Compre Special Project (PPE &IPE)

1. Calculate the energy transfer rate across 6 in.
wall of firebrick with a temperature

difference across the wall of 50 C. The thermal conductivity of the firebrick is 0.65 BTU/hrft-F at the temperature interest.
a. 285 W/m2
c. 112 W/m2
b. 369 W/m2
d. 429 W/m2
Solution

Q
A
kAT
x
kT
x
Where:
T
50
6in.
0.65
90F
0.5 ft.
BTU
hr ft F
Then:
Q
A
0.65 90
BTU
0.50
hr ft F
Q
A
W
3.153
BTU
m
117

hr ft 1 BTU ft
hr
Thus:
b
Q
A
368.90
W
m
2. At an average temperature of 100C, hot air flows through a 2.5 m long tube with an
inside diameter of 50 mm. The temperature of the tube is C along its entire length.
Convective film coefficient is 20.1 W/m2-K. Determine the connective heat transfer from
air to the tube.
a. 900 W
c. 624 W
b. 909 W
d. 632 W
Solution
Q
h AT
Where:
A
d L
0.050 2.5
0.3927m
20.1 0.3927 100
Thus;
d Q
20
631.46 W
3. Steam, initially saturated at 2.05 MPa, passes through a 10.10 cm standard steel pipe for
a total distance of 152 m. The steam line is insulated with 5.08 cm thickness of 85%
magnesia. For an ambient temperature of 22C, what is the quality of the steam which
arises at its destination if the mass flow rate is 0.125 kg steam per second?
Properties of Steam:
Pressure=2.05 MPa,
Temperature=213.67 C
Enthalpy: hf=914.52 kJ/kg
hfg=1885.5 kJ/kg
hg=2800.00 kJ/kg
Note: k for 85% magnesia is 0.069 W/m-K and ho for still air is 9.36 W/m2-K
a.93 %
c. 84 %
b.98 %
d. 76 %
Solution
From figure,
r1=5.05 cm
r2=10.13 cm
Q
t
r
ln
r
2kL
t
1
A h
Where:
A
2r L
2 10.13 152
96.746 m
213.67
ln 0.1013/0.0505
2 0.069 152
16,427.4 W
m h
22
1
96.746 9.36
16.43 kW
From:
16.43
0.125 2800
2668.6kJ/kg
Where:
2668
x
x
914.52
x 1885.5
0.92998 or 93 %
Thus;
a 93%
4. The sun generates 1 kW/m2 when used as a source for solar collectors, a collector with
an area of 1 m2 heat water. The flow rate is 3.0 liters per minute. What is the temperature
rise in the water? The specific heat of water is 4,200 J/kg C.
a. 4.8 C
c. 0.50 C
b. 0.48 C
d. 0.84 C
Solution
Q
mC T
Where:
Q
kW
1m
m
Li
min
4200
1kW
kg
Li
J
kgC
60
4.2
min
s
0.05
kg
s
kJ
kgC
Thus;
1
0.05 4.2 T
a T
4.76 C
5. The hot combustion gases of a furnace are separated from the ambient air and its
surrounding which are a 25C, by a brick wall 0.15 m thick. The brick has a thermal
conductivity of 1.2 W/m-K and a surface emissivity of 0.80. Under steady state conditions
and outer surface temperature of 100 C is measured. Free convection heat surface
temperature of 100 C is measured. Free convection heat transfer to the air adjoining this
surface is characterized by a convection coefficient of 20 W/m2-K. What is the inner
temperature in C?
a. 623.7
c. 461.4
b. 352
d. 256.3
Solution
Let, Q
heat transmitted by convection

Q
h t
20 100
1500
t
25
W
m
heat transmitted by radiation

Q
20,408.4
10 Fe T
20,408.4
10
1,872,793
520
0.80
T
100
J
hr
273
m
25
273
J
hr
W
m
Then;
Q
2020
W
m
1500
520
Thus;
Q
kA t
2020
b t
t
x
1.2 t
100
0.15
352.5 C
6. A 6 in. x 20 ft. uninsulated B.I. pipe conveys steam at 385 F wit han average ambient
temperature of 85F. If the cost of the fuel is P 250.00 per 106 BTU with the net energy
conversion efficiency of 75%, what is the annual cost of the heat lost?
a. P 60, 482.00
c. P 70, 482.00
b. P 65, 482.00
d. P 75, 482.00
Solution
For 6 in. pipe schedule 80
D
6.625 in.
5.761 in.
For iron;
k
52
30
W
m
BTU
hr ft F
For the surface coefficients;

h
1000
BTU
F
hr ft
BTU
ft
F
hr
Solving for Q:
t
D
ln
D
2kL
t
Q
1
Ah
1
A h
Where:
A
D L
D L
30.16ft
34.69ft
385
Q
1
30.16 1000
85
6.625
ln
5.761
2 30 20
Q
Then the annual cost of heat lost:
20,713
1
34.69 2
BTU
hr
20,713 8760 250

0.75 10
Thus;
a C
P 60,482.00
7. What is the external heating area in square feet of a tube with the following dimensions:
tube inside diameter = 5 in. wall thickness = in. length = 18 ft.
a. 26.5
c. 19.25
b. 24.25
d. 28.26
Solution
A
D L
Where:
D
6
12
18
1
2
6 in.
Thus;
A
d A
28.27ft
8. Determine the vacuum efficiency of a surface condenser which operates at a vacuum

of 635 mm Hg and exhaust steam enters the condenser at 45.81 C, the barometric pressure
is 760 mm Hg and the saturation pressure at 45.81 C is 0.010 MPa.
Solution
a. 80.4%
c. 92.7%
b. 85.2%
d. 98.3%
P
P
Vacuum Efficiency
P
P
Where:
P
101.325 kPa
0.010 MPa
760
125 mmHg
16.665kPa
10kPa
635
101.325kPa
760mmHg
Then;
Vacuum Efficiency
101.325 16.665
101.325 10
Vacuum Efficiency
0.9270
Vacuum Efficiency
92.70%
Thus;
9. A heat exchanger was installed purposely to cool 0.50 kg of gas per second. Molecular
weight is 28 and k=1.32. The gas is cooled from 150 C to 80 C. Water is available at the rate
of 0.30 kg/s and at a temperature of 12 C. Calculate the exit temperature of the water.
a. 48
c. 46
b. 42
d. 44
Solution
Q
m C
m C
m C T
12
m C
150
80
Where:
C
1.2247
4.187
kJ
kg K
kJ
kg K
Then;
0.30 4.187 t
12
0.50 1.2247 150
80
Thus;
c t
46.125 C
10. An uninsulated steam pipe passes through a room in which the air and walls are at
25 C. The outside diameter of the pipe is 70 mm, and its surface temperature and emissivity
are 200 C and 0.80 respectively. If the coefficient associated with free convection heat
transfer from the surface to the air is 15 W/m2-K, what is the rate of heat loss from the surface
per unit length of pipe?
a. 997.84 W/m
c. 797.84 W/m
b. 897.84 W/m
d. 697.84 W/m
Solution
Q
Q
Where:
Q
heat transmitted by convection
h A t
15 0.07 L 200
577.27
heat transmitted by radiation
t
25
W
m
20,408.4
10 A Fe T
20,408.4
10
1,514,032.40 L
42,057 L
0.07 L 0.8
J
hr
473
298
J
hr
W
m
Thus;
Q
577.27L
997.84 L
Q
L
420.57L
997.84
W
m
11. A heat exchanger is to be designed for the following specifications:

H2O gas temperature = 1145 C
CO2 gas temperature = 45 C
Unit surface conductance on the hot side = 230 W/m2-K
Unit surface conductance on the cold side = 290 W/m2-K
Thermal Conductivity of the metal wall = 115 W/m-K
Find the maximum thickness of the metal wall between the hot gas and cold gas so that
the maximum temperature of the wall does not exceed 545 C.
a. 10.115 mm
c. 17.115 mm
b. 13.115 mm
d. 20.115 mm
Solution
Q
A
t
1
h
t
x
k
1
h
Solving for Q/A:

Q
A
h t
Q
A
230 1145
Q
A
138,000
545
W
m
Then;
138,000
1.45 45
x
1
1
230 115 290
Thus;
d x
20.115 mm
12. Calculate the heat transfer per hour through a solid brick wall 6m long, 2.9 m high, and
225 mm thick, when the outer surface is at 5 C and the inner surface 17 C, the coefficient
of thermal conductivity of the brick being 0.5 W/m-K.
a. 2,004.48 kJ
c. 2,400.48 kJ
b. 3,004.48 kJ
d. 3,400.48 kJ
Solution
Q
kAT
x
0.60 6 2.9 17
0.225
556.8 W
556.8
2,004.48
J 360J0s
hr
s
kJ
hr
kJ
1000J
Thus;
a The heat transfer per hour is 2,004.48 kJ
13. A vertical furnace wall is made up of an inner wall of firebrick 20 cm thick followed by
insulating brick 15 cm thick and an outer wall of steel 1 cm thick. The surface temperature
of the wall adjacent to the combustion chamber is 1200 C while that of the outer surface
of steel is 50 C. The thermal conductivities of the wall material in W/m-K are: firebrick, 10;
insulating brick 0.26; and steel, 45. Neglecting the film resistances and contact resistance
of joints, determine the heat loss per sq. m. of wall area.
a. 1.93 W/m2
c. 1.55 W/m2
b. 2.93 W/m2
d. 2.55 W/m2
Solution
Q
A
Q
A
x
k
x
k
x
k
1200 50
0.20 0.15 0.01
10
0.26
45
1.93
Thus;
a
Q
A
1.93
W
m
14. A composite wall is made up of an external thickness of brickwork 110 mm thick inside
which is a layer of fiberglass 75 mm thick. The fiber glass is faced internally by an insulating
board 25 mm thick. The coefficients of thermal conductivity for the three are as follows:
Brickwork
1.5 W/m-K
Fiberglass
0.04 W/m-K
Insulating board
0.06 W/m-K
The surface transfer coefficients of the inside wall is 3.1 W/m2-K while that of the outside
wall is 2.5 W/m2-K. Take the internal ambient temperature as 10 C and the external
temperature is 27 C. Determine the heat loss through such wall 6 m high and 10 m long.
a. 330.10 W
c. 430.10 W
b. 230.10 W
d. 530.10 W
Solution
Q
AT
R
Where;
R
1
h
x
k
x
k
1
3.1
0.110
1.5
3.09
m C
W
x
k
0.075
0.04
1
h
0.025
0.06
1
2.5
Then;
Q
6 10 27
3.09
10
Thus;
a Q
330.10 W
15. One insulated wall of a cold-storage compartment is 8m long by 2.5 m high and
consists of an outer steel plate 18 mm thick. An inner wood wall 22.5 mm thick, the steel
and wood are 90 mm apart to form a cavity which is filled with cork. If the temperature
drop across the extreme faces of the composite wall is 15 C, calculate the heat transfer
per hour through the wall and the temperature drop across the thickness of the cork. Take
the coefficients of thermal conductivity for steel, cork and wood as 0.45, 0.045, and 0.18
W/m-K respectively.
a. 408.24 kJ, 12.12 C
c. 608.24 kJ, 13.12 C
b. 708.24 kJ, 11.12 C
d. 508.24 kJ, 14.12 C
Solution
Q
AT
R
x
k
0.018
45
2.125
x
k
x
x
0.09
0.045
0.09
0.045
0.0225
0.18
Then:
Q
8 2.5 15
2.125
141.176 W or
508.24
J
s
kJ
hr
Thus, the heat transfer per hour is 508.24 kJ

Solving for the temperature drop across the cork:
Q
AT
x
k
141.176
20 T
0.09
0.045
14.12 C
Thus;
d 508.24 kJ, 14.12C
16. A cubic tank of 2 m sides is constructed of metal plate 12 mm and contains water at
75C. The surrounding air temperature is 16 C. Calculate the overall heat transfer coefficient
from water to air. Take the coefficient of thermal conductivity of the metal as 48 W/m-K,
the coefficient of heat transfer of water is 2.5 kW/m2-K and the coefficient of heat transfer
of the air is 16 W/m2-K.
a. 15.84 W/m2-K
c. 16.84 W/m2-K
b. 14.84 W/m2-K
d. 13.84 W/m2-K
Solution
Let U
overall heat transfer coefficient

U
1
R
Where:
1
x
x
h
1
2.5
1
h
0.012
48
10
1
16
0.063m C /W
R
Then:
U
1
W
0.063 m C
15.84
W
m C
Thus;
a U
15.84
W
m C
17. Calculate the quantity of heat conducted per minute through a duralumin circular
disc 127 mm diameter and 19 mm thick when the temperature drop across the thickness
of the plate is 5 C.Take the coefficient of thermal conductivity of duralumin as 150 W/m-K.
a. 30 kJ
c. 35 kJ
b. 40 kJ
d. 45 kJ
Solution
Q
kAT
x
150
0.127
4
0.019
500.04 W
30 kJ/min
Thus;
a the quantity of heat conducted per minute is 30 kJ
18. A cold storage compartment is 4.5 m long by 4 m wide by 2.5 m high. The four walls,
ceiling and floor are covered to a thickness of 150 mm with insulating material which has
a coefficient of thermal conductivity of 5.8 x 10 -2 W/ m-K. Calculate the quantity of heat
leaking through the insulation per hour when the outside and inside face temperatures of
the material is 15 C and -5 C respectively.
a. 2185.44 kJ
c. 3185.44 kJ
b. 1185.44 kJ
d. 4185.44 kJ
Solution
Q
kAT
x
Where:
A
2 4.5 2.5
78.50 m
4 2.5
4.5 4
Then:
Q
5.8
10
78.5 15
0.15
607.07 Wor
2185.44
J
s
kJ
hr
Thus:
a the quantity of heat leaking through the insulation per hour is 2185.44
kJ
.
hr
19. Supplementary Problem

A blower operating at 15000 rpm compresses air from 20C and 1 atm to 1.6 atm.
The design flow is 38 m3/min and at this point the power input is 60 kW. Determine
the blower efficiency at the design flow.
a. 65%
c. 59.81%
b. 64.91%
d. 60.01%
Solution:
Blower Efficiency =
Isentropic Power
Power Input
Solving for Isentropic Power:

Pisen =
k-1
k -1
kPV (rp )
k-1
1.4-1
1.4(101.325)(38/60) (1.68/1) 1.4 -1

k-1
= 35.89 kW
then;
eblower =
35.89
= 0.5981
60
thus;
(a) eblower = 59.81%
A small blower handles 43.33 m3 of air per minute whose density is 1.169 kg/m3.
The static and velocity heads are 16.38 and 1.22 cm WG (at 15.6C) respectively.
Local gravity acceleration is 9.741 m/s2. Find the power input to the air from the
blower.
a. 1.64 kW
c. 1.76 kW
b. 1.91 kW
d. 1.24 kW
Solution:
P = Qh
Where:
h = 16.38 + 1.22
= 17.6 cm = 0.176 m
Q = 43.33 m3/min
= 0.72 m3/s
then;
P = 9.741(0.72)(0.176)
thus;
(d) P = 1.24 kW
A fan can developed a static pressure head or 350 mm water gage through
standard air condition. What is the new static pressure head if fan can operate at
95C and 729 mm of Hg?
a. 265 mm WG
c. 274 mm WG
b. 270 mm WG
d. 263 mm WG
Solution:
h2
= 2
h1
1
Solving for 2 :
2 =
720(101.325/760)
0.287(95 + 273)
then;
h2
0.91
=
350
1.2
thus;
(a) h2 = 265.09 mm WG
A two-stage radial-type airplane supercharger is designed to deliver 4535 kg of
air per hour at a pressure of 800 mmHg abs when operating at an altitude of 4570
m where the temperature is -15Cand the pressure is 429 mm Hg abs. it rotates at
18,000 rpm and is to have an adiabatic over-all efficiency of 72 percent. It is to
be tested at sea level (762 mm Hg abs and 26.67C at a speed of 14,000 rpm.
Considering that the efficiency at the design point does not change, determine
for the design point under test conditions the volume of air taken m3/s.
a. 2.27 m3/s
b. 2.12 m3/s
Solution:
Q2
N2
=
Q1
N1
c. 1.27 m3/s
d. 1.12 m3/s
Solving for Q1:

Q1 =
mRT1
P1
(4535/3600)(8.314/29)(-15 + 273)
429(101.325/760)
= 1.63 m3/s
Then;
14,000
Q2
=
1.63 18,000
Thus;
(a) Q2 = 1.27 m3/s
The fan has a total head of 190 m and a static pressure of 20 cm WG. If the air
density is 1.2 kg/m3, what is the velocity of air flowing?
a. 16.21 m/s
c. 16.67 m/s
b. 17.21 m/s
d. 17.766m/s
Solution:
hv =
v2
2g
Solving for hv:

h = hs + hv
190 = 0.20(1000/1.15) + hv
hv = 16.09 m
then;
16.09 =
V2
2(9.81)
V = 17.766 m/s
thus;
(d) 17.766 m/s
A sewerage aeration blower rotating at 3500 rpm is designed to deliver 567
m3/min of air from 20C and 1 atm to a discharge of 158 kpa (abs) with an
adiabatic efficiency of 65 %. During a summer the atmospheric temperature rises
to 43C but the barometric pressure does not change. It is desired to vary the
blower speed to maintain the same discharge pressure. Determine the discharge
volume of standard air with the new speed.
a. 8.20 m3/s
c. 8.64 m3/s
b. 9.10 m3/s
d. 9.74 m3/s
Solution:
2
Q1
h1
=
h2
Q2
h1
:
h2
Solving for
T2
h1
=
h2
T1
=
43 + 273
20 + 273
= 1.08
Then;
567
Q2
1.08 =
Q2 = 545.98 m3/min
Thus;
(b) Q2 = 9.10 m3/s
A 40 in. diameter fan rated at 160,000 cfm standard air at 16 in. starting pressure is
operating at 1200 rpm. Solve for the specific speed.
a. 386,845.18 rpm
c. 384,845.18 rpm
b. 380,125.20 rpm
d. 392,865.28 rpm
Solution:
Ns = specific speed
Ns =
=
NQ
3
12000 160,000
3
( 43 )
Thus;
(a) Ns = 386,845.18 rpm

A boiler requires 75,000 m3/hr of standard air. What is the motor power if it can
deliver a total pressure of 145 mm or water gage. The mechanical efficiency of
fan is 64 %.
a. 40.30 kW
c. 42.45 kW
c. 46.30 kW
d. 43.69 kW
Solution:
Pmotor =
Pair
Pfan
Solving for Pair:

Pair =QH
Where:
h = 0.145
1000
1.2
= 120.83 m
Then;
Pair = [(1.2)(0.00981)]
75000
(120.83)
3600
= 29.63 kW
Thus;
Pmotor =
29.63
0.64
(a) pmotor = 46.30 kW

Calculate the required motor capacity needed to drive a forced-draft fan
serving a stoker fired boiler using coal as fuel.
Combustion data includes the following:
Atmospheric Air
101.3 kPa ; 20 C
Weight of fuel burned per hour
10 tons
Ultimate Analysis of fuel :

C = 78 %
S=1%
H=3%
A=8%
O=3%
M=7%
ExcessAir
30 %
Fuel bed and air heater resistance
18 cm WG
Fan Efficiency
60 %
a. 87.84 kW
c. 84.87 %
b. 82.87 kW
d. 88.72 %
Solution:
Pmotor =
Pair
Pfan
Solving for Pair:

Theoretical Air required for the combustion of coal.
Wt =11.5C + 34.5
H -
O
+ 4.3S
8
= 11.5(0.78) + 34.5
0.03 -
0.03
+4.4(0.01)
8
= 9.92 kg air / kg fuel

Actual weight of air supplied into the boiler:
Wa = (1 + e) Wt
= (1 + 0.30)(9.92 kg air/kg fuel)[(10(1000)kg fuel/hr]
= 128,942 kg air/hr
Volume of air demanded by the boiler from the forced draft fan:
Q=
128,942 kg/hr
=107,451.77 m3 /hr
1.2 kg/m3
= 29.85 m3/s
Then;
Pair = 1.2 0.00981
29.85 (0.18)
1000
1.2
= 52.71 kW
Thus;
Pmotor =
52.71
0.60
(a) Pmotor = 87.84 kW

The motor power needed to drive the fan is 75 kW and the volume flow of air
delivered by fan is 23 m3/s and 20 cm water gage. The density of air is 1.2kg/m3.
What is the fan efficiency?
a. 60 %
c. 64 %
b. 62 %
d. 65 %
Solution:
efan =
Pair
Pmotor
Solving for Pair:

Pair =[1.2 0.00981 23 ] 0.20
1000
1.2
= 45.126 kW
Then;
efan =
45.126
75
= 0.60168
Thus;
(a) efan = 60.168 %
In a certain installation, a fan when driven by a 7.5 Hp motor at a speed of 600
rpm delivers 510 m3 of air per minute at a total pressure of 5 cm WC. If in the same
installation, 6.5 cm WC pressure is required. What power and motor speed willthe
fan be driven?
Solution:
Fan Drive Speed:
N2
N1
h2
h1
N2
=
N1
N2
=
600
h2
h1
6.5
5.0
N2 = 684.11 rpm
Motor Power required:

P2
N2
=
P1
N1
P2
=
7.5
684.11
600
P2 = 11.12 Hp
Thus;
(c) N2 = 684.11 rpm, P2 = 11.12 Hp

A fan is supplying forced draft into a boiler has the following specifications on its
name plate:
Capacity
280 m3/min
Air temp.
25C
Total pressure
4 cm WC
Motor Rating
5 Hp; 1200 rpm
A tabular air heater is installed in line with the boiler, and the fan is now required
to supply heated air for combustion at 90C. What drive power is required and
the new total pressure that this fan will operate if it is going to deliver the same
volume of heated air at 1200rpm?
a. 3.28 cm WC, 4.10 Hp
c. 3.28 cm WC, 5.50 Hp
b. 3.95 cm WC, 5.5 Hp
d. 3.95 cm WC, 4.10 Hp
Solution:
Solving for the new head;
h1
= 2
h2
1
Solving for
2
:
1
2 T1
=
1 T2
=
Then;
25+273
90+273

h2
= 0.82
4
h2 = 3.28 cm WC
thus;
(a) h2 = 3.28 cm WC & P2 = 4.10 Hp
Note: when air is heated, its density decreases and the pressure needed to move the air
to the combustion chamber will be lesser resulting to the decrease in the fan power
requirement.
A fan has a suction pressure of 5 cm water vacuum with air velocity of 5m/s. the
discharge has 20 cm WG and discharge velocity of 10 m/s. Find the total head of
the fan.
Solution:
h = hs + hv
where:
hs =
(hdw - hsw )w
a
[(0.20 - (-0.05)]1000
1.2
= 208.33 m
hv =
Vd 2 - Vs 2
2g
(10) - (5)
2(9.81)
= 3.28 m
Thus;
h = 208.11 + 3.28
(a) h = 212.15 m
An Air Handling Unit (AHU) for an air conditioning system has a centrifugal fan
with backward curved blades mounted on a scroll housing driven by a motor at
750 rpm. The fan delivers 2000 cfm of air against 3 in. WC static pressure (including
resistance of ducts, elbows, cooling coils, and outlet grills) and 0.80 in. WC
velocity pressure. Calculate the tip speed of the wheel.
a. 3100 fpm
c. 3586 fpm
b. 3000 fpm
d. 3500 fpm
Solution:
The speed of the wheel:
V=
2ghv
Solving for hv:

a hva = hvw w
(0.075)hva = (0.8/12)(62.4)
hva = 55.46 ft of air
Note:
air =0.075 lb/ft
water = 62.4 lb/ft
Then;
V=
2(32.2)(55.46)
= 59.77 ft/s
Thus;
(c) V = 3586 ft/min.
A steam generator supplies 180,000 kg of steam per hour at 5.5 Mpa abs and
540C with feedwater at 176C. At this output, the thermal efficiency is 85% when
burning 42,456 kJ/kg fuel oil at 15% excess air. The products of combustion with an
average molecular weight of 30 are removed from the unit by a pair of duplicate
induced-draft fans operating in parallel and the flue gas temperature for each
fan suction is 150C. Estimate the capacity of each fan using the following rule:
7.5 kg air required for perfect combustion for each 23,200 kJ per kg heat value
of oil. The fan differential pressure is 190 mm WC.
a. 69.57 kW
c. 59.75 kW
b. 89.75 kW
d. 76.57 kW
Solution:
air = Qh
= [1.2(0.00981)] Q h
Where;
h = 0.190 (1000/1.2)
= 158.33 m
Solving for Q:
Mass of fuel burned per hour:
mf =
ms hs - hv
eboiler Qh
180,000(3520-746)
0.85(42,456)
= 13 836.33 kg/hr
Total mass of air used per kg of fuel burned:

ma = 7.5
42,456
23,200
1.15 =17.58 kg air/kg fuel
Mass flow rate of air for combustion:

ma = (15.78)(13,836.33) = 218,839.17 kg/hr
Mass flow rate of gases leaving the boiler:
mg = ma + mf
= 218,839.17 + 13,836.33
= 232,225.5 kg/hr or mg = 116,112.75 kg/hr (each fan)
Mass flow gas handled by each fan:
Q=
(116,112.75)(8.314/30)(150 +273)
101.3
= 134,369.16 m3/hr
= 37.32 m3/s
Thus;
(a) Pair = [ 1.2(0.00981) ] (37.32) (158.33) = 69.57 kW
34. ME Board Problem
A fan listed as having the following performance with standard air:
Volume discharge = 120 m3/s
Speed = 7 rps
Static pressure = 310 mm water gage
Brake power = 620 kW
The system duct will remain the same and the fan will discharge the same volume
of 120 m3/s of air at 93C and a barometric pressure of 735 mm Hg when its speed
is 7 rps. Find the brake power input and the static pressure required.
a. 482 kW, 241 mm WG
c. 482 kW, 256 mm WG
b. 492 kW, 241 mm WG
d. 492 kW, 256 mm WG
Solution:
Brake power input=620
Solving for 2 :
2 =
735(101.325/760)
0.278(93+273)
= 0.9329 kg/m3
2
2
=620
1
1.2
Solving for the static pressure, h2:

0.9392
h2
=
310
1.2
h2 = 241 mm of water gage
then;
Brakepower input= 620
0.9329
=482 kW
1.2
thus;
(a) h2 = 241 mm & Pbrake = 482 kW
35. ME Board Problem
Local coal with higher heating value of 5,500 kCal/kg is burned in a pulverized
coal fired boiler with 25% excess air at the rate of 25.9 M.T. per hour when the
steam generated is 220 M.T. per hour. This boiler is served by 2 forced-draft fans of
equal capacity delivering the air at 305 mm of water to the furnace. Calculate
the capacity of each fan in m3/hr if fan capacity is to be 110 percent of the
maximum requirement. Ambient air is 100 kPa and 30C.
a. 31.75 m3/s
c. 28.87 m3/s
b. 63.50 m3/s
d. 57.73 m3/s
Solution:
The theoretical weight of air to burn the fuel is given in an appropriate formula when the
heat value ot the fuel is given:
Wta =
=
A
HHV, kCal/kg
=
F t
745
5,500
745
= 7.382 kg air / kg fuel

Mass flow rate of air required for combustion
ma = (7.382)(1.25)[25.9(1000)]
= 238,992.25 kg/hr
Volume of Air needed at 100 kPa and 303 K
ma Ra Ta
P
238,992.25(0.287)(303)
=
100
Q =
= 207 830.05 m3/hr

Capacity of each fan at 110% of the maximum requirement:
Q1 = Q2 =
207,830.05(1.10)
2
= 114,306.53 m3/hr = 31.75 m3/s

Thus;
(a) Q = 31.75 m3/s
Find the motor size needed provide the forced-draft service to a boiler that burns
coal at the rate of 10 to per hour. The air requirements are 59,000 cfm, air is being
provided under 6 in. water gage (WG) by the fan which has mechanical
efficiency of 60 percent. Assume fan to deliver the total pressure of 6 in WG.
a. 90 Hp
c. 97 Hp
b. 93 Hp
d. 99 Hp
Solution:
Pmotor =
Pair
0.60
Solving for Pair:

Pair = Qh
Where:
Q = 59,000 ft3/min. = 27.84 m3/s
h = 6(1000/1.2) in. = 127 m
then;
Pair = [ 1.2(0.00981)) ](27.84)(127)
Pair = 41.62 kW
Thus;
Pmotor =
41.62
=69.37 kW
0.60
(a) Pmotor = 92.99 Hp

A turbo-generator, 16 cylinder, Vee type diesel engine has an air consumption
of 3000 kg/hr per cylinder at rated load and speed. This air is drawn in thru a filter by
a centrifugal compressor direct connected to the exhaust gas turbine. The
temperature of the air from the compressor is
145C and a counterflow air cooler
reduces the air temperature to 45 C before it goes to the engine suction header.
Cooling water enter air cooler at
mean temperature difference.
30C and leaves at 38C. calculate the arithmetic
a. 41 C
b.
c. 61 C
51C
d,
71C
Solution:
t max 145 38 107C

t min 45 30 15C
t max t min
AMTD
2
107 15
AMTD
2
Thus;
(c) AMTD 61 C

A pond is covered by a sheet of ice 2 cm thick (thermal conductivity
1.68W / mC ). The temperature of the lower surface of the ice is 0C and that of the
upper surface is
of the ice?
10C. At what rate is heat conducted through each square meter
a. 840 W
c . 940 W
b. 740 W
d. 640 W
Solution:
kAt
x
(1.68)(1)(0 10)
Q
(0.02)
Q
Thus;
t max
t min
(a) Q 840W

How much heat is
conducted
through
sheet
of
plates
glass,
k 0.0024Cal / s m C which is 2m by 3m and 5mm thick, when the
temperatures of the surfaces are 20 C and 10 C.

