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PREFACE
TABLE OF CONTENTS
This book is designed to give more emphasis in
solving problems in Power Plant Engineering subjects due to
the increase number of problem solving on recent board
examinations.
The main purpose is to show the way by presenting
subject matter and material that have appeared in many board
examinations, together with worked-out solutions that are
acceptable to board examiners. By following a rather simple
pattern of self-discipline and working out many problems can
help you pass the jist time.
The contents of each topic are also well designed for
familiarization of expected questions that will come out in the
actual board examinations.
The author can guarantee a good passing grade in
Power Plant Engineering subjects if mastery of topics
including the principles behind in this reviewer were done.
Description
Thermodynamics
Fuels & Combustion
Variable Load Problems
Steam cycles
Boilers
Steam engine
Steam Turbine
Geothermal Plant
Diesel Plant
Gas Turbine Plant
Hydro-electric plant
Chimney
Machine Foundation
Heat Transfer
Air-compressor
Pumps
Fans & Blowers
Fluid Mechanics
Past Board Examination Elements
Refrigeration
Air-conditioning
Conversion of Units-
Page
1
72
95
97
123
135
142
149
158
188
199
219
226
237
253
276
304
314
331
390
422
453
,,"0_
Thermodynamics
THERl\10DYNAMICS
THERMOD YNAI~lICS PROPERTIES
Thermodynamics - 1 (Math-ME Ed Oct. 1998)
What is the pressure 8,000 ft (2000 m) below the surface of the ocean?
Neglect the compressibility factor, in Sl units.
Ao 21.4 Mpa
C. 21.0 Mpa
B. 20 I Mpa
D. 22.3 Mpa
SOLUTION:
Sea water
p = '>'1 h
U sing typical SG of sea water equal to 1.03
p = (1.03 x 98 J )(2000)
p = 20,208 Kpa
P c 20.21 Mpa
~~o~=r:2:~m
1
P=wh
Thermodynamics - 2 (Math-ME Ed Oct. 1998)
What is the temperature at which water freezes using the Kelvin scale'!
C. 278
A. 373
B. 273
D. 406
SOLUTION:
Freezing temperature of water is
"K = °C + 273
"K = 0 + 273
"K = 273
Thermodynamics - 3 (Math-ME Ed Oct. 1998)
The SI unit of temperature is:
A. of
B. oK
411.\', B
C. BTU
D. oR
o-c.
::'
~
Thermodynamics
Thermodynamics
.J
St lllJTlON
Thermodynamics - 4 (Math-ME Bd Oct. 1998)
iii
~~
The pressure reading of 35 psi in kpa is:
A. 342.72 kpa
B. 724.00
h = u + Pv
9500
u + 900(58)
II ~c 4280/t-lh/lh
-r-
C. 273.40
D. 42730
fi']
i:~'
;
Thermodynamics - 8 (Math-ME Bd Oct. 1997)
SOL UIION'
~
Pg
Pg
P abs
35 (101325 14.7)
241.25 kpag
=
Pabs
Pabs
Pg
+
Palm
24l.2" + 10132"
342.57 kpaa
The barometer reads 29.0 inches (737 mm)
absolute pressure if a vacuum gage reads 9.5
A. 3202 kpa
C
B. 3304 k p a .
D.
of mercury. What is the
psi (66 kpa) in 51?
31.36 kpa
31.86 kpa
SOLUTION:
101.325
Patrn
Thermodynamics - 5 (Math-ME Bd Apr. 1999)
1 torr is equivalent to pressure
A. 1 atm
C 2 mm Hg
Paun
r.,
_
C 14.7
D. 1/760 atm
An5. D
p abs
PaLs
(29)(---)
29.92
98.2 kpa
Pgage + Pann
-66 ~ 98.2
32.2 kpa
Thermodynamics - 9 (Math-ME Bd Oct. 1997)
Thermodynamics - 6 (Math-ME Bd Apr. 1999)
What is the standard temperature in the US?
A. Fahrenheit
B. Rankine
C Celsius
D. Kelvin
Ans. A
A fluid with a vapor pressure of 0.2 Pa and a specific gravity of 12 is
used in a barometer. If the fluids column height is 1 rn, what is the
atmospheric pressure?
A. 150.6 kpa
C 144.4 Kpa
B. 115.5 Kpa
D. 117.7 Kpa
SOLUTION
Pressure = (specific weight)(height)
Pressure
(12 x 9.81)(1 m)
Pressure c= 117.72 Kpa
rr;
Thermodynamics - 7 (Math-ME Bd Apr. 1999)
Given steam pressure of 900 Ib/fe, temperature of 300°F, specific
volume of 5.8 fellb. If the specific enthalpy is 9500 ft-lb/lb, what is the
internal energy per Ib of the system?
A. 4400
C. 3600
B. 3900
D. 4280
Thermodynamics - 10 (Math-ME Bd Apr. 1997)
What is the atmospheric pressure on a planet if the pressure is 100 kpa
and the gage pressure is 10 kpa?
A. 10 kpa
C. so kpa
B. 100 kpa
D. 90 kpa
Thermodynamics
Thermodynamics
4
';
SOLUTION'
Thermodynamics - 13
Pabs
P atm
+
P gage
If the
100
=
P atm
PanTI
=
90 kpa
temperature inside the
corresponding reading in OF?
A. 700.60
B. 750.60
+ 10
furnace
is 700 oK, what
is
the
C. 860.60
D. 800.60
SOLUTION:
Thermodynamics - 11 (Math-ME Bd Apr. 1997)
OK
= "C + 32
700= "C + 273
"C ~ 427
of = 9/5 De +- 32
of = 9/5 (427) + 32
°F= 800.6
A column of water 200 cm high will give a pressure equivalent to:
A. 9810 dyne/em 2
C. O. JO bar
2
B 0.1 atrn
D. 19,620 N/m
SOLl:TTON:
h= 200 em
h~ 2 m
3
w = 9810 N/m
P = wh
P = (9810)(2 m)
P == 19,620 Nlm 3
r
1
200cm
r:"
Thermodynamics - 14
11' the of scale is twice the °C scale, what will be the corresponding
reading in each scale'?
A. zz-c and 44°F
e. 40
and 80°F
B. 160 0e and 320°F
D. 1oo-c and 200°F
0e
lher mudynamics - 12
What is the equivalent "R of 400
A n0600R
B. 851.15°R
SOLUTION:
0K?
C. 670.2rR
D 344.25°R
9/5 -c + 32
z-c
2°e = 9/5 <C + 32
"C = 160
SOLUTION
°K- -c + 273
WI)
"C + 273
°e = 127
of
9/5°e + 32
of
9/5 (127) + 32
OF
260.6
oR = of + 460
oR
260.6 + 460
oR
720.6
OF
of
OF
=
of
=
2(160)
320
Thcrmodynamics - 15
Water enters the condenser at 25°C: and leaves at 40°C. What is the
temperature difference in of'?
6
D. 28
SOLUTION:
SOLUTION:
6°C
C. 150.9 KN/rn J
D. 82.2 KN/m J
A 102.3 KN/m]
B. 132.9KN/m J
en
A. 25
B. 26
7
Thermodynamics
Thermodynamics
=
Specific weight of mercury
Specific weight of mercury
Specific weight of mercury
°C 2 - °C I
,:'.°C =40-25
6°C = 15
6°F
9
6°C
5
6°F
9
15
5
t,
=40°C
(Specific gravity) (Density of H 20)
(1355)(9.81)
132.9 KlV/m
J
Thermodynamics - 18 (Math-ME Bd Apr. 1996)
An iron block weighs 5 N and has a volume of 200 em]. What is the
density of the block?
A. 988 kg/m '
C. 2550 kg/rrr'
B. 1255 kg/rn'
D. 800 kg/rn '
----
-------
SOLUTION:
6°F
27°F
Mass = 5/9.81
Mass = 0.5097 kg
Volume = 200 cm ' (1/100 3 )
Volume = 2 x 10.4 m'
Thermodynamics - 16
Mass
Wafer enters the heater with 28°C and leaves at 75°C. What is the
temperature change in OF?
A. 7480
C. 84.80
B 38.29
D. 57.36
w
w
w
SOLUT10N:
6°C
6°C
6°C
=
=
~c
°C 2 - °C l
75 - 28
47°C
i\OF/6°C = 9/5
6°F/47 = 9/5
6°F = 84.6 of
~
HEATER
Thermodynamics - 17 (Math-ME Bd Apr. 1996)
The specific gravity of mercury relative to water is 13.55. What is the
specific weight of mercury? The specific weight of water is 62.4 lb per
cubic foot.
Volume
0.5097
2xl0- 4
2548.42 kg/m'
Thermodynamics - 19
The suction pressure of a pump reads 540 mm Hg vaccum. What is the
absolute pressure in Kpa?
A. 40
C. 60
B. 3.3
D. 29.3
SOLUTION:
Pa b ,
Pabv
1\\"
1'.11"
1',,1,"
=
Pg + Paull
-540 mrn Hg + 760 rnrn Hg
220 mm Hg
220 X (101.325/760)
29.33 Kpa
#
rr®]
p.= -540 mm Hg
J)
Thermodynamics
8
SOltiTlON:
Thermodynamics - 20
A boiler installed where the atmospheric pressure is 755 mm Hg. Has
a pressure 12 kg/ern", What is the absolute pressure in Mpa?
A. 1.350
C. 1.200
B. 1.277
D. 1,700
Weight of water ~ !S.'7481)(62A)
Weight of water
4L7 lb
::200 + 350
Volume of sand and gravel
2.65(62.4)
Volume of sand and gravel
Patm=755 mm Hg
r.,
r.,
~­
12 x (101325/1.033) + 755 x (101.325/760)
127769 Kpa
Total weight
41.'7 -i- 200
Total weight - 685.7 lbs
2
1.277 Mpa
BOILER
Thermodynamics - 21
SOLUTION.
~G"0'd ,'om
.r.
350 ,. 9..+
Total volume
5!7.4S! - 33::26 ' CdS6
Total volurne- 4.48 jf
Weight of concrete per
A storage tank contains oil with specific gravity of 0.88 and depth of 20
m. What is the hydrostatic pressure at the bottom of the tank in
kg/ern"?
A. 1.76
C. 60
D. 3.0
R. 2.0
P - w x h
J
P - (0.88 x 981 KN/m ) ( 20 m)
P ~- 172.66 Kpa
P
J72.66 x (1.033/101.325)
P - 1.76 kg/em'
3.1 O(62..4)
Volume of cement -; 0.486 flO
p ~ -'- Parm
P'b,
3.326 ft'
cc
94
Volume of cement
SOLUTION:
Pabs
9
Th crmodynamics
CLl.
Weight of concrete/cu.
it. concrete
n. concrete
c-.
68".'7,4.48
153.0S8 lb.ft \
Thermodynamics - 23 (Math-ME Bd Oct. 19(5)
A batch of concrete consisted of 200 Ibs fine aggregate, 350 lhs (031 'it
aggregate, 94 Ills, cement, and 5 gallons water. The specific gr:n ity cf
the sand and gnwel may be taken as 2.65 and that of 011' <:,~m~-m :b
3.t O. Hew m uch by weight of cement is req uired to produce one cubic
yard'.'
A. 765
C. 675
13. 657
D. 567
SOLLTIC.JN:
Thermodynamics - 22 (Math-ME Bd Oct. 1997)
A batch of concrete consisted of 200 Ibs fine aggregate, 350 Ibs coarse
aggregate, 94 Ibs, cement, and 5 gallons water. The specific gravity of
the sand and gravel may be taken as 2.65 and that of the cement as
3.10. What was the weight of concrete in place per cubic foot?
A. 1721b
C. 1621b
B. 236 Ib
D. 153 Ib
Volume of water
~ '7 .48l
Volume of water C~ 0668 ft
]no+3:<J
Volume of sand and gravtl
':' t,5( 62.4)
J
3.~,:'6 n
Volume of sand ;mj zravel
94
Volume of cement
.1. i ()~ 62 A)
Thermodynamics
Thermodynamics
10
Volume of cernent > 0.486 ft°
Total volurne > 0.668 + 3.326 -i- 0.486
Total volume ~ 448 ft3
Weight of cement per ft° of concrete mixture
~ 94/4.48
3
3
2098 Ibs/ft 3 (27ft /yd )
II
Total volume
289.9 cm '
Specific volume
.~ Total volume
Specific volume
Total mass
289.9
250
566.5 lbs or 567lbs
Specific volume
1.1596 cm'rgr
Thermodynamics - 24
A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific
gravity of oil is 0.9, what is the mass of oil in the tank?
A 8482 kg
C. 1800 kg
B. 4500 kg
D. 7000 kg
o .,
SOLUTION:
Volume
Volume
Volume
Mass .~
Mass =
Mass ~
Mass =
n/4 0' h
n/4 (2)2 (3)
of cylinder = 9.425 m'
Density x Volume
w x V
3
(09 x rooo kg/m )(9 .425)
8482.3 kg
of cylinder
ofc~linder
=
=
_----
~_'!1_
Thermodynamics - 26 (Math-ME Bd. Oct. 1997)
100 g of water are mixed with 150 g of alcohol (w = 790 kg/m'). What
is the specific gravity of the resulting mixtures, assuming the fluids
mixed completely?
A. 0.96
C. 0.82
B 0.63
D. 0.86
SOLUTION
Total mass
Total mass
__
O. J 00 + 0.1 SO
0.250 kg
0.100
o.isu
Total volume = - - + ----1000
790
Total volume
2.899 x io' m 3
Thermodynamics - 25 (Math-ME Bd Apr. 1998)
Density of mixture
3
100 g of water arc mixed with 150 g of alcohol (w = 790 kg/m ) . What
is the specific volume of the resulting mixtures, assuming the fluids
mixed completely?
3
C. 0.63 cmvkg
A. 0.82 cm /kg
D. 1.20 crrr'zkg
B. 0.88 cm]/kg
- Density of mixture
0100
Total volume
0.250
2899 x 10-
4
Density Of mixture = 862 kg/rn:
Specific Gravity = 86211000
Specific gravity = 0.862
SOLUTION:
Total mass = 100 -t- 150
Total mass = 250 grams
Total mass
Thermodynamics - 27
0.150
Total volume
----+---
Total volume
1000
790
2.899 x 10.4 m 3
A spherical tank is full of water that has a total mass of 10.000 kg. If
the out side diameter of the tank is 2722 mm, how th ick is the wall of
the tank?
I
12
Thermodynamics
nerrnouynumus
1,)
SOLUTION'
A. 50 rnrn
B. 25 mm
C. 30 mm
D. 35 mm
SO}"-,UTI()r~:
v
rn/w
V
10,000 kgll,OOO kg/rrr'
V = 10 m 3
V = 4/3 it [3
10 = 4/3 it [3
r = 1336 m
r = 1336 mm
1
2722/2-1336
~
1 =
"' t l
O
i,
.
i
r,
+-
Q=
Q =
Q =
Q ~
volume flow of water flowing
A x vel
n/4 D 2 x vel
n/4(1)2(10)
Q
7.85jr/sec
=
~
:r;
I
..-- ; -- t+
1ft
i
J
'lS
EI%1§
I
I
IV
Thermodynamics - 30
A certain fluid is flowing in a 0.5 m x 0.3 m channel at the rate of 3
m/sec and has a specific volume of 0.0012 m 3/kg. Dctermine the mass
of water flowing in kg/sec.
A 380 kg/sec
C. 375 kg/sec
i3 390 kg/sec
D.370kg/sec
24.49 mm
Thermodynamics - 28
a.3m
A cylindrical t~WK is filled with water at the rate of 5.000 gal/min. The
height of water ill the tank after minutes is 20.42 ft. What is the
diameter of the tank'?
A 30 ft
C. 20 ft
B. 25 ft
D.9m
Q
=
Q
Q
Q
SOLUTION:
m
20.42
~-_Q --+
Thermodynamics - 29
Water is flowing through a 1 foot diameter pipe at the rate of W It/sec,
What is the volume flow of water flowing?
A. 7.50 ft3/sec
C. 7.85 m 3/sec
D. 0.22 mzsec
B. 7.95 ft3/sec
A x vel
(05 X 0.3)(3)
OA5 m ' /sec
=
OA5
After J5 minutes,
V c, 5000(15) ~, 75,000 gal
V = 75,000x(l fe/7ASI gal)
V = 10,025.39 ft3
V c. n'4 DC h
2
10,025.39 = n:/4 D (20.42)
D = 25ft.
P-O
3m/s r = c _
~
SOLUTION:
=
fiX v
rn (0.0012)
375 kg/sec
=
O.5m
J.
............ ..
IICI ".uuy"u,"'U.. . ...,
_'_~,.a
......
v
UWS OF THERMODYNAMICS
Thermodynamics - 33
Thermodynamics - 31 (Math-ME Bd. Apr. 1998)
One useful equation used is the change of enthalpy of compressible
liquid with constant specific heat is:
hs ub2 - h,u b l = c(T, ub1 - T.u b l ) + V(P.ub2 - P.ub l )
where: T. ubu = temperature at state n
p. u bn = pressure at state n
v = specific volume of liquid
Water with enthalpy with C,ubp = 4.187 KJIkg_oK and v = 1.00 x 10 to
rd
the _3 power cu.mIkg has the following states:
State I: T su b l = 19°C
P. ub l = 1.013 x 10 to the 5th power Pa
State II: T, ub2 = 30°C P.u b2 = 0.113 Mpa
What is the change in enthalpy from state I to state II?
A. 46.0 Kpa/kg
C. 46.0 KJlkg
B. 56.0 KJlkg
D. 46.0 KNlkg
SOLUTION:
h 2 - h.
h z - hi
h z - h,
The flow energy of 124 IiImin of a fluid passing a boundary to a
system is 2 KJ/sec. What is the pressure at this point.
A. 100 Kpa
C. 1,000 Kpa
B. 140.39 psi
D. 871 Kpa
SOLUTION:
W = Pressure x Volume
2 KJ/sec x 60sec/min = P(0.124 m 3/min)
P = 967.74 Kpa x 14.7/101.325
P = 140.39 psi
Thermodynamics - 34 (Math-ME Bd Apr. 1996)
Cp(Tz-T 1) + v(P 2-P t )
4.187(30 - 19) + 0.001(113 - 101.3)
46KJ/kg
Thermodynamics - 32
Steam at 1000 Ib/fr pressure and 300 0 R has a specific volume of 6.5
fe/lb and a specific enthalpy of 9800 ft-Ib/lb. Find the internal energy
per pound mass of steam.
A. 5400
C. 6400
D. 2500
B. 3300
SOLUTION:
What is the potential energy of a 500 kg body if it is dropped to a
height of 100 m?
A. 490.50 KJ
C. 560.50 KJ
B. 765.50 KJ
D. 645.48 KJ
h=u+Pv
9800 = u + 1000(6.5)
u = 3300 ft-lbfllb m
SOLUTION:
Potential
Potential
Potential
Potential
Energy
Energy
Energy
Energy
=
=
=
=
mxz
500 x 100
50,000 kg.m x 0.00981 KNlkg
490.50 KJ
Thermodynamics - 35
Air and fuel enter a furnace used for home heating. The air has an
enthalpy of 302 KJ/kg and the fuel has an enthalpy of 43,207 KJ/kg.
The gases leaving the furnace have an enthalpy of 616 KJ/kg. There
Thermodynamics
Thermodynamics
re 17 kg air/ kg fuel. The house requires 17.6 KW of heat, What is the
lei consumption per day?
A 85 kg
C. 45 kg
B 41 kg
D. 68 kg
SOLUTION:
By mass balance:
rn, + m, = m g
m/mr = 17
m, = 17mr
i 7mr + m, = mg
mg
18mr
C.C
gas
~
--.
air
.......
rna
EY-RJ*.A~e
---
heat
......
17.kw
fuel
mf
Thermodynamics - 3 7 (ME Rd. Apr. 1995)
The enthalpy of air is increased by 139.586 KJ/kg in a compressor,
The ~ate of air flow is 16.42 kg/min. The power input is 48.2 KW.
Which of the following values most nearly equals the heat loss from the
compressor in KW?
C. -9.95
A. -10.0
B. +10.2
D. +9.95
SOLUTION
U.42k
W
By heat balance:
rn, h,
mrhr= m g h g
(l7rnrJ(302) -t mt<43,207) = (l 8mf)(6 16) + \7.6
m, = 4.7244 x \0,4 kg/sec x 3600 x 24
m.
40.819 kg/day
.i..
Co
17
mh.
1
mh, -"- Q
=
Q
h.
Q .~ W + m(h i - h1 )
Q
W - mth, - hi)
Q = 48.2 - (16.42/60)(139.586)
I..W=~8.2kW
/
cc
COMPRESSOR
Q= 10 KW
Q = -10 KJV ( heat is rejected)
t
h
,
Thermodynamics - 36
Thermodynamics - 38 (ME Bd, Oct. 1982)
The power plant furnace burns coal at the rate of 108,200 kglhr. Air at
100.8 Kpa, 28°C is supplied at the rate of 13.8 kg/kg coal. Determine
the volume flow rate of air flow in mJ/min.
3/min
3
A. 21,327.64 rn
C. 20,435.26 rn /m in
B. 19,41462 m3/mim
D. 24,535.54 m3/min
A steam turbine receives 70 pounds of steam per minute with an
enthalpy of 1600 Btu per pound and a velocity of 100 It/sec. It 11:'3\'l:s
the turbine at 900 It/sec and 1320 Btu/lb enthalpy. The radiation loss is
84.000 Btu/hr. Find the horsepower output.
h.v,
SOLUTION:
A/F
AiF
rna
m,
rn,
rna
m/ml
13.8
=
C~
=
=
PY
~.
13.8mt
13.8(\08,200)
1493160 kglhr
24886 kg/min
B
100.8kPa
.
..
28°Cma_
.
:;~t FUR~~fE.
By heat balance:
mh,
t
KE,
co
VJ
mh- + KE 2 +Q+ W
W ~.. nuh, - h:) , m2
[VIC -
v/] - Q
kg/h
mRT
Q:::B4.000Btu!h
SOLUTION'
W
.~
(IOO.8)(Y) = 24886(0.287)(28 + 273)
Y
21,327.64 m 3/min
70( 1600
42
-z-
W
.c
1320)
70/60
- . - - - - +- - - - -
4()J Hp
~
2(32.2)
)
7
100 - - 900-
84,000
55!)
2~.t"
J ---------1
V;
Thermodynamics
Thermodynamics
19
(v ? i\]
(80)800
ierrnodynamics 39 (M.E Bd. Oct 1986)
= 750
T
-+-
2(9.81)(427)
---.,
2(9.81)(427)
h,=800Kcal/kg
earn enters a turbine stage with a enthalpy of 3628 KJ/kg at 10 m/sec
d leaves the same stage 'with an enthalpy of 2846 KJ/kg and a
locity of 124 m/sec. Calculate the work done by steam.
A. 77676 KJ/kg
C. 567.23 KJ/kg
D. 923.34 KJ/kg
B 873.45 KJ/kg
Y2
652.14 m/sec
---<>----,
V,=80m/s
~
TURBiNE
W
SOLUTION:
For
ITl ~c
V, Ih, =750kcal/kg
0=0
1 kg (basis)
~
By heat balance:
mh.! KE[
W
w
= mh 2+KE 2 + Q -rW
m (11:
- h 2 ) + -- (v: - v 2 )
.
ill
. 2
2
2g
-
""----1'
Q------
h,=2846KJlkg
Il
~,
t,
w
f,
w~
1(3628 - 2846,
-t
_ 1 _ [(70)2 -(24)2)(0.00981)
2(9.81)
VI
I
\
£,
v,
776.762 Kllkg
t;,
"~.
1
Thermodynamics - 41 (Power-MEEd Apr. 1998)
A volume of 450 cc of air is measured at a pressure of 740 rnm Hg
absolute and a temperature of 20°C. What is the volume in cc at 760
mm Hg absolute and O°C?
A. 516.12
C. 620.76
B. 408.25
D. 375.85
SOLUTION:
CD
~
I,
~
!
I
er ruodynamics - 40 (Math-ME Bd Apr. 1998)
am with all enthalpy of 800 Kcal/kg enters a nozzle at a velocity of
rn/sec, Find the velocity of the steam at the exit of the nozzle if its
~
Ii
r
PlY'
P 2Y]
TI
T2
-----
740( 450)
(760)(Y 2 )
halpy is reduced to 750 Kcal/kg, assuming the nozzle is hortzontal
! disregarding heat losses. Take g(9,81) m/sec and J = 427 kg-
(20+273)
(0 + 273)
(cal
Y 2 = 408.25 cc
A. 56124 m/s
B 142.5 m/s
II
@
V,= 450c:c:
V,
P, = 470
t, =20°C
P,=760
t, O°C
=
C 52.41 m/s
D. 652.14 mls
Thermodynamics - 42 (Power-ME Bd \pr. 1998)
SOLUTION:
V 2
I
hi-+---- = h 2 +
2gJ
V 2
2
-=-LgJ
Assuming compression is according to the law PV = constant.
Calculate the initial volume of gas at a pressure of 2 bar which will
occupy a volume of 6 cubic meters when it is compressed to a pressure
of 42 bar.
A. 126 m '
C. 130 m'
D. 136 mY
B. 120 m'
Th ermadynumics
Thermodynamics
20
.!l
SOLUTION
Thermodynamics - 45 (Math-ME Bd Apr. 1998)
PI V;
P2 V 2
2 (VI) = 42 (6)
CD
Q)
P, = ·2bar
P,=42 bar
V,=?
V,=6m'
VI
= 126 m
The mass of air in the room 3 m x 5 m x 20 m is known to be 350 kg.
Find its density.
. A 1617kg/m J
C. 1.167kg/m J
B.1.716kg/m'
D I.176kg/m '
3
SOLUTION
~
V
Thermodynamics - 43 (Power-ME Bd Apr. 1998)
3(5)(20)
,meso" [J
V - 300m'
How much heat , KJ must be transferred to 20 kg of air to increase the
temperature from 20 degrees C to 280 degrees C if the pressure is
maintained constant.
A 2500
C. 5200
B. 2050
D. 5500
SOLUTION:
1,=20°C
HEATER
Q = m cp (t 2 - t.)
Q = 20 (1.0) (280 - 20)
Q = 5200 KJ
280°C=t,
-------.
m=20kg
v
t
r
20
5m
350
Density -
300
\
f!
1.167 kg/m'
Density
r
~
Q
If air is at pressure, p, of 3200 Iblft\ and at a temperature, T, of 800
what is the specific volume, v?
C. 11.2 ftJlIb
A. I 4.2 fe/lb
B. 13.3 ft 3 /1b
D. 9.8 fellb
I
Thermodynamics - 46 (Math-ME Bd Apr. 1998)
I
A transportation company specializes in the shipment of pressurized
gaseous materials. An order is received for 100 liters of a particular
gas at STP (32°F and 1 atm).
What minimum volume tank is
necessary to transport the gas at 80°F and a maximum pressure of 8
I
Thermodynamics - 44 (Math-ME Bd Apr. 1998)
SOLUTION
m
Density
oR,
f.
a trn?
A
B
16 liters
14 liters
C. 10 liters
D. 12 liters
SOLUTION
PV = mRT
V
v
m
RT
PI VI
T\ V 2
1'[
1',
v
v
P
(53.3)(800)
1(100)
(8)(V 2 )
- - -
-~---
(32
t
4(0)
(80 + 460)
c~
3200
v = 13.375/f/lb
V.'
/4 liters
CD
@
P, = 1atm
t, = 32°F
V,= 1001i
P,=8atm
t, = 80°F
V 2 = ?•
22
Thermodynamics
Thermodynamics
23
Thermodynamics - 47 (Math-ME Ed Apr. 1997)
Thermodynamics - 49 (Math-ME Ed Apr. 1998)
A bicycle has a volu me of 600 em:'. It is inflated with carbon dioxide to
pressure of 80 psi at 20"e. How many grams of carbon dioxide are
contained in the tire?
C 4.63 g
A 5.98 g
D 3.83 g
B 6.4:-1 g
Air compressed in a diesel engine from an initial pressure of 13 psia
and a temperature of 120°F to one-twelfth of its original volume.
Calculate the final temperature assuming compression to be adiabatic.
A. 987
C. 981
B. 980
D. 1010
SOLUTION
SOLUTION:
M
M
M
R
R
V
V
P
P
T
T
=
C.C
=
=
8.314/44
=
0.189 KJ/kg- oK
600 em' 1 (100)'
0.0006 m 3
80 psi x (101.325/14.7)
551.43 kpa
20 I 273
293 uK
=
=
=
=
=
=
~
m
~
r,
=
1'2
m(0.189)(293)
~'
f'
~
V,-
>
._-~~
v,
....,.-.v
The compression ratio of an Otto cycle is 6:1, P sub l is 14.7 psia, T su b 1 is
68°F. Find the pressure and temperature at state 2.
A. 180.6 psia, 1081 OF
C. 180.6 psia, 139°F
B. 180.6psia,139°F
D. 180.6psig, 1081°R
p
SOLUTION:
~'
I
;:
i~
~,
SOLUTION:
Pv=nR1'
liter - atm
=
~' 120"F
L
VJ
I
r,
(~
n (0.0821
.
'~=c
Thermodynamics - 50 (Math-ME Bd. Oct. 1997)
~fe.
An Ideal gas at 0.60 atmospheres and 87"C occupies 0.450 liter. How
many moles are in the sample? (R = 0.0821 atm/mole K)
A. 0.0002 mole
C. 00198 mole
B 0.0378 mole
D. 0.0091 mole
n
4-1
120+460
V I / 12
1'2 = 1567°R
t 2 co 1567 - 460
t2 = 1101'F
I
=
V2
--l-
0.00598 kg
5.98 grams
Thermodynamics - 48 (Math-ME Bd Apr. 1997)
(060 atm)(0.450 li)
\
y-I
=l~j
1'2
(
PV=mR1'
(551.43 )(0.0006)
m
(
molecular weight of CO 2
12 +2(16)
44
2
p
P,
Compression ratio
Compression ratio
(P2/P\) = (V/V 2) k
(P 2fl4.7) =_ (6)14
P 2 = 180.6 psia
VIN]
6
2
~t'
P. ,=14.7 PSia
S=c
t,=68
0F
•
-I
)(87 + 273)K
1'2fT I
mole- K
0.009135 mole
=.
(V /V 2)k.\
1'2/(68+460) = (6)14.1
1'2
1081.11'R
l
+
!v
v,
1
~
4
Thermodvnamics
Thermodvnamics
25
'hermodynamics - 51
Thermodynamics - 53
automobile tire is inflated to 30 psig pressure at 50°F. After being
riven, the temperature rise to i5°F. Determine the final gage pressure
ssuming the volume remains constant.
A. 32. J9 psig
c. 0 psig
D. 38.9 psig
B 55 psig
If 8 Ihs of a substance receives 240 Btu of heat at constant volume and
undergo a temperature change of 150°F. What is the average specific
heat of the substance during the process?
,II
A 0.30 Btu/lb-oF
B. 025 Btu/Ib-oF
C G50 Btu/lb-oF
D 0.20 Btu/lb· OF
SOLUTION:
SOLUTION
P2
PI
T2 / T 1
Q
P2
(75 + 460)
(30+147)
(50+460)
~.'
m c, VH)
m = Bibs
aF= 150
--------
240 = 8(c,)( 150)
c, = 0.2 Btu/lb- OF
P= = 46 89 psia
Peo 46.89 - 14.7
P2
32.19 psig
0=240 Btu
cc
Thermodynamics - 54 (ME Bd, Oct. 1995)
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _T
.crmodynamics - 52
flO 111} of atmospheric air at zero degree centigrade temperature are
.ompressed to a volume of 1 m) at 100°C, what will be the pressure of
ur in Kpa?
C. 2,000
A. J 500
B. J,384
D. JOOO
A certain gas at 101.325 Kpa and 16 cC whose volume is 2.83 m J are
compressed into a storage vessel of 0.3] m' capacity, Before admission,
the storage vessel contained the gas at a pressure and temperature of
137.,8 Kpa and 24°C; after admission the pressure hasmcreased to
l1il.S Kpa. What should be the final temperature of the gas in the
vessel in Kelvin'!
A. 298.0
B. 319.8
C. [800
[) 4200
SOLUTION:
SOLUTION
PlY]
P
2Y .
2
----_
T1
P,
T2
=
101.325 Kpa (atmospheric air)
(101.325)(\0)
(p} )(\)
(0 + 273)
(\ 00 + 272\
P2 = 1384.4 Kpa
Solving for the mass of gas which is to be compressed:
PV -r- mRT
JOJ 32~(283) = m ,R( J6 + 2 7J )
m , ~ o 9922/R
Solving fur the mass of gas Initially contained in the
vessel:
PV
mRT
137.8(031) = m2R(24 + 273)
m=- 01433/R
Solving for the final temperature:
Thermodynamics
Thermodynamics
p]v, ~ rnJ RT 3
m, = rnlT rn2
m, ~. 0.9922/R+01438/R
m, cc 1.13 6/R
1171.8(0.31) = (l.136/R)RT
1'3 = 319.8°[(
27
Thermodynamics - 57
What is the specific volume of an air at 30°C and 150 Kpa?
A. I ml/kg
C. 12 m 3 /kg
3/kg
B. 0.5 m
D. 0.579 mJ/kg
SOLUTION:
ermodynamics - 55
PV = rnRT
v = Vim
v = RTfP
e temperature of an ideal gas remains constant while the absolute
.ssure changes from 103.4 Kpa to 827.2 Kpa. If the initial volume is
liters, what is the final volume?
A. 100 Ii
c. 8 Ii
B. 10 Ii
D. 1000 Ii
0.287(30 + 273)
v
ISO
SOLUTION
3
v = 0.579 m / kg
PI VI
~c
P2 V 2
1034(80)
.~
827.2(VJ)
Thermodynamics - 58
V, •• 10 liters
-rmodynarnics - 56
at is the density of air under standard condition:
A. j ] kg/m '
C. 1.2 kg/rn 3
B. I kg/rn '
D. I KN/m'
SOLUTION
P •.• 14.7 psi
P = 101.325 Kpa
t c. 70°F
t ~ 21.11 °C
PV
rnRT
w - m/R
w
P/RT
--.--
0.287(2 LlI + 273)
w
.c
1.2I,g/m 3
~
SOLUTION
M
~.
P1V 1 = P"V,/'
~.
PI V/ ~ P 2 V 2
P 2 / PI - (V I/V 2 )k
P2 / P I ~ (rK/
P2 / 100 = (5)14
P 2 .~ 951.82 Kpa
k
P,=100kPa
_v,-
v,
v
Thermodynamics - 59
101J25
V\,';::--
If
The compression ratio of an Otto cycle is 5. If initial pressure is 100
Kpa, determine the final pressure.
A. 1000 Kpa
C. 300 Kpa
B. 952 Kpa
3
D. 100 Kpa
P
How much work is done when 20 fe of an air initially at a pressure of
15 psia and a temperature of 40°F experience an increase of pressure
to 80 psi while the volume remains constant.
A 1000 Btu
C zero
B 3000 Btu
0 2000 Btu
Thermodynamics
Thermodynamics
28
~
SOLUTION:
For constant volume process, W
L=J
0
=
Thermodynamics - 6C
A perfect gas has a value of R = 58.8 ft-lb/lb-oR and k = 1.26. If 20 Btu
are added to Sibs of this gas at constant volume when initial
temperature is 90°F, find the final temperature.
A. 100°F
C. 154°F
B. 104°F
D. 185°F
29
Thermodynamics - 62
During the polytropic process of an ideal gas, the state changes from
20 psia and 40°F to 120 psia and 340°F. Determine the value of n.
A. 1.233
C. 1.355
B. L255
D. 1.400
SOLUTION:
T 2/T 1
=
(PZ/Pit-i/n
n-I
(340 -+- 460)
j
.!2.~J ~­
(40 + 460) -l20
SOLUTION:
6 n- l i n
n - I In 1.6
1.6
m c, (t 2 - t 1)
Q
Y,='{
c, =, R / k-I
58.8
c
(1.26-1)778
c, = 0.29086 Btu/lb-vf
m = 51bs
20
t
n
In6
n- 1
0.2623n
1.355
n
t
y
=
=
=
1
0=20 Btu
5(0.29086)(t2 - 90)
103. 76°F
Thermodynamics - 63 (ME Bd, Apr. 1984)
=
=
Thermodynamics - 61
While the pressure remains constant and 689.5 Kpa, the volume of air
changes from 0.567 m 3 to 0.283 m'. What is the work done?
A. -196 KJ
C. 206 KJ
B. 204 KJ
D. -406 KJ
A group of 50 persons attend a secret meeting in a room which is 12
meters wide by 10 meters long and a ceiling height of 3 m. The room is
completely sealed off and insulated. Each person gives off 150 Kcal per
hour of heat and occupies a volume of 0.2 m', The room has an initial
pressure of 101.3 Kpa and temperature of 16°C. Calculate the room
temperature after 10 minutes.
A. 36.35
C. 23.24
B. 33.10
D. 76.32
12m
SOLUTION:
SOLUTION:
W
689.5(0.283 - 0.567)
Volume
Volume
Volume
Volume
W
-195.82 KJ
Q
Q
W
=
P(V 2
-
VI)
m
of room = 12 x 10 x 3
of room = 360 m'
of air = 360 - (02 x 50)
of air = 350 m 3
i
50 x 150
7,500 Kcal/hr
PV/RT
m = (101.3)(350)/(0.287)(16~273)
0 =7500K
A
ca
I/h
Thermodynamics
30
Thermodynamics
427.46 kg
0 171 Kcal/kg-OC
After 10 minutes:
7500(10/60) ~- 427.46(0.171)(tz- 16)
m
c,
t2
33.1
=
cr
SOLUTION.
Since the molecular weight of ammonia is 17, then
R
8.3143/M
R = 8.3143/17
R
0.489 KJlkg-OK
T1
Thermodynamics - 64 (ME Bd. Oct. 1994)
1244.5
12445
1244.5
1244.5
Kpa.
Kpa,
Kpa,
Kpa,
T2
T2
599.96°K
gage, 60°C
60°C
599.96°C
Compression ratio
Compression ratio
(P/P j ) = (V/V,;'
(P/I013) = (6)14
.C
=
'.~
mRT I
P2 V 2 = mRT 2
(413)(V 2 ) = 22(0.489)(373)
V 2 c 9.716 rrr'
.
V IIV
6
= (V IV ) k. 1
I
2
T 2/(20 + 273)= (60)14.
T/T
=
PIV 1
W
=
P(V,-V 2)
W == 413(9.716 - 8.1(1)
W = 667 KJ
/244.5 Kpa
=
100 + 273
373°K
=
413(V I) = 22(0.489)(311)
VI = 8.101 m'
SOLUTION:
P2
38-1-273
311°K
T!
The compression ratio of an ideal Otto cycle is 6:1. Initial conditions
are 101.3 Kpa and 20 c C. Find the pressure and temperature at the end
of adiabatic compression.
A
B
C
D.
31
J
Thermodynamics - 66 (Power-ME Bd Oct. 1997)
T 2 = 599.96 oK
Thermodynamics - 65 (ME Bd. Apr. 1996)
Determine the average C, value in KJ/kg-K of a gas if 522 KJ/kg of
heat is necessary to raise the temperature from 300 OK to 800 0 K
making the pressure constant:
A. 1.440
C. 1038
B. 1.044
Ammonia weighing 22 kgs is confirmed inside a cylinder equipped
with a piston has an initial pressure of 413 Kpa at 38°C. If 2900 KJ of
heat is added to the ammonia until its final pressure and temperature
are 413 Kpa and 100°C, respectively, what is the amount of work done
by the fluid in KJ?
A. 630
C 420
B
304
D. :'102
D. 1026
SOLUTION
For constant pressure process,
Q .• m c p (t 2 - td
522 = I (c p ) (800 - 3(0)
cpc
UJ44 IU/kg-" J(
32
33
Thermodynamics
Thermodynamics
Thermodynamics - 67 (ME Bd. Oct. 1993)
Thermodynamics - 69 (Power-ME Bd Oct. 1(97)
A tank contains 80 fe of air at a pressure of 350 psi; if the air is cooled
until its pressure and temperature decreases to 200 psi and 70°F
respectively, what is the decrease in internal energy?
A. +4575
C. 5552
B. -5552
D. 0
A large mining company was provided with a 3 m of compressed air
tank. Air pressure in the tank drops from 700 kpa to 180 kpa while
the temperature remains constant at 28°C. What percentage has the
mass or air in the tank been reduced?
A. 74
C. 76
8.72
D. 78
SOLUTION:
3
SOLUTION:
m
m
m
PV/RT
(200 + 14.7)(144)(80)/(53.3)(70 + 460)
87.55 lbs
=
=
=
For constant volume process:
PI/T] = P2/T2
T 2 = 70 + 460
T 2 = 530 0 R
(350 + 14.7) (200 + 14.7)
T1
TI =
6U =
6U =
6U =
G)
Solving for m.,
PI VI = m, R T I
700(3) = ill] (0.287)(28 + 273)
ill I = 24.31 kg
PI =700Kpa
V, =3m'
1,=
28°1£
ill,
Before use
Q)
P,=180K;~
V, =3m'
_ t.,;:: 2S°t:
I
m,
I
I'
.
Afterl1se
Solving for rn-;
P2 V 2 = m R T 2
180(3) = m2 (0.287)(28 + 273)
rn, = 6.25 kg
530
0
900 R
mc v (T 2 - T I )
87.55(0.171)(530-900)
-5544 Btu
Percent of mass reduced:
24.3 i -- 6.25
- - - - -
24.31
74.29%
Thermodynamics - 70
Thermodynamics - 68 (ME Bd. Oct. 1993)
If 10 Ibs of water are evaporated at atmospheric pressure until a
volume of 288.5 fe is computed, how much work is done?
A. 1680 Btu
C. -610,000 ft-lb
B. no work
D. 550.000 ft-lb
SOLUTION:
In a diesel cycle, the air is compressed to one-tenth of its uriginal
volume. If the initial temperature of the air is 27"C, what is the final
tern peratu re?
C. 460~C
A. 420°C
D. 480"C
B. 440°C
SOLUTION:
VI = 10/62.4
V I = 0 . 1 6 f t3
W = P(V 2 - VI)
W cc (14.7 x 144)(288.5 - 016)
W -r- 610,358 ft-lb
V2 '=1/10 V I
Vi/Vz=lO
T: / T I = (VI / V2) k. l
Thermodynamics
Thermodynamics
36
37
I"
f,
IY
A 14.33 Kl/kg-OK
B. 2.34 Btullb-OR
C. 13.23 Kl/kg-OK
I
Thermodynamics - 76
D. 10.76 Btu!lb·oK
While the pressure remains constant at 689.5 Kpa the volume of a
system of air changes from 0.567 m 3 to 0.283 m3 • Find the change of
internal energy.
A 389.68 KJ
C. 678.68 KJ
8. 493.68 KJ
D. 245.68 Kl
SOLUTION:
t>.W r
t>.Wr
=
t>.W t
=
=
Pz VZ-P j V,
620.36(0.017) - 103.4(0.0566)
4.69 K1
SOLUTION:
t>.h = t>.U + t>.W r
16.35 = t>.U + 4.69
t>.U = 1 1.65 K.!
t>.h
mc p (T 2
t>.U
mC
y
-
=
t>.U
=
T))
m c, (Tz - T, )
P V
P V
me, (_2_2 mR
mR
_1-')
Cv
t>. U -- -(P2V2
- F;V1 )
(T2 - T,)
R
16.35
cp
----11.65 10.217
t>.U
14.33 KJ/kg- OK
Cp =
t>.U
=
0.716
--[(689.5)(0.567) - (689.5)(0.283)J
0.287
t>.U = 488.52 KJ
Thermodynamics - 75
Thermodynamics - 77 (ME Bd. Oct. 1996)
A perfect gas has a value of R = 58.8 ft-lbllb-oR and k = 1.26. If 20 Btu
are added to 5 Ib of this gas at constant 'volume when the initial
temperature is 90°F, find the final temperature.
A. 100.76°C
C. 10376°F
B. 167.76°C
D. 145.76°F
~.
SOLUTION:
c;
~
c, =
Ri(k -I)
58.8/778
c,
=
1.26- 1
0.2906 Btu/lb-oR
Q
20
=
mcvCt z - t l )
tz =
If initial volume of an ideal gas is compressed to one-half its original
volume and to twice its original temperature, the pressure.
A. doubles
C. remains constant
B. quadruples
D. halves
=
5(0.2906)(t z - 90)
103.76 OF
<D
t,=900F
..
VI
10
e
GJ
SOLUTION:
I
I
P2 T2
PIVI
----
T2
where:
Vz
r,
T,
=
V,/2·
=
2T,
P\TI
f
I
=
=
v,
P2(V\/2)
TI
PI
P2
2T}
Pz/4
4 PI
ITJ
~
T,
o
P,
V,=V,I2
T,=2T,
q.;
Thermodynamics
Thermodvnami. .,
59.7
Thermodynamics - 78 (ME Rd. Oct. 1996)
Thermodynamics - 80 (IVlE Bd. Oct. 1990)
SOLUTION:
Air is compressed polytropically so that the quantity PV I .4 is constant,
If O.D2 m ' of air at atmospheric pressllre(101.3 Kpa) and 4°C are
compressed to a gage pressure of 405 KN/m 2 , determine the fTI"]lli
temperature of the air in "C,
A. 123.23°C
C. 165.70°C
B. 187.23°C
D. 28J4SoC
For isothermal process:
=
PIY I In(P/P 2)
W
=
(14.7 x 144)(800) In(14.7/120)
W
=
-355562 i .557 ft-Ib/min
W
=
P2
540
590
P2 = 65.23 psia
What horsepower is required to isothermally compress 800 fe of air
per minute from 14.7 psia to 120 psia?
A. 13,900 Hp
C. 256 Hp
B. 28 Hp
D. 108 Hp
W
\ 1....;
SOLUTION
n-j
~~ ~~J~
3555621.557 ft - Ib I min
---------
T1
,33,000
W = 107.746Hp
-
101.3kPa
PI)
I' t,
~
r- --J-i
t,=4'C
14-J
-.21._=(405+ 101.3J-j-4
4+273
~-~
c........•.I
IOU
Thermodynamics .. 79 (Power-ME Bd Apr. 1997)
T2
All ideal gas at 45 psig and 80°F is heated in a closed container to
LWoF. What is the final pressure?
A. 54 psia
C. 75 psia
B. 65 psia
D. 43 psia
~
t2
=
t2
=
405kPa
~--?
438.7°K
438.7 - 273
/65.70 OC
Thermodynamics .. 81 (ME Rd. Oct. 1996)
SOLUTION
Pi
=
PI
=
T[
T[
T2
T2
Pj
=
T\
i
I
l
=
=
=
45 + 14.7
59.7 psia
80 + 460
540 0 R
130+460
590 0 R
P2
T2
CD
P, = 45psig
t, = BO°F
A refrigeration plant is rated at 20 tons capacity. How many pounds of
air per hour will it cool from 70 to 90°F at constant pressure.
A. 50,000 Ib/hr
C. 52,000 lb/hr
D. 45,000 lb/hr
B. 47,0001blhr
o
P,=?.
1, =130°F
SOLlJrION:
I
l
Tons of Refrigeration
mC p
(t 2 - t \ )
12,000
~ ()
Til ermodyn ([1/1 ics
m(0.24)(90 ·70)
20
,n
=
TIJ erun II lvnmnic-;
THERMODYNAlVIICS
4i
CYCLE
12 OOl)
50, (JOO lb/hr
Thermodynamics - 83 (Math-ME Bd Oct. 1998)
Fhermodvnarnics - 82 (ME Bd. Oct. 1996)
constant temperature, closed system process, 100 Btu of heat is
transferred to the working fluid at 100°F. What is the change of
entropy of the working fluid?
c: 0.25 KJiOK
A 0 18 KJ/oK
D. 034 KJ/oK
B. 0.57 KJ/oK
In
3
A steam engine operation hetween 150 0 e and 550°C. What is the
theoretical maximum thermal efficiency?
A. 99'%
C. 49%,
D. 73%
B. 27%
SOLUTION
SOLUTION:
Q
Q
TH
TH
=
=
100(1.055)
1055KJ
T= 5/9 (100-32)' 273
T ~- 310.78°K
.6.s
.6.s
==
QIT
=
0.3395 KJlkg- D!(
=
T[
=
TL
=
C =
105.5/31075
=
=
c =
e
=
550 T 273
823°K
150 T 273
423°K
TH
-
TL
---
TH
823 - 423
823
48.60'%
Thermodynamics - 84 (Math-ME Bd Apr. 1999)
An engine has a bore of 15 cm and stroke of 45 em. If the volumetric
compression is 2000 crn ', find the engine efficiency.
A. 46.2
C. 45.4
B 44.2
D. 40.3
SOLUTION:
v , = 2000 em"
V D =c (IT/4)(15i~ (45)
V D ~ 7952.156 em'
V,==V 2 + V D
V, = 2000 + 7952.156
V, - 9952.156em'
rl.
V:,"y',
42
Thermodynamics
Thermodynamics
rk
rk
e
e
=
=
=
=
\V
9952.156/2000
4.976
1
(4.976) 14-1
47.37%
=
e =
26 + 273
299°K
W
TH
QA
15
35
T]-j =
t]-j =
tH =
TH
-~
(.1
TL
~
.~.
~
s
523.25 -x
523.25 - 273
250.25"C
.\
2
s
The maximum thermal efficiency possible for a power cycle operating
between 1200°F and 225°F is:
A. 58~o
C. 57.54%
B. 58.73%
D. 57.40%
T;,
TH
TL
=
h
=
~.
~c
1200 + 460
1660 0R
225 + 460
685°R
T'4[J'
12~OF f
T H -TL
e = ----Ttl
1660- 685
e=~--~--
e
= 700 + 460
T H = 1160 0R
W = QA - QR
W = 130 - 49
(0;
tu
Thermodynamics - 87 (Power-ME Bd Apr. 1997)
Thermodynamics - 86 (Power-ME Bd Apr. 1997)
1H
~
3 QR=49
SOLUTION:
26°C
TH
SOLUTION:
1
,i;
'1
.-;)
W=15kw
299
A Carnot engine receives 130 Btu of heat
and rejects 49 Btu of heat. Calculate
reservoir.
A. -21.9°F
B. -24.2°F
•
700°F
W
130
1160
T L = 437.23°R
t L = 437.23 - 460
t L = -22.77' F
TH
-
.
-~--_.-
T "4' Q.=35KJ/s,
-
QA=130Btu
:;,
TH
1160 - T L
81
SOLUTION:
=
4
QA
1-----
A Carnot engine requires 35 KJ/sec from the hot source. The engine
produces 15 kw of power and the temperature of the sink is 26°C.
What is the temperature of the hot source?
A. 245.57
C. 250 18
B. 210.10
D. 260.68
TL
81 Btu
W
T H -T L
=
e =
Thermodynamics - 85 (Power-ME Bd Apr. 1997)
TL
43
=
.
3
22SoF
.
2
s
1660
58.73%
from a hot reservoir at 700°F
the temperature of the cold
C. -20.8°F
D. -22.7°F
Thermodynamics - 88 (Power-ME Bd Oct. 1997)
A heat engine is operated between temperature limits of ]370 DC and
260 DC. Engine supplied with 14,142 KJ per KWH. Find the Carnot
cycle efficiency in percent.
A. 70.10
C. 67.56
B. 6505
D. 69.32
45
Thermodynamics
Thermodynamics
44
SOLUT10N
Thermodynamics - 90 (Power-ME Bd Apr. 1998)
T
=
1370 + 273
T
I
Te
1643 "K
T-I = ?60
_ + 273
33"](
T;
5
~
-
o-c
•1
Q137
2
W
260"C
~
1 -
4
-
3
:
e T,
e zr: I - (53311643)
e - 6756%
.
•
2
S
A closed vessel contains air at a pressure of 160 KN/m gauge and
temperature of 30°C. The air is heated at constant volume to 60°C
with the atmospheric pressure as 759 mm A g • What is the final gauge
pressure?
A. 174
C 167
B. 169
D. 172
SOLUTION:
Patm = 759 (101.325/760)
Patm
Thermodynamics - 89 (Power-ME Bd Oct. 1997)
PI
--
TI
An Otto engine has clearance volume of 7%. It produces 300 kw
power. What is the amount of heat rejected in KW?
A. 170
C 152
B. 160
D 145
p'
rk
---
T2
P
- - -2
(30 + 273)
(60 + 273)
P2c~ 287 Kpa
P2 = 287 - 101.2
P2 = 185.8 Kpag (No exact answer in choices)
3
Thermodynamics - 91 (Power-ME Bd Apr. 1998)
c
An air standard engine has a compression ratio of 20 and a cut-off
ratio of 5. If the intake air pressure and temperature are 100 kpa and
27°C, find the work in KJ per kg.
A. 2976
C. 2437
B. 2166
D. 2751
14- 0.Q7
rk
=
---
rk
=
0.07
15.286
e
I0 1.2 kpa
P2
(160 + 101.2)
SOLUTIONI+c
=
I
=
1--r kk-1
V
SOLUTION:
I
e = 1----(15.286) 14-1
0.664
W/QA
0.664 = 300lQA
QA = 452 kw
W = QA - QR
300 ~. 452 - QR
QR = 152 kw
e
=
e
=
p
IrK - I
}
e = l--{ c
r k k-I k(r c -I)
I
e
2
,3
,4 '
I (5) I 4 _ I l
\ - (20FlI.4(5 - I)
e = 54.10%
T, = 27 + 273
J
27'C
100Kpa
V
4()
Thermodynomics
Thermodynamics
T, = 300 0 K
Process I to 2 is Isentropic process:
lk
T 1'--J
,
V,
47
4320
0.35
QA
1
QA
I
~.
12,342 K.J
2 -
\/2
T 2 = 300 (20)14-1
T, = 994.J4°K
Process 2 to J is constant pressure process:
T,
V"
_ c . = _ - =r
T
V
c
2
1
To = 994.34 (5)
To
4971.T'K
QA = m c p (T" - T 2)
QA
=
QA
=
In an air standard Otto cycle, the clearance volume is 18% of the
displacement volume. Find the compression ratio and or thermal
efficiency.
A. 0.52
C. 0.53
B. 0.55
D. 0.60
SOLUTION-
(1)(49717-994.34)
4001.3 KJ/kg
1+ c
rk
c
W
e
W
W
Thermodynamics - 93 (ME Bd, Oct. 1993)
1 + 0.18
QA
=
(0541)(4001.3)
=
21651U/kg
rk
rk
018
6.556
e
1-----
I
(6.556) 14-1
Thermodynamics - 92 (Power-ME Bd Oct. 1997)
The thermal efficiency of a particular engine operating on an ideal
cycle is 35'1<,. Calculate the heat supplied per 1200 watt-hr of work
developed in KJ.
A. 12,343
C. 14,218
B. 10.216
D 11,108
SOLUTION:
W
=
W
W
=
W
.~
e
=
1200 w-hr
1.20 kw-hr
1.2 (3600)
4320 KJ
w
()\
G.S3
e
Thermodynamics - 94
The clearance of a
determine the final
A.
B.
diesel cycle is 10%. If initial temperature is 27 DC,
tern peratu reo
e
5lO D
t
610De p
3
C. 540 D e
D. 1000 D e
2
SOLUTION:
I+c
Vj
rk --.
C
V2
t,=27°C
] + 0.1
1
rk
0.1
rk
=
] ]
lV,~
--
--- R>
V
V,
---.J-
J Ii Ie rmotlynamics
"'\'"
T'
.i
~,
'1
(V
I
lh crnuulynumics
.v.r'
49
1+0.15
l'\
T,
~
12:
273)
=
(]
015
If'
7.667
1'\
r: -
782 85°K
c
1. 2
78=.,85 -173
t2 ~
509.85r
c
c:
1-
IlLI
(7.667)'
47.89%
Thermodynamics - 97
T!l.:rmody!umll" - 95
hn n n ideal diesel cycle with overall value of k = 1.33, compression
is 15 and cu-off ratio of 2.1 ~ determine the cycle efficiency.
f\
50.62%
C 46.00';;0
B. "290";0
D. 4900%
;'1 [iO
'\11 ideal Otto cvcle. operating in hot air with k = 1.34 has compression
r a t ro of 5. Determine the efficiency of the cycle.
A 52 45%
C 64,27%
B. 64.27%
D. 36.46%
SOLUTION:
SOL UTION.
1
1
1--- ---- { - '----}
r l, kl kir-1)
~ c
e
(21):
t:
" '-'
(. 15)
I
1 ""
~
.~.
0.529
e
=
52.9%
I 1
- I
---- ---_._-----
1,';,7[-1
1.---, ••' \L., .
.)
ihcrmodYnarnics' 96
~o
ideal Otto cycle with 15"". clearance operates on 0.227 kg/sec of air
= 1.32. Determine the efficiency of the cycle.
/\. ';065%
C 4 \32%
f3 4384"/0
D. 47.89%
\ it h I,
e
I-~
r \..
e
1---(5) 134-1
e
42.14%
Thermodynamics - 98
An engine operates on the air standard Otto cycle. The cycle work is
900 KJ/kg, the maximum cycle temperature is 3000°C and the
temperature at the end of isentropic compression is 600°C. Determine
the engines compression ratio.
A 6.388
C. 867
B. 10.45
D. 7.87
SOLUTION:
SOUJTIClN
0\
r,
I
I
c
"
IllqT; - T 2 )
I;
,000 + 273
1;3273°](
I.
600 1- 273
so
Thermodynamics
T 2 = 873°K
QA/m = ciT) - T 2)
QA/m = (0.716)(3273 - 873)
QA/m o~ 1718.4 KJ/kg
e = W/QA
e = 900/1718.4
P
e = 0.5237
1
e = 1--rk
I 'he rtnodyn am ie"
Tlll:rlllol!v 11<1 III ics - I (Ill (I\U: Bd. Apr. 1(92)
-\ tIll',el cil~illl' is o[Jeraling on a -l-str okc cycle. has a heat rate of
II.JI5.6 KJ,K\V-hr brake. The compression ratio is 13. The cut-off
r.ui» IS 2. [i S lI 1g K = 1.32. what is the brake engine efficiencv.
i\ I)-;.~,
c. 735
3
=
:' 1=
r~
D~5.3
k-l
')()!!.
1
0.5237 = 1 - -..
(r k )
rk
; I ( ) '"
r r,
______
t,=600°C
6.338
v•
r, "
.s:
.. I
(
k (r c
-
1')
(1) 13" _ 1
n an air standard diesel cycle, compression starts at 100 Kpa and
OO°K. The compression ratio is 16 to 1. The maximum cycle
emperature is 2031 "K, Determine the cycle efficiency.
C. 65.98%
A. 60.34%
D. 45.45%
B. 56.23%
e =
1------
c
0:>010
,
(13)';2'
132(~> 1)
(1 !<v,- hr) (3600)
e:
1 Ul:5.6KJ
0:; 18
e,)
'>
~,
e, ,"
ec
0.318/0.5010
c..
63.5%
SOLUTION:
T 2/T 1 = (v/vd- '
T 2/(300+273) = (16/1)14
T 2 = 1737.01 10 K
I __
,I
I'hermodynamics - 99 (ME Bd. Apr. 1995)
0
1
P
rocess 2 to 3 is constant pressure:
V)V 2 = T)/T 2
r, = V)N 2 = 2031/1737.011
r, cc 1.169
1
r k -I
e = I---{ c
}
r k k-l k(r -I)
c
2
3
4
T,=2031°K
T,=3000K
P,=100okPa
1
Therrnodvna nucs - 101 (ME Bd. Apr. 1991)
Determine the air-standard efficiency of an engine operating on the'
diesel cvcle with clearance of 8'Yo when the suction pressure is 99.9:
Kpa and the fuel is injected for 6'Yo of the stroke. Assume K = 1.4.
SOLUTION
v
r
1
(1.169) J ( - 1
e=I---{
}
(l6)IH 1.4(1.169-1)
I', k- I
\I, - V;
e = 65.98%
5I
"
=
k -
kr r
'
i
1)
(
(J06V,)
v.
008V"
\I, - 008V il v 014 V)
()
Il() \"
r
Thermodynamics
52
rc
=
0.06V
v.v.
P·2·
(0).
VJ
C;;
Thermodynamics
3
PURE SUBSTANCE
=
0.14V O
4
0.08V o
1.175
i + 0.08
=
,1 '
IA~
0.08
13.5
1
=
(13.5)14-1
60.02%
{
Find the enthalpy at 100 psi and 97% quality, h r = 298.55 Btu/lb;
h rg = 889.119 Btu/lb.