318,400 Cal/min
b. 418,400 Cal/min
a.
c. 940 Cal/min
d. 618,400 Cal/min
Solution:
kAt
x
Where:
k 0.0024 Cal/s- cm C
A 2(3) 6 m 2
60,000 cm 2
t 20 10 30 C
L 5 mm 0.50 cm
Then;
0.002460,00030
0.50
= 8640 Cal/s
Thus;
(c)
Q 518,400 Cal/min
40. Supplementary problem

A cooper rod whose diameter is 2 cm and length 50 cm has one end in
boiling water, the other end in a jacket cooled by flowing water which enters at
10 C . The thermal conductivity of copper is 0.102 kCal / m s C. If 0.20 kg of

water flows through the jacket in 6 min, by how much does the temperature of the
water increase?
10.38 C
b. 9.38 C
a.
Solution:
Q
Where:
kAt
x
11.38 C
d. 12.38 C
c.
x
0.022
4
3.14 10 4 m 2
A
Then;
0.1023.14 10 4 100 10
0.50
= 0.005765 kCal / s
= 0.346 kCal / min 6 min
= 2.705 kCal
Heat required to raise the water temperature:
Q m C t
2,075 Cal 200g (1 Cal / g C )(t )
Thus;
(a) t 10.38 C

The thermal insulation of wooden glove may be regarded as being essentially
kCal / m s C. How
2
much heat does a person lose per minute from his hand of area 200 cm and skin
temperature 35 C on a winter day at 5 C.

a layer of quiescent air 3 cm thick, of conductivity 5.7 10
6.12 C
b. 7.12 C
9.12 C
d. 8.12 C
a.
c.
Solution:
kAt
x
Where:
k= 5.7 10
kCal / m s C
cm 2
2
= 0.02 m
A= 200
x=3 cm = 0.03 m
t 35 5 40 C
Q
5.7 10 .0240
6
0.03
1.52 10 4 kCal / s
Thus;
(c) Q 9.12 Cal / min

The temperature directly beneath a 3 in. concentrate road is
5 F and the air
temperature is 20 F . Calculate the steady flow per square foot through the
concrete. The thermal conductivity of the concrete is 0.50 Btu / ft hr F .

a. 30 Btu / hr ft
c. 50 Btu / hr ft
b. 40 Btu / hr ft
d. 60 Btu / hr ft
Solution:
Q kt
x
A
0.520 5
3 / 12
Thus;
(a)
Q
Btu
30
A
hr ft 2

At what rate does to sun lose energy by radiation? The temperature of the sun
is about 6000 K and its radius is 6.95 10 km. .
5
3.48 10 26 W
26
b. 4.48 10 W
5.48 10 26 W
26
d. 6.48 10 W
a.
c.
Solution:
P A T 4
Where:
5.7 10 12 W / cm 2 K 4
A 4R 2
4 6.95 10 5
6.07 1012 km 2
6.07 10 22 m 2
Then;
P 5.7 10 12 6.07 10 22 6000
Thus;
(b)
P 4.48 10 26 W

How many watts will be radiated from a spherical block body 15 cm in
diameter at a temperature of 800 C .

a. 5.34 kW
b. 4.34 kW
c. 6.34 kW
d. 3.34 kW
Solution:
P A T4
Where:
5.7 10 12 W / cm 2 K 4
A 4 (7.5) 2
706.86 cm 2
T 800 273
1073K
Then;
P 5.7 10 12 706.861073
Thus;
(a) P 5,340 W 5.34 kW
Calculate the radiation in watts per square centimeter from a block of copper
at 200 C and at 1000 C . The oxidized copper surface radiates at 0.60 the rate of a
black body.
a. 0.17
b. 0.27
c. 0.37
d. 0.07
Solution:
Q e AT4
Q
e T4
A
0.60(5.7 10 12 )(300 273) 4 0.17 W / cm 2
Thus;
(a)
Q
0.17 W / cm 2
A

A surface condenser serving a 50,000 kW steam turbo-generator unit receives
exhaust steam at the rate of 196,000 kg/hr. Vacuum in condenser is 702 mm. Hg. Sea
water for cooling enters at 29 and leaves at 37.5 C . For steam turbine condenser,
manufacturers consider 950 Btu/lb of steam turbine condensed as heat given up to
cooling water. Calculate the logarithmic mean temperature difference.
a. 4.57 C
c. 6.57 C
5.57C
b.
d.
7.57 C
Solution:
Let: LMTD Log mean temperature difference
LMTD
t max t min
t max
ln
t min
Where:
101.325kPa
P( condenser ) 101.325 702 psi

760 psi
7.733 kPa(t sat 40.86 C )
(t ) max 40.86 29.5 11.36C

(t ) min 40.86 37.5 3.36C
Then;
LMTD
11.36 3.36
11.36
ln
3.36
Thus;
(c)
LMTD 6.57C

The stack gas from a chemical operation contains noxious vapors that must
315C to 35C . The gas flow rate is

0.70 m 3 / s . Water is available at 10 C at 1.26 kg / s . A two shell and 4 tube pass,
be condensed by lowering its temperature from
conterflow heat exchanger will be used with a water flowing through the tubes. The
gas has a specific heat of 1.10
kJ / kg K and a gas constant of 0.26 kJ / kg K .
Calculate the logarithmic mean temperature difference.
102.8 C
b. 110.9 C
120.8 C
d. 118.9 C
a.
c.
Solution:
Average gas temperature
Density of gas
( ) :
315 35
2
175 C
P
RT
101.325
(0.26)(175 273)
0.867kg / m 3
Mass flow rate of gas:
mg (0.7 m 3 / s )(0.867 kg / m 3 ) 0.607 kg / s

Heat gained by cooling water = Heat Lost by the gasses
mw C pw t w m g C pg t g
(1.26)(4.187)(t 10) 0.60(1.10)(315 35)
t 45.5 C
Solving for
(t ) max and (t ) min :
(t ) max 315 45.5 269.5 C

(t ) min 35 10 25 C
Then;
LMTD
(t ) max (t ) min 269.5 25
(t ) max
269.5
ln
ln
25
(T ) min
Thus;
(a)
LMTD 102.8 C

Exhaust steam at 7 kPa at the rate of 75 kg/s enters a single pass condenser
2
containing 5,780 pcs copper tubes with a total surface area of 2950m . The steam
has a moisture content of 10 % and the condense leaves saturated liquid at steam
temperature. The cooling water flow rate is 4,413 liters per second entering at 20 C .
Size of tubes, 25 mm O.D. by 3 min thick wall. Find the overall heat transfer coefficient.
a. 5275 W / m
b. 2275 W
2 K
c. 4274 W / m
/ m 2 K
d. 3225 W
Solution:
Ethalpy of steam entering the condenser:
h1 (h f 1 xh fg ) 7 kPa 163.4 0.90(2572.5)

2478.65 kJ / kg
Ethalpy and temperature of condensate:
2 K
/ m 2 K
h2 h f @ 7 kPa 163.4 kJ / kg , Tsat @ 7 kPa 29 C

Ethalpy and temperature of the condensate:
Qwater Qsteam
m w C pw t w m s (h1 h2 )
4,413L / s(1kg / L)(4.187)(t 20) 75(2478.65 163.4)

t 29.40 C
(t ) max 39 20 19 C
(t ) min 39 29.4 9.6 C
LMTD
19 9.6
13.77 C
19
ln
9 .6
From:
Q A U (LMTD )
Where:
Q m s (h1 h2 )
75(2478.65 163.4)
173,643.75 kW
Then:
173,643.75 2950 U 13.77

U 4.275
kW
m2 K
Thus;
(a) U 4275
W
m K
2

What is the heat flow per hour through a brick and mortar wall 9 in. thick if the
coefficient of thermal conductivity has been determined as 0.40 Btu / ft hr F

and the wall is 10 ft high by 6 ft wide, the temperature on one side of the surface
being 330 F and on the other 130 F .

a. 6400 Btu / hr
c. 5400 Btu / hr
b. 7400 Btu / hr
d. 8400 Btu / hr
Soution:
kAt
x
(0.4)(10)(6)(330 130)
6
12
then;
(a) Q 6,400 Btu / hr
50. Supplementary problem

Water is flowing in a pipe with radius of 25.4 cm at a velocity of 5 m/s at the
temperature in the pipe. The density and viscosity of the water are as follows:
density 997.9 kg / m 3 and vis cos ity 1.131Pa s . What is the Reynolds number for
this situation.
a. 2241
b. 96.2
c. 3100
d.1140
Solution:
N RE
VDp
Where:
V 5 m/s
D 2(25.4) 50.8 cm
0.508 m
997.9kglm 3
1.131
Then;
N RE
5(0.508)(9977.9)
1.131
Thus;
(a) N RE 2241.08

A heat exchanger has an over-all
coefficient
of heat
transfer of
900W / m K . The mean temperature difference is 20 C and the heat loss is

15,000 W. Calculate the heat transfer area.
2
a. 0.833 m
c. 0.933 m
b. 0.733 m
d. 0.633 m
Solution:
Q AU
15,000 A(900)(20)
A 0.833m 2
Thus;
(a) A 0.833m

A complete furnace wall is made up of a 12 in. lining of magnesite refactory
brick, a 5 in. thickness of 85% magnesia, and a steel casing 0.10 in. thick. Flue gas
temperature is 2200 F and the boiler room is at 80 F .Gas side film coefficient is 15
Btu / hr ft 2 F air side is 4. Determine the thermal current Q / A .

a. 187.41 Btu / hr ft
c. 200.62 Btu / hr ft
b. 197.41 Btu / hr ft
d. 250.46 Btu / hr ft
Solution:
Q t
A RT
Where:
t 2200 80
2120 F
hi 15 Btu / hr ft 2 F
ho 4 Btu / hr ft 2 F
k12 20.5 Btu / hr ft 2 F (magnesite)
k 23 0.04 Btu / hr ft 2 F (magnesia)
k 34 25 Btu / hr ft 2 F ( steel )
RT
1 x12 x 23 x34 1
hi k12 k 23 k 34 h0
RT 11.312
Thus;
2120
Q
A 11.312
(a)
Q
187.41 Btu / hr ft 2
A
0.72 Btu / hr ft F ) 5
in. of insulating brick ( k 0.08) and 75 in. of red brick ( k 0.5) . The inner and outer
A wall of a furnace is made up of 9 in. firebrick ( k
surface temperatures t1 and t 4 of the wall are
1500 F and 150 F respectively.
Neglecting the resistance of the mortar joints, compute the rate of heat flow through
1 ft 2 of the wall.
a. 80 Btu/hr
b. 180 Btu/hr
c. 100 Btu/hr
d. 200 Btu/hr
Solution:
At
RT
Where:
RT
x12 x 23 x34
k12 k 23 k 34
RT
9 / 12 5 / 12 2.5 / 12
0.72 0.08
0 .5
Then;
1(1500 150)
7.5
Thus;
(d) Q = 180 Btu/hr
What is the heat transfer in the glass surface area of 0.70
m 2 having an inside
temperature of of 25 C and 13 C outside temperature. The thickness of the glass

surface is 0.007 m. The thermal conductivity is 1.8 W/m-K.
a. 4.16 kW
b. 3.16 kW
Solution:
kAt (1.8)(0.7)(25 13)
0.007
x
2160 W
Q
Thus;
(c)
Q 2160 W
c. 2.16 kW
d. 1.16 kW

The interior of an oven is maintained at a temperature of 1500F by means of a suitable
control apparatus. The walls of the oven are 9 in. thick and constructed from a material
having thermal conductivity of 0.18 Btu/hr-ft-F. Calculate the heat loss for each square
foot of wall surface per hour. Assume that the inside and outside wall temperatures are
1500F and 400F respectively.
a. 264 Btu/hr-ft2
c. 164 Btu/hr-ft2
b. 364 Btu/hr-ft2
d. 64 Btu/hr-ft2
Solution:

0.18 1 1500
9
12
400
264
Thus;
(a)
264
56. Past ME Board Problem

Compute the amount condensate formed during 10 minutes warm-up of 150 m pipe
conveys saturated steam with enthalpy of vaporization hfg = 1,947.8 kJ/kg. The minimum
external temperature of pipe is 2C and the final temperature is 195C. The specific heat
of pipe material is 0.6 kJ/kg-K and the specific weight is 28 kg/m.
a. 249.69 kg
c. 124.85 kg
b. 499.38 kg
d. 62.42 kg
Solution:

ms (1947.8) = 4200(0.60)(195-2)
Thus;
(a) ms = 249.69 kg
A high pressure steam generator is to be fitted with convection type superheater having
72 elements in parallel. Steam at the rate of 70,000 kg/hr from the boiler drum enters the
superheater inlet header at 8.3 Mpa and 485oC. Combustion products at 980oC enters
the superheater proper at the rate of 160,000kg/hr. Superheater elements are made of
60 mm O.D. by 8 mm thick tubing of 30 m length. Assume that the flue gas has the same
thermal properties of air. Calculate the heat transferred to the superheater tubes.
a. 12,152 kW
c. 10,152 kW
b. 11,512 kW
d. 13,152 kW
Solution:
From mollier Chart:
h1 = 2,715 kJ/kg @ 8.3 Mpa & 98% quality
h2 = 3,340 kJ/kg @ 8.0 Mpa & 485oC
Heat lost by flue gas = Heat gained by steam
mgCpg tg = ms (h1 h2)
(160,000)(1.0)(980 to) = 70,000 (3340 2715)
to = 705oC
The rate of heat transferred to the superheated tubes:
Q = ms (h2 h1)
= 70,000 (3340 2715)
= 43,750,000 kJ/hr
= 12,152,000 kJ/S
Thus:
(a) Q = 12,152 kJ/s or kW
In a hot water heating system, water heated to 95oC and then is pumped at the rate of 4
L/min through a radiator where it is cooled to 35oC. If the water arrives at the radiator at
at temperature of 85oC, how much heat does the radiator release each hour?
a. 50,244 kJ/hr
c. 55,344 kJ/hr
b. 45,422 kJ/hr
d. 65,244 kJ/hr
Solution:
Q = mCp t
= 4(4.187)(85-35)
= 837.4 kJ/min
= 50,244 kJ/hr
Thus:
(a) Q = 50,244 kJ/hr
Brine enters a cooler at the rate of 50 m3/hr at 15oC and leaves at 1oC. Specific heat and
specific gravity of brine are 1.07 kJ/kg-K and 1.1 respectively. Calculate the heat
transferred in kW.
a. 158.21 kW
c. 258.21 kW
b. 228.86 kW
d. 128.86 kW
Solution:
Q = mCp t
Where:
m = pV
= 1.1(1000 kg/m3)(50m3/hr)(1hr/3600s)
= 15.28 kg/s
Thus:
Q = (15.28)(1.07)(15-1)
(b) Q = 228.86 kJ/s or kW
A metal rod is 10 cm long and has a diameter of 2 cm one end is in contact with steam
at 100oC while the other end contacts a block of ice at 0oC. the cylindrical surface of the
rod is carefully insulated so heat flows only from end to end. In a time of 20 minutes, 320
grams of ice melts. What is the thermal conductivity of metal?
a. 0.28266 kJ/kgoC
c. 0.38266 kJ/kgoC
b. 0.18266 kJ/kgoC
d. 0.48266 kJ/kgoC
Solution:
Qrod = Qice
0.01 100
0.10
20 60
0.320 3.3 10
K = 282.66 J/kgoC
K = 0.28266 kJ/kgoC
Thus:
(a) k = 282.66 J/kgoC
A wall with an area of 10 m2 is made of 2 cm thickness of white pine (k = 0.113 W/moC)
followed by 10 cm of brick (k = 0.649 W/moC). The pine is on the inside where the
temperature is 30oC while the outside temperature is 10oC. Assuming equilibrium

conditions exist, what is the temperature at the interface between the two metals?
a. 15.65oC
c. 18.21oC
b. 17.64oC
d. 19.31oC
Solution:

Where:
0.02
. 113
0.10
0.649
RT = 0.331 m2oC/W
Then;

10 30 10
0.331
Q = 604.23 W
Solving for the temperature at the interface between the two materials:
Q = Q12
604.23
604.23
Thus;
(a) t2 = 19.31oC
In a hot air heating system, the furnace heats air from 60oF to 160oF. If the air is then
circulated at the rate of 330 ft3/min by the blower, how much thermal energy is
transferred per hour.
Note: the specific heat of air at constant pressure is 0.250 Btu/lboF, the density of air is
0.0806 lb/ft3 at atmospheric pressure.
a. 39,000 Btu/hr
c. 36,805 Btu/hr
b. 46,800 Btu/hr
d. 39,900 Btu/hr
Solution:
Q = mCpt
Where:
m = pV
= (0.0806 lb/ft3)(330 ft3/min)
= 26.60 lb/min
Then;
Q = (26.60 lb/min)(0.250 Btu/lboF)(160-60)oF
Q = 665 Btu/min
Thus;
(d) Q = 39,900 Btu/hr
If 1000 liters of air at 27oC and pressure of 1 atm has a mass of 1.115kg and a specific
heat at constant pressure of 1 x 103 J/kg K, how much heat is required to raise the
temperature of this gas from 27oC to 177oC at constant pressure?
a. 176.25 kJ
c. 167.25 kJ
b. 157 kJ
d. 175 kJ
Solution:
Q = mCPt
= (1.115)(1)(177-27)
Thus;
(b) Q = 167.25 kJ
Calculate the amount of energy required to heat the air in a house 30 by 50 by 40 ft from
10 to 70oF. The density of air is about 0.08 lb/ft3, and its specific heat at constant pressure
0.24 Btu/lboF.
a. 49,120 Btu
c. 69,120 Btu
b. 59,120 Btu
d. 79,120 Btu
Solution:
Q = mCpt
Where:
m = pV
= (0.08 lb/ft3)[(30)(40)(50)]ft3
= 4800 lb
Then;
Q = 4800(0.24)(70-10)
Thus;
(c) Q = 69,120 Btu

Water enters the condenser at 20oC and leaves at 35oC. What is the log mean
temperature difference if the condenser temperature is 40oC.
a. 16.37oC
c. 15.37oC
b. 13.37oC
d. 17.37oC
Solution:

Where:
(t)max = 45 20
= 25oC
(t)min = 45 35
= 10oC

25
10
25
10
Thus;
(a) LMTD = 16.37oC
When 200 grams of aluminum at 100oC is dropped into an aluminum calorimeter (k =
0.909 x 103 J/kg-K) of mass 120 grams and containing 150 grams of kerosene at 15oC the
mixture reaches a temperature of 50oC, what is the specific heat of kerosene?
a. 1004 K/kg-k
c. 1110 K/kg-k
b. 1050 K/kg-k
d. 1080 K/kg-k
Solution:
Heat loss by aluminum = heat gained by the kerosene and calorimeter
QA = Qk + Qc
maCata = mkCktk + mcCctc
(0.20)(0.909x103)(100-50) = (0.15)(Ck)(50-15) + (0.12)(0.909x103)(50-15)
Thus;
(a) Ck = 1004.23 J/kg-K
A calorimeter contains 66 kg of turpentine at 10.6oC. When 0.147 kg of alcohol at 75oC is
added, the temperature rises to 25.2oC. the specific heat of turpentine is 1.95x103 J/kg-oC
and the calorimeter is thermally equivalent to 30 grams of water. Find the specific heat
of alcohol.
a. 2.81745 kJ/kgoC
c. 0.81745 kJ/kgoC
b. 3.81745 kJ/kgoC
d. 1.81745 kJ/kgoC
Solution:
Heat loss by alcohol = heat gained by the turpentine and calorimeter
maCata = mtCttt + mcCctc
(0.147)(Ca)(75-25.2) = (0.66)(1.95x103)(25.2-10.6) + (0.03)(4.187)(25.2-10.6)
Ca = 2817.45 J/kgoC
Thus;
(a) Ca = 2.81745 J/KgoC
The temperature of a sample of molten lead near its temperature of solidification is falling
at the rate of 6 K/min. If the lead continues to lose heat at this same rate and takes 35
min. to solidify completely, what is the heat of fusion of the lead? The specific heat of
molten lead is 0.126 kJ/kg-K.
a. 16.46 kJ/kg-K
c. 36.46 kJ/kg
b. 26.46 kJ/kg-K
d. 46.46 kJ/kg
Solution:
Q = mCt = mLf
Lf = Ct
= (0.126 kJ/kg-K)(6K/min)(35min)
Thus;
(b) Lf = 26.46 kJ/kg
A counterflow heat exchanger is designed to heat fuel oil from 45oC to 100oC while the
heating fluid enters at 150oC and leaves at 115oC. Calculate the arithmetic mean
temperature difference.
a. 40oC
c. 60oC
b. 50oC
d. 70oC
Solution:

Where:
(t)max = 115 - 45
= 70oC
(t)min = 150 - 100
= 50oC

70
50
2
Thus;
(a) AMTD = 60oC
A fuel oil of 20oAPI is to be heated in a heater which makes two passes thru heater tubes
and the heating fluid makes one passes but the flow is cross flow through the heater due
to baffles inside the shell.
Quantity of oil to be heated
3000L/hr
Temperature of oil entering heater tubes
21oC
Temperature of oil leaving heater tubes
95oC
Heating fluid, steam enter at 05oC and leaves as condesate at 105oC. Assume specific
heat of oil to be 2.093 kJ/kg-K. Find the heating surface area if the over-all coefficient of
heat transfer is taken as 140 W/m2oC.
a. 24.76 m2
c. 23.75 m2
b. 30.75 m2
d. 32.54 m2
Solution:
Q = UAs(LMTD)
= UAs()
Where:

84
10
84
10
= 43.77oC
Q = moCpto
Solving for mo:
. .
S.G. = 0.934
mo = (3000L/hr)(1kg/L)(0.934)
141.5
131.5
141.5
20 131.5
= 2802 kg/hr
= 0.778 kg/s
From:
Q = (0.778)(2.093)(95-21)
Q = 120.55 kW
Q = 120,550 W
Then:
120,550 (140)(As)(34.77)
Thus;
(a) 24.76 m2
A 30 cm thick wall has an inside and outside surface temperatures of 300oC and 50oC
respectively. If the thermal conductivity of the wall is 8 W/m-K. Calculate the heat
transferred in kW/m2.
a. 6.67
c. 7.67
b. 5.67
d. 8.67
Solution:

8 300 50
0.30
= 6,666.67 W/m2
Thus;
(a) Q/A = 6.67 kW/m2
A 4-pass low-pressure surface type feedwater heater is designed to heat 92,730 kg/hr of
feedwater from 40oC initial to 80oC final temperature using steam bleed at 70 kPa abs.
containing 2,645 kJ/kg enthalpy. Assume no subcooling of condensate, determine the
effective length of 19 mm O.D. x 2 mm thick muntz metal tubes to be installed, if the
water velocity inside the tubes is 1.22 m/s and U = 3000 W/m2K based on the external
surface of the tubes.
a. 2m
c. 3m
b. 4m
d. 5m
Solution:
A=
. .
A=
0.019
Solving for LMTD or :

50
10
50
10
= 24.85oC
Solving for A:
By: energy balance:
ms(h1 h2) = mwCpw(t2 t1)
ms(2645 376.7) = 92,730(4.187)(80 40)
ms = 6846.72 kg/hr
Q = ms(h1 h2)
= 1.90(2645 376.7)
Q = 4,314.01 kW
Q = 4,314,010 W
Then;

4,314,010
3000 24.85
= 57.87 m2
Solving for n:
.
4
92,730
1000 3600
n = 477.91 pcs say 480 pcs
Thus;
A = 0.019
57.87 = 0.019
480
(a) L = 2.02 m tube length
0.015
4
1.22

Find the thermal conductivity of the 500 cm thick material with an area of 50,000
cm2 and a temperature difference of 10 K if the heat transmitted during 2 hours
test is 2000 KJ.
a. 0.014 W/m-K
c. 0.126 W/m-K
b. 0.025 W/m-K
d. 0.214 W/m-K
Solution:
Q =
kAt
x
10,000
k
10
2000
100 2

2 3600
0.500
Thus;
(a) k = 0.014 W/m-K

A pipe with an outside diameter of 2.5 in. is insulated with a 2 in. layer of asbestos
(ka = 0.396 Btu-in./hr-ft2-oF), followed by a layer of cork 1.5 in. thick (kc = 0.30 Btuin./hr-ft2-oF). If the the temperature of the outer surface of the cork is 90oF, calculate
the heat lost per 100 ft of insulated pipe.
a. 847.64 Btu/hr
c. 2847.42 Btu/hr
b. 3847.51 Btu/hr
d. 1847.14 Btu/hr
Solution:
Q
Thus;
t 1 t 3
r
r2
ln 3
ln
r1
r2
2k a L 2k c L
29090
4.75
3.25
ln
ln
3.25
1.25
0.396
0.30
100 2
100
2
12
12
c Q 2847.42 Btu/hr

At $ 0.25 per kW-hr, how much will it cost to maintain a temperature of 95oF for 24
hours in a box 2ft square on each side if the outside temperature is 72oF and the
over-all heat transfer coefficient for the box is 0.10 Btu/hr-ft2-oF ?
a. P 0.10
c. P 0.15
b. P 0.20
d. P 0.25
Solution:
Area of One side:
A 2 2 4 ft 2
Heat transferred through one side:
Q AU t 1 t 2
Q 4 0.1 9672
Q 9.6
Btu
hr
Total Heat Transferred through 6 sides in 24 hours:

Q 9.6 6 24
Q 1382.4 Btu
The Cost to maintain:
C
1382.4 $ 0.25
3412.75
Thus;
a C $ 0.10
A steam pipe having a surface temperature of 300oC passes through a room where
the temperature is 25oC. The outside diameter of pipe is 100 mm and emissivity
factor is 0.60. Determine the radiated heat loss for a 5 m pipe length.
a. 5.34 kW
c. 3.34 kW
b. 4.34 kW
d. 6.34 kW
Solution:
Q 20,408.4 x 108 AFe T1 4 T2 4
where:
A DL
T1 300 273 573 K
A 0.10 5
T2 25 273 298 K
A 1.57 m2
Then;
Q 20,408.4 x 108 1.57 0.60
Q 19,208,138
573 4 298
1hr
J
hr 3600s
J
Q 5,335.59 or W
s
Thus;
a Q 5.336 kW

An air-cooled condenser has an expected U value of 30 W/m2-K based on the air
side area. The condenser is to transfer 60 kW with an airflow rate of 15 kg/s entering
at 35oC. If the condensing temperature is to be 48oC, what is the required air-side
area?
a. 184 m2
c. 174 m2
b. 194 m2
d. 164 m2
Solution:
Q A U
Solving for :

max
t
ln
t
max
from:
min
min
Q mCp t
60 15 1 t
t 4 K
t 2 t 1 4 K
t 2 39 K
t
max
48 35 13C
min
48 39 9C
Then;

13 9
10.88C
13
ln
9
60,000 A 30 10.88
Thus;
a 183.82 m2

An air-cooled condenser is to reject 40 kW of heat from a condensing refrigerant
to air. The condenser has an air-side area of 210 m2 and a U value based on this
area of 0.037 kW/m2-K; it is supplied with 6.6 m3/s of air which has a density of 1.15
kg/m3. If the condensing temperature is to be limited to 55oC, what is the maximum
allowable temperature of the inlet air?
a. 30.7 oC
c. 50.7 oC
b. 40.7 oC
d. 20.7 oC
Solution:
max
t
ln
t
max
min
min
Solving for :
Q A U
70 210 0.037
9.018 K
from:
Q m Cp t
70 6.6 1.15

1.02 t 2 t 1
9.04
Then;
9.01
9.01
9.04
55 t 1
ln
55 t 2
9.04
55 t 1
ln
55 9.04 t 1
Thus;
b t 1 40.7 C

Two walls are composed of 150 mm thick insulating material at the outer layer (k =
0.139 W/m-K) and 300 mm thick material at the inner layer (k = 1.111 W/m-K).
Calculate the heat transmitted per m2 if the surface temperature of the cold side
and hot side are 25oC and 300oC respectively.
a. 203.79 W/m2
c. 254.65 W/m2
b. 303.79 W/m2
d. 354.65 W/m2
Solution:
t
Q

RT
A
where:
RT
0.30
0.15

1.11 0.139
R T 1.35
Q 30025

1.35
A
Thus;
Q
a 203.79 W/m2
A

The temperature of the flame in a furnace is 1277oC and the temperature of its
surrounds is 277oC. Calculate the maximum theoretical quantity of heat energy
radiated per minute per square meter to the surrounding surface area.
a. 19,321.65 kJ
c. 17,321.65 kJ
b. 18,321.65 kJ
d. 16,321.65 kJ
Solution:
Q 20408.4 x 108 Fe T1 4 T2 4
where:
Fe = 1 (if not given)
T1 = 1277 + 273 = 1550 K
T2 = 277 +273 = 550 K
Q 20408.4 x 108 1
1550 4 550
Q 1.1593 x 109 J/m2-K

thus;
a Q 19,321.65 kJ/m2-min

A small sphere has a radius of 3.50 cm and is maintained at a temperature of
360oC. Assuming it to be a black body surrounded by empty space, how much
energy does it radiate each second?
a. 30 J
c. 40 J
b. 35 J
d. 45 J
Solution:
Q 20408.4 x 108 Fe T1 4 T2 4
where:
T1 = 360 + 273 = 633 K
A = ( 0.035)2
A = 0.003848 m2
Q 20408.4 x 108 1 0.003848 633
Q 126,083.68 J/hr
Q 35.02 J/s
Thus;
b Q 35 J

The inner wall of a thermos bottle is at 0oC while the outer at 37oC. The space
between the walls is evacuated and the walls are silvered so the emissivity is
reduced to 0.10. It each wall has an area of 700 cm2, how much energy is
transformed by radiation between the walls each second?
a. 1.46 J
c. 16.5 J
b. 1.04 J
d. 17.03 J
Solution:
Q 20408.4 x 108 Fe T1 4 T2 4
where:
Fe = 0.10
A = 700 cm2 = 0.07 m2
T1 = 37 + 273 = 310 K
T2 = 0 + 273 = 273 K
Substituting Values:
Q = 5,258.116 J/hr
Q = 1.46 J/s
Thus;
(a) Thus; the energy transferred by the radiation between the walls each
second is 1.46 J

The hot gas temperature in a heat exchanger is 350oC (ho = 220 W/m2-K). What is
the surface temperature on the wall if the heat transferred is 1500 W/m2?
a. 3500C
c. 3430C
b. 3380C
d. 3580C
Solution:
Q
ho t
A
1500 = 220 (350 t1)
Thus;
c t 1 343.18o C
84 ME Board Problem
An oil heater heats 100 kg per minute of oil from 35oC to 100oC in a counterflow
heat exchanger. The average specific heat of the oil is 2.5 kJ/kgoC. Exhaust gases
used for heating enter the heater with an average specific heat of 1 kJ/kgoC, a
mass flow rate of 250 kg/min and an initial temperature of 200oC. The over-all heat
transfer coefficient is 75 W/m2oC. Determine the heating surface in m2.
a. 36.110C
c. 32.720C
b. 41.720C
d. 25.340C
Solution:
Q A U
Solving for total heat transferred Q and :
Qoil = Qgas
moCpo to = mgCpg tg
100(2.5)(100 35) = 250(1)(200 t)
t = 1350C
(t)max = 135 - 35 = 1000C
(t)min = 200 100 = 1000C
If (t)max = (t)min , then = average value = 1000C
Q = mCpt
Q
100
2.5 10035
60
Q 270.83 kW
Thus;
270.83 = A (0.075) (100)
a A 36.11 m2
If total resistance to heat flow of a composite wall is 3.0875 m2-K/W. What is the
over-all transfer coefficient of the wall?
a. 0.324 W/m2-K
c. 0.243 W/m2-K
b. 0.423 W/m2-K
d. 0.243 W/m2-K
Solution:
U
1
1

3.0875
RT
Thus;
(a) U = 0.324 W/m2-K

In a composite vertical furnace wall, the resistance due to insulating brick is 0.5769
oC/W.
What is the total resistance to heat flow if the percent of the total resistance
due to insulating brick is 96.80%.

a. 0.597 0C/W
c. 0.975 0C/W
b. 0.795 0C/W
d. 0.957 0C/W
Solution:
% brick
0.9680
R brick
RT
0.5769
RT
Thus;
(a) RT = 0.597 0C/W

A counterflow bank of boiler tubes has total area of 900 ft2 and its over-all efficiency
of heat transfer is 13 Btu/hr-ft2-oF. Calculate the heat transferred if the log mean
temperature difference is 1380oF.
a. 16,146,000 Btu/hr
c. 18,148,000 Btu/hr
b. 17,147,000 Btu/hr
d. 15,145,000 Btu/hr
Solution:
Q A U
Q 900 13 1380
Thus;
(a) Q = 16,146,000 Btu/hr

Calculate the quantity to be transferred to 3.25 kg of brass to raise its temperature
from 30oC to 250oC taking the specific heat of the brass as 0.394 kJ/kg-K.
a. 182 kJ
c. 151 kJ
b. 282 kJ
d. 251 kJ
Solution:
Q m Cp t
Q 3.25 0.394 25030
Thus;
(b) Q = 281.712 kJ

The mass of the copper calorimeter is 0.28 kg and it contains 0.4 kg of water at
15oC. Taking the specific heat of copper as 0.39 kJ/kg-K, calculate the heat
required to raise the temperature to 20oC?
a. 6.92 kJ
c. 8.92 kJ
b. 7.92 kJ
d. 9.92 kJ
Solution:
Water equivalent of calorimeter:
m 0.28
0.39
4.187
m 0.026 kg
Heat received by the water and calorimeter
Q m mw Cp t
Q 0.026 0.40 4.187 2015
Thus;
(c) Q = 8.92 kJ
(d)
In an experiment to find the specific heat of lead, a 0.50 kg of lead shot at a
temperature of 51oC is poured into an insulated calorimeter containing 0.25 kg of
water at 13.5oC and the resultant temperature of the mixture is 15.5oC. If the water
equivalent of the calorimeter is 0.020 kg, find the specific heat of the lead.
a. 0.1278 kJ/kg-K
c. 0.01389 kJ/kg-K
b. 0.0278 kJ/kg-K
d. 0.0389 kJ/kg-K
Solution:
Heat lost by the lead = Heat gained by the water and calorimeter
mL CL t L mw Cw t w mC CC t C
0.5 CL 51 15.5 0.25 0.02 4.187 15.513.5
thus;
a CL 0.1278 kJ/kg-K
max
55 t 1
min
55 t 2
max
min
t 2 t 1 9.04

The load on a water-cooled condenser is 90,000 Btu/hr. If the quantity of water
circulated through the condenser is 15 gpm, determine the temperature rise of the
water in the condenser.
a) 12F
c) 16F
b) 14F
d) 18F
Solution:
Q = mCpt
90,000
Btu
hr
=m
Btu
lb F
Where:
m = V
= 8.33
lb
gal
= 7497
15
gal
60 min
min
hr
lb
hr
Thus;
90,000 = 7497 1 t
(a) t = 12F

Thirty-six gallons of water per minute are circulated through a water-cooled
condenser. If the temperature rise of the water in the condenser is 2F, compute the
load on the condenser in Btu/hr.
a) 216,000 Btu/hr
c) 217,000 Btu/hr
b) 215,000 Btu/hr
d) 218,000 Btu/hr
Solution:
Q = mCpt
Where:
m = 36
gal
min
8.33
= 17992.8 lb/hr
Thus:
60 min
hr
Q =17992.8(1) (12)
(a) Q = 215,913.6 Btu/hr

The load on an air-cooled condenser is 121,500 Btu/hr. If the desired temperature of
the air in the condenser is 25F, determine the air quantity in cfm that must be
circulated over the condenser.
a) 4500 cfm
c) 5500 cfm
b) 3500 cfm
d) 3000 cfm
Solution:
Q = mCpt
Where:
Q = 121,500 Btu/hr
Cp= 0.24 Btu/F
t = 25F
Then:
121,500 = m(0.24)(25)
m = 20,250 lb/hr
the volume flow rate is:
V=
20,250 lb/hr
0.075 lb/ft
V = 270,000 ft3/hr = 4500 ft3/min

Thus;
(a) V = 45000 cfm

Three thousand cubic feet per minute of air are circulated over an air-cooled
condenser. If the load on the condenser iss 64,800 Btu/hr , compute the temperature
rise of the air passing over the condenser.
a) 16F
c) 20F
b) 18F
d) 22F
Solution:
Q = mCpt
Where:
Q = 64, 800 Btu/hr
m = V
m = 0.075 lb/ft3 (3000 ft3/min) (60min/hr)
m = 13,500 lb/hr
Cp = 0.24 Btu/lb F
Thus;
6400 = 13,500 (0.24) t
(c) t = 20F

The weight of ammonia circulated in a machine is found to be 21.8 lb/hr. If the vapor
enters the compressor with a specific volume of 9.6 ft3/lb , calculate the piston
displacement, assuming 80% percent volume efficiency.
a) 261.6 ft3/hr
c) 281.8 ft3/hr
b) 271.6 ft3/hr
d) 291.6 ft3/hr
Solution:
Actual volumetric efficiency =
Volume flow rate at suction

Piston displacement
eva = V1/VD
0.80 =
21.8(9.6)
VD
VD = 261.6 ft3/hr
Thus;
(a) VD = 261.6 ft3/hr

A single-stage ammonia compressor is producing 10 tons of refrigeration and the
power consumed is 15 Hp. Suction pressure is 25 psi, condensing pressure is 180 psi.
Brine temperature is 20F off brine cooler. Determine the actual coefficient of
performance.
a) 10.14
c) 12.14
b) 11.14
d) 13.14
Solution:
COP =
=
Refrigeration Capacity
compressor power
10 3.516
15 0.746
= 13.14
Thus;
(d) COP = 13.14

In an ammonia condensing machine (compressor plus condenser) the water used for
condensing is 55F and the evaporator is at 15F. Calculate the ideal COP.
a) 11.875
c) 10.875
b) 12.875
d) 13.875
Solution:
COP =
T1
T2 T1
Where:
T1 = 15 + 460 = 475 R
T1 = 55 + 460 = 515 R
Then;
COP =
475
515 475
Thus;
(a) COP = 11.875
Calculate the specific volume of an air-vapor mixture in cubic meters pre kilogram of
dry air when the following conditions prevail : t = 30C, w = 0.015 kg/kg, and Pt = 90
kPa.
0.99 m3/kg
c) 0.79 m3/kg
b) 0.89 m3/kg
d) 0.69 m3/kg
a)
Solution:
v=
RaT
Pt Pv
Solving for Pv:

w = 0.622
Pv
Pt Pv
0.015 = 0.622
Pv
90 Pv
Pv = 2.12 kPa
Thus;
v=
(0.287) (30+273)
90 - 2.12
(a) v = 0.99 m3/kg
Compute the Humidity Ratio of air at 62 % relative humidity and 34C when the
barometric pressure is 101.325 kPa.
a) 0.021 kgvapour/kgdry air
c) 0.041 kgvapour/kgdry air
b) 0.031 kgvapour/kgdry air
d) 0.051 kgvapour/kgdry air
Solution:
w = 0.622
Pv
Pt Pv
Solving for Pv:

Psat @ 34C = 5.32 kPa
Pv = (RH) Psat
= (0.62) (5.32)
= 3.30 kPa
Then:
w = 0.622
3.30
101.325 3.30
Thus;
(a) w = 0.021 kgvapour/kgdry air
A sample of air has dry-bulb temperature of 30C and a wet-bulb temperature of
25C. The barometric pressure is 101.325 kPa. Calculate the enthalpy of the air if it is
adiabatically saturated.
a) 75.94 kJ/kg
c) 79.54 kJ/kg
b) 70.94 kJ/kg
d) 74.09 kJ/kg
Solution:
Enthalpy of air if adiabatically saturated:
h = Cpt + whg
Solving for humidity ratio, w:
w = 0.622
Pv
Pt Pv
From steam table, at 25C :

Pv = 3.17 kPa, hg = 2547.2 kJ/kg
w2 = 0.622
3.17
101.325 3.17
= 0.02 kgvapour/kgdry air

Thus;
h2 = 1.0 kJ/kg-C (25C) + (0.02) (2547.2 kJ/kg)
(a) h2 = 75.94 kJ/kg
An air-vapor mixture has dry bulb temperature of 30C and a humidity ratio of 0.015.
Calculate the enthalpy at 85 kPa barometric pressure.
a) 68.34 kJ/kg
c) 72.45 kJ/kg
b) 54.35 kJ/kg
d) 67.45 kJ/kg
Solution:
h = Cpt + whg
From steam table, at 30C :
h = 2556.3 kJ/kg
then,
h = (1 kJ/kg-C) (30C) + (0.015) (2556.3 kJ/kg)
Thus;
(a) h = 68.34 kJ/kg
In an air conditioning unit, 3.5 m3/s of air 27C dry-bulb temperature 50 % relative
humidity and standard atmospheric pressure enters the unit. The leaving condition of the
air is 13C dry-bulb temperature and 90% relative humidity. Using the properties from the
psychrometric chart, calculate the refrigerating capacity in kW.
a) 87.57 kW
c) 57.87 kW
b) 77.57 kW
d) 58.77 kW
Solution:
Refrigerating Capacity QA:
QA = m (h2 - h1)
Solving for m:
m=
=
Volume Flow Rate

Ave. Specific Volume
3.5
1/2(0.866 0.822)
= 4.15 kg/s
Thus;
QA = 4.15 (55.3 - 34.2)
(a) QA = 87.57 kJ or kW
A stream of outdoor air is mixed with a stream of return air in an air conditioning
system that operates at 101 kPa pressure. The flow rate of outdoor system air is 2 kg/s
and its condition is 35C dry-bulb temperature and 25C wet-bulb temperature. The
flow rate of return air is 3 kg/s and its condition is 24C and 50% relative humidity.
Determine the enthalpy of the mixture.
a) 91.56 kJ/kg
c) 91.56 kJ/kg
b) 91.56 kJ/kg
d) 91.56 kJ/kg
Solution:
By Energy Balance:
m1h1 + m2h2= m3h3
m1h1 + m2h2 = ( m1+ m2 ) h3
h3 =
2 (75.9) 3 (48)
2+3
Thus;
(d) h3 = 59.16 kJ/kg
What is the specific volume of an air-vapor mixture at 30C and a relative humidity of
45C at 101.325 kPa.
a) 0.578 m3/kg
c) 0.875 m3/kg
b) 0.785 m3/kg
d) 0.758 m3/kg
Solution:
v=
RaT
Pt Pv
Solving for Pv :
Psat @ 30C = 4.24 kPa
Pv = (RH) Psat @ 30C
= 0.45 (4.24)
= 1.91 kPa
Thus;
v=
0.287(30+273)
101.325 - 1.91
(c) v = 0.875 m3/kg
A mixture of dry-air and water vapor is at temperature of 21C under a pressure of 101
kPa. The dew point temperature is 15C. Calculate the relative humidity.
a) 68.56 %
c) 56.68 %
b) 65.68 %
d) 58.66 %
Solution:
RH =
Pv
Psat @ 21C
Where:
PV = Psat @ 15C
= 1.7044 kPa
Psat @ 21C = 2.4861 kPa
Thus;
RH =
1.7044
2.4861
= 0.6851
(b) RH = 68.56%
The density of air at 35C and 101 kPa is 1.05 kg/m3. The humidity ratio is:
a) 0.036 kgvapour/kgdry air
c) 0.36 kgvapour/kgdry air
b) 0.063 kgvapour/kgdry air
d) 0.63 kgvapour/kgdry air
Solution:
PV = mRT
P = mRT/V
Pair = 1.08 ( 0.287 )( 35 + 273 )
Pair = 95.48 kPa
Pt = Pair + Pvapor
101 = 95.48 + Pv
Pv = 5.53 kPa
Then;
w = 0.622
= 0.622
Pv
Pt Pv
5.53
101 5.53
Thus;
(a) w = 0.036 kgvapour/kgdry air
If the sensible heat ratio is 0.80 and the cooling load is 100 kW, what is the amount of
sensible heating?
a)
80 kW
c) 125 kW
b) 60 kW
d) 100 kW
Solution:
SHR = sensible heat ratio
=
Qs
Qs QR
0.80 = Qs/100
Thus;
(a) Qs = 80 kW

A 4m x 4m x 4m room has a relative humidity ratio of 80 %. The pressure in the room is
120 kPa and temperature is 35 C (Psat = 5.628). What is the mass of vapor in the room.
Use Rvapor = 0.4615 kN-m/kg-K.
a) 3.03 kg
c) 4.03 kg
b) 2.03 kg
d) 5.03 kg
Solution:
PvV = mvRvT
Solving for Pv :
RH = Pv / Psat
0.08 = Pv / 5.628
Pv = 4.5024 kPa
Thus;
4.5024 [(4)(4)(4)] = mv (0.4615) (35 + 273)
(a) mv = 2.027 kg
45. The bore and stroke of an air compressor are 276 mm and 164 mm respectively. If the
piston displacement is 0.039 m /s, what is the operating speed of the compressor?
a. 238.49 rpm
c. 338.49 rpm
b. 261.54 rpm
d. 361.54 rpm
Solution:
Vd =
d L N
4
0.039 =
(0.276) 0.164 N
4
N = 3.975 rev/s = 238.49

Thus;
(a) N = 238.49 rpm
46. A turbo-compressor is a gas turbine plant is used to compress 10 kg/s air from an initial
pressure of 102 kpa to a discharge pressure of 622 kpa, with inlet and discharge
temperature measured at 297 K and 527 K respectively. The compressor inlet pipe is 50
cm ID and the discharge pipe is 20 cm ID. Find the inlet discharge velocities of air.
a. 42.56 m/s, 77.40 m/s
c. 45.45 m/s, 57.45 m/s
b. 34.72 m/s, 76.56 m/s
d. 54.45 m/s, 56.7 m/s
Solution:
Velocity of air at suction:
V
Q
s= s
As
Velocity of air in at discharge:

V
d=
Qd
Ad
solving for Q:
Qs= V
mRT1
P1
s =
10 (.287)(297)
102
=8.357 m3 /s
Qd= V
d=
mRT2
P2
10 (.287)(527)
622
2.43 m3 /s
then;
V
s=
8.357 m3 /s
=42.56 m/s
2
(.5)
4
2.43 m3 /s
d=
=77.40 m/s
2
(.20)
4
Thus;
(a) Vs=42.56m/s ,
Vd=77.40 m/s
47.
The initial condition of air in an air compressor is 100 kpa and 25
and
discharges air at 450 kpa. The bore and stroke are 276 mm and 186 mm
respectively with 8 % clearance running at 6 rev per second. Find the volume of
air at suction.
a. 203.39 m /hr
c. 261.25 m /hr
b. 303.39 m /hr
d. 361.25 m /hr
Solution:
Solving for e :
ev =1+c-c (
P1 1/n
)
P2
=1+0.08-0.08 (
450 1/1.4
100
=0.84575
Vd= D2 LN= (0.276)2 (0.186)(6)
4
=0.0668 m3 /s
Then;
V1=0.84575(0.0668 )
=203.39 m3 /hr
48. If the power to drive shaft is 7 hp and the mechanical efficiency is 75 %, what is the
actual compressor power?
a. 5 hp
c. 2 hp
b. 3 hp
d. 4 hp
Solution: e
Compressor power
Power to drive the shaft
0.74 =
Compressor power
7
Thus;
(a) Compressor Power = 5.18 hp
49.A two stage compressor receives 0.35 kg/s of air at 100 kpa and 629 K and delivers it
at 1000 kpa. Find the heat transferred in the intercooler?
a. 70.49 kw
c. 90.49 kw
b. 80.49 kw
d. 100.49 kw
Solution:
Q
mCp (Tx - T1
Solving for Tx :
Px =
Tx
T1
=(
Tx
269
100 5000 = 707.11 kpa
Px
P1
=(
k-1
k
707.11
100
1.4-1
1.4
Tx =470.40 K
Thus;
Q=0.35(1)(470.40 K- 269 K)
Q=70. 49 kw
50.An air compressor which operates at 900 rpm has a piston displacement of 4500cm .
Determine the mass flow rate of air standard density considering that the volume
efficiency is 77 %.
a. 224. 53 kg/hr
c. 314. 57 kg/hr
b. 324.35 kg/hr
d. 137.54 kg/hr
Solution:
ev =
V1 '
VD
0.77=
V1 '
4500
V'1 =3465 cm3

Then;
V'1 =3465 cm3 900 =3,118,500 cm3 /min
At standard temperature and pressure: air =1.2 kg/m3
m=1.2
kg
cm3
1m3
60 min
3118500
3
min (100)3 cm3
hr
m
Thus;
(a) m=224.53 kg/hr
1.A fuel is delivering 10 gallons per minute of oil with a specific gravity of 0.83. The total
head is 9.14m; find how much energy the pump consumes in KJ per hour.
a. 169
c. 189
b. 199
d. 179
Solution;
P= QH
Where:
=9.81 .083 =8.14 KN/m3
Q=10 gal/ min =2.27 m3 /hr
H=9.14 m
Then;
P=8.14 2.27 9.14
Thus;
(a) P=168.89
kj
hr
2. A pump receives 8 kgs of water at 220 kpa and 110
and discharges it at 1100 kpa.
Compute the power required in kw.