A. 1,170 Btu/lb
C. 1,734 Btu/lb
B. 1,161 Btu/lb
D. 1,803 Btu/lb
LV
vp
$.08 Vp
e = 1e
P,=99.97
Thermodynamics - 103 (Math-ME Bd. Oct. 1999)
rk = - -
r,
53
V2
rc =
rc
r
1
p
(1.75)~4 - 1
}
1.4(1.75-1)
SOLUTION:
h
=
he + x h eg
Thermodynamics - 102 (Math-ME Bd Apr. 1996)
h
=
298.55 + 0.97(889.119)
A heat engine (Carnot cycle) has its intake and exhaust temperature of
157°C and 100°C, respectively. What is its efficiency?
A. 12.65%
C. 15.35%
B. 14.75%
D. r3.7'>%
h
=
1161 Btu/lb
Thermodynamics - 104 (Power-ME Bd. Oct. 1999)
SOLUTION:
T H = 157 + 273
T H = 430 0K
T L = 100 + 273
T L = 373°K
Efficiency
=
T'l~l
-c
TH -TL
TH
430- 373
Efficiency
,v
=
430
Efficiency = 13.25%
'1-
6J.
•
j
180 grams of saturated water of temperature 95°C undergoes
evaporation process until all vapor completely vaporized. Determine
the changed in volume.
At 95°C, Vr = 0.00 I 0397 mvkg, V g = 1.9819m3/kg
3
C. 0.2565 rrr'
A. 0.1656 m
B. 0.4235 m'
D. 0.3656 m 3
2
SOLUTION:
T
s
Volume =
Volume =
Volume =
Volume =
Specific Volume x mass
(v g - vr) x m
(1.9819 - 0.0010397)(0.18 kg)
0.3565 m 3
Thermodynamics - 105
s
Five kilograms of saturated liquid at 120 Kpa is heated until its
moisture content is 5%. Find the work done for this process.
54
Thermodynamics
Thermodynamics
A. 813.59 KJ/kg
B. 643.23 KJ/kg
SOLUTION:
C. 542.34 KJ(kg
D. 753.12 KJ/kg
Thermodynamics - 107
For constant pressure process,
W
=
P(V2 -
55
VI)
From steam table: At 120 kpa
VI = vr at 120 Kpa (sat. liquid)
v g = 1.4284 roJ/kg
VI = 0.0010473 mJ/kg
V2 = vr + x Vrg
x = 1- Y
x = I - 0.05
x ~ 0.95
V2
0.00J0473 + 0.95(1.4284 - 0.0010473)
V, = 1.357
Vi = 120(1.357 - 0.0010473)
W = 162.73 KJ/kg (5 kg)
W = 813.59 KJ
5
Steam at 2 Mpa and 250°C in a rigid cylinder is cooled until the
quality is 50%. Find the heat rejected from the cylinder.
At 2 Mpa and 250°C
3
V = 0.11144 m /kg
u = 2679.6 KJ/kg
At 2 Mpa, (saturated)
vr = 0.0011767 rrr'zkg
v g = 0.09963 mJ/kg
Ur = 906.44
Urg = 1693.8
A. -432.23 KJ/kgC. -834.55 KJ/kg
B -926.26 KJ/kgD. 1082.34 KJ/kg
SOLUTION
Q = (U 2
Thermodynamics - 106
Twenty kilograms of water at 40°C is confined in a rigid vessel. The
heat is supplied until all the water is completely vaporized. Find the
heat added in KJ.
C. 45,252 KJ
A. 45,422 KJ
D. 65,233 KJ
B. 43,122 KJ
-
UI)
U 1 = 2679.6
U. = U, + x Urg
U 2 = 906.44 + 0.5(1693.8)
U2
1753.34 KJ/kg
Q
Q
=
CC
(175334 - 2679.6)
-926.26 KJ/kg
Thermodynamics - 108
SOLUTION:
V=c
For rigid vessel,
Q
UI
U2
=
v,)
(VI =
m (U 2
Ud
= U, (saturated liquid)
= U g (saturated vapor)
Q == m (U rg )
Q = 20 (2262.6)
=
t = 40°C
-
Q = m (U, - U r)
Q
me = 20kg
45,252 KJ
1
Q
Find the entropy in KJ/kg-K at 90% moisture of a IMpa steam-water
mixture?
At 1 Mpa:
Sg = 6.5865
Srg = 4.4478
A. 4.87
C. 2.583
B. 6.34
D. 4.36
SOLUTION:
x
x
I - 0.9
0.10
Thermodynamics
')6
Thermodynamics
S ~' Sf + X Sfg
Sfg = Sg - Sf
44478 = 6.5865 - Sf
Sf = 2.1387
SOLUTION:
For isothermal process, t)=
Thermodynamics - 109
SOLUTION:
S = Sf + X Sfg
4 = 2.2525 + x (6.4953 - 2.25 IS)
x = 0.412
h
hf X hfg
h = 814.93 + 0.412(1972.7)
h
Thermodynamics - I I I
A tank contains exactly one kilogram of water consisting of liquid and
vapor in equilibrium at I Mpa. If the liquid and vapor each occupy
one-half the volume of the tank, what is the enthalpy of the contents of
the tank?
/>. 644.40 KJ/kg
C. 8331i0 KJ/kg
B. 774.40 KJ/kg
D. 435.2lJ KJ/kg
At I Mpa
Vf= 00011273
Vfg
hIe 762.81
hfg " 20 I 53
~c
=
Let V
.--(Sat. Vapor)
Sf' 1.8607
Sfg = 4.9606
At 300 Kpa and 151.86°C
S = 70888 KJ/kg
A 652.34 KJ/kg
B. 535.16 KJ/kg
~.
total volume of tan-,
in! == Vl/Vl.
1/2 V
mi.
Mixture with 80% quality at 500 Kpa is heated isothermally until its
pressure is 300 Kpa. Find the heat added during the process.
At 500 Kpa
mL
00011273
443.54 V
m,
v«.
1" :2 V
m.
01944
.' 572 V
m
m,
x
-r-;
I'; \
C. 983.44 KJ/kg
D. 765.34 KJ/kg
0.19444
=
SOLLJTlON:
1627.71 KJ/kg
Thermodynamics - 110
t2
Q = T (S2 - S) )
SI = Sf + X Sfg
SI = 18607 + 0.3(49606)
SI = 5.829
S2 = 7.0888
Q = (151.86 + 273)(7.0888 - 5.829)
Q = 535.16 KJ/kg
= 2.1387 + 0.10(44478)
S = 2.583 KJ/kg-"K
At 1.3 Mpa, mixture steam and water has an entropy of 4 KJ/kg_°K.
Find the enthalpy of the mixture.
At 13 Mpa:
sf~22515
hf = 8 14 .93
Sg = 6.4953
hfg = 1972.7
A. 1627.71 KJikg
C. 1234.45 KJ/kg
B. 1533.33 KJ/kg
D 1734.45 KJ/kg
5'7
+- ,n L
L::::
, rn, = 1 kg
(Sat. Liquid]
r -3
-
--
m.
me
'2_
iV'I.,v,'
r
.1..
v=;v
2
'
-
.4-_
58
Thermodynamics
2.572V
x
c. ].672
D. 3.230
A. 156
B. 2.12
2.572V + 443.542V
SOLUTlON:
0.005765
= h, + xh rg
h = 762.81 + 0.005765(2015.3)
h = 774.43 KJ/kg
x
h
59
Thermodynamics
=
From steam tables:
At 70 bar(7 Mpa) and 65°C
VI = 0.001017 mJ/kg
At 50 bart S'Mpa) and 700
V2 = 0.06081 rrr'zkg
0K(427°C)
Thermodynamics - III (ME Bd. Oct. 1996)
ml
A vessel with a volume of 1 m contains liquid water and water vapor
in equilibrium at 600 Kpa. The liquid water has a mass of 1 kg. Using
stea m tables, calculate the mass of water vapor.
A.3.16kg
C. 1.57kg
B. 0.99 kg
D. 1.89 kg
m2
=
01 I
J
v) = Q2 I V2
AxV\
A(]OO)
0.00 I0 17
0.06081
---
VI
=
1.672 m/sec
SOLUTlON:
From steam tables, at 600 Kpa:
~
~i~~~
-'m
v.
vg
>
=
0.001 ]01 mJ/kg
0.3157 m 'zkg
1L
m,
Volume ofliquid
Volume of liquid
Volume of liquid
=
=
=
mL VL
](0.001101)
0.001101 m'
m
__
Thermodynamics - 114 (ME Bd. Oct. 1991)
Vv
t,
(Sat. Vapor)
Water substance at 70 bar and 65°C enters a boiler tube of constant
inside diameter of 25 mm. The water leaves the boiler tube at 50 bar
and 700
at velocity of 100 m/s. Calculate the inlet volume flow(li/sec)
A. 0.821
C. 0.344
B.1.561
0.2.133
0K
6S'C
v
7Ob~
600Kpa
Volume of vapor
Volume of vapor
=
=
1-0.001101
J
0.998899 m
(Sat. Liquid)
Mass of vapor = 0.998899/0.3157
Mass of vapor = 3.164 kg
Thermodynamics - 113 (ME Bd. Oct. 1991)
Water substance at 70 bar and 65°C enters a boiler tube of constant
ins'ide diameter of 25 mm. The water leaves the boiler tube at 50 bar
and 700 0 K at velocity of 100 m/s. Calculate the inlet velocity(m/sec)
SOLUTION:
V,
hom steam tables:
At 70 bar(7 Mpa) and 65°C
3
VI = 0.001017 m /k g
Ef
'
-
:
At 50 bar(5 Mpa) and 700 0 K ( 4 2 7 ° C )
3
V2 = 0.06081 m /k g
mt
QI
=
I
VI
m2
=
Q2
I
V2
700 0K
~ba'
V,=100m/s
Thermodynamics
Thermodynamics
60
AxV j
A(lOO)
--0.001017 0.06081
V I = 1.672 mJsec
Q\ = A x vet,
QI = rc/4 (0.025/(1.672)
Q\ = 0.8207li/sec
61
Thermodynamics - 120 (ME Bd. Apr. 1989)
Steam at the rate of 600 kg/hr is produced by a steady flow system
boiler from feedwater entering at 40°C. Find the rate at which heat is
transformed in Kcal per hour if enthalpy of steam is 660 Kcal/kg and
of the feedwater at 40 Kcal/kg.
C. 345,200
A 372,000
B. 387,000
D. 312.444
SOLUTION
Thermodynamics - 115 (ME Bd. June 1990)
One Ib (0.455 kg) of a mixture of steam and water at 160 psia(l 103.2
Kpa) is in rigid vessel. Heat is added to the vessel until the contents are
at 560 psia (3861.2 Kpa) and 600°F (315.55°C). Determine the quantity
of heat in KJ added to the water and steam in the tank.
A. 1423.70
C. 1562.34
B. 1392.34
D. 1294.45
Rate at which heat is transformed
~
ms(h, - h-)
Rate at which heat is transformed
=
600(660 - 40)
372,000 Kcal/hr
Rate at which heat is transformed
Thermodynamics - 12] (ME Rd. Oct. 1988)
SOLUTION:
For a rigid vessel, the volume is constant:
From steam tables:
At 1103.2 Kpa
3/kg
vr ,~ 0.0011332 m
J/kg
v g = 0.17704m
U. = 780.65 KJ/kg
U rg = 1805.8 KJ/kg
At 3861.2 Kpa and 315 .55°C,
3/kg
V2 = 0.06378 m
U 2 = 2761.3 KJ/kg
Solving for the quality of mixture:
Vj = V2
vr+ XVrg = V2
0.0011332 + x(0.17704-0.00 11332) = 0.06378
x = 0.3561 = 35.61%
Solving for U\:
U = U r + xUfg
U, = 780.65 + 0.3561(1805.8)
U\ = 1423.70 KJ/kg
Heat added = m(U 2 - U I)
Heatadded = 0.455(2761.3 - 1423.7)
Heat added = 608.6 KJ
Steam leaves an industrial boiler at 827.4 Kpa and 171.6°C.·A portion
of the steam is passed through a throttling calorimeter and is
exhausted to the atmosphere when the calorimeter pressure is 10].4
Kpa. How much moisture does the steam leaving the boiler contain if
the temperature of the steam at the calorimeter is] ]5.6°C?
A 3.78%
C. 456%
B. 308%
D. 2.34%
SOLUTION:
I
~
E
At 827.4 Kpa (171.6°C):Jl
I
r
Calorimeter
~ 101.4kPa
115.6°C
h; = 727.25 KJ/kg
<i.i
h fg = 20432 KJ/kg
From table 3:
At 101.4 Kpa and 115.6°C:
827.4kPa
h 2 = 2707.6 KJ/kg
171.6°C
Let x = quality of steam entering the throttling
calorimeter.
hi = h2
hr + xh., = h2
727.25 + x(2043.2) = 2707.6
x = 0.9692
62
Thermodynamics
y = moisture content
Thermodynamics
63
SOLUTION:
y = I - 0.9692
V=O.058m'
y = 0.0308
y = 3.08%
Thermodynamics - 122 (ME Bd. Oct. 1995)
Steam enters a throttling calorimeter at a pressure of 1.03 Mpa, The
calorimeter downstream pressure and temperature are respectively
0.100 Mpa and 125°e. What is the percentage, moisture of the supply
steam?
Properties of steam:
P,Mpa
h,
htg
hg
103
2010.7
2779.25
Note At 0.100 Mpa and 125°C
h ~ 27266 Kl/kg
A. 2.62
C. 3. J 5
D. 198
8 5.21
SOLUTION.
VI = V 2 = 0.058/2
VI = V2 = 0.029m3
T] ~ 27 + 273
T I = 300 0 K
T 2=I77f273
T 2 ~ 450 0 K
=
ill I =
m, =
ill 2 =
m, =
m, =
ill I
I
I
I
/il'
v,
V,
137.8kPa :413.4kPa
2r>C
I
17r>C
PIV/RT 1
(I 37.80)(0.029)/(0.287)(300)
00464 kg
P 2V 2/RT 2
(413.4)(0.029)/(0.287)(450)
0.0928 kg
Heat loss = Heat gain
rn, C v2 (t 2 - tf) = rn, cv l (t f - t,)
0.OlJ2S(0.7I 6)(1 77 - tf) = 0.0464(0.7I6)(tf - 27)
tf = 127°C
T, = 127 + 273
T r = 400
L\.s = m c, In(T tiT1)
L\.SI = 0.0464(0.716) In(400/300)
L\.SI = 0.00956
L\. S 2 = 0.0928(0.716) 1n(400/450)
L\.SI = -0.00783
L\.s = 0.00956 - 0.00783
L\.s = 0.00173 KJ/"C
0K
h n = 2779.25 - 2010.7
1.03,MPa :f. C.10MPa, 125°C
h n ~ 768.55 KJ/kg
For throttling process:
hi = h2
h, + xh., = h2
76855 -+- x(20 10.7) ~ 2726.6
x
09738
x
97.38%
y .~ 100 - 97.38
Y = 2.62%
Thermodynamics - 125
Thermodynamics - 123 (ME Rd. Apr. 1996)
A vessel of 0.058 m 3 capacity is well insulated and is divided equally by
a rigid conducting diaphragm. Initially both halves contain air at
pressure of 137.8 Kpa and 413.4 Kpa and temperature of 27°C and
177°C respectively. What is the increase of entropy of the system in
Using steam table, find the enthalpy of steam at 250 kpa if its specific
volume is 0.3598 m 3/kg
A. 1625.86 KJlkg
C. 1543.45 KJlkg
B. 1785.34 KJlkg
D. 1687.55 KJlkg
KJ/OC?
A 1.002
B. 0.5080
C. 0.00173
D. 0.1080
SOLUTION:
At 250 kpa:
Thermodynamics
t\~
h[= 535.37 Kl/kg
h,~ = 2181.5 KJ!kg
v,' = 0.0010672 m]/kg
v g = 0.7187 m]/kg
v = v, -+- XVlp
03598 = 0.0010672 + x(0.7187 - 0.(010672)
x r--. 0.49988
Solving for h:
h
=
h,
-+-
xhfg
h = 53537 + 0.49988 (2181.5)
n
/625.86 KJ/kg
z;
Thermodynamics
65
Thermodynamics - 127
Steam enters an isothermal compressor at 400°C and 100 kpa,
exit pressure is 10 Mpa, determine the change of enthalpy.
A. 198 KJlkg
C. 187 KJ/kg
B. 178 KJ/kg
D. 182 KJ/kg
The
SOLUTION:
At 100 kpa and 400 oe:
h = 3278.2 KJ/kg
For isothermal process,
t 2 = t\ ~~ 400°C
Thermodynamics - 126
A throttling calorimeter is connected to the desuperheated steam line
supplying steam to the auxiliary feed pump on a ship. The line
pressure measures 2.5 Mpa. The calorimeter pressure is 110 kpa and
150"C: Determine the entropy of the steam line.
A. 6.8 KJ!kg-OK
C. 6.6 KJlkg-OK
8. 7.2 KJ/kg-OK
D. 7.5KJlkg-OK
At 400"C and 10 Mpa:
h = 3096.5 Kl/kg
/"h ". h, - h2
,)h= 3278.2 - 3096.5
L\h = ] 81. 70 lU/kg
SOLUTION:
At 110 kpa and 150 oe:
h2 = 2775.6 Kl/kg
At 2.5 Mpa:
h. = 962.11 Kl/kg
hfg = 1841 KJ/kg
Sf = 2.5547
Sfg = 3.7028
For throttling process: (h I = h2 )
hi = h2
h, + xh.,
2775.6 + 962.11 + x(1841)
x = 0.985
s I - Sf + XSfg
s\- 25547 + 0.985(37028)
s\
6.202 KJ/kg-"K
Cc
l
Thermodynamics - 128
Stearn enters an adiabatic turbine at 300°C and 400 kpa, It exits as a
saturated vapor at 30 kpa, Determine the work done.
A. 476.34 KJ/kg
C. 436.33 KJlkg
B. 441.50 Kl/kg
D. 524..34 Kl/kg
SOLUTION:
At 300°C and 400 kpa:
b, .= 3066.8 Kl/kg
At 30 kpa and saturated vapor:
h 2 =. h g = 2625.3 KJ/kg
W= hi - h2
W = 3066.8 - 2625.3
W
44/.51U/kg
n~
J
_1
). ._~~,~J.l~_ .~\-"'.__
..
300°0
400Kpa
----------toS
66
Thermodynamin'
Thermodynamics
67
Thermodynamics - 129
Thermodynamics - 131
A 0.5 m ' tank contains saturated steam at 300 kpa. Heat is transferred
until pressure reaches 100 kpa. Find the final temperature.
A I kg steam-water mixture at 1.0 Mpa is contained in an inflexible
tank. Heat is added until the pressure rises to 3.5 Mpa and 400°C.
Determine the heat added.
A. 1378.64 KJ
C. 1456.78 KJ
B. 1532.56KJ
D.1635.45KJ
A. 934S"C
B ~3::'3uC
C 99.63°C
D 103.2'C
SOLl;TJO,\:
SOLUTION:
At 300 kpa:
() 6058 m ' kg
\
VI
-
At 100 kpa, ~ 0 00 I 0432 m'/kg
Yf
J' 694 m3/kg
v ~ ,.
Since v I is in between Vr and vb at I ()o kpa. then the
temperature is equal to the saturation temperature at
I no kpa which is equal to 99.631!C.
At 3.5 Mpa and 400°:
V2 = 0.08453 KJ/kg-OK
U 2 = 2926.4 KJ/kg
At 1 Mpa:
vr = 0.0011273 m3/kg
vg = 0.19444 m3/kg
U, = 761.68 m 3 lk g
U rg = 1822 KJ/kg
for inflexible tank,
VJ
=
V2 =
VI
Vr
=
V2
+
XVrg
Thermodynamics - 130
0.08453 = 0.0011273 + x(0.1944 - 0.001127)
x = 0.4314
A 500 Ii tank contains a saturated mixture of steam and water at
300"C. Determine the mass of vapor if their volumes are equal.
A. 1154 kg
C 1345 kg
UJ =
+ x U rg
U I = 761.68 + 0.4314(1822)
U I = 1547.76 KJ/kg
Q=m(U 2 - U r)
8.]034kg
SOLUTION
D.1634
J,g
u,
Q = 1(2926.4 - 1547.76)
Q = 1378.64 KJ
At 300°C
vb =
mv
-r-
002167 m'/kg
V[
~.
Thermodynamics - 132 (Math-ME Bd. Oct. 1999)
v
"
500/2
VI
---
VI
111 v
1000
025 m'
0.25/0.02167
mv
11.54 kg
Atmospheric pressure boils at 212°F. At the vacuum pressure at 24 in
Hg, the temperature is 142"F. Find the boiling temperature when the
pressure is increased by 45 psia.
A. 342.34°F
C. 479.13"F
B. 526.34°F
D. 263.45°F
Thermodynamics
Thermodynamics
68
HEAT CAPACITY
SOLUTION:
'"'
P2
P2
PI
PI
=
=
14.7 + 45
59.7 psia
=
-24(14.7/29.92) + 14.7
=
2.908
Thermodynamics - 132 (Power-ME Bd Apr. 1998)
By interpolation:
t2
59.7
212
14.7
142
2.908
t,-212
59.7-14.7
142
59.7 - 2.908
t2
-
69
T,
212"F
142°F
t2 -212 = 0.7923tr 112.515
t 2 = 478.98°F
14.7+45psi
14.7psi(atmospheric)
-24"Hg
What is the temperature in degree C of 2 liters of water at 30°C after
500 calories of heat have been added to?
A. 35.70
C. 38.00
D. 39.75
B. 30.25
SOLUTION:
Q
Q
Q
30
0
e
HEATER
t,
m c p (t 2 - t.)
500 cal
Q=500cal
= 0.50 Kcal
0.50 x 4.187 = (2 x I) (4. I 87)(t 2 - 30)
t2 = 30.25"C
=
=
Thermodynamics - 133 (ME Bd. June 1990)
A mass of 0.20 kg of metal having a temperature of lOODC is plunged
into 0.04 kg of water at 20 DC. The temperature of the water and metal
becomes 48 DC. The latent heat of ice at ODC is 335 KJ/kg-DK, and the
specific heat capacity of water is 4.19 KJ/kg-DK. Assuming no heat loss
to the surroundings, determine the specific heat capacity of the metal
in KJ/kg-DK.
A. 0.234
C. 0.754
water
Metal
B. 0.564
.---"----p. OA51•r - - -•
SOLUTION:
0.2kg
0
100 e
0.04 kg
20°C
Mixture
Q
Heat loss by metal = heat gam by water
(m c p flt)metaJ = (m c p flt)water
70
Thermodynamics
0.20( c m )( 100 - 48)
c.,
=
=
Thermodynamics
0.04(4.19)(48 - 20)
Heat loss by iron
0.451 KJ/kg- oK
m, cp,(t I
Thermodynamics - 134 (ME Bd. Apr. 1996)
What is the total energy required heating in raising the temperature of
a given amount of water when the energy applied is 1000 KWH with
heat losses of 25%?
A. 1000
C. 1333
B. 1500
D. 1250
Q
--.
HEATER
=
=
Heat gain by water
- t )
2
rn., c pw (t 3
where: t, = equilibrium temperature after mixing
30(0.4)(220 - t) = 14.33(4.1 87)(t) - 10)
220 - t) = 5(t) - 10)
t) = 45°C
i\.s = Q,
T]
11000~Wh
Q - 0.25(Q) = 1000
=
t3 )
i\.s ~ m, c p, (t 3 ~tl)
SOLUTION:
Q
-
1
1333 KWH
Loss=O.25Q
Thermodynamics - 135 (Power-ME Bd Oct. 1999)
A 30 kg iron was put in a container with 14.33 kg water. The water is
at looe and the iron has an initial temperature of 493°K until the iron
was in thermal equilibrium with water. Find the change in entropy.
(c, for iron = 0.4 KJ/kg°K)
A. -12.56 KJf'K C. -25.78 K!f'K
B. -43.58 KJf'K D. -6.6 KJf'K
Iron
SOLUTION:
t,
Water
30 kg
220°c
1,
'\:
t,
t,
-r-
=
493 -273
220°C
Mixture /
D
14.33kg
20 0 e
t3
+ 273
30(0.4)(45- 220)
.6.5
= ------
i\.s
=
45 + 273
-6.6 KJlkg
71
Fuels & Combustion
Fuels & Combustion
72
SOLLTION
FUELS AND COMBUSTION
Fuel
0,
CIj!];O +
Fuels & Combustion - 1
C1JI!;u
A diesel power plant consumes 650 liters of fuel at 26°C in 24 hours
with 28°API. Find the fuel rate in kg/hr.
A 23.83
C. 22.85
D. 26.85
B. 24.85
~
~
~. A .
~
3.76N z
Product of Combustion
~
CO 2
-i-
7e
~
)(
o'
i
21.5(3.76)N 2
-- ---
Theoretical A/f
I
=
\02.34
")
'X" ""'" . 2.2:-:\_·~~ActuaJ
~
14CO:.; 15H 20
21.5 + 21.5(376)
r
t) lv\t1.~7"7 0t'1-~'j.~
H20 +- 3.76 he
71.- :"J') Theoretical A.T
0" )( l''J
f/'l I
~"-!1
SOLlJT!ON:
131.5+° API
141.5
Air
2150 2 + 21.5(376)N 7
i:"-Y\-\
141.5
+
.L
f\ M~k~ Nt'; 1'.,
:7
Sg.t 156°C
73
Y))
102.34 (1.15)
117.69 mol air/mol fuel
A/F
Actual AlF
7._--=~"-":i.2..
~~
Fuels & Combustion - 3
Sgat1560C
Sg.t156"c =
SG a1 W C
SG at 26"C
=
=
131.5+28
0.887
0.887[ I - 0.0007(26 - 15.6)]
0.88
Density of fuel = 0.88(1 kg/Ii)
Density of fuel = 0.88 kg/Ii
w
m
v
V = 650/24
V = 27.0833 li/hr
0.88 = m/27.0833
m = 23.83 kg/hr
A diesel power plant uses fuel that has a density of892.74 kg/m}
at 15.67°C. Find the heating value of fuel.
SOLUTiON:
Density of fuel
SG
- - - - - _.. _----
Density of water
892.74 /1000
SG
SG
0.89274
"API
o.h,PI
Fuels & Combustion - 2
A boiler burns fuel oil with 15% excess air. The fuel oil may be
represented by C I4 H 30 • Calculate the molal air fuel ratio.
A 14
C. 102.34
B 117.69
D.17.14
C. 43,000 K.J:'kg
D. 43562 KJ/kg
A. '(4.911() KJ/kg
B. \9,301 Btu/lb
=
1..J 1.5
------ - 131.5
0.89274
27
Q ~. 41.130
Q =1-],130
Q
cc
r-t-
139PAPJ
!39.6(27)
44,899.31 KJ/kg
Fuels & Combustion
74
.e
Flld\
7~
Combustion
Fuels & Combustion - 4
Fuels & Combustion - (,
A certain coal has the following ultimate analysis:
C = 69%
N2 =5%
H 2 == 2.5%
S = 7%
Moisture = 8%
Ash = 5%
O 2 = 3.5%
Determine the heating value of fuel used.
C. 25,002.4 KJ/kg
A. 26,961.45 KJ/kg
8. 45,256 KJ/kg
D. 26,000 KJ/kg
A diesel engine consumed 945 liters of fuel per day at 30°C. If the fuel
was purchased at 15.5°C and 30° API at P5.00/li, determine the cost of
fuel to operate the engine per day.
C P4888.90
A P5677.50
. O. P5000.0Q
B. P4677.50
SOUTlON
SOLUTION:
SG 13 6 ' [
131.5 + 30
SG I 3 6 C = 0.87616
SG,QcC = 0.87616 [I - 00007(30 - 156)J
SG w c = 0.8673
v ;O'C
SG IS 6'C
o
Qh = 33,820C + 144,2l2(H --) + 9,304S
8
0.035
33,820(0.69) + 144,212(0.025 - - - ) + 9,304(0.07)
8
= 26,961.45 KJlkg
Oh =
Oh
1415
=
VI' 6C
945
SG JO'C
0.87616
-------
VIS 6"C
Vt3 6
Cost
Cost
Fuels & Combustion - 5
A diesel power plant uses fuel with heating value of 45,038.8 KJ/kg.
What is the density of fuel at 30°C?
A. 0.900 kg/Ii
C. 0.850 kg/Ii
B. 0.887 kg/li
D. 0.878 kg/Ii
SOLUTION:
Oh
= 41,130 + 139.6 °API
45,038.8 = 4],130 + 139.6(oAPI)
API = 28
C
=
=
0.8673
= 935.44 li
P500/li (935 44 Ii)
P4,677.20
Fuels & Combustion - 7
\. cylindrical tank 3 m long and 2 m diameter is used for oil storage.
How many days can the tank supply the engine having 27° API with
fuel consumption of 60 kglhr?
3m
A.484
C. 7.84, B 5.84
O. 8.84 ~
2m
SOLUTI.ON
SG I5 6, C =
]41.5
---
131.5+0 API
SG I5 6, C = 0.8872
SG.1 30, C = 0.8872[ 1 - 0.0007(30 - 15.6)]
SG ot 30'C = 0.8782
Density of fuel = 0.8782(1 kg/Ii)
Density of fuel = 0.8782 kglli
V
~ ;v4 0 2h
V - "1412/(3)
V.
4.42478 m '
1415
'-,( J
t,
f. (
1]1.5~·27
~,"'Okglh
I~~
·
..
~
'..
~
/0
Fuels & Combustion
SG l 5 6
0
C
2.
Fuels & Combustion
0.8927
77
SOLUTION:
Density of fuel ~. 0.89274(1000 kg/m3)
Density of fuel = 892.74 kg/nr'
W = m/V
892.74 = 60IV
V = 0.0672 m 3/hr
A/F = 11.5C + 34.5(H - 0/8) + 9.3S
A/F = 11.5(0.7) + 34.5(0.03 - 0.04/8) + 4.3(0.06)
A/F ~~ 9.1705(1.25)
A/F = 11.46 kg air/kg fuel
Number of days = 9.42478/0.0672
Number of days = 140.23 hrs
Number of days = 5.843 days
Fuels & Combustion - 10 (ME Bd. Apr. 1991)
:it A 650 Bhp diesel engine uses fuel oil of 28° API gravity, fuel
Fuels & Combustion - 8
Determine the minimum volume of day tank in
nr' of 28° API fuel
having a fuel consumption of200 kg/hr.
A. 10.43 m'
B. 5.41 m'
l
C. 6.87 rn'
(
D. 7.56 m'
consumption is 0.65 lb/Bhp-hr. Cost of fuel is P7.95 per liter. For
continuous operation, determine the minimum volume of cubical day
tank in em), ambient temperature is 45°C.
3
A. 5,291,880 em
C. 5,491,880 em]
B. 5,391.880 crn'
D. 5,591,880 crn'
SOLUTION:
SOLUTION:
1415
SG 15 6°e
=
SG 15 6°e
=
141.5
131.5 + 28
0.887
SG\S6'C
=
0
0
5.4/ m3
Solving for fuel consumption'
m. = 0.65(650)
m. = 422.5 lb/hr
m- = 191.61 kg/hr
V r = 191.61/0.869
V f = 220.495 li/hr
Fuels & Combustion - 9
Given the following ultimate analysis:
C = 70%
O 2 = 4%
N2 = 5%
~ = 6%
Moisture
H 2 = 3%
Ash = 5%
Using 25% excess air, determine the actual air fuel ratio
A. 11.46
B. 24.85
C. 23.85
D. 26.85
--
131.5-t-28
SG!56'C = 0.887
SG 45 C = SG1WC[l - 0.0007(t - 15.6)]
SG 4 5 C = 0.887[1 - 0.0007(45 - 15.6)]
SG 4 5, C = 0.86Q
Density of fuel =2 0.869(1 kg/li) ~
Density of fuel = 0.869 kg/li
~
Density of fuel = 0.887(1000)
Density offue1 = 887 kg/rn '
W = m/V
887 = 200N
V = 0.22548 m 3/hr x 24 hrs/day
V
=
=
8%
Volume
Volume
Volume
Volume
of dav
of day
of day
of day
tank
tank.
tank.
tank.
=
=
=
=
220.495 x 24 hrs
3/1000li
5,291.88 Ii x Im
3
5.291.88 m' x (100)3 e m 3 1m
5,291,880 em'
Fuels & Combustion
Fuels & Combustion
78
S( l[l
79
iT!ON:
Fuels & Combustion - 11 (ME Bd. Apr. 1991)
A 650 Bbp diesel engine uses fuel oil of 28°API gravity, fuel
consumption is 0.65 lblBbp-br. Cost of fuel is P7.95 per liter. For
continuous operation, determine tbe cost of fuel per day at 45°C.
A. P42,870.45
C. P42,570.45
B. P42,070.45
D. P42,170.45
SOLUTION:
SG 15.6o C =
141j
---
131.5 + 28
SG 1W C = 0.887
SG w c = SG 1W C[1 - 0.0007(t - 15.6)]
SG w c = 0.887[1 - 0.0007(45 - 15.6)]
SG w c = 0.869
Density offuel = 0.869(1 kg/li)
Density of fuel = 0.869 kg/li
Solving for fuel consumption:
m, = 0.65(650)
m, = 422.5 lb/hr
IDr= 191.61kglhr
v, '" 191.61/\).869
V r = 220.4951i/hr
Volume of day tank = 220.495 x 24 hrs
Volume of day tank = 5,291.88 li
Cost of fuel per day = 5,291.88 li x P7.95/li
Cost of fuel per day = P42,070.45
60°F
80°F
=
=
SG 15 6"C
15.6°C
26.6°C
1415
= ---
131.5 + 30
0.876
= 0876[1 - 00007(2667 - 1556)]
= 0869
250gal/24hrs x 3.7851i/gal
39.431lilhr
39.431(0.869/0.876)
39.1161ilhr
2700/24
112.5 KW
39.l16xP3.00/li
Cost per KW-hr =
112.5
Cost per KW-hr = Pl.043/KW-h
SG I5 6"C
SG 26 6 "C
SG 2 6 6 , C
At 26.6°C. m, =
At 26.6°C. m. =
At 15.6°C, m. =
At 15.6°C,mr =
Load =
Load =
=
Fuels & Combustion - 13 (ME Bd. Oct. 1981)
A logging firm in Isabela operates a Diesel Electric Plant to supply its
electric energy requirements. During a 24 hour period, the plant
consumed 250 gallons of fuel at 80°F and produced 2700 KW-hrs.
Industrial fuel used is 30° API and was purchased at P3.00/li at 60°F.
Determine the overall thermal efficiency of the plant.
A. 26.08%
C 43.12%
B. 34.23%
D. 18.46%
SOLUTION:
Fuels & Combustion - 12 (ME Bd, Oct. 1981)
A logging firm in Isabela operates a Diesel Electric Plant to supply its
electric energy requirements. During a 24 bour period, tbe plant
consumed 250 gallons of fuel at 80°F and produced 2700 KW-hrs.
Industrial fuel used is 30° API and was purchased at P3.00m at 60°F.
Determine cost of fuel to produce one Kwb
A. P3.043/KW-h
C. P1.043/KW-h
B. P4.043/KW-h
D. n.043/KW-h
I
Qh
Qh
Qh
=
=
=
60°F
80°F
41,130 + 139.6 x °API
41,130 + 139.6(30)
45,3 18 KJ/kg
=
=
15.6°C
26.6°C
Fuels & Combustion
ISU
SG 1S6 o C
141.5
=
81
Fuels & Combustion
Mass ot Iuel
--
-r-
24.933 kg/hr
Qh ~, 41,1:10 + l396CO API)
Qh = 41,13 0' 139.6(2 8)
Qh = 45,039 KJlkg
131.5+ 30
SG 1W C = 0.876
SG 26 6 C = 0.876[1 - 0.0007( 26.67 - 15.56)]
SG 26 6 , C = 0.869
At 26.6°C, IDr = 250gaV 24hrs x 3.7851i/gal
At 26.6°C, IDr = 39.431l ilhr
Load == 2700/24
Load = 112.5 KW
IDr = 39.431l ilhr x 0.869 kg/li x Ihr/360 0sec
IDr = 0.00952 kg/sec
Power' Output
Overall Efficiency = ----- ---=0
Overall Efficien cy
Power Output
-----mrQ h
82.5
Overall efficien cy
Overall efficien cy
(24.933 ! 3600)(4 5.039)
26.47%
IDfQ h
Overall efficiency =
Overall efficiency
=
112.5
0.00952(45,318)
26.08%
Fuels & Combu stion -14 (ME Bd, Oct. 1981)
A diesel electric plant in one of the remote provinces in the South
utilizes diesel fuel with an °API of 28 at 15.6°C. The plant consum
es
680 liters of diesel fuel at 26.6°C in 24 hrs, while the power genera
ted
for the same period amoun ts to 1,980 KW-hr s. Determ ine overall
therma l efficiency of the plant
A. 26.47%
B. 12.34%
C. 23.45%
D. 34.34%
SOLUT ION:
SG 15 6 , C =
141.5
---
131.5+ 28
SG I5 .6 , C = 0.887
SG 26 6, C = 0.887[ 1 - 0.0007( 26.6 - 15.6)]
SG 26 6, C = 0.88
Density = 0.88 x 1 kg/Ii
Density = 0.88 kglli
Mass of fuel = 680 (0.88)/2 4
-----_._---
Fuels & Combu stion - 15 (ME Rd. Oct. 1991)
A circula r fuel tank 45 feet long and 5.5 feet diamet er
is used for oil
storage . Calcula te the numbe r of days the supply tank
can hold for
continu ous operati on at the followi ng conditi ons:
Stea m now = 2000 Ibs!h r
Steam dry and saturat ed at 200 psia
Fecdw ater temper ature = 230°F
Boiler Efficien cy = 75%
Fuel oil = 34° API
C 13.45
A. 1234
D. 2344
B 1758
SOLUT ION:
45ft
From steam tables:
At 200 psi( 1380 Mpa), h, =c 2789.6 KJ1<g
At 2JO°F( 1 10°C), h F = 461.3 KJ/kg
Qh
OJ,
c
In,
111, =
41,130- + 139.6(3 4)
45.876 KJlkg
2000/2 205
lJ07 kg/hr
1ll,(h s - h F )
'1l> = - - - - mfQIJ
rr-""" ""'''''
5.5ft
~
m,=2000 Iblh
,---
J2~Opsia
In,
•BOILER
, 75%
·
Y
+;
-'.
230°F
,1\
Yv~-
~
FURNAC E
82
Fuels & Combustion
Fuels & Combustion
907(2789.6 - 461.3)
0.75 = - - - - - - - ' m f(45,876)
me = 61.376 kg/hr
141.5
SG J5 .6 , C = - - 131.5+ 34
SG J5 6"C = 0.855
Density = 0.855(1000 kg/nr')
Density = 855 kg/m?
83
Fuels & Combustion - 17 (ME Bd. Apr. 1'987)
A steam generator burns fuel oil with 20% excess air. The fuel oil may
be represented by C 14 H lOo The fuel gas leave the preheater at 0.31
Mpa. Determine partial pressure of H10.
A. 23.34 kpa
B. 29.34 kpa
C. 35.7 kpa
D. 32.34 kpa
SOL,UTION:
Volume of tank
Volume of tank
=
=
1[/4 (5.5/3.28i)2(45/3.281)
30.297 m3 .
Total weight of fuel
Total weight offuel
= 30.297 m 3 x 855 kg/m"
= 25,904 kg
25,904
Number of Days = - - - 61.376(24)
Number of Days = 17.58 days
C 14 H30 + O2 + 3.76N2 - ; CO 2 + H20 + 3.76 N z
C I4H30 + 21.50z + 21.5(3.76)Nz -; 14 COz + 15HzO + 21.5(3.76)Nz
Combustion reaction with 20% excess air:
C 14H30 + 1.20(21.5)02 + 1.2(21.5X3.76)Nz -;
14 CO 2 + 15 HzO
+ 1.2(21.5)(3.76)N z + 0.20(21.5)Oz
A steam generator burns fuel oil with 20% excess air. The fuel oil may
be represented by C 14 HJ(Io The fuel gas leave the preheater at 0.31
Mpa. Determine the actual air-fuel ratio in kg air per kg fueL
Partial pressure of HzO
Partial pressure of H20
Partial pressure ofHzO
Fuel + Air
•
C 14H30 + O 2 + 3.76N2
=
=
=
(15/130.3 08)( 0.3 1)
0.0357 Mpa
35.7 kpa
C. 12.34
D. 19.45
Fuels & Combustion - 18 (ME Bd. Oct. 1985)
Product of combustion
- ; CO2 + H20 + 3.76 N 2
C J4H30 + 21.502 + 21.5(3.76)N 2 -; 14 C~ + 15H20 + 21.5(3.76)N2
..
21.5(32) + 21.5(3.76X28)
Thoeritical A/F =
14(12) + 30(1)
Thoeritical A/F = 14.91 kg air/kg fuel
Actual AIF = 14.91(1.2)
Actual AfF = 17.89 kg airikg fuel
t
Product of combustion
Total mols of product = 14 + 15 + 1.2(21.5X3.76) + 0.2(21.5)
Total mols of product = 130.308 mols
SOLUTION:
i
-;
Fuels & Combustion - 16 (ME Bd. Apr. 1987)
A. 17.89
B. 15.67
\
Fuel + Air
~xit temperature
above the dew point. Estimate the dew point temperature of the flue
gas produced by combustion having the gravimetric analysis of:
In the boiler design, it is desired to have the flue gas
N 2 = 71.84%
CO 2 = 20.35%
O 2 = 3.61%
H20 = 4.20010
A. 32°C
B. 23°C
C. 45°C
D. 39°C
84
Fuels & Combustion
Fuels & Combustion
85
SOU 'II< Ii'-..
Fuel and Combustion -20 (ME Rd. Apr. 1984)
Converting the given analysis to volumetric:
N2
0.7184/28
0.02565714
CO;
= 0.2035/44
0004625
O2
= 00361/32
= 000112812
H/)
= 0.042118
= 0.00233333
The dry exhaust gas from oil engine has the following gravimetric
analysis:
CO 2 = 21.6% O 2 = 4.2%
N 2 = 74.2%
Specific heats at constant pressure for each component of the exhaust
gas in Kcal/kg °C are:
CO 2 = 0.203
O 2 = 0.2]9
N 2 ;= 0.248
Calculate the gas constant in J/kg_°K.
A. 272
C. 274
C. 276
D. 278
-r-
Total mols of product
Total rnols or product
0.025657 +- 0.00462 -t- 0.00 I] 28 + 000233
003374359
Partial Pressure ofH,O ~. (0.0023333/0.0337436)(10\.325)
Pallial Pressure ofH·O= 7.006 Kpa
.
From steam table:
At 0.007006 Mpa, tsar -~ 39°C
Dew po mt temperature = 39 '-'C
SOLUT]ON
'.
Fuels & Combustion - ]9 (ME Bd. Oct. 1995)
CO 2
O2
N2
A flue gas has the following volumetric analysis:
CH 4 = 68%
C 2 H 6 = 32%
Assume a complete combustion with ]5% excess air at 101.325 Kpa,
what is the partial pressure of water vapor in Kpa.
A 15.95
C. 12.45
B.23.12
D.27.34
~
R
R
R
R
T
T
Considering 15% excess air:
I Fuel
«
1.15(24)0 2
1.15(2.48)(3.76)N 2 ~ ] 32C0 2 -t- 2.32H 20 "
1.I5(2.48)(376)N 2 +- 0.] 5(2.4)0 2
[
Total rnols of product
Total mots of product
=
=
Partial pressure of H 20
Partial pressure of H 20
1.32 + 2.32 i 2.48(3.76)(115) + 0.15(248)
14.735 mols
(2.321] 4.73 5)( 10 I.325 )
15.95 Epa
0.216/44
0042132
0.742/28
0.004909
0.00]3]2
~QI6500
=
=
=
=
8.3 ]4/M
8.3 ]4/30.56
0.27206 KJ/kg-OK
272.06 Jlkg- OK
Product of Combustion
1360, +- 136(376)N, ~0.68CO" + I 36H 2 0 i \.36(3.76)N)
(U::'C,!-fc,' 112Q~ +LI2(376Y~" ~ 0.6~_CO~ +- Q.96H 2Q+-1.12(3.76)N"
i lULL
2480, + 248(3.76)N, ~\.32CC}, +232H 20+2.48(376)N 2
(}(lSCfI.j
0.2\6
0042
0.742
0.03272] mols/kg-mol
Molecular Weight ~ 1/003272]
Molecular Weight = 30.56 kg/kg-mol
SOLUTION
Fuel +- Air
Converting the gravimetric analysis to volumetnc:
,r
Fuel and Combustion - 2] (ME Rd. Apr. 1984)
The dry exhaust gas from oil engine has the following gravimetric
analysis:
CO 2 = 21.6% O 2 = 4.2%
N 2 = 74.2%
Specific heats at constant pressure for each component of the exhaust
gas in Kcal/kg °C are:
CO, = 0.203
O 2 = 0.219
N 2 = 0.248
Calculate the specific gravity if the molecular weight of air is
28.97 kg/kg-mol
A. 0.98]
CLOSS
B ] .244
D. 0.542
86
Fuels & Combustion
Fuels & Combustion
SOLUTION:
A. 2.870
B. 3.120
87
C. 2.274
D. 6.233
Converting the gravimetric analysis to volumetric:
SOLUTION
CO,
O2
N,
0.216
0.042
0.742
0216/44
0042/32
0.742/28
0.004909
0.001312
0.026500
0.032721 mols/kg-rnol
Molecular Weight = 1/0032721
Molecular Weight = 3056 kg/kg-mol
SG = 30.56/2897
SG ~ 1.055
Thea. A/F
Thea. A/F
Thea AIF
11.5C + 34.5(H - 0/8) + 4.3S
11.5(0.715) + 34.5(0.05 - 0.07/8) + 4.3(0.036)
9.8 lb air/lb coal
O 2 in air by weight
=
232%
Therefore:
Theoretical weight of O 2
Theoretical weight of O,
0.232(9.8)
2.274 1Mb coal
Fuel and Com bustion - 22 (ME Bd. Apr. 1984)
The dry exhaust gas from oil engine has the following gravimetric
analysis:
CO 2 = 21.6% O 2 = 4.2%
Nz = 74.2%
Specific heats at constant pressure for each component of the exhaust
gas in Kcal/kg °C are:
CO 2 = 0.203
O 2 = 0.219
N2 = 0.248
Calculate the specific heat ofthe gas in KJ/kg_°K.
A. 0.872
C. 0.452
B. 0.992
D. 0673
=
=
=
There are 20 kg of flue gases formed per kg of fuel oil burned in the
combustion of a fuel oil C 12 H 26 • What is the excess air in percent?
A. 26.67%
C. 12.34%
B. 18.34%
D. 20.45%
SOLUTION:
C I , H 26 + 18.5 O 2 + 18.5(3.76)N 2 --+ 12 CO 2 + 13H 20 + 18.5(3.76)N z
SOLUTION:
cp
cp
cp
Fuel and Combustion - 24 (ME Bd. Oct. 1996)
0216( ruo]) +- 0.042(0.219) + 0.742(0.248)
0.237 K(;,.' ll~b-oC x 4.187
O. Q92 KJ/kg- 't:"
Thea. A/F
18.5 + 3.76( 18.5)
I
Thea. A/F
88.06 mol/mol
Thea. A/F in kg/kg
A bituminous coal has the following composition:
C = 71.5%
H = 5.0%
0 = 7.0%
N
S = 3.6°/;,
Ash = 8.2%
W = 3.4%
Determine the theoretical weight of Oxygen in lb/lb of coal
= 1.3%
-----
12(12)+26(1)
15 kg air/kg fuel
= 20 kg flue gas - I kg fuel gas
= 19 kg air
= 19 kg air/kg fuel
= Thea. A/F ( I + e)
+ e)
Thea. A/F in kg/kg
Fuel and Combustion - 23 (ME Bd. Oct. 1986)
88.06(28.97)
=
Mass of air
Mass of air
Actual A/F
Actual A/F
19 = 15( I
e = 0.2667
e = 26.67%
=
Fuel and Combustion ~ 25 (ME Bd. Apr. 1984)
A gaseous fuel mixture has a molal analysis:
H 2 = 14%
CRa =: 3%
CO = 2Tlfo
O 2 ~ 0.6%
CO2 = 4.5%
N2 = 50.9%
Determine the air~fuel ratio for complete combustion on molal basis.
A. 2.130
B. 3.230
C. 1.233
D. 1.130
89
Fuels & Combustion
Fuels & Combustion
88
Combustion reaction with 125% theoretical air:
7108C -+- 7050H z -+ 0.0165 -t- O.004N 2 ~. 1.25(l0649)02 +
l.25(l0649)(376)N?·c 7108C0 2 + 7.050H 20 + 0.0 16S0 2
+ 0.25(10.649)0 2
Ill! = total 1110ls in product
rn- = 7108 i 7.050 + 0.016 + 50.054 + 2.662
m- = 66.89 mols
Partial pressure of H2 0 = (7.050/66.890)( 170)
Partial pressure of H 20 = /7.92 Kpa
i-
50054N 2 .
SOLUTION:
Fuel and Combustion - 27
Chemical reaction with Oxygen:
0.14H2 + 0.0700 2 = 0.14H20
O.03C}-4 + 0.0600 2 = 0.03C02 + 0.06H20
0.27COz + 0.1350;\ = 0.27C02
0.265 O2
Actual O 2 in product = 0.265 O 2 - 0.006 O 2
Actual O2 in product = 0.259 Oz
0.259 + 0.259(3.76)
Molal AJF
1
Molal NF = 1.233 mots air/mol a/fuel
Calculate the theoretical air needed for the complete combustion of
ethane C 2 H 6 for 20 kg fuel.
C. 234.45 kg
A. 432.23 kg
D. 320.32 kg
B. 28745 kg
SOLUTION:
C 2 Hc, + O 2 + (3.76)N" -,) CO 2
-t-
I-hO -i- (3.76) N?
Balancing the equation:
C 2 Hr. " 350 2 -L 3.5(3.76)N z --~ 2C0 2 + 3HP + 3.5(3.76) N z
Fuel and Combustion - 26 (ME Bd, Apr. 1995)
A steam generator burns fuel oil that bas the foUowing chemical
analysis by mass in percent:
C = 85.3
H 2 = 14.1
S = 0.5
N 2 = 0.1
A. 19.85
B. 11.l4
C. 17.93
.,
(2xI2) + (lx6)
Theoretical A/F = 16.016 kg airlkg fuel
Mass of air needed ~ 16.016(20)
Mass of air needed = 320.32 kg
Fuel and Combustion - 28
Converting the given mass analysis to molal analysis:
Hz
S
Nz
3.5(32) + (3.5)(3.76)(28)
=
D. 14.20
SOLUTION:
C
Theoretical A/F
85.3/12
14.1/2
0.5/32
0.1/28
=
7.108
7.050
0.016
0.004
A fuel oil is burned with 50 percent excess air, and the combustion
characteristics of the fuel oil are similar to en H 2• • Determine the
volumetric (molal) analysis of CO 2 in the product of combustion.
A. 9.34°'u
C. 6.34%
B. 8.66'%
D. 7.45%,
l)O
Fuels & Combustion
Fuels & Combustion
SOLUTiON
13/12
83.5
c/b=376
83.5/b = 3.76
b
222
2d = (13/]2)(26)
d
]4.08
N) balance
Oo-N,ratlo
C"H i 6
"
0::' (376l N 2 ~ CO 2 +
n.o
t
91
(3.76) N 2
it
-
C
-c-
-r;
Balancing the equation:
Hi balance:
r-r-
C,l-l!(, + (I 85)0 2 + 18.5(3.76)N 2 ~ 12C0 2 + 13IhO + 18.5(3.76) N 2
Considering the 50% excess air,
Cloth" (15)(185)0 2 + (1.5)(185)(376)N 2 ~ :12C02 + 13H 20
. (1.5)(18.5)(37 11 ) N 2 ~ (0.5)( 185 )0 2
= total mols of product
rn- -z: 12 -t- 13 + 1.5(18.5)(3.76)
mT = 138.55 mols
% CO 2 = 12/138.55
% CO 2 = 0.0866
°lr, CO 2 - 8.66%
Divided the equation by a to determine the combustion equation for I mole
of fuel.
C. 11", + 20.500 ~ 7708N 2 '-c> I J .8C07 +- 0 I &eO + 3.230 2 T 7708N 2
~ 13H 2 0
(20.5)(32)
Actual A/F
011
T
(0.5)(18.5)
Actual AT
- - -
=
T
(77.08)(28)
._-- ---"---------
..
][] 2 ( J 2) + 26(1)]
16 5 kg air/kg fuel
The balance equation for 100 percent theoretical air is
C, Hi,~' 1850 2 + 18.5(3.76)N 2 ~ ]2CO, + 13H 20 + 18.5(3.76)N 2
Theoretical .'\/F- (18.5)(32) + (! 8.5)(376)(28)
Fuel and Combustion - 29
Theoretical A.T'
Fuel oil, C l 2 H 2o , is burned in air at atmospheric pressure. The Orsat
analysis of the products of combustion yields
: 12.8%
CO 2
: 3.5%
O2
CO
: 0.2%
: 83.5%
N2
Determine the percent excess air.
A. 12.34?/o
C. 10.52'}"
B. 8.34%
D. ! 8.45~o
SOLUTION
Applying the conservation of mass on each reactants:
aC t2 HOb
i
b0 2 + eN, -) 12.8C0 2 + 0.2eO + 83.5N 1 + dH 2 0
C balance:
12a
12.8
T
0.2
E~Cess
16.5 - J4.93
air
Excess air
1[12(12) + 26(I)J
]4.93 kg air/kg fuel
=
14,93
lO.52%
Fuel and Combustion - 30
An unknown fuel has the following Orsat analysis:
; 12.5%
CO 2
CO
: 0.3%
: 3.1 '%
O2
: 84.1 %
N2
Determine the actual air-fuel ratio in kg air per kg fuel.
A 17.13
C i923
B. 1234
D. 23.23
Fuels & Combustion
92
Fuels & Combustion
SOLUTION:
C.Ht. + C02 + dN 2
~
Theoretical O?/F ~ Weight of Oxygen
Weight of fuel
6.5(32)
Theoretical 0 21 F = - - - 12(4)+ 1(10)
Theoretical Oj/F = 3.586
12.5C02 + 0.3CO + 3.102 + 84.1N2 + eR20
C balance:
N 2 balance:
O 2 - N 2 ratio:
O 2 balance:
12.5 + 0.3
12.8
= 84.1
die = 3.76
84.llc = 3.76
c = 22.36
2236 = 12.5 + 0.3/2 + 3.1 + el2
e = 13.2
a
=
a
d
=
b
(22.36 + 84.1X28.97)
=
1[12(12.8) + 1(26.4)J
Actual AIF
Mass of O 2
Mass of O 2
=
3.586(1000)
= 3586 kg
Fuel and Combustion - 32
2e
=
b = 2(13.2)
b = 26.4
C l2 .s H26.4 + 22.3502 + 84.1N2 ~ 12.5C02 + O.3CO + 3.102 + 84.1 N 2 +
13.2H20
Actual AfF
93
=
17.13 kg airlkgfuel
A volumetric analysis of a gas mixture is as follows:
: 12%
CO 2
: 4%
O2
N2
: 82%
CO
: 2%
What is the percentage of CO on a mass basis?
A. 1.0%
C. 1.2%
B. 1.5%
D. 1.9%
Converting to mass basis:
CO 2
o?
Fuel and Combustion - 31
What mass of liquid oxygen is required to completely burned 1000 kg
of liquid butane, C 4 H 10• on a rocket ship?
A. 4568
B. 2746
C. 3586
D. 6345
N?
CO
~
0.12 x 44
= 5.28
0.04 x 32
= 1.28
~ 0.82 x 28
= 22.96
~ 0.02 x 28
0.56
Total mass of product = 5.28 + 1.28 + 22.96 + 0.56
Total mass of product = 30.08 kg
% mass of CO = 0.56130.08
% mass of Ct) = 1.9%
=
-c-
SOLUTION:
Fuels and Combustion - 33 (ME Rd. Apr. 1998)
C 4 RIO + O 2 + (3.76)N2
~
CO 2 +J{20 + (3.76) N 2
Balancing the equation:
C 4 RIO + 6.502 + 6.5(3.76)N2 ~ 4C02 + 5H20 + 6.5(3.76) N 2
What is the percent theoretical air for a combustion process to which
the fuel and combustion gas analysis are known as follows:
Fuel:
% by volume
CO 2
: 12.4%
CO
: 27%
VAR IABL E LOA D PRO BLE MS
Fuels & Combu stion
94
11.,
N-~
Combu stion ga~
CO z
Oz
Hz
: 2.2%
: 58.4%
% by volume
24.6%
: 1.0%
: 74.4%
A. III
B. 121
Variab le Load Problem s - ]
a load
A 50 MW fL"er plant has an averag e load of31,50 0 KW and
factor of 70%. Find the reserve over peak.
C. 5 MW
A. 4 MW
D.6M W
B. 3 MW
C 116
D. 126
SOLUT ION
SOLUT ION:
Combu stion reaction with theoreti cal air:
Ave. Load
Load Factor = - - - Peak Load
31,500
O. 70 C~
Peak Load
Peax Load = 45,000 Kw
Peak Load = 45 Mw
Reserve over peak = 50 - 45
Reserve over peak = 5 Mw
N;o
() 124CO: + I) 27CO + 0022H : -r o 584N 2 + 0.1460" O. J 46(3.76)
76)N?
O.146(3
;584N
2
--> C:i94CO o 0022H 2 0 ~ o
T
Combu stion with excess air:
1460,
0.] 24CO;o + onCO t 0022H 2 + o 584N 2 + (I +x)O
(I ix)OI4 6(176) N 2 --> o 394CO , -t 0.022H 20 + o 584N 2 +
0146(3 .76)(1 +-x)N 2 + x(O 14\))0,
~
water:
Express ing the percent age of oxygen in the product s excludi ng the
Variab le Load Problem s - 2
0.01
x = 0.11
1 + 0.11
Percent age Theoret ical air
Percent age Theoret ical air = III %
0=
KWThe daily energy produc ed in a certain power plant is 480,000
load?
e
hrs, What is the daily averag
A. 10 MW
C. 25 MW
B. 15 MW
D. 20 MW
SOLUT ION:
Averag e Load
Energy Pr oduced
A verage Load
No. of hours
480,000 /24
A verage Load
20,000 Kw
A verage Load
20Mw
~ 'ariable
96
Load Problems
STEAM CYCLE
Variable Load Problems - 3
The annual energy produced in a 100 MW power plant is 438,000,000
KW-hrs. What is the annual capacity factor of the plant?
A. 40%
C. 35~'o
B. 50%
D. 60%
SOLUTION:
Annual Energy Pr oduced
Annual Capacity Factor
=
Annual Capacity Factor
=
Annual Capacity Factor
~
Plant Capacity x 8760
438,000,000
100,000x 8760
50%
Steam cycle - 1 (ME Bd Oct. 1999)
In a Rankine cycle steam enters the turbine at 2.5 Mpa (enthalpies &
entropies given) and condenser of 50 Kpa (properties given), what is
the thermal efficiency of the cycle?
At 2.5 Mpa: h g = 2803.1 KJ/kg Sg = 6.2575
At 50 kpa:
Sr = 1.0910
Srg = 6.5029
h, = 340.49
h rg = 2305.4
Vr= 0.0010300
A. 25.5~~~~
C. 34.23%
B. 45.23%
D. 12.34%
SOLUTION:
s = Sr + x Srg
6.2575 = 1.0910 + x(6.5029)
x = 0.7945
h z = h, + xh rg
h z = 340.49 + 0.7945(2305.4)
L-·---------hz = 2172.13
h, = 340.49 KJ/kg
h, = h r + vr(P z - PI)
h, = 340.49 + 000 I 03(2500 - 50)
h, = 342.98
A power plant has a use factor of 50% and capacity factor of 44%.
How many hours did it operate during the year?