a. 8.126
c. 7.014
b. 5.082
d. 6.104
Solution:
Let : P=power in kw
P=Q(Pd - Ps )
Where:
Q= 8
kg
s
0.001
m3
= 0.008 m3 /s
kg
Pd =1,100 kpa
P=(0.008)(1100-220)
Thus;
(b) P=7.04 kw
3.A pump lifts water at a rate of 283 liters per second from alake and force it into a tank 8
m above the level of the water at a pressure of 137 kpa. What is the power required in
kw?
a. 71
c. 61
b. 41
d. 51
Solution:
P= QH
Where:
=9.81 kN/m3
Q=
283L
=0.283 m3 /s
s
H=8+
137
9.81
H=21.97
Thus;
P=9.81 0.283 21.97
(c) P=60.99
4.A pump discharges 150 liters per second of water to a height of 75 m. if the efficiency is
75 % and the speed of the pump is 1800 rpm, what is the torque in the N-m to which the
drive shaft is subjected?
a. 771
c. 791
b. 781
d. 681
Solution:
Let: ep =pump efficiency
P=
QH
2TN
or P=
ep
60
Where;
N=1800 rpm
P=
QH 9.81(0.150)(75)
=
=147.15 kw
ep
0.75
Then;
147.15=
2T(1800)
60
T=0.781 kN-m
Thus;
(b) T=781 N-m
5.A centrifugal pump delivers 80 liters per second of water on test suction gauge reads
10 mm hg vacuum and 1.2 m below pump center line. Power input is 70 kw. Find the
total dynamic head in meters.
a. 66
c. 62
b. 60
d. 64
Solution:
=9.81 kN/m3
Q=0.80 m3 /s
P=0.74 70 =51.80
Note: 74% is the usual pump efficiency used if not given.
Thus;
51.80=9.81 0.80 H
(a) H=66 m
6.A pump with a 400 mm diameter suction pipe and a 350 mm diameter discharge pipe
is to deliver 20,000 liters per minute of 15.6
water. Calculate the pump head in meters
if suction gage is 7.5 cm below the pump centerline and reads 127 mm Hg vacuum and
discharges gage is 45 cm above the pump centerline and reads 75 kpa.
a. 15 m
c. 20 m
b. 5 m
d. 10 m
Solution:
H=total dynamic head
H=
Ps -Ps (Vd 2 -Vs 2 )

+
+Zd -Zs
2g
Q=20,000
Vs =
Q
2
li
=0.33 m/s
min
=
0.33
2
d /4 (0.4) /4
=2.63 m/s
Vd =
0.33
2
(0.35) /4
=3.43 m/s
Thus:
H=
75-(-16.93) (3.432 -2.632 )

+
+0.45+0.075
9.81
2(9.81)
(d) H=10.14 m
7.A centrifugal pump delivers 300,000 liters per hour of water to a pressurized tank whose
pressure is 284 kpa. The source of water is 5 meters below the pump. The diameter of the
suction pipe is 300 mm and the discharge pipe is 250 mm. calculate the kw rating of the
driving motor assuming the pump efficiency to be 72%.
a. 41.75 kw
c. 43.28 kw
b. 35.23 kw
d. 38.16 kw
Solution:
Let: Pbrake =brake input power
Pbrake =
QH
ep
Where:
Q=300,000
li
=0.0833m3 /s
hr
Solving for H:
From: Bernoullis Equation:
H=

+
+Zd -Zs
2g
where:
Vs =
0.0833
2
(0.3) /4

=1.18 m/s
0.0833
0.25 /4
1.7 m/s
Then;
2
280-0 (1.70) -(1.18)

+
+5-0
H=
2(9.81)
9.81
H=33.62 m
Thus;
Pbrake =
9.81(0.0833)(33.62)
0.72
(d) Pbrake = 38.16 kw

8.A pump delivers 500 gpm of water against a total head of 200 ft and operating at 1770
rpm. Changes have increased the total head to 375 ft. at what rpm should the pump be
operated to achieved the new head at the same efficiency.
a. 2800 rpm
c. 3434 rpm
b. 3600 rpm
d. 2424 rpm
Solution:
H1 N1 2
=( )
H2 N2
200
375
1770
N
Thus;
(e) N2 =2423.67 rpm
9.The rate of flow of water in a pump installation is 60.6 kg/s. the intake static gage is 1.22
m below the pump centerline and reads 68.95 kpa gage ; the discharge static gage is
0.61 m below the pump centerline and reads 344.75 kpa gage. The gages are located
lose to the pump as much as possible. The areas of the intake and discharge pipes are
0.093 m and 0.069 m respectively. The pump efficiency is 74 %. Take the density of
water equals 1000 kg/m . What is the hydraulic power in kw?
a. 17.0
c. 31.9
b. 24.5
d. 15.2
Solution:
Pwater =Phydraulic =QH
Where:
Q=
60.6 kg/s
1000 kg/m3
Q=0.0606 m3 /s
Vs =
0.0606
Vd =
0.0606
0.093
0.069
=0.65 m/s
=0.88 m/s

H=

+
+Zd -Zs
2g
H=
344.75-68.95 (0.88) -(0.65) )

+
+(-0.61+1.22)
9.81
2(9.81)
H=28.74 m
Thus the hydraulic power is:
Phydraulic = 9.81 (0.0606) (28.74)
Phydraulic =17.09 kW
10.It is desired to deliver 5 gpm at a head of 640 ft in a single stage pump having a
specific speed not to exceed 40. If the speed is not exceeding 1352 rpm how many
stages are required?
a. 3
c. 5
b. 4
d. 2
Solution:
Let; n=no. of stages
h=head per stage
then,
h=
640
n
from;

Ns =
40=
NQ
h3/4
13525
640 3/4
( n )
Thus, n=2 stages

11. The power output is 30 Hp to a centrifugal pump that is discharging 900 gpm and
which operates at 1800 rpm against a head H = 120 ft, 220 V, 3 phase, 60 Hertz. If this
pump is modified to operate 1200 rpm, assuming its efficiency remains constant,
determine its discharge in gpm, the theoretical head it imparts to the liquid and the
power input to the pump.
a. 600 gpm, 53.33 ft, 8.89 Hp
c. 500 gpm, 50.33 ft,7.89 Hp
b. 700 gpm, 63.33 ft, 9.89 Hp
d. 650 gpm, 53.33 ft, 8.95 Hp
Solution:
Solution for the discharge, Q :
Q1 N1
=
Q2 N2
900 1800
=
Q2 1200
Q2 =600 gpm
Solution for the theoretical head, H :
H1 N1 2
=( )
H2 N2
120 1800 2
=(
)
1200
H2
H2 =53.33 ft
Solving for the Power input, P :
P1 N1 3
=(
)
P2 N2
3
30 1800
=(
)
P2 1200
P2 =8.89 Hp
Thus;
(a) Q2 =600gpm , H2 =53.33 ft , P2 =8.89 Hp
12. A pump operating at 1750 rpm delivering 500 gal/min against a total head of 150 ft.
Changes in the piping system have increased the total head of 360 ft. At what rpm
should the pump be operated to achieve this new head at the same efficiency?
a. 2730 rpm
c. 2711 rpm
b. 2740 rpm
d. 2600 rpm
Solution:
H1 N1 2
=( )
H2 N2
2
150 1750
=(
)
360
N2
Thus;
N2 =2711.09 rpm
13. Water in the rural areas is often extracted from underground water source whose free
surface is 60 m below ground level. The water is to be raised 5 m above the ground by a
pump. The diameter of the pipe is 10 cm at the inlet and 15 cm at the exit. Neglecting
any heat interaction with the surroundings and frictional heating effects. What is the
necessary power input to the pump in kW for a steady flow of water at the rate of 15 li/s?
Assume pump efficiency of 74 %.
a. 9.54
c. 7.82
b. 5.54
d. 12.90
Solution:

Input Power=
Input Power=
where:
Water Power
Pump Efficiency
QH
ep
Q = 15 li/s
Q = 0.05 m3
Vs =
0.015
2
(0.10) /4
Vs =1.91 m/s
Vd =
0.015
2
(0.015) /4
Vd =0.85 m/s
H=
Vd 2 -Vs 2
+Zd -Zs
2g
H=
(0.85) -(1.91)
+ 5-(-60)
2(9.81)
H = 64.85
Thus;
Input power =
9.81(0.015)(64.85)
0.74
(d) Input Power = 12.896 kW
14. Past Board Problem

Find the hydraulic horsepower and the mechanical efficiency of a rotary pump direct
connected to a 5 Hp electrical motor operating at full load under the following
conditions:
Fluid handled
oil
Temperature
21C
Specific gravity
0.85
Volume flow rate
20 li./sec
Total Head
175 kPa
a. 4.69 Hp , 94%
c. 5.69 , 74%
b. 3.69 Hp , 84%
d. 6.69 , 78%
Solution:
Hydraulic Power of the Pump :
Phydraulic = Q H
where:
= 9.81(0.85) = 8.34 kN/m3
Q = 20 li/s = 0.020 m3/s
H = =
H = 20.98 m
then ;
Phydraulic = (8.34) (0.020) (20.98)
= 3.50 kW
Phydraulic = 4.69 Hp
ep =
ep = 0.94 = 94%
thus;
(a) Phydraulic = 4.69 Hp , ep = 94%
15. Past Board Problem
Water from an open reservoir A at 8 m elevation is drawn by a motor driven pump to an
open reservoir B at 70 m elevation. The inside diameter of the suction pipe is 200 mm and
150 mm for the discharge pipe. The suction line has a loss of head three times that of the
velocity head in the 200 mm pipe. The discharge line has a loss of head twenty times that
of the velocity head in the discharge pipeline. The pump centerline is at 4 m. Overall
efficiency of the system is 78 %. For the discharge rate of 10 li/s, find the power input to
the motor and the pressure gage readings installed just at the outlet and the inlet of the
pump in kPag.
a. 7.82 kW, 39 kPa, 650 kPa
c. 6.82 kW, 35 kPa, 550 kPa
b. 8.82 kW, 40 kPa, 680 kPa
d. 5.82 kW, 30 kPa, 500 kPa
Solution:
Power input of the motor:
Pinput =
where:
Q = 0.010 m3/s
Vd =
.
.
Vd = 0.565884242 m/s
.
Vs =
Vs = 0.318309886 m/s
H=
+Z
H=0+
+ 20
]+3
] + 64 4
H = 62.3530768 m
thus;
.
Pinput =
Pinput = 7.842098505 kW
Pressure Gages Readings:
Ps = Hs = 9.81(3.99)
Ps = 39.14 kPa
Pd = Hd = 9.81(66.34)
Pd = 650.80 kPa

A pump adds 167.6 m of pressure head to 45.43 kg/s of water. What is the hydraulic
power in kW.
a. 64.69 kW
c. 66.54 kW
b. 74.69 kW
d. 76.54 kW
Solution:
P=QH
where:
Q=
/
/
= 0.04543 m3/s
P = 9.81 (0.04543) (167.6)

thus;
(b) P = 74.69 kW

A pump driven by an electrical motor moves 25 gal/min of water from a reservoir A to B,
lifting the water to a total head of 245 ft. The efficiency of the pump is 64%. Neglecting
velocity head, friction, and minor losses. What size motor is required?
a. 2.64 Hp
c. 1.55 Hp
b. 2.55 Hp
d. 1.64 Hp
Solution:
P=Qh
= (8.33 lb/gal) (25 gal/min.) (245 ft)
= 51,021.25
thus;
(a) P = 1.55 Hp
A centrifugal pump is powered by a direct drive induction motor is needed to discharge
150 gal/min against a total head of 180 ft when turning at fully loaded speed of 3500
rpm. What type of pump should be selected?
a. Radial
c. Mixed Flow
b. Francis
d. Propeller
Solution:
Ns =
=
= 872.286 rpm
thus;
(a) Radial turbine

A boiler feed pump receives 40 L/s at 4 MPa and 180C. It operates against a total head
of 900 m with an efficiency of 60%. Determine the power output of the driving motor.
a. 450.21 kW
c. 500.21 kW
b. 459.64 kW
d. 523.26 kW
Solution:
Pmotor =
Solving for Pwater:
From Steam table:
At 4 Mpa and 180C
h1 = 764.76 kJ/kg
v1 = 0.00112484 m3/kg
=
= 889.015 kg/m3
= 8721.24 N/m3
= 8.721 kN/m3
Pwater = Q h
= (8.721 kN/m3) (0.040 m3/s) (910 m)
= 313 956 kW
thus;
Pmotor =
.
.
(a) Pmotor = 523.26 kW

A submersible, multi-stage, centrifugal deep well pump 260 gpm capacity is installed in a
well 27 ft below the static water level. Drawdown when pumping at rated capacity is 10
ft. the pump deliverd that water into a 25,000 gallons capacity overhead storage tank.
Totaldischarge head developed by pump including friction in piping is 243 ft. Calculate
the brakepower required to drive the pump if pump efficiency is 70%.
a. 15.86 Hp
c. 30.16 Hp
b. 21.22 Hp
d. 10.52 Hp
Solution:
Pbrake =
Solving for Pwater:
Q = 260
= 34.76 ft3/min.
H = 243 (27 10)
= 226 ft
Pwater = (62.4
) (34.76
= 490,199.42
Pwater =
) (226 ft)

thus;
Pwater = 21.22 Hp
A boiler feed pump receives 45 li./s of water at 190C and enthalpy of 839.33 kJ/kg. It
operates against a head of 952 m with efficiency of 70%. Estimate the water leaving
temperature assuming that the temperature rise is due to the inefficiency of the input
energy.
a. 190.96 C
c. 199.71 C
b. 194.66 C
d. 200 C
Solution:
m C
- m h

4.187 t 190) =
(h 839.33)
Solving for h :
m (h - h ) = 0.00981 (952)
h = 848.67 kJ/kg
then;
.
4.187 (t 190) =
(848.67 839.33)
(a) t = 190.96 C
A pump is driven by an electric motor moves 25 gal/min of water from reservoir A to
reservoir B, lifting the water to a total of 245 ft. The efficiency of the pump and motor are
64% and 84% respectively. What size of motor in Hp is required?
a. 3 Hp
c. 2 Hp
b. 5 Hp
d. 7 Hp
Solution:
Pmotor =
Solving for Pwater:
Q = 25 gal/min = 0.0557 ft /s
Pwater = (62.4
= 851.54 ft-lb/s
) (0.0557

) (245 ft)
Pwater = 1.55 Hp
Pbrake =
.
.
= 2.42 Hp
thus;
(a) Pmotor = 2.88 Hp or 3 Hp
A vacuum pump is used to drain a flooded mine shaft at 20C water. The pump pressure
of water at this temperature is 2.34 kPa. The pump is incapable of lifting the water higher
than 10.16 m. What is the atmospheric pressure?
a. 90.21 kPa
c. 102.01 kPa
b. 96.02 kPa
d. 108.01 kPa
Solution:
From Bernoullis Theorem:

=

.
+Z =
+
.
.
+Z
+ Z
+ 0 + 10.16
(c) thus; P = 102.01 kPa

Water reservoir is pumped over a hill through a pipe 45 mm in diameter and a pressure of
1 kg/cm (98.08 kPa) is maintained at the summit. Water discharge is 30 m above the
reservoir. The quantity pumped is 0.50 m /s. Frictional losses in the discharge and suction
pipe of the pump is equivalent to 1.5 m head loss. The speed of the pump is 800 rpm,
what amount of energy must be furnished by the pump in kW?
a. 206 kW
c. 250 kW
b. 210 kW
d. 245 kW
P=QH
Solving for H:
H=
.
/ .
+Z

.
Z + h
+ 30 +1.5
h = 42 m
then;
P = 9.81 (0.50) (42)
thus;
(a) P = 206 kW
A pump is to deliver 80 gpm of water at 140F with a discharge pressure of 150 psig.
Suction pressure indicates 2 in. of mercury vacuum. The diameter of suction and
discharge pipes are 5 in. and 4 in. respectively. The pump has efficiency of 70%, while the
motor efficiency is 80%. Determine the power input to the drive motor.
a. 12.59 Hp
c. 15.590 Hp
b. 10.59 Hp
d. 20.59 Hp
Solution:
Pmotor =
Pbrake =
Solving for Pwater:
Q = 80 gal/min = 0.1782 ft /s
Vs =
Vd =
.
/ /
.
/ /
= 1.307 ft/s
= 2.043 ft/s
From steam table:

At 150 psig (164.7 psi) and 140F:
= 61.424 lb/ft
h=
.
.
= 354 ft
then;
Pwater = 61.424 (0.1782) (354)
= 7.05 Hp
= 3874.80
=
= 10.07 Hp
Pmotor =
.
.
thus;
(a) Pmotor = 12.59 Hp
Determine the water horsepower of a centrifugal water pump which has an input of 3.5
Hp if the pump has an 8 in. nominal size suction and 6 in. nominal size discharge and
handles 150 gpm of water of 150F. The suction line gage shows 4 in. Hg vacuum and the
discharge gage shows 26 psi. The discharge gage is located 2 ft above the center of the
discharge pipe line and the pump inlet and discharge lines are at the same elevation.
a. 2.52 Hp
c. 2.78 Hp
b. 3.52 Hp
d. 3.78 Hp
Solution:
P=QH
Solving for h:
Q = 150 gal/min = 0.334 ft /s

Vs =
= 0.957 ft/s
Vd =
= 1.701 ft/s
From steam table:

At 150F:

=
H=
= 61.2 lb/ft
.
.
= 67.83 ft
thus;
P = (61.2) (0.334) (67.83)
= 1386.50 ft-lb/s
(a) P = 2.52 Hp
Water from an open reservoir A at 8 m elevation is drawn by a motor-driven pump to an
open reservoir B at 70 m elevation. The inside diameter of the suction pipe is 200 mm and
150 mm for the discharge pipe. The suction line has a loss of head three times that of the
velocity head in the 200 mm pipe. The discharge line has a loss of head 20 times that of
the velocity head of the discharge pipeline. The pump centerline is at 4 m. Overall
efficiency of the system is 78%. For a discharge rate of 10 li./s, find the power input to the
motor.
a. 10.06 kW
c. 6.12 kW
b. 4.80 kW
d. 7.85 kW
Solution:
Pinput =
:
Solving for P

Vs =
.
.
= 0.318 m/s
Vd =
.
.
= 0.566 m/s
.
=3
= 0.01546 m
20
= 20
+ Z
= 0.32642 m

H=
=
.
.
= 62.35 m
+ h
+ (66 4) + (0.01546 + 0.32642)
then;
= 9.81 (0.010) (62.35)
= 6.12 kW
thus;
Pinput =
.
.
(d) Pinput = 7.85 kW

A double suction, single stage, cenyrifugal pump delivers 900 m3/hr of seawater (S.G. =
1.03) from a source where the water level varies 2 m from high tide to low tide level. The
pump centerline is located 2.6 m above the surface of the water at high tide level. The
pump discharges into a surface condenser, 3 m above pump centerline. Loss of head
due to friction in suction pipe is 0.80 m and the discharge side is 3 m. Pump is directly
coupled to a 1750 rpm, 460 V, 3 phase, 60 Hz motor. Calculate the specific speed of
pump in rpm.
a. 5,149.20 rpm
c. 5,500 rpm
b. 6,149.20 rpm
d. 6,500 rpm
Solution:
Ns =
Solving for H:
hs = 2 + 2.6 + 0.8
= 5.4 m
= 17.72 ft
hd = 3 + 3 = 6 m
= 19.69 ft
H = 17.72 + 19.69
= 37.41 ft
Q1 = Q2 = 900/2
= 450 m3/hr
= 1981 gal/min
then;
Ns =
thus;
(a) Ns = 5,149.20 rpm
A DC driven pump running at 100 rpm delivers 30 liters per second of water at 40C
against a total pumping head of 27 m with a pump efficiency of 60%. Barometer pressure
is 758 mm Hg abs. What pump speed and capacity would result if the pump rpm were
increased to produce a pumping head of 36 m assuming no change in efficiency.
a. 115.47 rpm , 34.64 L/s
c. 110.51 rpm , 34.46 L/s
b. 115.47 rpm , 38.68 L/s
d. 110.51 rpm , 38.68 L/s
Solution:
New Speed required:
N
N
H
H
N
100
36
27
N = 115.47 rpm
New Capacity Required:
Q

Q
N
N
Q

30
115.47
100
Q = 34.64 L/s
thus;
(a) N = 115.47 rpm , Q = 34.64 L/s
A centrifugal pump discharged 20 L/s against a head of 17 m when the speed is 1500
rpm. The diameter of the impeller was 30 cm and the brake horsepower was 6.0. A
geometrically similar pump 40 cm in diameter is to run at 1750 rpm. Assuming equal
efficiencies, what brake horsepower is required?
a. 51.55 HP
c. 40.15 HP
b. 50.15 HP
d. 45.15 HP
Solution:
New brake horsepower required:
P
D N
P
D N
6

0.30 1500
P
0.40 1750
thus;
(c) P = 40.14 Hp
A two-stage centrifugal pump delivers 15,000 kg/hr of 110C water against 76 m head at
3500 rpm. What is the specific speed of the pump?
Solution:
Ns =
Solving for Q:
,
Q=
=
Q = 0.004383 m3/s
H = 76/2 = 38 m
then;
Ns =
.
.
= 780.39 rpm
thus;
(a) Ns = 780.39 rpm

Calculate the impeller diameter of the centrifugal pump that requires 15m head
to deliver if pump speed is 1500 rpm.
a. 218.43 mm
c. 345.75 mm
b. 300.75 mm
d. 276.45 mm
Solution:
V=
Solving for V:
V= 2
= 2 9.81 15
= 17.15 m/s
then;
17.5 =
D = 0.21843 m
thus;
(a) D = 0.21843 m

A large centrifugal pump has a 254 mm diameter inlet and a 127 mm diameter
outlet pipe. The measured flow rate is 51.6 L/s of cold water. The measured inlet
pressure is 127 mmHg above atmospheric and discharge pressure measured at a
point 1.22 m above the pump outlet is 212 kPa abs. The pump input is 10 Hp. Find
the pump efficiency.
a. 78.51 %
c. 74.54 %
b. 70.62 %
d. 76.77 %
Solution:
epump = Pwater / Pinput
Solving for Pwater :
Vs = Q / As = 0.0516 / /4 0.254
Vs = 1.018 m/s
Vd = Q / Ad
=
.
.
= 4.073 m/s
H=
212101.325 127
101.325
760
9.81
4.073 2 1.018 2
2 9.81
1.22
= 11.57 m
Pwater =9.81(0.516)(11.57)
=5.86 Kw=7.85Hp
thus;
epump = 7.85/10
(a) epump = 0.785 = 78.51%
A closed tank contains liquefied butane gas whose specific gravity is 0.60. The
tank pressure us 1.7 Mpa gage which is also the equilibrium vapor pressure of
butane at the pumping temperature. Suction line losses is 1.5 m of gas and the
static elevation gain is 4m. What is the Net Positive Suction Head available
(NPSH)?
a. 2.5 m
c. 3.5 m
b. 1.5 m
d. 1.7 m
Solution:
Use (+) for static elevation gain.

NPSH =
= 0 + 4 1.5
Thus;
(a)
NPSH= 2.5 m available

A condensate pump at sea level take water from a surface condenser where the
vacuum is 15 in. of mercury. The friction and turbulence in the piping in the
condenser hot well and the pump suction flange is assumed to be 6.5 ft. If the
condensate pump to be installed has a required head of 9 ft, what would be the
centerline of the pump to avoid cavitation?
a. 2.5 ft
c. 18 ft
b. 15.5 ft
d. 5.5 ft
Solution:
The Net Positive Suction Head required by the pump is 9.
then;
NPSH =
9 = 0 + S 6.5
Thus;
(b) S = 15.5 ft
A boiler feed pump receives 130 cfm of water with specific volume of 0.0025 m3
at a head of 800 m. If the pump efficiency is 64%, what is the output of the driving
motor?
a. 299 kW
c. 250.34 kW
b. 350.16 kW
d. 299.64 kW
Solution:
epump = Pwater / Pinput

Solving for Pwater;
Q = 130 ft3 = 0.061 m3/s
=
1
0.0025
0.00981
3.9245 kN/m3
Pwater = 3.924(0.061)(800)
= 191.49 Kw
Then;
Poutout = 191.49/0.64
Thus;
a
Poutout = 299.21 kW

A dearator heater supplies 150 L/min of daerated feedwater into a booster pump
at 115
pumping temperature, The heater pressure is maintained at 100 kPag by
bleed steam. Pump centerline is located 1m above the floor level. Suction line
losses is 0.60 m. Determine the minimum height of water level in the heater that
must be maintained above the centerline of the pump to avoid cavitation, if the
pump to be installed has a reuired suction head 5.8 m.
a. 9.05 m
c. 3.25 m
b. 2.25 m
d. 5.09 m
Solution:
NPSH =
where;
P = 100 kPag
Pa = 101.325 kPa
PV = Psat at 115
Pv = 172 kPa
Vf at 15.6
= 0.001 m3 / kg
Vf at 11.5
= 0.001055 m3/ kg
S.G. =
0.001
0.001055
= .948
then;
5.8 =
100 101.325172
9.81 .948
+ S 0.60
thus;
(c) S = 3.25
A pump running at 1000 rpm delivers water against a head of 300 m. If the pump
speed will increased to 1500 rpm, what is the change in head.
a. 575 m
c. 675 m
b. 375 m
d. 475 m
Solution:
H2
H1
=( )
N
H2 = 300(1500/1000)2
H2 = 675 m
thus;
(b) H = 675 300 = 375 M

Oil is being pumped from a truck to a tank 10 ft higher than the truck through a 2
in. galvanized pipeline 100 ft long. If the pressure of the discharge side of the
pump is 15 psi, at what rate in gpm is oil flowing through the pipe? The oil has a
specific gravity of 0.92 at the temperature in the pipe.
a. 542.22 gpm
c. 642.44 gpm
b. 442.44 gpm
d. 742.44 gpm
Solution:
Q = AV
H=P/
15(144)
+ hs = 62.4(.92)
10
= 47.66 ft
2gh =
V=
2 32.2 47.66
= 55.40 fps
A=
22 /
= 0.0218 ft2
then;
Q = 0.0218 (55.40)
= 1.209 ft3/s (7.489 gal/1ft3)(60 s/min)
thus;
(a) Q = 542.44 gal / min or gpm

A pump is operated by motor developing 30 Kw and is delivering water at a
pressure of 200 psig. If the pump is drawing water from a lake 20 ft below the
centerline and the mechanical and hydraulic efficiency may be assumed to be
97% and 65% respectively; estimate the amount of water discharge in gallons per
minute.
a. 358.26 gpm
c. 208.26 gpm
b. 258.26 gpm
d. 308.26 gpm
Solution:
Pwater = QH
Solving for Pwater;