A. 7700 hrs
C. 7709 hrs
B. 7800 hrs
D. 7805 hrs
s""·
SOLUTION:
Annual Energy Pr oduced
Annual Capacity Factor
Plant Capacity x 8760
Annual Enerzv Pr oduced
~-
=
Plant Capacity x 8760
Energy Produced = 3854.4(Plant Capacity)
Use Factor
0.50
T
h, = 2803.1 KJ/kg
Solving for hz:
Variable Load Problems - 4
0.44
97
Steam Cycles
Efficiency
Efficiency =
Energy Pr oduced
=
=
(hi -h z)-(h 4 -h,)
-------
(h j - h 4 )
(2803.1 - 2172.11)- (342.98 - 340.49)
(2803.1 - 342.98)
-----'=-------
Plant Capacity x t
Energy Pr oduced
Efficiency = 25.55%
~
Plant Capacity x t
Energy Produced = 050(Plant Capacity)t
0.50(Plant Capacity)t = 3854.4(Plant Capacity)
t = 7708.8 hrs.
Steam cycle - 2
,
In an ideal Rankine cycle, the steam throttle condition is 4.10 Mpa and
440°C. If turbine exhaust is 0.105 Mpa, determine the pump work in
KJ per kg.
98
Steam Cycles
A. 6.34
B. 5.34
C. 4.17
D. 2.12
SOLUTION:
Solving for h.:
h, =h fatO.105Mpa
h- = 423.24 KJ/kg
V3 = 0.0010443 mvkg
Solving for h.:
Using pump work equation:
h,
V3(P4 - P3) + h 3.
3/kg
V3 = 00010443 m
h, = 0.0010443(4100 - lOS) + 423.24
h, ~ 427.412 KJ/kg
W p = h4 - h3
W p = 427.412 - 423.24
W p = 4.172 KJ/kg
-c;
Steam cycle - 3
A thermal power plant generates 5 MW has also 300 KW power
needed for auxiliaries. If the heat generated by fuel is 13,000 KJ/sec,
determine the net thermal efficiency.
A. 35.78%
C. 30.56%
B. 36.15%
D. 3367%
SOLUTION:
5,000 - 300
11 net
Steam Cycles
SOLUTION:
Solving for h.:
At 410 Mpa and 440°C (Table 3)
hi = 3305.7 KJ/kg
s\ = 6.8911 KK/kg-OK
Saving for h 2:
At 0.105 Mpa(Table 2)
sf=1.3181
h r=423.24
Sfg = 6.0249
h rg = 2254.4
s I = S2 = Sr + XSfg
6.8911 = 1.3 I 81 + x(6.0249)
x = 0.925
h2
=
11 net = 36.15%
Steam cycle - 4
In an ideal Rankine cycle, the steam throttle condition is 4.10 Mpa and
440°C. If turbine exhaust is 0.105 Mpa, determine the thermal
efficiency of the cycle.
A. 20.34%
C. 34.44%
B. 27.55%
D. 43.12%
CD
@
G»)
•
1
h, + xh.,
h 2 = 423.24 + 0.925(2254.4)
h 2 = 2508.54 KJ/kg
Solving for h.;
h, = h-at 0.105 Mpa
h, = 423.24 KJ/kg
Solving for h.:
Using pump work equation:
h, = Vi(P 4 - P 3 ) + h,
3/kg
V3 = 0.0010443 m
h4.= 0.0010443(4100 - 105) + 423.24
h, = 427.412KJ/kg
QA = h. - h,
QA = 3305.7 - 427.412
QA = 2878.29 KJ/kg
=
13,000
99
WT =
Wr =
WT =
Wp =
Wp =
w, =
hi - h2
3305.7 - 2508.54
797.16 KJ/kg
h, - h,
427.412 - 423.24
4.172 KJ/kg
W ne t = W T - W p
W net = 797..16 - 4.172
W n e t = 79299 KJ/kg
11, = W nc,lQA
11, = 792.99/2878.29
11t = 27.55%
s
100
101
Steam Cycles
Steam Cycles
Steam cycle - 6 (ME Bd. Oct. 1989)
1--
Steam cycle - 5 (ME Bd. Oct. 1991)
In a Rankine cycle, saturated liquid water at 1 bar is compressed
isentropically to 150 bar. Fir..t by heating in a boiler, and then by
superheating at constant pressure of 150 bar. the water substance is
brought to 750°K. After adiabatic reversible expansion in a turbine to
1 bar, it is then cooled in a condenset to saturated liquid. What is the
thermal efficiency of the cycle (%)?
A. 23.45%
C. 34.24%
D. '18.23%
B. 16.23%
SOLUTION:
SOLUTION:
CD
At 150 bar(l5 Mpa) and 750 0 K ( 4 7 7 ° C )
h, = 3240.5 KJ/kg
SI = 6.2549 KJ/kg-OK
At I bar(O.IO Mpa)
Sf = 1.3026
Sfg = 6.0568
vr = 0.001043
A steam generating plallt has two 20 MW turbo-generators. Steam is
supplied at 1.7 Mpa and 320°C. Exhaust is at 0.006 Mpa. Daily
average load factor is 80%. The steam generating units operate at
70% efficiency when using bunker fuel having a heating value of
31,150 'KJlkg and an average steam rate of 5 kg steam/K'W-hr.
Calculate the Mtons of fuel oilfbunker fuel required per 24 hours.
A. 515
C. 6.17
B. 432
D. 762
Load Factor
Peak Load
Av/;;. Load
= S2 = Sf + xSsg
6.2549 = 1.3026 + x(6.0568)
x = 0.8176
x = 81.76%
hz = 417.46+0.8176(2258)
hz = 2263.6 KJ/kg
14 = V3(P 4 - P 3) + h3
14 = 0.0010432(15,000-100)+417.46
14 = 433 KJ/kg
Wp = 14 - h,
Wp = 433 -417.46
Wp = 15.54 KJ/kg
Wr = h, - h z
Wr = 3240.5 - 2263.6
Wr = 976.9 KJ/kg
SI
Efficiency
976.9 - 15.54
=
=
I
~
0.8 = - - - 20,000 x 2
Ave. Load = 32,000 KW m,
h f = 417.46
h fg = 2258.0
Efficiency
Ave. Load
=
(3240.5 - 433)
34.24%
s
I
I
®
@
From Steam Tables:
PM
.J
h, = 3077 KJ/kg
Pw
h, =, 151.53 KJ/kg
3/kg
V3 = '0.0010064 m
Solving for 14:
14 = V3(P 4 - P 3) + h)
h, = 0.001 0064( 1700 - 6) + 151.53
h, = 153.23 KJ/kg
m, = 5(32,000)
m, = 160,000 kg/hr
m s (h l - h 4 )
llb =
mfQ h
160,000(3077 - 153.23)
0.70 = ------'------'m f(31,150)
m, =.21,454 kg/hr
21,454(24)
Fuel needed for 24 hours operation =
1000
Fuel needed for 24 hours operation = 514.9 Mtons
102
Steam cycle - 7 (ME Bd. Oct. 1994)
A back pressure steam turbine of 100,000 KW serves as a prime
mover in a cogeneration system. The boiler admits the return water at
a temperature of 66°C and produces the steam at 6.5 Mpa and 455°C.
Steam then enters a back pressure turbine and expands to the
pressure of the process, which is 0.52 Mpa. Assuming a boiler
efficiency of 80% and neglecting the effect of pumping and the
pressure drops at various location, what is the incremental heat rate
for electric?
The following enthalpies have been found; turbine entrance = 3306.8
KJ/kg, exit = 2700.8; boiler entrance = 276.23 KJ/kg, exit = 3306.8.
A. 21,504 KJIKW-hr
C. 23,504 KJIKW-hr
B. 22,504 KJIKW-hr
D.24,504 KJIKW-hr
WT = turbine work
WT = m(h, - h2)
WT = m(3306.8 - 2700.8)
WT 606 m KW
QA
(m x 3600)(h) - h 4)
(m x 3600)(3306.8 -
...........
L.71.J0.~)
m
•
,
0.8
13,637,565m KJ/hr
Heat rate
Heat rate
13,637,565m
=
=
SOLUTION:
®
Net Output = 1000 - 0.09( 1000)
Net Output = 910 MW
Net Output = 910,000 KW
Heat generated = m, Qh
9800(907)
Heat generated =
Qh
m
I
(6,388.9 x 4.187)
pw
J
24 x 3600
Heat generated = 2,752,001 KW
Steam Cycle - 9 (ME Bd. Oct. 1995)
llbo
QA
value of 6,388.9 Kcal/kg and the steam generator efficiency is 86%.
What is the net station efficiency of the plant in percent?
A. 30%
C. 33%
B. 25%
D.38%
Station efficiency = Net output/Heat input
Station efficiency = 910,000/2,755.00 I
Station efficiency = 33.07%
SOLUTION:
QA
103
Steam Cycles
Steam Cycles
~l
I
•
•
606m
m
22,504 KJIKW-h,
Steam Cycle - 8 (ME Bd. Oct. 1994)
A coal-fired power plant has a turbine-generator rated at 1000 MW
gross. The plant required about 9% of this power for its internal
operations. It uses 9800 tons of coal per day. The coal has a heating
A superheat steam Rankine cycle has turbine inlet conditions of 17.5
Mpa and 530°C expands in a turbine to 0.007 Mpa. The turbine and
pump polytropic efficiencies are 0.9 and 0.7 respectively, pressure
losses between pump and turbine inlet are 1.5 Mpa. What should be
the pump work in KJ/kg?
C. 37.3
A. 17.3
D. 47.3
B. 27.3
SOLUTION:
»;>
V 3 (P4
- P3)
IIp
where: Using density of water = 1000 kg/rrr'
V3 = 1/1000
3/kg
V3 = 0.001 m
P4=17.5+1.5
P4 = 19 Mpa
P4 = 19,000 Kpa
P3 = 0.007 Mpa
104
Steam Cycles
P3
IIp
=
7 Kpa
=
0.70
0.001(19,000 - 7)
QA
Steam properties:
At 1.70 Mpaand 370°C: h = 3187.1 KJ/kg S = 7.1081
At 0.17 Mpa:
h f = 483.20
Sf = 1.4752
hfg = 2216.0
Sfg = 5.7062
At 65.soC: h, = 274.14
C. 91.24
D. 69
SOLUTION:
h,
=
3187.1 KJ/kg
Solving for h2 :
SI
=
S2
=
Sr
h) - h 4
3187.1- 274.14
QA
=' - - - - -
QA
=
0.8
3641.2 KJfkg
Cogeneration efficiency
:.;- Steam Cycle - 10 (ME Bd, Oct. 1995)
A. 78
B. 102.10
=
T]b
W = -----p
0.70
Wp = 27.1 KJ/kg
A steam plant operates with initial pressure of 1.70 Mpa and 370°C
temperature and exhaust to a heating system at 0.17 Mpa. The
condensate from the heating system is returned to the boiler at 65.5°C
and the heating system utilizes from its intended purpose 90% of the
energy transferred from the steam it receives. The T]T is 70%. If boiler
efficiency is 80%, what is the cogeneration efficiency of the system in
percent. Neglect pump work.
+ XSrg
7.1081 = 1.4752+x(5.7062)
x .~ 0.9871
h z = h, + xh rg
hz = 483.20 + 0.9871(2216)
hz = 2670.60 KJ/kg
h, = h, = 274.14 KJ/kg
WT = (h, - hz)lh
W 1 = (3187.1-2670.60)(0.70)
WT = 361.55 KJ/kg
OR = 0.90(hi - h 3)
QR = 0.90(2670.6 - 274.14)
QR = 2156.81 KJ/kg
105
Steam Cycles
=
QT+Q R
QA
361.55 + 2156.81
Cogeneration efficiency
=
Cogeneration efficiency
=
3641.2
69.16%
Steam Cycle - 11 (ME Bd. Apr. 1996)
In a cogeneration plant, steam enters the turbine at 4 Mpa and 400°C.
One fourth of the steam is extracted from the turbine at 600 Kpa
pressure for process heating. The remaining steam continues to
expand to 10 Kpa, The extracted steam is then condensed and mixed
with feed water are constant pressure and the mixture is pumped to the
boiler pressure of 4 Mpa. The mass flow rate of steam through the
boiler is 30 kg/sec. Disregarding any pressure drops and heat losses in
the piping, and assuming the turbine and pump to be isentropic, how
much process heat is required in KW?
Steam properties:
At 4 Mpa and 400°C: h = 3213.6 KJlkg, s = 6.7690
At 600 Kpa: hr = 670.56
Sf = 1.9312
h rg = 2086.3 Sfg = 4.8288
A. 15,646.8
C. 1.9312
B. 2,468.2
D. 1,027.9
SOLUTION:
®
52 = St· + xSrg
6.7690 = 1.9312 +x(4.8288) .-.
..
•
1I J.
®
x = 1.00(saturated vapv"
- ~
51
hz
hz
=
=
=
h r+ xh rg
670.56 -I. 1.00(2086.3)
6bj'i,Ij.;1
®
p)..
•
y
Steam Cycles
106
Steam Cycles
h2 == 275/,)."7
h, == h-at 600 Kj .
h, == 670.56
Q == m p (h, - h 3 )
Q == (30/4)(2756.9 - 670.56)
Q == 15,647.5 KW
Steam Cycle - 12
In an ideal Reheat cycle, the steam throttled condition is 8 Mpa and
480°C. The steam is then reheated to 2 Mpa and 460°C. If turbine
exhaust is 60°C, determine cycle efficiency.
A. 38.3%
C. 34.3%
B. 24.3%
D. 45.2%
SOLUTION:
At 8 Mpa and 485°C(Table 3)
hi == 3348.4 KJ/kg
51 == 6.6586KJ/kg-OK
8ft1Pa
4l10°C<D@
2MPa
- ' I ' - 4 l1
At 2 Mpa(Table 3),
(51
==
53
l
60~C
==
54
11 5 ~c 251.13 KJ/kg
vr == 0.0010172 m3lkg
h, == V5(P6 - P 5) + hs
h, == 0.0010172(8000-19.94)+25I.13
h, == ·259.25 KJ/kg
QA == (hi - h 6) + (h, - h2)
QA =' (3348.4 - 259.25) + (3379.5 - 2963.145)
QA == 3505.501 KJ/kg
WT · == (hi - h 2) + (h, - ~)
WT == (3348.4-2963.145)+(3379.5-2411.41)
WT == 1353.345 KJ/kg
QR == h, - h,
2.8MPa
QR == 2411.41 - 25I.13
540°C
QR == 2160.28 KJ/kg
W p == h, - h,
w, =259.25-25I.13
W p == 8.12KJlkg
W net = W T - Wp
W net == 1353.345 - 8.12
W net == 1345.225 KJ/kg
Tj == Wne/QA
Tj == 1345.225/3505.505
®
Tj == 38.37%
52)
6.6388
2952.3
6.6586
h2
6.6828
2976.4
By interpolation, h 2 == 2963.145 KJ/kg
At 2 Mpa and 460°C (53 == 54)
h 3 == 3379.5 KJ/kg
53 == 7.3147
At
460°C
(19.94 kpa) -Table 1
5r == 0.8312
h r == 25I.13
5fg == 7.0784
h fg == 2358.5
== 5r + X5rg
7.3147 == 0.8312+x(7.0784)
x == 0.916
h, == 25I.13 + 0.916(2358.5)
h, == 2411.41 KJ/kg
h, == h, at 60°C
107
P)Oll
®
•
y
5,=5,
Steam Cycle - 13 (ME Bd. Apr. 1991)
5
A reheat steam cycle has 13,850 kpa throttle pressure at the turbine
inlet and a 2800 Kpa reheat pressure, the throttle and reheat
temperature of the steam is 540°C, condenser pressure is 3.4 Kpa,
engine efficiency .of high and low pressure is 75% find the cycle
thermal efficiency
SOLUTION:
At 13.85 Mpa and 540°C,
hi == 3434 ..1 KJ/kg(interpolated)
51 == 6.53553 KJlkg-OK(interpolated)
At 2.8 Mpa(Table 3) and 51 == 52 == 6.53553
h 2 == 2974.914 KJ/kg
At 2.8 Mpa and 540°C,
108.
Steam Cycles
h, = 3548.5 KJlkg
5] = 7.3810 KJlkg-OK
At 0.0034 Mpa:
hr = 109.84
5f = 0.384
Sfg ~~ 8.1488
hlg = 2439.5
v; = 0.0010032
53 = 54 = Sr + XSrg
7.3810 = 0.384 + x(8.1488)
x = 0.8586
14 = 109.84 + 0.8586(2439.5)
h, = 2204.426 KJlkg
s
h s = h, at 0.0034 Mpa
h, = 109.84 KJlkg
h, = vs(P 6 - Ps) + h,
h, = 0.0010032(13850 - 3.4) + 109.84
h, = 123.73 KJlkg
Considering the engine efficiency:
W T = (h, - h 2 )T] stl + (h, - 14)T]st2
WT = (3434.1 - 2974.9)(0.75) + (3548.5 - 2204.5)(0.75)
WT = 1,352.4 KJlkg
Wp = h, - h,
Wp = 123.73 - 109.84
Wp = 13.89 KJlkg
QA = (hi - ~) + (h, - h2)
QA = (3434.1 - 123.73) + (3548.5 - 2974.9)
QA = 3,883.97 KJlkg
WOe l = 1352.4 - 13.89
Wnet = 1338.51 KJlkg
Efficiency = 1338.51/3883.97
Efficiency = 34.46%
Steam Cycle - 14
Steam is delivered to turbine at 5.4 Mpa and 600°C. Before
condensation at 31°C, steam is extracted for feedwater heating at 0.6
Mpa. For an ideal regenerative cycle, find the thermal efficiency.
A. 23.45%
C. 28.34%
B. 34.34%
D. 44.14%
SOLUTION:
At 5.4 Mpa and 600°C(Tab1e 3)
h, = 3663.3 KJlkg
Steam Cycles
109
51 = 7.2206 KJlkg-OK
At 0.60 Mpa and s, = s- (Table 3)
SI
=
52
7.1816 _ _.2957.2
7.2206 _ _ h2
7.2214
2978.2
h2 = 2977.78 KJlkg
At 31°C (Table I)
Sf = 0.4507
hr = 129.97
Srg = 7.9822
hrg = 2428.1
Vr == 0.0010046
SI = S2 = S3 = Sf + XSfg
7.2206 = 0.4507 + x(7.9822)
x = 0.848
h, = 129.97 + 0.848(242~.l)
h, == 2189.30 KJlkg
14 == hr at 31°C
14 = 129.97 KJlk 6
h, == V4(PS -P 4) + 14
P, = Psat at 31°C
P, == 0.004496 Mpa
P, = 4.496 Kpa
h, = 0.0010046(600 - 4.496) + 129.97
h, == 130.56 KJlkg
hs == hrat 0.6 Mpa
h, == 670.56 KJlkg
V6 == 0.0011006 m3lkg
h- = V6(P 7 -P 6) + h6
h7 = 0.0011006(5400 - 600) + 670.56
h- = 675.84 KJlkg
By heat balance in Regenerative heater
mh, + (l-mjh, = 1~
h 6 - hs
m=
h2 - hs
67056 - 130.56
m
2977.78-13056
m == 0.1896
QA == 1(h1 - h-)
QA == 1(3663.3 - 675.84)
s
QA == 2987.46 KJlkg
WT = 1(h1 - h2) + (I-m)(h2 - h3)
WT == 1(3663.3 - 2977.78) + (I - 0.1896)(2977.78 - 2189.30)
e
110
Steam Cycles
WT = 1324.504 KJ/kg
QR = (1 - m)(h 3 -14)
QR = (1- 0.1896)(2189.30 - 129.97)
QR = 1668.88 KJ/kg
W PT = W P J + Wrz
W PT = (1 - m)(h s -14) + l(h? - 11,;)
W PT = (1 - 0.1896)(130.56 - 129.97) + 1(675.84 - 670.56)
W PT = 5.758 KJ/kg
W Oe l = W T - W TP
W Oe l = 1324.504 - 5.758
W Oe l = 1318.746 KJ/kg
TJ = Woell QA
TJ = 1318.746/2987.46
TJ = 44.14%
III
Steam Cycles
Calculate the mass flow rate of subcooled liquid if steam flow rate is
0.865 kg/s.
Steam Properties are:
At 7 bar, saturated vapor:
h g = 2763.5 KJ/kg
At 7 bar and 25°C:
h r = 105.5 KJ/kg
At 7 bar, saturated liquid:
h, = 697.22 KJ/kg
7 bar
A. 2.725
C. 2.286
m,=O.865kg/s
B. 3.356
D. 3.948
(Sat. liquid)
SOLUTION:
m,
Efficiency
=
Efficiency
=
•
•
HEATER
7 bar
Heat Absorbed
Heat Supplied
Steam Cycle - 15 (ME Rd. Apr. 1998)
A steam condenser receives 10 kg/s of steam with an enthalpy of 2570
KJ/kg. Steam condenses into a liquid and leaves with an enthalpy of
160 KJ/kg.
Cooling water passes through the condenser with
temperature increases from 13 degrees C to 24°C. Calculate the water
flow rate in kg/s.
A. 533
C. 523
D. 528
B. 518
i
m L (h 3 - h 2 )
m (h. - h
s
)
3
m L (697.22 -105.5)
0.90 = --"'--------0.865(2763.5 - 697.22)
mL = 2.725 kg/s
SOLUTION:
By heat balance in the condenser:
13°C
m=10kg/s
_
24°C
Heat rejected by steam = Heat absorbed by water
m s (hI - h 2) = mw c p (t 2 - t.)
10(2570 - 160) = m w (4.187)(24 - 13)
m., = 523.2 kg/s
Steam Cycle - 16 (ME Rd. Apr. 1998)
In an open feedwater heater for a steam plant, saturated steam at 7
bar is mixed with subcooled liquid at 7 bar and 25°C. Just enough
steam is supplied to ensure that the mixed steam leaving the heater
will be saturated liquid at 7 bar when heater efficiency is 90%.
Steam Cycle - 17 (ME Rd. Oct 1997)
Steam expands adiabatically in a turbine from 2000 kpa, 400°C to 400
kpa, 250°C. What is the effectiveness of the process in percent
assuming an atmospheric pressure uf 15°C. Neglect changes in kinetic
and potential energy.
Steam Properties are:
At 2000 Kpa and 400°C.
h = 3247.6 KJ/kg
s = 7.1271 KJ/kg-K
At 400 Kpa and 250°C.
h = 2964.2 KJ/kg
s = 7.3789 KJ/kg-K
A. 82
C. 80
B. 84
D. 86
7 bar,
112
Steam Cycles
113
Steam Cycles
SOLUTION:
Q =
Q =
Q =
Qs =
Qs =
Qs =
2000Kpa
400°C
(h, - h 2 )
3247.6 - 2964.2
283.4 KJlkg
T (S2 - s.)
('" + 273)(7.3789 - 7.127
400Kpa
250°C
72.5 KJlkg
283.4
Effectiveness
=
Effectiveness
=
283.4 + 72.5
79.60%
s
Steam Cycle - 18 (ME Rd. Oct. 1997)
Steam enters the superheater of a boiler at a pressure of 25 bar and
dryness of 0.98 and leaves at the same pressure at a temperature of
370°C. Calculate the heat energy supplied per kg of steam supplied in
the superheater.
Steam Properties:
At 25 bar and 370°C:
h = 3]71.8 KJIkg
At 25 bar
hf = 962.11 KJIkg
hfg = 1841.0 KJ/kg
A. 407.46
C. 405.51
B. 408.57
D. 406.54
370°C
SOLUTION:
h
=
Steam enters the turbine of a cogeneration plant at 7.0 Mpa and
500°C. Steam at a flow rate of7.6 kg/s is extracted from the turbine at
600 Kpa pressure for process heating.• The remaining steam continues
to expand to 10 kpa. The recovered condensates are pumped back to
the boiler. The mass flow rate of steam that enters the turbine is 30
kg/so Calculate the cogeneration efficiency in percent.
Steam properties:
At 7.0 Mpa and 500°C:
h = 3410.3
7Mpa
s = 6.7976
500°C
At 600 Kpa:
CD
h r = 670.56
m.=30kg/s
h rg = 2086.3
Sf = 1.9312
Srg = 4.8288
At 10 Kpa:
h, = 191.83
Q)
h rg = 2392.8
Sr = 0.6493
s
Srg = 7.5009
A. 60
C. 65
D. 5'5
B. 50
e
SOLUTION:
WT
J 25~ar
hr + xhrg
Steam Cycle - 19 (ME Rd. Oct. 1997)
51
=
mj(h l
= 52 = 5r+
6.797
=
-
h 2) + mih 2 - h,)
X5rg
1.9312 +
X2
(4.8288)
= 1.0
h2 = hr + X2 hig
X2
hi = 962.11 + 0.98(1841.0)
BOILER
Q
h, = 2766.3 KJlkg
Q
=
h2 - h,
Q
=
3171.8 - 2766.3
Q
=
405.5 KJ/kg
25bar.
(x=98%)
..
h- =.670.56 + 1.0(2086.3)
h 2 = 2756.86 KJ/kg
Sj = 52 = :>3 = Sf + X3 Sig
6.7976 = 0.6493 + x3(7.5009)
x3=0.8196
h, = hfJ + xh fg
h, = 191.83 + 0.8196(2392.8)
h, = 2152.96 KJlkg
WT = 30(3410.3 - 2756.89) + (30 - 7.6)(2756.86 ·2152.96)
WT = 33,129.66 kw
114
Steam Cycles
Steam Cycles
QR = (rn. - m2l (h, - 14)
h, = hc at 10 Kpa
14 = 191.83 KJ/kg
QR = (30 - 7.6)(2152.96 - 191.83)
QR = 43,929.312 Kw
QA = ml(h l-hc4)
QA = 30(3410.3 - 191.83)
QA = 96,554.1 Kw
33,129.66+43,929.312
Cogeneration efficiency = - - - - - - - 96,554.1
Cogeneration efficiency = 79.81%
(No exact answer in the choices)
115
Steam Cycle - 21
Pump work of Rankine cycle is 15 KJ/kg. Density of water entering
the pump is 958 kg/rn", If condenser pressure is 100 Kpa, what is the
pressure at the entrance of tlie turbine?
A. 14.47 Mpa
C. 15.67 Mpa
B. 20.48 Mpa
D. 17.77 Mpa
SOLUTION:
W p = v(P l - P 4 )
W p = (l/w)(P 1 - P4 )
15 = (I/958)(P[ - 100)
P = 14,470 Kpa
P = 14.47 Mpa
Steam Cycle - 20 (ME Bd. Oct. 1997)
A heat exchanger was installed purposely to COOl U.50 kg of gas per
second. Molecular weight is 28 and k = 1.32. The gas is cooled from
150°C to 80°e. Water is available at the rate of 0.30 kgls and at a
temperature of rz-c, Calculate the exit temperature of the water in
"C.
A. 48
C. 46
B. 42
D. 44
SOLUTION:
R
R
=
=
r~
8.314/28
0.2969 KJ/kg-K
kR
cp
=
cp
=
k-I
1.32(0.2969)
1.32- 1
cp = 1.2247 KJ/kg-K
11t =
180°C
I
•
~
t,,=12 c
t
~
1000-13
2800
11t = 35.25%
HEAT
t:XCHANGE
Q
By heat balance:
In a Rankine cycle the turbine work is 1,000 KJ/kg and pump work of
13 KJ/kg. If heat generated by generator is 2800 KJ/kg, what is the
efficiency of the cycle?
A. 35.25%
C. 38.65%
D. 30.25%
B. 40.75%
SOLUTION:
14.
::l
Steam Cycle - 22
1SOoC
Steam Cycle - 23
m w=O.3kgls
Qgain = Qloss
m; cp (t, - to) = fig cpg (t2 - t l )
(0.30)(4. 187)(tb - 12) = (0.5)( 1.2247)(150 - 80)
t b = 46.125"C
In a Reheat power plant the difference in enthalpy at the entrance and
exit is 550 KJ/kg for first stage and second stage is 750 KJ/kg. If both
stages has an efficiency of92% and heat added to boiler is 3,000
KJ/kg. Determine the plant cycle efficiency neglecting the pump work.
A. 30%
C. 40%
B. 35%
D. 45%
water required because of high pressure exit velocity if the steam flow
rate is 38 kg/s and the cooling water temperature rise is iz'c with an
inlet condition of 30°C.
SOLUTION:
llT
=
QA
llT =
llT
=
SOLUTION:
3000
39.87%
Steam Cycle - 24
An adiabatic feed pump in a steam cycle delivers water to a steam
generator at a temperature of 200°C and a pressure of 10 Mpa. The
water enters the pump as a saturated liquid at 180°C. If the power
supplied to the pump is 75 kw, determine the mass flow rate.
A. 6.23 kg/s
B. 8.34 kg/s
C. 7.39 kg/s
D. 9.12 kg/s
p...
200"C ~ 180"C
~.
=
=
10 Mpa
P sat
h2
h2
h2
=
=
=
763.22 KJ/kg
3
0.0011274 m /kg
= 1.0021 Mpa
If the exit velocity is not considered:
At 15 Kpa:
h, = 225.94 KJ/kg
h fg = 2373.1 KJ/kg
hi = h f + X h fg
h, = 225.9 + 0.9(2373.1)
hi = 2361.73 KJ/kg
h2 = h, at 15 kpa
h2 = 225.94 KJ/kg
Qgain =
38 kg1s
15Kpa (90%quality)
30°C
h,
m.
Qloss
m; (4.187)(12) = 38(2361.73 - 225.94)
m; = 1615.38 kg/s
SOLUTION:
vr
C. 13.23 kg/s
D. 21.78 kg/s
A. 11.23 kg/s
B. 17.23 kg/s
W n + Wn
0.92(550) + 0.92(750)
At 180°C:
h,
117
Steam Cycles
Steam Cycles
116
75 Kw
If exit velocity is considered:
(1)(240)2
h, = 2361.73 + [(1/ 2)
]
1000
h, = 2390.53 KJ/kg
IDw (4.187)(12) = 38(2390.53 - 225.94)
m; = 1637.1 kg/s
Mass difference
Mass difference
vr (P2 - PI) + hi
0.0011274(10,000 - 1002.1) + 763.22
773.36 KJ/kg
W p = m s (h2 - hi)
75 = m s(773.36 - 763.22)
rn, = 7.39 kg/s
Steam Cycle - 25
A condenser receives steam from a turbine at 15 kpa, 90% quality,
and with a velocity of 240 m/s. Determine the increase in circulating
=
=
1637.1 - 1615.38
21.78 kg/s
Steam Cycle - 26
A steam generator has an exit enthalpy of 3195.7 KJfkg at the rate of
10 kg/s. The enthalpy available at the turbine inlet is 3000 KJfkg.
Determine the heat lost between boiler outlet and turbine inlet.
A.1957kW
B. -1957 kw
C.1873kw
D. -1873 kw
118
Steam Cycles
Steam Cycles
SOLUTION:
119
h, = 289.23 + 0.0010223 (2000 - 30)
h, = 291.24 KJ/kg
By mass balance in the heater:
Assume supply steam = Ikg
m\(h z) -t (I - m.jh, = I h,
h, = h-at 2 Mpa
h6 = 908.79 KJ/kg
ml(2902.5) + (I - m l)(291.25) = 1(908.79)
m, = 0.2364 kg extracted steam/kg supply
Q = m (h z - h.)
Q = 10(3000 - 3195.7)
Q=-1957kw
Steam Cycle - 27
A Rankine cycle has a turbine unit with available enthalpy of 800
KJ/kg. The pump has also 10 KJ/kg energy available. Find the net
cycle output of the plant if mass now rate is 5 kg/so
A. 2619 kw
C. 8745 kw
B. 3950 kw
D. 4234 kw
SOLUTION:
W net = m, (W T - W p)
W net = 5(800 - 10)
W net = 3950 kw
Steam Cycle - 29
An adiabatic turbine in a steam generating plant receives steam at a
pressure of 7.0 Mpa and 550°C and exhausts at 20 kpa. The turbine
inlet is 3 m higher than the turbine exit, the inlet steam velocity is 15
m/s and the exit is 300 m/s. Calculate the turbine work in KJ/kg.
A. 1297.45
C. 1093.45
B. 1197.10
D. 1823.45
SOLUTION:
At 7.0 Mpa and 550°C:
h = 3530.9 KJ/kg .
s = 6.9486 KJ/kg-OK
At 20 kpa:
Steam Cycle - 28
In a Regenerative cycle, the steam is extracted from the turbine at 2
Mpa and 250°C for feedwater heating and it is mixed with condenser
exit at 30 kpa after pumping. Find the fraction of vapor extracted
from the turbine.
C. 0.5632
A. 0.23464
B. 0.19338
D. 0.3855
SOLUTION:
h, ~~ 251
I C"-550°C
V'==15mi~ ~
,4
w
.~ L'-"il • •
~""
hhg - 23:J8.3
1
Sr = 0.8320
Srg = 7.0766
s = Sr + XSrg
6.9486 = 0.8320 + x(7.70766)
x = 0.864
h, = 251.4 + (0.864)(2358.3)
h2 = 2288.9 KJ/kg
~
\61
20 Kpa
V,= 300 m/s
2 Mpa
At 2 Mpa and
hz =
At 30 kpa,
h, =
Vr =
hs =
250°C:
2902.5 KJ/kg
m, T 250°C
-
30 Kpa
289.23 KJ/kg
0.0010223 m3/kg
14 + V4(PS - P4)
1
HEATER
W T = (hi - h2 ) + 1/2 m (v/ - v/) + (PEl - PEz)
50 Kpa
1-m,
2-300 2
W T = (3530 -2288.9) +
WT
=
1197.IOKJ/kg
15
2(1000)
+
3(1 x9.81)
1000
Steam Cycles
120
121
Steam Cycles
Steam Cycle - 30
Steam Cycle - 31
A steam power plant operates on the Rankine cycle. The steam enters
the turbine at 7 Mpa and 550°C with a velocity of 30 m/s. It discharges
to the condenser at 20 kpa with a velocity of 90 m/s. Calculate the net
work in kw for a flow of 37.8 kg/s.
C. 34.22 Mw
A. 23.23 Mw
D. 46.54 Mw
B. 53.34 Mw
A Carnot cycle uses steam as the working substance and operates
between pressures of. 7 Mpa and 7 kpa. Determine the cycle thermal
efficiency.
A. 44.17%
C. 34.23%
B. 54.23%
D. 59.44%
SOLUTION:
SOLUTION:
At 7.0 Mpa and 550°C:
. h = 3530.9 KJ/kg
S = 6.9486 KJ/kg-OK
At 20 kpa:
hf = 251.4
hhg = 2358.3
Sf = 0.8320
Sfg = 7.0766
At 7.0 Mpa:
t = 285.88°C
At 7 kpa:
t = 39°C
T H = 285.88 + 273
T H = 558.88°K
T L = 39 + 273
TL = 312
w
e =
e
6.9486 = 0.8320 + x(7.70766)
x = 0.864
h 2 = 251.4+(0.864)(2358.3)
= 2288.9 KJ/kg
= h f at 20 kpa
= 251.4 KJ/kg
= 0.001017 m3 /k g
= h, + V3 (P4 - P 3 )
= 251.4 + 0.0010 17(7000 - 20)
= 258.5 KJ/kg
30 2 _ 90 2
WT = (3530.9 ·2288.9) + - - 2(1000)
WT = 1238.4 KJ/kg
WOe l = 1238.4 - (258.5 - 251.4)
WOe l = 1231.3 KJ/kg (37.8 kg/s)
WOel = 46,543.19 kw
WOe l = 46.54 Mw
2
285.B80C
1
[;]
3
5,-5,
4
TI-l -TL
5
TI-l
S=Sf+XSfg
h2
h,
h]
v]
h,
h,
h,
T
558.88 - 312
=
558.88
e = 44.17%
Steam Cycle - 32
A supercritical power plant generates steam at 25 Mpa and 560"C.
The condenser pressure is 7.0 kpa.. Determine the exit quality of steam
if it expands through a turbine in this power plant.
A. 45.66%
C. 56.56%
rI D. 74.26%
'1'
B. 68.45%
+
W25MPa
560°C
SOLUTION:
At 25 Mpa and 580°C:
h = 3430.5
S = 6.2897
/
/-
7Kpa
r
@
5
Steam Cycles
122
At 7 kpa:
Sc =
SCg =
BOILERS
0.5592
7.7167
S = Sc
123
Boilers
+ XSCg
6.2897 = 0.5592 + x(7.7167)
x = 74.26%
Boiler - 1
The heating surface area of water tube boiler is 200 m2 , what is the
equivalent rated boiler horsepower?
A. 217 Hp
C. 200 Hp
B. 2365.93 Up
D. 219.78 Hp
Steam Cycle - 33
Steam enters a turbine at 1.4 Mpa and 320?C. The turbine internal
efficiency is 70%, and the total requirement is 800 kw. The exhaust is
to the back pressure system, maintained at 175 kpa. Find the steam
flow rate.
C. 3.23 kg/s
A. 2.62 kg/s
D.
5.34 kg/s
B. 4.23 kg/s
SOLUTION:
Rated Boiler horsepower = RS.!0.91
Rated Boiler horsepower = 200/0.91
Rated Boiler horsepower = 219.78 Hp
SOLUTION:
Boiler - 2
W=800kw
--.
At 1.4 Mpa and 320°C:
h, = 3084.3
SI = 7.0287
At 175 Kpa:
Sc = 1.4849
SCg = 5.6868
hc = 489.99
V,= 90 m/s
hcg = 2213.6
Solving for the quality:
7.0287 = 1.4849 + x (5.6868)
x = 0.9748
h 2 = 489.99 + 0.9748(2213.6)
h 2 = 2647.93 KJ/kg
WT = IDs (hi - h 2)(llT)
800 = ills (3084.3 - 2647.93)(0.70)
ills = 2.62 kg/s
The rated boiler horsepower of a fire tube boiler is 500 Up. What is
the heating surface area of the boiler?
2
A. 500 m2
C. 400 m
2
2
B. 300 m
D. 550 m
SOLUTION:
Rated Boiler horsepower = H. S:! 1.1
500 = H.S.! 1.1
U.S, = 550m 2
Boiler - 3
A water tube boiler has a heating surface area of 500 m2 • For a
developed boiler hp of 825. Determine the percent rating of the boiler.
A. 120.15%
B. 160.15%
C. 150.15%
D. 300.15%
Boilers
Boilers
124
125
Rated boiler Hp = 281.32 Hp
Dev. Boiler Hp
xIOO%
%R =
Rated Boiler Hp
Dev. Boiler Hp
2 = ----281.32
Developed Boiler Hp = 562.64 Hp
ills (2257 x 1.08)
562.64 = - - - - - " - - - - 35,322
ID, = 8153.02 kg/hr
SOLUTION'
Rated boiler horsepower = 500/0.91
Rated boiler horsepower = 549.45 hp
Dev. Boiler Hp
xIOO%
%R =
Rated Boiler Hp
%R = 825/549.45 x 100%
%R = 150.15%
Boiler - 4
Boiler - 6
The factor of evaporation of a boiler is 1.1 and a steam rate of 0.79
kg/sec. What is the developed boiler horsepower?
A. 300
C. 869
B. 200
D. 250
SOLUTION:
FE
=
h, - h,
The actual specific evaporation of a certain boiler is 10. Factor of
evaporation is 1.05. If the heating value of fuel is 30,000 KJ/kg, find
the boiler efficiency.
A. 60%
C. 70%
B.65%
D.79%
SOLUTION:
h, - h F
--=
2257
2257 x 1.1
Developed Boiler hp
Developed Boiler hp
Developed Boiler hp
l1b =
ills(hs-h F )
ills(hs-h F )
IDrQ h
IDs
35,322
(0.79 x 3600)(2257 x 1.1)
10
ID r
35,322
199.89 Hp
l1b
1O(2257xI.05)
=
30,000
l1b = 79%
Boiler - 5
The percent rating of water tube boiler is 200%, factor of evaporation
2
of 1.08 and heating surface area is 256 m • Find the rate of
evaporation.
B. 7,200 kg/hr
A. 8,153.02 kg/hr
D. 8,500.46 kg/hr
B. 5,153.02 kg/hr
SOLUTION:
Rated boiler Hp = 256/0.91
Boiler - 7
The AS ME evaporation units of h boiler is 24.827,000 KJ/hr. The
boiler auxiliaries consumes 1.5 MW. What is the net boiler efficiency if
the heat generated by the fuel is 30,000 KJ/hr?
A. 64.75%
C. 62.76%
B. 68.94%
D. 68.54%
127
Boilers
Boilers
lLO
SOLUTiON:
SOLUTiON.
ms(h s - h F )
11ner =
-
Boiler Aux.
mrQ"
(24,827 ,000 I 3600) - (L5xl 000)
llne'
(30,000,000 I 3600)
11net ,=
Theo A/F =
Theo. NF =
Theo. A/F =
Actual A/F =
Actual A/F =
64.75%
11.5C + 345(H - 0/8) T 4.3S
11.5(0.705) + 34.5(0.045 - 0.06/8) + 4.3(0.03)
9.53 kg air/kg fuel
9.53(1.3)
12.389 kg air/kg fuel
=
h, - h F
-2257
(h, - h F ) = FE x 2257
(h, - h f ) = 1.1 x 2257
33,820C + 144,212(H - 0/8) + 9304S, Kl/kg
33,820(0.705) + 144,212(0.045 - 0.06/8) + 9,304(0.03)
=
29,930 KJ/kg
FE~
Boiler - 8
A 100,000 kg of coal supplied two boilers. One has a capacity of 150
kg/hr. How many days to consume the available fuel if the other boiler
consumes 200 kg/hr?
A. S days
C. 15 days
D. 12 days
B. 7 days
Qh
Qh
Qh
=
m s (h s
llb =
0.70
-
hF)
mrQ h
175,000( 1.1 x 2257)
=
------
m r (29,930)
SOLUTION:
m,
m
=
m
=
mrl + mf2
m = ISO + 200
350 kg/hr
No. of days = 100,00/350
No. of days = 285.71 hrs
No. of days = 11.9 days
=
rna =
rna =
PV =
101.325(V) =
V
=
21,018.456 kglhr
21,018456(12389)
260,397.651 kg/hr
mRT
260,397.651(0.287)(15.6 ~ 273)
3/hr
212,830 m
Boiler - 10 (ME Bd. Apr. 1981)
Boiler - 9 (ME Bd. Apr. 1981)
The following coal has the following ultimate analysis by weight:
C = 70.5%
Hz = 4.5%
O, = 6.0%
N z = 1.0%
S = 3.0%
Ash = 11 %
Moisture = 4%
A stocker fired boiler of 175,000 kg/hr steaming capacity uses this
3/hr
coal as fuel. Calculate volume of air in m
with air at 60°F and 14.7
psia pressure if boiler efficiency is 70% and FE = 1.10.
3/hr
A. 212,830 m
C. 213,830 rrr'rhr
D.. 214.830 m'zhr
B.. 2 J 5,830 mY/hr
The following coal has the following ultimate analysis by weight:
C = 70.5%
Hz = 4.5%
Oz = 6.0%
N z = 1.0%
S = 3.0%
Ash = 11 %
Moisture = 4%
A stocker fired boiler of 175,000 kglhr steaming capacity uses this
3/hr
coal as fuel. Volume of air in m
with air at 60°F and 14.7 psia
pressure. Weight in metric tons of coal needed for 24 hours operation
at rated capacity if boiler efficiency is 70% and FE = 1.10.
A. 503.443 Mtons
C. 502443 Mtons
B. 508.443 Mtons
D. 504,443 Mtons
128
Boilers
(h, - h-)
Theo. A/F
Theo. A/FTheo AiF
ActuaiA/F =
Actual A/F·~
129
Boilers
SOLUTION:
1I5e i 34.5(H - 0/8) + 4.3S
115(0.705) j 34.5(0.045 - 0.06/8) + 4.3(0.03)
953 kg air/kg fuel
9.53(13)
12.389 kg air/kg fuel
h, - h F
FE = - 2257
(h, - hr ) = FE x 2257:
(h, - hr) = 1.1 x 2257
On = 33.820C + 144,212(H - 0/8) + 9304S, KJrKg
Oh = 33.820(0.705) -i- 144,212(0.045 - 0.06/8) + 9,304(0.03)
Oil = 29.930 KJ/kg
ms\h s - h F )
llb
=
0.70 =
mrQ h
175,000(1.1x2257)
rn r (29,930)
rn, = 21,018.456 kg/hr
Coal needed in 24 hrs =
21,018.456(24)
1000 .
Coal needed in 24 hrs = 504.443 Mtons
Boiler - 11 (ME Bd. Oct. 1982)
Two boilers are operating steadily on 91,000 kg of coal contained in a
bunker. One boiler is producing 1591 kg of steam per hour at 1.2
factor of evaporation and an efficiency of 65% and another boiler
produces 1364 kg of steam per hour at 1.15 factor of evaporation and
an efficiency of 60%. How many hours will the coal in the bunker run
the boilers if the heating value of coal is 7590 Kcal/kg?
C. 230.8 hrs
A. 220.8 hrs
B. 256.2 hrs
D. 453.3 hrs
SOLUTION:
For Boliler No. I:
(h, - h r) = FE x 2257
llb =
=
1.2 x 2257
m s ( h s - h f- )
mrQ"
1591( 1.2 x 2257)
0.65 = - - - - - rn f ( 7590 x 4.l 87)
mfI= 208.605 kg/hr
For Boiler No.2:
(hs-hF ) = FE x 2257
(h, - hr)= 1.15 x 2257
ms(hs-h F )
llb
1591 kg/h
911,)00 kg
.m
Bunker
=
mrQ h
1364(1.15 x 2257)
0.60 = - - - ' - - - - m r (7590 x 4.187)
mf2 = 185.673 kg!hr
rn- = total fuel consumed
mT = rnfl + rnf2
m- = 208.605 + 185.673
m- = 394.278 kg!hr
No. of hours = 91,000/394.278
No. of hours = 230.8 hrs
B1
Boiler#1
B2
1364kgfh
Boiler#2
Boiler - 12 (ME Bd. Oct. 1984)
A steam generating plant consisting of a boiler, an economizer and
superheater generates superheated steam at the rate of 50 tons /hr,
Feed water enters the boiler at 5 Mpa and 120°C. Steam leaves the
superheater at 4.5 Mpa and 320°C. If the coal used has a heating
value of 30,000 KJ/kg, calculate the number of tons of coal fired per
hour for a gross efficiency of 85%.
A.4.89
C. 5.34
B. 6.34
D. 45.5
SOLUTION:
At 4.5 Mpa and 320°C(Table 3),
h, = 3000.6 KJ/kg
130
coal that could be used in order to ensure the generation of required
At 5 Mpa and 120°C(Table 4),
h F = 507.09 KJ/kg
llb
=
steam.
11kg/c:m'
_
h,
SOLUTION:
h.
(50)(3000.6 - 507.09)
p
_,I r (30,000)
m, = 4.889 tons/hr
C. 23,556
D. 30,976
A. 28,464
B. 29,977
ms(hs-h F )
mrQ
h
0.85 =
131
Boilers
Boilers
a.,..30000kJ/k
m,
11 kg/cm 2 x 101.325/1.033
1079.1 Kpa
1.0791 Mpa
=
P
=
p
=
From Table 2,
h,
From Table 1,
Boiler - 13 (ME Bd. Oct. 1986)
A water tube boiler has a capacity of 1000 kg/hr of steam. The factor
of evaporation is 1.3, boiler rating is 200%, boiler efficiency is 5%,
heating surface area is 0.91 m 2/boiler Up, and the heating value of
fuel is 18,400 Kcal/kg. The total coal available in the bunker is 50,000
kg. Determine total number of hours to consume the available fuel
A. 533.45
C. 634.34
B. 743.12
D. 853.26
llb
=
0.85
o,
hs
=
h,
=
hF
hF
m, =50000kglhr
BOILER
h g at 1.0791 Mpa
2781 KJ/kg
= hr at 80°C
= 334.91 KJ/kg
ms(h s - h F )
mrQ h
50,000(278-1 - 334.91)
= -------
4,800Q h
=
29,977 KJ/kg
SOLUTION:
(h, - h F) = FE x 2257
(h, - h F)= 1.3 x 2257
ms(hs-h F )
llb =
mrQ h
1000(1.3 x 2257)
0.65 = ----'--------'---m r (18,400x4.l87)
m, = 58.592 kglhr
No. of hours
No. of hours
=
50,000/58.592
=
853.36 hrs
Boiler - 14 (ME Bd. Apr. 1984)
A boiler operating at 11 kgicm 1 is required to generate a minimum of
50,000 kg/hr of saturated steam. Feed water enters the boiler at 80°C.
The furnace is designed to fire coal at an average rate 4,800 kg/hr and
boiler efficiency is 85%. Compute the minimum heating value of local
Boiler - 15 (ME Bd. Apr. 1984)
A boiler operating at 11 kg/cm 2 is required to generate a minimum of
aC.
50,000 kg/hr of saturated steam. Feed water enters the boiler at 80
The furnace is designed to fire coal at an average rate 4,800 kg/hr and
boiler efficiency is 85%. Compute the developed boiler horsepower.
A. 3462.56 hp
C. 2345.67 hp
D. 4233.34 hp
~'m.
B. 1234.56 hp
SOLUTION:
P
=
P
=
11 kg/crni x 101.325/1.033
1079.1 Kpa
P = 1.0791 Mpa
From Table 2,
h, = h g at 1.0791 Mpa
h, = 2781 KJlkg
From Table 1,
h F = h, at 80°C
132
Boilers
Boilers
At 145°C: h,
610.63
A.65
B. 95
133
z;
hF
334.9i
-
50,000(2781- 334.91)
Developed Boiler Hp
Developed Boiler Hp
~.J!h:g
==
C. 88
D. 78
SOLUTION:
35,322
3,462.56 Hp
Tlb
ms(hs-h F )
mfQ h
Boiler - 16
Tlb =
A steam boiler generating 7.1 kg/s of 4.137. Mpa, 426.7°C steam is
continuously blown at the rate of 0.31 kg/sec. Feed water enters the
economizer at 148.9 0C. The furnace burns 0.75 kg coal/sec of 30,470.6
KJ/kg higher heating value. Calculate the overall thermal efficiency of
steam boiler.
A. 76.34%
C. 82.78%
B. 84.23%
D. 88.34%
SOLUTION:
From Steam Table:
h, = 629.87 KJ/kg
h, = 3274.1 KJ/kg
h, = 109702 KJ/kg
7.1(3274.1) + 0.31(1097.02) - 7.41(629.87)
Tlb
0.75(30,470.6)
Tlb == 82.78%
23.5(3195.7 - 610.63)
2.75(25,102)
Tlb = 88%
'*
Boiler - 18 (ME Bd. Apr. 1997)
A steam boiler on a test generates 885,000 Ib of steam in a 4-hollr
period. The average steam pressure is 400 psia, the average steam
temperature is 700°F, and the average temperature of the feedwater
supplied to the boiler is 280°F. If the boiler efficiency for the period is
82.5%, and if the coal has a heating value of 13,850 Btu/lb as fired,
find the average amount of coal burned in short tons per hour.
At 400 psia and 700°F, h, == 1362.7 Btu/lb
At 2BO°F, hI = 249.1 Btu/lb
A. 9.84 short tons per hour
B. 10.75 short tons per hour
C. 12.05 short tons per hour
D. I 1.45 short tons per hour
SOLUTION:
Boiler - 17
23.5 kg of steam per second at 5 Mpa and 400°C is produced by a
steam generator. The feedwater enters the economizer at 145°C and
leaves at 205°C. The steam leaves the boiler drum with a quality of
98%. The unit consumes 2.75 kg of coal per second as received having
a heating value of 25,102 KJ/kg. What would be the overall efficiency
of the unit in percent?
Steam properties:
At 5 Mpa and 400°C: h = 3195.7 KJ/kg
At 5 Mpa: h; = 1154.23
hfg = 1640.1
At 205°C: h, = 875.04
rn, = 885,000/4
m, = 221,250 lb/hr
Tlb =
0.825
ms(hs-h F )
mfQ h
221,250(1,362.7 - 249.1)
I400
r.-....:..--.
m r (13,850)
mf= 21,563 Ib/hr
Q. = 13,850 Btull~
mf= 2 1,563/2000
mf- 10.78 short tons per hr
p.sia
700~F m, = 885,0001b
BOILER
I
280°F
134
Steam Engine
Boilers
135
STEAM ENGINE
Boiler-19
A steam boiler generates 401,430 kg of steam in a 4-hour period. The
steam pressure is 2750 kpa and 370°C. The temperature of the water
supplied to the steam generator is 138°C. If the steam generator
efficiency is 82.5% and the coal has a heating value of 32,200 KJ/kg,
find the average amount of coal burned per hour.
A. 9771
C. 9563
B. 8734
D. 7354
SOLUTION:
Steam Engine - 1
A steam engine have 10% brake thermal efficiency and delivers 750
kglhr steam. The enthalpy of engine entrance is 2800 KJ/kg and
condenser exit is 450 KJ/kg. Determine the brake power of the engine.
A. 46 KW
C. 49 KW
B. 47 KW
D. 48 KW
SOLUTION:
At 2750 kpa and 370°C:
h, = 3166.9 KJ/kg
At 138°C:
hF = hf at 138°C
hF = 580.54 KJ/kg
401,430
Brake Power
lltb =
ID s ( h s
0.10
--------
(750 /3600X2800 - 450)
Brake Power = 48.96 KW
IDs
IDs =
=
- hf2 )
Brake Power
4 x3600
27.877 kg/s
IDs(hs-h F )
llb =
IDFQ h
0.825
ID F (32,200)
IDF =
IDF =
Steam Engine - 2
27.877(3166.9 - 580.54)
2.714 kg/s (3600)
9770.77 kg/hr
The indicated efficiency of a steam engine is 60%. The engine entrance
is 2700 KJ/kg and exit is 2000 KJ/kg. if steam consumption is 800
kg/hr and mechanical efficiency is 90%, what is the brake power of the
engine?
A. 55 KW
C. 65 KW
B. 84 KW
D.70KW
SOLUTION:
Indicated Power
llei
ffi s
0.60
( h \ - h2 )
Indicated Power
=
(800/3600)(2700 - 2000)
Steam Engine
136
Steam Engine
Indicate d Power 93.33 KW
11m = BP/IP
0.90 = BP/93.3 3
BP = 84 KW
137
V 0 = 2[11:/4 (030):' (045) (220/60 )]
V o = 0.23326 rnzsec
Indicate d Power .~ 392.4(0 .233263 3)
Indicate d Power = 91.53/0 .746
Indicate d Power = 122.7 Hp
=
91.53 KW
Steam Engine - 3
g at
A steam engine has bore and stroke of 300 mm x 420 mm runnin
Kpa.
400
is
engine
the
of
e
pressur
ed
250 rpm has mean indicat
Determ ine the indicat ed power.
C. 65 KW
A. 100 KW
D. 99 KW
B 50KW
SOLUT ION
2
2[11:/4 D L N]
= 2[rc/4 (0.3)2 (0.42)(2 50/60)]
V::J = 0.247 m'
Indicate d Power = P mi X V 0
Indicate d Power = 400 x 0.247
Indicate d Power = 98.96 KW
VD
VD
Steam Engine - 5
1034.25
A steam engine develop s 60 Bhp with dry saturat ed steam at
736.36
is
ption
consum
Kpa absolut e and exhaus t at 124.11 Kpa. Steam
90%
on
based
cy
efficien
engine
kg/hr. Calcula te the indicat ed
cy.
efficien
ical
mechan
A. 34.23%
C. 45.23%
B. 54.23%
D. 66.74%
=
SOLUT ION:
hi = h g at 1.03425 Mpa
hi = 2779.4 KJ/kg (interpo lated)
S I = Sg at 1.03425 Mpa
5\
6.5748 KJlkg-O K(interp oiated)
=
At 0.12411 Mpa:(B y interpol ation)
h f = 443.43
Sf = 1.17165
h fg = 2241.56
Sfg = 5.9152
Steam Engine - 4
220 rpm.
The crank shaft of a double acting steam engine rotates at
and the
mm,
450
x
mm
300
is
engine
steam
the
of
stroke
and
The bore
the
Find
,
mean effectiv e pressur e acting upon the piston is 4 kg/ern"
indicat ed horsep ower develop ed in the cylinde r.
C. 143.2 hp
A. 122.7 hp
D. 176.3 hp
B. 110.3 hp
SI
=.
S2 =
Sf
+
XSfg
6.5748 = 1.37165 j x(5.915 2)
x = 0.8796
h2 = 443.43 + 0.8796( 2241.56 )
h 2 = 241.5.16 Kl/kg
Indicate d Power
Indicate d Power
SOLUT ION:
Indicate d Power
Pm'
Pm'
VD
=
=
=
Pm'
4 kg/em" x 101.325 /1.033
2
3924 KN/m
2[11:/4 [)2 L N]
X
VD
11el
(60/0.9 ) x 07'+6
49.733 KW
Indicate d Power
=
m,(h j
11"
rtei
=
=
=
-0
h2 )
49.733
(736.36 /3600)( 2779.4 - 24\51)
66.74%
138
Steam Engine
Steam Engme - 6
A steam engine develops 60 Bhp with dry saturated steam at 1034.25
Kpa absolute and exhaust at 124.]] Kpa. Steam consumption is 736.36
kg/hr. Calculate the thermal efficiency of equivalent Rankine engine.
A. 15.59%
C. 12.45%
B 34.23%
D. 21.34%
Steam Engine
indicated mean effective pressure is 600 Kpa, determine brake thermal
efficiency.
A. 23.34%
C. 14.66%
B. 18.34%
D. 27.34%
SOLUTION:
VD
VD
SOLUTION:
hi
hi
=
=
SI
=
SI
=
S 1"= S:
=
Sf
+ XSfg
6.5748 = 1.37165 + x(5.9152)
x = 0.8796
h,
~
hz
hf2
=
=
1,'2
=
Y]R
=
Y]R
Y]R
=
2779.4 - 443.43
15.59%
A 350 mm x 450 mm engine running at 260 rpm has an entrance steam
condition of 2 Mpa and BO°C and exit at 0.] Mpa. The steam
consumption is 2,000 kg/hr and mechanical efficiency is 88%. If
=
At 0.1 Mpa:
s,
=
Sfg
=
hf2
=
1.3026
6.0568
h, = 417.46
h rg = 2258
417.46 KJ/kg
Brake Power
Y]tb
m s (h J - h f2 )
m s (h l - h f 2 )
2779.4 - 2415.1
-----
Prru X V D
600 x 0.37522
225.13 KW
=
=
At 2 Mpa and 230°C(Table 3)
h, = 2849.6
SI = 6.4423
m s (hI - h 2)
=
2[71:/4 (0.35)2 (0.45) (260/60)]
0.37522 m 3 /sec
Brake Power = Indicated Power(Y]rn)
Brake Power = 225.13(0.88)
Brake Power = 198.11 KW
443.43 + 0..8 796(2241.56)
2415.16 KJ/kg
h, at 0.12411 Mpa
443.43 KJ/kg
:!/.:: Steam Engine - 7
l
=
=
Indicated Power
Indicated Power
Indicated Power
hg at 1.03425 Mpa
2779.4 KJ/kg (interpolated)
Sg at 1.03425 Mpa
6.5748 KJlkg-OK(imerpolated)
At O. ] 2411 Mpa:(By interpolation)
Sf = 1.37165
h, = 443.43
Sfg = 5.9152
h rg = 2241.56
139
198.11
Y]tb
=
Y]tb
=
(2000/3600)(2849.6 - 417.46)
14.66%
Steam Engine - 8
A 350 mm x 450 mm engine running at 260 rpm has an entrance steam
condition of 2 Mpa and BO°C and exit at 0.1 Mpa. The steam
consumption is 2,000 kg/hr and mechanical efficiency is 88%. If
140
Steam Engine
Steam Engine
indicated mean effective pressure is 600 Kpa, determine indicated
thermal efficiency.
A. 16.66%
C. 12.34%
B. 34.23%
D. 21.23%
SOLUTION:
V D = 2[rc/4 (0.35)2 (0.45) (260/60)]
V D = 0.37522 m 31sec
Indicated Power
Indicated Power
Indicated Power
=
Prru
=
600 x 0.37522
225.13KW
=
X
VD
14'1
SOLUTION:
VD
=
piston displacement
VD
=
2[(71:/4)D 2LN]
VD
=
2[(71:/4)(10/12)2(12/12)(300)]
VD
=
327.25 felmin
P rru V D
Indicated Power
Indicated Power = (120)(144)(327.25)
33,000
At 2 Mpa and 230°C(Table 3)
hi
=
Sl
=
2849.6
6.4423
Indicated Power
At 0.1 Mpa:
Sf =
Sfg =
hf2
=
1.3026
6.0568
hf = 417.46
hfg = 2258
417.46 KJ/kg
Indicated Power
11tl
m s ( h J - h f2 )
225.13
11 ti
116
-t
(2000/3600)(2849.6 - 417.46)
16.66%
Steam Engine - 9 (ME Bd. Apr. 1997)
Steam is admitted to the cylinder of an engine in such a manner that
the average pressure is 120 psi. The diameter of thepiston is 10" and
the length of stroke is 12". What is the engine when it is making 300
rpm?
A. 171.5
C. 173.2
D. 174.4
B. 175
I
=
171.40 hp
1-.+2
.\·1t'1If11
Turbine
STEAM TURBINES
Stca m Turbine - 2 (ME Rd. Apr. 1987)
An industrial power plant requires 1,5 kg of dry saturated steam per
second at 165°C for heating purposes. This steam may be supplied
from an extraction turbine which receives stearn at 4 Mpa and 380°C
and 'is exhaust to condenser at the rate of 0.8 kg steam per second at
0.0034 Mpa while rejecting 1400 KW to the cooling water. If
mechanical- electrical efficiency is 95% and that the he-at loss in the
turbine casing is 10 KW, calculate the power generated by 'he plant.
A. 2,126.44 Kw
C. 3,123.34 Kw
B. 1,556.5 Kw
D. 4,344.33 Kw
Steam Turbine - 1
A steam turbine receives 5,000 kglhr of steam at 5 Mpa and 400°C and
velocity of 25 m/sec. It leaves the turbine at 0.006 Mpa and 85%
quality and velocity of 20 rn/sec. Radiation loss is 10,000 KJ/hr. Find
the KW developed.
A. 1273.29
C. 1373.29
C. 2173.29
D. 7231.29
SOLUTION:
SOLUTION:
At 5 Mpa and 40QoC
hi = 3195.7 KJ/kg
Sl = 6.6459
---
5MPa.400oC
m7=5000kg/h
At 0.006 Mpa
h r = 15 I. 53 and hrg = 2415.9
hz = hr +- xhrg
hz = 151.53 +- 0.85(2415.9)
hz = 2205.045 KJ/kg
w
TURBINE
KE 1 = l/2 m yZ
KE 1 = 1/2 (5,OOO/3600)(25)z
KE 1 = 434.03 W
KE 1 = 0.43403 KW
KE z = 1/2 m yZ
KE z = l!2 (5,000/3600)(20)z
KE z = 277.78 W
KE z = 0.2778 KW
W
=
W
=
143
Steam Turbine
By energy balance:
KE, -t- mh I = KE z -i- mh, +- Q -i- W
W = (KE 1 - KE z) +- m(h, - h z) - Q
5000
10,000
(0.43403 - 0.2778) -i- ( - - )(3195.7 - 2205.045) - - 3600
3600
1373.29 KW
From Steam tables:
h, = 3165.9 KJ/kg
10i,{'.~
hz = hg at 165°C
4--~Gen
hz = 2763.5 KJ/kg
! ~~) OutP':lt
h, = h, at 0.0034 Mpa
h) = 109.84 KJ/kg
° . ~<;> @~.
By mass balance:
~,~~~5~gIS
. 1Tl,=O.8kgls
rn, = 1.5 + 0.8
f
1,400kw
m, = 2.3 kg/sec-:
h
By heat balance:
m.h, = 1.5h z +- O.Sh) + 10 +- 1400 +- \'1
2.3(3165.9) = 1.5(2763.5) + 0.8(109.84) +- 10 -t- 1400" W
W = ]638.45 KW
Generator Output = 1638.45(0.95)
Generator Output = 1,556.5 KW
r».
't
7=20m/s
(x=85%)
O.006MPa
-------~------------
Steam Turbine - 3
A steam turbine with 90% stage efficiency receives steam at 7 Mpa
and 550°C and exhausts as 20 Kpa. Determine the turbine work.
A.. 117 KJ/kg
C. 123 K.l/kg
.
B. 132 KJ/kg
D. 143 KJ/kg
SOLUTION:
At 7 Mpa and 550°C
hi = 3530.9 KJ/kg
s\ = 6.9486
)44
Steam Turbine
.~
At 20 Kpa(0.020 Mpa)
h, = 251.40
h,g = 2358.3
s I = S2 = Sf + XSfg
6.9486 = 0.8320 + x(7.0766)
x = 0.8643
h2 = 251.40 + 0.8643(2358.3)
h2 = 2289.78 KJ/kg
p
hi - h Za
llsT =
h, - h z
j~30.9 - h
Za
0.90 =
3530.9 - 2289.78
h2• = 2413.89 Kl/kz
WT = hi - h2 •
W T == 3530.9 - 2413.89
W T = ]]7.01 KJ/kg
.
145
Steam Turbine
Sr = 0.8320
Srg = 7.0766
090
3530.9 - h
2a
----
~
3530.9 - 2289.78
h 2> .= 2413.89 KJlkg
h2> = h, + xhrg
2413.89 = 25.\.40 + x(2358.3)
x
=
0.9167
x == 91.67%
Steam Turbine - 5
A small steam turbine power plant of 5,000 KW capacity has a full
load steam rate of 6 kg steam per KW-hr. No load steam consumption
may be taken as 10% of the full load steam consumption. Write the
equations of WILLANS LINE of this turbine and at 60% of rated
load, calculate the hourly steam consumption of this unit.
C. 19200 kg/hr
A. 19,100 kg/hr
B. 19,300 kg/hr
D. 19,400 kg/hr
5
Steam Turbine - 4
SOLUTION:
A steam turbine with 90% stage efficiency receives steam at 7 Mpa
and 550°C and exhausts as 20 Kpa. Determine the quality at exhaust.
C. 82.34%
A. 87.45%
B. 76.34%
D. 91.690/,
Full load steam consumption == 6(5000)
Full load steam consumption = 30,000 kg/hr
No load steam consumption
No load steam consumption
p
0.10(30,000)
3,000 kg/hr
SOLUTION:
By Two point slope formula:
At 7 Mpa and 550°C
h, = 3530.9 KJIkIl
SJ = 6.9486
At 20 Kpa(0.020 Mpa)
Sf = 0.8320
h, = 251.40
Sfg = 7.0766
hfg = 2358.3
SI
==
S2 =
sr+ XSrg
6.9486 = 0.8320 + x(7.0766)
x = 0.8643
h2 = 251.40 + 0.8643(2358.3)
h2 = 2289.78 KJ/kg
hl-h Za
llsT =
hi - h,
m.
(kg)
ms
5
-
3,000 ==
30,000 - 3,000
5.4(0.6 x 5000)
rn, at 60% load
m, = 19,200 kg/hr
2
,(sooo,
(L - 0)
5,000 - 0
ms-3,000 == 5.4L
rn, = 5.4 L + 3,000
FL
,
Nl~~30-0-00)-- -+
3000
,,
,
:
L(kw)
Steam Turbine - 6
Steam flows into a turbine at the rate of 10 kg/sec and 10 KW of heat
are lost from the turbine. Ignoring elevation and kinetic energy effects,
30000)
146
Steam Turhine
147
Steam Turbine
calculate till' power output from the tur hine. Inlet enthalpy is 2739
KJ/kg ami exhaust enthalpy is 2300.5 KJ/kg.
A. 4605 KW
C 4375 KW
B. 4973 KW
D. 4000 KW
rn,
=
55(42ll0)
rn,
~
26,100 kg/hr
I
3.000
SOLUTION:
W =
ffi(h 1 -
W
=
10(2739 - 2300.5) - 10
W
=
4375 KW
h2 ) - Q
Steam Turbine - 9
A steam turbine has an entrance enthalpy of 3050 KJ/kg. The exit has
2500 KJ/kg. Determine the actual enthalpy after isentropic expansion
if stage efficiency is 90%.
A 1255 KJ/kg
C. 2500 KJ/kg
B. 2555 KJ/kg
D. 2000 KJ/kg
Steam Turbine - 7
SOLUTION:
Steam entering the turbine has a rate of 10 kg/sec with enthalpy of
3400 KJ/kg and 2600 KJ/kg at the exhaust. If 100 KW of heat is
rejected from turbine casing, what is the turbine work?
A. 7900 KW
'c. 5600 KW
B. 7700 KW
D. 5400 KW
h, - h 1 a
"lsr
h, - h 1
3050 - h 1 a
0.90
SOLUTION:
3050- 2500
W
=
W
W
=
=
W
=
(h, - h-) - Q
10 (3400 - 2600) - 100
8000 -100
7900KW
h1a
ffi s
=
2555 KJ/kg
Steam Turbine - 10
Steam Turbine - 8
A steam turbine of 6 MW capacity has a Willan's line equation of rn,
=
5.5L + 3,000, kg/hr. Determine the steam consumption at 70% load.
A. 3564 kg/hr
C. 26,100 kg/hr
B. 3546 kg/hr
D. 58,000 kg/hr
SOLUTION:
At 70% load,
L
=
0.7(6,000)
L
=
4200 KW
Steam enters the turbine at the rate of2.5 kg/sec with enthalpy of 3200
KJ/kg and exhaust enthalpy of 1100 KJ/kg. Steam is extracted from
the turbine at the rate of 1 kg/sec for heating purposes with enthalpy
of 2750 KJ/kg. What is the turbine work.
A. 2000 KW
C. 3000 KW
B. 2500 KW
D. 3600 KW
SOLUTION:
ffi\
=
2.5
rn-
=
=
+ ffi3
1+ m,
1.5 kg/s
ffi2
Stemn Turbine
Ic\X
W = 2~(1:()()) -
W
=
i(2750)-1.5(1100)
Geothermal Power Plant
149
GEOTHERMAL POWER PLANT
36()() 1\W
Geothermal Power Plant - 1
Steam Turbine - 11
A steam turbine has an entrance enthalpy of 3400 KJ/kg and 2500
KJ/kg at exit. If generator generates 2430 KW and has 90% efficiency,
what is the mass of steam entering the turbine?
A 10,400 kg/hr
C. 10,700 kg/hr
B. 10,600kglhr
D.l0,800kg/hr
SOLUTION:
Mass flow rate of ground water in a geothermal power plant is
1,500,000 kg/hr and the quality after throttling is 30%. Determine the
brake power of turbine if the change of enthalpy of steam at inlet and
outlet is 700 KJ/kg.
C. 64.5 MW
A. 68.5 MW
B. 87.5 MW
D. 89.5 MW
SOLUTION:
W T = m, (hi - h2 )
243010.90 = TIl, (3400 - 2500)
m, = 3 kg/sec x 3600sec/hr
m, = }O,800 kg/hr
ID, =
m,
0.3(1,500,000)
450,000 kg/hr
ID, = 125 kg/sec
WT = m, (h, - h4 )
W T = 125(700)
W T = 87,500 KW
W T = 87.5 MW
ID, =
Steam Turbine - 12
Steam turbine in Rankine cycle has an exhaust enthalpy of 2650 KJ/kg
and delivers 0.8 kg/sec of steam. Determine the heat rejected from the
condenser if enthalpy at exit is 200 KJ/kg.
A. 1960 KW
C. 1995 KW
D. 1909 KW
B. 1940 KW
SOLUTION:
OR
=
mlh 2 - h3 )
OR = 0.8(2650 - 200)
QR = 1960 KW
x mg
=
Geothermal Power Plant - 2
Ground water of geothermal power plant has an enthalpy of 700
KJ/kg and at turbine inlet is 2,750 KJ/kg and enthalpy of hot water in
flash tank is 500 KJ/kg. What is the mass of steam flow entering the
turbine if mass flow of ground water is 45 kg/sec?
A. 3.27 kg/sec
C. 4.27 kg/sec
B. 2.27 kg/sec
D. 9.27 kg/sec
SOLUTION:
h2
=
700
x
=
hi + x(h, - hf)
500 + x(2750 - 500)
0.0888
=
!:'()
Geothermal Power Plant
0.888(45)
The enthalpy entering the turbine of a geothermal power plant is 2750
KJ/kg and mass rate of 1 kg/sec. The turbine brake power is 1000 KW
condenser outlet has enthalpy of 210 KJ/kg. If temperature rise of
. cooling water in condenser is 8°C, what is the mass of cooling water
requirement?
C. 46 kg/sec
A. 44 kg/sec
B. 45 kg/sec
D. 47 kg/sec
SOLUTION:
W T = ill,(h 3 - h 4 )
1000 = 1(2750 - h4 )
h, = 1750 KJ/kg
=
Qw
ills(~
- h s) = ill w Cp(t2 - t.)
1(1750 - 210) = ill w(4.187)(8)
m., = 45.97 kg/sec
A 16,000 KW geothermal plant has a generator efficiency and turbine
efficiency of 90% and 80%, respectively. If the quality after throttling
is 20% and each well discharges 200,000 kg/hr, determine the number
of wells are required to produce if the change of enthalpy at entrance
and exit of turbine is 500 KJ/kg.
A. 4 wells
C. 6 wel1s
B. 5 wells
D. 8 wells
SOLUTION:
W T = ill s(h3 - h 4 )
16,000
- - = i l l s(500)
0.9(0.8)
ills = 44.44 kg/sec
ills = 160,000 kg/hr
160,000 = 0.20 illg
illg = 800,000 kg/hr
No. of wel1s = 800,000/200,000
No. of wells = 4 wells
Geothermal Power Plant - 6 (ME Bd. Apr. 1988)
Geothermal Power Plant - 4
In a 12 MW geothermal power
the turbine is 26 kg/sec. The
enthalpy of ground water is
efficiency of the plant.
A. 7.4%
B. 9.6%
plant, the mass flow of steam entering
quality after throttling is 25% and
750 KJ/kg. Determine the overall
C. 5.4%
D. 15.4%
SOLUTION:
ills = x illg
26 = 0.25 illg
illg = 104 kg/sec
12,000
f]()
Geothermal Power Plant - 5
4 kg/sec
Geothermal Power Plant - 3
f]o
151
x fIl g
ills
ill,
ill,
QR
Geothermal Power Plant
104(750)
15.38%
A geothermal power plant draws pressurized water from a well at 20
Mpa and 300°C. To produce a steam water mixture in the separator,
where the un flashed water is removed, this water is throttled to a
pressure of 1.5 Mpa. The flashed steam which is dry and saturated
passes through the steam collector and enters the turbine at 1.5 Mpa
and expands to 1 atm. The turbine efficiency is 85% at a rated power
output of 10 MW. Calculate overall plant efficiency
A.7.29%
C. 9.34%
B. 12.34%
D. 19.45%
SOLUTION:
At 1.5 Mpa (Table 2)
h, = 2792.2 KJ/kg
S3 = 6.4448
At I arm (100°C)
152
Geothermal Power Plant
Sf
Sfg
= 1.3069
=
h,
hf.~ -
6.048
4 I9.04
2257
S3 = S4 = Sf -t X4Sf~
64448 = 1.3069 + X4 (6.0480)
X4 ~ 0.8495
h, = h r + xh fg
.h, = 419.04 + 0.8495(2257)
h, = 2336.4 KJ/kg
Wr = ms(h 3 " h4)Tlr
10,000 = m,(2792.2 - 2336.4)(0.85)
rn, = 25.81 kg/sec
At 20 Mpa and 300°C(Table 4)
h,= 1333.3 KJ/kg
At 1.5 Mpa:
hr = 844.89
hfg = 1947.3
h, = h2 = h, + X2 hfg
1333.3 = 844.89+X2(1947.3)
X2 = 0.25
m, = Xl fig
(25.81 X 3600) = 0.25(m g \
fig = 37 I ,664 kg/hr
10,000
hr '= 640.23
hrg = 2108.5
m.
h, = h g at 0.50 Mpa
1500kPa
h, = 2748.7 KJ/kg
h, = n2 '= hr + xzhrg
.
697~22 = 640.23 + Xz (2~108.5)
® I" I
X2 - 0.027
I
:
I
I
m.m, = xm g
\-'
m, = 0.027(29.6)
,29.6 kg/s
m, = 0.80 kg/sec
-IJG)
6t;"10°C
From Mollier Diagram: j ~
14 = 2211 KJ/kg
Power produced = m s(h 3 - ~)
Power produced = 0.8(2748.7 - 22 I I)
Power produced = 430.16 KW
[d:>
'!®
+"""
=
+@
Geothermal Power Plant - 8 (ME Bd. Oct. 1985)
A flashed steam geothermal power plant is located where
underground hot water is available as saturated liquid at 700 Kpa.
The well head pressure is 600 Kpa. The flashed steam enters a turbine
at 500 Kpa and expands to 15 Kpa, when it is condensed. The flow rate
from the wdl is 29.6 kg/sec. Determine the cooling water flow in kg/sec
if water is available at 300 e and a lODe rise is allowed through the
s
(371,644/3600)(1333.3)
110vel'all
153
Geothermal Power Plant
7.26%
condenser.
SOLUTION:
Geothermal Power Plant - 7 (ME Bd. Oct. 1985)
A flashed steam geothermal power plant is located where
underground hot water is available as saturated liquid at 700 Kpa.
The well head pressure is 600 Kpa. The flashed steam enters a turbine
at 500 Kpa and expands to 15 Kpa, when it is condensed. The flow rate
from the well is 29.6 kg/sec. Determine the power produced in KW
! .. 430.13 kg/s
C. 370.93 kg/s
B. 540.23 kg/s
D. 210.34 kg/s
h,
h,
=
=
h-at 0.70 Mpa
697.22 KJ/kg (i)
e-t><:I--+-
1#.':
® ll.t=10°C
i!
SOLUTION:
h,
=
h f at 0.70 Mpa
m,
m,
h,- 697.22 KJ/kg
At 500 Kpa:
It
=
=
xm g
0.027(29.6)
154
m,
=
155
Geothermal Power Plant
Geothermal Power Plant
P4 .~ 0.016932 Mpa
From Mollier diagram:
14 = 2085 KJ/kg
Shaft Power = (480,915/3600)(2796 - 2085)( I - 0.015)(0.75)
Shaft Power = 70,167 K W
Plant Output = 70,167(0.97)
Plant Output = 68,062 kw
Plant Output = 68.075 MW
0.80 kg/sec
h, = hrat 15 kpa(0.015 Mpa)
h, = 225.94 KJ/kg
m s(14 - h s) = m; c p (At)
0.8(2211-225.94) = m w(4.187)(10)
= 37.93 kg/sec
m;
Geothermal Power Plant - 9 (ME Rd. Oct. 1981)
Geothermal Power Plant - 10
In a certain geothermal area, studies show .that 1,500,000 kglhr of
pressurized ground water is available at 2,500 psia and 620°F. The
water will be throttled to 250 psia to produce wet steam and this
mixture will be passed through a water separator to remove the water
droplets so that saturated steam at 250 psia is available at the entrance
of steam turbine for the proposed power plant. Other power plant
data are as follows:
Discharge pressure of turbine = 25 in Hg vacuum
Turbine engine efficiency = 75%
Mechanical Loss = 1.5% of shaft power
Generator Efficiency = 97%
Assume atmospheric pressure to be 30 in Hg. Determine the maximum
amount of power in Kilowatts that the plant can generate.
A. 60 MW
C. 66 MW
B. 68 MW
D. 74 MW
SOLUTION:
At 2500 psi(17.232 Mpa) and 620°F(326.667°C)
hi = 1490.9 KJ/kg
At 250 psi(I.7232 Mpa)
hr = 875 KJ/kg
h rg = 1921
h, = h g at 1.7232 Mpa
h, = 2796 KJ/kg
h, = h 2 = h r + X2hrg
.
~il2?.
...
1490.9 = 875 + X2(1921)
X2 = 0.3206
;j
rn, = X2 m,.g
'1'-'
'I
m, = 0.3206(1 ,500,000) ~
rn, = 480,915 kg/hr
'J 2500 psi
P 4 = (30 -25) x 0.101325/29.97
I®
+
@25 in Hg
vacuum
A geothermal power plant has an output of 16,000 KW and combined
efficiency of 80%. The pressurized ground water at 175 bar and 280°C
leaves the wells to enter the nash chamber maintained at 14 bar. The
steam collected enters the turbine at 14 bar and exhaust at 1 atm. If
one well discharges 180,000 kg/hr of hot water, how many wells are
required?
A. 2
C. 3
B. 4
D. 5
SOLUTION:
~
"
l~
h g at 14 bar(1.4 Mpa) m,
@
h, = 2790 KJ/kg
I
~-.§.,
S3 = Sg at 1.4 Mpa H<D
S3 = 6.4693
~175bar
At I atm(100°C):
Sr = 1.3069
h r = 419.04
Srg = 6.0480
h rg = 2257
S3 = S4 = Sr + X2Srg
6.4693 = 1.3069 + ~(6.0480)
X4 = 0.8536
14 = 419.04 + 0.8536(2257)
14 = 2345.55 KJ/kg
16,000/0.80 = m.(h3 -14)
20,000 = m s(2790 - 2345.55)
m, = 45 kg/sec
At 1.4 Mpa:
h r = 830.30
h rg = 1959.7
At 175 bar(17.5 Mpa) and 280°C(Table 4)
hi = 1231 KJ/kg
hi = h 2 = hr + X2hrg
h,
=
+®
156
Geothermal Power Plant
1231
X2
=
830.30
t-
Geothermal Power Plant
xl(l959.?)
hi
h2
he + xhcg
870<; I - 770.38 + xz(2009.2)
Xl
0049836
rn. .~ xm~
121.8 = 0.049836m g
m g = 2,444 kg/sec
= 0.2046
m, = Xl m g
(45 X 3600) = 0.2046mg
m g = 791699.25 kg/hr
No. of wells = 791699.25/180,000
No. of wells = 4.3984
No. of wells « 5 wells
Geothermal Power Plant - 11 (ME Rd. Oct. 1995)
A liquid dominated geothermal plant with a single flash separator
receives water at 204°C. The separator pressure is 1.04 Mpa. A direct
contact condenser operates at 0.034 Mpa. The turbine has a polytropic
efficiency of 0.75. For a cycle output of 50 MW, what is the mass flow
rate of the well-water in kg/s?
A. 2871
C. 186
D. 2444
B. 2100
At 204°C:
he = 870.51 KJ/kg
At 1.04 Mpa:
hcg = 2009.2
h c = 770.38
5 g = 6.5729
h g = 2779.6
At 0.034 Mpa:
hcg = 2328.8
hc = 301.40
Srg = 6.7463
Sc = 0.9793
SOLUTION:
h, = h. at 1.04 Mpa
2779.6 KJ/kg
1m•
Solving for h.t:
h,
S3 =
S4
=
<i>
-
=' Sr + xSrg
6.5729 = 0.9793 + )4(6.7463)
~<D
)4 = 0.829
h, = 301.4 + 0.829(2328.8) U 0c
h, = 2232.3 KJ/kg
204
W T = m, (h, - h.t)
50,000 = m..(2779.6 - 2232.3)0.75
m, = 121.8 kg/sec
Solving for X2: (hi = h2)
@
157
158
Diesel Power Plant
Diesel Power Plant
A 1000 KW
B. 775 KW
DEISEL POWER PLANT
159
C. 968 KW
D. 588 KW
SOLUTION:
Diesel Power Plant - I
Indicated Power
110 =
Determine the indicated mean effective pressure of an engine in psi
having a brake mean effective pressure of 750 Kpa and 80%
mechanical efficiency.
A 136 psi
C. 138 psi
B. 137 psi
D. 140 psi
mrQ h
0.44
Indicated Power
=
0.05(44.000)
Indicated Power = 968 KW
SOLUTION:
11m = PmWPml
0.80
Diesel Power Plant - 4
750/P m t
14.7
PIT" = 937.5--101.325
Pm' = 136 psi
=
A 750 KW diesel electric plant has a brake thermal efficiency of 34%.
If the heat generated by fuel is 9,000,000 KJ/hr, what is the generator
efficiency?
A 85.33%
8. 65.88'%
C. 75.55%
D. 88.23%
SOLUTION:
Diesel Power Plant - 2
Brake Power
11tb
Determine the friction power of an engine if the frictional torque
developed is 0.3 KN.m running at 1200 rpm.
A. 40.6 KW
C. 36.5 KW
B. 37.7 KW
D. 50.3 KW
0.34
mrQ h
Brake Power
9,000,000/3600
SOLUTION:
Friction power
Friction power
Friction power
=
=
=
Brake Power = 850 KW
110 = 750/850
110 = 88.23%
2 IT T N
2 n(O.3)(l200/60)
37.70 KW
Diesel Power Plant - 5
Diesel Power Plant - 3
What is the power developed in the cylinder if indicated
thermalefficiency is 44%, the engine uses 0.05 kg/sec fuel with heating
value of 44,000 KJ/kg?
A ) 6-cylinder V-type diesel engine is directly coupled to a 5000 KW
AC generator. If generator efficiency is 90%, calculate the brake
horsepower of the engine,
A 7447 Hp
C. 8542 Hp
B. 6468 Hp
D. 7665 Hp
160
Diesel Power Plant
Diesel Power Plant
°AP1
27
141.5
SG.= - - 131.5+°API
141.5
SG
131.5+ 27
SG = 0.8927
SOLUTION:
llCi -
161
(icn. Output
Brake Power
0.9 = 5000/Brake Power
Brake Power = 5555.55 KW (1/0.746)
Brake Power = 7447.12 Hp
=
Diesel Power Plant - 8
Diesel Power Plant - 6
Determine the brake power of the engine having a brake thermal
efficiency of 35% and uses 25° API fuel with fuel consumption of 40
kg/hr.
A. 165.84 KW
C. 173.52 KW
B. 173.52 KW
D. 160.67 KW
A 500 KW diesel engine operates at 101.3 Kpa and 27°C in Manila. If
the engine will operates in Baguio having 93 Kpa and 23°C, what new
brake power will developed if mechanical efficiency is 85%?
A. 600 KW
C. 459 KW
B. 754 KW
D. 971 KW
SOLUTION:
SOLUTION:
Qh
Qh
Qh
11tb
WI
=
P/RT
WI
=
WI
=
101.325/(0.287)(27 + 273)
1.1765 kg/rrr'
41,130 + 139.6°API
= 41,130+ 139.6(25)
= 44,620 KJ/kg
Brake Power
W2
=
W2
=
mfQ h
Brake Power
0.35 = - - - - - (40/ 3600)(44,620)
Brake Power = 173.52 KW
Diesel Power Plant - 7
Determine the specific gravity of fuel oil having a heating value of
44,899.2 KJ/kg.
A. 0.90
C. 0.877
D. 0.893
B. 0.80
W2 =
P/RT
= 93/(0.287)(23 + 273)
= 1.0947 kg/rrr'
llm = BP / IP
0.85 = 500/ind. Power
Ind. Power = 588.23 KW
Friction Power = IP - BP
Friction Power = 588.23 - 500
Friction Power = 88.23 KW
1P2
w2
IP\
wj
IP2
1.0947
----58823 1.1765
SOLUTION:
Qh = 41,130 + 139.6°API
44,899.2
=
41,130 + 139.6°API
IP 2 = 547.336 KW
BP 2 = 547.336 -8823
BP 2 = 459.106 KW
Diesel Power Plant
16:::'
Diesel Power Plant
163
Diesel Power Plant - 9
Diesel Power Plant - I I
What is the displacement volume of 300 nun x 400 mm, 4-stroke, 1200
rpm, 8 cylinder diesel engine?
C. 5.75 kg/sec
A. 0.243 m1/sec
D. 1.25 m] /sec
B. 2.262 m]/sec
Determine the output power of a diesel power plant if the engine and
generator efficiency is 83% and 95%, respectively. The engine uses
25° API fuel and has a fuel consumption of 0.08 kg/sec.
A. 2795 KW
C. 9753 KW
B. 8642 KW
D. 2815 KW
SOLUTION:
SOLUTION:
Yo
=
Yo
n/4D 2 L N c
Qg
Qg
Qg
1200
(n/4)(0.3)2(OA)(--) (8)
2 x60
=
mrQh
0.08[41,130 + 139.6(25)]
= 3569.6 KW
Generator Output = 35696(0.83)(0.95)
Generator Output = 2814.63 KW
Yo = 2.262 m'rsec
=
Diesel Power Plant - 12
Diesel Power Plant - 10
What is the friction horsepower of a 300 KW diesel engine having a
mechanical efficiency of 86%.
C. 90.5 Hp
A. 86.5 Hp
D. 65.5 Hp
B. 87.5 Hp
Determine the piston speed of a 250 mm x 300 mm diesel engine
running at 1200 rpm.
A. 6 m/sec
C. 18 rn/sec
B. 12 mlsec
D. 5 m/sec
SOLUTION:
SOLUTION:
Piston Speed
Piston Speed
Piston Speed
2LN
2(030)( 12(0/60)
J 2 rn: sec
Brake Power
11m
Indicated Power
300
0.86
Indicated Power
Indicated Power
=
348.84 KW
Friction Power = 348.84 - 300
Friction Power = 48.84 KW
Friction Power = 65.46 Hp
Diesel Power Plant - 13
Determine the speed of a 6-pole generator of 60 hz.
A. 600 rpm
C. 1200 rpm
B. 1000 rpm
D. 3600 rpm
SOLUTION:
N =
120 f
P
Diesel Power Plant
Diesel Power Plant
164
120(60)
N
Diesel Power Plant - 16
165
.
6
N
1200 rpm
An engine-generator rated 9000 KV A at 80% power factor, 3 phase,
4160 V has an efficiency of 90%. If overall plant efficiency is 25%,
what is the heat generated by the fuel.
A. 18,800 KW
C. 7500 KW
B. 28,800 KW
D. 20,000 KW
Diesel Power Plant - 14
A 50 Bhp blast furnace engine uses fuel with 10 felBhp-hrs. The
heating value of gas is 33,500 KJ/m 3 • Determine the brake thermal
efficiency.
A.80%
25.76%
B. 50%
D. 28.31%
SOLUTION:
C:
Gen. Output = pfx KVA
Gen. Output = 0.8 x 9000
Gen. Output = 7200 KW
Gen Output
11 overall
SOLUTION:
Qg
lOft]
Qg
=
Qg
=
Qg
=
Qg
=
[
1
1
x 50Bhp x - - x
, ](33,500)
Bhp - hrs
3600 (3.281)-
0.25 = 7200/Qg
Qg = 28,800 KW
131.73 KW
50 x 0.746
Diesel Power Plant - 17
131.73
28.31%
A 4-stroke, 8 cylinder diesel engine with bore and stroke of 9 in x 12
in, respectively and speed of 1000 rpm has brake mean effective
pressure of 165 psi. Determine the engine brake horsepower.
A. 1753.34 Hp
C. 1950.34 Hp
B. 1850.34 Hp
D. 1272.34 Hp
Diesel Power Plant -15
The heat generated by fuel is 2500 KW If the jacket water loss is 30%.
Determine the mass of water circulated in the engine if the
temperature rise is soc.
A. 20.5 kg/sec
C. 58.5 kg/sec
B. 22.4 kg/sec
D. 12.3 kg/sec
~
SOLUTION:
Vo
v:
V0
SOLUTION:
Jacket
Jacket
Jacket
Jacket
750 =
row =
water loss = 0.3 Qg
water loss = 0.3(2500)
water loss = 750 KW
water loss = row c p (At)
m., (4.187)(8)
22.39 kg/sec
1t/4D2LNc
= 1t/4(9)2 (12) (1000/2)(8)
= 3053628.059 in3/roin
=
Brake
Brake
Brake
Brake
Brake
power =
Power =
Power =
Power =
Power =
Pmb X V0
165(3053628.059)
503848629.8 in-lb/min x 1/12
41987385.82/33,000
1272.34 hp
I ()()
Diesel Power Plant
Diesel Power Plant
167
SOLUTION:
Diesel Power Plant - 18
SG
What torque is developed by the JOn KW engine running at 900 rpm?
A. 2.65 KN.m
C. 3.18 KN.m
B. 6.85 KN.m
D. 5.65 KN.m
SG
141.5
131.5+ 25
0.904
Density of fuel
Density of fuel
SOLUTION:
=
300
.~
T
3.18 KN.m
=
Time to run the engine
Time to run the engine
Time to run the engine
211: TN
KW
211: T (900/60)
Diesel Power Plant - 19
What is the mechanical efficiency of a diesel engine if friction power is
30 KW and brake power of 150 KW?
A. 85.55%
C. 65.44%
B. '83.33%
D. 75.88%
SOLUTION:
BP/IP
BP + FP
=
150 + 30
= 180 kw
11m = 150/180
11m = 83.33%
11m
IP
IP
IP
0.904( I)
0.904 kg/Ii
v
=
240/1080
V
=
0.2221i1sec
m
m
=
0.904(0.222)
0.2 kg/sec
=
Diesel Power Plant - 21
The indicated thermal efficiency of a two stroke diesel engine is 50%.
If friction power is 3% of heat generated, determine the brake thermal
efficiency of the engine.
A.33%
C. 36%
B. 34%
D.37%
=
=
Diesel Power Plant - 20
A 3MW diesel engine consumes 240 liters of 25° API fuel and generates
900 KW-hr. Determine the rate of fuel consumed by the engine.
A. 0.2 kg/sec
C. 0.6 kg/sec
B. 0.4 kg/sec
D. 0.8 kg/sec
90013000
0.3 hr
1080 sec
SOLUTION:
1111 = IP/Qg
0.5 = IP/Qg
IP = 0.5 Qg
BP = IP - FP
BP = O.5Qg - O.13Qg
BP = 0.37Qg
11tb = BP/Qg
11tb = 0.37QglQg
11tb = 37%
168
Diesel Power Planl
Diesel Power 1'/11"1
169
Dlcsel Power Plant - 24
Diesel Power Plant - 22
During the dynamometer test of an engine for 1 hr steady load, the
engine consumes 40 kg fuel having 43,000 KJ/kg heating value. If the
torque developed is 2.5 KN-m during the test at 600 rpm, what is the
brake thermal efficiency of the engine?
A. 31.22%
C. 55.77%
B. 32.88%
D. 25.99%
What maximum power that can be delivered by the 2000 KW engine
at JOOO ft elevation considering the pressure effect alone?
A. 1600 KW
C. 1900 KW
B. 1700 KW
D. :300 KW
SOLUTION:
Solving for the pressure at higher elevation:
B = 29.92 - (3000/1000)
B = 26.92 in Hg
SOLUTION:
m.
=
40/3600
m; = 0.0 III kg/sec
BP
BP
BP
=
=
=
2nTN
2n(2.5)(600/60)
157.08 KW
157.08
11tb
11'b
=
B
P,
=
P,
=
r,
=
Diesel Power Plant - 25
0.0 IIII( 43,000)
32.88%
A 1000 KW diesel engine operates at an altitude of 1500 m elevation.
Considering the pressure effect alone, find the power developed by the
engine at higher elevation?
C. 345 KW
A. 753 KW
D. 983 KW
B. 823 KW
Diesel Power Plant - 23
A waste heat recovery boiler generates 250 kg/hr with h, -h F =
23GOKJ/kg. What is the exhaust loss from the engine if the actual heat
developed is 35%?
C. 456.4 KW
A. 492.6 KW
B. 365.7 KW
D. 845.4 KW
SOLUTION:
T
=
T
=
SOLUTION:
Pe
Waste heat recovered
Waste heat recovered
Waste heat recovered
Exhaust Loss
Exhaust Loss
Ps ( - - )
29.92 .
2000(26.92/29.92)
1799.46 KW
=
=
=
rn, (h, - hF)
25W3600 (2300)
159.72 KW
159.72/0.35
456.35 KW
'.,
=
3.6
520 - - - x (l500x3.281)
1000
502.28
Ps~ 5:0
J502.28
1000 - 520
P e = 982.82 KW
P,
=
: "jl I
/)/I'S/'!
171
Diesel Power Plant
Power Plant
SOLUTION:
Diesel Power Plant - 26
BP = 2 rt TN
BP = 2 rt (5)( 1800/60)
BP = 94248 KW
The piston speed of an engine running at 1200 rpm is 12 m/s. Find the
stroke of the engine in inches.
A. 15.74 mm
C. 300 mrn
B. 16 in
D. 15.75 in
= BP/Qg
0.31 = 942A8/Qg
Qg = 3040.25 KW
Tllb
SOLUTION:
Piston Speed = 2 L N
12- 2L(l200/60)
L ~ 0.3 rn
L ~ 30n mm
Diesel Power Plant - 29
In a diesel engine
clearance volume is
A.
B.
the fuel is injected at 6.5% of the stroke, the
10% of the stroke. Find the cut-off ratio.
1.80
C. 1.70
1.65
D. lAO
Diesel Power Plant - 27
The density of air entering the engine is 1.176 kg/m'' has a volume flow
rate of 0.375 m 3/sec. If the air fuel ratio is 21, find the mass flow rate of
fuel.
A. O. J 0 kg/sec
C. 0.30 kg/sec
D. OAO kg/sec
B. 0.02 kg/sec
SOLUTION:
Y3 I V 2
Cut-off ratio
Cut-off ratio
=
YI +0.065Y D
YI
SOLUTION:
YI
m, == W Y
m, =." 1.176(0.375)
m. = 0441 kg/sec
A/F = m, / m,
21 = OA41/mr
=
O.IOYD
Cut-off ratio
0.10Y D + O.065Y D
O.IOY D
Cut-off ratio = 1-, 65
m, = 0.0211 kg/sec
Diesel Power Plant - 30 (ME Bd. Oct. 1989)
Diesel Power Plant - 28
A diesel engine develops a torque of 5 KN-m at 1800 rpm. If the brake
thermal efficiency is 31 %, find the heat generated by the fuel.
A. 3050.25 KW
C. 3000.25 KW
B. 304025 KW
D. 5000 KW
A 373 KW (500 Hp) internal combustion engine has a brake mean
effective pressure of 551.5 Kpa at full load. What is the friction power
if mechanical efficiency is 85%?
C. 56.34 hp
A. 88.23 hp
D. 76.23 hp
B. 97.33 hp
Diesel Po wer Plant
17.'
S()I>UTlON:
Tlm
Brake Power/Indicated Power
0.85
SOO/Indicated Power
Indicated power = 588.23 Hp
Friction Power = Ind. Power - Brake power
Friction Power = 588.23 - 500
Friction Power = 88.23 Hp
Diesel Power Plant - 31 (ME Bd. June. 1990)
A 4-stroke 394 mm bore and 534 mm stroke single acting diesel engine
with four cylinders is guaranteed to deliver 350 Bhp at 300 rpm. The
engine consumed 66.8 kg/hr of fuel with a heating value of 44,251
K.J/kg. Calculate indicated mean effective pressure in Kpa if
mechanical efficiency is 89%.
A. 450.8 kpa
C. 345.6 kpa
B. 234.5 kpa
D. 643.2 kpa
SOLUTION:
Vo
=
rr/4 D 2 L N c
300
Vo = rr/4 (0.394)2(0.534)(--)( 4)
2x60
0.651 m 3/sec
Brake power = Pmb X V 0
350(0.746) = P mbxO.651
Pmb = 40I.2 Kpa
Tlm = Brake Power/Indicated Power
11m = P mb / P ml
0.89 = 401.2/P m,
P mt = 450.8 Kpa
Vo
C 45.34%
A. 31.53%
B. 27A5%
o
54.23%
SOLUTION:
m,
rn,
rn,
0.26 kg/KW-hr x 150 KW
39 kg/hr
= 0.010833 kg/sec
Brake Power
Brake thermal efficiency
mrQ h
150
Brake thermal efficiency
0.0 I0833( 43,912)
31.53%
Brake thermal efficiency
=
=
Diesel Power Plant - 33 (ME Bd. Oct. 1986)
A four-stroke, 8 cylinder Diesel engine with bore and stroke of 9
inches and 12 inches, respectively and speed of 950 rpm has a brake
mean effective pressure of 164 psi. The specific fuel consumption is
0.39 Ib/bhp-hr and the fuel heating value is 18,500 Btu/lb. Determine
thermal efficiency.
A. 35.27%
C. 38.23%
B. 45.23%
D. 54.23%
=
SOLUTION:
PmbLANc
Bhp
=
A
=
A
=
33,000
n/4 (9/12/
OA418ft 2
Bhp
----"'-''-------
(164x144)(l2/ 12)(0.4418)(950/2)(8)
33,000
Diesel Power Plant - 32 (ME Bd. Apr. 1990)
Bhp
A 305 mm x 457 mm four stroke single acting diesel engine is rated at
150 KW at 260 rpm. Fuel consumption at rated load is 0.26 kg/KW-hr
with a heating value of 43,912 KJ/kg. Calculate brake thermal
efficiency.
rn;
m,
=
=
1,201AHp
0.39 lb/Bhp-hr x 120 I A Bhp
468.546 lbs/hr
174
Diesel Power Plant
Diesel Power Plant
Thermal efficiency
=
120 l~i2545)
468.546( 18,500)
175
C. 242
D. 292
A. 234
B. 873
SOLUTION:
Thermal efficiency = 35.27%
Note: 1 Hp = 2545 Btu/hr
Diesel Power Plant - 34 (ME Bd. Apr. 1988)
A dynamometer test was done for one hour-at steady load on a 6
cylinder diesel engine. It was found to use 42 kg of fuel having Qh =
42,000 J/g. Cylinder is 22.8 em x 27.2 em, 4-cycle type. Speed, 550 rpm
arid dynamometer torque at 27000 kg-em. Determine the brake
thermal efficiency.
C. 35.34%
A. 31.13%
B. 43.22%
D. 45.32%
SOLUTION:
T
T
=
=
27,000 icg-cm x 0.00981 x 1/100
2.6487 KN-m
Brake Power = 2 rc TN
Brake Power = 2rc(2.6487)(550/60)
Brake Power = 152.552 KW
VD
=
VD
=
280
(rc/4)(0.4)z(0.60)(--)(8)
2x60
1.4074 mJ/sec
Indicated Power = (130 x 101.325/14.7)(1 4074)
Indicated Power = 1,261.16 KW
Indicated Power = 1,690.56 Hp
mr = 0.6(1,690.56)
m, = 1,014.3361bslhr
Heat supplied = m-Q, = 1,014.336(19,100)
Heat supplied = 19,373,817.6 Btu/hr
Heat gained by water = 0.25(19,373,817.6)
Heat gained by water = 4,843,454.4 Btu/hr
Heat gained by water = 80,724.24 Btu/min
Heat gained by water = m w cp (t z - t.)
80,724.24 = rn., (1)(40)
m; = 2,018.106 Ib/min
2,018.106 x 7.481
V
62.4
V = 241.9 gal/min
152.554
Brake thermal efficiency
=
Brake thermal efficiency
=
(42/ 3600X42,000)
31.13%
Diesel Power Plant - 35 (ME Bd. Oct. 1992)
A certain diesel engine with the following classification, 8 cylinder, 400
mm x 600 mm, four stroke cycle has a fuel consumption of 0.6 Ibs/hphr based on 19,100 Btu/lb. Engine speed is 280 rpm with an indicated
mean effective pressure 130 psi. If the jacket water carries away an
estimated 25% of the heat supplied, find its capacity(gpm) required if
the allowable rise is 40°F.
Diesel Power Plant - 36
~ME
Bd. Oct. 1985)
When the pressure is 101.325 Kpa and temperature is 27°C, a diesel
engine has the full-throttle characteristics listed:
Brake power = 275 KW
Brake specific fuel consumption = 0.25 kg/KW-hr
Air fuel ratio = 22
Mechanical efficiency = 88%
If the engine is operate at 84.5 Kpa and temperature of 15.5°C, find
brake specific fuel consumption kg/kwh.
A. 0.294
C. 0.423
B. 0.862
D. 1.08
Diesel Power Plant
176
Diesel Power Plant
36
T = S20-(----XI9ln2x3.181)
1000
T = 496.6°R
rBIT
SOLUTION:
At 103.3 Kpa and 2JOC:
Density = P/RT
Density = 101.325/(0.287)(27 + 273)
Density = 1.1765 kg/rrr'
Indicated power = 275/0.88
Indicated power = 312.5 KW
Friction power = 312.5 - 275
Friction power = 37.5 KW(constant)
mn = 0.25(275)
mn = 68.75 kg/hr (constant)
mal = 68.75(22)
3/hr
mal = 1512.59 m
Val = 1512.5/1.1765
Val = 1285.59 m 3/hr (constant)
At 84.5 Kpa and 15.5°C:
Density = P/RT
Density = 84.5/(0.287)(15.5 +273)
Density = 1.0205 kg/rrr'
Indicated power = 312.5(1.0205/1.1765)
Indicated power = 271.06 KW
Brake power = 271.06-37.5
Brake power = 233.56 KW
Brake spec. fuel consumption = 68.75/233.56
Brake spec. fuel consumption = 0.294kgIKWh
Find the power which a 2.5 MW natural gas engine can develop at an
altitude of 1981.2 meters taking into consideration the pressure and
temperature change
C. 2.56 MW
A. 2.34 MW
D. 1.67 MW
B. 1.912 MW
SOLUTION:
B
=
I
29.29---(1981.2x3.281)
1000
B
=
23.42 in Hg
P,
=
1-)
23.42 .. !496.62.5(-j(,/"-)
P,
=
29.92 V 520
1.912 KW
Po
=
P, (--)(
29.92
~ 520
Diesel Power Plant - 38 (ME Rd. June 1990)
A 1119 KW, six cylinder, 589 mm x 711 mm, 225 rpm, four stros,e
diesel engine has a fuel consumption of 0.23 kg/kWh based on 44,fJi,H i
KJ/kg fuel (heating value). A waste heat recovery boiler recovers 3S'''-o
of the exhaust loss. Jacket water 105s are estimated at 30%. Assume
losses due to friction, etc. at 8%. Calculate tilt: quantity of 136 KP3
steam that can be produced in kg/hr, if jacket water at 70"C is used for
boiler feed.
. C. 543.32 kg/hr
A. 439.4 kg/hr
B. 623.4 kg/hr
D. 984.45 kgihr
SOLUTION:
m,
m,
Diesel Power Plant - 37 (ME Rd. Apr. 1992)
J77
0.23(1119)
257.37 kg/hr
Heat generated by file! = m, Qh
Heat generated by fuel = (257.37/3600)(44,099)
Heat generated by fuel > 3152.7 K\V
By heat balance ill the engine:
Heat generated = Jacket water less + Brake power -I- Exhaust
loss + friction and radiation loss
3152.7 cc 0.3(3152.7)-+ 1119+Exhaustloss+O.08(3152.Ti
Exhaust loss = 835.7 KW
Heal utilized in the boiler = 0.35(835.7)
Heat utilized ;n the" 292.495 KW
Using Steam table:
h, '.~ h, at 70°C
hr
292.98 KJ/kg
h,
h,. at i36 Kpa
h,
=
=
"IiR91 K.Lkg
178
ills = mass of steam
ills(hs - hd = heat utilized in the boiler
ills(2689.1 - 292.98) = 292.495
ills = 0.12207 kg/sec
ills = 439.4 kg/hr
Diesel Power Plant - 39 (ME Rd. Oct. 1995)
A 2000 KW diesel engine'unit uses I bbl oil per 525 KWH produced.
Oil is 25°API. Efficiency of generator 93%, mechanical efficiency of
engine 80%. What is the thermal of engine based on indicated
power(%)?
C. 39.60
A. 31.69
D. 35.60
B. 29.47
SOLUTION:
1 bbl
42 gallons
=
141.5
SG =
179
Diesel Power Plant
Diesel Power Plant
Diesel Power Plant - 40 (ME Rd. Oct. 1995)
A waste heat recovery boiler produces 4.8 Mpa(dry saturated) steam
from 104°C feedwater. The boiler receives energy from 5 kg/sec of
954~C dry air. After passing through a waste heat boiler, the
temperature of the air is has been reduce to 343°C. How much steam
in kg is produced l'er second? Note: At 4.80 Mpa dry saturated, h =
2796.
A. 1.3
C. 2.1
B. 0.92
D. 3.4
SOLUTION:
hr
=
(Steam)
4.8MPa
approximate enthalpy offeedwater
T
~
I
i
hr = cp t
h r = 4.187(104)
I ENGINE
BOILER
h r = 435.45 KJlkg
~." .."'. »». .•, ,••.•
x'·' '.' ',' '.' '.' '.'
Heat loss = Heat gain
.... '" '" '"104°C
~t
illg Cp (t 1 - t z) = ills (hs - h F)
\~eeCJ water)
.- .
1343°C
5( 1.0)(954 - 343) = ills(2796.0 - 435.45)
ills = 1.3 kg/sec
fOry
I
25 + 131.5
SG = 0.904
Diesel Power Plant - 41 (ME Rd. Oct. 1995)
w = 0.904(1 kg/Ii)
w = 0.904 kg/li
rn,
= V
m,
m,
=
=
xw
(42 x 3.785)(0.904)
143.724 kg
Qh = 41,130 + 139.6(25)
Qh = 44,620 KJlkg
A diesel electric plant supplies energy for Meralco. During a 24-hour
period, the plant consumed 200 gallons of fuel at 28°C and produced
3930 KW~hr. Industrial fuel used is 28°API and was purchased at
P5.50 per liter at 15.6°C. What is the cost of fuel be to produce one
KW-hr?
C. PI.069
A. Pl.05
D. P1.00
B. PLIO
SOLUTION:
Indicated work
Indicated work
=
=
525/(0.93)(0.80)
705.645 KW-hr
SG 15 6C =
705.645(3600)
TJti
TJti
143.724( 44,620)
39.6%
141.5
131.5+ 28
SG I5 6C = 0.887
Density at 15.6°C = 0.887(1 kg/li)
Density at 15.6°C = 0.887 kg/li
SG Z8°C = 0.887[ I - 0.0007(t - 15.6)]
sat.)
Diesel Power Plant
180
SG 28C = OJ!7<)
Density at 28°('
Density at 28')C
o 879( I kg/li)
0.879 kg/li
V 280C / V \5 6 .~ SG 15 6'C I SG 28 0e
200 I V IWC = 0.887/0.879
cC
V 15 e-c
V 15 6C
198.196 gallons x 3.785 Ii/gal
750.171861i
(5.5)(750.172)
Cost
3930
P1.05/Kw-hr
Cost
Diesel Power Plant
Diesel Power Plant - 43 (ME Bd. Apr. 1995)
.\ supercharged six-cvlinder four stroke cycle diesel engine of 10.48 em
bore and 12.'7 cm stroke has a compression ratio of 15. When It IS
tested on a dynamometer with a 53.34 ern arm at 2500 rpm. the scale
reads 81.65 kg. 2.86 k.g of fuel of 45,822.20 KJ/kg heating value are
burned during a 6 min test, and air metered to the cylinders at the rate
of tl.I 82 kg/sec. Find the brake thermal efficiency.
A. U327
C 0.307
B 0.367
D. 0.357
SOLUTION
T~·
T=
Diesel Power Plant - 42 (ME Bd. Apr. 1996)
A single-acting, four-cylinder, 4 stroke cycle diesel engine with a bore
x stroke of 21.59 x 27.94 cm, operating at 275 rpm, consumes 8.]89
kg/h of fuel whose heating value is 43,961.4 KJ/kg. The indicated mean
effective pressure is 475.7 Kpa. The load on the brake arm, which is
93.98 cm is ] 13.4 kg. What is the brake mean effective pressure in
Kpa?
C. 319.95
D. 645.33
A. 415.20
B. 124.17
VD
=
275
n/4 (0.2159)2 (0.2794)(--)(4)
2x60
3
0.09376 m /sec
2nTN
2n(0.42725 )(2500/60)
111.854 KW
2..86
mj
m,
=
6x60
0.00794 kg/s
I \ 1.854
SOLUTION:
=
(81.65xO.0098J)(05334)
04~72S KN-m
Brake Power
Brake Power
Brake Power
llib
VD
181
lltb
(0.00794)( 45,822.20)
3070%
Diesel Power Plant - 44 (M E Bd. Oct. 1998)
T = (1 13.4 x 0.00981)(0.9398)
T = 1.045 KN-m
Brake power = 2nTN
Brake power = 2n(1.045)(275/60)
Brake power = 30 KW
BP=PmbXV D
30 = P mb X 0.09376
Pmb = 319.97 Kpa
In a double acting, 2 stroke compression ignition engine. 8-q linder,
the diameter of the cylinder is 700 mm, stroke is 1350 mm and the
piston rod diameter is 250 mrn. When running at 108 rpm, the
indicated mean effective pressure above and below the pistons are 5.80
bar and 4.90 bar respectively. Calculate the brake power of the engine
with a mechanical efficiency of 80% in kilowatts.
A. 6050
C. 6010
B. 6030
D. 6070
Vw
Vw
SOLUTION:
At head end:
(11:/4)0 2 LNc
= (n/4)(070)ZC1.35)(108/60X8)
3
= 7.481 m /s
Indicated Power = Pmi X V0
Indicated Power = (5.80 x 100)(7.481)
Indicated Power = 4,339 Kw
At crank end:
2)LNc
Vo = (1t/4)(02 - d
V D = (11:/4)[(0.
(0.25)2]( 1.35)(1 08/60)(8)
V D = 6.527 m 3/s
Indicated Power = (4.9 x 100)(6.527)
Indicated Power = 3,198 Kw
Total Indicated Power = 4,339 + 3,198
Total Indicated Power =' 7,537 Kw
Brake Power = 7,537(0.8)
Brake Power = 6,030 kw
V»
Vo
Vn
=
=
3.0455(7.481)
22.80 gpm
Diesel Power Plant - 46 (ME Bd, Apr. 1997)
=
Brake Power
7i -
A six cylinder, four stroke diesel engine with 76 mm bore x 89 mm
stroke was run in the laboratory at 2000 rpm, when it was found that
the engine torque was 153.5 N-m with all cylinders firing but 123 N-m
when one cylinder was out. The engine consumed 12.2 kg of fuel per
hour with a heating value of 54,120 KJIkg and 252.2 kg of air at 15.6'C
per hour. Determine the indicated power.
A. 32.1kw
C. 23.3kw
B. 38.4 kw
D. 48.3 kw
SOLUTION:
Brake Power = 2 11: r N
Brake Power = 21t (0.1535)(2000/60)
Brake Power = 32.15 kw
Friction power per cylinder = 32.15(5/6) - 21t(0.123)(2000/60) Friction
power per cylinder = 1.031 kw
Indicated Power
Diesel Power Plant - 45 (ME Bd. Apr. 1997)
In a test laboratory, it was found out that of the 80 Bhp developed by
an engine on test, 45 Hp are absorbed by the cooling water that is
pumped through the water jacket and the radiator. The water enters
the top of the radiator at 200°F. At that temperature, enthalpy of the
water is 168.07 Btu/lb. Water leaves the bottom of the radiator at
190°F and with an enthalpy of 158.03 Btu/lb. What is the water flow
rate for a steady-state operation?
C. 23 gpm
A. 25 gpm
D. 24 gpm
B. 20 gpm
SOLUTION:
80 bhp
Q = m., (hi - h2 )
45(42.4) = rn., (168.07 - 158.03)
m; = 190.04 Ib/min
V w ,~ 190.04/62.4
VYo = 30455 ft3/ min
183
Diesel Power Plant
Diesel Power Plant
182
Qh
190°F
200°F
Total friction power
Total friction power
Indicated power
Indicated power
=
=
=
=
1.031 (6)
6.19 kw
32.15 + 6.19
38.34 kw
Brake
I
,
I
Friction Power
Diesel Power Plant - 47
A waste heat recovery boiler receives energy from 10 kg/s of 950'C hot
gases from a diesel engine. The exit temperature of hot gases (c p =
1.0) has been reduced to 250°C. Steam is produced at 5 Mpa (dry
saturated) from 95°C feedwater. How much steam in kg is produced
per hour? At 5 Mpa, h g = 2794.3 KJIkg
A. 10,515
C. 11,055
B. 10,155
D. 11,515
] 8·1
Diesel Power Plan'
Diesel Power Plant
18"
(Exhaust)
d~
SOLUTION:
Using approximate value of hr:
h r = cp t
Engine
h, = 4.187(95)
h, = 397.765 KJ/kg
By heat balance in recovery boiler:
95°C
m c p (t 2 - t 1) = m, (h, - h F )
Feedwateri
(10)(1)(950 - 250) = m, (2794.3 - 397.765)
ills = 2.92 kg/s (3600)
rn, = 10,515 kg/hr
After 15 minute test:
10 kg/s
Boiler
1
rn.
=
fir
=
SG
=
SG
=
v
I
131.5+28
0.887
6.048138
(0.887 x I)
6.82 liters
250°C
V
6.72 x 10'3 (IS x 60)
6.048 kg
141.5
=
Diesel Power Plant - 48
Diesel Power Plant - 49
A SIX cylinder, four stroke cycle diesel engine has an engine thermal
efficiency of 34%. The engine was tested on a dynamometer with a 23
in arm, running at 1800 rpm, the scale reads 210 Ibs. During the 15
min test, the engine uses fuel with 28° API and air metered to the
cylinder at the rate of 0.201 kg/so Find the fuel consumption during
rhe 15 min test.
A. 6.33 liters
C. 6.83 liters
D. 8.97 liters
B. 5.97 liters
A 3 MW diesel electric power plant uses 3700 gallons in a 24 hours
period. Oil is 25°API. What is the thermal efficiency of the engine
based on indicated power if the generator is 90%. efficient and 95°/~)
mechanical efficiency is assumed?
A.55%
C. 65%
B.60%
D.70%
SOLUTION:
SOLUTION:
m,=O.201kg/s ~
F x distance
210 (23!l2)
402.5 ft-lb
T
T
T
27t (402.5)(1800)
p
31,000
137.94 hp
= 102,906 kw
Qh = 41,130 + 139.6(28)
Qh = 45,038,8 KJikg
Brake Power
P
P
=
TIth
=
0.34
mr
=
mrQ h
102.906
=
m r (45,038.8)
6.72 x 10,1 kg/s
,.....
141.5
SG
SG
=
131.5 + 25
0.904
Density of fuel = 0.904(1000)
Density of fuel = 904 kg/m?
rn,
m,
Pi
3700x3.785
=
904(
=
1000
12,661AI kg
=
Pi
)
power input
3,000
(0.9)(0.85)
3921.57 kw
Qh = 41,130+ 139.6(25)
Qh c= 44.620 KJ/kg.
PI
=
1111 -
Ill,
3921.57
12,661.41
(
)(44,620)
24 x 3600
60%
187
Diesel Power Plant
Diesel Power Plant
IXh
SOLUTION:
.
VD
2000
(n/4 )(0.076)L(0.089)(--)( 6)
2 x60
V D = 0.0404 m 3/s
(137.4 ,I 3600)(0.287)( 15.5... 273)
Diesel Power Plant - 50
Va
A single acting, 8 cylinder, 4 stroke cycle diesel engine with a bore to
stroke of 15.24 em x 22.86 ern, operates at 1200 rpm. The load on the
brake arm which is 101.60 ern is 120 kg. What is the brake mean
effective pressure in kpa?
C. 445.5 kpa
A. 4505 kpa
C. 495.5 kpa
B. 455.5 kpa
SOLUTION:
J200
V D = (n/4)(0.1524)2(0.2286)(--)(8)
2 x60
V D = 0.3336 m 3/s
T = Force x distance
T = (120 x 0.00981)(1.016)
T ~~ 1.196 KN-m
P
P
P
=
~c
2rrTN
2 7t (I. 196)( 1200/60)
=
ISO.298Kw
P =, Pmb X V D
150.298 = Pmb (0.3336)
Pmb = 450.5 Kpa
Diesel Power Plant - 51
A six cylinder four stroke diesel engine with 76 mm bore and 89 mm
stroke was run in the laboratory at 2000 rpm. The engine consumed
137.4 kg of air at IS.SoC per hour. Determine the volumetric efficiency
of air only.
A. 69.84%
C. 92.54%
B. 88.32%
D. 77.23%
101.325
0.0312 m3/s
Va
Va
11"
VTJ
11"
=
0.0312/0.0404
11,
=
77.23%
188
189
Gas Turbine Plant
Gas Turbine Plant
] 123
583 =(f p )
GAS TURBINE POWER PLANT
14
= 9.9]9
fp
e
14-1
=
I
] _
Gas Turbine - 1
(9.919)
e = 48.08%
An air-standard Brayton cycle has a pressure ratio of 12. Find the
thermal efficiency of the cycle.
C. 56.32%
A. 34.23%
B. 50.&3%
D. 65.23%
14-1
14
Gas Turbine - 3
SOLUTION:
f"
= 1-
An air-standard Brayton cycle has a pressure ratio of 8. The air
properties at the start of compression are 100 kpa and 25°C. The
maximum allowable temperature is 1l00°C. Determine the net work.