Pwater = Pbrake (epump)
= (30)[(.97)(.65)]
= 18.915 Kw
Solving for H;
H=
200 144
62.4
20
= 481.54 ft
= 146.766 m
Then;
18.915 = 9.81 Q (146.766)
Q = 0.01314
thus;
(c) Q = 208.26 gpm
A 30 Hp centrifugal pump is used to deliver 70 cfm water. Calculate the number
of stages needed if each impeller develops a 38 ft head.
a. 4
c. 8
b. 6
d. 10
Solution:
No. of stages = Total head / head per stage
Solving for the total head, H;
Q = 70 ft3/min
Q = 0.033 m3/s
P=
QH
30 / 0.746 = 9.81 (0.033) H
H = 124.22
n = 124.22 / 38
Thus;
(a) n = 3.26 say 4 stages
In a test of a centrifugal pump driven by an electric motor, the suction pipe is 10
in. in diameter and its gage indicates a partial vacuum of 2.5 ft of water. The
discharge pipe is 5 in. in diameter, is 2 ft higher than the suction gage and shows
a pressure of 50 ft of water. if the pump efficiency is indicated assuming motor
efficiency of 85%?
a. 60%
c. 75%
b. 70%
d. 65%
Solution:
Electrical Power Input = Water power / (epump)(emotor)
Solving for water power:
Vs = Q/AS = 1.6 / /4 10/12 2 = 2.935 fps
Vd = Q/Ad = 1.6 / /4 5/12 2 = 11.74 fps
H = (50 + 2.5) + 2 + [(11.74)2 (2.935)2]/ 2(32.2)
H = 56.51 ft
Pwater = (62.4)(1.6)(56.51)
= 5,641.96 ft-lb /s
= 10.26 Hp
then;
12/ .746 = 10.26 / (epump)(0.85)
thus;
(a) epumo = 0.7504 = 75.04%
43 Supplementary Problem
A centrifugal pump is designed for 2000 rpm and head of 70 m. What is the speed
if the head id increased to 100 m.
a. 2100 rpm
c. 3000 rpm
b. 2390 rpm
d. 3010 rpm
Solution:
)2
100/70 =(N2/2000)2
Thus;
(b) N2 = 2,390.46 rpm
A water pump develops a total head of 200 ft. The pump efficiency id 80% and
the motor efficiency is 87.5%. If the power rate is 1.5 cents per Kw-hr, hat is the
power cost for pumping 100 gal?
a. 34 cents per hour
c. 2.34 cents per hour
b. 1,34 cents per hour
d. 3,34 cents per hour
Solution:
P=
QH
= (62.4 lb/ft3)[(1000 gal/hr)(1 ft3/ 7.48 gal)(1 hr / 16000 s)](200 ft)

= 436.46 Hp
P motor = 0.84 / 0.80(.875)
Power Cost = 0.895 Kw ($ 0.015 / Kw- hr)
Thus;
(b) Power Cost = $ 0.0134 per hr or 1.34 cents per hr
A test on a centrifugal pump operating at 1150 rpm showed a total of 37.6ft at a

capacity of 800 GPM. Estimate the total head and capacity if the pump were
operated at 1750 rpm. Assume normal operation at point of maximum efficiency in
each case.
a. 87.07 ft., 1217.4 gpm
c. 97.07 ft., 1217.4 gpm
b. 87.07 ft., 1517.4 gpm
d. 97.07 ft., 1517.4 gpm
Solution:
New Head Required:
H2 = 87.07 ft
Q2 = 1217.4 gpm
Thus;
(a) H2 = 87.07 ft; Q2 = 1217.4 gpm

A centrifugal pump operating at 1800 rpm develops total head of 200 ft at a
capacity of 2500 gpm. What is the specific speed?
a. 1590 rpm
c. 1650 rpm
b. 1690 rpm
d. 1550 rpm
Solution:
Ns =
Thus:
(a) Ns = 1690 rpm

A centrifugal pump operating at 1150 rpm showed a total head of 40 ft at a
capacity of 600 gpm. The impeller diameter is 10.5 in. Estimate the total head and
capacity of a geometrically similar pump at 1150 rpm with an impeller diameter of 10
inches.
a. 30. 25 ft.,513.8 gpm
c. 36.28 ft., 618.3 gpm
b. 36.28 ft., 518.3 gpm
d. 30.25 ft., 618.3 gpm
Solution:
New Head Required:
H2 = 36.28ft
Q2 = 518.3gpm
Thus;
(b) H2 = 36.28 ft; Q2 = 518.3 gpm

A pump delivers 20 cfm of water having a density of 62 lb/ft3. The suction and discharge
gage reads 5 in. Hg vacuum and 30 psi respectively. The discharge gage is 5 ft above
the suction gage. If pump efficiency is 70%, what is the motor power?
Solution
Pwater
0.70
Pmotor
30
Z Z
5 14.7/29.920 144
62
80.38
Pwater
62
ft3
20
3
min
80.38
99674.87
Pwater
1
33, 000
3.02 Hp
Thus;
Pwater
3.02
0.70
4.31

A centrifugal pump-motor unit draws 100 li/min of water from a supply which has a level
at the centreline of the pump. The discharge pressure is 28 kg/cm2 and the over-all unit
efficiency is 67 %. What could be the required input to the electric motor in kW if the
head is 280 m.
a. 6.83 kW
c. 7.83 kW
b. 5.83 kW
d. 8.83 kW
Solution
Pinput
9.81
Pwater
0.67
280
0.67
Pinput
6.83 kW

A double-suction centrifugal pump delivers 3 m3/s of water at a head of 15 m and
running at 1200 rpm. Calculate the specific speed of the pump. Ns
a. 9958.46 rpm
c. 16, 000 rpm
b. 12, 110.64 rpm
d. 17, 000 rpm
Solution
Ns
Where:
3
47, 556.14
2

15
Ns
23, 778.07
49.215
120023, 778.07
49.215 /
9958.56
Llamera, Kristine Joyce D.
10 19388
ME 5206
Problems in Industrial Plant Engineering
Part 4: Fans & Blowers
1. What horsepower is supplied to air moving at 20 fpm through a 2 x 3 ft duct under a
pressure of 3 in. water gage?
a. 0.786 Hp
c. 0.642 Hp
b. 0741 Hp
d. 0.0566 Hp
Solution:

Solving for :
2 3
2
Solving for :

then,
2

31.2
Thus,
(d)
0.0567
2. A fan whose static efficiency is 40% has a capacity of 60,000
per hour at 60
and barometer of 30 in. Hg and gives a static pressure of 2 in. of water column on
full delivery. What size electric motor shall be used to drive this fan?
a. Hp
c. 2 Hp
b. 1 Hp
d. 1 Hp
Solution:

Brake (input) Power =
Solving for
where;
,
16.67
then;
.
16.67
173.37
0.315
thus;
.
0.788
(b) Use 1 Hp
3. Air is flowing in a duct with a velocity of 7.62 m/s and a static pressure of 2.16 cm
water gage. The duct diameter is 1.22 m, the barometric pressure 99.4 kPa and the
gage fluid temperature and air temperature are 30 . What is the total pressure of air
against which the fan will operate in cm of water?
a. 3.25
c. 3.75
b. 2.50
d. 1.25
Solution:
Solving for velocity head,
.
.
2.959
Solving for velocity in terms of cm of water:

1000
2.959
where:
.
.
1.143
then;

1000
2.959 1.143
0.0034 m of water
0.34 cm of water
thus;
h
2.16
(b) h
0.34
2.50 cm of water
of air per second through a duct 91 cm in diameter
4. A ventilating fan discharge 4.4
against a static pressure of 22 mm water gauge. Barometric pressure is 730 mm Hg,

the temperature of air is 29.44
and the gauge fluid density is 994 kg/
. If the power
input is 2.65 kW, determine the static efficiency.

a. 35.62 %
c. 45.62 %
b. 25.62 %
d. 55 %
Solution:
.
= 0.3562
thus;
(a) e = 35.62 %
5. A ventilation system includes a fan of 570
per minute. A capacity discharging thru
a 122 cm x 91 cm air duct against 30 mm static pressure. Air temperature is at 21

and barometer pressure is 730 mm Hg. (97 kPa). What input power will be required to
give the fan an efficiency of 44.3%?
a. 5 Hp
c. 3 Hp
b. 10 Hp
d. 7 Hp
Solution:
Solving for
:
@
= 995.85 kg/
3
= 1.150 kg/
Static pressure in cm WG converted into m of air:
0.30
1.150
995.85 0.30
25.98 m of air
Velocity of air at the fan outlet:

/
.
8.56 /
Velocity pressure:
.
.
= 3.73 m of air
Total Pressure created by Fan:
+
= 5.98 + 3.73
= 29.71 m of air
then;

= [1.150(0.00981)(570/60)(29.71)]
= 3.18 kW
= 4.27 Hp
thus;
.
.
9.64
10
6. The mechanical efficiency and static pressure of a fan are 40% and 20 m of air
respectively. What is the static efficiency if the total pressure created by fan is 25 m of
air?
a. 30.44%
c. 35.44%
b. 33.44%
d. 37.44%
Solution:
= (0.443)(20/25)
thus;
(a)
0.3544 or 35.44%
7. Air enters a fan through a duct at a velocity of 6.3 m/s and an inlet static pressure of
2.5 cm of water less than atmospheric pressure. The air leaves the fan through a duct
at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above
the atmospheric pressure. If the specific weight of the air is 1.20 kg/
delivers 9.45
and the fan
/s, what is the fan efficiency when the power input to the fan is 13.75
kW at the coupling?
a. 71.81%
c. 52.34%
b. 61.81%
d. 72.34%
Solution:
Fan efficiency =
Solving for
:
+
1000
.
.
= 88.761 m of air
then;

= [1.2(0.00981)](9.45)(88.761)
= 9.874 kW
thus;
.
.
(a)
= 71.81%
8. The forced draft fan is in parallel with a capacity if 73.2
per second each supplying
combustion air to a steam generator. Air inlet is at 43.33 , a static pressure of 254
mm water gage is developed and the fan speed is 1200 rpm. The fan input is 257 kW
each. Calculate the capacity of the fan for a speed increase of 20 percent.
a. 77.86
/s
c. 89.46
/s
b. 87.84
/s
d. 59.49
/s
Solution:
1.21
= 1.21 (73.2)
thus;
(b)
87.84
/s
9. Calculate the air power of a fan that delivers 1200
/min of air through a 1 m by 1.5
m outlet. Static pressure is 120 mm WG and density of air is 1.18.

a. 20.45 kW
c. 30.45 kW
b. 25.64 kW
d. 35.64 kW
Solution:

Solving for h:
0.120
101.695
from:
Q=AV
1 1.5
V = 13.33 m/s
.
.
= 9.06 m of air
+
= 101.695 + 9.06
= 110.756 m of air
thus;
1.18 0.00981
(b)
10. A fan delivers 4.7
110.756
25.64 kW
/s at a static pressure of 5.08 cm of water when operating at a
speed of 400 rpm. The power input required is 2.963 kW. If 7.05
/s are desired in the
same fan and installation, finds the pressure in cm of water.

a. 7.62
b. 17.14
c. 11.43
d. 5.08
Solution:
Solving for
600 rpm
then;
.
thus;
(c)
11.43 cm of water
11. A fan described in a manufacturers table is rated to deliver 500
/min at a static
pressure (gage) of 254 cm of water when running at 250 rpm and requiring 3.6 kW. If
the fan speed is changed to 305 rpm and the air handled were at 65
instead of
standard 21 , find the power in kW.

a. 3.82
c. 4.66
b. 5.08
d. 5.68
Solution:
Power Required at 305 rpm and 65 :
Solving for the Power required at 305 rpm and 21 :
6.5 kW
then;
.
thus;
(d)
5.68 kW
7
12. What is the set efficiency of a fan if the fan efficiency is 45% and motor efficiency is
90%?
a. 40.50%
c. 30.41%
b. 35.65%
d. 40.94%
Solution:
Set of Efficiency = Fan Efficiency x Motor Efficiency
= (0.45)(0.90)
thus;
0.405 or 40.50%
(d)
13. A fan draws 1.42
/s of air at a static pressure of 2.54 cm of water through a duct 300
mm diameter and discharges it through duct of 275 mm diameter. Determine the

static fan efficiency if total fan mechanical efficiency is 70% and air is measured at
25
and 760 mm Hg.

a. 60%
c. 30%
b. 50%
d. 40%
Solution:
Solving for
= 1.18 kg/
Solving for
.
.
= 21.52 cm of air
.
.
.
.
20.09
23.9
.
.
.
8.54 m of air
Solving for h:
+
= 21.52 + 8.54
8
= 30.06 m of air
thus;
0.70 21.52/30.06
(b)
0.5011 or 50.11%
14. A fan manufacturer rates his fans at 152 mm water gage static pressure for 10
of
air per second at 21 , 1200 rpm and static efficiency of 69%. At what speed would
these fans operate to develop 130 mm water gage when the temperature is 316 ?
a. 1570 rpm
c. 1770 rpm
b. 1670 rpm
d. 1470 rpm
Solution:
At 21 ,
1.52
A change in temperature will effect a change in density:
With the same pressure;
1.52 2 2
13.09
At 21
thus;
(a)
1570.8 rpm
15. A 12 Hp motor is used to drive a fan that has a total head of 20 m. If the fan
efficiency is 70%, what is the maximum capacity in
/s?
a. 26.63
/s
c. 24.43
/s
b. 25.53
/s
d. 27.73
/s
9
Solution:
0.70 12
8.4
= 6.27 kW

6.27 = [(1.2)(0.00981)]Q(20)
thus;
/s
(a) Q =26.63
16. Find the air horsepower of an industrial fan that delivers 26

by 1.2 m; pressure is 127 mm of water; air temperature is 21
/s of air through a 1 m
and barometric pressure
is 760 mm of mercury.
a. 53.35 Hp
c. 46.45 Hp
b. 43.33 Hp
d. 56.45 Hp
Solution:

Solving for h:
21.667 /
.
.
= 1.2 kg/
@ 101.325
998.6
2
1.20 21.62
998.6 2 9.81
0.029
127
0.127
h = 0.127 + 0.029 = 0.156
then;
P = (9.81)(26)(0.156)
= 39.79 kW
thus;
(a) P = 53.34 Hp
10
17. A large forced draft fan is handling air at 1 atm, 43.3
under a total head of 26.6
cm WG (at 43.3 ). The power input to the fan is 224 kW and the fan is 75 percent
efficient. Compute the volume of air handled per minute. Local gravity of
acceleration is g = 9.81m/ .
a. 3,862.87
c. 2,862.87
b. 4,862.87
d. 4,567.97
Solution:
Power Input to fan = Shaft Power
P=
224 =
Q = 64.38
= 3,862.87
/s
/
thus;
(a) 3,862.87
18. A fan develops a brake power of 150 kW at 1.2 kg/

brake power of the fan if it operates at 100 kPa and 30
air density. What is the new

at the same speed?
a. 163.75 kW
c. 133.86 kW
b. 143.75 kW
d. 173.86 kW
Solution:
Solving for
:
.
= 1.15
then;
.
.
thus;
(b)
143.75 kW
11

A cold storage compartment is 4.5 m long by 4 m wide by 2.5 m high. The four walls, ceiling
and floor are covered to a thickness of 150 mm with insulating material which has a
coefficient of thermal conductivity of 5.8 x 10-2 W/m-K. Calculate the quantity of heat
leaking through the insulation per hour when the outside and inside face temperatures of
the material is 15C and -5C respectively.
a.
2185.44 kJ
c.
3185.44 J
b.
1185.44 kJ
d.
4185.44 kJ
Solution
kAt
x
where:
A
2 [ (4.5)(2.5) + (4)(2.5) + (4.5(4) ]
78.5 m2
then:
(5.8x10-2)(78.50)(15+5)
607.07 W or J/s
2185.44 kJ/hr
0.15
thus;
(a) the quantity of heat through the insulation per hour is 2185.44 kJ
19 Supplementary Problem
A thin square steel plate, to 10 cm on a side, is heated in a blacksmiths forge to a
temperature of 800C. If the emissivity is 0.60, what is the total rate of radiation of energy?
a.
900 Watts
c.
300 Watts
b.
400 Watts
d.
700 Watts
Solution
Q
A e T4
(0.020m 2 ) (0.60) (5.67 x 10 -8
thus;
(a) Q = 900 Watts
W
m2K4
) (1073)4 K4

A furnace wall consist of 35 cm firebrick (k= 1.557 W/m-K), 12 cm insulating refractory
(k=0.346) and 20cm common brick (k=1.692) covered with 7 cm steel plate (k=45). The
temperature at the inner surface of the firebrick is 1230 degree C and at the outer face of
the steel plate is 60 degree C. Atmosphere 27 degree C. What is the value of combined
coefficient for convection and radiation from the outside wall?
a. 31.13 W/m2-K
c. 41.3 W/m2-K
b. 30.13 W/m-K
d. 40.13 W/m2-K
Solution
Q
A
t
RT
where:
RT
k12
k23
X12
x23
0.35
x34
0.12
1.557
k34
0.346
k45
x45
0.2
0.692
m2K
0.862
then;
Q
A
Q
A
(1230-60)K
0.862
Q5-0
m2K
t5 - to
1
h0
1357.15
60-70
1
h0
h0
41.13
W
m2K
1357.15 W/m2
0.07
45

A dry ice storage chest is a wooden box lined with glass fiber insulation 5cm thick. The
wooden box (k= 0.069) is 2 cm thick and cubical 60 cm on an edge. The inside surface
temperature is -76 degree C and the outside surface temperature is 18 degree C. Use k=
0.035 for fiber glass insulation. Determine the heat gain per day.
a. 10211 kJ
c. 12211kJ
b. 11195 kJ
d. 9185 kJ
Solution
Q
At
RT
where:
A
6[(0.60)(0.60)]
=
RT
=
=
2.16m2
k12
x12
k23
x23
0.05
0.035
0.02
0.069
0.1718
then;
Q
=
=
2.16(18+76)
1.718
(118.18)
10211.092
118.18
J
S
(3600s)
(24hrs)
(1kJ)
(hr)
(Day)
(1000J)
kJ
day
thus;
(a) the heat gain per day is 10211.092kJ

One side of the refrigerated cold chamber is 6 m long by 3.17 m high and consists of
168mm thickness of cork between outer and inner walls of wood. The outer wood wall is
30 thick and its outside face temperature is 20 degree C, the inner wood wall is 35 mm
thick and its inside face temperature is -3 degree C. Taking the coefficient of thermal
conductivity of cork and wood as 0.42 and 0.20 W/m-K respectively, calculate the heat
transfer per second per sq. m of surface area.
a. 5.138 J
c. 6.318 J
b. 4.138 J
d. 3.318 J
Solution
Q
A
t
RT
where:
RT
0.03
0.2
0.168
0.042
0.035
0.2
4.325
then;
Q
A
=
=
20+3
4.325
5.318
5.918W
J
s
thus;
(a) the heat transfer per second per sq.m of the surface is 5.318 J

Hot gases at 280 degree C flow on one side of a metal plate of 10 mm thickness and air
at 35 degree C flows on the other side. The heat transfer coefficient of the gases is 31.5
W/m-K and that of the air 32 W/m-K. Calculate the overall transfer coefficient.
a.
15.82 W/m2-K
c.
14.82 W/m2-K
b.
15.82 W/m2-K
d.
17.82 W/m2-K
Solution
U
1
RT
where:
RT
=
=
F
h1
k12
x12
1
x12
0.0632
thus;
U
1
0.06032
(a) U=15.82
15.82
W
m2-K
1
31.5
0.01
50
1
32

The surface temperature of the hot side of the furnace wall is 1200 degree C. It is desired
to maintain the outside of the wall at 38 degree C. A 152 mm of refractory silica is used
adjacent to the combustion and 10 mm of steel covers the outside. What thickness of
insulating bricks is necessary between refractory and steel, if the heat loss should be keep
at 788 W/m2? Use k= 12.84 W/m-K for refractory silica; 0.15 for insulating brick, and 45 for
steel.
a. 220 mm
c. 260 mm
b. 240 mm
d. 280 mm
Solution
RT
RT
x12
k12
0.152
31.84
x23
k23
x34
k34
x23
0.15
0.01
45
Solving for RT
788
788
RT
t
RT
(1200-38)
RT
1.475
then;
1.475
0.152
13.48
thus;
(a)
x23
0.22mm
x23
220mm
x23
0.15
0.01
45

How much heat will flow in 24 hours through a plaster wall that is 0.50 in thick and 8 ft x 14
ft in area if the temperature is 80 degree F on one side and 40 degree F on the other? Use
= 3.25 Btu-in/hr-ft2-degree F
a. 5.99 x 105 Btu
c. 7.99 x 105 Btu
b. 6.99 x 105 Btu
d. 4.99 x 105 Btu
Solution
Q
kAT
x
3.25[(8)(14)](80-40)(24)
thus;
(b) Q = 6.99 x105 Btu
0.5

A hollow sphere has an outside radius of 1 m and is made of polystyrene foam with a
thickness of 1 cm. A heat source inside keeps the inner surface 5.20 degree C hotter that
the outside surface. How much power is produced by the heat source? Thermal
conductivity of polystyrene foam is 0.033 W/M degree C.
a. 200 W
c. 300 W
b. 216 W
d. 316 W
Solution
Q
kAT
x
where:
A
=
=
4(1)2
12.56 m2
(0.033)(12.56)(5.20)
thus;
(b) Q= 216 W
0.01
216 W

A glass window has an area of 1.60 m2 and a thickness of 4mm. If one side is at a
temperature of 6.80 degree C and other is at -5 degree C, how much thermal energy flows
through the window in a time of 24 hours? The thermal conductivity of glass is 1.89 x 10-4
Kcal/ m-s-degree C
a. 26200 kCal
c. 40700 kCal
b. 58000 kCal
d. 77100 kCal
Solution
Q
=
=
=
kAT
x
(1.89 x 10-4)(1.60)(6.80+5)(3600)(24)
0.004
77100 kCal
thus;
(d) Q = 77100 kCal

The wall of a cold room consist of a layer of cork sandwiched between outer and inner
walls of wood, the wood walls being each 30 mm thick. The inside atmosphere of the room
is maintained at -20 degree C when the external atmospheric temperature is 25 degree
C, and the heat loss through the wall is 42 W/m2. Taking the thermal conductivity of wood
and cork as 0.20 W/m-K and 0.05 W/m-k respectively, and the rate of heat transfer
between each exposed wood surface and their respective atmospheres as 15 W/m2 K,
calculate the thickness of the cork.
a. 31.90 mm
c. 41.90 mm
b. 21.90 mm
d. 51.90 mm
Solution
Q
kT
where:
Q
A
42
W
m2
W
0.05
30mm
m-K
=
0.30m
from the heat through each wood wall:

42
0.20t
0.03
6.30C
Interface temperature of outer wood and cork:

=
22.2 - 6.3
15.6C
Interface temperature of inner wood and cork:

=
-17.2
10.9C
Temperature difference acros cork:

=
15.9-(-10.9)
26.8C
From heat flow through cork:

42
0.05(26.8)
x
0.0319m
thus;
(a) x = 31.90 mm

A slab of material has an area of 2m2 and is 1mm thick. One side is maintained at a
temperature of 0 degree C while the other is at 12 degree C. It is determined the 6820 J of
heat flows through the material in a time of 10 minutes. What is the thermal conductivity
of the material?
a. 4.74 x 10-4 W/mC
c. 2.66 x 10-4 W/mC
b. 5.74 x 10-4 W/mC
d. 9.79 x 10-4 W/mC
Solution
Q
kAT
x
6820 J
10(60) s
k
k(2)(12-0)
0.001
4.74 x 10- 4
thus;
(a) k = 4.74 x 10-4
W
mC
J
s-m-C

An insulated steam pipe located where the ambient temperature is 32C, has an inside
diameter of 50 mm with 10 mm thick wall. The outside diameter of the corrugated
asbestos insulation is 125 mm and the surface coefficient of still air, h0 = 12 W/m2-K.
Inside the pipe is steam having a temperature of 150C with film coefficient hi = 6000
W/m2-K. Thermal conductivity of pipe and asbestos insulation are 45 and 0.12 W/m2-K
respectively. Determine the heat loss per unit length of pipe.
a. 110 W
c. 130 W
b. 120 W
d. 140 W
Solution:
Q=
Where:
RT =
RT =
RT = 0.98345 / L
Then;
Q=
Thus;
(b) Q/L = 120 W per meter length

A pipe 200 mm outside diameter and 20 m length is covered with a layer, 70 mm thick
of insulation having a thermal conductivity of 0.05 W/m2-K and a thermal conductance
of 10 W/m2-K at the outer surface. If the temperature of the pipe is 350C and the
ambient temperature is 15C, calculate the external surface temperature of the
lagging.
a. 32.6C
c. 42.6C
b. 22.6C
d. 53.6C
Solution:
Q(for lagging) = Q(for surface film)

.
0.340 10
T2 = 32.6C
15

Dry and saturated steam at 6 Mpa abs. enters a 40 m length of 11.5 cm O.D. steel pipe
at a flow rate of 0.12 kg/s. The pipe is covered with 5 cm thick asbestos insulation (k=
0.022 W/m-k). The pipe is located in a tunnel with stagnant air temperature of 27 degree
C. The unit outside convective coefficient is 10 W/m2-K. Neglecting steam film and pipe
wall resistances, determine the mass of steam.
a. 4.86 kg/hr
c. 5.86 kg/hr
b. 3.86 kg/hr
d. 6.86 kg/hr
Solution:
The temperature of the outer surface of the pipe is equal to that of the steam since the
resistance of metal pipe and vapor are negligible. Vapour temperature t1 =t2 =tsat @
6Mpa = 275 degree C.
RT
=
=
=
R2
R0
ln(10.75/5.75)
2(40)(0.022)
0.11687
t
RT
2112 W
2.122kW
1
(0.215)(40)(10)
C
275-27
0.11687
The heat necessary to condense steam at 6Mpa is hfg @ 6Mpa is equal to 1571 kJ/kg.
Since there are 2.122 kW of heat lost from the steam, then
Steam
Condensed
=
=
=
2.122kJ/s
1577kJ/s
0.00135 kg/s
4.86 kg/hr

Calculate the heat loss per linear ft from 2 in. nominal pipe (2.375 in. outside diameter )
covered with 1 in. of an insulating material having an average thermal conductivity of
0.0375 Btu/ hr- ft- degree F. Assume that the inner and outer surface temperatures of the
insulation are 380 degree F and 80 degree F respectively.
a. 110 Btu/ hr-ft
c. 120 Btu/ hr-ft
b. 116 Btu/ hr-ft
d. 126 Btu/ hr-ft
Solution
Q
t
RT
t
ln( d2/d1)
2kL
380
ln(4.375/2.375)
2(0.0375)
116
Btu
hr-ft

Calculate the heat loss per linear foot from a 10 in. normal pipe ( outside diameter = 10.75
in. ) covered with a composite pipe insulation consisting of 1 in of insulation I placed next
to the pipe and 2 in. of insulation II placed upon insulation I. assume that the inner and
outer surface temperatures of the composite insulation are 700o F and 100oF respectively,
and that the thermal conductivity of material I is 0.05 Btu/hr-ft-oF and for material II is 0.039
Btu/hr-ft-oF.
a. 423.13 Btu/hr-ft
c. 120 Btu/hr-ft
b. 123.13 Btu/hr-ft
d. 126 Btu/hr-ft
Solution
Q =
Where:
.
RT =
/
.
= 1.826
Then:
Q =
Thus;
(d) Q = 323.13
/
.