A. 373.24 KJ/kg
C. 321.34 KJ/kg
B. 283.45 KJ/kg
D. 398.23 KJ/kg
k-j
(f p )
k
I
e = I -
--1-4-1
(12)
e = 50.83%
SOLUTION:
14
P2
-=8
PI
T 1 = 25+273
T 1 = 298°K
Gas T.urbine - 2
k-I
An air-standard Brayton cycle has an air leaving the high-temperature
heat exchanger at 850°C and leaving the turbine at 310°C. What is the
thermal efficiency?
C. 48.08%
A. 42.21%
B. 23.34%
D. 56.34%
SOLUTION:
T3
T3
T4
T4
=
=
=
=
850 + 273
I ]23°K
310+273
583 OK
r.
2
c:J0
T
- 3 = (r )
T4
P
1 °
T
_1
2c1311000C
14-1
=(8r1.4
298
T 2 = 539.81°K
•
Wnet
° 4
. 1 °100Kpa,2SoC
s·
k-I
3 850°C
~: =[:J'
T 3 = 1100 + 273
T 3 = 1373°K
Wnet
k-I
~:t:r
r.
° 4 310°C
k
5
14-1
1373
._--=(8) 14
T4
T 4 = 757.95°K
Gas Turbine Plant
190
Gas Turbine Plant
We = m Cp (1'2 - Tj )
We = m(1 )(539.81 - 298)
We 1m = 241.81 KJ/kg
Wr = m Cp (1' J - 1'4)
Wr = m (1)(1373 - 757.95)
W r 1m = 615.05 KJ/kg
Wiler = 615.05 - 241.81
Wnet = :; 73 .24 KJ/kg
191
Gas Turbine - 5
The compressor for an actual gas rurbine requires 300 KJ/kg of work
to quadruple the inlet' pressure. ..,he inlet air temperature is 100°C
Determine the compressor air exit temperature.
A. 234°K
C. 653°K
B. 542°K
D. 673°K
SOLUTION:
Gas Turbine - 4
The air-standard Brayton cycle has a net power output of 100 kw. The
working substance is air, entering the compressor at 30°C, leaving the
high-temperature heat exchanger at 750°C and leaving the turbine at
300°C Determine the mass now rate of air.
A. 1698 kg/hr
C. 1543 kg/hr
B. 1756 kg/hr
. D. 2344 kg/hr
1', = 100+273
1'1 = 373°K
We = m Cp (1'2 - Tj)
We 1m = Cp (1'2' - Tj)
300 = 1 (1'2' - 373)
1'2' = 673°K
SOLUTION:
1'1 = 30 + 273
1'1 = 303°K
1'3 = 750+273
1'3 = 1023°K
1'4 = 300 + 273
1'4 = 573°K
1'2
1'3
Gas Turbine - 6
Power
---
Output
1'1
1'2
100kw
1'4
1023
---303
573
1'2 = 540.96°K
We =, m Cp (1'2 - 1'1)
We = m (1)(540.96 - 303)
We = 237.96 m
W r = mc p (1' J - 1' 4 )
W r = m(1)(1023 - 573)
W r = 450 m
WOe l = Wr - We
100 = 450 m - 237.96 m
m = 0.4716 kg/r (3600)
m = 1697.79 kg/hr
The compressor for an actual gas turbine requires 300 KJ/kg of work
to quadruple the inlet pressure. The inlet air. temperature is 100°C
Determine the compressor efficiency.
A. 34.56%
C. 60.42%
B. 53.23%
D. 76,34%
SOLUTION:
P 2 = 4 PI
rp = P2 I PI
rp = 4P 1IP j
1'3 = 100+273
1'3 = 373 OK
i~OKJlkg
k-l
rp
=
1'2
4
l'
2
1'1
(
=l~] k
14-1
373 =(4)--;-4
P
1
Gas Turbine Plant
192
Gas Turbine Plant
T 2 = 554.27°K
We/m = cp(T 2 - T 1)
We/m = 1(554.27 - 373)
We/m = 181.27 KJ/kg
11e = 181.27/300
l1e = 60.42%
A. 20 kgls
B. 25 kg/s
193
C. 30 kg/s
D. 35 kg/s
SOLUT~ON:
T, = 280 0 K
Gas Turbine - 7
PI = 100 kpa
P 3 = 1000 kpa
T 3 = 1167
W T = 11,190 kw
An ideal gas turbine operates with a pressure ratio of 8:1 and
temperature limits of 20°C and 1000°C. The energy input in the high
temperature heat exchanger is 200 kw. Determine the air flow rate.
A. 560 kg/hr
B. 970 kglhr
W T = ill cp (T 3 - T4 )
P4 = PI = I 00 kpa
r p = P3/P4
rp = 1000 I 100
P31P4 = 10
C. 873 kg/hr
D. 453 kg/hr
SOLUTION:
rp
~
8
T3
P,
T4 =
I
l'P
QA = 200 kw
QA = m cp (T 3 - T 2)
(
T2
I
k-J
( 1--
k-I
1167
'\P2 I k
_--=2,---
k
.
O
3
2
~~~~!<:a
11,190Kw
•
Wnet
•
. 4
1 ·100Kpa, 280 0K
4
-=(10)
14-1
•
S
14
T4
~=lI\ j
T
j
T
T 4 = 604.44°K
14-1
= (8) ~
20+ 273
T 2 = 530.75°K
T] = 1000+273
T 3 = 1273°K
200 = m (I) (1273 - 530.75)
m = 0.26945 (3600)
m = 970 kg/hr
Gas Turbine - 8
In an air-standard Brayton cycle the compressor inlet conditions are
oK.
100 kpa and 280
The turbine inlet conditions are 1000 Kpa and
1167°K. The turbine produces 11,190 kw. Determine the air flow rate.
11,190 = ill (1)(1167 - 604.44)
m = 19.89 kg/s
\.was
1 urbine
-9
An air-standard Brayton cycle has air enter the compressor at 27°C
and 100 kpa, The pressure ratio is 10 and the maximum allowable
temperature in the cycle is 1350°K. Determine the cycle efficiency per
kilogram of air.
A. 48.20%
C. 45.23%
B. 51.34%
D. 65.23%
194
Gas Turbine Plant
Gas Turbine Plant
SOLUTION:
85% and there is a 27 kpa drop between the compressor discharge
and the turbine inlet. Determine thermal efficiency of the cycle per
kilogram of air
T 1 = 27+273
T} = 300 K
I'
P2
= - =
P
p
A 34,23~'o
B 43,23%
10
1
PI
TI
T]
T 2 = 579.6 K
P}=P2=100kpa
T 3 = T",ax = 1350 OK
T
4
1.4-1
1350
--=(10)
1.4
T4
T 4 = 698.8°K
We = ill Cp ( T2 - T 1)
We = 1(1)(579.6-300)
We = -280.9 KJ/kg
Wt = ill Cp (T 3 - T 4 )
WT = 1(1)(1350 - 698.8)
W T = 651.2 KJ/kg
W net = 651.2 - 280.9
W net = 370.3 KJ/kg
QA
=
QA
QA
=
e
=
~
~.
(1~.
_
Wnel
•
0
-
~ 1 3 500 K
_
r p=10
41
7 C, 100Kpa,
S
(T 3 - T 2)
1(1)(1350 - 579.6)
ill C p
770.4
Wnel
=--
Q"
e = 370.41770.4
e = 48.06%
Gas Turbine - 10
An air-standard Brayton cycle has air enter the compressor at 27°C
and 100 kpa. The pressure ratio is 10 and the maximum allowable
oK.
temperature is 13S0
The compressor and turbine efficiencies are
~.
100 kpa
300 K
300(10)14-1/14
=
=
=
T]
k-I
4
C 3050%
D. 47.23%
SOLUTION:
T 2 = T 1 (fpik-I)1k
T 2 = 300 (10)(14-1)/14
2i=[~]k
T
P
195
579.6 K
579,6 - 300
=
0.85
=
1, - 300
628,goK
10(100) - 27
973 kpa
1350(100/973)(14-1)/14
704,3°K
=
0,85
1c
P.
P3
T4
T4
T'lt
.~
=
=
=
=
1350-T4
------'
T4
=
1350 -704.1
801.0 OK
We
We
We
=
mCp(T] - T 1)
=
1(1)(628.9 - 300)
328,9 KJ/kg
W T = m cp(T 4 ' - T,)
W T = I ( I )(1350 - 80 I)
W T = 549 KJ/kg
QA = I1lC p (T 3 - T]')
QA = 1(1)(1350-628,9)
QA = 724.5 KJ/k~
W ne, = 549 - 328.9
W nc ! = 220.1 KJ/kg
e = 220.11724,5
e = 30.38%
=
Gas Turbine - 11 (ME Bd Oct. 1997)
A gas turbine working on an air standard Brayton cycle has air enter
into the compressor at atmospheric condition and 22°C. The pressure
197
Gas Turbine Plant
Gas Turbine Plant
196
ratio is 9 and the maximum temperature in the cycle is 1077"e.
Compute for the cycle efficiency per kg of air in percent.
A. 44.85%
C. 41.65%
B. 43.92%
D. 46.67%
785.9
Combustor Efficiency
Combustor Efficiency
984.7
79.8%
SOLUTION:
Gas Turbine - 13 (ME Rd. Oct. 1995)
Efficiency
1--k-I
(r p)
0K
T
310noc
k
r.=9
Efficiency
4
1 -
14-1
(9)
Efficiency
(1
14
s
46.62%
Air enters the compressor of a gas turbine at 100 Kpa and 300
with
a volume flow rate of 5 m 3/sec. The compressor pressure ratio is 10
and its isentropic efficiency is 85%. At the inlet to the turbine, the
pressure is 950 Kpa and the temperature is 1400°K. The turbine has
an isentropic efficiency of 88% and the exit pressure is 100 Kpa. On
the basis of air standard analysis, what is the thermal efficiency of the
cycle in percent?
C. 31.89
A. 42.06
D. 25.15
B. 60.20
SOLUTION:
Gas Turbine - 12 (ME Rd. Apr. 1996)
In a gas turbine unit, air enters the combustion chamber at 550 kpa,
277°(' and 43 m/s. The products of combustion leave the combustor at
511 kpa, 1004°C and 140 m/s. Liquid fuel enters with a heating value
of 43,000 KJ/kg. For fuel-air ratio of 0.0229, what is the combustor
efficiency of the unit in percent?
Air
SOLUTION:
43kg/s
Fuel
-;OKpa
27°C
Heat supplied by fuel = m, Qh
Heat supplied by fuel = 0.0229(43 00)
Heat supplied by fuel = 984.7 KJ/k air
0,=43,000 K Jlkg
(1.0)(1004-227) + 1/2
Q/m
=
785.9 KJ/kg air
":l
Combustor
T 2 = 579°K
Solving for T 4:
T}/T 4 = (PiP 4)k.11k
1400/T4 = (950/100)14.1/14
Pro uct
5111Kpa
100 4°C
[(140)2 _(43)21
.
1000
J
950Kpa
0K
1400
T
T 4 = 736°K
Q = heat absorbed by fuel
2
Q/m = cp (T 2 - T I) + 1/2 (V 2 - V 12)
Q/m
Solving for the mass flow rate:
PV = mRT
100(5) = m(0.287)(300)
m = 5.81 kg/s
Solving for T 2 :
T 2/T 1 = (p/PI)k.11k
T 2/300 = (10)'4.1/14
We = mC p(T2 - T 1)
We = 5.81(1.0)(579 - 300)
We = 1621 KW
We' = 1621/0.85
We' = 1907 KW
WT = mCp(T} - T 4 )
W T = 5.81(1.0)(1400 -736)
W T = 3858 KW
WT ' = 3858(0.88)
W T ' =3395 KW
100Kpa
s
198
Gas Turbine Plant
WI'<'
c
WT '
-
We'
WN '
3395-1907
W N ' ~ 1488 KW
QA = mCp(T3 - T z)
QA = 5.81(1.0)(1400 - 579)
QA = 4770 KW
Efficiency = 1488/4770
Efficiency = 31.19%
Hydro-electric Power Plant
199
HYDRO-ELECTRIC PLANT
Hydro-electric Plant - 1
A hydro-electric power plant consumes 52,650,000 KW-hrs per
annum. Expected flow is 1665 m3/min and overall efficiency is 65%.
What is the net head?
C. 32 m
A.30m
B. 31 m
D. 34 m
SOLUTION:
Gen. Output
11 net
0.65
=
=
Water Power
52,650,000/8760
------
Water Power
Water Power = 9246.575 KW
Water Power = wQh
9246.575 = 9.81 (1665/60)h
h = 33.966 m
Hydro-electric Plant - 2
In a hydro-electric power the tail water level fixes at 480 m. The net
head is 27 m and head loss is 4% of the gross head. What is the head
water elevation?
A. 500.34 m
C. 456.34 m
B. 508.12 m
D. 567.34 m
SOLUTION:
h
27
hg
~ h~
+ hL
hg + 0.04h g
28.125 m
200
Hydro-electric Power Plant
Hydro-electric Power Plant
h g = H.W.Elev - T.W.r::lev.
28.125 = H.W.Elev - 480
H.W.Elev = 508.J25 ill
h = 33.25 ill
Water Power
Water Power
Water Power
=
=
=
201
wQh
9.81(10)(33.25)
3261.825 KW
Hydro-electric Plant - 3
Hydro-electric Plant - 5
The available flow of water is 25 illJ/sec at 30 m elevation. If a hydroelectric plant is to be installed with turbine efficiency of 85% and
generator efficiency of 90%, what maximum power that the plant
could generate?
A. 4658.5 KW
C. 5628.5 KW
B. 3478.5 KW
D. 4756.5 KW
SOLUTION
The flow of a river is 20 mJ/sec and produces a total brake of 6,000
KW. If it is proposed to install two turbines each has 85% efficiency,
what is the available head?
C. 39 ill
A. 35 ill
D. 36 ill
B.37m
SOLUTION
Water Power
Water Power
Water Power
Gen. Output
Gen. Output
=
=
=
wQh
9.81(25)(30)
7357.5 kw
7357.5 (0.85)(0.9)
5628.49 kw
Water Power
~
6,000/0.85
Water Power
=
7058.82 KW
Water Power
=
wQh
7058.82
h
=
=
9.81(20)h
35.98 m
Hydro-electric Plant - 4
For a proposed hydro-electric plant, the tail water and head water
elevation is 160 m and 195 m, respectively. If available flow is 10
nr'zsec and head loss of 5% of water available head. What is the water
power?
A. 3261.8 KW
C. 5874.5 KW
B. 4254.6 KW
D. 2456.5 KW
SOLUTION:
Hydro-electric Plant - 6
Two turbines generates a total brake power of 5000 KW. If one unit is
thrice the capacity of the other, find the capacity of smaller unit.
A. 1250 KW
C. 2456 KW
R 3450 KW
D 5763 KW
SOLUTION'
hg = H.W.E1ev - T.W.Elcv
hg = 195 - 160
h, ~, 35 ill
h = hg - h L
h = 35 - 0.05(35)
WT
WI + W,
W:= 3W,
5000 c. \\/,t 3\\.
-r-;
WI
=
1250 KW
202
Hydro-electric Power Plant
Hydro-electric Power Plant
203
Hydro-electric Plant - 7
Hydro-electric Plant - 9
In a hydro-electric plant the brake power is 1800 KW running at 450
For a generator running at 300 rpm and 60Hz, find the number of
generator poles.
C. 18 poles
A. 24 poles
B. 8 poles
D. 20 poles
rpm and net head of 30 m. Determine the specific speed of the turbine.
A 6029 rpm
C. 75.29 rpm
R 65.29 rpm
D. 71.29 rpm
SOLUTlON:
SOLUT£ON:
Hp
cc
n,
=
120 f
N
NJHP
N,
N.
1800 x I Hp/O.746KW
24]2.87 Hp
p
h 5/4
=
450J2ill.S7
(30x3.281)5!4
300
N, = 71.29 rpm
Hydro-electric Plant - 8
120(60)
--
p
24 poles
P
Hydro-electric Plant - 10
The specific speed of turbine is 75 rpm and running at 450 rpm. If the
head is 20 m and generator efficient) is 90%, what is the maximum
power delivered by the generator.
A. 4505 KW
C. 650.5 KW
B. 354.5 KW
D. 7805 KW
The penstock of hydro-electric plant is 0.5 x 0.5 m with velocity of 5.5
mlsec has a head of 20 m. What is the output of the turbine if turbine
efficiency is 87%?
A. 845.32 K\V
C. 654.56 KW
B. 789.34 KW
D. 234.56 KW
SOLUTION
SOLUTION:
'"
-
"'s'
N.JHP
h 5/ 4
=
Q
=
A xv
(0.5 x 0.5)(5.5)
Q = 1.375 m 3/sec
450!HP
Generator Output
Generator Output
Q
Water Power = w Q h
Water Power = 981(1.375)(20)
Water Power = 269.775 KW
(%X92 x 0.746)(0.9)
65():,\ KW
Turbine Output - 269.775(0.87)
Turbine Output
234.70 KW
204
Hydro-electric Power Plant
Hydro-electric Power Plant
205
Hydro-electric Plant - 13
Hydro-electric Plant - II
A 410 x 0.510 channel has water velocity of 5 m/sec. If the head is 100
m, what is the annual energy produced if overall efficiency is 70%?
A. 40,456,000 KW-hrs
B. 34,456,345 KW-hrs
C. 60,154,920 KW-hrs
D. 54,867,234 KW-hrs
'" turbine has a mechanical efficiency of 95%, volumetric efficiency of
and total efficiency of 80%. If effective head is 40 rn, find the
total head.
C. 46.72 m
A. 48.72 m
D. 34.72 m
B 4072 m
l)"1('l'"
SOLU'ION
SOLUTION:
Q
Q
Q
TlT = llmllhllv
Ax v
(4 x 0.5)(5)
3
= 10 m (sec
Water Power c , w Q h
Water Power 0' 9.81(10)(100)
Water Power" 9810 KW
Annual Energy Produced = 981O(8760hrs)(0.7)
Annual Energy Produced = 60,154,920 KW-hrs
=
'0.8
0.9S(Ylil)(0.97)
0868
Total head = hllh
Total head ~. 40(0.868)
Total head 'c 3472 m
=
Tlh
=
=
Hydro-electric Plant - 14
Hydro-electric Plant - 12
In Francis turbine, the pressure gage leading to the turbine casing
reads 380 Kpa and center of spiral casing is 3 m above the tailrace. If
velocity of water entering the turbine is 8 rn/sec, what is the net head
of the turbine?
A. 30 m
C. 40 m
B. 35 m
D. 45 m
A Pelton type turbine has 30 m head friction loss of 4.5 m. The
coefficient of friction head loss(from Moorse) is 0.00093 and penstock
length of 80 m. What is the penstock diameter?
A. 1234 mrn
C. 677 mm
B. 3476 mm
D. 1686 rnm
SOLUTION:
y
h
h
SOLUTION:
y
h
I
h
i
h
P
=
=
I
380
8)
-+3+-9.81
2(981)
i
1
I
l
=
v
-+z+w
2g
45 m
y
hl =
~2gh
30 - 4.5
25.5
~2(9.81)(25.5)
2237 m/sec
2 f L y2
gD
2(
0.00(93)(80)(223 7) 2
4.5
9.81 D
D
D
I .(,l~6 IT1
I M{h mITI
206
Hydro-electric Power Plant
Hvdro-electric Power Plant
A. 58.960 KW-hl's
B. 60,960 KW-hl's
Hvdro-electric Plant - 15
From a height of 65 m water flows at the rate of 0.85 m 3/s and is
driving a turbine connected to 160 rpm generator. If frictional torque
is 540 N-m, calculate the turbine brake power.
A. 533 KW
C. 455 KW
B. 677 KW
D. 488 KW
C. 65,960 K).V-hrs
D 70,960 Kw-hrs
SOLUTION:
Plant Capacity
Plant Capacity
8,000(0.88 )(24i
168,960 KW-hrs
Secondary Power
Secondary Power
SOLUTION:
WT
=
168.960 - 110.000
58.960 K\N-Im.
w Q h - Friction power
Hydro-electric Plant - 18 (ME Bd. Oct. 1986)
Friction Power = 2 IT T N
Friction power = 2 IT (0.054)(160/60)
Friction power = 9.047 kw
WT = 9.81(0.85)(65) - 2n(0.54)(160/60)
WT = 532.95 KW
Hydro-electric Plant - 16
What is the turbine output of 100 m head and delivering 1 m
hydro-electric power plant if turbine efficiency is 88%?
A. 863.28 KW
C. 167.34 KW
B. 734.28 KW
D. 492.34 KW
3/sec
in a
At a proposed hydroelectric plant site, the average elevation of
headwater is 600 m, the tailwater elevation is 480 ill. The average
annual water flow is determined to be equal to that volume flowing
through a rectangular channel 4 m wide and 0.5 m deep and average
velocity of 5.5 m/sec. Assuming that the plant will operate 350
days/year, find the annual energy in Kwh that the plant site can
develop if the hydraulic turbine that will be used has an efficiency of
80% and generator efficiency of 92 % • Consider a headwork loss of 4%
of the available head.
A. 76,854,851
C. 75,234,45'5
B. 65,234,556
D. 82,456,677
SOLUTION:
SOLUTION:
WT
=
207
(wQh)(llT)
WT
=
[(9.81)(I)(l00)J(0.88)
WT
=
863.28 KW
Hydro-electric Plant - 17
In an 8,000 KW hydro-electric plant the over-all efficiency is 88% and
the actual power received by the customer is 110,000 KW-hrs for that
day. What is the secondary power could this plant deliver during the
entire day?
hg
hg
hg
H.W.Elev - T.W.Elev
600 - 480
= 120 m
h = 120 - 0.04(120)
h = 115.2 In
Q = Axv
Q = 4(0.5)(5.5)
Q = II mJ/sec
Generator Output = (w Q h)IlTf)C;
Generator Output = (9.81 x I i x 1152)(OR)(092)
Generator Output = 9,149.39 KW
Annual Energy
9,149.39(24)(350)
. Annual l.ncrgy = 76,854,851 KW-hrs
=
=
-c;
208
Hydro-electric Power Plant
v
Hydro-electric Plant - 19 (ME Rd. Apr. 1992)
=
2.368 m/sec
2f Lv
hL =
In Maria Cristina Hydroelectric Project, the available head is 140 10;
the water flow is one cubic meter per second; efficiency of the turbine
is 95%; efficiency of the generator is 95%, three phase, 60 cycles,
voltage delivered is 4160 V. Determine yearly income of the plant if the
cost of the generated electric energy is PO.60 per Kwh.
C. P6.514,778
A. P5,234,567
B. P7,385,678
O. P9,354,677
o
=
2( 0.00093)(80)(22.368)
9.810
1.686 m
Axv
Q =
Q = [n/4 (1.686)2](22.368)
Q = 49.94 rrr'zsec
Power = w Q h = 9.81(49.94)(25.5)
Power
Annual Energy = 1239.49(8760)
Annual Energy = 10,857,963.06 KW-hrs
Yearly Income = 10,857,963 x PO.60
Yearly Income = P6,514,777.80
2
gO
4.5 =
SOLUTION:
Generator Output = (w Q h)llT llG
Generator Output = (9.81 x 1 x 140)(0.95)(0.95)
Generator Output = 1239.49 KW
209
Hydro-electric Power Plant
=
12,492.74 kw
Hydro-electric Plant - 21 (ME Rd. Apr. 1989)
3/sec
From a height of 65 10, water flows at the rate of 0.85 m
and is
driving a water turbine connected to an electric generator revolving at
160 rpm. Ca lculate the power developed by the turbine in KW if the
total resisting torque due to friction is 540 N.m and the velocity of the
water leaving the turbine blades is 4.75 m/sec.
A. 753
C. 635
B. 523
O. 833
Hydro-electric Plant - 20 (ME Rd. Apr. 1987)
A Pelton type turbine was installed 30 10 below the head gate of the
penstock. The head loss due to friction is 15 percent of the given
elevation. The length of penstock is 80 10 and coefficient of friction is
0.00093. Determine the power output in KW.
A. 12,273
C. 12,345
B. 13.2:3'1
O. 12,493
SOLUTION:
0.15(30)
= 4.5 m
h = 30 - 4.5
h = 25.5 m
v = ~2gh
hL
hL
v
=
=
~2(9.81)(25.5)
1~:-
r
Om
L JQ
SOLUTION:
(4.75)2
65---2(9.81)
h = 63.85 m
Water Power = w Q h
Water Power = 9.81(0.85)(63.85)
Water Power = 532.41 kw
h
~~
Friction Power
Friction Power
Friction Power
Turbine Power
Turbine Power
r;
~~
=
=
=
2nTN
2 rt (0.54 )(160/60)
9 KW
532.4 - 9
532.40 kw
210
211
Hydro-electric Power Plant
Hydro-electric Power Plant
Hydro-electric Pian! - 22 (ME Bd, Oct. I 985)
Hydro-electric Plant - 23 (ME Bd, Apr. 1990)
A proposed hydro-electric power plant has the following data:
Elevation of normal headwater surface = l 94 m
Elevation of normal tailwater surface = 60 m
Loss of head due to friction = 6.5 m
Turbine discharge at full gate opening = 5 m 3/sec
Turbine efficiency at rated capacity = 90%
Turbine is to be connected to a 60 cycle AC generator.
Find the num ber of poles of the generator
A. 6
C. 8'
B. 10
D. 12
A Francis turbine is installed with a vertical draft tube. The top of the
draft is 1.0 m below the center line of spiral casing while the tail race
water level) is 3.0 m from the top of the draft tube. There is no velocity
of whirl at the top or bottom of the draft tube and leakage are
negligible. The elevation of water from the reservoir to the center line
of the turbine spiral casing is 50 m, and water velocity at the inlet is 5
m/sec. Discharge is 2.5 m 3/sec, hydraulic efficiency is 86%, overall
efficiency of 83%. Determine the reading of a pressure gage(in psi) if
one is installed at the penstock just before the water inlet to the
turbine.
C. 56.34
A. 69.35
B. 74.33
D. 92.45
SOLUTION:
h
h
h
=
=
=
SOLUTION:
(194 - 60) - 6.5
127.5mx3.281
418.2 feet
Brake Power
Brake Power
Brake Power
=
=
=
P
h
33
718.5
1I,!-
i
1m+=:
<;2
50 :.' - ' - + --.-9.81 2(9.81)
P
P
=
=
~~r;;y
'= s
- om
r: _
I
2g
P
i)lt,
4--
'"_~._
_
-
478 Kpa x 14.7/101.325
69.35 psi
NJHP
h 5/4
Hydro-electric Plant - 24 (ME Bd. Apr. 1990)
N.J5628.5 I 0.746
_.~~~--
(418.2)5/4
H.W. Elev. = 194m
N = 718.5 rpm
N
yl
-+W
(w Q h}rh
(9.81 x 5 x 127.5)(09)
5,628.5 KW
From MRII charts and tables:
For 418.2 feet net head, using Francis Turbine
N, = 33
Ns
=
Brake
Power
120f
p
120(60)
=
P
P = 10.02 poles say 10 poles
T.W. Elev.
=60m
A Francis turbine is installed with a vertical draft tube. The top of the
draft is 1.0 ill .below the center line of spiral casing while the tail race
water level) is 3.0 m from the top of the draft tube. There is no velocity
of whirl at the top or bottom of the draft tube and leakage are
negligible. The elevation of water from the reservoir to the center line
of the turbine spiral casing is 50 m, and water velocity at the inlet is 5
m/sec. Discharge is 2.5 m 3/sec, hydraulic efficiency is 86%, overall
efficiency of 83%. Determine the mechanical efficiency of the turbine
A 76.23%
C. 96.51 %
B. 83.45%
D. 64.34%
Hydro-electric Power Plant
212
213
Hydro-electric Power Plant
SULUTION
Hvdro-electric Plant - 26 (ME Rd. Oct. 1989)
A hydro-electric plant has a 20 MW generator with an efficiency of
96%. The generator is directly coupled to a vertical Francis type
hydraulic turbine having an efficiency of 80%. The total gross head on
the turbine is 150 m while the loss of head due to friction is 4% of the
gross head. The runaway speed is not to exceed 750 rpm. Determine
the flow of water through the turbine in cfs,
A. 651
C. 763
B. 534
D. 827
111
llh 11m 11v
where: 11v = 1.0 (ifnot given)
0.83 = 0.86( 11rn)( 1.0)
11m = 96.51%
Hydro-electric Plant - 25 (ME Rd. Apr. 1984)
SOLUTI0N:
A remote community in Mountain Province plans to put up a small
h = hg - h L
h = 150 - 0.04(150)
h = 144 m
Generator Output = (w Q h) 11T 11G
20,000 = (9.81 x Q x 144)(0.8)(0.96)
3
Q = 18.435 m3/sec x 35.3 I ft 1l m'
Q = 650.94 cfs
hydro-electric plant to service six closely -located barangays estimated
to consume 52,650,000 KW-hrs per annum. Expected flow of water is
1665 m 3/min. The most favorable location for the plant fixes the tail
water level at 480 m. The manufacturer of turbine generator set have
indicated the following performance data: turbine efficiency - 87%;
generator efficiency is 92%; loss in head work not exceed 3.8% of
available head. In order to pinpoint the most suitable area for the
dam, determine the head water elevation
A.234m
C. 345 m
C. 842 m
D. 509 m
Hydro-electric Plant - 27 (ME Rd. Oct. 1989)
SOLUTION:
Annual Energy Output = (w Q h)l1T11o x 8760
52,650,000 = (9.81 x 1665/60 x h)(0.87)(0.92)(8760)
h = 27.583 m
h = hg - h L
h = hg - 0.038h g
h = 0.962 h g
27.583 = 0.962 hg
h g = 28.67 m
Head water elevation
Head water elevation
A hydro-electric plant has a 20 MW generator with an efficiency of
96%. The generator is directly coupled to a vertical Francis type
hydraulic turbine having an efficiency of 80%. The total gross head on
the turbine is 150 m while the loss of head due to friction is 4% of the
gross head. The runaway speed is not to exceed 750 rpm. Determine
the rated speed of the turbine.
C. 400
A. 300
D.600
B. 500
SOLUTION:
480 + 28.67
508.67 m
h = h g - hL
h = 150 - 0.04(150)
h = 144 m
From MRII charts and tables, for 144 m(472 ft) head, is
2J4
Hydro-electric Power Plant
within the range
or francis
turbine with N,
Hydro-electric Power Plant
29
20,000
HP
096(0746)
27,927 hp
HP
Ns =
29
NM
h 5/4
NJ27j27
=
215
(472)5/4
N = 382 rpm, use 400 rpm (standard)
Hydro-electric Plant - 29 (ME Bd. Oct. 1995)
A hydroelectric generating unit station is supplied from a reservoir of
capacity 6,000.000 m'' at a head of 170 m. Assume hydraulic efficiency
of 80% and electrical efficiency 90%. The fall in the reservoir level
after a load of 15 MW has been supplied for 3 hours, if the area of the
reservoir is 2.5 sq. Km is closet to:
A S39cm
C 5.98 em
B. 4.32 em
D. 4.83 em
SOLUTION
A = 2.5 krn
r-.
-:
Ac :2.5 x 10 m '
Generator Output = (w 0 h)11 r 11G
15,000' (981 x Q x 170)(0.80)(U90)
Q .zz: 12.492 mJisec
A her 3 hours,
Q ~ I:2A92(3x3600)
C) = 134.914 m 3
Volume =c Area x Height
134,914 = (2.5 x 10") H
H = 0.0539 m
H = 5.39 ern
Hydro-electric Plant - 28 (ME Bd. Oct. 1995)
Water flows steadily with a velocity of 3.05 mls in a horizontal pipe
having a diameter of 15.24 em. At one section of the pipe, the
temperature and pressure of the water are 21°C and 689.3 Kpa,
respectively. At a distance of 304.8 m downstream, the pressure is
516.9 Kpa. What is the friction factor?
A. 0.134
e.0.0189
B. 0.0050
D. 0.641
SOLUTiON:
hL
p - P
~
Hydro-electric Plant - 30 (M E Bd. Oct. 1995)
_d_ _s
~
w
689.3-516.9
hL =
9.81
h L=17.574m
f Lv
hL =
2
2gD
A pelton wheel is to be designed to run at 300 rpm under an effective
head of 150 m. The ratio of the nozzle diameter of the pitch circle is
1/12. Assuming efficiency of 84%, what is the size of the wheel in
meters. Assume a speed ratio of 0.45.
A. 1.05
C. 1.55
B 2.00
D. 2.86
SOLUTION:
q)=
7.574
f
=
~2gh
[(3048)(3.05/
=
2(9.81)(0.1524)
0.0 185
nDN
~-~
OA'i
I)
n D (300 / 60)
)2(9.81)(150)
I"''i In
216
---_.-
--
Hydro-electric Plant - 31 (ME
217
Hydro-electric Power Plant
Hydro-electric Power Plant
na. Oct. 1995)
275 (035)2
-+--9.81 2(9.81)
h
A Francis turbine is installed with a vertical draft tube. The total head
to the center of the spiral casing at the inlet is 38 m and velocity of
water at the inlet is 5 m/sec, The discharge is 2.1 m3isec. The hydraulic
efficiency is 0.87 and overall efficiency is 0.84. The top of draft tube is
1 m( water) below the center line of the spiral casing while the tailrace
(water) level is 3m from the top of the draft tube. Neglected velocities
of whirl and leakage losses. What is the total head on the turbine in
meters?
A 34.72.
C.,55.20
B. 43.27
D 48.12
h
=
28.039 m
~iJJI'
nON
<1>=
J2gh
1[
Turbine
Tail race
(0.55)(520/60)
<1> = ----;=====
J2(9.81)(28.039)
«1> = 0.638
SOLUTION:
Hydro-electric Plant - 33 (ME Bd, Apr. 1998)
h
=
h
=
total head
p
V
- + Z+
W
A hydraulic turbine receives water from a reservoir at an elevation of
LOO meters above it. What is the minimum water now in kg/s to
produce a steady turbine output of 50 Mw?
A. 50,247
C. 50,672
B. 50,968
D. 50,465
B
2g
5 2 - 02
38 + (1 + 3) + 2(9.81)
h
h
2_V 2
A
=
43.27 m
Hydro-electric Plant - 32 (ME Bd, Apr. 1998)
SOLUTION:
Water Power
wQh
Reservoir
.~.
A hydro-electric plant discharges water at the rate of 0.75 m3/s and
enters the turbine at 0.35 mps with a pressure of 275 Kpa. Runner
inside diameter is 550 mm, speed is 520 rpm and the turbine efficiency
is 88%. Find the turbine speed factor.
A. 0.638
C. 0.368
B. 0.386
D. 0.836
50Mw
100m
Q = 50.968 m 3/s
50.968 (lOOJ)
m
=
m
= 50,968 kg/s
SOLUTION:
h
P
v2
-+W 2g
Hydro-electric Plant - 34 (ME Bd, Apr. 1997)
A hydro-electric plant having 50 sq. km reservoir area and 100 m head
is used to generate power. The energy utilized by the consumers
whose load is connected to the power plant during a five-hour period is
218
Hydro-dearie Power Plant
Chimney
13.5 X 10 kwh. The overall generation efficiency is 75%. Find the fall
in the height of water in the reservoir after the 5-hour period.
A. 2.13 m
C. 3.21 m
B. 1.32 m
D. 0.53 m
CHIMNEY
219
6
SOLUTION:
Chimney - I
Energy Output = Power x time
Energy Output = (w Q h)11 x time
6
13.5 x 10 = 9.81(Q)(100)(0.75)(5)
Q = 3669.725 m 3/s
A boiler uses 2500 kg of coal per hour and air required for combustion
is 16 kg per kg coal. If ash loss is 10%, determine the mass of gas
entering the chimney.
C. 85,452 kg/hr
s: 42,250 kg/hr
D. 33,800 kg/hr
B 78,300 kg/hr
h I.:. .... _.. _.. __ ..
Reservoir
Volume after 5 hrs = 3669.725(5 x 3600)
Volume after 5 hrs = 66,055,050 m '
SOLUT10N:
rng
Volume = A x height
66,055,050 = (50 x 106 ) h
h = 1.321 rn
rna
1-
m, - mash
A/F= rna / m16 =- rna / m.
m a = 16 rnf
Hydro-electric Plant 35 - (Math-ME Bd. Oct. 1997)
mg
=
l6mr+mr-O.lInf
mg
=
16.9 m,
16.9(2500)
42,250 kg/hr
rn g =
A reaction turbine develops 500 bhp. Flow through the turbine is 50
cfs, Water enters at 20 fps with a 100 ft pressure head. The elevation
of the turbine above the tailwater level is 10 ft. Find the effective head.
A. 130 ft
C. 110 ft
B. 120 ft
D. 116.2 ft
mg
=
Chimney - 2
The gas density of chimney is 0.75 kg/m'' and air density of 1.15 kg/rn ',
. If the driving pressure is 0.25 Kpa, determine the height of chimney,
A. 54.6 m
C. 74.6 In
B. 63.7 m
D. 68.5 In
SOLUTION:
h
h
=
P
V 2_V 2
-+z+ A
B
W
2g
100+ 10 + (20)2_ 0 2
2(32.2)
h
116.21 ft
10ft
SOLUT10N:
h.,
=
H(d. - dg )
0.25
=
H
63.7l
=
H( 1.15 - 0.75)(0.00981)
In
220
22\
( 'h nnncy
Chimney
Chimney - 3
The actual velocity of gas entering in a chimney is 8 m/sec, The gas
temperature is 25°C and pressure of 98 Kpa with a gas constant of
0.287 KJ/kg_°K. Determine the chimney diameter if mass of gas is
50,000 kg/hr.
A. 1.57 m
C. 3.56 m
B. 1.81 m
O. 1.39m
rn, TIn,
InA~Jl t In l',
46,500 . 3000
o 11(3000) +- mg
m~ = 49,170 k~hr
PV = mR'1
98.2(V g)
(49.170/3600)(0.277)(285
V g = 21.498 m 'zsec
Let 0 = diameter of chimney
Q = Axv
Q = ( rc/4 02)V
21.498 = (rc/4 0 2)(7 5)
D
SOLUTION:
=
-t- 273)
19\ m
Pg V s = mg Rg T g
98(V g) = (50,000/3600)(0.278)(25 + 273)
Vg
=
A power plant situated at an altitude having an ambient air of 96.53
Kpa and 23.88°C. Flue gases at a rate of 5.0 kg/sec enter the stack at
200°C and leaves at 160°C. The flue gases graVimetric analysis are
18% COz, 7% O, and 75% N z. Calculate the height of stack necessary
for a- driving pressure of 0.20 Kpa.
12.12m3/sec
Vg = A x v
11.74 = rc/4 0 2 (8)
D
=
Chimney - 5 (ME Bd, Apr. 1990)
SOLUTION:
1.39 m
Solving for the molecular weight and gas constant of the flue gas:
CO 2
Chimney - 4 (ME Bd. Apr. 1981)
O2
A coal fired steam boiler uses 3000 kg of coal per hour. Air required
for combustion is 15.5 kg per kg of coal at barometric pressure of 98.2
Kpa. The flue gas has temperature of 28SoC and an average molecular
weight of ~O. Assuming an ash loss of 11 % and allowable gas velocity
of 7.S mlsec, find the diameter of chimney.
SOLUTION:
Amount of air required = 15.5(3000)
Amount of air required = 46,500 kglhr
R = 8.3141M
R = 8.314/30
R = 0.277 KJlkgOK
By mass balance:
mf
FURNACE
m(ash)=11 %mf
•
o
285°C
N2
18%
7%
75%
Mg = 110.03306
M, = 30.25
R g = 8.314/30.25
R g = 0.275
T g = (200 + 160)/2
T g = 180°C
d g = P/RT
96.53
dg =
(0.275)(180 + 273)
d g = 0.775 kg/m'
d. = P/RT
0.18/44
0.07/32
0.75/28
=
=
=
0.00409
0.00219
0.02678
0.03306
-,"'-:.
Chimney
Chimney
l)().'i 3
h., = 26.30
do
(02S7)(23.88 + 273)
d•.= I.! 33 kg/m'
Draft
lIZ d. - d g )
0.20 = H(1I33 - 0.775)(0.00981)
H = 56.95 m
0
Chimney - 6 (ME Bd. Apr. 1990)
A power plant situated at an altitude having .an ambient air of 96.53
a
Kpa and 23.88 C. Flue gases at a rate of 5.0 kg/sec enter the stack at
a
200 C and leaves at 160 a C. The flue gases gravimetric analysis are
18% CO z, 7% O 2 and 75% N z• Calculate the diameter of stack in
meters for a driving pressure of 0.20 Kpa.
SOLUTTON:
Solving for the molecular weight and gas constant of the flue gas:
CO 2
18%
7%
75%
O2
N2
0.18/44 = 0.00409
0.07/32 = 0.00219
0.75/28 = 0.02678
0.03306
01
223
of flue gas
v. = 'ihl9
'")
-\..8\ V'\ 76
- .~.
v
cc
22.716 m/sec
Actual velocity
40% v
Actual velocity
0.40(22.716)
Actual velocity
9.1 In/sec
Q = Ax v
2)(91)
(5/0.775) = (11'/4 D
D = 0.95 m
Chimney - 7 (ME Ed. Apr. 1995)
A steam generator with economizer and air hearer has an O'\"IC)'g H draft
loss of 21. 78 em of water. If the stack gases are at l7i"C and jf the
atmosphere is at 101.3 Kpa and 26°C, what theoretical height of stack
in meters is needed when no draft fan arc used? Assume that th~' gas
constant for the flue gases is the same as that for air.
A. 565
C. 545
B. 535
D. :'50
SOLUTION:
M g = 1/0.03306
M g = 30.25
R g = 8.314/30.25
R, = 0.275
200+ 160
YV
:.-=-
P'"E.T
101.325
do
Tg
Z0.287)(26 + 273)
d. = I.! 80 kgn-: 3
2
T g = 180°C
dg = P/RT
dg
96.53
dg
(0.275)(180 + 273)
d g ~~ 0.775 kg/rn '
v = theoretical velocity
v
= ~2gh",
hw
=
0.20
0.775(OOO9RI)
dg
101.3
(0.28 7 )0 T'
0.784 kg im3
-i-
2n)
Draft
O.2178( lOOO)
Draft == 217.8 Kg/rn'
Draft = H(d a - dwl
217.R
H(Ll8 - 0.78"+)
5~(\ m
!1
224
C. 40
D. 56
A.46
B. 50
Chimney - 8 (ME Bd, Oct. t 995)
If the actual draft required for a furnace is 6.239 em of water and the
frictional losses in the stack are 15% of the theoretical draft, calculate
the required stack height in meters. Assume that the flue gas have an
average temperature of 149°C and molecular weight of 30. Assume air
temperature of 21°C
C. 220
A. 215
D. 210
B. 230
225
Chimney
Chimney
SOLUTION:
t o =12°C
m,~9,OOOkg/hr
i
I
I
i
I
IH
I
---+
SOLUTION
hw
"
h,
=
total draft
6.239 ~. O.ISh"
Ill' =, 7.34 em water
h; = 00734(9.81)
h, = 0.72 Kpa
d'l -c. PRT
101325
d,
d,
dg
=
~
(0.287)(21 +273)
1.2 kg/m '
P/RT
d =
101325
g
(8.314/30)(149+273)
d g = 0.867 kg/m '
h",= H(do - d g)
0.72 = H(1.2 - 0.867)(0.00981)
H ~c 220 m
Chimney - 9 (ME Bd, Apr. 1998)
A steam boiler plant consumes 9,000 kg of coal per hour and produces
20 kg of dq flue gases per 'kg of coal fired. Outside air temperature is
32"C, average temperature of the flue l::as entering the chimney is
343°C and average temperature of dry flue gas in the chimney is
260"C. The gage fluid density is 994.78 kg per m' and the theoretical
draft of 2.286 em of H 20 at the chimney base is needed when the
barometric pressure is 760 mrn Hg. P .terrnine the height of the
chimney,
h; = hxw
h; = (0.02286)(994.78)
h; = 22.74 kg/rrr'
p
d =-•
RaTa
P = 760 mm Hg
P = 101.325 kpa
101.325
da - (0.287)(32 + 273)
d, = 1.157 kg/rrr'
P
ds =
R g Tg
ds
=
101.325
--
(0.287)(260 + 273)
d g = 0.662 kg/rrr'
h... = H(d a - d g )
22.74 = H(!.!57 - 0.662)
H = 46m
226
Machine Foundation
Machine Foundation
227
MACHINE FOUNDATION
Machine Foundation - 5
The machine foundation must have a factor of safety of
A.3
C.5
B.4
D.6
Machine Foundation - I
of sufficient
mass and base area to prevent or minimize the transmission of
vibrations.
A. steel foundations
C. solid foundations
B wooden foundations
D. soil foundations
_
All heavy machinery should be supported on
The answer is: C
IS:
C
Machine Foundation- 6
Foundations should be isolated from floor slabs or building footings at
least _ _ mm around its perimeter to eliminate transmission of
vibration.
A. 20
C. 30
B 25
D. 35
Machine Foundation - 2
Foundations mass should be from
machinery it is supposed to support.
A. 3 to 5
B. 5 to 7
The answer
times the weight of the
The answer is: B
C. 4 to 6
D. I to 2
The answer is: A
Machine Foundation - 7
Machine Foundation - 3
Foundations are preferably built of concrete in the proportion of
If the unbalanced inertia forces produced by the machine can be
calculated, a mass weight equal to
times the forces should be
used to dampen vibration.
A. 2 to 4
C. 10 to 20
B. 5 to 6
D.l0tol5
A.I:2:3
B. 1:3:3
C. 1:2:5
D. 1:2:4
The answer is: D
The answer is: C
Machine Foundation - 8
Machine Foundation - 4
For stability, the total combined engine, driven equipment, and
foundation center of gravity must be kept _ _ the foundation's top.
C. I m
A. above
B. below
D. none of these
The answer is: B
The machine should not be placed on the foundation until 7 days have
elapsed or operated until another "_ days have passed.
A 6
B.7
The answer is: B
C. 8
D. 10
"'2X
111acllllll'
Machine Foundation
loundation
Machine Foundation - 9
Machine Foundation - 13
Concrete foundations should have steel bar reinforcements placed
both vertically and horizontally, to avoid
.
A. damaging
C. superheating
8 thermal cracking
D. superposing
No foundation bolts shall be less than
A. 10
B. 12
22Y
mm diameter
C. 14
D. 16
The answer is: B
The answer is 8
Machine Foundation - 14
Machine Foundation - 10
The weight of reinforced steel in a foundation should be from
the weight of the foundation.
A. l/2% to 1%
C. 2% to 3%
B. 1% to 2%
D. 3% to 4%
of
Machine should be leveled by driving wedges between the machine's
base and concrete foundation and with the aid of a _ _
A. hose and water
C. spirit level
B. try square
D. level hose
The answer is: C
The answer is: A
Machine Foundation - 15
Machine Foundation - 11
Foundation bolts of specified size should be used and surrounded by a
pipe sleeve with an inside diameter of at least _ _ times the diameter
of the anchor bolt.
A. I
C.3
B. 2
0.4
Grouting all spaces under the machine bed with a thin mixture of one
part cement and _ _ part sand.
A. I
C. 3
D. 4
B. 2
The answer is: A
The answer is: C
Machine Foundation - 16
Machine Foundation - 12
Foundation bolts length should be at least
times the diameter of
the anchor bolt.
A. 12
C. 16
B. 14
D. 18
The answer is: 0
For foundation of stacks, the maximum pressure on the soil is _ _ to
the pressure due to the weight and the wind movement.
A. more than
C. equal
B. less than
D. none of the above
The answer is: C
230
Machine Foundation
Machine Foundation
231
Machine Foundation - 17
Machine Foundation - 22
Guyed stacks seldom exceed 1.83 m diameter and _ _ m high.
A. 32.45
C. 43.23
B. 36.34
D. 30.48
The answer is: D
The steam turbine foundation should be designed to support the
machine load plus _ _ for impact, condenser load, floor loads and
dead loads.
C. 25%
A.10%
D. 50%·
B. 15%
The answer is: C
Machine Foundation - 18
The angle betwee.i the stacks and the guy wire is usually _ _ degrees.
A. 45
C. 75
B. 60
D. 90
The answer is: B
Machine Foundation - 19
The angle between wires in a set is _ _ degrees for a set of three.
A. 90
C. 110
B. 100
D. 120
The answer is: D
Machine Foundation - 23
For diesel engines foundation the concrete mixture must be !:2:4, and
the broken stone or gravel must have a size of
mm maximum.
A. 30
C. 50
B. 40
D. 60
The answer is: C
Machine Foundation - 24
In pouring a concrete mixture for foundation of diesel engine, it should
be poured _ _ time
A. 1
B. 2
Machine Foundation - 20
The maximum unit pressure of turbine and generator on the
reinforced concrete should not exceed _ _ kg/cm 2 •
A. 17.62
C. 19.34
B. 15.34
D. 21.34
The answer is: A
Machine Foundation - 21
For steam turbine foundations the concrete mixture should be
A. 1:2:3
C. 1:3:5
B. 1:2:4
D. 1:2:2
The answer is: B
.
C.3
D.4
The answer is: A
Machine Foundation - 25
After pouring the foundation of diesel engine, the tops should be
covered and wet down twice daily until the forms are removed at the
end of _ _ day.
A. third or fourthC. fourth or fifth
B. second or third
D. none of these.
The answer is: A
232
233
Machine Foundation
Machine Foundation
Machine Foundation - 26
Machine Foundations - 30 (ME Bd. Oct. 1995)
The diesel engine should not be place on the foundation until 10 days
have elapsed, nor operated until after another _ _ days.
A. 7
C. 9
B. 8
D. 10
The answer is: 0
What is the required base area of the foundation to support an engine
with specified speed of 1200 rpm and weight of 9,000 kg? Assume
bearing capacity of soil as 47.867 kpa. Use e = 0.11.
o
2
A 5.57 m"
C. 7.75 m
2
B. 8.87 m 2
D. 10.5 m
SOLUTION:
Machine Foundation - 27
The safe soil bearing pressure of diesel engine foundations is
A. 4,600 kg/ern'
C. 5,633 kg/ern"
D. 2600 kg/ern"
B. 4,890 kg/ern"
WF =
weight of foundation
WF
exWex.,fN
0.1 1(9,000).J1200
34,295 kg
=
WF
Wr
The answer is: B
Sb =
We + W F
A
Machine Foundation - 28
(9,000+ 34,295)(0.00981)
47.867
The foundation depth may be taken as a good practical rule, to be
_ _ times the engine stroke.
A. 2.3 to 4.5
C. 3.2 to 5.2
D. 5.2 to 6.2
B. 3.2 to 4.2
The answer is: B
A
=
8.87 m2
A
Machine Foundations - 31
Machine Foundation - 29
Anchor bolts should be embedded
the bolt diameter.
A. 10
B. 20
The answer is: C
;'1
concrete of at least
C. 30
0.40
times
A foundation measures 3 m x 4 m x 5 m. Find the yd:' of stone needed
for a concrete mixture of 1:2:4.
C. 78
A. 65
D. 69
B. 56
SOLUTION:
v
V
V
=
3x4x5
60 m3
(3.281)3
60---(3)3
V ~ 78.48 yd'
For 1 yu 3 of concrete, it needs 0.88 yd'' of stone.
234
Machine Foundation
Therefore.
Machine Foundation
SOil :T10N:
Volume of stone
Volume of stone
088
yd 'stone
78 .48 _vd ' concrete .. _. - _.J -yd concrete
A =
69 yd' of stone
A =
A
V
V
V
V
Machine Foundations - 32
A foundation measures 10 ft x I2 ft x 15 ft. Find the number of sacks
of cement needed for 1:2:4 mixture.
A 300
C. 350
B. 400
D. 450
=
=
=
=
=
(b
l+b 2)h
2
(2 + 3)( 1.5)
2
3.75 m 2
A L
3.75(4)
15 rrr'
19.622 yd 3
For I yd'' concrete, it needs 0.44 yd' of sand. Therefore,
SOLUTION:
Volume of sand
Volume of sand
V
235
=
10 x 12 x 15
1 d3
v
0.44(19.622)
8.633 yd 3 of sand
1800 ft' (---t-)
3 ft J
Machine Foundations - 34
j
v
=
66.67 yd' of concrete
For every 1 yd ' concrete, it needs 6 sacks of cement.
Therefore,
No. of sacks = 6 (66.67)
No. of sacks = 400 sacks
Machine Foundations - 33
A machine foundation has a trapezoidal cross-section with bases of 2
m and 3 m. The height is 1.5 m and foundation length of 4 m. Find
J
the required yd of sand to be used for 1:2:4 mixture.
A. 7.33
C. 9.34
B 12.4
D. 863
A foundation has a weight equal to 4 times the weight of the engine.
The safe bearing pressure of soil is 60 kpa and the foundation has a
base dimension of 2 m x 4 m. Find the maximum weight of the engine
to be mounted.
A 8,234 kg
C. 9,736 kg
B 6,455. kg
D. 12,344 kg
SOLUTION:
F
F
PA
60 (2 x 4)
F = 480 KN
=
=
IF, -;-; 0
We +- We
=
F
but W f = 4 We
We j 4 We = 480
W,
96 KN(l/O 00981)
W,
().7:1631 ,,!C
Heat Transfer
Machine Foundation
236
237
HEAT TRANSFER
Machine Foundations - 35
A rectangular foundation cross-section has a bed plate dimension of 4
ft x 8 ft. The uniform clearance on each side is 1 ft. The height of
foundation is 2.5 ft. If the weight of the steel bar reinforcements
needed is 1/2% of weight of foundation, find the weight of steel bars.
lise concrete density of 2400 kg/m 3 •
C. 65 kg
A. 51 kg
D. 76 kg
B. 45 kg
SOLUTION
A
A
=
(4+2)(8+2)
60 ft2
V
=
Ah
=
=
A 15 em tbick wall bas a tbermal conductivity of 5 W/m-°K. If inside
and outside surface temperature of tbe wall are 200°C and 30°C,
respecnvely, Determine tbe beat transmitted.
A. 5.67 KW/m 2
C. 8.87 KW/m 2
2
B. 4.68 KW/m
D. 6.87 KW/m 2
SOLUTION:
V = 60(2.5)
V=150fe
V = 4.247 nr'
W
Heat Transfer - 1
Q
=
kA(t z - t])
x
wV
W = (2400)(4,247)
W = 10,192.57 kg
Weight of steel bars
Weight of steel bars
Weight of steel bars
(1/2%) W F
=
=
0.005(10,192,57)
50.96 kg
Q
5(200 - 30)
A
0.15
200
o
e .l.
.l. 30 e
0
Q
Q/A
=
5666.67 W/m 2
K,
Q/A
=
5.6667 KW/m 2
Jscm).
Heat Transfer - 2
Two walls of cold storage plant are composed of an insulating material
(k = 0.25 KJlhr-m·oC), 100 mm thick at the outer layer and material
(k = 3.5 Kl/hr-m-e'C), 15 em thick at inner layer. If the surface
temperature at the cold side is 30°C and hot side is 250°C, find the
heat transmitted per square meter.
2
2
A. 0.138 KW/m C. 0.025 KW/m
2
B. 0.450 KW/m D. 0.065 KW/m 2
238
Heat Transfer
Heat Transfer
239
SOLUTION:
Q
Heat Transfer - 4
A(t 2 -tI)
=
Xl
X
ki
k2
{,=Z50·C
2
--+-
Q/A
t t,=30'C
Q
250 - 30
=
f
(0.15 I 3.5) + (0.1 I 0.25)
Q/A
Q/A
=
=
496.77 KJ I hr - m
Determine the thermal conductivity of a material that is use a 2 m 2 test
panel, 25 mm thick with a temperature difference of 10.8 OF between
the surfaces. During the 5 hours test period, the heat transmitted is
200 ~J.
k 0.045 W/m-oK
C. 0.023 W/m-oK
oK
D. 0_370 W/m-oK
B. 0.560 W 1m-
2
JX,J
3600
0.138 KW/m 2
SOLUTION:
X,
~oC
I
~oF =
519
~oC/lO.8 =
Heat Transfer - 3
~oC
Sea water for cooling enters a condenser at 27°C and leaves at 37°C.
The condenser temperature is 45°C, what is the log mean temperature
difference?
A. 50. 56°C
C. 37.82°C
B. 12.33°C
D. 80.54°C
Q
~tA =
~tB =
~tB =
~t1l1e.n
45 - 27
18°C
45 - 37
8°C
45'C
I
~t A - ~t B
~tA
In - ~tB
~tll1eon
~t1l1e.n
18-8
~7'~
27'C
kA(t 2 -t I)
--=----''-
x
Heat Transfer - 5
45'C
At,
= 6°C
200
k(2)(6)
--0.D25
5 x 3600
k = 2.3148 X 10- 5 KW/m-oC
k = 0.023148 W/m-oC
SOLUTION:
~tA =
=
5/9
.6.t.
A steam pipe having a surface temperature of 200°C passes through a
room where the temperature is 27°C. The outside diameter of pipe is
80 mm and emissivity factor is 0.8. Calculate the radiated heat loss for
3 m pipe length.
A. 1434.47 W
C. 2756.57 W
B. 3746.35 W
D. 3546.45 W
27°C
SOl UTION:
In 181 8
12_33°C
A, = 7tDL
Ao = 7t(0.08)(3)
Ao = 0.7539 m 2
Solving for heat due to radiation:
24\
Heat Transfer
240
lIeul Transfer
20,408.4 x 10-8AJ:,('l14 - 1'21 ) , J/hr
200 + 273
OR
OR
OR
OR
=
=
=
T1
=
T1
=
473°K
T 2 = 27 + 273
T7 = 300 0 K
20,408.4 x 10-8 (0.8)(0.7539) [(473)4 - (300)4]
5164079.866 JIhr x Ihr/3600sec
1434.47 W
C. 56.80 ft2
D. 15.80 ft2
A. 51.80 ft2
B. 37.30 ft2
{i
I,
~
·1
SOLUTION:
o=
11
=
AU L1t
A(0.5)(l5)
2 2!lm2
A = 1.467 m 2 x 3.281 ft
A = 15.79 m
2
I
Heat Transfer - 6
A counter flow heat exchanger is designed to heat fuel oil from 30 0 e to
90 0 e while the heating fluid enters at 140 0 e and leaves at 105°C.
Determine the arithmetic mean temperature difference.
A. n.5°e
c. 45.5°C
B. 62.5°e
D.67.5°C
SOLUTION:
L1t A
L1tA
L1t B
L1tB
105-30
75°C
= 140 - 90
= 50 0 e
A
_
L\t A - L\tB
,-,tnean L\
tA
In-L\tB
75- 50
L1tnean =
In75/50
L1t",ean = 6I.66°e
=
=
\
I
Heat Transfer - 8
,1
3/hr
SOLUTION:
Density of brine = 1.1(1000 kg/m')
Density of brine = 1100 kg/rrr'
11OO( 60)
m=
m
=
3600
18.33 kg/sec
o m c L\t
o =18.33(1.072) [-10 -(-16)1
o = 117.92 KW
=
p
TR = 117.92/3.516
TR = 33.54 Tons of refrigeration
Heat Transfer - 7
A heat exchanger has an overall coefficient of heat transfer of 0.50
KW/m 1 _ 0 C. Heat loss is 11 KW and the mean temperature difference
is 15°C. What is the heat transfer area in fe?
i\i
Brine enters a circulating brine cooler at the rate of 60 m
at
lOoe and leaves at -16°C. Specific heat of brine is 1.072 KJ/kg-OK and
specific gravity of 1.1. Determine the tons of refrigeration.
A. 53.5 TR
C. 33.5 TR
B. 65.3 TR
D. 44.5 TR
,~
'i~
,'11
I
242
Heat Transfer
243
Heat Transfer
Q/A = 230(1145 - 545)
Heat Transfer - 9
t,=1145°C t~ t,=45°C
QIA = 138,000
A heat exchanger has a hot gas temperature of 300°C and surface
A(t! -t 2 )
conductance on hot side is 200 W/m 2. oK. If heat transmitted is 1000
W/m 2• what is the surface temperature on the wall at hot side?
A. 295°C
C. 234°C
B 465°C
D. 354°C
Q
= 1
x
Q
1
--+-+h.
k h2
Q/A
h,
(t l - t 2 )
1
x
1
-+-+h,
k h,
SOLUTION:
138,000 = ----'--------'----(1 1230) + (x I 115) + tl 1290)
Q/A = (ho)(!!t)
=
LJ
(1145 - 45)
Q = A (ho)(!!t)
1000
h,
k
x = 0.020115 m
x = 20.115 mm
200 (300 -tt)
t l = 295°C
Heat Transfer - 11 (ME Rd. Oct. 1985)
Heat Transfer - 10 (ME Bd, Oct. 1986)
A heat exchanger is to be designed for the following specifications:
Hot gas temperature; 1145°C
Cold gas temperature. 45°C
Unit surface conductance on the hot side, 230 W/m 2 .oK
Unit surface conductance on the cold side. 290 W/m 2. oK
Thermal conductivity of the metal wall, 115 W/m-oK
Find the maximum thickness of metal wall between the hot gas and the
cold gas, so that the maximum temperature of the wall does not
exceed 545°C.
C. 20 mm
A. 10 mm
D.40mm
B 30mm
SOLUTION:
Q = hi A (t] - ta)
Q/A = hi (t, - (2)
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/m 2 -oK , what is the rate of heat loss from the
surface per unit length of pipe?
2
2
C. 762 w/m
A. 998 w/m
2
2
D. 422 w/m
B 872 w/m
SOLUTION:
Qc = heat transmitted by convection
Qc = h, Ao (t, - t 2 )
Qc '" 15[1t (0.07) LJ(200 - 25)
Qc I L = 577.268 W/m
QR = heat transmitted by radiation
4
QR = 20,408.4 X 10,8 Ao Fe (T ]4 - T z ) , Jihr
where: Ao = n(0.07)L
T I '" 200+273
T. = 473°K
T 2 = 25 + 273
Tz '" 298°K
244
Heat Transfer
245
Heal Transfer
8
OR = 20,408.4 x 10- [n(0.07)L](0.8)[(473)4 _ (298)41
OR/L = 1,514,032 J/hr x 1/3600
OR/L = 420.564 Wlm
OIL = Oc + OR
OIL = 577.268 + 420.564
OIL = 997.832 Wlm
the compressor is 145°C and a counter flow air cooler reduces the air
temperature to 45°C before it goes to the engine suction header.
Cooling water enters air cooler at 30°C and leaves at 38°C. Calculate
the quantity of cooling water in mJ/h,. required to cool the total air
requirements of the engine at rated load and speed.
A. 144
B. 136
C. 123
D. 165
Heat Transfer - 12 (ME Bd. Oct. 1981)
SOLUTION:
A turbo-charged, 16 cylinder, Vee-type dieset engine has an air
consumption of 3,000 kg/hr per cylinder at rated load and speed. This
air is drawn in through a filter by a centrifugal compressor directly
connected to the exhaust gas turbine. The temperature of the air from
the compressor is 145°C and a counter flow air cooler reduces the air
temperature to 45°C before it goes to the engine suction header.
Cooling water enters air cooler at 30°C and leaves at 38°C. Calculate
the log mean temperature difference.
45 - 30
=
~tA =
Ave. temp of water = (30 + 38)/2
Average temperature = 34°e
~tB
145 - 38
~tB
iov-c
~t
'mean
=
~tA - ~tB
~tA
In-~tB
~t.nean
~t.ne.n =
4~
~8.C
30'C
107 -- 15
In(l07 I 15)
46.82°e
Heat Transfer - 13 (ME Bd. Oct. 1981)
A turbo-charged, 16 cylinder, Vee-type diesel engine has an air
consumption of 3,000 kg/hr per cylinder at rated load and speed. This
air is drawn in through a filter by a centrifugal compressor directly
connected to the exhaust gas turbine. The temperature of the air from
=
0.0010056 mJ/kg
Volume flow
Volume flow
145°C
15°e
(m cp ~t).il
m w (4 .187)(38 - 30) = 48,000(1.0)(145 - 45)
= 143,301 kg/hr
From steam table,
vr at 34°e
SOLUTION:
=
m.,
e. 3~e
D. 65°e
A. 4re
B. 87°e
MA
Heat gained by water = heat lost by air
(m cp ~t)w.ter
143,301 (0.001 0056)
144.1 m 3 fhr
Heat Transfer - 14 (ME Bd. Apr. 1983)
An oil heater heats 100 kg per minute of oil from 35°C to tOO°C in a
counter flow heat exchanger. The average specific heat of the oil is 2.5
KJ/kg_°C. Exhaust gases used for heating enter the heater with an
average specific heat of 1 KJ/kg-OC, a mass flow rate of 250 kg/min
and an initial temperature of 200°C. The overall heat transfer
coefficient is 75 W/m 2 _ 0 C. Determine heating surface in square meters
A. 30
B. 63
C. 40
D. 36
SOLUTION:
Heat gained by oil
=
Heat lost by gas
(m c p ~t)"ll= (m c p ~t)g.s
246
Heat Transfer
Heat Transfer
100(2.5)(100 - 35) = 250(1)(200 - to)
to = 135°C
Heat transferred = (100/60)(2.5)( I 00 - 35)
Heat transferred = 270.83 K W
t.t A = 135 - 35
t.t A = 100°C
M B = 200 - 100
L\tB = 100°C
Since f.t A = t.tB, use the av~rage value
100+ 100
t.t m
2
t.t m = 100°C
Q = U A t.t m
270.83 = (0.075) (A) (100)
A = 36.11 m 2
tmean
tmean
tmean
247
11.36 - 3.36
In( 11.36/3.36)
6.56°C x 1.8
11.82°e
Heat Transfer - 16 (ME Bd. Apr. 1983)
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 29.5°C and
leaves at 37.5°C. For steam turbine condenser design, manufacturers
consider 950 Btullb of steam turbine condensed as heat given up to
cooling water. Calculate the required quantity of cooling water in
cubic meters per hour.
A. 10,374
C. 11,345
B. 12,445
D. 13,509
SOLUTION:
Heat Transfer - 15 (ME Bd, Apr. 1983)
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 29.5°C 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 logarithmic mean temperature difference in
of.
A. 10
C. 12
B. 14
D. 16
SOLUTION:
Condenser pressure = 101.325 - (702 x 10 1.3251760)
Condenser pressure = 7.733 Kpa
From steam table, at 0.007733 Mpa, t Sa l = 40.86°C
t.tA = 40.86 - 29.5
t.tA = 11.36°e
t.t B = 40.86 - 37.5
t.tB = 3.36°e
Heat absorbed by cooling water
Heat absorbed by cooling water
=
950(196,000)(2.205)
= 410,571,000 Btulhr
AverageSG of sea water is 1.03 and cp of gas is 0.93 Btu/lb-vl­
cp t.t = 410,571,000
m(0.93)(37.5 - 29.5)(1.8) = 410,571,000
ill = 30,657,930.11 Ibsfhr
ill
v =
30,657,930.11/(62.5)(1.03)
V = 477,003 ft3fhr
V = 477,003/35.31
V = 13,509 m 3lhr
Heat Transfer - 17 (ME Bd. Oct. 1994)
Calculate the energy transfer rate across 6" wall of firebrick with a
temperature differences across the wall of 50°C. The thermal
248
Heat Transfer
Heat Transfer
conductivity of the firebrick is 0.65 Btu/hr-It-Pf" at the temperature
interest.
A. 285 W/m 2
C. 112 W/m 2
B. 369 W/m 2
D. 429 W/m 2
v
=
NR
NR
249
0.0011334 m2/sec
0.508(5)
=
0.0011334
= 2241
SOLUTION:
t a - tb = 50(9/5)
t. - tb
Q
Heat Transfer - 19 (ME Rd. Oct. 1995)
90°F'
kA(t -t b )
=
=
a
x
0.65(90)
Q/A
Q/A
Q/A
Q/A
117 Btu/hr-ft'
369 W/m 2
=
X
(6/12)
117 Btu/hr-fr'
1055J/Btu X Ihr13600sec
X
10.76ft2/m2
Heat Transfer - 18 (ME Bd. Oct. 1994)
Water is flowing in a pipe with radius of 25.4 em at a velocity of 5
mlsec at the temperature in the pipe. The density and viscosity of the
water are as follows: density = 997.9 kg/sec
viscosity = 1.131
Pa-s. What is the Reynolds Number for this situation?
A. 2241
C. 3100
B. 96.2
D. 1140
SOLUTION:
nv,
R
v
where:
D
D
D
D
=
=
=
=
Vo =
Vo
v
v
=
=
=
diameter
2(25.4)
50.8 em
0.508 m
velocity
5 m/sec
kinematics viscosity
1.131/997.9
The hot combustion gases of a furnace are separated from the ambient
air and its surrounding, which are at 25°C, by a brick wall 0.15 m
thick. The brick has a thermal conductivity of 1.2 W/mfK and surface
emissivity of 0.8. Under steady state conditions and outer surface
temperature of lOO°C is measured. Free convection heat transfer to
the air adjoining this surface is characterized by a convection
coefficient of 20 W/m 2 -oK. What is the brrck inner surface
temperature in DC?
A. 623.7
C. 461.4
B. 352.5
D. 2563
SOLUTION:
Qc = convection heat transfer
Qc = Ah, (t, - t2)
Qc 1A = 20(100 - 25)
2
= 1500 W/m
Qr = radiated heat loss
.
4
2
8
Qr = 20,408.4 X 10 Fe (T 1 - T/) ,1/hr-m
8
Qc = 20,408.4 X 10. (0.8)[(100+273)4 - (25+273)4]
Qc = 1,872,793 l/hr-m 2 X (1/3600)
c
Qr = 520 W/m
Q = Qc + Q,
Q = 1500 + 520
Q = 2020 W/Ill'
k(t a -thY
o,
Q
X
2020
1.2(t a
-
0.15
t, = 352.5°C
lOa)
Heat Transfer
L.JV
2668.6 = 914.52 + x(l885.5)
x = 93%
Heat Transfer - 20 (ME Bd. Oct. 1996)
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 a 5.08 cm thickness of 85% magnesia. For an ambient
temperature of 22°C, 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 Temperature
Enthalpy
2.05
213.67
hf~ 914.52
hfg = 1885.5
hg = 2800.0
k for 85% magnesia = 0.069 W/m-oK
f, for still air = 936 W/m-oK
A.93%
C. 84%
B.98%
0.76%
Heat Transfer - 21(ME Bd. Apr. 1999)
Compute the amount of condensate form during 10 minutes warm-up
of 150 meter pipe conveys the saturated steam with enthalpy
vaporization h rg = 1,947.8 LJ/kg. The minimum external temperature
of pipe is. 2°C. The final temperature of pipe is 195°C. The specific
heat of pipe material is 0.6 KJ/kg_°C. The specific weight is 28 kg/m.
A. 249.69 kg
C. 294.54 kg
B. 982.45 kg
D. 42345 kg
SOLUTION:
SOLUTION:
rl
'=
rl
'=
10.10/2
5.05
(Steam~
~
r2 '= 5.08 + 5.05
•
rz'=10.13cm
A, = 21trzL
A, = 21t(O.I013)(152)
A, '= 96.746 m Z
Q
1o=2°C
m, = mass of pipe
m, = 28(150)
rn, = 4200 kg
Heat loss by steam = Heat loss from pipe
m, (h, - h.) = mp cp (tz - t.)
m,(l947.8) '= (4200)(0.6)(195 - 2)
m, = 249.69 kg
jQ
C.
"I"
~ h,
~m.
15h
(t , -to)
In( r 2 I r l )
2' I
Heat Transfer
Heat Transfer - 22 (ME Bd. Oct. 1999)
2
I
---+-­
21tkL
Aoh o
(213.67 -22)
In(O.lOI3/0.0505)
I
+
21t(0.069XI52)
96.746(9.36)
Q = 16,427.4 W
Q = 16.427 KW
Q = m,(h l - hz)
Q
16.4274 = 0.125(2800 - hz)
hz = 2668.6
h '= hr + xhfg
What is the heat transfer in the glass surface area of 0.7 m having an
inside temperature (room) of 2SoC and 13°C outside temperature
(surrounding). The thickness of glass surface is 0.007 m. The thermal
eonductivity is 1.8 W/m-°K.
C. 6.2 kw
A. 5.8 kw
D. 2.34 kw
B. 3.6 kw
SOLUTION:
252
Heat Transfer
Q=
Air Compressor
253
AIR COMPRESSOR
kA(t 2 -t 1 )
x
(1.8XO.70)(25 - 13)
Q = --'-----'------'---------'-0.007
Air Compressor - 1
An air compressor takes air at 100 Kpa and discharges to 600 Kpa. If
the volume flow of discharge is 1.2 mJ/sec, determine the capacity of
air compressor.
3/sec
A. 432 m 3/sec
C. 6.85 m
3
B. 3.33 m /sec
D. 7.42 m3/sec
Q = 2160 watts
Q = 2.16 kw
SOLUTION:
PI V l n = P2 V 2n
25°e
13°e
a
n
-h-+
1.4 (for standard air)
=
IOO(V I)L4
VI
=
=
(600)(1.2)14
4.315 m 3/sec
Air Compressor - 2
The discharge pressure 01 an air compressor is 5 times the suction
pressure. If volume flow at suction is 0.1 mJ/sec, what is the
compressor power assuming n = 1.35 and suction pressure is 98 Kpa?
A. 21.67 KW
C. 25.87 KW
B. 19.57 KW
D. 10.45 KW
SOLUTION:
W
=
nP V
P
n-I
PI
n-I
-~
_ 1 - 1 [(_2) n
_
I]
135-1
W -
1.35(98)(0.1)[(5P / PI)
I
1.35-1
W
19.57 kw
=
135
-
1]
Air Compressor
2:l4
255
Air Compressor
1
'.,
Air Compressor - 5
Air Compressor - 3
3/sec
A 10 Hp motor is use to drive an air compressor. The compressor
efficiency is 75%. Determine the compressor work.
A. 5.0 KW
C. 7.6 KW
B.6.5KW
D.5.6KW
SOLUTION:
at 97 Kpa and
An air compressor has a suction volume of 0.25 m
discharges to 650 Kpa. How much power saved by the compressor if
there are two stages?
A. 8.27 KW
B. 6.54 KW
C. 3.86 KW
D. \0.0 KW
SOLUTION:
W
YJc =
For single stage:
W
0.75
W
Brake Power'
=
=
P
--­
10xO.746
5.59 KW
n-I
-~-
n PI VI ((_2)
W
n -1
n
-
l]
PI
I~
\.4(97)(0.25) ((650 197)
W~
1.4 - \
W= 6\.28 KW
Air Compressor - 4
The initial condition of air in an air compressor is 98 Kpa and 27°C
and discharges air at 400 Kpa. The bore and stroke are 355 mm and
381 mm, respectively with percent clearance of 5% running at 300
rpm. Find the volume of air at suction.
3
A. 600 rrr'zhr
C. 620 m /hr
3
D.630m3/hr
B. 610 m /hr
Px
=
r,
=
P,
=
W
YJy
=
1 + C _ c(.!2)I!n
PI
YJy = 1 -i- 0.05 - 0.05(400/98)1114
YJy = 0.913
VD
= nl4
VD
VD
=
D2 L N
n/4 (0.355/(0.381 )(300160)
0.1885 m3/sec
VI = 0.1885(0.913)
3!sec
VI = 0.17215 m
3
VI = 619.75 m /hr
=
-lJ
For two stages:
~
~97(650)
251.097Kpa
p
SOLUTION:
-1
14
n-l
­
2nP1Vl((~)n _I]
n- 1
PI
14-1
2(\.4)(97)(0.25) [(25\.097 /97)
\.4 - 1
W= 53 KW
W=
power Saved = 61.28 - 53
power Saved = 8.27 KW
14
-
11
256
and 0.2
The suction conditi on of an air compre ssor is 98 Kpa, 27°C
3/sec.
frce air
the
ine
determ
20°C,
and
Kpa
100
nding air is
If surrou
m
3/sec.
capacit y in m
A. 0.15
B. 0.]9
C. 0.25
D. 0.23
D. 11%
I
SOLUT ION:
11v
=
0.87
1 +c _c(pz/P j)!fn
=
1 + c - C(3P 11P1) l f 14
c
SOLUT ION:
P,V s
Pr,-V
- -F - - Ts
IF
100(V F )
1i
C. ]5%
A.5%
R 7%
Air Compr essor - 6
257
Air Compressor
Air Compressor
=
10.9]%
Air Compr essor - 9
98(0.2)
(20 + 273) (27 + 273)
3isec
V F = 0.]914 m
piston
The compre ssor work of an air compre ssor is 100 KW. If the
3/min,
e.
pressur
e
determ ine the mean effectiv
speed is 15 m
C. 400 Kpa
A. 200 Kpa
D. 600 Kpa
B. 300 Kpa
SOLUT ION:
Air Compr essor - 7
W=Pm xV o
x 381 mm air compre ssor has a piston displac ement of
A 355 mm
3/sec.
Determ ine the operati ng speed of the compre ssor.
0.1885 m
C 350 rpm
A. 250 rpm
600 rpm
D.
B. 300 rpm
100
Pm
=
Pm ( ] 5/60)
= 400
Kpa
Air Compr essor - 10
SOLUT ION:
2
Yo = nl4 D L N
0.1885 = rr/4 (0.355) 2(0.38] )N
N = 5 rev/sec x 60sec/m in
N ~ 300 rpm
A double acting air compre ssor has 16 in x 7 in, 600 rpm has what
volume displac ement?
3
C. 488 /min
A. 688 ft /min
3
D. 977 ft /m i n
B. 'i55 ft3/m in
fe
SOLUT ION:
Air Compr essor - 8
87%
Determ ine the percen t clearan ce of an air compre ssor having
the
thrice
be
to
e
pressur
air
ssor
compre
and
cy
'Ivolumetric efficien
suction pressur e.
V o=2(r rD
v,
Vo
=
=
2LN]
2(rr/4 (16/12)2 (7112) (600)]
3
977.38 ft /m in
258
Air Compressor
Air Compressor - 11
Air Compressor
A two-stage air com pressor has a suction pressure of 14 psi and
discharge pressure of 130 psig. What is the intercooler pressure in
Kpag.
A. 209 Kpag
C. 477 Kpag
B. 600 Kpag
D. 300 Kpag
~
13
The piston speed of an air compressor is
displacement of 0.2 m 3/sec. Determine
cylinder,
A. 500 mm
B. 358 mm
VD =
Piston
140 =
LN =
]30+]4.7
]47.5 psia
P2
P2
Jp
Px
Px
j
P2
=
=
140 mlmin and has a volume
the diameter of compressor
C. 467 mm
O. 246 mm
2
rei4 0 L N
Speed = 2 L N
2 LN
70 m/min
0.2 = 7[ / 4 0 2 (70/60)
D = 467,19 mm
J14(l447)
=
259
SOLUTION:
SOLUTION:
P,
P,
P,
Air Compressor
45 psi x 1Ol.325/14.7
310.24 Kpaa - ]0l.325
208.91 Kpag
Air Compressor - 14
Air Compressor - 12
A two stage air compressor has an intercooler pressure of 3 kg/ern".
What is the discharge pressure if suction pressure is 1 kg/cm/?
A. 3 kg/ern"
C. 12 kg/ern"
B. 9 kg/ern'
D. 15 kg/ern:'
An air compressor piston displacement is 5000 em:' when operates at
900 rpm and volumetric efficiency of 75%. Determine the mass now
of air at standard density.
.
C. 386.4 kg/hr
A. 365,3 kg/hr
D. 465,2 kg/hr
B. 243,5 kg/hr
SOLUTION:
VI
11v
SOLUTION:
Px=~
p/= P (P2)
1
32
=
1(P 2 )
P 2 = 9 kg/em/
Vo
0.75
VI
V,
5000
3750 cm 3 (900)
3375000 cmvrnin
w
m
m
m
1,2 kg/m' (at standard)
1.2(3375000/1 00 3 )
4.05 kg/min
243 kg/hr
V,
=
260
Air Compressor
Air Compressor
Air Compressor - 15
0.4
A two-stage compressor air at 100 Kpa and 22°C discharges to 690
Kpa, If intercooler intake is 105°C, determine the value of n.
A 1400
C. 1.345
8. 1.358
D. 1.288
=
261
2[7(/4 (D)2(D)(300/60)]
D = 0.37067 m
L = 0
L = 370.67 mm
SOLUTlON:
Px
=
P,
P,
=
l
n-)
(p
x
An air compressor takes air at 97 Kpa at the rate of 0.5 m/sec and
discharge 500 Kpa, If power input to the compressor is 120 KW,
determine the heat loss in the compressor,
A 26.85 KW
C. 30.45 KW
B. 18.55 KW
D. 22.36 KW
JlOO(690)
262.68 Kpa
C~
T
Air Compressor - 1i
~PI P2
r-;;­
x I
~= ~j
n-I
(l 05 + 273)
(22 + 273)
= (
l
SOLUTION:
262.68.J--;;
100
P
n-I
1.281
n-- I
=
(2.6268)
In 1.28\
n
n
In 2.6268
n - I = 0.2564n
n = 1.345
W
n PI VI [(_2)
n -I
n-]
-­
n
_
I]
PI
14-1
W
W
1.4(97)(0.5) [(500 / 97)
1.4 - 1
101.45 KW
Heat Loss
Heat Loss
14
-I]
120-101.45
18.55 KW
Air Compressor - 16
The piston displacement of a double acting compressor running at 300
rpm is 0.4 mJ/sec. If bore and stroke are unity, determine the length of
stroke.
A. 350 mm
C. 371 mm
8. 380 mm
D. 400 mm
SOLUTION:
L
o
V D =2(nI4D 2LN)
(for unity)
Air Compressor - 18 (ME Bd. Apr. 1983)
A single acting air compressor has a volumetric efficiency of 87%,
operates at 500 rpm. It takes in air at 100 Kpa and 30°C and
discharges it at 600 Kpa. The air handled is 6 mJ/min measured at
discharge condition. If compression is isentropic, find mean effective
pressure in Kpa
A ]82
C. ]98
B. 973
D. 204
262
Air Compressor
Air Compressor
SOLUTION.
0.45 m 3/min
PI = 90 - 5
P,=85kpa
P2 = 600 -t- 10
P2 = 610 kpa
Tlv = I + c - c(P 2/p,)lfn
1']v = 1+0.10-0.10(610/85)11128
1']v = 0.633684
V I = 0.45(0.633684)
V I = 0.285 mJ/min
m = PV/RT
m = 85(0.285)/(0.287)(29.3 + 273)
m = 0.2792 kg/min
m ~ 16.76 kg/hr
Vo
PI V l
k
= P2 V/
IOO(V 114) = 600(6)14
VI = 21.58 rrr'zrnin
V0
=
21.58/0.87
Vo = 24.8 m3/min
W =
nP V
.
n~l
P ­
[(_2) n
n- I
PI
_'_I
_
I]
I
1.4(100)(21.58) [(600/ 100)
W =
1.4-1
4~
14
=
I
_
I]
w=
5,049 KJ/min
W = Pm X Vo
5,049 = Pm X 24.8
Pm = 203.6 Kpa
Air Compressor - 19 (ME Bd. Oct. 1984)
A single acting reciprocating air compressor has a clearance volume of
10%. Air is received at 90 Kpa and 29.3°C and is discharged at 600
Kpa. The compression and expansion are polytropic with n = 1.28. The
pressure drop is 5 Kpa at suction port and 10 Kpa at the discharge
port. The compressor piston displacement is 500 cnr' when operating
at 900 rpm. Determine the mass of compressed air in kg/hr
A. 16.76
C. 98.33
B. 20.45
O. 28.23
Air Compressor - 20 (ME Bd, Apr. 1986)
A single acting air compressor operates at 150 rpm with an initial
condition of air at 97.9 Kpa and 27°C and discharges the air at 379
Kpa to a cylindrical tank.. The bore and stroke are 355 mm and 381
mm, respectively, with 5% clearance. If the surrounding air is at 100
Kpa and 20°C while the compression and expansion process are PVI.J
= C, determine free air capacity, m 3/sec
A. 0.0818
C. 1.23
B. 2.13
O. 4.23
SOLUTION:
2LN
Vo =1t/40
Vo = 1t/4 (0.355)2 (0.381) (150/60)
3/sec
Vo = 0.094278 m
11n
1']v = I + c - c(P 21Pd
1'], = 1 + 0.05 - 0.05(379/97.9)1113
1']v = 0.908
V I = 0.908(0.09427tn
VI = 0.085604 mJ/sec
SOLUTION:
v,
= (rr/4 n 2 L) N
VD = (500)(900)
Vo = 450,000 crrr'zmin
263
Solving for free air capacity:
PFV
PlV l
- -F - - TF
T]
264
Air Compressor
Air Compressor
I OO( VF )
D
D
D
97 .9( 0.085604)
(20 + 273)
(27 + 273)
3/sec
= 0.081851 m
VF
=
=
=
265
0.45 m
450 mm
L = 450 mm
Air Compressor - 23 (ME Bd. Nov. 1983)
Air Compressor - 21 (ME Bd. Apr. 1987)
The piston displacement of a double acting compressor is 0.358 m 3/sec,
delivers gas from 101.325 Kpa and 300 0 K to 675 Kpa at the rate of
0.166 m 3/sec at 150 rpm. Value of n for compression and expansion is
1.33. Find the compressor percent clearance..'
C. 12.34
O. 18.44
k 16.96
B. 14.23
SOLUTION:
A 2-stage, double acting, L-type air compressor 16" x 10" x 7", 600
rpm, has a free air unloader at each end for capacity control. It is
driven through V-belts by a 150 Hp electric motor, 460 V, 3 phase, 60
Hz, 1200 rpm. Barometric pressure is 125 psi gage. Calculate piston
displacement in m 3/hr
3
A. 562 m 3/hr
C. 649 m !hr , 1661 m 3 /hr
3/hr,
3
3/hr
B. 762 m
O. 833 m
724 m !hr
SOLUTION:
llv=V}/V D
nv = 0.166/0.358
llv = 0.4637
Piston displacement of first stage:
VD
=
2[11:/40 2 L N]
llv = 1 + c - c(pZ/Pj)l/n
0.4637 = 1 + c - c(675/10 1.325)1/133
VD
=
2[11:/4 (16/12)2 (7/12)(600)]
VD
=
977.384 cfm
VD
VD
=
977.384(60)/35.31
1,661 m 3/hr
c
c
=
=
0.1696
16.96%
=
Piston disp lacement of second stage:
Air Compressor - 22 (ME Bd. Apr. 1987)
The piston displacement of a double acting compressor is 0.358 m3/sec,
delivers gas from 101.325 Kpa and 300 0 K to 675 Kpa at the rate of
0.166 m3/sec at 150 rpm. Value of n for compression and expansion is
1.33. Find the bore and stroke assuming bore = stroke.
C. 350 mm
O. 450 mm
A. 300 mm
B. 400 mm
SOLUTION:
VI)
=
0.3 58
2(11:/4 D 2 L N)
=
2[11:/4 (0)2 (D) (150/60)]
V D = 2[11:/4 (10/12)2(7/12)(600)]
V D = 381.791 cfm
V D = 381.791(60)/35.31
3
V D = 648.75 m /hr
Air Compressor - 24 (ME Bd. Apr. 1984)
A two-cylinder single-acting air compressor is directly coupled to an
electric motor running at 1000 rpm. Other data are as follows:
a. size of each cylinder = 150 mm x 200 mm
b. clearance volume = 10% of displacement
c. exponent (n) for both compression and re-expansion
266
process = 1.6
d. air constant = 1.4
e. air molecular mass M = 29
Calculate the volume rate of air delivery in terms of standard air for a
delivery pressure 8 times ambient pressure under ambient air
conditions of 300
and 1 bar.
3
3
A. 2 m /hr
C. 3 m /hr
3
B. 4 m /hr
D. 5 m 3/hr
0K
SOLUTION:
71:/4 D2 L N
= 71:/4 (0.15)2 (0.2) (1000)
3/min
Vn = 7.068 m
11, = 1 + c - c(P2/P I ) 11n
11, = 1 +0.10-0.10(8P/P\)IIL6
11, = 0.733
V I = 0.733(7.068)
VI = 5.181 m 3/min
Standard air is at 70°F(21.11 "C) and 14.7 psi(10 1.325 Kpa)
r; v. PI Vi
Vo
V0
=
-----
r,
r,
101.325(V s )
100(5.181)
(21.11 + 273)
(300)
V, = 5.013 m 3/min
267
Air Compressor
Air Compressor
SOLUTION:
VOl = 94,390 cml/sec
Vol = 0.09439 m 3/sec
V I = Vo l(11,) = 0.09439(0.85)
3/sec
VI = 0.0802315 m
PI = 14.5 psi x 101.325/14.7
PI = 99.946 Kpa
P, = (30 + 14.7)(101.325/14.7)
P, = 308.11 Kpa
P 2 == (100 + 14.7)(101.325/14.7)
P 2 = 790.61 Kpa
Solving for the mass flow rate:
PI VI = m R r,
99.946(0.0802315) = m(O.287)(22 + 273)
m = 0.09471 kg/sec
Solving for polytropic exponent n:
(T, 1 T 1) = (P, 1 Pj)",lfn
n-l
105 + 273 =(308.11 )---;;-­
22 + 273
99.946
1.2813 = (3.08277)",l/n
n -1
In 1.2813
n
In3.08277
n -~ 1.2824
Compressor power of the first stage:
n-l
W
n(mRT)
=
n- 1
P~ ­
[(--) n
-
1]
PI
L28-1
Air Compressor - 25 (ME Bd. Apr. 1991)
A two stage compressor with first stage piston displacement of 94390
cnr'rsec is driven by a motor. Motor output is 35 Hp, suction
temperature 22°C, volumetric efficiency is 85%. Mechanical efficiency
is 95%, the intercooler pressure is 30 psi gage. Air temperature in and
out of the intercooler are lO5°C and 44°C. Final discharge pressure is
100 psi gage, suction estimated 14.5 psi. Find the compression
efficiency
A. 77.60%'
C. 87.34%
B. 63.34%
D. 98.23%
1.2824(0.0947)(0.287)(22 + 273) [(308.11/99.946) -'28
1.2824-1
W =
W = 10.245 kw
_
1]
Compressor power of the second stage:
128-1
1.2824(0.0947)(0.287)(44 + 273) [(790.611/308.11) 128
1.2824-1
W =
W = 8.956 kw
WT == 10.245 + 8.956
W T = 19.20 kw
_ 1]
268
Air Compressor
Compression efficiency
=
Compression efficiency
=
Air Compressor
269
19.20
Air Compressor - 27 (ME Rd. Apr. 1995)
(35 )(0.746 )(0.95)
77 AI %
. Air Compressor - 26 (ME Rd. Apr. ~995)
An air compressor is to compress 8.5 mJ/min from 98.56 Kpa to 985.6
Kpa, Assuming conditions ideal, and with n ;=: 1.3, what will be. the
saving in work due to two staging?
A. zero
C. 5.6 KW
B. 4.6 K'W
D. 3.5 KW
A single stage air compressor handles 0.454 mJ/sec of atmospheric
pressure, 27°C air, and delivers it to a receiver at 652.75 Kpa. Its
volumetric efficiency on an isothermal basis is 0.85 and its mechanical
efficiency is 0.90. If it operates at 350 rpm, what power in KW is
required to drive it?
C. 120
A. 95
B 112
D. 100
SOLLTIO~
For Isothermal process.
SOLUTION:
W
=
Pi VI In(P 2fP 1)
\V
=
101.3(0.454) In(652.75/l01.3)
\\
=
85.685 KW
For single stage compressor:
n-1
W
=
nP V
P. _'_I
[(_2)
n -I
0
IJ
_
Pj
13-1
1.3(93.56)(8.5 / 60)
W
W
=
=
[(985.6/98.56)
1.3 - I
42043 KW
13
85.685
Power Input
---
085(0.90)
112KW
IJ
Power Input
For two stage-compressor:
r,
=
P,
=
P
w
2nP J VI [(.2_)
n- I
W
W
Air Compressor - 28
.J98.56(985.6)
311.67 Kpa
=
A two stage compressor receives 0.50 kg/s of air at 120 kpa and 300"K
and delivers it at 7 .\1 pa, Find the heat transferred in the intercooler.
A. 118]4 kw
C. 233.23 kw
B. 134.55 kw
D ) 87.34 kw
0-1
-
n
-1]
P,
2(1.3)(98.56)(8.5/60) [(311.67 / 98.56)
1.3 - 1
36.83 KW
Power Saved
=
42.43 - 36.83
Power Saved
=
5.6 KW
13-1
1.3
-1]
SOU;T10~
Pi
Pi
P,
P
:::L
P,
7 Mpa
T
7f)()() kpa
m=O.Si'ljlc,
intercooler pre~SUTI
j(T2(Jj7~(Kif!)
c,
P,
91n515 kpa
P,
r-'
rl
c,
Air Compressor
Air Compressor
270
and 300
and discharged it at 377.1 kpa. Determine the isothermal
efficiency.
C. 81.89%
A 67.34%
D. 67.48%
B. 76.34%
0K
n-l
Tx
T]
271
=r.~]-;\ PI
1.4-1
~=(916.515)~
SOLUTION:
(300) \ 120
= 536.28 "K
Q = m cp (T x - T I)
Q = 0.5( I)(536.28 - 300)
Q = 118.14kw
r,
For isothermal compression:
Air Compressor - 29
SOLUTION:
pII
n-I
P -
n PI VI [(_2)
n-I
n
_
I]
PI
]4-1
w
w
=
PI V , ln(P2 / Pd
W
=
101.4(0.189) In(377.11101.4)
W = 25.17kw
An air compressor is tested and it is found that the electric motor used
37.3 kw when the compressor handled 0.189 m3/s of air at 101.4 kpa
and 300 0 K and discharged it at 377.1 kpa, Determine the adiabatic
efficiency.
A. 67.34%
C. 81.89%
B. 76.34%
D. 92.34%
w
W
1.4(101.4)(0.189) [(377.1/101.4)
1.4 - I
30.54 kw
L4
-1]
Isothermal Efficiency = 25.17/37.3
Isothermal Efficiency = 67.48%
Air Compressor - 31
Calculate the volumetric efficiency of a single-cylinder, double acting
compressor with a bore and stroke of 0.45 x 0.45 m. The compressor
is tested at 150 rpm and found to deliver gas from 101.3 kpa and
3/s
300 0 K to 675 kpa at a rate of 0.166 m when n = 1.33 for expansion
and compression process.
C. 48.34%
A. 46.39%
D. 74.23%
B. 56.23%
SOLUTION:
Adiabatic Efficiency = 3054137.3
Adiabatic Efficiency = 81.89%
Vo
V0
Vo
Air Compressor - 30
n.
An air compressor is tested and it is found that the electric motor used
37.3 kw when the compressor handled 0.189 m 3/s of air at 101.4 kpa
llv
n,
2
2(nI4)d L N
= 2[(nI4)(0.45)2(0.45)(l50/60)]
3/s
= 0.3578 m
= V [I v«
= 0.166/0.3578
= 46.39%
=
,~
r
272
V D = 0.1227
V I = 0.1227 (0.5289)
3/s
VI = 0.0649 m
Air Compressor - 32
A reciprocating compressor with a 3% clearance receives air at 100
kpa and 300 0K and discharges it at 1.0 Mpa. The expansion and
compression are polytropic with n = 1.25. There is a 5% pressure
drop through the inlet and outlet valves.
Find the volumetric
efficiency.
A. 76.23%
C. 98.33%
B. 82.50%
D. 65.33%
100kPa
SOLUTION:
PI
PI
P2
P2
=
~
=
~
T]v =
n,
=
T]y =
273
Air Compressor
Air Compressor
100 (1 - 0.05)
95 kpa
1,000 (1 + 0.05)
1050 kpa
P2
n-I
nP V
P
--
n-l
PI
_1_1 [(_2) n
=
-
1]
1 3-1
1.3(95)(0.0649)
W
=
W =
[(2000/95)
--"
1-,
-1]
1.3 - I
27.25 kw
1000kPa
Q---t=f
1 + c - c(-)
PI
W
Air Compressor - 34
3/s
I/n
Compressor
1+ 0.03 - 0.03(1050/95)1/125
A water-jacketed air compressor handles 0.143 m of air entering at
965 kpa and 21°C and leaving at 480 kpa and 132°C; 10.9 kg/h of
cooling water enters the jacket at ISoC and leaves at 21°C. Determine
the compressor power.
..
C. 34.44 kw
D. 19,33 kw
A. 26.163 kw
D. 17.23 kw
SOLUTION:
82.50%
~
W
Air Compressor - 33
A reciprocating compressor has a 5% clearance with a bore and
stroke of 25 x 30 em. The compressor operates at 500 rpm. The air
enters the cylinder at 27°C and 95 kpa and discharges at 2000 kpa. If
n = 1.3, determine the compressor power.
A. 12.34 kw
C. 27.25 kw
B. 18.45 kw
r 'j'-;
=
480
21+273
\96.5
n-llnl.377
n
n
SOLUTION:
W
Compressor r
I
1.249(96.5)( 0.143)
[( 480/96.5)
1.249 - 1
W = 26.087 kw
Q -z: heat loss
Q = me p(t 2 - t.)
Q = (10.9/3600)(4.187)(21 - 15)
=
1-S oC
-
Compressor Power
In4.974
1.249
=
Radiator
-""Q
n
PI
132 + 273
r-;
-..I
n-I
D, 34.23 kw
P2 lin
Ilv = 1 + c - c(-)
PI
T]y ~ 1 + 0.05 - 0.05(2000/95)1111
Tly = 52.89%
VD ~ (11:/4)(0.25)2(0.3)(500/60)
'1-
=l~j
2
T
T\
21°C
n-I
(
i 249-1
._._1.249
-1]
274
Q
275
Air Compressor
Air Compressor
0.076 kw
Compressor power = W + Q
Compressor power =, 26.087 + 0.076
Compressor power = 26.163 xw
% Increase
%Increase
,'.
'.~
41.63-37.45
=
.
.;'.
=
=
37.45
//./6%
'il
}!
h
~
:.1",
d
Air Compressor - 36 (ME Bd Oct. 1997)
Air Compressor - 35 (ME Bd. Apr. 1998)
A 2-stage compressor operates between constant pressure limits of
98.6 kpa and 1.103 M pa. The swept volume of the low pressu re piston
is 0.142 m'', Due to failure of the cooling. water supply to tbe
intercooler , air is passed to the high pressure cylinder without
reduction in temperature. Using PV I. 2 = C, determine the percentage
increase in power.
A. 26
B. 21
C. 11
An ideal-single stage air compressor without clearance takes in air at
100 kpa with a temperature of 16°C and delivered it at 413 Kpa after
isentropic compression. What is the discharge work done by the
compressor in KJ/kg?
SOLUTION:
D. 16
c,
c,
=
~PIP2
=
';'-98-.6-(-11-03-)
2nP,V\
n- 1
W =
W
=
~ --;;--1
[()
PI
2(1.2)(98.6)(0.142)
98.6Kpa y
i
c,
]
=
_1_'
n -1
W
W
l PI /
r
[(329.8/98.6)
­
1.2-1
12
-
n-I
P ­
[(_2 ) n
_
1]
PI
1.2(98.6 )(0.142)
­
-----'-'- -[(1103/98.6)
1.2 - 1
41.63 KJ
1.2-1
1.2
-
I]
m
1- 1.4
1]
4 - J. pe
~
LJ
I
14-'
1
l100
W/m = -/45 KJ/kg (no answer in the given choices)
Single stage
Solving for power due to single stage:
W
1- k
Jk
I
I
1.2 - 1
37.45 KJ
nP V
kmRTilP2
k-I
W = -1.4(0.287)(16+273)1(413)-;-4-_
n-I
=
W
,-­
Two-stage
P,
P, = 329.8 kpa
Solving for power due to two staging:
W
100Kpa
r'(
SOLUTION:
P,
P,
C -54.75
o -5613
A. -59.22
B. -52.43
276
Pumps
Pumps
~.
h,
h,
PUMPS
=
277
:2.5 + 2 + 0.8
5.3 m
Pumps -1
Pumps - 3
A double suction centrifugal pumps delivers 70 fe/sec of water at a
head of 12 m and running at 1250 rpm. What is the specific speed of
the pump?
A.5014rpm
C. 2345 rpm
B. 6453 rpm
D. ~968 rpm
A centrifugal pump requires 40 ft head to deliver water from low level
to higher level. If pump speed is 1600 rpm, determine the impeller
diameter of the centrifugal pump.
C. 154 mm
A. 185 mm
D. 176 mrn
B. 160 mm
SOLUTION'
SOLUTION:
N,
v
J2gh
---_.
v = J2(9.81(40/3.281)
v = 15.466 m/sec
v=nDN
15.466 = nD(l600/60)
D = 0.]8461 m
D = 184.61 mrn
N.JQ
=-~
Q =
Q =
h =
h =
h
70/2 fe/sec x 7.481 gal/1 ft3 x 60sec/l min
15710.10 gal/min
70ft'/s
12 x 3.281
39.37 ft
~--
Ns
=
1250.J15710.10
(39.37)3/4
N,
=
9968.4 rpm
35ft'/s
-----.,
~
....... I
ill
=
s~ft'/S
Pumps - 4
Pumps - 2
The pump centerline of a centrifugal pump is located 2.5 m above
from the high tide level. The sea. water varies two meters from high
tide to low tide level. If friction loss at the suction is 0.8 m, determine
the total suction head.
A. 5.30 m
C. 6.30 m
B. 2.30 m
D. 8.23 m
r: .... rr®D= 4
SOLUTION:
h,
=
total suction head
H~~~ Tide Level
The suction pressure of a pump reads 2 in. of mercury vacuum and
discharge pressure reads 130 psi is use to deliver 100 gpm of water
with specific volume of 0.0163 fellb. Determine the pump work..
A. 4.6 KW
C. 7.4 KW
B. 5.7 KW
D. 8.4 KW
SOLUTION:
PI = -2 in Hg x 101.325/29.9:­
PI = - 6.773 Kpa
P2 = 130 psi x 101.325/14.7
P 2 = 896.071 Kpa
w = l/v
w = 1/.0163
w = 61.35 Ib/ft; x 9.81/62.4
w = 9.645 KN/m>
278
Pumps
P2
h
h
=
-
279
Pumps
PI
w
896.071- (-6.773)
T\combtned
0.85(0.7)
11combined
59.50%
-----'----....:...
9.645
h = 93.075 m
Q = 100 gal/min x 3.785Ii1lgal x Im 3/1000li x 1/60
Q = 0.006308 m 3/sec
P = wQh
P = 9.645(0.006308)(93.075)
P=5.69KW
Pumps - 7
In a boiler feed pump, the enthalpy at the entrance is 765 KJ/kg. If
pump has a head of 900 m, what is the exit enthalpy of the pum p.
A 897 KJ/kg
C. 774 KJlkg
B. 465 KJ/kg
D. 864 KJlkg
Pumps - 5
SOlCTION
A pump is to deliver 150 gpm of water at ahead of 120 m. If pump
efficiency is 70%, what is the horsepower rating of motor required to
drive the pump?
A. 40.44 Hp
C. 38.44 Hp
B. 25.66 Hp
D.2111Hp
m(h:? - hi) = m x h x 0.00981
h:?-765 = 900 x 0.00981
h:? = 773.83 KJlkg
h,
h,
~~.\
h=900m
SOLUTION:
W, = wQh
W, = 9.81 (150gal/min x 0.003785m 3/lgal x 1/60)(120)
w, = 11.139 KW
BP = 11.139/0.7
BP = 15.913 KW
BP = 21.33 hp
Pumps - 8
A submersible pump delivers 350 gpm of water to a height of 5 ft from
the ground. The pump were installed 120 ft below the ground level and
a draw down of 8 ft during the operation. If water level is 25 ft above
the pump, determine the pump power.
C. 7.24 KW
A. 7.13 KW
B. 4.86 KW
D. 864 KW
Pumps - 6
SOLUTION:
A motor is used to drive a pump having an efficiency of 85% and 70%
respectively. What is the combined efficiency of pump and motor?
A. 59.50%
C. 62.50%
B. 61.50%
D. 65.50%
SOLUTION:
11combtned =
11p 11m
h = 5+120-(25-8)
h = 108/3.281
h = 32.916 m
Q = 350 gal/min x 0.003785m 3/gal x Imin/60sec
Q = 0.02246 m 3/sec
W, = wQ h
DiU
Pumps
WI'
Pumps
9.81 (0.02246)(32.916)
725 KW
w,
BP
BP
=
=
281
32846/0.65
505.32 KW
Pumps - 9
Pumps -11
Determine the number of stages needed for a centrifugal pump if it is
used to deliver 400 gal/min of water and pump power of 15 Hp. Each
impeller develops a head of 38 ft.
A. 6
C. 8
f}. 4
D.7
What power can a boiler feed pump can deliver a mass of 35 kg/s
water at a head of 500 m?
C. 456.64 KW
A. 356.56 KW
D. 171.67 KW
B. 354.54 KW
SOLUTION:
SOLUTION:
W,
15 x 0.746
=
h
h
=
=
P
=
mxhxO.00981
P
=
35 x 500 x 0.00981
p
=
171.675KW
w Q h
9.81 (400 galimin x 0.00785m J /gal x 1/60)h
45.20 m x 3.281 film
148.317 ft
=
Number of stages ~ 148.317/38
Number of stages = 3.903 stages
N umber of stages "" 4 stages
Pumps - 12
A pump running at 100 rpm delivers water against a head of 30 m. If
pump speed will increased to 120 rpm, what is the increase in head?
A. 43.2 m
C. 34.6 m
B. 13.2 m
D. 56.3 m
Pumps - 10
SOLUTION:
A boiler feed pump receives 50 Ii/sec of water with specific volume of
0.00112 mJ/kg at ahead of750 m. What is the power output of driving
motor if pump efficiency is 65%?
A. 505.32 KW
C. 785.56 KW
B. 643.54 KW
D. 356.45 KW
SOLUTION:
W,
w,
w,
=
=
=
w Q h
(1/0.001 12x 0.00981)(0.050)(750)
32846 KW
h,
N2
2
-=(-)
hi
N1
h,
120 2
-" = ( - )
30
100
h 2 = 43.2 m
Increased = 43.2 - 30
Increased = 13.2 m
~I
II
,.~
·1
.Jv·
-,·r
282
Pumps
283
Pumps
Pumps - 13
Pumps - 15
A pump is used to deliver 50 Ii/sec of sea water at a speed of 120 rpm.
If speed will increased to 135 rpm, determine the increase in pump
A certain pump is used to deliver 150 gpm of water having a density of
61.2 Ib/fe. The suction and discharge gage reads 4 in Hg vacuum and
25 psi, respectively. The discharge gage is 2 ft above the suction gage.
What is the brake power ofthe motor if pump efficiency is 75%?
A. 3.24 Hp
C. 5.45 Hp
B. 2.67 Hp
D. 6.89 Hp
capacity.
A. 56.25 li/sec
B. 34.56 li/sec
C. 87.54 Ii/sec
D. 6.260 IiIsec
SOLUTION:
SOLUTION:
Q
N
2-2
-
Q1
N\
Q
135
h
=
Pd
-
P,
+z
W
Ps = - 4 in Hg x 14.7/29.92
P, = -1.965 psi
Pd = 25 psi
25 - (-1.965)
h = [
](144) + 2
61.2
h = 65.45 ft
BP = w Qh
(61.2)(150/7.481)(65.45)
BP = ...:...----'--'----...:....:...--....:...
33,000(0.75)
BP = 3.24 Hp
2- -
50
120
Q2 = 56.25 Ii/sec
Increased = 56.25 - 50
Increased = 6.25 Ii/sec
Pumps - 14
A 15 KW motor running at 350 rpm is used to drive a pump. If speed
will changed to 370 rpm, what is the increase in power?
A. 2.72 KW
C. 56.45 KW
B. 17.72 KW
D. 5.67 KW
SOLUTION:
P2
No
3
p\
P2
N,
370
3
]5
350
- = ( - ")
-=(-)
P2 = 17.72 KW
Increased = 17.72 - IS
Increased = 2.72 KW
Pumps - 16
The discharge pipe of a pump is 400 mm in diameter delivers 0.5
mJ/sec of water to a building which maintains a pressure of 100 Kpa at
a height of 30 m above the reservoir. If equivalent head is 2 m, what
power must be furnished by the pump?
A.21IKW
C.340KW
B. 480 KW
D. 240 KW
SOLUTION:
Q = Axv
0.5 = (re/4 x 0.4 2)
V = 3.9788 mlsec
V
284
A 1265 KW
B. 23.54 KW
P
v
h = --+-+Z
2g
w
(3.9788) 2
h =
2(8.81)
43 m
285
Pumps
Pumps
100
+-+(30+2)
9.81
r
SOLUTION:
-+Pd
.=~~~_.
110m
I
n =
W p = wQh
W p = 9.81(0.50)(43)
W p = 210.92 KW
v,
C. 14.17 KW
D. 45.35 KW
A
12m
Wp = w Q h
h = (72 - 10) + 0.1
h = 62.15 ill
W p = 9.81(0.015)(62.15)
w, = 9.145 KW
Power input = 9.145/0.65
Power input = 14.07 KW
+-
Pumps -17
A centrifugal pump is designed for 1800
Determine the speed if impeller diameter is
254 mm.
A. 1000 rpm
C.
B. 1250 rpm
D.
rpm and head of 61 m.
reduced from 305 mm to
1500rpm
1600 rpm
SOLUTION:
h,
D2
2
h)
h,
D1
254
2
-=(-)
-=(-)
42.3
N2
The elevation of suction reservoir is 5 m above the pump centerline
and delivers to 85 m elevation tank which maintain 150 Kpa. If 1.5
mJ/sec of water is used to deliver a total head of 3m, determine the
power needed by the pum p.
A. 1446 KW
C. 4675 KW
B. 2567 KW
D. 3456 KW
h
h
P
-=(-)
hI
N]
-=(-)
Pumps - 19
SOLUTION:
61
305
h 2 = 42.30 m
h2
N2 2
2
P
P
=
=
(85 - 5) + 3 + 150/9.81
98.29 m
=
w Qh
=
9.81(1.5)(98.29)
1446.34 KW
=
61
1800
N 2 = 1499 rpm
Pumps - 20 (ME Bd. Oct. 1989)
Pumps - 18
Water from a reservoir A 10 m elevation is drawn by a motor driven
pump to an upper reservoir B at 72 m elevation. Suction and discharge
head loss are 0.15 rn, respectively. For discharge rate of 15 li/sec, find
the power input to the motor if overall efficiency is 65%.
I
151i/s
Water from a reservoir is pumped over a hill through a pipe 900 mm
in diameter and a pressure of one kg/crrr' is maintained at the pipe
discharge where the pipe is 85 m from the pump centerline. The pump
have a positive suction head of 5 m. Pumping rate of the pump at 1000
rpm is 1.5 mJ/sec. Friction losses is equivalent to 3 m of head loss.
What amount of energy must be furnished by the pump in KW?
I
286
Pumps
A. 1372 kw
B. 1523 kw
Pumps
C. 1234 kw
D. 1723 kw
0.5
Yd =
v.
SOLUTION:
•Yd
=
I,
Q/A
1.5
(1t 14)(0.45)2
:41
5 II
1.em'
85m
Yd
Yd =
(1t/4)(0.9)2 I
2.358 mlsec
':=-=0
Pd = 1 kg/ern" x 101.325/1.033 .
P d = 98.088 Kpa
P 5 = O(open to atmosphere)
p_p
h = (zd -zs)+(
d
w
h
=
h
=
Yd
= 3.144 mlsec
Pd
=
Pd
~~
1.0 kg/ern' x 101.325/1.033
98.088 Kpa
h
p_p
(zd-zs)+( d
s)+(hLs+h Ld)+
h
(30) + (
h
42m
w
98.088-0
9.81
2_y 2
y
s)+(hLs+h Ld)+
----c:-
_------.J
Q=1.5m'/s
5m
s
d
2g
.
98.088 - 0
(2.358)2 - (0)2
(85 - 5) + (
) + 3 + ...:....-------:---=---..:.9.81
2(9.81)
93.28 m
Water Power = w Q h = 9.81(1.5)(93.28)
Water Power = 1372.6 KW
Pumps - 21 (ME Bd. Oct. 1986)
. Water from a reservoir is pumped over a hill through a pipe 450 mm
in diameter and a pressure of 1 kg/cm 2 is maintained at the summit.
Water discharge is 30 m above the reservoir. The quantity pumped is
0.5 m'/sec. Frictional losses in the discharge and suction pipe and
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, KW?
A. 202
C. 204
B. 206
D. 208
= Q/A
d
s
2g
(3.144)2 -(0»)
) + 1.5 + ...:....------'-------'-2(9.81)
Pumps - 22 (ME Rd. Apr. 1988)
Water from an open reservoir A at 8 m elevation is drawn by a motordriven pump to an open reservoir R 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 a
discharge rate of 10 Ii/sec, find the power input to the motor and the
pressure gage readings installed just at the outlet and inlet of the
pump in Kpag.
A. 3.34 kw
C. 6.59 kw
B. 5.45 kw
D. 7.84 kw
v.;SOLUTION:
v,
Yd
2_y 2
y
Water power = w Q h
Water power = 9.81(0.5X42)
Water power = 206 kw
y, = Q/A
SOLUTION:
287
=
0.010,
v, = 0.31831 mlsec
Yd = Q/A
r~ ~
8m
10lils
®
~J
:
--
I
288
0.010
v,
Vd
(re /4 )(0.15)2
= 0.566 m/sec
Yd
h Ls
=
3(
(0.31831)2
Ys
(Zd~Zs)+(
Pd
-
Ps
w
0.1782
(rr / 4)( 5 / 12) 2
1.307 fils
0.1782
Yd =
(n I 4)(4 / 12) 2
= 2.043 fils
From Steam Table, at 150 psig(l64.7 psi) and 140°F,
w = 61 .424 Ib/fi l
p_p
y 2_y 2
h=(d
s)+(d
s)
w
2g
)
2(9.81)
hLs = 0.01549 m
(0.566)2
h Ld = 20(
)
2(9.81)
h Ls = 0.32642 m
h
289
Pumps
Pumps
Yd
Y/
- Ys2
)+(hLs+hLd)+·~-~-
2g
h =(
150(l44)-(-2xI4.7xI44)
61.424
+[
(2.043)2 -(1.307)2
]
2(32.21)
h = 354 ft
h = (66_4)+0+(0.01549+0.3264)+[(0.566)2 -(0.31831)2
2(9.81)
]
h = 62.35 m
Water Power = w Q h
Water Power = 0.010(9.81)(62.353)
WaterPower = 6.12KW
20wer Input = 6.12/0.78
Power Input = 7.84 KW
Water Power
(61.424)(80 17.481)(354)
33,000
Water Power = 7.05 hp
Brake horsepower = 7.05/0.7
Brake horsepower = 10.07 Hp
Power Input of motor = 10.07/0.80
Power Input of motor = 12.59 Hp
Power Input of motor = 9.39 KW
Pumps - 23 (ME Rd. Apr. 1982)
Pumps - 24 (ME Bd. Apr. 1986)
A pump is to deliver 80 gpm of water at 140°F with a discharge
pressure of 150 psig. Suction pressure indicates 2 inches of mercury
vacuum. The diameter of suction and discharge pipes are 5 inches and
4 inches, respectively. The pump has a efficiency of 70%, while the
motor efficiency is 80%. Determine power input to the drive motor
A. 7.23 kw
C. 8.34 kw
B 2.34 kw
D. 9.39 kw
SOLUTION:
Q
Q
=
=
(80)/(7.481 x 60)
0.1782 fills
Determine the water horsepower and the mechanical efficiency of a
centrifugal water pump which has an input of 3.5 Up if the pump has
an 8 inches nominal size suction and 6 inches nominal size discharge if
it handles 150 gpm of water at 150°F. The suction line gage shows 4"
Ug vacuum and the discharge gage shows 26 psi. The discharge gage is
located 2 feet 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. 4.23 hp
B. 6.33 hp
D. 8.34 hp
290
Pumps
291
Pumps
SOLUTION:
Pumps - 26 (ME Bd. Oct.l984)
150 gal/min x I fel7.48gal x IminJ60sec
0.334 ft3/ sec
P, = - 4 in Hg x 14.7/29.92
P, = -1.965 psi
Vs = velocity at suction
v. > Q/A
v, = 0.334/[n/4 (8112)2]
v, = 0.957 ft/sec
Vd = Q/A
Vd = 0.334/[n/4 (6!12i]
Vd = 1.701 ft/sec
From steam table, at 150°F,
w = 1/0.01634
w = 61.2 lb/ft'
h=(
Q
=;
Q
=
(26xI44)-(-1.965xI44)
+2+[
=
SOLUTION:
(1.701)2 -(0.957)2
61.2
h
A boiler feed pump receives 40 liters per second at 180°C. It operates
against a total head of 900 m with an efficiency of 60%. Determine
power output of the driving motor in KW.
A. 453.23
C. 983.45
C. 623.34
D. 523.27
From steam table(table 4), at 4 Mpa and 180°C,
h, = 764.74 KJ/kg
3/kg
VI = 0.00112484 m
Density = 1/0.00112484
3/1000kg)
Density = 889.015 kg/rrr' (Im
Density = 0.889 kg/li
Waterpower = (40xO.889015)(900)(0.0098!)
Water power = 313.964 kw
Power output of motor = 313.964/0.60
Power output of motor = 523.273 kw
]
2(32.2)
67.83 ft
Water Hp
(61.2)(0.334)(67.83)
=
33,000
Water Hp
=
2.521 .Hp
Pumps - 27 (ME Rd. Oct. 1984)
Pumps - 25 (ME Rd. Oct. 1984)
A boiler feed pump receives 40 liters per second at 4 Mpa and 180°C.
It operates against a total head of 900 m with an efficiency of 60%.
Determine the enthalpy leaving the pump in KJ/kg
A. 783.45
C. 756.23 .
B. 773.57
D. 765.23
SOLUTION:
180°C, -4MPa
-401ils
----.....'
Pump Work = m(h z - hi) = m x h x 0.00981
(h- - 764.74) = 900 x 0.00981
hz = 773.57 KJ/kg
'"
•
,1 '
p
SOLUTION:
2
~®_
.
A boiler feed pump receives 40 liters per second at 180°C. It operates
against a total head of 900 m with an efficiency of 60%. Determine
discharge pressure in Kpa for a suction pressure of 4 Mpa.
A I J ,850 kpa
C. 12,566 kpa
B. 13,455 kpa
D. 14,233 kpa
"
h, - h.. gOOm h, .
Pump Work = m(hz-h\) = mxhxO.00981
(h z - 764.74) = 900 x 0.00981
h 2 = 773.57 KJ/kg
h 2-h l
=
Vl(P 2
- P \)
773.57 - 764.74
P 2 = 11,850 Kpa
=
0.001 12484(P2 - 4000)'
29::'
Pumps
293
Pumps
Water power
Pumps - 28 (ME Bd, Apr. 19(0)
(260/7.481)(62.4)(226)
33,000
Water power
A boiler feed pump receives 45li/sec of water at 190°C and enthalpy of
839.33 KJ/kg. It operates against a head of 952 m with efficiency of
Brake Hp = 14.85/0.70
Brake power = 21.21 Hp
70%0 Estimate the water leaving temperature assuming that the
temperature rise as due to the inefficiency of the input energy
A. 191°C
C. 123°C
B. 143°C
D. 165°C
SOLUTION:
Let m,
mass flow rate. kg/sec
Pump work = rn., x h x 0.00981
mw (h 2 - h.) = m; x h x 0.00981
h- - 839.33 = 952 x 0.00981
11 2 = 848.67 KJ/kg
m
- h )
_,,~.",----_I -m",(h 2 -hl)=m c
-t
14.85 Hp
Pumps - 30 (ME Bd. Apr. 1985)
Cc
w
1"1 p
P(t 2
(84867 - 839.32>__ (848.67- 839.33) = (4.187)( t 2
0.70
12 = 191°C
-
l
)
190)
A submersible, multi-stage, centrifugal deep well pump 260 gpm
capacity is installed in a well 27 feet below the static water level and
running at 3450 rpm. Drawdown when pumping at rated capacity is
10 feet. The pump delivers the water into a 25,000 gallons capacity
overhead storage tank. Total discharge head developed by pump,
including friction in piping is 243 feet. Calculate the diameter of the
impeller of this pump in inches if each impeller diameter developed a
head of 38 ft.
A 3.28
C. 4.23
B. 5.33
D. 6.34
SOLUTION:
Pumps - 29 (ME Bd, Apr. 1985)
A submersible, multi-stage, centrifugal deep well pump 260 gpm
capacity is installed in a well 27 feet below the static water level.
Drawdown when pumping at rated capacity is 10 feet. The pump
delivers the water into a 25,000 gallons capacity overhead storage
tank. Total discharge head developed by pump, including friction in
piping is 243 feet. Calculate brake horsepower req uired to drive the
pump if pump efficiency is 70%.