A steam pipe carrying a steam at 380 kPa pressure for a distance of 120 m in a chemical
plant is not insulated. Estimate the saving in steam cost that would be made per year if this
8 cm steam line were covered with 85% magnesia pipe covering 5 cm thick. Take room
temperature to be 25C, the cost of steam is 65 cents per 1000 kg. thermal conductivity of
magnesia k = 0.0745 W/m-K, unit convective coefficient of room air, ho = 12 W/m2-K.
a. $ 305
c. $ 505
b. $ 405
d. $ 605
Solution
Steam temperature = 142o C
Latent heat hfg = 2139.4 kJ/kg
Heat Loss from the bare pipe:
Q1 =
Where:
Ro =
=
=
0.00276 oC/W
Then;
Q1 =
= 42,343.64 W
=
42,344 kW
Total Resistance from the insulated pipe:

R1 = R2 + Ro
=
/
.
R1 = 0.01113 C/W
Heat loss from the insulated pipe:
Q2 =
= 10,512.13 W
Heat Saved = Q1- Q2
= 42,34364 10.51213
= 31,831 kW
Amount of steam saved due to condensation (m2):
m =
/
. /
= 0.014878 kg/yr

A liquid to liquid counter flow heat exchanger is sued to heat a cold fluid from
120 310 . Assuming that the hot fluid enters at 500F and leaves at 400F, calculate the
log mean temperature difference for the heat exchanger
a. 132F
c. 332F
b. 232F
d. 432F
Solution
LMTD = log mean temperature difference

LMTD =
Where:
Thus, LMTD =
= 232
400
120
280
500
310
190

A blower with the inlet open to the atmosphere delivers 300 cfm of air at a pressure of 2in.
WG trough a duct 11 in. diameter, the manometer being attached to the discharge duct
at the blower. Air temperature is 70F, and the barometer pressure is 30.22 in Hg. Calculate
the horsepower.
a. 1.54 Hp
c. 3.54 Hp
b. 2.54 Hp
d. 0.75 Hp
Solution:
Pair = gh
where:
Q = 3000 cfm
Solving for h:
Air density at the following condition by correcting the standard density.
air = 0.075 (
30.2
29.92
hs =
hv =
62.4
12 0.076
V
2g
=[
) = 0.076 lb/ft3
) = 136.84 ft of air
3000/60
11 2
)
12
/4(
2(32.2)
] = 89.13 ft
then;
P = 0.075 (3000/60(136.84 + 89.13)
P = 847.39
ft - lb
s
thus;
(a) P = 1.54 Hp
1 Hp
550 ft-lb/s

A certain fan delivers 12,000 cfm at a static pressure of 1in. WG when operating at a speed
of 400 rpm and requires an input of 4 Hp. If the same installation 15,000 cfm are desired,
what will be the new speed, and the new power needs?
a. 450 rpm, 6.81 Hp
c. 500 rpm, 6.81 Hp
b. 500rpm, 7.81 Hp
d. 450 rpm, 7.81 Hp
Solution:
New Speed Required:
N2
N1
N2
400
The New Power Required:
Q2
P2
Q1
P1
15000
P2
12000
N2 = 500 rpm
= (
= (
N2
N1
500
400
P2 = 7.81 hp
thus;
(b) 500 rpm, 7.81 Hp

A certain fan delivers 12,000 cfm at 70F and normal barometric pressure at a static
pressure of 1 in. WG when operating at 400 rpm and requires 4 Hp. If the air temperature is
increased to 200F (density 0.06018 lb/ft3) and the speed of the fan remains the same, what
will be the new static pressure and power?
a. 0.81 in. WG, 3.21 Hp
c. 0.71 in. Wg, 3.24 Hp
b. 0.81 in. WG, 2.24 Hp
d. 0.71 in. WG, 2.24 Hp
Solution:
New Static Pressure Required:
h2
h1
New Power Required:
P2
P1
2
1
h2
1
0.06018
P2
0.075
h2 = 0.80 in. WG
0.06018
0.075
P2 = 3.21 Hp
thus;
(a) 0.81 in. WG, 3.21 Hp

If the speed of the fan in problem 39 is increased so as to produce a static pressure of 1
in. WG at 200F. What will be the new speed and new capacity?
a. 446.54 rpm, 13,396.33 cfm
c. 457.45 rpm, 12,457.45 cfm
b. 454.34 rpm, 15,345.17 cfm
d. 745.54 rpm, 11,345.34 cfm
Solution:
New Speed Required:
N2
400
0.075
Q2
0.06018
12000
N2 = 446.54 rpm
0.075
0.06018
Q2 = 13,396.33 cfm
thus;
(a) 446.54 rpm, 13,396.33 cfm

If the speed of the fan of the previous examples (Problem 39-40) is increased so as to
deliver the same weight of air at 200F as at 70F. What will the new speed, new capacity,
new static pressure and new power?
a. 498.50 rpm, 14,955.14 cfm, 1.25 in. WG, 6.21 Hp
b. 646.54 rpm, 15,396.33 cfm, 2.25 in. WG, 7.21 Hp
c. 464.54 rpm, 15,396.33 cfm, 3.25 in. WG, 5.21 Hp
d. 546.54 rpm, 12,396.33 cfm, 4.25 in. WG, 4.21 Hp
Solution:
New Speed Required:
N2
400
New Static Pressure Required:
0.075
h2
0.06018
N2 = 498.50 rpm

Q2
12000
0.075
0.06018
h2 = 1.25 in. WG
New Power Required:
0.075
P2
0.06018
Q2 = 14,955.135 cfm
=(
0.075
0.06018
P2 = 6.21 Hp
thus;
a. 498.50 rpm, 14,955.14 cfm, 1.25 in. WG, 6.21 Hp

A fan discharges 10,000 cfm of air through a duct 2 ft by 2 ft against a static pressure of
0.90 in. of water. The gage fluid density is 62lb/ft3, air temperature is 85F and the
barometric pressure is 28.7 in. Hg. If the power input to the fan is measured as 3.6 Hp, what
is the over-all mechanical efficiency of the fan?
a. 50 %
c. 60 %
b. 56 %
d. 65 %
Solution:
em =
Pair
3.6
Solving for Pair :

V=
Q
A
h = hs + hv
h = 19.05 + 8.22 = 27.27 m
Q = 10000 ft3/min
V = 2500 ft/min = 41.67 ft/s
Q = 4.72 m3/s
V=
then;
2gh
41.67 =
2(32.2) hv
hv = 29.96 ft = 8.22 m
hs = hw (
1000
1.2
Pair = [1.2(0.00981)](4.72)(27.27)
Pair = 1.515 kW
air
hs = 0.9 (
Pair = gh
) = 750 in.
hs = 19.05 m
Pair = 2.03 Hp
thus;
(a) em =
2.03
3.6
= 0.564 = 56.4 %

A ventilation system includes a fan with a mechanical efficiency of 45% against a static
pressure of 30 cm WG. If the total pressure created by fan is 300m of air, what is the static
efficiency?
a. 37.5 %
c. 40.5 %
b. 35 %
d. 45 %
Solution:
es = em
thus;
hs
h
(a) es = 0.375 or 37.5 %
Solving for hs:

hs = 0.30
1000
1.2
hs = 250 m
then;
es = 0.45
250
300

It requires 55 Hp to compress 1000 cfm of air at 60F and 14.7 psi to a pressure of 10 psig.
The temperature of the air leaving the blower is 184F. What is the flow in cfm from the
blower discharge?
a. 852.64 cfm
c. 737.06 cfm
b. 801.62 cfm
d. 700.91 cfm
Solution:
P1 V1
T1
P1 Q
T1
P2 V
=
=
T2
P2 Q
T2
(14.7)(1000)
60 + 460
(10 + 47)Q2
184 + 460
thus;
(a) Q2 = 737.06 cfm

A blower draws 3000 cfm of air through a duct of 12 in. in diameter with a suction of 3 in.
of water. The air is discharged through a duct 10 in. in diameter against a pressure of 2in.
of water. The air is measured at 70F and 30.2 in. Hg. Calculate the air horsepower. Use
specific weight of 62.34lb/ft3.
a. 2.82 Hp
c. 3.87 Hp
b. 2.87 Hp
d. 1.75 Hp
Solution:
P = gh
Solving for h:
h = Zd Zs +
Pd - Ps
Vd2 Vs2
2g
where:
3000
Vs = 6010
ft /s
2
( ) ft
= 91.67 ft/s
4 12
Vd =
3000 3
ft /s
60
2 2
(1) ft
4
= 63.56 ft/s
Ps = h = (62.34)(2/12) = 0.072 psi

Pd = (62.34)(-3/12) = -0.108 psi
Zs = Zd
The density of Air @ 30.2 in. Hg and 70F
P
RT
14.7
30.2(29.92)(144)
53.34(70 + 460)
= 0.0756 lb/ft3
then;
h=
(0.072 + 0.108)(144)lb/ft3
0.0756 lb/ft3
(91.67) -(63.66)
2(32.2)
h = 410. 42 ft of air
thus;
Pair = gh = 0.0756(3000/60)(410.42) ft-lb/s
Pair = 1551.39 ft-lb/s (
1 Hp
550 ft-lb/s
(a) Pair = 2.82 Hp

A blower operating at 15,000 rpm, compresses air from 68F and 14.7 psia to 10 psig. The
design flow is 1350 cfm and at this point the BHp is 80. Determine the efficiency of the
blower at the design point.
a. 48.41 Hp
c. 55.62 Hp
b. 40.54 Hp
d. 57.65 Hp
Solution:
Pair = gh
Solving for the density of air at 68F and 14.7 psia
=
P
RT
(14.7)(144)
53.34 (68 + 460)
= 0.07516 lb/ft3
Solving for h:
h=
h=
kRT1
k - 1
k-1
[( 2 ) k -1
P1
1.4(53.34)(68 + 460)
1.4 - 1
24.7 1.4-1
[(
14.7
) 1.4 -1
h = 15,743.7 ft
then;
Pair = 0.07516(
1350
60
(15,743.7)
Pair = 26,624.17 ft-lb/s

thus;
(a) Pair = 48.41 Hp

A fan running at 2000 rpm delivers 16,000 cfm against 3 in. static pressure, thereby
consuming 15BHp. If the rpm is increased to 2200 rpm so that the rpm ratio is 1:1:1. What
is the new cfm?
a. 17,500
c. 17,600
b. 16,600
d. 16,500
Solution:
N2
N1
Q2
Q1
1.1 =
Q2
16000
thus;
(a) Q2 = 17,600 cfm

A 0.70 m vane axial fan is running at 2000 rpm delivers 7.5 m3/s against 0.08 m static
pressure thereby consuming 12 BkW. If the fan wheel diameter is increased from 0.70 m
to 0.76 m, so that the diameter ratio 1.10:1, what is the new static pressure?
a. 0.10 m
c. 0.13 m
b. 0.09 m
d. 0.15 m
Solution:
h2
h1
h2
0.8
D2
D1
= (1.10
thus;
(a) h2 = 0.097 m

At standard air density a fan delivers 8m3/s against 0.05 m static pressure consuming 12
BkW. If it will operate in Baguio City where due to high altitude the air density is only 82.4%
of the standard air density, what is the new BkW?
a. 7.86 BkW
c. 9.89 BkW
b. 15 BkW
d. 13 BkW
Solution:
P2
P1
P2
12
2
1
= (0.824)
thus;
(a) P2 = 9.89 BkW

What horsepower is supplied to air moving at 20 fpm through a 2x3 ft duct under a
pressure of 3in. WC?
a. 0.057 Hp
c. 0.123 Hp
b. 0.043 Hp
d. 0.241 Hp
Solution:
Pair = gh
where:
Q = AV
Q = (20 ft/min)[ (2)(3) ft2 ]
Q = 120 ft3/min = 2 ft3/s
h = 3 in. (
62.4
0.075
h = 2496 in.
h = 208 ft of air
then;
Pair = 0.075(2)(208) = 31.2 ft-lb/s
thus;
(a) Pair = 0.057 Hp

A fan whose static efficiency is 40% has a capacity of 60,000 ft3/hr at 60F and barometer
of 3 in. Hg and give a static pressure of 2 in. WC on full delivery. What size of electric
motor be used to drive this fan?
a. Hp
c. 2 Hp
b. 1 Hp
d. 1.5 Hp
Solution:
Pair
efan =
Pmotor
Solving for Pair:

Q = 60,000 ft3/hr
Q = 16.67 ft3/s
h = 2(
62.4
0.075
h = 1664 in. of air

h = 138.67 ft of air
then;
Pair = gh
Pair = 0.075(16.67)(138.67)
Pair = 173.37 ft-lb/s
Pair = 0.315 Hp
thus;
0.40 =
0.315 Hp
Pmotor
(b) Pmotor = 0.788 Hp say 1 Hp

A fan delivers 4.7 m3/s at a static pressure of 5.08 cm of water when operating at speed
of 400 rpm. If the power input required is 2.963 kW and if on the same installation 7.05
m3/s is desired, find the static pressure in cm of water.
a. 12.43 cm of H2o
c. 10.43 cm of H2o
b. 9.43 cm of H2o
d. 11.43 cm of H2o
Solution:
h2
h1
Solving for N2:

N2 2
N2
N1
N1
=(
h2
5.08
N2 2
N2
400
400
=(
Q2
Q1
7.05
4.7
N2 = 600 rpm
thus;
h2
5.08
600 2
=(
400
(d) h2 = 11.43 cm of H2o

What horsepower is supplied to air moving at 7m/min through a 70 cm x 90 cm duct
under pressure of 8 cm of H20?
a. 0.049 Hp
c. 0.077 Hp
b. 0.058 Hp
d. 0.066 Hp
Solution:
Pair = gh
where:
Q = AV
Q = [ (0.70)(0.90) ](7/60) m3/s
Q = 0.0735 m3/s
h = 0.08(1000/1.2)
h = 66.67 m of air
then;
Pair = [ 1.2(0.00981) ] (0.0735) (66.67)
Pair = 0.058 kW
thus;
(c) Pair = 0.077 Hp

What horsepower is supplied of air moving at 25 fpm through an air duct 2 ft x 3 ft? Fan
pressure is 4 in. of water column.
a. 0.0945 Hp
c. 0.0495 Hp
b. 0.495 Hp
d. 0.849 Hp
Solution:
Pair = gh
where:
Q = [ (2)(3) ](25/60) ft3/s = 2.5 ft3/s
h = 4(62.4/0.075) = 3328 in. of air = 277.33 ft of air
then;
Pair = (0.075)(2.5)(277.33)
Pair = 52 ft-b/s
thus;
(a) Pair = 0.0945 Hp

A fan delivers 1.42 m3/s air at static pressure head of 2.54 cm of water through a duct 300
mm in diameter and discharge it through a duct 275 mm in diameter. Determine the
static fan efficiency if the total fan mechanical efficiency is 70% and air is measured at
25C and 760 mm of Hg.
a. 60 %
c. 30 %
b. 40 %
d. 50 %
Solution:
estatic = efan (
hs
h
Solving for h:
V1 = Q/A1 = 1.42/( /4)(0.3)2
V1 = 20.089 m/s
V2 = Q/A2 = 1.42/( /4)(0.275)2
V2 = 23.907 m/s
hv =
V22 V12
2g
(23.907)2 (20.089
2(9.81)
hv = 8.562 m of air
hs = 0.0254 (1000/1.2)
hs = 21.17m of air
h = hs + hv
h = 21.17 + 8.562
h = 29.73 m of air
thus;
estatic = 0.70 (21.17/29.73)
estatic = 0.4984
(d) estatic = 49.84 %

A 95 tons refrigeration system has a compressor power of 90 Hp. Find the coefficient
of performance, COP.
a. 3.85
c. 4.77
b. 2.77
d. 1.99
Solution:
where:
RE = 95 tons = 334.02 kW
= 90 hp = 120.64 kW
thus;
334.02
120.64
(b) COP = 2.77
A refrigeration system operates on the reversed Carnot cycle. The minimum and
maximum temperatures are -25
and 72 , respectively. If the heat rejected at the
condenser is 6000 kJ/min, find the power input required.

a. 1 663.38 kJ/min
c. 1 686.83 kJ/min
b. 1 888.38 kJ/min
d. 1 886.83 kJ/min
Solution:

where:
T
72
25
273
345
273
345K
3
248
solving for :
= 17.39 kJ/min-K
thus,
345
(c)
248 17.39
1 686.83 /
248K

In a
0 refrigerating plant, the specific enthalpy of the refrigerant as it leaves the
condenser is 135 kJ/kg and as it leaves the evaporator it is 320 kJ/kg. If the mass flow the
refrigerant is 5 kg/min, calculate the refrigerating effect per hour.
a. 55 500 kJ/hr
c. 65 500 kJ/hr
b. 35 500 kJ/hr
d. 45 500 kJ/hr
Solution:

where:
m = 5 kg/min = 300 kg/hr
320 /
135 /
thus;
RE = 300 ( 320 135 )
(a) RE = 55 500 kJ/hr
A refrigerator is 2 m high, 1.2m wide and 1m deep. The over-all heat transfer
coefficient is 0.532 W/
. How many kilograms of 0
temperature is maintained at 10
ice will melt per hour if the inside
while the surrounding air temperature is at 35 ?
a. 1.60 kg
c. 2.60 kg
b. 1.80 kg
d. 2.80 kg
Solution:
Q=m
Solving for Q:
Q=UA
where:
A = 2 2 1.2
1.2 1
2 1
A = 11.2
COP = 0.532
11.2
Q = 148.96 W = 0.14896 kW
)(35 10)
then;
0.14896 = m ( 335 )
m = 4.4466 x 10
kg/s
thus; ice melted per hour is:

(a) m = 1.60 kg/hr
The power requirement of a Carnot refrigerator in maintaining a low temperature
region at 300 K is 1.5 kW per ton. Find the heat rejected.
a. 4.02 kW
c. 5.02 kW
b. 7.02 kW
d. 6.02 kW
Solution:
where:
=
+1
Solving for COP:

.
COP =
COP = 2.34
then;
=

+ 300
427.99
.
.
0.012
thus;
427.99 0.012
(c)
5.02

A simple saturated refrigeration cycle for R-12 system operates at an evaporating
temperature of -5
and a condensing temperature of 40 . Determine the volume
flow rate for a refrigerant capacity of 1 kW.

Properties of R-12:
At -5 ,
349.3
0.06496
At 40 ,
238.5
a. 0.0005866
c. 0.05865
b. 0.005866
d. 0.00005866
Solution:

Solving for m:

1 /
349.3 238.5 /
m = 0.00903 kg/s
thus;
= (0.00903)(0.06496)
(a)
0.0005866
25. Past ME Board Exam

A vapor compression refrigeration system is designed to have a capacity of 150 tons
of refrigeration. It produces chilled water from 22
to 2 . Its actual coefficient of
performance is 5.86 and 35% of the power supplied to the compressor is lost in the
form of friction and cylinder cooling losses. Determine the condenser cooling water
required for a temperature rise of 10 .
a. 14.75 kg/s
c. 18.65 kg/s
b. 15.65 kg/s
d. 13.76 kg/s
Solution:
By Energy Balance:
where:
RE = 150 ( 3.516 )
RE = 527.4 kW
from:
COP =
.
5.86 =
= 90 kW
then:
527.4
90
617.4 =
617.4
617.4 =
417.4 10
thus;
(a)
14.75

An air conditioning system of a high rise building has a capacity of 350 kW of
refrigeration, uses R-12. The evaporating and condensing temperatures are 0
35
respectively. Determine the mass of refrigeration 12 circulated per second.
Properties of R-12
At 0
At 35
0.05339
200 /
233.5 /
351.48 /
a. 2.97 kg/s
c. 4.57 kg/s
b. 3.57 kg/s
d. 1.97 kg/s
Solution:
m=
where:
35
350
0
351.48 /
and
35
233.5 /
thus;
m=
(a)
m = 2.97 kg/s

Cold salt brine at an initial temperature of 0
from 40
to 5
is used in a packing plant to chill beef
in 18 hours. Determine the volume of brine in liters per minute required
to cool 1000beeves of 250 kg each, if the final temperature of brine is 3 . Specific

heat of brine is 3.76 kJ/kg
and S.G. = 1.05. Specific heat of beef is 3.15 kJ/kg .
a. 37.59 kg/s
c. 38.79 kg/s
b. 39.67 kg/s
d. 35.67 kg/s
Solution
Volume of brine circulated:
=
where:
= 1
1.05
Solving for the
(3.14
=
424
then;

1.05
424
3.76 3
37.59
thus;
(a) 37.59 kg/s
424
)(40 5)

A simple vapor compression cycle develops 13 tons of refrigeration. Using ammonia
as refrigerant and operating at a condensing temperature of 24
and evaporating
temperature of -18 ; and assuming that the compressions are isentropic and that the
gas leaving the condenser is saturated. Find the power requirement.
a. 13 kW
c. 12 kW
b. 8.79 kW
d. 9.79 kW
Properties of R-12:
At 24 :

312.87 /
h @ 974 kPA ( P at 24
and
974 kPA
1657 /
At -18 :

-18
1439.94 /
0.5729
Solution
where:
@ 947
&
= 1657 kJ/kg
@
18
= 1439.94 kJ/kg
from:
13tons ( 3.516 kW/ton) = m ( 1439.94 312.87 ) kJ/kg
m = 0.0415 kg/s
thus;
0.0405 1567
(b)
24
= 8.79 kW
1439.94

A belt driven compressor is used in refrigeration system that will cool 10 liter per
second of water from 13
to 1 . The belt efficiency is 98%, motor efficiency is 85%
and the input of the compressor is 0.7 kW per ton of refrigeration. Find the actual
coefficient of performance of over-all efficiency is 65%.
a. 4.44
c. 6.44
b. 5.44
d. 3.44
Solution
COP =
where:
= [ 10(1)] ( 4.187 ) ( 13 1 )
= 502.44 kW
= 142.90 TOR
Solving for compressor work,
.
=
0.65 =
.
.
.
.
78.055
thus;
COP =
.
.
(c) COP = 6.44

Four thousand liters per hour of distillates are to be cooled from 21
to 12
of wax by weight is separated out at 15 . The specific heat of oil is 2 kJ/kg
and 12%
and S.G. is
0.87. The specific heat of the wax is 2.5 and the latent heat of fusion is 290 kJ/kg. The
specific heat of the wax is 2.5 and the latent heat is 290 kJ/kg. Allow 10% for the losses,
find the capacity of the refrigerating machine.
a. 20 TOR
c. 40 TOR
b. 51.08 TOR
d. 31.08 TOR
Solution
Distillate
Oil
21
15
Oil
-12
Wax
15
Wax
-12
where:
m = mass of distillate chilled per hour
( 0.87 ) 1
m = 400
m = 3,480 kg/hr = 0.97 kg/s

and;
= 0.97
(21 15)
= 11.64 kW
= (0.12)(0.97)
(290)
= 33.76 kW
= (0.12)(0.97)
(2.5)
(15+12)
= (0.88)(0.97)
(2)
(15+27)
33.76
7.86
0.10 11.64
= 7.86 kW
= 46.09 kW
46.09
9.935
thus;
109.285
(d)
31.08

A 50 ton vapor compression system using Ammonia as refrigerant operates between
20
condenser and -16
evaporator temperature. If simple saturation cycle with
isentropic compression is assumed, determine the piston displacement of the

reciprocating compressor to be used in the system operating at 600 rpm.
Properties of Ammonia:
At 20
At -16
274.9 /
1424.4 /
@ 857.12
&
@20
0.5296 /
1640 /
a. 5000
c. 7000
b. 6000
d. 8000
Solution
Piston displacement of the compressor:
=
Where:

m=
974 kPA
20
m=
-16
m=
m = 0.513 kg/s
then;
=
0.0081
8000
thus;
(a)
8000

A simple vapor compression cycle develops 15 tons of refrigeration using Ammonia
as refrigerant and operating at condensing temperature of 24
temperature of -18
and assuming compression are isentropic and that the gas
leaving the condenser is saturated, find the power per ton.

Properties of Ammonia
At 24
At -16
312.87 /
1439.94 /
1665 /
and evaporating
a. 0.702 kW/ton
c. 0.602 kW/ton
b. 0.802 kW/ton
d. 0.502 kW/ton
Solution

=
=
where:
Solving for m:
15(3.516) =
1439.94
312.87
M = 0.04688 kg/s
then;
0.04688 1665
1439.94
10.531
thus;

(a)
0.702

In an Ammonia refrigerator the pressure in the evaporator is 267,58 kPa and the
ammonia at entry is 0.12 dry while at exit is 0.91 dry. During compression the work
done per kg of ammonia is 17,033 kg-m. Calculate the coefficient of performance. If
the rate of ammonia circulation is 5.64 kg, calculate the volume of vapor entering
the compressor per minute. The compressor is single acting, its volumetric efficiency is
80% and it runs at 120 rpm. The ratio of stroke to bore is 1. Calculate the bore and
stroke. It is given that the latent enthalpy and specific volume of ammonia a 267.58
kPa are 320 kCal/kg and 0.436
/kg respectively.
a. 2.235
c. 1.457
b. 4.357
d. 3.567
Solution
267.58
@ 267.58
320
0.436
0.91 320
291.2
0.12 320
38.4
Solving for COP:

COP =
where:
RE =
201.2 -
38.4
= 252.8 kCal/kg = 1058.47 kJ/kg

= 17 033
( 0.00981 kN/
then;
COP =
.
.
= 6.335
Solving for the volume of vapor entering the compressor per minute:
where:
m = 5.64 kg/min
= 0.001527 + 0.91(0.436)
= 0.3983
then;
= 5.61
0.3983
= 2.235
Solving for the bore and stroke:

from:
=
LN
=
.
.
LN
(D)(120)
Note:
L=D
thus;
(a) L = D = 0.31 m = 31 cm
Freon-12 leaves the condenser of a refrigerating plan as a saturated liquid at 5.673
bar. The evaporator pressure is 1.509 bar and the refrigerant leaves the evaporator at
this pressure and at a temperature of -5 . Calculate the refrigerating effect per kg.
a. 132.88 kJ/kg
c. 160.91 kJ/kg
b. 123.77 kJ/kg
d. 123.86 kJ/kg
Solution
Properties of Freon-12
At 5.673 bar
At 1.509 bar ( t = - 20
54.87 /
17.82 /
178.73 /
160.91 /
Since, the saturation temperature at 1.509 is -20
and the refrigerant at this pressure
leaves the evaporator at -5 , it is superheated by -5 , it is superheated by 15

@ 1.509 bar superheated 15
then;
RE =
RE = 187.75 54.87
thus;
(a) RE = 132.88 kJ/kg
= 187.75 kJ/kg

An ammonia compressor operates at an evaporator pressure of 316 kPa and a
condenser pressure of 1514.2 kPa. The refrigerant is subcooled 5
and is superheated
8 . A twin cylinder compressor with bore to stroke ratio of 0.85 is to be used at 1200
rpm. The mechanical efficiency is 76%. For a load of 87.5 kW, determine the size of the
driving motor.
a. 24.26 kW
c. 34.26 kW
b. 25.26 kW
d. 35.26 kW
Solution
Properties of Ammonia
@ 316
& 0
1472 /
@ 316
& 0
0.41
@ 34
361.2 /
@ 1514.2
=
=
=
1715 /
34
1514 kPa
316 kPa
Solving for m:
Q=
87.5 = m (1472 361.2)

m = 0.079 kg/s
=
thus;
(b)
= 25.26 kW

A refrigeration system having a 30 kW capacity requires 10 Hp compressors. Find the
COP of the system.
a. 2.78
c. 4.02
b. 3.78
d. 5.02
Solution
COP =
=
thus;
(c) COP = 4.02
37.SupplementaryProblem
Arefrigeratingmachineusesammoniaastheworkingfluid.Itleavesthecompressoras
drysaturatedvaporat8.57barpassesthecondenseratthispressureandleavesassaturated
liquid.Thepressureintheevaporatoris1.902barandtheammonialeavestheevaporator0.96
dry.Iftherateofflowoftherefrigerantthroughthecircuitis2kg/min,calculatethevolume
takenintothecompressorinm/min,andtherefrigeratingeffectinkJ/min.
a. 1.198m/min,2,183kJ/min
b. 2.198m/min,3,183.38kJ/min
c.1.198m/min,3,183kJ/min
d.2.198m/min,2,183kJ/min
Solution:
PropertiesofAmmonia:
At8.57bar
Solvingforthevolumetakenintocompressorperminute:
SolvingfortheRefrigeratingeffectinkJ/min:
RE=m(h1h2)
89.8 0.96 1330.2

1336.79 /
275.1 /
Then;
At1.902bar
=275.1kJ/kg
=1462.6kJ/kg
=1420kJ/kg
=0.6237m/kg
=1330.2kJ/kg
Where:
/ min 1366.79
2,183.38
275.1 /
Thus;
(a)
=1.198m/min
=89.8kJ/kg
38.SupplementaryProblem:
Thewaterentersthecondenserat30Candleavesat50C.Iftheheatrejectedinthecondenser
is500kW,determinethevolumeofwaterneededtocooltherefrigerant.
a. 5.69kg/s
b. 4.69kg/s
4.187 50
30
c.6.69kg/s
d.7.69kg/s
Solution:
500
5.97
Then;thevolumeofwateris:
.
Thus;
.
A500kWrefrigerationsystemisusedtoproducecooledwaterfrom24Cto3C.Calculatethe
massflowrateofwaterinkg/s.
a. 5.69kg/s
b. 4.69kg/s
c.6.69kg/s
d.7.69kg/s
Solution:
500
4.187 24
(a)
Thus;

Avaporcompressionrefrigerationsystemisdesignedtohaveacapacityof100TOR.Itproduces
chilledwaterfrom22Cto2C.Itsactualcoefficientofperformanceis5.86and35%ofthe
powersuppliedtothecompressorislostintheformoffrictionandcylindercoolinglooses.
DeterminethesizeoftheelectricmotorrequiredtodrivethecompressorinkWandthevolume
flowrateofchilledwaterisL/s.
a. 92.31kW,4.199L/s
b. 90.71kW,5.277L/s
c.93.75kW,5.724L/s
d.91.75kW,7.575L/s
Solution:
5.86
.35
0.65
60
Thus;

92.31
Solvingforthevolumeflowofchilledwater:
100 3.516
4.187 22
4.199
Thus;
.
/
/
4.199 /
Thus;
(a)
Afourcylinder,singleacting,Vtypecompressorwith8cmand10cmstrokeoperatesat600
rpm.ItusedinaFreon12vaporcompressionsystemwithcondenserandevaporatorpressure
of725.5kPaand189.5kParespectively.Ifthecompressionisdryandisentropic,theclearance
is2percentandthethereisnosubcoolingorsuperheating(beforecompression)ofthe
refrigerant,determinetherefrigeratingcapacityofthecompressorintons.
a. 7.31TOR
b. 8.54TOR
c.7.54TOR
d.8.31TOR
Solution:
PropertiesofFreon12
14
345.365 /
14
0.0878951
@ 29
@ 725.5
227.557 /

@29
368 /
And
345.365
SolvingfortherefrigeratingCapacity
227.557
Solvingform:

0.0878951
0.10 4
Where;
1.02
0.9544
0.02
Then;
0.0878951
0.2183
0.9544
/
0.10 4
Thus;
0.2183 345.365

(a)
25.72
227.557
Thedrynessfractionsofthe
enteringandleavingtheevaporatorofarefrigeratingplantare
of
atthe
0.28and0.92respectively.Ifthespecificenthalpyoftheevaporation(
evaporatorpressureis290.7kJ/kg,Calculatethemassoficeat5Cthatwouldtheoreticallybe
throughthemachineis0.5kg/s.
madeperdayfromwaterat14Cwhenthemassflowof
Note:
Specificheatofwater=4.2kJ/kgK
Specificheatoffusionofice=2.04kJ/kgK
Enthalpyoffusion=335kJ/kgK
a. 17.89tons/day
b. 18.89tons/day
c.19.89tons/day
d.20.89tons/day
Solution:
throughevaporator:
Specificenthalpygainof
0.92
0.28
0.64 290.7
186.05 /
0.92
0.28
Heattobeextractedfromwatertomake1kgofice:
1 4.2 14
404 /
0.50
335
Then;
1 2.04 0
404
0.5 186.05
0.23
Thus;themassoficeintonsperday:
0.23
(b)
43.PastMEBoardProblem
Avaporcompressionrefrigerationsystemhasa30kWmotordrivingthecompressor.The
compressorinletpressureandtemperatureare64.17kPaand20Crespectivelyanddischarge
pressureof960kPa.Standardliquidenterstheexpansionvalve.UsingFreon12asrefrigerant,
determinethecapacityoftheunitintonsoftherefrigeration.
a. 17.145TOR
b. 18.145TOR
c.19.145TOR
d.20.145TOR
Solution:
PropertiesofFreon12
345 /
398 /
238.5 /
SolvingfortheRefrigeratingCapacity:
345
238.5
Solvingform:
30
398
0.566
0.566 345
60.28
345
Thus;
238.5
17.145
(a) 17.145TOR
ArefrigeratingsystemoperatesonthereversedCarnotCycle.Thehighertemperatureofthe
refrigerantinthesystemis120Fandtheloweris10F.Thecapacityis20tons.Neglectlosses.
DeterminethenetworkinBtu/min.
a. 935.21Btu/min
b. 457.57Btu/min
c.745.71Btu/min
d.765.81Btu/min
Solution:
Where;
120
10
460
580
460
470
Solvingfor :
8.511
Thus;
580
470 8.511
.
(a)
Whatisthecoefficientofperformanceofavaporcompressionrefrigerationsystemwiththe
followingproperties:Enthalpyatsuctionis190kJ/kg;enthalpyaftercompressionis210kJ/kg.
Theenthalpyaftercondensationis60kJ/kg.
a. 4.5
b. 5.5
Solution:
c.6.5
d.3.5
Thus;
(c)
Arefrigeratingmachineisdrivenbyamotorofoutputpower2.25kWand2.5tonsoficeat7C
madeperdayfromwaterat18C.Calculatethecoefficientofperformanceofthemachineand
expressitscapacityintermsoftonsoficeper24hoursfromand0C,takingthefollowing
values:
Specificheatofwater=4.2kJ/kgK
Specificheatoffusionofice=2.04kJ/kgK
Enthalpyoffusion=335kJ/kgK
a. 5.476,3.17tons/day
b. 4.476,3.17tons/day
c.5.476,4.17tons/day
d.4.476,4.17tons/day
Solution;
Where;
2.25
Solvingfor
4.2 18
0.029 424.88
12.32
Then;
.
.
5.476
Solvingformintonsperday;
2.5
424.88
335
2.04 0
2.5
335
424.88
Then;
3.17
Thus;
.
(a)
&
Determinetheheatextractedfrom2000kgofwaterfrom25Ctoiceat10C.
a. 621,150kJ
b. 721,150kJ
c.821,150kJ
d.921,150kJ
Solution:
Where;
209,350
2000 335
2000 4.187 25
Thus;
670, 000
2000 2.09 0
10
41,800
209, 350
(d)
670, 000

41, 800
Asingleacting,twincylinder,ammoniacompressorwithboreequaltostrokeisdrivenbyan
engineat250rpm.Themachineisinstalledinachillingplanttoproduce700kWofrefrigeration
at18Cevaporatingtemperature.Atthistemperaturethecoolingeffectperkgmassis1160kJ.
Thespecificvolumeofthevaporcompressoris0.592mperkilogram.Assume85%volumetric
efficiency,determinetheboreinmm.
a. 400mm
b. 300mm
c.450mm
d.500mm
Solution:
6.545
Solvingforthepistondisplacement,
.
.
From:

700
1160
0.603
0.42 /
Then;
Thus;
0.42
6.545
(a) D=0.40m=400mm
SaturatedvaporFreon12refrigerantat219.12kPaleavestheevaporatorandentersthe
compressorat5C.Therefrigerantleavesthecondenserassaturatedliquidat25Candenters
theexpansionvalveat22C.Heatrejectedfromthecondenseramountto74kW.Theworkto
thecompressoris55.5kJ/kgwhiletheheatlostfromthecompressoris4.2kJ/kg.If1.15kJ/kg0f
heatarelostinthepipingbetweenthecompressorandcondenser,calculatetherefrigeration
capacityintons.
a. 15.06TOR
b. 17.76TOR
Solution:
c.14.57TOR
d.12.75TOR
PropertiesofFreon12
@219.12
@ 25
223.65 /
@ 22
4.2
401.30 /
347.13 /
Byenergybalanceinthecompressor:
220.75 /
@ 219.12
350 /
55.5
Byenergybalanceinthepipingfromthecompressortocondenser;
1.15
400.15 /
Byenergybalanceinthecondenser;
223.65
400.15
176.5 /
Solvingforthemassflowrate:
0.419
Thus;therefrigeratingcapacity:

(a)
0.419 347.13
.
220.75
.
Themassflowofwaterenteringthecondenseris20kg/s.Ifthetemperaturedifference
betweentheentranceandexittemperatureis20C,determinetherejectedheatinthe
condenser.
a. 1,674.80kW
b. 1,774.80kW
Solution:
c.1,574.80kW
d.1,884.80kW
20 4.187 20
Thus;
(a)
Thecoolingloadofasmallwalkinfreezerhasbeencalculatedtobe1.10tonsofrefrigerationat
30C.Acompressorandmotormustselectedtohandletheload.Thefollowingconditionsare
given:
Refrigerant
F12
Compressor,rpm
600
Motor,rpm
1800
Compressordischargepressure
800kPa
Liquidreceivertemperature
20C
Assumedryandisentropiccompression,compressorvolumetricefficiencyof80%,mechanical
efficiencyof85%,andpowertransmissionefficiencyof90%.Calculatethedisplacementofthe
compressorincm.
a. 641.40
b. 651.40
c.661.40
d.671.40
Solution:
PropertiesofFreon12
@ 800
33

338.143 /
375 /
218.321 /
0.159375 /
Solvingfor
Where:
.
. 322
Then:
0.006414
0.0006414
Thus;
(a)
TherefrigerantleavesthecompressorandentersthecondenserofaFreon12refrigerating
plantat5.673barand50 andleavesthecondenserassaturatedliquidatthesamepressure.
Atcompressorsuctionthepressureis1.826barandtemperature0 .Calculatethecoefficient
ofperformance.
a. 3.09
b. 4.09
c.5.09
d.6.09
Solution:
PropertiesofFreon12
At5.673bar,Sat.temp=20
Thus,At50 refrigerantissuperheatedby30 ,andthecompressor

216.75 /
discharge,
At1.826bar,Sat.temp.=15
thus,At0 refrigerantissuperheatedby15
190.15 /

54.87
/

54.87 /
Then;
Thus;
(c)
Anindustrialplantrequires10kg/stocoolwaterfrom30 to1 .Findthetonsofrefrigeration
required.
a. 345.34
b. 245.34
c.145.34
d.445.34
Solution:
10 4.187 30
1, 214.23
345.34
Thus;
.
(a)
Airflowingatarateof2.5kg/sisheatedinaheatexchangerfrom10 to30 .Whatisthe
rateofheattransfer?
a. 100kW
b. 150kW
c.200kW
d.50kW
Solution:
2.5
100
1.0
30
10
Thus;
(a)

To cool farm products, 300 kg of ice at -4.4 0C are placed in bunker. Twenty four
hours later, the ice have melted into water at 7.2 0C. What is the average rate of
cooling provided by the ice in kJ/hr?
a. 2679.28 kJ/hr
c. 3679.8 kJ/hr
b. 5679.8 kJ/hr
d. 4679.28 kJ/hr
Solution

2.09 0
4.4
335
4.187 7.2
thus;
(d)
4679.28 /

The combined loads of an ice and cold storage are 25 tons of ice per day and
137,000 kJ/hr, respectively. Refrigeration required per ton of ice is 1.925. Ammonia
compressor carrying these combined loads operates between -14 0C and 42 0C
liquefaction. Determine the number of units of 7 pass multiple tube condenser each
unit made up of 200 mm pipe shell where there are 7 pieces extra strong 50 mm
tubes inside each pipe shell. Length is 6 m condensing water enters at 29 0C and
leaves at 38 0C. U = 539 W/m2 K, LMTD = 39.22 0C and cross flow factor = 0.75.
a. 3 units
c. 5 units
b. 4 units
d. 6 units
Solution
h1 = 1427.7 kJ/kg
h2 = 1714.0 kJ/kg
h3 = h4 = 383.5 kJ/kg
Let: n = no. of units

Q = heat gained by cooling water
Q1 = heat transferred per unit
then;

Solving for Q:
kW refrigeration for cold storage = 137,000 kJ/hr (1hr/3600s)
= 38.06 kW
kW refrigeration for ice plant
= 25(3.516)(1.925)
= 169.21 kW
QT = Total kW = 38.06 + 169.21

QT = 207.27
From:
m=
m=

.
.
m = 0.1985 kg/s
then;
Q = m (h2 h1)
= 0.1985 (1714 383.5)
= 264.12 kW
Solving for Q1:

Area of each unit, A:
A = (O.D.) L N
= (0.06) (6) (7)
= 7.92 m2
then;
Q1 = A U F (LMTD)
= (7.92)(539)(0.75)(39.22)
= 125,569.11 W
= 125.57 kW/units
thus;
n=
.
.
n = 2.10 units ( or n must be 3 units )

(a) n = 3

Calculate the power required by a system of one compressor serving two
evaporators. One evaporator carries a load of 35 kW at 10 0C and the other a load
of 70 kW at -5 0C. a back pressure valve reduces the pressure in the 10 0C evaporator
to that of the -5 0C evaporator. The condensing temperature is 37 0C. if the refrigerant
is ammonia , then what is the COP.
a. 4.33
c. 6.33
b. 5.33
d. 3.33
h3 = h4 = h7 @ 37 0C = 375.9 kJ/kg
h5 = h6 = hg @ 10 0C = 1471.6 kJ/kg
h8 = hg @ -5 0C = 1456.2 kJ/kg
h2 = h @ 1432 kPa (Psat @ 37 0C) and S2 = S1 = 1665 kJ/kg
then;

COP =
COP =
Solving for WC:
m4 =
= 0.0319 kg/s
m7 =
= 0.0648 kg/s
m1 = m6 + m8
= m4 + m7
= 0.0967 kg/s
By Energy Balance:
m1h1 = m6h6 + m8h8
0.0967h1 = 0.0319(1471.6) + 0.0648(1456.2)
h1 = 1461.3 kJ/kg
then;
WC = m (h2 h1)
= 0.0967(1665 1461.3)
= 19.7 kW
thus;
COP =
(a) COP = 5.33

Twenty pounds of water at an initial temperature of 80 0F are heated until the
temperature is increased to 190 0F. Compute the quality of heat energy supplied.
a. 2200 BTU
c. 2400 BTU
b. 2300 BTU
d. 2500 BTU
Solution
Q = mCpt
= (20lb)(1
thus;
(a) Q = 2200 Btu
)(190-80)0F

Suppose that 30 gpm of water are removed from 60 0F to 40 0F. Calculate the heat
energy removed in Btu per hour.
a. -299,880 Btu/hr
c. -199,880 Btu/hr
b. -399,880 Btu/hr
d. -499,880 Btu/hr
Solution
Q = mCpt
= m(1)(40-60)
Solving for m:
m = V
= (8.33
)(30
)(60
= 14,994
thus;
Q = (14,994)(1)(-20)
(a) Q = -299,880

If the latent heat of water is 144 Btu/lb, determine the quantity of latent heat given
up by 10 lb of water at 32 0F when it freezesinto ice at 32 0F.
a. 1550 Btu
c. 2880 Btu
b. 1440 Btu
d. 3100 Btu
Solution
Q = mLf
= 10 lb (144
thus;
(b) Q = 1440 Btu

Compute the cooling rate (energy flow rate in Btu/hr) produces by ice melting at the
rate of 150 lb/hr.
a. 30000 Btu/hr
c. 21,600 Btu/hr
b. 10,530 Btu/hr
d. 15,000 Btu/hr
Solution
Q = mLf
= (150 )(144
thus;
(c) Q = 21,600 Btu/hr

Twenty kilograms of water at initial temperature of 25 0C are heated until the
temperature is increased to 80 0C. Compute the quantity of heat energy supplied.
a. 4,605.7 kJ
c. 2,000.1 kJ
b. 4,000.2 kJ
d. 2,302.85 kJ
Solution
Q = mCpt
= (20 kg)(4.187
thus;
(a) Q = 4,605 kJ
)(80-25) K

One-tenth m3 of water is cooled from 39 0C to 2 0C. Determine the quantity of heat
energy rejected by the water.
a. 15,491.90 kJ
c. 17,321.90 kJ
b. 14,591.90 kJ
d. 18,231.80 kJ
Solution
Q = mCpt
=
1000
4.187 35
10
thus;
(a) Q = 15,491.90 kJ

Suppose that 30 kg/s of water are cooled from 35 0C to 10 0C. Compute the required
energy flow rate in kJ/s.
a. 3140.25 kW
c. 3457.75 kW
b. 3240.25 kW
d. 3567.25 kW
Solution
Q = mCpt
= 30
4.187
35
10
thus;
(a) Q = 3140.25 kJ/s or 3140.25 kW

Compute the cooling rate produced by ice melting at the rate of 150 kg/hr.
a. 737.5 kW
c. 937.5 kW
b. 837.5 kW
d. 637.5 kW
Solution
Q = mLf
= m 335
where:
m = 1.50 kg/hr
= 2.50 kg/s
Q = 2.5
335
thus;
(b) Q = 837.5 kJ/s or kW

Seventy-five hundred pounds of fresh beef enter a chilling cooler at 102 0F and are
chilled to 45 0F each day. Compute the product load in Btu per 24 hours. The specific
heat of beef above freezing is 0.75 Btu/lb-0F.
a. 320,600 Btu
c. 220,600 Btu
b. 420,600 Btu
d. 520,600 Btu
Solution
Production Load = (7500)(0.75)(102-45)
= 320,600 Btu per 24 hrs
(a) Product Load = 320,600 Btu per 24 hrs

Calculate the piston displacement of a two cylinder compressor rotating at 1450 rpm
if the diameter of the cylinder is 2.5 in. and the length of stroke.
a. 16.48 ft3/min
c. 14.48 ft3/min
b. 15.48 ft3/min
d. 17.48 ft3/min
Solution
VD =
2 1450 2
= 28,470.68 in.3/min
thus;
(a) VD = 16.48 ft3/min

Calculate the compression ratio of an R-12 compressor when the suction
temperature is 20F and the condensing temperature is 100F.
a. 4.68
c. 2.68
b. 1.68
d. 3.68
Solution:
Compression =
R=
Where:
At 20F, Ps = 35.75 psi
At 100F, Pd = 131.6 psi
Thus;
R=
.
.
(d) R = 3.68

Determine the shaft power required by the compressor if the theoretical power is 2.66
Hp and the overall efficiency of the compressor is 80%.
a. 2.22 Hp
c. 4.44 Hp
b. 3.33 Hp
d. 1.11 Hp
Solution:
Let eo = overall efficiency
Wc = compressor theoretical
Ws = shaft power
Then:
eo =
0.80 =
Thus;
(b) Ws = 3.33

A refrigeration compressor having a 10 in. flywheel is driven by a four-pole,
alternating current motor. If the diameter of the motor pulley is 4 in., determine the
speed of the compressor.
a. 700 rpm
c. 600 rpm
b. 500 rpm
d. 800 rpm
Solution:
N1D1 = N2D2
Where:
N1 = speed of the compressor
D1 = diameter of the compressor flywheel
N2 = speed of the compressor driver
D2 = diameter of the driver pulley
Note: If the compressor driver is a four-pole, alternating current motor operating
on 60 cycle power, the approximate driver speed is 1750 rpm. For a two-pole, alternating
motor the approximate speed is 3500 rpm.
N(10) = (1750)(4)
Thus;
(a) 700 rpm

Determine the estimated condenser load for an open-type compressor having a
cooling capacity of 16,500 BTU/hr and a heat rejection factor of 1.32.
a. 22,280 BTU/hr
c. 21,780 BTU/hr
b. 20,780 BTU/hr
d. 19,780 BTU/hr
Solution:
Condenser Load = Compressor Capacity x Heat rejection factor
= (16,500)(1.32)
= 21,780 BTU/hr
Thus;
(c) Condenser Load = 21,780 BTU/hr

If the load on the water-cooled condenser in 150,000 BTU/hr and the temperature rise
of the water in the condenser is 10F, what is the quantity of water circulated in gpm.
a. 30
c. 20
b. 40
d. 50
Solution:
Q = mCpt
150,000 = m(1)(10)
m = 15,000 lb/hr
Solving for V in gpm:
V=
,
.
/
/
V = 1800.72 gal/hr = 30.01 gal/min

Thus;
(a) V = 30 gpm
CHAPTER VI

A coil has an inlet temperature of 60 and outlet of 90 . If the mean temperature of the
coil is 110 , find the bypass factor of the coil.
a. 0.20
b. 0.30
c. 0.40
d. 0.50
Solution:

Bypass factor =
BF =
Thus;
(a) BF = 0.40

If the latent and sensible heat loads are 20 kW and 80 kW respectively, what is the
sensible heat ratio?
a. 0.80
b. 0.60
c. 0.70
d. 0.90
Solution:
Let: SHR = sensible heat ratio
SHR =
Thus;
(a) SHR = 0.80
14. A room being air conditioned is being held at 25 dry bulb and 50% relative
humidity. A flow rate of 5 of supply air at 15 dry bulb and 80% RH is being
delivered to the room to maintain that steady condition at 100 kPa. What is the sensible
heat absorbed from the room air in kW?
a. 50.8
b. 60.8
c. 40.5
d. 70.9
Solution:
=m
Solving for m:
PV = mRT
100(5) = m (0.287)(15 + 273)
m = 6.049 kg/s
thus;
= (6.049)(1.003)(25 25)
60.80 kW
(a)

Compute the pressure drop of 30 air flowing with a mean velocity of 8 m/s in a circular
sheet-metal duct 300 mm in diameter and 15 m long. Use a friction factor, f = 0.02 and
1.1644 / .
a. 37.26 Pa
b. 25.27 Pa
c. 29.34 Pa
d. 30.52 Pa
Solution:
=
=
Thus;
.
.
(a)
37.26 Pa

A pressure difference of 350 Pa is available to force 20oC air through a circular sheetmetal duct 450 mm in diameter and 25 m long. At 20oC,
1.204 /
and take
friction factor, f = 0.016. Determine the velocity.
a. 25.57 m/s
b. 27.55 m/s
c. 28.54 m/s
d. 24.85 m/s
Solution:
=
350 =
.
.
Thus;
(a) V = 25.57 m/s

A duct 0.40 m high and 0.80 m wide suspended from the ceiling in a corridor, makes a
right angle turn in the horizontal plane. The inner radius is 0.2 m and the outer radius is
1.0 m measured from the same center. The velocity of air in the duct is 10 m/s.
Compute the pressure drop in this elbow. Assuming; f = 0.3,
1.204 /
and L =
10 m.
a. 341 Pa
b. 441 Pa
Solution:
=
Where:
=
(for rectangular duct)
c. 143 Pa
d. 144 Pa
. .
= 0.53 m
Thus;
=
(a)
.
.
341 Pa

A rectangular duct has a dimension of 0.25 m by 1 m. Determine the equivalent
diameter of the duct.
a. 0.40 m
b. 0.80 m
c. 0.70 m
d. 0.30 m
Solution:
=
=
.
.
Thus;
0.40 m
(a)

A 0.30 x 0.40 m branch duct leaves a 0.30 x 0.60 main duct at an angle of 60. The air
temperature is 20 . The dimensions of the main duct remain constant following branch.
. What is the pressure downstream in the main duct.
The flow rate upstream is 2.7
Note: at 20 ,
1.2041 / .
a. 346 Pa
b. 436 Pa
Solution:
Pressure Loss in the main duct:
c. 634 Pa
d. 643 Pa
0.4 1
and
Solving for
1.3
2.7 =
= 1.4
= 7.78 m/s

.
.
= 15 m/s

.
.
= 10.43 m/s
Then;
=
= 3.38
From: Bernoulli Equation:

2
9.81 1.2041
Thus;
250 3.38
9.81 1.02041
15
7.78
2 9.81
345. 64
(a)