A. 3l.31
C. 21.21
B. 41.41
D. 51.51
Let 0 = diameter of impeller
v
=
nDN
V
=
~2gh
1t
0 (3450/60) == ~2(322)(3'6)
D
~c
D
=
0.2738 ft
3.28 inches
SOLUTION:
Pumps - 31 (ME Bd. Oct. 1981)
Total dynamic head
Total dynamic head
243 - (27 - 10)
226 ft
A double suction, single stage, centrifugal pump delivers 900 m 3/hr of
sea water(SG = 1.03) from a source where the water level varies two
meters from high tide to low tide level. The pump centerline is located
294
295
Pumps
Pumps
Water Power
2.6 meters 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 0.8 m and that in 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. 3131 rpm
C. 4141 rpm
B. 5151 rpm
D. 6161 rpm
wQh
62.4(0.0557)(245)
Water Power
550
1.548 Hp
Water Power
=
Motor size
Motor size
1.548/0.64
2.42 Hp
Therefore, use 3 Hp motor
SOLUTION:
Total suction head = 2 + 2.6 + 0.8
Total suction' head = 5.4 m
Total suction head = 17.71
0=900m'/h
Total discharge head = 3 + 3
Total discharge head = 6 m
I,
Total discharge head = 19.686 ft
,I·:
Q = 900/2 (double suction)
Q = 450 m 3/hr
+1--Q = 1981 gal/min
0/2 Fl~ 012
h = 17.71 + 19.68
h = 37.392 ft
2.6m
I750 v"l98!
Ns
I iHighT'Ide level
(37.3 92) 3/4
cbts=.._:r
Ns 5151 rpm
2m
r
I
I ilow Tide level
Pumps - 33 (ME Bd. Oct. 1996)
Water is pumped at I mJ/sec to an elevation of 5 m through a flexible
hose using a 100% efficiency pump rated at 100 KW. Using the same
length of hose, what size of motor is needed to pump I mJ/sec of water
to a tank with no elevation gain? In both cases both ends of hose are at
atmospheric pressure. Neglect kinetic energy.
A. 51 KW
C. 43 KW
B. 18 KW
D. 22 KW
n
SOLUTION:
At 5 m elevation:
Water Power = w Q h
100 = 9.81(1)(h)
h = 10.194 m
~
Pumps - 32 (ME Bd. Oct. 1996)
A pump 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 feet. The
efficiency of the pump and motor are 64% and 84% respectively.
What size of motqr(HP) is required?
A. 5 Hp
C. 4 Hp
B. 3 Hp
D. 7.5 Hp
SOLUTION:
Q
Q
=
=
25 gal/min x 1min/60sec x 1ft317.481 gal
0.0557 ft3/sec
-'
Ifthere is no elevation:
h = 10.194 - 5
h = 5.194 m
Power
Power
9.81(1)(5.194) .
51 KW
Pumps - 34 (ME Bd. Apr. 1996)
A vacuum pump is used to drain a flooded mine shaft of 20°C water.
The pump pressure of water at this temperature is 2.34 Kpa. The
pump 1\ incapable of lifting the water higher than 10.16 m. What is the
atmospheric pressure?
L'Jb
Pumps
A. 1o»
B. 112
C. 98
D. 101.9
Pumps - 36 (ME Bd. Apr. 1985)
!:.
SOLUTION:
Using Bernoulli's Theorem:
2
P2 V22
PI VI
-+--+Zl =-+--+Z2
W
w
2g
2g
PI
P 2 V2 2-V t 2
-=-+
+(Z2- ZI )
w
w
2g
~
2.34
9.81 = 9.81 + 0 + 10.16
The rate of flow of water in a pump installation is 60.6 kg/sec. The
intake static gage is located 1.22 m below the pump center line 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
close to the pump as much as possible. The area of the intake and
discharge pipes are 0.093 m 2 and 0.069 m 2, respectively. The pump
efficiency is 70%. Take density of water equals 1000 kg/rrr', What is
the hydraulic power in KW?
A. 17.0
C. 31.9
D. 15.2
B. 24.5
SOLUTION:
,
Q = 60.6/l 000
Q = 0.0606 m3/sec
PI == 101.9 Kpa
0.0606/0.093
0.652 m/s
0.0606/0.069
0.878 m!s
==
=
Vct =
Yct ==
Ys
Ys
Pumps - 35 (ME Bd. Oct. 1995)
It is desired to deliver 5 gpm at a head of 640 m in a single stage pump
having specific speed not to exceed 40. If the speed is not to exceed
1352 rpm, how many stages are required? .
A. 3
C. 5
B.4
D.2
p_p
h
h ==(
SOLUTION:
(
d
s)
2_y 2
y
N == N.jQ
h 3/4
40 = 1352..[5
h 3/4
h = 319.54 ft (head per stage)
Number of stages == 640/319.54
Number of stages == 2 stag.es
s
d
+z+(--~-)
w
344.75- 68.95
2g
)+(-0.61+1.22)+[
9.81
h == 28.742 m
s
297
Pumps
(0.878)2 - (0.652)2
]
2(9.81)
Hydraulic power == w Q h
Hydraulic power == 9.81(0.0606)(28.742)
Hydraulic power == 17.10 KW
Pumps - 37 (ME Bd. Apr. 1996)
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
298
Pumps
299
Pumps
10 em at the inlet and 15 em at the exit. Neglecting any heat
interaction with the surroundings and frictional heating effects, what
is the necessary power input to the pump for a steady now of water at
the rate of 15 Ii/sec in KW if pu mp efficiency is 85%?
A. 9.54
C. 7.82
B. 5.343
D. 11.23
SOLUTION:
H = total head
P
H = z + ­
w
137
H = 8 +­
9.81
H = 21.96 ill
Power = w Q H
Power = (0.283)(9.81)(21.96)
Power = 61 kw
SOLUTION:
Q
Q
15 Ii/sec
0.015 rrr'zsec
0.015
v,
v,
8m
Q=28Jlps
(n /4)(0.10)2
1.91 m1s
0.015
Vd
Vd
Pumps - 39 (ME Bd. Apr. 1998)
(n /4)(0.15)2
0.85 m/s
V
h
)+ (
(Zd- Z
s
.
2 _ V 2
d
2g
s)
A pump receives 8 kg/s of water at 220 Kpa and 110·C and discharges
it at llOO kpa. Compute for the power required in kilowatts.
A. 8.126
C. 7.041
B. 5.082
D. 6.104
SOLUTION:
h
= 5-(-60)+ (0.85)2 -(1.91)2
2(9.81)
h = 6485
Power =
ill
Water Power = w Q h
Water Power = 9.81(0.015)(64.85)
Water Power = 9.54 KW
Power input = 9.54/0.85
Power input = 11.22 KW
h =
ill X
h x 0.00981
1100-220
m=8kgls
9.81
h = 89.704 m
Power = (8)(89.704)(0.00981)
Power = 7.04 kw
Pumps - 38 (ME Bd. Apr. 1998)
Pumps - 40 (ME Bd. Apr. 1998)
A pump lifts water at a rate of 283 Ips front a lake 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 kilowatts,
'
A. 71
C. 61
B. 41
D. 51
A fuel pump is delivering 10 gallons per minute of oil with a specific
gravity of 0.83. The total head is 9.14 m, find how much energy does
the pump consumes in KJ per hour.
A. 169
C. 189
B. 199
D. 179
301
Fans & Blowers
300
Pumps
SOLUT ION:
SO!JJT ION:
3.785(60)
Q = 10x--1000
3/hr
Q = 2.271 m
Power
Power
=
=
20,000
Q
=
Q
=
127 (101.325/760)
16.93 kpa
0.33
PI
PI
(0.83 x 9.81)(2.271)(9.14)
169 KJ/hr
Yj
Pumps - 41 (ME Bd, Apr. 1998)
of 75
A pump dischar ges 150 liters per second of water to a height
1800
is
pump
the
of
speed
the
and
75%
is
cy
efficien
meters. If the
ed?
subject
is
shaft
drive
the
which
to
N-m
in
torque
the
is
what
rpm,
C. 791
A. 771
681
D.
B 781
YI
(rr 14)(0.40)2
= 2.626
0.33
Y2
=
Y2
=
(rr 1 4)(0.35)2
3.429 m/s
y 2_y 2
p_p
s)
s)+Z+ (d
(d
2g
W
h
75-(-1 6.93)
h =[
SOLUT ION:
h
= w Q h
Power = (0.150)(9.81)(75)
Power = I] 0.4 kw
Brake power = 110.4/0.75
Brake power = 147.2 kw
1000(60)
3/s
0.33 m
=
]+(0.45 +0.075 )+[
9.81
10.05 m
(3.429)2 -(2.626 )2
]
2(9.81)
Power
Brake Power
Pumps - 43 (ME Bd. Oct. 1997)
Power = wQh
Brake Power = 2 rt T N
147.200 = 2rrT(l8 00/60)
T = 781 N-m
.Pumps - 42 (ME Bd. Oct. 1997)
er
A pump with a 400 mm diamet er suction and a 350 mm diamet
water.
15.6°C
of
minute
per
dischar ge pipe is to deliver 20,000 liters
below
Calcul ate the pump head in meters if suction gage is 7.5 em
is
gage
ge
dischar
and
vacuum
Hg
mm
127
pump centerl ine and reads
kpa.
75
reads
and
ine
centerl
pump
the
45 cm above
C. 20 m
A. 15 m
D. 10m
B. 5 m
to a
A centrif ugal pump deliver s 300,000 liters per hour of water
is 5
water
of
pressur ized tank whose pressur e is 280 kpa. The source
and
mm
300
is
pipe
suction
m below the pump. The diamet er of the
driving
the dischar ge pipe is 250 mm. Calcula te the kw rating of the
72%.
be
to
cy
motor assumi ng the pump efficien
C. 43.28%
A. 41.75 kw
D.38.1 6kw
B. 35.75k w
SOLUT ION:
Q =
300,000
1000(3600)
3/s
Q= 0.0833 m
302
0.0833
VI
VI
Pump Efficiency
70
(9.81)(0.8)h
=
------'--'-----'-
0.74
h = 66 m
(rr /4)(025)2
1.697 mls
p_p
v 2_v 2
s)+Z+(d
5)
h=(d
2g
W
280 - 0
(1.697)2 -(1.178)2
h =(--) + 5 + [
.
]
9.81
2(9.81)
h=33.62m
Water Power
Pump Efficiency
Brake Power
Water Power = w Q h
Water Power = (9.81)(0.0833)(33.62)
Water Power = 27.473 kw
27.473
0.72
Water Power
Brake Power
(rr /4)(0.3)2
1.178 mls
00833
V2
Vl
303
Pumps
Pumps
Pumps - 45 (ME Bd, Apr. 1997)
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 achieve the new head
at the same efficiency?
A. 2800 rpm
C. 3434 rpm
B. 3600 rpm
D. 2424 rpm
SOLUTION:
=
Brake Power
Brake Power = 38.16 kw (motor rating)
hI
NI
h,
200
N2
1770
-==(-)
Pumps - 44 (ME Bd. Oct. 1997)
A centrifugal pump delivers 80 liters per second of water on test.
Suction gage reads 10 mm Hg vacuum and 1.2 meters below pump
centerline. Power input is 70 kw. Find the total dynamic head in
meters.
A. 66
C. 62
B. 60
0.64
SOLUTION:
Q
Q
=
=
80/1000
0.08 mJ/s
Using the typical pump efficiency of 74%.
2
L
-=(-)
375
N2
N 2 = 2424 rpm
304
SOLUTION:
FANS AND BLOWERS
P
wa
-
Fans & Blowers - 1 (ME Apr. 1997)
wa
=
A fan whose static efficiency is 40% bas a capacity of 60,000 fe/hr at
60°F and barometer of 30 in Hg and gives a static pressure of 2 in of
water column on full delivery. What size of electric motor should be
used to drive the fan?
C. 2 Hp
A. 1/2 Hp
D. 1 '1/2 Hp
B. 1 Hp
Wa
SOLUTION:
h,
RT
101.325
0.287(25 + 273)
1.18 kg/m'
h w «;
=
Wa
(00254)( I 000)
hs =
U8
hs = 2 1.52 m
1.42
v =
(n/4)(OJ)2
s
h=
s
hs
~
h
W
~
(2 112)(62.4)
=
wa
1
f
~s:;;:~: ';:{!;)~ =~~~~
2in
wa
30 in Hg
v,
=
Vd
=
20.09 mls
1.42
1
(n
23.9 mfs
(239)-1 - (20.09)"'
=
Vd
4)(0.275)-
h = -----v
2(9.81)
h, = 8.5--1 m
300mmO
h, =' 10Alw.
wa Q h
W a (60,00 160)(10.4 I
305
Fans & Blowers
Fans & Blowers
Air Power
~
Air Power
=
-''------------=---
Air Power
=
0.315 Hp
W
a)
33,000
h = h, + h,
h = ~IS~ + 8.54
h = 30.0C' 111
Therefore: Use I hp motor (standard)
Hi>
(1.18 x 0.00981XI.42X30.06)
0.70
A fan draws 1.42 m J per second of air at a static pressure of 2.54 cm of
water through a duct 300 mm diameter and discharges it through a
duct of 275 mm diameter. Determine tbe static fan efficiency if total
fan mechanical is 70% and air is measured at 25"C and 760 mm Hg.
A. 50.1 1%
C. 65.67%
B. 54.34%
D. 45.34%
760rnrnHg
25°C
,
BP
= 0.7058 kw
BP
I]
=
wnQ h,
= ---
BP
p.1 S:\ O.00981)(l.42X21.52)
1], -=
1],
c
0.7058
50113%
a
L:"'=J-4T~-'----+
w.\Qh
'11
Fans & Blowers - 2 (ME Bd. Oct. 1997)
~
-c
~}54,m
Fans & Blowers
Fans & Blowers
306
Fans & Blowers - 3
Find the air horsepower of an industrial
a 900 mm by 1200 mm outlet.
~lIll'r gal-:l' and air density is 1.18 kg/nr',
A. 65.35 Hp
B. 52.35 Hp
Ihruu~h
fan that delivers 25 rolls of air
Static pressure is 127 mm of
C. 60.35 Hp
D. 70.35 Hp
307
b, = 125 rn
Air power = wQh
Air Power = (1.2 x 0.00981)(90,000/3600)(125)
Air Power = 36.78 KW
Size of motor = Brake Power
Size of motor = 36.78/0.65
Size of motor = 56.59 KW
,t II LIT/ON:
Fans & Blowers - 5
Q=Axv
25 = (0.9 x 1.2) v
v = 23.15m1sec
h, = v 2/2g
h, = (23.15)2/2(9.81)
h, = 27.315 m
h, = hw(dw/d a )
h, = 0.127(1000/1.18)
h, = 107.63 m
h = h, + h,
h = 107.63 + 27.315
h = 134.94 m
Air Power = w Q h
Air Power = (1.18 x 0.00981)(25)(134.94)
Air Power = 39.052 KW
Air Power = 52.35 Hp
At 1.2 kg/m 3 air density
operates at 98 Kpa and
brake power of the fan?
A. 68.4
B. 36.7
SOLUTION:
For Standard air, w = 1.2 kg/m '
h, = hwCdw/d.)
h, = 0.15(10001l.2)
C. 67.5 KW
D. 93.3 KW
KW
KW
SOLUTION:
WI
W2
W2
W2
1.2 kg/rn'
P/RT
98/(0.287)(32 + 273)
1.1195 kg/m'
=
=
=
=
w2
BP 2
----
BPI
WI
BP1
J.I195
---_.--
Fans & Blowers - 4
A boiler requires 90,000 ml/hr of standard air. The mechanical
efficiency of fan to be installed is 65%. Determine the size of driving
motor assuming fan can deliver a total pressure of 150 mm of water
gage.
A. 56.6 KW
C. 45.5 KW
B. 78.5 KW
D. 23.5 KW
a fan develops a brake power of 100 KW. If
32°C with the same speed, what is the new
100
BP2
=
1.2
93.29 KW
Fans & Blowers - 6
A fan has a suction pressure of 30 mm water vacuum with air velocity
of 3 m/sec. The discharge has 150 mm of water gage and discharge
velocity of 7 m/sec, Determine the total head of fan if air density is 1.2
3
.
kg 1m .
C. 154 rn
A. 150 m
D.156m
B. 152 m
I
*
",;
301l
Falls & Blowers
SOLUTION:
Fans & Blowers
A. 150 mm water gage
B. 180 mm water gage
h,
hs
(h w 2
-
h W1 )d w
da
[0.15 - (-0.03)J(I 000)
150 m
v
hv
WI
2g
h,
7
2
_
32
2(9.81)
h, = 2.038 m
h = 150 + 2.038
h = 152.038 m
1.2 (standard air density)
Wz = PIRT
735( 10 1.325/ 760)
Wz
0.287(93 + 273)
0.933 kg/rn '
2 _ V 2
_ct_ _
s
C. 24 J mm water gage
D. 456 mm water gage
SOLUTION:
1.2
hs
309
Wz
h2
w2
hi
h,
WI
0.933
- - - -
310
1.20
11 sz = 24 J mm water gage
Fans & Blowers - 9
Fans & Blowers - 7
A 50 KW motor is used to drive a fan that has a total head of 110m. If
fan efficiency is 70%, what is the maximum capacity of the fan using
standard density of air?
J
A. 27 m /sec
C. 3 I rrr'/sec
B. 29 mJ/sec
D. 33 m'zsec
The total head offan is 185 m and has a static pressure of 210 mm of
water gage, what is the velocity of air flowing if density of air is 1.15
kg/rn''?
A. 6.85 m/sec
C. 4.76 m/sec
B. 3.45 m/sec
D. 8.54 m/sec
SOLUTION:
SOLUTION:
h, =0.21(1000/].J)
h, = 182.6J m
h = h, + h,
J85 = 182.61 + h,
h, = 2.39J m
hv = y 2/2g
2.3J = YZ/2(9.8J)
v =6.85 m/sec
Air power = 50(0.7)
Air power = 35 kw
Air power = w Q h
35 = (1.2 x 0.00981)(Q)(110)
Q = 27.02 m'zsec
Fans & Blowers - 8
Fans & Blowers - 10
A fan using standard air condition can developed a static pressure
head of 310 mm water gage. If fan will operate at 93°C and 735 mm of
Hg, find the new static pressure required.
The volume flow of air delivered by fan is 20 mJ/sec and 180 mm water
gage. The density of air is 1.185 kg/rn'' and the motor power needed to
drive the fan is 44 KW. What is the fan efficiency?
Fan.\ .e Blowers
j/u
A. 70.26%
B. 80.26%
C. 75.26%
O. 90.26%
Fans & Blowers
Static pressure
Static pressure
31 ]
(0.9329/J .2)(310)
24] mm water gage
SOLUTION:
h, = 0.18(1000/1.185)
h, = ]51.89m
Air power = w Q h
Air power = (1.185 x 0.00981)(20)(151.89)
Airpower = 35.316 KW
T] = 35.3 ]6/44
T] = 80.26%
Fans & Blowers - 11 (ME Bd. Apr. 1984)
A fan is listed as having the following performances with standard air:
Volume discharged - 120 m3/sec
Speed - 7.0 rps
Static pressure - 310 mm water gage
Brake power required - 620 KW
The system duct will remain the same and the fan will discharge the
3/sec
same volume of 120 m
of air at 93°C and a barometric pressure of
735 mm Hg when its speed is 7.0 rps. Find the brake power input and
the static pressure required.
A. 23] rum water
C. 24] mm water
B. 251 mm waterO. 261 mm water
Air enters a fan through a duct at a velocity of 6.3 mls and an inlet
static pressure of 2.5 em of water less than atmospheric pressure. The
air leaves the fan through a duct at a velocity of 11.25 rn/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/m' and the fan
delivers 9.45 m3;sec. what is the fan efficiency when the power input to
the fan is 13.75 KW at the coupling?
C. 81.34%
A 7181%
D. 61.34~·{'
B. 91.23%
SOLUTION:
h = h, + h,
p_p
h=(d
For standard air, w
=
1.2 kg/rrr'
Solving for the density at 93°C and 735 mm Hg
w = P/RT
73S(101.J25 I 760)
y 2 _ y2
s)+(d
w
h =[
h
SOLUTION:
w
Fans & Blowers - 12 (ME Bd. Oct. 1994)
=
0.0762 - (-0.025)
s)
2g
](1000)+[
1.20
88.76 m
(11.25) 2
-
(6.3)2
]
2(9.81)
Air power ,~ w Q h
Air power = (1.2 x (\.0098])(9.45)(88.761)
Air power = 9.874 KW
Fan Efficiency = 9.874/13.75
Fan Efficiency c= 71.81 %
0.287(93 + 273)
w
= 0.9329 kg.nr'
Using fan laws:
Brake power input = (0.9329/] .2)(620)
Brake power input = 482 kw
Fans & Blowers - 13 (ME Bd. Apr. 1995)
A fan delivers 4.7 m 3/sec at a static pressure of 5.08 em of water when
operating at a speed of 400 rpm. The power input required is 2.963
KW. If 7.05 m3/sec are desired in the same fan and installation, find
the pressure in cm of water.
l
i
'J
312
Fans & Blowers
Fans & Blowers
A. 7.62
B. 17.14
BPI
C. 11.43
D. 5.08
BPz
w2
T]
6.5 65 + 273
N,
P2
P2
-1 - Q2 N 2
4.7
400
7.05 N z
N 2 = 600 rpm
h,
NJ z
-==(-)
h,
N2
5.08
400 2
-==(-)
hz
600
h1 = 11.43 em of water
Fans & Blowers - 14 (ME Bd. Apr. 1995)
=
2 1+ 273
5.68 KW
Fans & Blowers - 15 (ME Bd. Oct. 1996)
Air is flowing in a duct with velocity of 7.62 m/s and static pressure of
2.16 em water gauge. The duct diameter is 1.22 m, the barometric
pressure 99.4 Kpa and the gage fluid temperature and air temperature
are 30°e. What is the total pressure against which the fan will operate
in em of water?
A. 3.25
C. 3.75
B. 2.5
D. 1.25
SOLUTION:
V
A fan described in a manufacturer's table is rated to deliver 500
m3/m in 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 air handled were at 65°C instead of standard 21°C, find the power
in KW.
A. 3.82
C. 4.66
B. 5.08
D. 5.68
SOLUTION:
At 305 rpm and 21 0C:
PI/P z = (N,/N 2 ) 3
3.6/P2 = (250/305)3
P1 = 6.5 KW
At 305 rpm and 65°C:
w = P/RT
WI
PI RT,
---\V
P/RT2
2
\V 1
T2
---
w2
T]
T2
--=-=-
SOLUTION:
Q
WI
313
hv
2
fa
~b
h
(7.62)2
=
---
2(9.81)
h, = 2.959 m
w = P/RT
w = 99.4/(0.287)(30+273)
w = 1.143 kg/nr'
Solving for the velocity head in terms of em of water
h, = 2.959(1.14311000)
h, = 0 0034 m of water
h, = 0 34 em of water
h = h, + h,
h = 216 + 0.34
h = 2.5cmofwater
v
'[I
I
11
314
Fluid Mechanics
Fluid Mechanics
315
FLUID MECHANICS
Fluid - 2 (Math-ME Bd Oct. 1997} .
What is the expecten nead loss per mile of a closedcircular pipe (17
in inside diameter), friction factor of 0.03 when 3300 gal/min of water
now under pressure?
A. 38 ft
C. 0.007 ft
B.3.580ft
D. 0.64 ft
Fluid - I (Math-ME Bd Oct. 1997)
A perfect venturi with throat diameter of 1.8 inches is placed
horizontally in a pipe with a 5 in inside diameter. Eighty (80) Ib of
water now through the pipe each second. What is the difference
between the pipe and venturi throat static pressure?
z
A. 29.91b/in
C. 5020lb/in z
z
B. 34.81b/in
D. n.3lb/in z
SOLUTION:
Velocity
~-
QIA
3300/7.481
SOLUTION
Velocity
(n ! 4 )(17 / 12) 2 (60)
~
VI
Velocity = 4.66 ft/s
L = 5280 ft (I mile)
QIA I
80/62.4
VI
CD
-..::c..~":.:c _ _
®
I
--."""::"-.... -:.~J.~::-:'-=--(rr 14)(5/12)2
m=80lb/s
V
9.4 fils
--+ \ 5" ................8"1 P ~
VI
QIA z
80/62.4
Vz
V,
p,(,L~i~:,:.~~.,.;"=·~l~_~< "-~ c
_
Head loss =
f L v
2
..Y.J
12:] Q=33~Ogal/min
L=1mile
2 g 0
)
Head loss
0.03(5280)(4.66) 2
2(32.2)(17/12)
Head loss = 37.7/1
Vz
(n 14)(1.8/12)2
= 72.549 fils
Vz
Fluid - 3 (Math-ME Bd Apr. 1997)
Apply 109 Bernoulli's Equabon:
2
VI
P2 V/
PI
-+--+Zt =-+--+Z
w
2g
. w
2g
For horizontal, z,
= Zz
PI - P2
2_V 2
V_2_
_1
w
2g
PI -
P2
62.4
(72.549)2 _(9.4)2
2(32.2)
PI - Pz = 5014.281b/tY x I1144
PI - P2 = 34.82 psi
A rigid container is closed at one end and measures 8 in diameter by
] 2 in long. The container is held vertically and is slowly moved
downward until the pressure in the container is 15.5 psia. What will
be the depth of the water surface measure from the free. water
surface'?
A. 22 in
C. 12 in
B 9.2 in
D. 9.8 in
Water Surface
SOLUTION:
Pabs
=
P~age + Patn.
15.5 = Pgage + 14.7
Pgage = 0.8 psi
Pgage = w h
0.8(144) = 62.4(h)
h = 1.846 ft x 12
h = 22.15 in
- - ---- -- --- -_
-.h
8"16
Air
15.5psia
-
12"¢
316
Fluid Mechanics
Fluid Mechanics
PI -P2
Fluid - 4 (Math-ME Bd Apr. 1997)
C. 1.05 hp
D. 2.54 hp
h
h
V
2
)
'+(Z2- Zj
V/-O
0.075
2(32.2)
+0
V z = 73.2 ft/sec (3600/5280)
V 2 = 49.9mph
P/w
25 x 144
~TUrblne~gpm
62.4
57.69 ft
P,-P,=25psi
Turbine power = w Q h
7.35( 62.4 )(57 .69)
Turbine power =
33,000
Turbine power = 0.802 hp
BBN
Let V = volume of stone
D. 62 mph
Rear
For the liquid
PI-Pz=wh
PI - P2 = (0.6 x 62.4)(2/12)
PI - Pz = 6.24 Ib/ft 2
By applying the Bernoulli's equation:
1
PI VI
P2
V2 - + - + Z , =-+--+Z2
W
2g
w
2g
w.s.
~
Two tubes are mounted to the roof of a car. One tube points to the
front of the car while the other point to the rear. The tube are
connected to a manometer filled with fluid of specific gravity 0.60.
When the height difference is 2 inches, what is the car's speed?
A. 46 mph
C. 50 mph
SOLUTION:
What is the density of a stone that weighs 19.9 Ib (88 N) in air and 12.4
Ib (55 N) in water?
--../::
SOLUTION:
Fluid - 5 (Math-ME Bd Apr. 1997)
B. 43.8 mph
Fluid - 6 (Math-ME Bd, Apr. 96)
A 2,651.2 kg/rn' C. 2,578.2 kg/rrr'
3
3D.
B. 2,612.5kg/m
26,700kg/m
Q = 5517.4ISl
Q = 7.35 ft 3/min
2
_
6.24
--=
SOLUTIOW
h
22
w
2g
w = density of air
3
w = 0.075 Ib/ft (standard)'
The pressure drop across a turbine is 25 psi. The flow rate is 55
gallons per minute. Calculate the power output of the turbine.
A. 0.802 hp
B. 0.41 hp
V
317
rront
r
CD
+v;-
BBN
t
If body floats. then the weight of the object is
equal to the Bouyant force.
W = BF
55N
BF is also the difference in weight of object in air
and in water.
BF = 19.9 - 12.4
BF = 7.51bs
BF = wV
7.5 = (62.4) V
V = 0.1202 ft3
Density of stone = 19.91b/0.1202 ft3
Density of stone = 165.561b/ft3 (1000/62.4)
Density of stone = 2,653.2 kg/m'
2"
Fluid
Fluid - 7 (Math-ME Bd. Oct. 1995)
The now rate of water through a cast iron pipe is 5000 GPM. The pipe
is 1 ft and the coefficient of friction f = 0.0173. What is the pressure
drop over a 100 ft length of pipe?
318
fluid Mechanics
A. 3] 7 26 lblin z
B 21078 lb/in z
C. 337.261b/ft z
D. 23.7801b/ft z
.
Turbine power
8.02(62.4)(69.23)
=
-----'--'------'--
33,000
Turbine power = 1.05 hp
SOLUTION
5000
o
o
=
7481(60)
11.139 ft3 Is
Fluid - 9 (ME Bd. Oct. 1996)
~:.D1ft
v =
,
)
a=50~Ogpm
L=100ft
v =
v =
h,
319
Fluid Mechanics
SOLUTION:
=
v
0.0173(100)(14.183)2
=
hi
=
@
P J - w h P,
PJ-P Z = wh
6.2(144) = (62.4 x 0.8)(h)
2gD
hr
The fluid in a manometer tube is 60% water and 40% alcohol (SC
0.8). What is the manometer fluid height difference if a 6.2 psi
pressure is applied across the two ends of a manometer?
A 15.5 in
C. 36 in
B.186in
D.215in
T
_
2(32.2)(1)
5.404 ft of water
h
17.88 ft
=
=
SG=O.80
214.6 in
Fluid - 10 (ME Bd. Apr. 1996)
Fluid - 8 (Math-ME Bd. Oct. 1995)
Pressure drop across a turbine is 30 psi, the flow rate is 60 gpm.
Calculate the power output of the turbine.
A 0.41 hp
C. 6,30 hp
D. 2.54 hp
B. 105 hp
An air bubble rises from the bottom ofa well where the temperature is
25°C, to the surface where the temperature 27°C. Find the percent
increase in volume of the bubble if the depth of the well is 5 m.
Atmospheric pressure is 101.528 Kpa,
A. 49.3
C. 56.7
D. 38.6
8413
SOLUTION:
SOLUTION:
h
=
o
o
t,=27"C
30 x 144
h
h
P/w
=
=
=
P,~~..... ~
~=60gpm
624
69.23 ft
60/7.481
8.02 ft3 /min
Turbine power
=
P,·P,=30psi
P2 T2
PlY]
T2
T]
----
, J:
wh+ 101.528
PI = 9.81(5) + 101.528
PI = 150.378Kpa
150.378(Y 1 )
101.528(Y2 )
PI
wOh
33,000
·_·-~'V2
=
(25 + 273)
(27 + 273)
5m
.. _ _' V,
t,=25°C
320
fluid Mechanics.
~
V2
321
Fluid Mechanics
1491V 1
Fluid - 12 (Math-ME Bd Oct. 1998)
%lncrease in volume
=
V2 -V1
---
'\ 24 inches long rod floats vertically in water. It has a I sq. in. cross
section and has a specific gravity of 0.6. What length L is submerged?
A. 14.4 in
C. 9.6 in
B. 24 in
D. 18.0 in
V1
%lnerease in volume =
%lnerease in volume
=
1.491Vj
VI
-
VI
49.10%
~w
SOLUTION:
Fluid - II
For floating object,
An empty, open can is 30 cm high with a Io-cm diameter. The can,
with the open end down, is pushed under water with a density of 1000
kg/m ', Find the water level in the can when the top of the can is 50 cm
below the surface.
W
=
w., v,
C. 4.20 em
D. 5.87 em
24"
vc;
Consider the air pressure:
r'IV, = P ZV 2
10 1.325(A x 0.3)
30.3795
Pz =
(0.3 - x)
Pw = Pz
109.173 - 9.81x
2
.
A
=
=
0.02l18m
2.12 em
':]m
I,. ~-- -.-----,1
O
.
80
I
P"P,
=
Pz[A (0.3 - x)]
80-x
bo . ,- __=- x
!
=
14.4 in
=
1000 [I xL)
--. ·~U
t
BF
10
I
-----=-------
Pw
Flow of water taking over in a pipe having a velocity of 10 m/s.
Determine the velocity head of the water.
A 50.1 m
C. 8.2 m
B. 5.1 ill
D. 100 m
SOLUTION:
h = vZ/2g
h = (10)z 12(9.81)
h = 5.1 ill
J.,
II . , . . ~10mJs
30.3795
(0.3- x)
112.1 16x + 2.3705 = 0
By qua' Iratie formula:
x
..A
v,
Fluid - 13 (Power-ME Bd Oct. 1999)
Consider the water pressure:
t'w = wh + 101.325
P w = (0.8 - x)(9.81) + 101.325
P; = 109.173 - 9.81x
9.81x
W.~L t=
I_: J
,I __
____
=!j=__
L
L
SOLUTION:
A=1 in'---.
BF
(0.6 x 1000)( 1 x 24)
A. 17.20 em
B. 2.12 em
R~
Fluid - 14 (Power-ME Bd Oct. 1999)
The length of pipe is 168 meters. If the pressure drop is 50 Kpa for
every 30 meters, what is the total oressure drop?
A. 260 kpa
C. 280 kpa
B. 300 kpa
D. 100 kpa
,')')
Fluid Mechanics
.)-­
"OU: IJON:
PI-p"
Total Pressure Drop
Total Pressure Drop
=
SO kpa (168 m!30 m)
=
280 kpa
VI
VI
Hvd ra ulic clficicncv S5%, find Q in Ii/s, Power developed 10,500 kw
C 327,1
D 3623
2g
(re / 4)(0.08)2
3.98 m/s
V2
=
z,
(re I 4)(0.04)2
15.91 mJs
p,-p)
9.81
PI - P?
11
I
15 m
-1.5 m
0.02
---' =
SOU lION
-z)
2
0.02
\/2
Ill.
A 3935
B 3e.j.52
,
?
V7'-V t ­
=--~--+(z
w
Z, - 7.1
Z2 - ZI
Fluid - 15 (I'O\~er-ME Bd Apr. 1999)
under head of 320
323
Fluid Mechanics
=
(15.91)
2
-(3.98)
2
+ (-1.5)
1-
Reference
2(9.81)
104.016 kpa
Brake Power
~-
Water Power
10,500
0.85 =
Water Power
Water Power = 12,352.94 kw
Water Power = w Q h
12,35294 = 981 (Q) (320)
Q = 3935 m 3/s (1000)
Q = 3935 li/s
Fluid - 17 (Power-ME Bd Apr. 1999)
A fluid that has a velocity of 18 m/s will have an equivalent head of:
A. 16.51 m
B. 12.44 rn
C 18.34 m
D. (0.34 m
SOLUTION:
h
v
2g
FIlJi,[ - 16 (l)o\\cr-i\lE Bd Apr. 1999)
(18)2
A c vl iud rira l pipe with water flowing downward at 0.02 mJ/s having
lOp d ia me t cr "I' O.OS, bottom diameter of 0.04 m and height of 1.5 m.
Find rhc prcxsur« between the pipe.
A 104 kpa
C 120 Kpa
B 97 kpa
0 143 kpa
h
2(9.81)
h
16.51 m
SUU.' IIO"-J.
Pi
VI"
P1
V,"
w
2g
-+--+zl = -+--+Z)
\V
~g
­
Fluid - 18
Determine the size of pipe which will deliver 5 liters of medium oil (v =
6.10 X 10,6 m 2/s) assuming laminar now conditions.
324
522 111m
A
B 454
C. 550 mm
D. 650 mm
!TIfT!
Fluid - 20
The type of flow occupying in a 30 em diameter pipe which water
6
2/s
flows at a velocity of 2.10 m/s. Use v = 1.13 X 10. m for water.
A turbulent
C. laminar
B. constant
D. none of these
SOLUTION
Q
v
325
Fluid Mechanics
Fluid Mechanics
A
0005
v
SOLUTION:
2
4)d
V ~c o 006366 d)
dV
Re ~(T[
dV
Re
v
For laminar flow, Re
2000
~
v
2000
=
d(0.006366 I d
2
_~
(0.30)(2.1 0)
)
Re
113
610 x 10
d
0522 m
d- 522 mm
Re
=
X
10-
0
577,52212
Since it is greater than 2000, then it is turbulent flow
Fluid - 19
The power available in ajet having a cross-sectional area of II.OOS m Z
with a velocity of 25.80 rn/s,
A. 34 kw
C 43 kw
B 49 k w
D. 23 kw
SOL!)TION
h
=
'/12g
(2580)2
h
----
~~'
~""
tIJijiiiI1J
..."
2(98 J)
h = 33926 m
Q .~ Av
Q ~ (0.005)(2580)
Q= 0129 mJ/s
P = wQh
P = (981)(0129)(33926)
P = 42.93 kw
..
Fluid - 21
A man weighing 64 kg causes a flat area 30 em thick to be just fully
submerged in a sea water (SG = 1.03). Neglecting weight the area
must be:
C. 0.173 m 2
A. 0.085 m '
2
D. 0.062 m 2
B. 0.756 m
S
p
---+ 25.80m/s
A"'O.005m
SOLUTION:
M
1
For floating object:
W
W
64
A
=
=
=
=
BF
wV
(1.03 x 1000)(0.3 x A)
0062 m)
~;i,~·
·f••
r
SF
326
Fluid Mechanics
327
Fluid Mechanics
o 778
V,
Fluid - 22
rn/s
0.055
V2
What force is exerted by water jet 50 mm diameter if it strikes a wall
at the rate of 15 m/s?
A. 342 N
C. 764 N
D. 5113 N
B. 442 N
,
(rt . 4)(0.075)~
V,= 12456 mis
o
)
(12456f - (0.778r
P, - P,
--+-0
2(981)
w
P, -. P,
79 111
SOLUTION:
w
F = wQv
Q = A v
Q = [(11/4)(0.05)2](15)
Q = 0.02945 m3/s
F
F
Flu id - 2-t
A jet of water 50 mm diameter with a velocity of 35 mls is being
eli,charged in a horizontal direction from a nozzle mounted on a fire
truck. The force required to hold the nozzle stationary is:
A 1.34 KN
C. 4.23 KN
D. 323 KN
B 240 KN
(1000)(0.02945)( I 5)
441.75 N
Fluid - 23
SOLUTION:
A 300 mm x 75 mm venturi meter is inserted in a 300 mm diameter
pipeline where water flows at 55 liters/s. Neglecting friction loss,
compute the drop in pressure head from the inlet to the throat.
A8m
C.6m
B. 10 m
D. 12 m
F = wQ v
Q = A x v
Q = [(11/4)(0.05/](35)
Q = 0.0687 m 3/s
\61
SOLUTION:
F
"
V 75mm
,
v,
2
PI
Vj
w
2g
P2
V/
w
2g
-+--+Zl = -+--+Z2
PI-PO
V)
---- w
2
-VI
2g
0.055
VI
(11 /4)(0.3)2
F
F
Z,
Z,
(1000)(0.0687)(35)
2404.5 ~
2.404 KN
Reference
Ftuid-25
2
+(Z2- Z j )
An open storage vessel has 3 m of oil (SG = 0.82) and 6 m of water.
The pressure at the bottom is:
C. 83 kpa
A 45 kpa
D. 92 kpa
B 65 kpa
328
Fluid Mechanics
Fluid Mechanics.
329
SOLlfTiON
Fluid - 27
F=--C:=-Oil--­
P = pressure at the bottom
p =
h, + Ww h.,
P = (082 x 9.81)(3) + 981(6)
P = 83 kpa
3m=h.
""0
Water
6m=h w
p
A 300 0101 diameter pipe discharges water at the rate of 200 Ii/s. Point
I on the pipe has a pressure of 280 kpa and 3.4 m below point 1 is
point 2 with a pressure of 300 kpa. Compute the head loss between
points 1 and 2.
A 4.2 m
C. 6.3 m
B. 2.5 m
D. 1.4 m
Fluid - 26
SOLUTION:
A 200 0101 diameter pipe gradually reduces to a 100 0101 diameter.
The 200 0101 diameter pipe is connected to another pipe having a
pressure of 600 kpa horizontally with a flow of 0.04 m 3/s. Find the
pressure at the 100 0101 diameter.
A 588 kpa
C 566 kpa
B 642 kpa
D. 598 kpa
PI
V
2
PI -
hL =
;:,OLUT10N:
hL
hL
PI V\2
P2 V2 2
-+--+ZI =-+--+Z2
W
2g
w
2g
Ffuid -
0.04
VI
VI
(n: /4)(0.2/
1.2732 m/s
0.04
Q=0.04m'/s
-----+
100mm
P2
j
v/
- + - + Z \ =-+--+Z2 +h L
W
2g
w
2g
P2
+(Z\-Z2)
w
280- 300
----+3.40
9.81
1.36 m
CD
I
p ,=280K Pa
3.4m
P,=300Kpa
2~
An object weighs 90 N in air and when immersed in water it weighs 50
N. Compute the specific gravity of the object.
A. 1.25
C. 2.25
B. 3.25
D. 4.25
V2
V2
(n: 14)(0.1/
509 m/s
SOLUTION:
-W'$''--T~~~-c.:
P7
(5.09)2
600 (12732)2
-+
+0=-- +
+0
9.81
2(9.81)
9.81 2(9.81)
BF = 90 - 40
BF = 50 N
Po = 587.85 kpa
BF = W Va
50 = 9810 v;
V o = 0.00408 rrr'
11F
56N
Wo
weight of object in air
330
Fluid Mechanics
Wo
90
w.,
SG
SG
c.
=
=
=
Past Board Examination Elements (1994-1999)
331
w, v;
"0 (0.00408)
22,058.82 N/m J
22,058.82/9810
2249
PAST BOARD EXAMINATION
ELEMENTS
Fluid - 29
Elements - 1 (ME Rd. October 1994)
A rectangular open box 7.6 m by 3 m in plan and 3.7 m deep, weighs
J5U KN and is launched in fresh water. If water is 3.7 m deep what
weight of stone placed in the box will cause it to rest at the bottom?
/\ 35034 KN
C. 498.34 KN
B 65345 KN
D. 477.57 KN
When a substance is gaseous state is below the critical temperature it is
called:
A. Vapor
B cloud
C. moisture
Do steam
Sl)J.l iTlON:
Elements - 2 (ME Rd. October 1994)
w.s.
Total weight
=
J."Oj W,
w V
=
BF
Is the condition of pressure and temperature at which a liquid and its vapor
are indistinguishable:
A. critical point
B. dew point
C. absolute humidity
D. relative humidity
Elements - 3 (ME Rd. October (994)
3"i()
W,
f
~
W, = 9.111 [(7.6)(3)(3.7)]
477.57 KN
If the temperature is held constant and the pressure is increased beyond the
saturation pressure, we have a:
A. saturated vapor
C. saturated liquid
B. compressed liquid
D. subcooled liquid
Elements - 4 (ME Rd. October (994)
A Francis turbine has what flow:
A. inwardflow reaction
B. outward flow impulse
C. outward flow reaction
D. inward flow impulse
332
Past Board Examination
/:'/1'1111'/11\
(1994-1999)
Past Board Examination Elements (1994-1999)
333
A. velocity of now only
B. pressure only
C. height above a chosen datum, density, internal energy,
Elements - 5 (ME Rd. October 1994)
pressure and velocity ojflow
The latent heat 01 vaporization in joules per kg is equal to:
2
A 5.40 x 10
3
B.4.13xl0
5
C. 22.6 X 10
5
D. 335 X 10
D. pressure, height above a chosen datum, velocity of f1ow,
density of fluid
Elements - 10 (ME Rd. October 1994)
Elements - 6 (ME Rd. October 1994)
Form of energy associated with the kinetic energy of the random motion of
large number of molecules:
A. internal energy
B. kinetic energy
C. heat of fusion
o
A type of water turbine:
A. Parson
B. Hero
C. Pelton
D. Banki
Elements - 11 (ME Rd. October 1994)
heat
Elements - 7 (ME Rd. October 1994)
In a P- T diagram of pure substance, the curve separating the solid phase
from the liquid phase is:
A. vaporization curve
B. fusion curve
C. boiling point
D. sublimation point
Elements - 8 (ME Rd. October 1994)
The number of protons in the nucleus of an atom of the number of
electrons in the orbit of an atom:
A. atomic. volume
B. atomic number
C. atomic weight
D. atomic mass
If the pressure of the confined gas is constant, the volume is directly
proportional to the absolute temperature:
A. Boyle
B. Joule
C. Charles
D. Kelvin
Elements - 12 (ME Rd. October 1994)
A theoretical body which when heated to incandescence would emit a
continuous light-ray spectrum:
A black body radiation
B. black body
C. blue body
D. white body
Elements - 13 (ME Rd. October 1994)
Elements - 9 (M E Bd. October 1994)
The energy off1uid flowing at any section in a pipeline
IS;I
Iuncuon of
Ignition of the air fuel mixture in the intake of the exhaust manifold:
A. backlash
B. backfire
334
Past Board Examination Elements (1994- / 999)
C. exhaust pressure
D. back pressure
Elemer..ts - 14 (ME Rd. October 1994)
Is the condition of pressure and temperature at which a liquid and its vapor
are indistinguishable:
A. relative humidity
B. absolute humidity
C. critical point
D. dew point
Elements - 15 (ME Rd. October 1994)
When a substance in gaseous state is below its critical temperature it is
called:
A. steam
B. cloud
C. moisture
D. vapor
Elements - 16 (ME Rd. October 1994)
.
Which of the following a set of standard condition
A. 1 atm, 255k, 22.41 m 3/kg mole
B. 101.325,273 'k, 22.4 m'rkg mole
C. 101.325, 273°k, 23.66 m3/kg mole
D. 1 arm, lOoC, 22.41 m 3/kg mole
Elements - 17 (ME Rd. October 1994)
Number of molecules in a mole of any substance is a constant called:
A. Rankine cycle
B. Avogadro's number
C. Otto cycle
D. Thompson constant
Pas: Board Examination Elements (/994-1999)
335
Elements - 18 (ME Rd. October 1994)
A simultaneous generation of-electricity and steam (or heat) in a single
power plant:
A. gas turbine
B. steam turbine-gas turbine plant
C. waste heat recovery
D. cogeneration
Elements - 19 (ME Rd. October 1994)
Is one whose temperature is below
corresponding to its pressure:
A. compression
B. condensation
C. constant volume process
D. subcooled liquid
the
saturation' temperature
Elements - 20 (ME Rd. October 1994)
Pump used to increase air pressure above normal, air is then used as a
motive power:
A. air cooled engine
B. air compressor
C. air condenser
D. air injection
Elements - 21 (ME Rd. October 1994)
If the temperature is held constant and the pressure is increased beyond the
saturation pressure, we have a:
A. compressed liquid
B. subcooled liquid
C. saturated %por
D. saturated liquid
Elements - 22 (ME Rd. October 1994)
The locus of elevations:
336
Past Board Examination Elements (1994-/999)
A.
B.
C.
D.
337
critical point
hydraulic gradient
energy gradient
friction gradient
Elements - 23 (ME Ed. October 1994)
In sensible cooling process, the moisture content:
A. does not change
B. decreases
C. indeterminate
D. increases
Elements - 24 (ME Bd. April 1995)
What is the
A.
B.
C.
D.
Past Board Examination Elements (1994-1999)
process that has no heat transfer?
reversible
isothermal
polytropic
adiabatic
Elements - 25 (ME Ed. April 1995)
The internal combustion engines never work on_ _cycle:
A. Rankine
B. diesel
C. dual combustion
D. Otto
Elements - 27 (ME Bd. April 1995)
What is the force which tends to draw a body toward the center about
which it is rotating?
A. centrifugal force
B. centrifugal in motion
C. centrifugal advance
D. centripetal force
Elements - 28 (ME Bd. April 1995)
A simultaneous generation of electricity and steam (or heat) in a single
power plant:
A. steam turbine - gas turbine
B. cogeneration
C. gas turbine plant
D. waste heat boiler
Elements - 29 (ME Bd. April 1995)
Percent excess air is the difference between air actually supplied and
theoretically required divided by:
A. tlle theoretically air supplied
B. the deficiency air supplied
C. gas turbine plant
D. waste heat boiler
Elements - 30 (ME Bd. April 1995)
Elements - 26 (ME Ed. April 1995)
The dividing point between the high-pressure arid low pressure sides of the
refrigeration cycle occurs at the:
A. expansion 'valve
B. compressor
C. condenser
D. cooling oil
What amount of air is required in a low bypass factor?
A. greater
B. lesser
C. indeterminate
D. does not change
Elements - 31 (ME Bd, April 1995)
Work done per unit charge when charge is moved from one point to
another:
331<
Past Board Examinution Elements (1994-11)99)
Past Board Examination Elements (1994-1999)
339
A. equipotential SlJrI~ICC
B. potential at II point
C. electrostatic unit
D. potential difference
Elements - 32 (ME Bd. April 1995)
A pressure of lrnillibar is equivalent to:
A. 1000 dynes/em'
B. 1000 em of Hg
C. 1000 psi
D 1000 kg/ern"
Elements - 33 (ME Bd. April 1995)
Heat transfer due to density differential:
A. convection
B. nuclear
C. conduction
D. radiation
Elements - 34 (ME Bd. April 1995)
When a system deviates infinitesimally from equilibrium at every instant
of its state, it is undergoing:
A. isobaric process
B. quasi-static process
C. isometric process
D. cyclic process
Elements - 35 (ME Bd. April 1995)
The ratio of the average load to the peak load over a designated period of
time is called:
A. load factor
B. reactive factor
C. diversity factor
D. plant use factor
Elements - 36 (ME Bd. April 1995)
Vv hat is the clockwork-operated device which records continuously the
humidity of the atmosphere?
A. hetograph
B. hygrometer
C. hydrodeik
D. hygrograph
Elements - 37 (ME Bd. April 1995)
What is an apparatus used in the analysis of combustible gases?
calorimeter differential
B. calorimeter gas
C. calorimetry
D. calorimeter
A.
Elements - 38 (ME Bd. April 1995)
If the fluid travels parallel to the adjacent layers and the paths of individual
particles do not cross, the flow is said to be:
A. turbulent
B. critical
C. dynamic
D. laminar
Elements - 39 (ME Bd. April 1995)
What equation applies in the first law of thermodynamics {or an ideal gas
in a reversible open steady-state system?
A. Q-W=Uz_U j
B. Q+VdP=Hz_H 1
C. Q - VdP = Hz_HI
D. Q-PdV=H2_H,
Elements - 40 (ME Bd. April 1995)
Is one whose pressure is higher than the saturation pressure corresponding
to its temperature:
.' ·tIJ
Past Board Examination Elements ( J fJfJ4-1999)
A.
B.
C.
D.
saturated Iiqu i d
compressed liquid
saturated vapor
compressed gas
Elements - 41 (ME Rd. April 1995)
The locus of elevation to which water will rise in the piezometer tube is
termed:
A. energy gradient
B. friction head
C. hydraulic gradient
D. critical path
Past Board Examination Elements (1994-1999)
341
C. continuous flow
D. turbulent flow
Elements - 45 (ME Rd. April 1995)
The hydraulic formula CA~2gh is used to find:
A.
B.
C.
D.
quantity ofdischarge through an orifice
velocity of flow in a closed conduit
length of pipe in a closed network
friction factor of a pipe
Elements - 46 (ME Rd. April 1995)
The sum of the energies of all molecules in a system, energies appear in
Elements - 42 (ME Rd. April 1995)
The total energy in a compressible or incompressible fluid flowing across
any section in a pipeline is a function of:
A. pressure and velocity
B. pressure, density and velocity
C. pressure, density, velocity and viscosity
D. flow energy, kinetic energy, height above datum and
several complex forms, is the:
A. kinetic energy
B. potential energy
C. internal energy
D. thermal energy
Elements - 47 (ME Rd. April 1995)
internal energy
The temperature at which its vapor pressure is equal to the pressure exerted
Elements - 43 (ME Bd. April 1995)
The ratio of the density of a substance to the density of some standard
substance is called:
A. relative density
B. specific gravity
C. specific density
D. relative gravity
Elements - 44 (ME Rd. April 1995)
At any instant, the number of particles passing every cross-section of the
stream is the same, the flow is said to be:
A. steady flow
B. uniform flow
on the liquid
A. absolute humidity
B. calorimeter
C. boiling point
D. thermal energy
Elements - 48 (ME Rd. April 1995)
Type of turbine that has high pressure and low pressure is called:
A. compound engine
B. gas turbine
C. impulse turbine
D. compound turbine
342
Past Board Examination Elements (1994-1999)
Elements - 49 (ME Bd. April 1995)
The design of an air supply duct of an air conditioning system:
A. adds moisture to the air
B. lowers the temperature of the air
C. does no affect the distribution of air
D. affects the distribution of air
Elements - 50 (ME Bd. April 1995)
The changing of solid directly to vapor, without passing through the liquid
state is called:
A. evaporation
B. vaporization
C. sublimation
D. condensation
Elements - 51 (ME Bd. April 1995)
The volume of a fluid passing a cross-section stream in unit time is called:
A. steady flow
B. uniform flow
C. discharge
D. continuous flow
Elements - 52 (ME Bd. April 1995)
Weight per unit volume is termed as
A. specific gravity
B. density
C. weight density
D. specific gravity
Elements - 53 (ME Bd. April 1995)
S.l. unit of force:
A. pounds
B. Newton
Past Board Examination Elements (1994-1999)
343
C. kilograms
D. dyne
Elements - 54 (ME Board Oct.l995)
Heat exchanger used to provide heat transfer between the exhaust gaser
and the air prior to its entrance to the combustor:
A. evaporator
B. combustion chamber
C. regenerator
D. heater
Elements - 55 (ME Board Oct.1995)
Heat normally flowing from a high temperature body to a low temperature
body wherein it is impossible to convert heat without other effects is called
the'
A. second law of thermodynamics
B. first law of thermodynamics
C. third law of thermodynamics
D. zeroth law of thermodynamics
Elements - 56 (ME Board Oct.1995)
What are the immediate undesirable products from the petroleum-based
lubricating oil when subjected to high pressure and temperature?
A. gums, resins and acids
B. sulfur
C. soots and ashes
D. carbon residue
Elements - 57 (ME Board Oct.l995)
The intake pipe to a hydraulic turbine from a dam is:
A. tailrace
B. spiral casing
C. surge tank
D. penstock
344
Past Board Examination Elements (1994-1999)
Past Board Examination Elements (1994-1999)
345
Elements - 58 (ME Board Oct.1995)
Elements - 63 (ME Board Oct.1995)
When 1 mol of carbon with 1 mol oxygen:
A. 2 mols carbon dioxide
B. 1 mol carbon dioxide
C. 1 mol carbon and 1 mol CO 2
D. 1 mol carbon monoxide
Percent excess air is the difference between the air actually supplied and
the theoretically divided by:
A. the deficiency air supplied
B the actually air supplied
C. none of these
D. the theoretically air supplied
Elements - 59 (ME Board Oct.1995)
Elements - 64 (ME Board Oct.1995)
A device for measuring the velocity of wind:
A. aneroid barometer
B. anemometer
C. anemoscope
D. anemograph
Mechanism designed to lower the temperature of air passing through is:
A. air cooler
B. air defense
C. air spillover
D. air cycle
Elements - 60 (ME Board Oct.1995)
Elements - 65 (ME Board Oct.1995)
Air standard efficiency ora diesel engine depends on:
A. speed
B. compression ratio
C. fuel
D. torque
Elements - 61 (ME BO?· _ O.l.l995)
Heavy water is:
A. B,O
B. H 20
C. W 20
D. D 20
The term "exposure" in radiological effects is used as a measure of a
Gamma ray or an X-ray field in the surface of an exposed object. Since
this radiation produces ionization of the air surrounding the object, the
exposure is obtained as:
A. x= no. of ions produced per mass ofair x coulombs per kg
B. x= mass.of air x surface area of an exposed object
C. xc~ mass of air over surface area of an exposed object
D. x= no.of ions produced per mass of air + coulombs per kg
Elements - 66 (ME Board Oct.l995)
Elements - 62 (ME Board Oct.1995)
The ratio of the sum of individual maximum demands of the system to the
overall maximum demand of the whole system:
A. demand factor
B. diversity factor
C. power factor
D. utilization factor
The viscosity of most commercially available petroleum lubricating oil
changes rapidly above:
A. 120°F
B
ISO°F
C. I SO°F
D. I :10°F
I
I
1
346
Past Board Examination Elements (1994-1999)
Elements - 67 (ME Board April 1996)
Past Board Examination Elements (1994-1999)
347
B, brew kelt le
C. cooler
D. starting tubs
A type of water turbine where ajet of water is made to fall on the blades or
buckets and due to the impulse of water, the turbine starts to move:
A Pelton wheel
B, Steam turbine
C. Francis turbine
D. reaction turbine
Elements - 68 (ME Board April 1996)
What condition exists in an adiabatic throttling process?
A. enthalpy is variable
B, enthalpy is constant
C. entropy is constant
D. specific volume is constant
Elements -69 (ME Board April 1996)
The specific gravity of a substance is the ratio of its density to the density
of:
A. mercury
R. gas
C. air
D. water
Elements - 70 (ME Board April 1996)
'''''-.,
Which is used as a moderator in certain types of nuclear reactors?
A. vapor
B. heavy water
C. hotwater
D. cold water
Elements - 71 (ME Board April 1996)
Yeast as raw material for beer making is added to the equipment called:
A. fermenters
Elements - 72 (ME Board April 1996)
What keeps the moisture from passing through the system?
A.
B.
C.
D.
dehydrator
aerator
trap
humidifier
Elements - 73 (ME Board April 1996)
What are the main components in a combined cycle power plant?
A. diesel engine and air compressor
B. gas engine and waste heat boiler
C. steam boiler and turbine
D. nuclear reactor and steam boiler
Elements - 74 (ME Board April 1996)
What do you call the changing of an atom of an element into an atcrn of a
different element with a different atomic mass?
A. atomization
B. atomic transmulation
C. atomic pile
D. atomic energy
Elements - 75 (ME Board April 1996)
What do you call the weight of the column of air above the earth's surface'!
A. air pressure
B, aerostatic pressure
C. wind pressure
D. atmospheric pressure
348
Past Board Examinatian Elements (1994-11)1)1))
Elements -76 (ME Board April 19(6)
C'-rnbined process of cooling and humidifying is also known as:
A. heating and humidifying
B. cooling tower
C. evaporative cooling process
D. moisture removal process
Elements - 77 (ME Board April 1996)
What is the fore- required to accelerate a mass of I gram at a rate of I
em/sec/sec?
A dyne
B. poundal
C. slug
D. kg force
Elements - 78 (ME Board April 1996)
What type of turbine has low head and high discharge?
A Pelton Wheel
B. Francis turbine
C. Jonval turbine
D. Kaplan turbine
Elements - 79 (ME Board April 1996)
What is a Bull Head Tee?
A a pipe tee with head shaped like a bull
B. a welded built-up tee
C. a pipe tee with its run larger than its branch
D. a pipe tee the branch of which is larger than tire run
Elements - 80 (ME Board April 1996)
What is the main power generating plant that produces the most electricity
per unit thermal energy in the fuel and has the greatest surplus of electricity
for most cogeneration systems')
Past Board Examination Elements (1994-1999)
A.
8.
C.
D.
349
steam engine
steam turbine
gas turbine
diesel turbine
Elements - 81 (ME Board April 1996)
What is the term as the ratio of the volume at the end of heat addition to
the volume at the start of heat addition?
A. compression ratio
B. air-fuel ratio
C. volumetric ratio
D. cut-off ratio
Elements - 82 (ME Board April 1996)
What is the
A.
B.
C.
D.
ideal cycle for gas turbine work?
Brayton cycle
Stag combined cycle
Bottom cycle
Ericson cycle
Elements - 83 (ME Board April 1996)
What do you call the passing of heat energy from molecuie to molecule
through a substance?
A conduction
B. radiation
C. conservation
D. convection
Elements - 84 (ME Board April 1996)
What is the lowest temperature to which water could possibly be cooled in
a cooling tower?