A sudden enlargement in a circular duct measures 0.20 m diameter upstream and 0.40
m downstream. The upstream pressure is 150 Pa, downstream pressure is 200 Pa.
What is the flow rate of 20C air through the fitting? Use p = 1.02041 kg/m3.
a. 0.49 m3/s
c. 0.38 m3/s
b. 0.83 m3/s
d. 0.94 m3/s
Solution:
Q = AuVu
Solving for Vu:
1
Ploss =
where:
=(
= 0.25
then:
(200-150) =
0.25
= 12.15 m/s
Thus;
Q=
(12.15)
(a)Q = 0.38 m3/s

Water at 55oC is cooled in a cooling tower which has an efficiency of 65%. The
temperature of the surrounding air is 32oC dry bulb and 70% relative humidity. The heat
dissipated from the condenser is 2,300,000 kJ/hr. Find the capacity in liters per second
of the pump used in the cooling tower.
a. 8.50 L/s
c. 7.60 L/s
b. 6.80 L/s
d. 6.70 L/s
Solution:
Pump Capacity = m f@t4:
Solving for m:
e=
from psychometric chart:
At 32oC and 70% RH:
= 27.40oC
0.65 =
= 37.06oC
Using energy balance is the condenser:
mCp(t3 t4) =
m(4.187)(55 37.06) =
m = 8.51 kg/s
From steam table at t4 = 37.06oC:
f = 1.0068 L/kg
Thus;
Pump Capacity = (8.51 kg/s) (1.0068 L/kg)
(a) Pump Capacity = 8.57 L/s

An atmospheric cooling tower is to provide cooling for the jacket water of a four stroke,
800 kW Diesel generator. The cooling tower efficiency is 60% at a temperature
approach of 10oC. If the ambient air has a relative humidity of 70% and dry bulb
temperature of 32oC, determine the cooling tower supplied to the diesel engine in liters
per hour. Generator efficiency is 97% useful work = 30% and cooling loss = 25%.
a. 39,800 L/hr
c. 45,700L/hr
b. 35,700 L/hr
d. 49,800 L/hr
Solution:
Volume of water = m f at t4 :
Solving for m and f :
At tdb1 = 32oC and RH = 70%
twb = 27.45oC
tapproach = t4 27.45
10 = t4 27.45
t4 = 37.45oC
Brake power of engine = Power input to generator

=
Heat supplied to Engine, QA:
= 824.74 kW
QA =
.
.
= 2749.14 kW
Heat absorbed by Cooling water, Qw:

Qw = 0.25 (2749.14)
= 687.285
mwCpw(t3 t4) = 687.285
mw(4.187)(52.4 37.45) = 687.285
mw = 10.98 kg/s
= 39,527.14 kg/hr
Specific Volume of water at 37.45oC, f = 1.007 L/kg
Thus;
Vol. of cooling water, Vw:
Vw = 39,527.14(1.007) L/hr
(a) Vw = 39,803.83 L/hr

Fifty gallons per minute of water enters a cooling tower at 46oC. Atmospheric air at 16oC
db and 55% RH enters the tower at 2.85 m3/s and leaves at 32oC saturated. Determine
the volume of water that leaves the tower.
a. 4.10 L/s
c. 2.10 L/s
b. 3.10 L/s
d. 5.10 L/s
Solution:
Volume water leaving the tower, V4:
V4 = m4(f at t4)
Solving for m4 and f:
At tdb1 = 16oC and 55%RH
1 = 0.828 m3/kg
At 32oC and 100% RH

2 = 0.0308 kg/kg
1 = 0.0056 kg/kg
h2 = 110.9 kJ/kg
h1 = 32 kJ/kg
h3 = hf at 46oC
ma =
= 192.62 kJ/kg
3 = 0.0010103 m3/kg
= 3.44 kg/s
m3 =
.
.
= 3.12 kg/s
By mass balance:
m3 m4 = ma(W2 W1)
3.12 m4 = 3.44 (0.0308 0.0056)
(a) m4 = 3.09 kg/s
By energy balance:
m3h3 m4h4 = ma(h2 h1)
3.12(192.62) 3.09h4 = 3.44(110.9 32)
h4 = 106.65 kJ/kg
From Steam table 1:
t4 = 25.42oC
4 = 1.0031 L/kg
Thus; volume of water leaving, Vw:
Vw = 3.09(1.0031)
(a) Vw = 3.10 L/s

A 250,000 kg/hr of water 35oC enters a cooling tower where it is to be cooled to 17.5oC.
The energy is to be exchanged with atmospheric air entering the units at 15oC and
leaving the unit at 30oC. The air enters at 30% RH and leaves at 85% RH. If all process
are assumed to occur at atmospheric pressure, determine the percentage of total water
flow that is make up water.
a. 2.22%
c. 4.44%
b. 3.335
d. 1.11%
Solution:
Percentage make-up water =
Solving for mass of make-up water:

At 15oC and 30% RH:
h1 = 23.02 kJ/kg
w1 = 0.0033 kg/kg
At 30oC and 85% RH:
h2 = 89.01 kJ/kg
w1 = 0.0233 kg/kg
Heat lost by water = Heat gained by air
mwCpw = ma(h2 h1)
250,000(4.187)(35 17.5) = ma(89.01 23.02)
ma = 277,589.41 kg/hr
then; the mass of make-up water, ms:
m5 = ma(w2 w1)
m5 = 277,589.41(0.0233 0.0033)
= 5,551.79 kg/hr
Thus; the percentage make-up water,
%make-up =
.
,
(a) %make up = 0.0222 or 2.22%

How much refrigeration capacity is required to cool 56.67 m3 of air per minute from
29oC to 21oC. Assume that the cooled air is saturated.
a. 4.76 TOR
c. 3.76 TOR
b. 1.76 TOR
d. 2.76 TOR
Solution:
Refrigeration Capacity, QA:
QA = ma(h1 h2)
Solving for ma:
From psychometric chart:
At 21oC db and 100% RH
h1 = 70 kJ/kg
1 = 0.875 m3/kg
At constant SH intersecting 29oC db:
h2 = 70 kJ/kg

ma =
=

.
.
ma = 64.766 kg/min
= 1.079 kg/s
Then;
QA = 1.079(70-61)
= 9.715 kW
Thus;
(a) 2.76 tons of refrigeration

Find the refrigeration capacity required to cool 29 cubic meter per minute from 29oC to
18oC if air from the outside has an RH of 90%.
a. 2.9 TOR
c. 4.9 TOR
b. 3.9 TOR
d. 5.9 TOR
Solution:
Qn = ma(h1 h2)
Solving for ma:
From psychometric chart:
h1 = 88.45 kJ/kg
h2 = 50.45 kJ/kg
1 = 0.886 m3/kg
ma =
= 32.73 kg/min
ma = 0.546 kg/s
then;
QA = 0.546(88.45 50.45)
= 20.75 kW
Thus;
(a) QA = 5.9 tons of refrigeration
The temperature of the air in a dryer is maintained constant by the use of steam coils
within the dryer. The product enters the dryer at the rate of one metric ton per hour. The
initial moisture content is 3 kg moisture per kg of dry solid and will be dried to moisture
content of 0.10 kg moisture per kg dry solid. Air enters the dryer with a humidity ratio of
0.016 kg moisture per kg of dry air and leaves with a relative humidity of 100% while the
temperature remains constant at 60 . If the total pressure of the air is 101.3 kPa,
determine the capacity of the forced draft far to handle this air in
.
a. 85.75
c. 55.87
b. 87.55
d. 58.75
Capacity of fan =
:
Solving for m and

At point 1:
.
0.16 =
2.54
1
101.3
2.54 1 = 0.287 (60 + 273)

1
0.968
At point 2:
= 19.94 kPa
= RH (
= (1)(19.94)
= 19.94 kPa
=
=
.
.
= 0.1524 kg/kg
At point 3:
= 250 kg
At point 4:
.
Moisture content =
= 0.0909 or 9.09%
= 0.0909
= 0.0909
= 275
250
/
By mass balance in the dryer:

=
=
Thus;
Capacity of Fan = 5315.25 (0.968)
= 5142.16
(a) Capacity of the Fan = 85.75
Alternate Solution:
Fan Capacity =
From psychrometric chart:
At 60
and 0.016 humidity ratio:
= 0.968
From steam table at 60 ,

=
0.1524
Moisture removed = m (
19.94
3 (350) (0.10)(250) = m (0.154- 0.1016)

M = 5315.25 kg/hr
Thus;
Fan Capacity = 5315.16 (0.968)
= 5145.16
(a) Fan Capacity = 82.75

Wet material containing 215% moisture (dry basis) is to be dried at the rate of 1.5 kg/s
in a continuous dryer to give a product containing 5% moisture (wet basis). The drying
medium consist of air heated to 373 K and containing water vapor equivalent to a partial
pressure of 1.40 kPa. The air leaves the dryer at 310 K and 70% saturated. Calculate
how much air will be required to remove the moisture.
a. 50 kg/s
c. 60 kg/s
b. 55 kg/s
d. 65 kg/s
Solution:
Amount of moisture removed = amount of moisture absorbed by air.
Let m = rate of flow of dried product
.
(1.5) = 0.95 m
m = 0.501 kg/s
Amount of moisture removed = 1.5 0.501

= 0.999 kg/s
Solving for W1:
W1 =
.
.
= 0.00871 kg/kg
From psychrometric chart, W2 = 0.0289 kg/kg
Then,
ma(0.0289 0.00871) = amount of moisture removed
ma(0.0289 0.00871) = 0.999
thus;
(a) ma = 49.48 kg/s

One hundred fifty cubic meters of air per minute at 35oC dry bulb and 25oC wet bulb
temperature are to be cooled to 21oC. Determine the refrigeration capacity.
a. 10 TOR
c. 12 TOR
b. 11 TOR
d. 13 TOR
Solution:
QA = maCp(t1 t2)
Solving for ma:
1
= 0.892 m3/kg,
ma =
then;
QA = 2.8 (1) (35 - 21)
= 39.24 kW
Thus;
(a) QA = 11.16 TOR
= 0.855 m3/kg
= 168.16 kg/min

In an auditorium maintained at a temperature not to exceed 24oC and relative humidity
not to exceed 60%, a sensible heat load of 132 kW and 78 kg of moisture per hour to be
remove. Air is supplied to the auditorium at 18oC. How many kilograms of air must be
supplied per hour?
a. 79,200 kg/hr
c. 72,900 kg/hr
b. 97,200 kg/hr
d. 92,700 kg/hr
Solution:
Qs = maCp(t2 t1)
132 = ma(1)(24 - 18)
ma = 22 kg/s
thus;
(a) ma = 79,200 kg/hr

Eleven thousand three hundred kilograms per hour of water enters a cooling
tower at 45oC. Atmospheric air at 16oC and 55 percent relative humidity enters
the tower at the rate of 10,200 m3/hr and leaves at 32oC and saturated.
Determine the mass of water evaporated per hour during the cooling process?
a. 2,912.53 kg/hr
c. 1,292.53 kg/hr
b. 2,219.53 kg/hr
d. 1,912.53 kg/hr
Solution:
Mass of water evaporated, mW:
mW = ma (SH2 SH1)
From Psychrometric chart:
Entering air at 16oC and 55% RH:
V1 = 0.83 m3/kg; SH1 = 0.07 kg/kg
Leaving air at 32oC and 100% RH (Saturated)
SH2 = 0.307 kg/kg
Mass of air entering the tower:
3
ma =
10,200 m /hr
0.83 m3 /kg
= 12,289.16 kg/hr
then;
mW = 12,289.16 (0.307 0.07)
thus;
(a) mW = 2,192.53 kg/hr
Water at 55 C is cooled in a cooling tower which has an efficiency of 65%. The
temperature of the surrounding air is 32 C dry bulb and relative humidity of 70%.
The heat dissipated from the condenser is 2,300,000 kJ/hr . Find the capacity in
liters per second of the pump used in the tower.
a. 8.66 L/s
c. 4.76 L/s
b. 8.76 L/s
d. 7.26 L/s
Solution:
From Psychrometric Chart:
twb1 = 27.5oC
The temperature of water leaving the tower can be determined by tower
efficiency equation:
Tower eff. =
=
0.65 =
Actual cooling range

Theoretical Cooling range
ta -tb
ta -twb1
55-tb
55-27.5
tb = 37.125 oC
By Energy Balance in the condenser:
QR = mw Cpw ( ta tb )
2,300,000 = mw (4.187) (55-37.125)
mw = 30,731.15 kg/hr
Density of water at 55oC:
w =
1
vf@55o C
1
0.0010146
= 985.6 kg/m3
Then; the capacity of the pump to be used in the cooling tower:
Pump capacity =
kg
hr
kg
935.6
hr
30,731
(1000
(3600
L
)
m3
s
)
hr
thus;
(a) Pump Capacity = 8.66 L/s
A dryer is to deliver 1000kg/hour of palay with a final moisture content of 10%. The
initial moisture content in the feed is 15% at atmospheric condition with 32 C dry
bulb and 21 C wet bulb. The dryer is maintained at 45 C while the relative
humidity of the hot humid air from the dryer is 80%. If the steam pressure supplied
to the heater is 2 Mpa, determine the air supplied to the dryer in m/hr.
a. 1332.25 m/hr
c. 1223.25 m/hr
b. 1233.25 m/hr
d. 1523.13 m/hr
Solution:
Amount of moisture removed = Amount of moisture absorbed;
Let; m = amount of palay in wet feed
Solid in wet feed = solid in dried product

0.85 = 0.90 (1000)
m = 1,058.83 kg/hr
Amount of moisture removed:
m = 1,058.83 -1000 = 58.28
W1 = W2 = 0.0111 kg/kg; v2 = 0.915 m3/kg
W3 = 0.0515 kg/kg
then; the amount of moisture absorbed;
= ma (W3 W2)
58.823 = ma (0.0515 0.0111)
ma = 1456.015 kg/hr
Va = 1456.015 kg/hr (0.915 m3/kg)
thus;
(a) Va = 1332.25 m3/hr
Copra enters a dryer containing 60% water and 40% of solids and leaves with 5%
water and 95% solids. Find the weight of water removed based on each pound of
original product.
a. 0.58 lb
c. 0.47 lb
b. 0.40 lb
d. 0.67 lb
Solution:
Let; m = weight of original product per lb of wet feed
Solid in wet feed = solid in dried product
0.95 = 0.40 (1)
m = 0.42 lb
thus;
Weight of water removed = 1 0.42
(a) Weight of water removed = 0.58
In an air conditioning system, If the re-circulated air is three times the outside the
mass of supply air is 20 kg/s, what is the mass of the outside air?
a. 3 kg/s
c. 5 kg/s
b. 4 kg/s
d. 6 kg/s
Solution:
mo + mt = ms
mo + 3mo = 20
4mo = 20
thus;
(a) mo = 5 kg/s
An auditorium is to be maintained at a temperature of 26 C dry bulb and 50%
RH. Air is to be supplied at a temperature not lower than 15 C dry bulb. The
sensible heat gain is 110 kW and the latent gain is 37.5 kW. Take ventilating air as
25% by weight of the air from the room, and is at 35 C dry bulb and 60% RH.
Determine refrigerating capacity in tons.
a. 43.45
c. 63.28
b. 54.23
d. 76.34
Solution:
Refrigeration Capacity = ms (h4 h1)
h3 = 90.49 KJ/kg
h2 = 53 KJ/kg
Solving for ms:
Qs = ms Cp (t2 t1)
110 = ms (1.0) (26 15)
ms = 10 kg/s
Solving for h3:
QT = ms (h2 h1)
110 + 37.5 = 10 (53 h1)

h1 = 38.25 KJ/kg
Solving for h4:
by mass balance:
mo + mr = ms
0.25mr + mr = 10
mr = 8 kg/s
by Heat Balance
moh3 + mrh2 = msh4
[0.25(8)] (90.49) + 8 (53) = 10 h4
h4 = 60.50 KJ/kg
thus;
Refrigerating Capacity = 10 (60.50 38.25)
= 222.48 kW
Refrigerating Capacity = 63.28 Tons of Refrigeration
An assembly hall was to have an air conditioning unit installed which would be
maintained at 26 C dry bulb and at 50% RH. The unit delivers air at 15 C dry bulb
temperature and the calculated sensible heat load is 150 kW and the latent heat
is 51.3 kW. Twenty percent by weight of extracted air is made up of outside air at
34 C dry bulb and 60% RH, while 80% is extracted by the air conditioner from the
assembly hall. Determine the air conditioners refrigeration capacity in tons of
refrigeration and its ventilation load in kW.
a. 83.22 TOR, 37.47 TOR
c. 89.56 TOR, 45.77 TOR
b. 76.43 TOR, 57.34 TOR
d. 56.78 TOR, 47.68 TOR
Solution:
QA = ms (h4 h1)
Ventilation load, QV:
QV = mo (h3 h1)
Solving for ms:
Qs = ms Cp (t2 t1)
150 = ms (1.0) (26 15)
ms = 13.64 kg/s
h3 = 86.5 KJ/kg
h2 = 53 KJ/kg
Solving for h1:
QT = ms (h2 h1)
150 + 51.3 = 13.64 (53 h1)
h1 = 38.24 KJ/kg
Solving for mr and mo:
mr = 0.80 (13.64)
mr = 10.91 kg/s
mo = 0.20 (13.64)
= 2.73 kg/s
By heat balance:
moh3 + msh4
2.736(86.5) + 10.91(53) = 13.64h4
h4 = 59.69 kJ/kg
thus;
QA = 13.64 (59.69 38.24)
= 292.61 kW
= 83.22 tons of refrigeration
Qv = 2.73(86.5 38.24)
= 131.75 kW
= 37.47 tons of refrigeration
(a) QA = 83.22 TOR, QV = 37.47 TOR
An air conditioned theater is to be maintained at 80 F dry bulb temperature and
50% RH. The calculated total sensible heat load in the theater is 620,000 BTU/hr,
and the latent heat load is 210,000 Btu/hr. The air mixture at 84 F and 59 F wet
bulb temperature by chilled water cooling coils and delivered as supply air to the
theater. Calculate the tons refrigeration required.
a. 100.65 TOR
c. 142.67 TOR
b. 124.67 TOR
d. 112.60 TOR
Solution:
Conditioner Capacity, QA:
QA = ms (h4 h1)
Solving for ms:
QT = ms (h2 h1)
h4 = 35.82 Btu/lb
h1 = 25.78 Btu/lb
h2 = 31.35 Btu/lb
then;
620,000 + 210,000 = ms (31.35 25.78)
ms = 149,012.57 lb/hr
thus;
QA =
149,012.57 35.82 - 25.78 Btu/hr

12,000 Btu/hr
TOR
(a) QA = 124.67 Tons of Refrigeration

Determine the quantity of heat required to raise 20 m /min of air 20 C and 80
percent relative humidity to 35 C.
a. 5 kW
c. 7 kW
b. 6 kW
d. 8 kW
Solution:
At tdb1 = 20oC and 80% RH
h1 = 50kJ/kg
v1 = 0.85 m3/kg
h2 = 65.5 kJ/kg
Q = m (h2 h1)
Solving for m:
m=
V
v
20
0.847
= 23.61 kg/min
= 0.394 kg/s
thus;
Q = 0.394 (65.5 50)
(a) Q = 6.10 kJ/s or kW
Determine the partial pressure of water vapor if the barometric pressure is 101.325
kPa and the humidity ratio is 0.05.
a. 7.54 kPa
c. 5.74 kPa
b. 4.75 kPa
d. 5.47 kPa
Solution:
W = 0.622
Pv
Pt -Pv
0.05 = 0.622
Pv
101.325-Pv
thus;
(a) 7.54 kPa
The evaporative condenser of an ammonia refrigeration plant has a water flow
rate 226 kg/s and enters a natural draft cooling tower at 40 C. The water is
cooled to 29 C by air entering at 38 C db and 24 C wb. The air leaves the
tower as saturated at 40 C db. Calculate the make-up water required in kg/hr.
Water properties:
At 49oC; hf = 167.48 kJ/kg
Air Properties:
At 38oC db and 42oC wb
At 29oC; hg = 121.43 kJ/kg
h = 72.5 kJ/kg
w = 0.013 kg/kg
At 40oC db saturated;
h = 166 kJ/kg, w = 0.0488 kg/kg
a. 8977
c. 8055
b. 8055
d. 8388
Solution:
m = ma (W2 W1)
Solving for mass of air, ma:
Heat absorbed by air = heat rejected by water
ma (h2 h1) = mw Cw t
ma (166 72.5) = 126 (4.187) (40 29)
ma = 62.07 kg/s
then;
m = 62.07 (0.0488 0.013)
= 2.22 kg/s
thus;
(a) m = 7999.08 kg/hr
Determine the absolute humidity (vapor density) of an air sample that has a dew
point temperature of 45 F if the value of the gas constant R for low pressure
water vapor is 85.66 ft-lbm R. The vapor pressure corresponding to a saturation
temperature of 45 F is 0.1475 psia.
a. 0.000491 lb/ft
c. 0.000149 lb/ft
b. 0.000941 lb/ft
d. 0.000194 lb/ft
Solution:
PV = mRT
m
V
=
==
P
RT
0.1475 (144)
85.66 (45+460)
thus;
= 0.000491 lb/ft3

A certain sample of the air has a temperature of 70 F (partial pressure of 0.36
psia) and a dew point temperature of 50F. The partial pressure of the water is
vapor corresponding to a 50 F dew point temperature is 0.178 psia. Determine

the relative humidity RH.
a. 49.44%
c. 39.44%
b. 59.44%
d. 69.44%
Solution
RH =
=
Actual partial pressure

partial pressure at saturation
0.178
0.36
x 100
x 100
thus;
(a) RH = 49.44 %
Air at normal atmospheric pressure has a temperature of 70 F and a dew point
temperature of 50 F. Determine the saturation ratio of the air. The humidity ratios
corresponding to dew point temperatures of 50 F and 70 F respectively are
0.00763 lb/lb and 0.01576 lb/lb respectively
a. 58.51%
c. 38.31%
b. 28.21%
d. 48.41%
Solution:
Saturation ratio =
=
Wactual
Wsaturation
0.00763
0.01576
x 100
x 100
thus;
(c) Saturation ratio = 48.41
Determine the sensible heat of 5 lb of air having a dry bulb temperature of 70 F
and a humidity ratio of 0.0092 lb/lb. The latter corresponding to a dew point
temperature of approximately 55F.
a. 94 BTU
c. 84 BTU
b. 48 BTU
d. 49 BTU
Solution:
Qs = m (0.24 DB)
= 5 (0.24)(70)
thus;
Qs = 84 BTU
If the total heat removed per pound of dry air is 0.10 Btu and the sensible heat
removed per pound of dry air is 6 BTU. Compute the sensible heat factor (SHR).
a. 0.50
c. 0.70
b. 0.60
d. 0.40
Solution:
SHR =
QR
QS
6
10
thus;
(a) SHR = 0.60
Determine the approximate load on a cooling tower if the entering and leaving
temperatures are 96 F and 88 F, respectively and the flow rate of the water over
the tower is 30 gpm.
a. 2500 Btu/min
c. 3000 Btu/min
b. 2000 Btu/min
d. 3500 Btu/min
Solution:
Tower load = 8.33 V t Btu/min
= 8.33 (30) (96 88)
thus;
(b) Tower load = 2000 Btu/min

Determine the equipment standard air volume for 150 m/s of air having a dry
bulb temperature of 15 C.
a. 150 m/s
c. 170 m/s
b. 160 m/s
d. 180 m/s
Solution:
Vs =
150 (21+273)
(15+273)
thus;
(a) Vs = 153 m3/s
Determine the quantity of water lost by bleed off if the water flow rate over the
tower is 30 gpm and the range is 10 F. Percent bleed-off required is 33%.
a. 0.077 gpm
c. 0.099 gpm
b. 0.088 gpm
d. 0.066 gpm
Solution:
The quantity of water lost by bleed-off,
Vw = 0.33 (30)
thus;
(a) Vw = 0.099 gpm

The mass of an outside air is 50C in an air conditioning unit is 60 kg. find the
temperature after mixing if the outside air mixed with 40 kg with recirculated air at
35 C.
a. 44 C
c. 52 C
b. 39 C
d. 47 C
Solution:
moto + mrtr = msts
60(50) + 40(35) = (60 + 40)ts
thus;
(a) ts = 44oC

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1. Calculate the energy transfer rate across 6 in.

wall of firebrick with a temperature

20.1 0.3927 100

Enthalpy: hf=914.52 kJ/kg

heat transmitted by convection

heat transmitted by radiation

For the surface coefficients;

Then the annual cost of heat lost:

20,713 8760 250

8. Determine the vacuum efficiency of a surface condenser which operates at a vacuum

0.50 1.2247 150

heat transmitted by convection

heat transmitted by radiation

11. A heat exchanger is to be designed for the following specifications:

Solving for Q/A:

c. 608.24 kJ, 13.12 C

b. 708.24 kJ, 11.12 C

d. 508.24 kJ, 14.12 C

Thus, the heat transfer per hour is 508.24 kJ

overall heat transfer coefficient

19. Supplementary Problem

Solving for Isentropic Power:

1.4(101.325)(38/60) (1.68/1) 1.4 -1

Solving for Q1:

Solving for hv:

26. Supplementary Problem

Solving for Pair:

(a) pmotor = 46.30 kW

Weight of fuel burned per hour

Ultimate Analysis of fuel :

Fuel bed and air heater resistance

Solving for Pair:

= 9.92 kg air / kg fuel

(a) Pmotor = 87.84 kW

Solving for Pair:

30. Supplementary Problem

5 Hp; 1200 rpm

c. 3.28 cm WC, 5.50 Hp

b. 3.95 cm WC, 5.5 Hp

d. 3.95 cm WC, 4.10 Hp

Solving for hv:

Total mass of air used per kg of fuel burned:

1.15 =17.58 kg air/kg fuel

Mass flow rate of air for combustion:

c. 482 kW, 256 mm WG

b. 492 kW, 241 mm WG

d. 492 kW, 256 mm WG

Solving for the static pressure, h2:

= 7.382 kg air / kg fuel

= 207 830.05 m3/hr

= 114,306.53 m3/hr = 31.75 m3/s

Solving for Pair:

(a) Pmotor = 92.99 Hp

37. Supplementary Problem

145C and a counterflow air cooler

30C and leaves at 38C. calculate the arithmetic

t max 145 38 107C

38. Supplementary Problem

10C. At what rate is heat conducted through each square meter

39. Supplementary Problem

k 0.0024Cal / s m C which is 2m by 3m and 5mm thick, when the

temperatures of the surfaces are 20 C and 10 C.

40. Supplementary problem

10 C . The thermal conductivity of copper is 0.102 kCal / m s C. If 0.20 kg of

Compre Special Project (PPE &IPE)

Compre Special Project (PPE &IPE)