A. the effective temperature
B. the temperature of adiabatic saturation
C. the wet bulb depression
D the dew point temperature of the air
350
Past Board Examination Elements (1994-1999)
Past Board Examination Elements (1994-1999)
A
Elements - 85 (ME Board April 1996)
The indicator used to determine the anti-knock characteristics of gasoline:
A. aniline point
B. Cetane No.
C. Octane No.
D. Diesel Index
Elements - 86 (ME Board April 1996)
Dew point is defines as:
A. the temperature to which the air must be cooled at
351
isotropic
B. adiabatic
C. isometric
D. isobaric
Elements - 90 (ME Board April 1996)
A receiver in an air compression system is used to:
A. avoid cooling air before using
B. increase the air discharge pressure
C. collect the water and grease suspended in the air
D. reduce the work needed during compression
constant pressure to produce saturation
B. the point where the pressure and temperature lines meet
C. the temperature which dew is formed in the air
D. the pressure which dew is formed in the air
Elements - 87 (ME Board April 1996)
What type of lubricating oils are produced entirely from the crudes chosen
through elimination of undesirable constituents by suitable refining
processes')
A. additives
B. inert
C. straight
D. premium
Elements - 88 (ME Board April 1996)
In a liquid-dominated geothermal plant, what process occurs when the
saturated steam passes through the turbine?
A. isobaric
B. polytropic
C. isometric
D. isentropic
Elements - 89 (ME Board April 1996)
Gas being heated at constant volume is undergoing the process of:
Elements - 91 (ME Board April 1996)
Foundations are preferably built of concrete in the proportion of what
measures of portland cement: sand: crushed stones?
A. 1:2:5
B. 2:4:6
C. 2:3:4
D. 1:2:4
Elements - 92 (ME Board April 1996)
How does the values for work per unit mass flow of air in the compressor
and turbine influenced by the addition of a regenerator?
A. slightly increased
B. unchanged
C. greatly decreased
D. greatly increased
Elements - 93 (ME Board April 1996)
The work done by a force of R newtons moving in a distance of L meters is
converted entirely into kinetic energy is expressed by the equation:
A. RL =' 2My2
B. RL =' RL N-m
352
Past Board Examination Elements (1994-1999)
C. RL = 1/2 MV1
D. RL= 1/2 MV
Past Board Examination Elements (1994-1999)
353
Elements - 98 (ME Board April 1996)
What is the suggested maximum permissible dose (MPD) of gamma ray
exposure for general individuals not working in a nuclear setting, by
Elements - 94 (ME Board April 1996)
In a steam generator with good combustion control, what occurs if the load
is increased?
A. air temperature leaving air heater decreases
B. air temperature entering heater increases
C. furnace pressure approximately constant
D. economizer gas outlet temperature decreases
Elements - 95 (ME Board April 1996)
The color of lubricating oil indicates:
A. does not indicated contamination
B. does not indicates qualities
C. qualities
D. viscosity
Elements - 96 (ME Board April 1996)
For design stability, the center of gravity of the total combined engine,
driven equipment and foundation should be kept:
A. anywhere
B. above the foundation top
C. in line with the surface of the foundation
D. below thefoundation top
Elements - 97 (ME Board April 1996)
Most commercially available petroleum lubricating oil deteriorates starting
from operating temperature of:
A. 150°F
B. 200°F
C. 300°F
D. 250°F
choice, in rem/year?
A I
B. 5
C. 1/2
D. 3
Elements - 99 (ME Board October 1996)
There are two broad types in the classification of lubricating oils, they are:
straight and
A.
B.
C.
D.
active
inactive
crocked
additives
Elements - 100 (ME Board October 1996)
Amount of air required in the low by-pass factor
A. does not change
B. greater
C. lesser
D. indeterminate
Elements - 101 (ME Board October 1996)
What is the
A.
B.
C.
function of the compression joint of pipes or tubes?
it is used to connect two pipes by welding
it is used to connect two pipes by pressing both ends
when tightened, compress tapered, sleeves so that they
form a tight joint on the periphery of the tubings they
connect
D. it connects two pipes with the use of threaded couplings
Elements - 102 (ME Board October 1996)
The components of a rotary pump;
A. gears
354
Past Board Examination Elements (1994-11)1)1))
Past Board Examination Elements (1994-1999)
R. piston
C impeller
D. screw
355
--------
Elements - 107 (ME Board October 1996)
Elements - 103 (ME Board October 1996)
An instrument commonly used in most Research and Engineering
Laboratories because it is small and fast among the other thermometers:
A. mercury thermometer
B. liquid-in-gas thermometer
C. gas thermometer
D. thermocouple
Elements - 104 (ME Board October 1996)
What is the term used in to express the ratio of specific humidities, actual
versus saturated')
A. relative hurrtidity
B. absolute humidity
C. degree ofsaturation
D. percent saturation
Elements - 105 (ME Board October 1996)
What is the process whereby a fissionable species utilized as a source of
neutrons to produce more nuclei of its own kind than are used up')
A. developing
B. culturing
C. multiplying
D. breeding
Elements - 106 (ME Board October 1996)
A process of heat transfer due to motion of matter caused by a change in
density:
A. absorption
B. radiation
C. conduction
D. convection
What is the most efficient thermodynamics cycle')
A. carnot
B. diesel
C. rankine
D. brayton
Elements - 108 (ME Board October 1996)
How do you treat a statement that is considered a scientific law')
A. We postulate to be true
B. Accept as a summary ofexperimental observation
C. We generally observed to be true
D Believe to be derived from mathematical theorem
Elements - 109 (ME Board October 1996)
The transmission of heat from one place to another by fluid circulation
between spots of different temperature is called:
A. convection
B. radiation
C. conservation
D. conduction
Elements - 110 (ME Board October 1996)
What is referred by volume control?
A. an isolated system
B. closed system
C. fixed region in space
D. reversible process only
.~]
Elements - III (ME Board October 1996)
Which of the following types of flow meters is most accurate?
A. venturi tube
B. pitot tube
356
Past Board Examination Elements (1994-1999)
C. flow nozzle
D. foam type
Elements - 112 (ME Board October 1996)
Pneumatic tools are powered by:
A. steam
B. water
C. natural gas
D. air
Elements - 113 (ME Board October 1996)
A graphical
A.
B.
C.
D.
representation between discharge and time is known as:
hectograph
monograph
hydrograph
topograph
Past Board Examination Elements (1994-1999)
Elements - 116 (ME Board October 1996)
The specific measurement of moisture content in air:
A. relative humidity
B. percent saturation
C. degree of saturation
D. specific humidity
Elements - 117 (ME Board October 1996)
Highest pressure drop in refrigeration cycle:
A. compressor
B. condenser
C. expansion valve
D. evaporator
Elements -118(ME Board October 1996)
Elements - 114 (ME Boa rd October 1996)
In a diesel engine, what elements in the fuel that make the work of the
lubricant more difficult?
A. water and ash content
B high octane number
C. high cetane number
D. sulphur and asphaltene content
Elements - 115 (ME Boa rd October 1996)
What is the
A.
B.
C.
D.
function of a radiation pyrometer?
boiler water weight
boiler pressure
furnace temperature
boiler drum pressure
What is an expansion loop?
A. a double loop radius elbow to minimize friction losses
B. a pipe bent to a loop to change direction
C. a pipe expander fitting
D. a large radius bend in pipe line to absorb longitudinal
expansion in the pipe line due to heat
Elements - 119 (ME Boa rd October 1996)
What is the
A.
B.
C.
D.
color code of steam pipe lines?
silver gray
green
red
yellow
Elements - 120 (ME Board October 1996)
What is absorbed by.sulphites in boiler water treatment?
A. oxygen
B. carbon dioxide
357
358
Past Board Fvaminntion Elements (1994-11)1)1))
C. impurities I "III,,£! ill mud drums
D. carbon diox idc and oxygen
Past Board Examination Elements (1994-/999)
C. sulfur dioxide
D methyl chloride
Elements - 121 (ME Board October 1996)
Elements - 125 (ME Board October 1996)
What is meant by choking in pipe flow?
A. the specified mass flow rate cannot OCcur
B. shock waves always occur
C. a valve is closed in a line
D. a restriction in flow area occurs
Water turbine converts:
A. mechanical energy into electrical energy
B. hydraulic energy into electrical energy
C. mechanical energy into hydraulic energy
D. hydraulic energy into mechanical energy
Elements - 122 (ME Board October 1996)
Elements - 126 (ME Board October 1996)
What is a check valve?
How do you differentiate surge from water hammer?
A. time for a pressure to traverse the pipe
B. the pressure of reservoir at the end of the pipe
C. rate of deceleration of flow
D. relative compressibility of liquid to expansion
A. a valve design to allow a fluid to pass through in one
direction only
B. a valve designed to release the excess pressure
C. a valve which allows flow of fluid in either direction
D. a valve used for checking the pressure of fluid
Elements - 123 (ME Board October 1996)
What is the purpose of providing the lubricating oil pre-heater in an
emergency stand-by diesel genset?
A. to keep the lube oil viscosity down under the cold
condition and enhance the starting of the cold engine
B. to avoid moisture condensation in the engine
C. to avoid corrosion to engine parts
D. to see to it that the lubrication system is functioning
properly
Elements - 127 (ME Board October 1996)
Throttling of the refrigerant throuzh the expansion valve
refrigeration cycle is:
A. reversible adiabatic process
8. constant entropy process
C. irreversible adiabatic process
D. isometric process
III
Elements - 128 (ME Board October 1996)
Assuming real process, the net entropy change in the universe is:
A. must be calculated
B. equal to zero
C. negative
D. positive
Elements - 124 (ME Board October 1996)
Which of the following refrigerants is most highly toxic?
A. ammonia
B. freon 12
359
Elements - 129 (ME Board October 1996)
What characterizes a reaction turbine?
a vapor
360
Past Board Examination Elements (1994-1999)
Past Board Examination Elements (1994-/999)
,..
B.
C.
D.
361
D. nitrogen dioxide
steam losses velocity as it leaves the diaphragm
steam strikes the blades at right angles
steam will react with a force in the diaphragm
steam is deflected
Elements - 134 (ME Board October 1996)
What takes place in a uniflow scavenging?
A. turbo blower in exhaust header to create vacuum in
cylinders
B. air reversing direction in cylinders
C. uses two blowers to purge cylinders
D. air travelling in one direction
Element's - 130 (ME Board October 1996)
The work done in an adiabatic process in a system:
A. is equal to the change in total energy in a closed system
B. is equal to the net heat transfer plus the entropy change
C. is equal to the change in total energy closed system
plus entropy change
D. is equal to the change in total energy of closed system
plus net heat transfer
or
Elements - 135 (ME Board October 1996)
The diagonal lines in the Psychrqmetric Chart represent:
A. Effective temperature
B. dry-bulb temperature
C. Wet-bulb temperature
D. dew-point temperature
Elements - 131 (ME Board October 1996)
Hr
do you increase the output of a centrifugal pump?
A. speed up rotation
B. install circulation line
C. increase the suction pipe area
D. increase the discharge pipe area
Elements - 136 (ME Ro~rrl Ol'tohpr 1996)
Elements - 132 (ME Board October 1996)
Based on the first law of thermodynamics, which of the following
wrong?
A. the heat transfer equals the work plus the energy change
B. the heat transfer cannot exceed the work done
C. the net heat transfer equals the net work of the cycle
D. the net heat transfer equals the energy change if no work
is done
Elements - 133 (ME Board October 1996)
The main cause of air pollution as a result of burning fuel oil is:
A. sulfur dioxide
B. silicon dioxide
C. hydrogen dioxide
IS
An ideal gas is compressed isothermally. The enthalpv change is:
A. sometimes negative
B. zero
C. sometimes positive
D. indeterminate
Elements - 137 (ME Board October 1996)
A system with paddle wheel work is irreversible, therefore, the change m
its entropy:
A.
B.
C.
D.
is zero
greater than zero
maybe negative
maybe positive, negative or zero
Elements - 138 (ME Board October 1996)
What is meant by brake horsepower?
362
Past Board Examination Elements (1994-1999)
Past Board Examination Elements (1994-1999)
A.
B.
C.
D.
power developed in the engine with cylinder
final horsepower delivered to the equipment
actual horsepower delivered to the engine drive shaft
work required to raise a weight of 33,000 pounds at a
height of one foot in one minute time
Elements - 139 (ME Board October 1996)
Enthalpy of an ideal gas is a function only of:
A. entropy
B. internal energy
C. temperature
D. pressure
Elements - 140 (ME Board October 1996)
When droplets of water are carried by steam in the boiler:
A. priming
B. foaming
C. carryover
D. embrittlement
Elements - 141 (ME Board October 1996)
Mechanical
A.
B.
C.
D.
energy of pressure transformed into energy of heat:
kinetic energy
enthalpy
heat exchanger
heat of compression
Elements - 142 (ME Board April 1997)
The ratio between the actual power and the apparent power in any circuit is
know as the of that circuit.
A. Measured Power
B. Capacity
C. Power Factor
D. KVAR
363
Elements - 143 (ME Board April 1997)
The products of complete combustion of gaseous hydro carbons.
A. Carbon dioxide and water
B. Carbon monoxide
C. Carbon monoxide, water and ammonia
D. Water, carbon monoxide and carbon dioxide
Elements - 144 (ME Board April 1997)
1
~\
The part that directs the flow of the refrigerant through the compressor:
A. wrist pin
B. valve
C. piston
D. connecting rod
.f'~.
...
ii
~.
·'··.·1.··
~il
~ij
Elements - 145 (ME Board April 1997)
An odorless refrigerant, its boiling point varies over a wide range of
temperatures:
A. Freon 22
B. Freon 12
C. Freon refrigerant
D.' Ammonia
Elements - 146 (ME Board April 1997)
The boiling point of Freon 22 is:
A. -41.04°F
B. 40.60°F
C. -38AO°F
D. -31AO°F
Elements - 147 (ME Board April 1997)
Medium pressure when applied to valves and fittings, implies they are
suitable for a working pressure of from:
A. 862 to 1200 kPa
I
'.1
.~
i~
11
.' :
i~
.1.:
!..
364
Past Board Examination
ElI.'IIIl'It(.\
(1994-1999)
B. 758 to I ooo kPa
C. 500 to 1000 kPa
D. 685 to 1050 kPa
Elements - ] 48 (ME Board April 1997)
A general term for a device that receives information in the form of one of
more physical quantities, modifies the information and/or its form. if
required, and produces a resultant output signal:
A. Converter
B. Transducer
C. Sensor
D. Scanner
Elements - 149 (ME Board April 1997)
In the process of pair formation, a pair cannot be formed unless the
quantum has an energy greater than:
A 2m sub o C
B.ll2mY'
C. 05MeY
D. hv/C
Elements - ]50 (ME Board April ]997)
The temperature of hot metals can be estimated by their color. For steel or
iron, the color scale at 2200°F is roughly:
A white
B orange
C. dark red
J
yellow
Elements - 15] (ME Board April ]997)
Mathematically, a thermodynamic property is which of the following?
A. a pflint function
B. discontinuous
C. a path function
D. exact differential
Past Board Examination Elements (1994-1999)
365
Elements - 152 (ME Board April 1997)
A device whose function is to pass an information in an unchanged form or
in some modified form:
A. relay
B. sensor
C. transmitter
D. transducer
Elements - ] 53 (ME Board April] 997)
A device whose primary function is to meter the flow of refrigerant to the
evaporator:
A. sniffer valve
B. equalizers
C. thermostatic expansion valve
D. crossover valves
Elements - 154 (ME Board April 1997)
The volume
stroke:
A.
B.
C.
D.
remaining when the piston reaches the end of the compression
air cell
combustion chamber
turbulence chamber
pre-combustion chamber
Elements - 155 (ME Board April ]997)
Specific heat capacity is anSI derived unit described as:
A. J/kg
B. W/moK
3
C. J/m
D. JlkgOK
Elements - ] 56 (ME Boa rd April] 997)
The fundamental difference between pipe and tubing is:
A. The dimensional standard to which each is manufactured
Past Board Examination Elements (1994-1 YYY)
366
jol
Past Board Examination Elements (J 994-1999)
B. Compression joints
C. The smoothness of the surface
D. Bell and spigot joint
Elements - 157 (ME Board April 1997)
One of the most popular types of compressor utilized for supercharging
engine is the:
A. Roots type hlower
B. Pulse turbocharger
C. Constant pressure turbocharger
D. Turbo compressor
Elements - 158 (ME Board April 1997)
Crankshaft of reciprocating type compressor is basically made of:
A. semi-steel
B. aluminum alloy
C. cast iron
D. steel forging
Elenlents - 161' (ME Board April 1997)
The temperature of the fluid flowing under pressure through a pipe is
usually measured by:
A. glass thermometer
B. electric-resistance thermometer
C. thermocouple
D. all of the above
Elements - 162 (ME Board April 1997)
An increase in the deposition of slag and ash on the surface for heating of
oil-fired boilers in both marine and stationary service has affected boiler
efficiency. The following are the causes except:
A. Low temperature corrosion of the cold section of air
heaters and duct works
B. Slagging of high temperature superheater surfaces
C. High temperature corrosion steel
D. Increase of heat transfer in the hoiler
Elements - 163 (ME Board April 1997)
Elements - 159 (ME Board April 1997)
A chemical
and sodium
A.
B.
C.
D.
method of feedwater treatment which uses calcium hydroxide
carbonate as reagents:
thermal treatment
lime soda treatment
The type of filter where the filtering element is replaceable:
A. paper edge filter
B. metal edge filter
C. pressure fi Iter
D. filter with element
demineralization process
ion exchange treatment
Elements - 164 (ME Board April 1997)
Elements - 160 (ME Board April 1997)
Engines using heavy fuels requires heating of the fuel so that the viscosity
at the injector is:
A. around 200 SSU
B. 100 SSU or less
C. 200 SSU±50
D. J50 SSU or slightly higher
Which does
A.
B.
C.
D.
not belong to the group?
air injection system
mechanical injection system
time injection system
gas admission system
· 368
Past Board Examination Elements (1994-1999)
Elements - 165 (ME Board April 1997)
Coaling water system consists of equipment to dissipate heat absorbed by
the, engine jacket water, lube oil and the heat to be removed from air
intercooler is measurable to keep the engine outlet water temperature
constant and the differential of the cooling water at a minimum preferably
not to exceed:
A. IOt030°F
B. 10 to 50°F
C. 10 to zo-r
D. IOt040°F
Past Board Examination Elements (1 YY4-1 YYY)
.)o~
A. direct expansion system
B. chilled water system
C. flooded system
D. multiple system
Elements - 170 (ME Board October 1997)
When four events take place in one revolution of a crankshaft of an engine,
the engine is called:
A. rotary engine
B. steam engine
C. two-stroke engine
D. four-stroke engine
Elements - 166 (ME Board October 1997)
In a water tube boiler, where is heat and gases of combustion passed?
A. through the combustion chamber only
B. through the tubes
C. away from the tubes
D. around the tubes
Elements - 167 (ME Board October 1997)
A pneumatic tool is generally powered by:
A. water
B. electricity
C. steam
D. air
Elements - 168 (ME Board October 1997)
The instrument used to measure atmospheric pressure is:
A. rotameter
B. manometer
C. venturi
D. barometer
Elements - 169 (ME Board October 1997)
A refrigeration system in which only part of the refrigerant passes over the
heat transfer surface is evaporated and the balance is separated from the
vapor and recirculated:
Elements - 171 (ME Board October 1997)
What occurs in a reversible polytropic process?
A. enthalpy remains constant
B. internal energy does not change
C. some heat transfer occurs
D. Entropy remains constant
Elements - 172 (ME Board October 1997)
In a deepwell installation or operation, the difference between static water
level and operating water level is called:
A. suction lift
B. drawdown
C. priming level
D. clogging
Elements - 173 (ME Board October 1997)
What characteristics an impulse turbine?
A. steam striking blades on angle
B. no steam reaction to velocity
C. steam striking blades at zero angle
D. steam reversing direction
-' 7()
Past Board Evamination Elements (/1)1)4-/999)
Past Board Examination Elements (1994-1999)
Elements - 174 (ME Hoard October 1997)
371
C. convection
D. conduction
Air receives in a compressed air plant must be:
A. without pressure gauges
B. vented to the atmosphere
C. rectangular in shape
D. installed with safety valve and drain valve
Elements - 175 (ME Board October 1997)
A gas which will not be found in the flue gases produced by the complete
combustion offuel oil is:
A. carbon dioxide
B. hydrogen
C. oxygen
D. nitrogen
Elements - 176 (ME Board October 1997)
Scale in boiler can:
A. create low steam quality
B. cause foaming
C. overheat blow-off line
D. inhibit circulation anti heat transfer
Elements -179 (ME Board October 1997)
Where is lithium bromide used in a refrigeration system')
A. condensate return lines
B absorbers
C. centri fugal compressors
D. ion exchangers
Elements - 180 (ME Board October 1997)
Amount of heat liberated by the complete combustion of a unit weight or
volume of fuel is:
A. heating value
B latent heat
C. sensible heat
D work or compression
Elements - 181 (ME Board October 1997)
A temperature above which a given gas cannot be liquified:
A. cryogenic temperature
B. vaporization temperature
C. absolute temperature
D. critical temperature
Elements - 177 (ME Board October 1997)
The effectiveness of a body as a thermal radiator at a given temperature.
A. absorptivity
B. emissivity
C. conductivity
D. reflectivity
Elements - 178 (ME Board October 1997)
In a cooling tower, the water is cooled mainly by:
A. condensation
B. evaporation
Elements - 182 (ME Board October 1997)
The ratio of the sum of individual maximum demands of the svstern to the
overall maximum demand of the whole system is:
A. diversity factor
B. utilization factor
C. power factor
D. demand factor
Elements - 183 (ME Board October 1997)
When fuel oil has a high viscosity, we mean that the fuel oil will:
372
Past Board Framination Elements (11)94-/999)
A
B.
C.
D.
evaporate easily
have a low specific gravity
bum without smoke
flow slowly through pipes
Elements - 184 (ME Board October 1997)
Percentage of excess air is the difference between the air actually supplied
and the theoretically required divided by:
A. actual air supplied
B. theoretical air supplied
C. theoretical less actual supplied
D. deficiency air supplied
Elements - 185 (ME Board October 1997)
In a refrigeration system, the heat absorbed in the evaporator per kg mass
of refrigerant passing through is:
A equals the increase in enthalpy
B. does not depend on the refrigerant used
C. is decreased if pre-cooler is used
D. equals the increase in volume
Elements - 186 (ME Board October 1997)
Air that controls the rate of combustion in the combustion chamber is
known as:
A. secondary air
B. excess air
C. control air
D. primary air
Elements - 187 (ME Board October 1997)
An aftercooler on a reciprocating air compressor is used primarily to:
A. cool the lubricating oil
B. condense the moisture in the compressed air
C. improve the compressor efficiency
D. increase the compressor capacity
Past Board Examination Elements (1994-1999)
373
Elements - 188 (ME Board October 1997)
In a hydro-electric plant using a Francis turbine with medium head, the
speed can be regulated by using the:
A. deflector gate
B. nozzle
C. wicket gate
D. weir
Elements - 189 (ME Board October 1997)
To protect adequately the engine bearings, what type and better
arrangement of lubricating oil filter is most practical?
A. full-flow type filter installed between the lubricating oil
pump and the bearings
B. duplex filter installed before the lubricating pump
C. by pass filter with cleanable and replaceable elements
D. splash lubricating system in the crankcase
Elements - 190 (ME Board October 1997)
In radiation, the heat transfer depends on:
A. temperature
B. heat rays
C. heat flow from cold to hot
D. humidity
Elements - 191 (ME Board October 1997)
The main purpose of a subcooler in a refrigerating system especially a 2stage system is to;
A. increase the heat rejection per ton and avoid system
shutdown
B. improve the flow of evaporator gas pel ton and increase
the temperature
C. reduce the total power requirements and return oil to the
compressor
D. reduce the total power requirements and heat rejection to
the 2nd stage
Past Board Examinauon Elements (1994-1999)
j
/4
375
Past Board Examination Elements (1994- 1999)
Elements - 192 (ME Board October 1997)
The performance of a reciprocating compressor can be expressed by'
A adiabatic work divided by adiabatic input
B. heat radiation
C condensate water level
D. air volume
Elements - 197 (ME Board April 1998)
8. adiabatic work divided by indicated work
C. isothermal work divided by indicated work
D. isothermal work divided by adiabatic work
Elements - 193 (ME Board October 1997)
A reciprocating pump is considered positive displacement pump because:
A. displacement ofthe liquid is affected by the displacement
Pres. F. V. Ramos approved on February 12, 1998 a Republic Act, which is
an act to regulate the practice of Mechanical Engineering in the
Philippines, otherwise known as the M.E. Law. What is this act?
A. RA No. 9845
B. RA No. 8495
C. RA No. 8594
D RA No. 8945
of the piston
B positive pressure is given to the liquid
C. liquid is discharge with positive pressure
D. liquid is lifted due to the vacuum created inside the
cylinder
Elements - 194 (ME Board October 1997)
A change in the efficiency of combustion in a boiler can usually be
determine by comparing the previously recorded readings with the current
readings of the:
A. stack temperature and CO
8. over-the-fire-draft and CO
C. Ringelman chart and CO 2
D. stack temperature and COl
Elements - 195 (ME Boa rd October 1997)
A boiler steam gauge should have a range of at least:
A one-half the working steam pressure
B. I 1/2 times the maximum allowable working pressure
C. the working steam pressure
D twice the maximum allowable working pressure.
Elements - 196 (ME Board April 1998)
----_._--~---~_._----
Elements - 198 (ME Board April 1998)
The relationship of water vapor in the air at the dew point temperature to
the amount that would be in the air if the air were saturated at the dry bulb
temperature
A.
B.
C
D.
is:
partial pressure actual at dew point
percentage humidity
reLative humidity
partial pressure of water
Elements - 199 (ME Board April 1998)
What is the color code of air pipelines?
A. light-bLue
8. red
C brown
D. violet
Elements - 200 (ME Board April 1998)
The CO (carbon dioxide) percentage in the flue gas of an efficiently tired
2
boiler should be approximately:
A.I%
B.12%
C 18%
D 20%
A manometer IS an instrument that is used to measure:
A air pressure
37(,
Past BOl/rd Evamination Elements (1')')4-1 '1'1'1)
Elements - 201 (M E Hoard April 1998)
An unloader is used on air compressor to:
A. to relief air pressure
B. start easier
C. stop easier
D. run faster
Elements - 202 (ME Board April 1998)
How many pounds of air are theoretically needed to burn one pound of
diesel fuel oil?
A. 28
B 14
C. 18
D.22
Elements - 203 (ME Board April 1998)
Which of the following is a great advantage of a fire-tube boiler?
A. steam pressure is not steady
B. contains a large volume of water and requires long
Past Board Examination 1:'1£'111£'111\ (I 994-1 999j
A.
B
C.
D.
cleaning the cup on a rotary cup burner
cleaning a completely clogged oil strainer
replacing a leaking valve
replacing a blown fuse
Elements - 206 (ME Board April 1
An air current in a confined space such as that in a cooling tower or
chimney is called:
A. variable flow
B. velocity profile
C. velocity gradient
D. draft
Elements - 207 (ME Board April 1998)
What kind of a heat exchanger where water is heated to a point that
dissolved gases are liberated?
A. evaporator
B. condenser
C. intercooler
D. deaerator
interval oftime to raise steam and not so flexible as to
changes in steam demand
C. can not use impure water
D. radiation losses are higher because fire is on the inside of
the boiler and is surrounded by water
Elements - 204 (ME Board April 1998)
What should be the temperature of both the water and steam whenever they
are present together?
A. saturation temperature for the existing pressure
B. boiling point of water at 101.325 kPa
C. superheated temperature
D. one hundred degree centigrade
Elements - 205 (ME Board April 1998)
One of the following tasks which is an example of preventive maintenance
IS:
377
Elements - 208 (ME Board April 1998)
What is the
A.
B.
C.
D.
function of steam separator?
trapping the steam and letting water through
throttling
changing direction of the steam flow
steam metering
Elements - 209 (ME Board April
Which of the following is not a main part ofa typical coal burner?
A. air registers
B. a nozzle
C. an atomizer
D. an ignitor
j/X
Past Board Examination Elements (1994-1999)
Past Board Examination Elements (1994-1999)
Elements - 210 (ME Board April 1998)
379
Elements - 214 (ME Board April 1998)
Which of the following types of air dryers works by absorbing moisture on
a solid dessicant or drying material such as activated alumina, silicon gel,
or molecular sieve?
A. Regenerative dryer
B. Deliquescent dryer
C. Spray dryer
D. Refrigerated dryer
In the processing section, there is an instrument frequently used to measure
the flow rate of fluids. What is the instrument consisting of a vertical
.passage with variable cross-sectional area, a float and calibrated scale"
A. rotameter
B. pitot-tube
C. rota-aire
D. manometer
Elements - 211 (ME Board April 1998)
Elements - 215 (ME Board April 1998)
A heat-transfer device that reduces a thermodynamic fluid from its vapor
phase to its liquid phase such as in vapor-compression refrigeration plant
or in a condensing steam power plant.
A. flash vessel
B. cooling tower
C. condenser
D. steam separator
How do you describe a non-flow process where in the volume remains
constant?
A. isometric
B. isentropic
C. isobaric
D. iscnthalpic
Elements - 216 (ME Board April 1998)
Elements - 212 (ME Board April 1998)
A goose neck is installed in the line connecting a steam gauge to a boiler
to:
A. maintain constant steam flow
B. protect the gauge element
C. prevent steam knocking
D. maintain steam pressure
A branched system of pipes to carry waste emissions away from the piston
chambers of an internal combustion engine is called:
A. exhaust nozzle
B. exhaust deflection pipe
C. exhaust pipes
D. exhaust manifold
Elements - 217 (ME Board April 1998)
Elements - 213 (ME Board April 1998)
The law that states entropy of all perfect crystalline solids
absolute zero temperature.
A. Newton Law
B. Third Law of Thermodynamics
C. First Law of Thermodynamics
D. Second Law of Thermodynamics
IS
zero at
Measure of ability of a boiler to transfer the heat given by the furnace
the water and steam is:
A. grate efficiency
B. stocker efficiency
C. furnace efficiency
D. boiler efficiency
TO
380
Past Board Examination Elements (/994-1999)
Elements - 218 (ME Board April 1998)
A major cause of air pollution resulting from the burning of fuel oils is:
A. nitrous
B hydrogen
C sulfur dioxide
D silicon
Past Board Examination Elements (1994-1999)
381
A. reduce the speed 01 the 1T10tor when the maximum
pressure is reached.
B. drain the condensate from the cylinder
C release the pressure in the cylinders in order to
reduce the starting load
D. prevent excess pressure in the receiver
Elements - 223 (ME Board April 1998)
Elements - 219 (ME Board April 1998)
An engine indicator is generally used to measure:
A. steam temperature
B heat losses
C steam cylinder pressure
D. errors in gauge reading
Fluids that
than water
A.
B.
C
are pumped in processing work are frequently more VISCOUS
Which of the following statements is corrects'!
Reynolds number varies directly as the viscosity
Efficiency of a pump increases as the viscosity increases
Increased fluid friction between the pump parts and
the passing fluid increases useful work
D. Working head increases as the viscosity increases.
Elements - 220 (ME Board April 1998)
Elements - 224 (ME Board April 1998)
The power required to deliver a given quantity of fluid against a given head
with no losses in the pump is called:
A. wheel power
B. brake power
C hydraulic power
D. indicated power
The size of a steam reciprocating pump is generally designated by a three
digit number size as 646. What would be the first number designate?
A. stroke of the pump in inches
B inside diameter of the steam cylinder measured in
inches
C percent clearance
D number of cylinders
Elements - 221 (ME Board April 1998)
Elements - 225 (ME Board April 1998)
A liquid whose temperature is lower than the saturation temperature
corresponding to the existing pressure.
A. subcooled liquid
B. saturated liquid
C pure liquid
D. compressed liquid
Peak load for a period of time divided by installed capacity is:
A. capacity factor
B. demand factor
C utilization factor
D load factor
Elements - 226 (ME Board April 1999)
Elements - 222 (ME Board April 1998)
The function of an unJoader on an electric motor-driven compressor is to:
At what temperature where in an oil at any grade becomes cloudy and it
freezes, thus in application is limited.
A. coLdpoint
382
Past Board Examination Elements (1994- J 999)
Past Board Examination Elements (1994-1999)
B. pour point
C. flash point
D. freezing point
A.
B.
C.
D.
Elements - 227 (ME Board April 1999)
critical temperature
dry bulb temperature
dew point temperature
wet bulb temperature
-
A turbine pipe determined its "nominal" size refers to:
A. inside diameter
Elements - 232 (ME Board April 1999)
B. outside diameter
C. approximate size
D. pipe wall thickness
For reciprocating compressor slip at negative displacement:
A. cd>}
B. cd-cl
C. cd
D. cd
Elements - 228 (ME Board April 1999)
If the exhaust lowered or the boiler is raised the moisture content of steam:
A. vaponzes
B. decreased
C. liquifies
D. increase
Elements - 229 (ME Board April 1999)
The relative humidity become [00% and where the water vapor starts to
condensate.
A. critical point
B. dew point
C. saturated point
D. cold point
Elements - 230 (ME Board April 1999)
For reciprocating compression slip at positive displacement.
A. cd = ]
=
0
= [
Elements - 233 (ME Board April 1999)
When the number of reheat stages in a reheat cycle is increased, the
average temperature:
A. increases
B. constant
C. decreases
D. zero
Elements - 234 (ME Board April 1999)
When the boiler pressure increases or when the exhaust pressure decreases,
the amount of moisture:
A. increases
B. constant
C. decreases
D. zero
Elements - 235 (ME Board April 1999)
B. cd<1
C. cd» l
D. cd = 0
Elements - 231 (ME Board April 1999)
A temperature measurement in an ordinary thermometer wh ich
constant specific humidity.
383
has
The purpose of the nozzle in a combustor of gas turbine plant is to:
A. increase the velocity
B. increase the pressure
C. increase the moisture
D. increase the power
384
Past Board Examination Elements (I <Jl)4-11)91))
-------
Elements - 236 (ME Board April 1999)
During the sensible heating, the absolute humidity remains constant but the
relative humidity.
A. increases
B. remains constant
C decreases
D. zero
Elements - 237 (ME Board April 1999)
For negative slip, the coefficient of discharge:
A. increases
B. zero
C decreases
D. constant
Elements - 238 (ME Board April 1999)
Past Board Examination Elements (1994-1999)
385
B Throttle valve
C Thermostatic expansion valve
D. Control valve
Elements - 241 (ME Board Oct. 1999)
A valve that sense the loss of ignition in a diesel engine.
A. combustion control
B. fire control
C flame detector
D. fire extinguisher
Elements - 242 (ME Board Oct. 1999)
If evaporator oil clogs in the evaporator, it cause:
A. increase in heat transfer
B. vaporized oil
C Low suction pressure
D. High pressure
For positive slip, the
A.
B.
C
D.
coefficient of discharge:
increases
zero
Elements - 243 (ME Board Oct. 1999)
decreases
constant
Elements - 239 (ME Board Oct. 1999)
Is an abrupt reduction in flow velocity due to sudden increase of water
depth in the down stream direction
A. Hydraulic energy
B Hydraulic jump
C Weirs
D. Hydraulic gradient
Elements - 240 (ME Board Oct. 1999)
Is a valve that regulates the flow ofrcfrigerants.
A. Direct expansion valve
In cooling, if humidity ratio remains unchanged,
A. sensible cooling
B sensible heating
C Pre-cooling
D. Latent cooling
Elements - 244 (ME Board Oct. 1999)
Property of lubricating oil that responds at very low temperature, the oil is
known as:
A. viscous
B. pour point
C cloud point
D. solid point
38!J
Past Board Examination Elements (/994-/999)
Past Board Examination Elements (1994-1999)
387
Elements - 245 (ME Board Oct. 1999)
Elements- 249 (ME Board Oct. 1999)
A turbine in which all available energy of the flow is converted by a nozzle
into kinetic energy before in contact to moving blades.
A. Kaplan turbine
B. Francis turbine
C. Propeller turbine
D. Impulse turbine
In a cooling tower the temperature of water is lower than the wet bulb
temperature of entering air and it is found that air cannot cool. What
temperature of water in cooling water.
A. above
B. lower
C. constant
D. none of these
Elements - 246 (ME Board Oct. 1999)
Elements - 250 (ME Board Oct. 1999)
A class of system of a refrigeration in which the wet bulb temperature is
not more than the temperature of air.
A. evaporator preload system
B. direct system
C. indirect system
D. chilling system
What is the critical temperature where water & vapor are in equilibrium to
the atmospheric pressure.
A. ice point
B. critical point
C. steam point
D. freezing point
Elements - 251 (ME Board Oct. ] 999)
Elements - 247 (ME Board Oct. 1999)
A type of throttle of air fuel ratio in constant charging.
A. quantitative
B. qualitative
C. hit and miss
D. none of these
Elements - 248 (ME Board Oct. 1999)
A company manager want to used comfort air, what is the most efficient
setting of the conditioning unit:
A most attainable value
B. moderate value
C maximum value
D. minimum value
In a pipe specification, schedule is used, when the pipe specified as "
schedule 80", the pipe corresponds to the
A. "extra standard" weight
B. extra strong
C. internal pressure
D. "old standard" weight
Elements - 252 (ME Board Oct. 1999)
The entropy of a substance at a temperature of absolute zer, is.
A. unity
B. infinity
C. zero
D. 100
Elements - 253 (ME Board Oct. 1999)
The ratio of cross-sectional area of flow to the wetted perimeter.
A. hydraulic lead
388
Past Board Examination Elements (/994-1999)
Past Board Examination Elements (1994-1999)
389
r,.
'.;,1·.·
B hydraulic radius
C hydraulic energy
,Ii;
Elements - 259 (ME Board Oct. 1999)
D. hydraulic gradient
Elements - 255 (ME Board Oct. 1999)
The effect of superheating the refrigerant is:
A. increased in COP
8. reduced in (lOP
C. high COP
D. constant COP
Elements - 256 (ME Board Oct. 1999)
The absolute zero in Celcius scale:
A. 100
The ice making capacity is always:
A. directly proportional to the refrigerating effect
B. less than the refrigerating effect
C. greater than the refrigerating effect
D. equal to the refrigerating effect
Elements - 260 (ME Board Oct. 1999)
When the air is saturated, the wet bulb depression is:
A. zero
B. unity
C. constant
D. 100%
B. -273
c.O
Elements - 261 (ME Board Oct. 111l)Q)
D. 273
Elements - 257 (ME Board Oct. 1999)
The water in the product of combustion is in vapor state.
A. Ultimate analysis
B. analysis
C. Lower heating value
D. Higher heating value
Elements - 258 (ME Board Oct. 1999)
The thermal efficiency of gas-vapor cycle as compared to steam turbine or
gas turbine.
A. greater than
B. lower than
C. less than
D. equal to
The process in which the heat energy is transferred to a thermal storage
device. It is known as:
A. adiabatic
B. intercooling
C. regenerator
D. isentropic
Elements -262 (ME Board Oct. 1999)
The liquid pressure in the surface per area in the surface at the bottom is:
A. magnitude pressure
B. cohession pressure
C. intensity pressure
D. adhession pressure
390
Refrigeration
Refrigeration
391
x = 0.1028
x = 10.28%
REFRIGERATION
Refrigeration - 3
Refrigeration - 1
The enthalpy at the entrance of the condenser is 1660 KJ/kg and exit is
315 KJ/kg. The compressor has an enthalpy of 1450 KJ/kg at entrance.
Determine the COP.
A. 4.2
C. 5A
R 4.0
D. 6.0.
The refrigerating effect of 100 cons refrigeration is 117 KJ/kg.
Determine the mass flow of refrigerant.
C. 3 kg/sec
A. I kg/sec
D. 4 kg/sec
B. 2 kg/sec
SOLUTION:
TR = ~(h I
SOLUTION:
-
h4 )
3.516
COP
COP
COP
Re frjgerating Effect
m (117 )
Compressor Power
hi - h 4
100
h, - hI
1450-315
ms
1660-1450
COP = 5AO
-s - -
3.516
3 kg/sec
Refrigeration - 4
The density of R-12 refrigerant at compressor suction is IlJ.81 kg/m'.
For mass flow rate of 2 kg/sec, determine the volume now at suction.
J
'
A.O.IOm/sec
C. 0.30 rrr'zsec
B. 0.25 mJ/sec
D. OAO rrr'zsec
Refrigeration - 2
An evaporator has a temperature of 3°C with entrance enthalpy of
352.75 KJ/kg. At 3°C, h r = 319.56 KJ/kg and h g = 642.45 KJ/kg. Find
the quality after expansion.
A. 16.27%
C. 15.67%
B. 21A8%
D. 10.27%
SOLUTION:
VI
VI
VI
VI
= mVI
= m (1/w))
= 2(1/19.81)
= 0.10 mJ/sec
SOLUTION:
h, = h, = h r + x(h g - hr)
Refrigeration -
352.75 = 319.56 +x(642A5-319.56)
A 100 tons refrigeration system has a COP of 5. Determine the
compressor horsepower.
;Ii.
'.'
392
Refrigeration
Refrigeration
A. 94.26 lip
B. 8676 Hp
C 8676 lip
D. 65.65 Hp
SOLUTION:
COP
Re frigerating Effect
Compressor Power
393
Refrigeration - 7
A 90 tons refrigeration system has a compressor input of 0.97 KW per
ton refrigeration. If compressor efficiency is 75%. determine the heat
rejected from the condenser.
A. 350 KW
C. 500 KW
B. 250 KW
D. 382 KW
SOLUTION:
5
100(3.516)
Compressor Power
RE
Compressor Power
=
70.32 KW x 1/0.746
Compressor Power
=
94.26 Hp
RE = 90x3.516
316.44 KW
Compressor Power = 097(90)(0.75)
Compressor Power = 65.475 KW
Heat Rejected = Ref. Effect + Camp. Power
Heat Rejected = 316.44 + 65.475
Heat Rejected = 381.915 kw
Refrigeration - 6
A 90 tons refrigeration system has a compressor input of 0.97 KW per
ton refrigeration and COP of 5. What is the efficiency of the
compressor?
A. 72.49%
C. 56.34%
B. 90.42%
D. 83.33%
SOLUTION:
Compressor Power
Refrigeration - 8
The mass flow rate of refrigerant entering the compressor is 0.25 kg/~
and the change of enthalpy between the inlet and outlet is 320 KJ/kg.
If 134 Hp motor is used to drive the compressor, determine the heat
loss from the compressor.
A. 15 KW
C. 25 KW
B. 20 KW
D. 30 KW
90(3.516)
=
5
Compressor Power
63.29 KW
Compressor input = 0.97(90)
Compressor input = 87.3 KW
Compressor Efficiency
Compressor Efficiency
63.29
87.3
72.497%
SOLUTION:
Compressor Power
Compressor Power
Compressor Power
Power input
Power input
Heat Loss
Heat Loss
mth, - hi)
0.25(320)
80KW
134 x 0.746
99.964 kw
99.964 - 80
19.964 kw
394
Refrigeration
395
Refrigeration
Refrigeration - 9
Refrigeration - 11
A 100 tons refrigeration system is used to produce chilled water from
22°C to 2°e. Determine the volume now of water in Ii/sec.
A. ).0
C. 4.2
The difference in temperature between the water leaving the
evaporator and evaporator temperature is SoC. [f the temperature of
water leaving is 3rF, what is the evaporator temperature?
B. 3.5
D. 5.5
C. -4°C
D. -SoC
A. O°C
B. _2°C
SOLUTION:
SOLUTION:
Ref Effect
100x3.516
m.,
=
m., cp (t 2 - t.)
=
=
m w(4.187)(22-2)
=
4.187 kg/sec (lIilkg)
Q
=
4.1871ils
An industrial plant requires to cool 120 gal/min of water from 20°C to
soc. Determine the tons of refrigeration required.
A. 100 TR
C. 145 TR
D. 135 TR
The heat rejected from condenser is 300 KW. The water required to
cool the refrigerant is S li/sec. Determine the temperature of water
leaving the condenser if water enters at 25°C.
A. 30°C
C. 45.33°C
B. 35°C
D. 39.33°C
SOLUTION:
QR
=
SOLUTION:
300
=
=
Ref Effect = rn., c p (t 2 - t.)
rn., = 120 gal/min x 3.785 liIgaI x Imin/60sec
7.57 Ii/sec x Ikg/Ii
7.57 kg/sec
Ref. Effect = 7.57(4.187)(20 - 5)
Ref. Effect = 475.43 kw
TR
=
0- 5
-5°C
Refrigeration - 12
Refrigeration - 10
m.,
m.,
5/9 (32 - 32)
=
Evaporator Temp.
Evaporator Temp.
4.187kg/sec
Q
B. 130 TR
°C
"C = O°C
TR = 475.43/3.516
135.22 ton of ref.
t2
~
m., c p (t 2 - tt)
(5
X
1)(4. I 87)(t 2
-
25)
39.33°C
Refrigeration .: t::
The mass now of refrigerant entering the compressor is 0.1 kg/sec with
change of enthalpy of 400 KJ/kg. For compressor efficiency of 70%
and motor efficiency of 80%, find the electrical energy needed for one
day.
,(if,
Refrigeration
R c] rr::';l'rl/tion
A 1524 KW-Ilt
B, 1685 KW-III
C 3455 KW-Iil
D 1714 KW-Iil
SOLUTION:
SOLUTION
TR
Compressor Power
Compressor Power
Compressor Power
Input Power of motor
397
~
Re frigerating Effect
3S 16
m(h: - hi)
o 1(400)
40 kw
TR
~
(12 / 60)( \ 000)
3.516
TR= 56,88 tons of ref
40
0,7(0,8)
Input Power of motor ,"'" 71,428 kw
Energy = 71,428 x 24' hrs
Energy = 1714,28 Kwh
Refrigeration - 16
The mass flow of water entering the condenser is 10 kg/sec. If the
temperature difference between inlet and outlet temperature is l8 F,
determine the heat rejected from the condenser.
A, 400 KW
C 500 KW
B. 419 KW
D, 324 KW
D
Refrigeration - 14
An 80 tons refrigeration system requires lOS KW from VECO, If
motor compressor efficiency is 60%, find the COP,
A. 5,84
C 4,46
B, 3,56
D, 3,78
SOLUTION
SOLUTiON:
.6. D C
Compressor Power = 105(0,6)
Compressor Power = 63 kw
RE
COP = - - - - - Compressor Power
COP
C~
5/9 OF
.6."C = (5/9)( 18)
.6. DC = lODC
QR = 10(4,187)(10)
QR = 418,70 KW
80(3,516)
Refrigeration - 17
63
COP
4,46
Refrigeration - 15
The change of enthalpy between the inlet and outlet of evaporator is
1000 KJ/kg and mass flow of refrigerant is 12 kg/min. What is the
capacity of plan!'?
A 15TR
C 57 TR
B, 25 TR
C, 60 TR
The change of enthalpy in the condenser is 1500 KJ/kg. The
temperature change of water is 8°C and the refrigerant flow is 0.13
kg/sec, Determine the gpm of water required for cooling.
A. 92 gpm
C 45 gpm
B, 88 gpm
D, 67 gpm
SOLUTION:
QR
m; cp (.6.t)
Refrigeration
398
0.13( 1500)
rn., (4.187)(8)
rn;
=
5.8216 kg/sec x Iii/kg
Ow
=
5.8216 Ii/sec x I gal/3.785 Ii x 60sec/min
Ow =
Refrigeration
T] = -25 + 273
T, = 248°K
OR = T2 (s, - S4)
6000 = 345(5 I - S4)
SI - 54 = 17.3913
92.28 gpm
Refrigeration - 18
The heat rejected from the condenser is 200 KW. The mass flow of
water entering is 5 kg/sec at 23°C. If the temperature between the
condenser cooling water outlet and condenser temperature is 5°C,
what is the condenser temp?
A. 30.56°C
C. 35.78°C
B. 39.94°C
D. 37.67°C
SOLUTION:
OR = rn., c p (t 2 -t.)
200 = 5( 4.187)(t2 - 23)
t2 = 32.55°C
t.t ,= teon - t 2
5 = t eond - 32.55
teond = 37.55°C
399
Hp
Hp
Hp
Hp
=
(T 2 - Td(s) - S4)
=
(345 - 248)(17.3913)
1686.96 KJ/min (Imin/60sec)
28.116 KW/0.746
37.69 Hp
=
Hp
=
=
Refrigeration - 20
The temperature difference between the minimum and maximum
temperature of Carnot cycle is 50°C. What is the minimum
temperature if COP is 5.5.
A. 0.31°C
C. 3°C
B. 2°C
D. 5°C
SOLUTION:
T2
T2
=
=
COP
50 + 273
323°K
T}
T2 -T}
\
5.5 =
Refrigeration - 19
T,
. --
232 - T)
323 - T 1 = 0.1818 T,
T) = 273.307°K
t[ = 273.307 - 273
t) = 0.307°C
The heat rejected from the condenser of Carnot refrigeration is 6000
KJ/min. The minimum and maximum temperature is -25°C and noc
respectively. Determine the horsepower required to drive the
compressor.
A. 30 Hp
C. 38 Hp
B. 35 Hp
D. 42 Hp
Refrigeration - 21
SOLUTION:
T 2=72+273
T 2 = 345°K
The power required to drive the compressor in a Carnot refrigeration
is 50 Hp. It operates between -SoC and 40°C. Determine the tons of
refrigeration.
400
Refrigeration
A, 57TR
55TR
D, 63 TR
C
B. 34TR
SOLUTION:
1'1 = -5+273
1'1 = 268°K
KJ/kg-OK and latent heat of fusion is 233 KJ/kg. If specific heat below
freezing is 1.68 KJ/kg-OK, find the freezing temperature.
A. -565°C
.
C -1.0 I°C
B. -219°C
D. -10°C
SOLUTION
T2=40+27~
1'2 = 313°K
268
COP
313 - 268
5,955
COP
o
m[cp,(tl-tr)f L+c p2(tr - t 2)]
01 6.4 37
=
331.24
I 10[3 .:23 (20 - tr) -+- 233 + I. 68{tr - (- 18) }]
64.6 - 3.23tr + 233 + 1.68tr + 30.24
RE
COP
155t(
Compressor Power
5,955
401
Refrigeration
RE
50xO,746
RE = 222,12 KW
TR = 222.12/3.516
TR = 63,17 tons of ref
Refrigeration - 22
Determine the heat to be remove from one ton of water at 26°C to an
ice at -6°C.
A. 215,765 KJ
C 345,654 KJ
B. 4D 956 KJ
D. 834,582 KJ
t( ,=
=
-3.4
-2.19°C
. Refrigeration - 24
A refrigeration compressor has a specific volume of 0.0482 mJ/kg
entrance and 0.017 mJ/kg at exit. If volum'etric efficiency is 90%,
determine the percent clearance of the compressor.
A. 5.45%
C 10%
B 6.35%
D. 12%
SOLUTION:
11v
=
I+
C - C(VI/V2)
SOLUTION:
0.90
Qremoved
Qremoved
m[cpl(t,- tr) + L + Cp2(tr - t2 )1
=
I +- c - c(0.0482/0.0 17)
0.90 = I + c - 2.835c
907[ 4.187(26-0) + 335 + 2.09( 0- (-6) )]
C =
5.45%
Qremoved = 413,956.514 KJ
Refrigeration - 23
Refrigeration - 25
The heat required to remove from beef 110 kg is 36,437 KJ which will
be cooled from 20°C to -18°C. The specific heat above freezing is 3.23
Milk must be received and cooled from 80°F to 38°F in 5 hrs. If 4000
= 0.935 Btu
gallons of fresh milk received having SG of 1.03 and
per Ib-oF, find the refrigeration capacity.
"p
Refrigeration:
402
A. 20 TR
B. 30 TR
h rg = 149.975
235.503 = 202.78 + x(l49.975)
x = 0.218
x = 21.80%
SOLUTION:
Ref Effect
w = rn/V
403
Refrigeration
C.·22 ..:'i TR
D. 34.6 TR
m cp (tz - t j )
Refrigeration - 27 (ME Bd. Oct. 1991)
m
1.03 (62.4)
An air-conditioning plant with a capacity of 400 KW of refrigeration
has an evaporating and condensing temperature of 3°e and 37°e
respectively. If it uses Refrigerant 12, what is the volumetric rate of
flow under suction condition?
C. 0.172 mJ/s
A. 0.272 m 3/s
D. 0.243 ml/s
B. 0.453 m 3 Is
4000 I 7.481
m = 34364.39 Ibs
Ref. Effect = (3436439/5x60)(0.935)(80~38)
Ref. Effect = 4498.298 Btu/min
Ref. Effect = 4498.298/42.42
Ref. Effect = 106.04 Hp
TR = (106.04 x 0.746)/3.516
TR = 22.50
p
SOLUTION:
2
Refrigeration - 26 (ME Bd. Oct. 1991)
An air-conditioning plant with a capacity of 400 KW of refrigeration
has an evaporating and condensing temperature of 3°e and 37°C
respectively. If it uses Refrigerant 12, what will be the mass of flash
gas per kg 'of refrigerant circulated?
C. 14.56%
A. 21.80%
D. 1834%
B. 12.45%
p
SOLUTION:
I,?",
From R-12 tables:
hi = h g at 3°e
hi = 352.755 KJ/kg
h r at 3°e = 202.780 KJ/kg
VI
= 0.05047 m3/kg
h 3 = 14 = hr at 37°e
h, = h, = 235.503 KJ/kg
Let x
=
mass
h, =
hrg =
h rg =
of flash gas or quality after expansion
h, = h r + xh rg
hg - hr
352.755 - 202.780
From R-12 tables:
hi = h g at 3°e
hi = 352.755 KJ/kg
h, at 3°e = 202.780 KJ/kg
3/kg
"I = 0.05047 m
h, = h, = h rat37°e
h, = h, = 235.503 KJ/kg
Refrigerating Effect = mth, -14)
400 = m(352.755 - 235.503)
m = 3.411 kg/sec
VI
VI
VI
v
v
=
m
=
3.411(0.05047)
=
0.I72m
VI
3/sec
Refrigeration - 28 (ME Bd. Apr. 1986)
An air-conditioning system of a high rise building has a capacity of 350
KW of refrigeration; uses R-12. The evaporator .and condenser
temperature are ooe and 35°e, respectively. Determine work of
compression in KW
404
Refrigeration
R efrigeration
A, 34 kw
B. 52 kw
'i
C. 43 kw
D,65kw
SOLUTION'
405
Refrigerating capacity = rruh. - h 4 )
Refrigerating capacity = 0,566(345 - 238,5)
Refrigerating capacity = 60,279 kw
Tons of refrigeration = 60,279/3,516
Tons of refrigeration = 17.144 tons ref.
p
From R -12 tables and P-h chart:
hi = h g at O°C
'f ,.~ ~
hi '= 351.477 KJ/kg
VI = vgatO°C
3
VI = 0,0553892 m /kg
At O°C:
h f = 200 KJ/kg
h fg = 351.477KJ/kg
From R-12 chart:
h z = 369 KJ/kg
h, = h, = 233.498 KJ/kg
Refrigerating Effect = rruh, - h 4 )
350 = m(35 1.477 - 233.498)
m = 2,967 kg/sec
Work of compression
m(hz-hl)
Work of compression
2,967(369-351.477)
Work of compression
52 kw
\
I
2,
Refrigeration· 30 (ME Bd. June 1990)
?"
•
v
Refrigeration - 29 (ME Bd. Apr. 1990)
A vapor compression refrigeration system has a 30 KW motor driving
the compressor. The compressor inlet pressure and temperature are
64.17 Kpa and -20°C respectively and discharge pressure of 960 Kpa.
Saturated liquid enters the expansion valve. Using Freon 12 as
refrigerant, determine the capacity of the unit in tons of refrigeration.
SOLUTION:
From R-12 tables and chart:
hi = 345 KJ/kg
h z = 398 KJ/kg
h, = h, = 238,5 KJ/kg
Compressor work = m(h z - hi)
30 = m(398 - 345)
m = 0,566 kg/sec
A simple vapor compression cycle develops 15 tons of refrigeration
using ammonia as refrigerant and operating lit condensing
temperature of 24°C and evaporating temperature of -18°C and
assuming compression are isentropic and that the gas leaving the
condenser is saturated, find the power per ton
A. 0.333 kw/ton
C. 0.452 kw/ton
B. 0.533 kw/ton
D. 0.702 kw/ton
p
SOLUTION:
.~
From ammonia tables and chart:
h, = 1439.94 KJ/kg
h z = 1665 KJ/kg
h, = h, = 312.87 KJ/kg
3/kg
VI = 0.572875 m
Refrigerating Effect = m(h 1 - h 4 )
15(3.516) = m(l439.94 - 312.87)
m ~o 0.0467 kg/sec
Power
Power
Power
Power
Power
v
Requirement = m(h z - hi)
Requirement = 0.0467(1665 - 1439.94)
Requirement = 10.531 lew
per ton = 10.531115
per ton = 0.702 K W Iton
Refrigeration - 31 (ME Bd. Apr. 1983)
A vapor compression refrigeration system is designed to have a
capacity of 100 tons of refrigeration. It produces chilled water from
-
Refrigeration
406
Refrigeration
2rc to 2°C. 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 size of the electric motor required tc
drive the compressor in kilowatts.
C. 87.23 kw
A. 92.31 kw
D. 98.23 kw
B. 76.34 kw
407
By heat balance in the system.
OR = We + RE
OR = 351.6 + 60
OR = 411.6 KW
mwC4.187)(10) = 411.6
low = 9.83 kg/sec
SOLUTION:
Refrigeration - 33 (ME Bd. Oct. 1984)
Ref. Effect = 100(3.516)
Ref. Effect = 351.6 kw
COP = Ref. Effect/Compressor power
351.6
5.86 = - - - - - EVAPORATOR
Compressor Power
4
Compressor Power = 60 kw
Compressor efficiency = 1 - 0.35
Compressor efficiency = 0.65
Motor rating = 60/0.65
22°F+
+2°F
Motor rating = 92.31 kw
low
A belt driven compressor is used in a refrigeration system that will
cool 10 Ii/sec of water from BOC to l°e. The nelt efficiency is 98%,
motor efficiency is 85%, and the input of the compressor is 0.7
KW/ton of refrigeration. Find the actual coefficient of performance if
overall efficiency is 65%.
A. 2.34
C. 3.45
B. 4.32
D. 6.44
SOLUTION:
Refrigeration - 32 (ME Bd. Apr. 1983)
A vapor compression refrigeration system is designed to have a
capacity Of 100 tons of refrigeration. It produces chilled water from
2rC to 2°C. 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 in kg/sec for a temperature rise of 100e.
A. 9.83
C. 12.23
B. 7.45
D. 4.34
Ref. Effect = 10 cp (t2 - t l )
Ref. Effect = (10 x 1)(4.187)(13-1)
Ref. Effect = 502.44 kw
TR = 502.4413.516
TR = 142.90
Compressor Input
0.7(142.90) Expansion
Valve
Compressor Input
100 kw
100(0.65)
Compressor work
(0.98)( 0.85)
Compressor work
78 kw
Actual COP
502.44178
Acw~COP
6.44
13°C 11Oli/s
mw
e=98%
11 -c
SOLUTION:
EVAPORATOR
Refrigerating Effect = 100(3 .516)
~,
Refrigerating Effect = 351.6 KW
COP = Ref. Effect/Compressor power
351.6
5.86 =
22°F
Compressor Power
rn,
Compressor Power = 60 KW
1
2°F
Refrigeration - 34 (ME Bd. Oct. 1984)
A belt driven compressor is used in a refrigeration system that will
cool 10 Ii/sec of water from 13°C to 1°C. The belt efficiency is 98%,
motor efficiency is 85%, and the input of the compressor is 0.7
KW/ton of refrigeration. Find the mass flow rate of condenser cooling
water warmed from 21°C to 32°C if overall efficiency is 65%.
Refrigeration
Refrigeration
408
A. 10.34 kg/s
B. 12.60kg/s
5.937 m 3/min
5.937/0.7
8A81 mvmin
2
11:/4 D L N c
= L (unity)
8A81 = 11:/4 (D)2 (D) (1200) (6)
D=0.114m
D = 114mm
VI =
VD =
VD =
VD =
For D
C. 23.23 kg/s
D. 15.34 kg/s
SOLUTION:
Ref Effect = m c p (t 2 - tt)
Ref Effect = (10 x 1)(4.187)(13-1)
Ref Effect = 502A4 kw
TR = 502A4/3.516
TR = 142.90
Compressor Input = 0.7(142.90) Expansion
Valve
Compressor Input = 100 kw
100(0.65)
Compressor work = - - - - - ' (0.98)(0.85)
Compressor work = 78 kw
QR
QR
QR
QR
=
RE + We
502A4 + 78
=
580A4 KW
=
13°C1101i/S
rn,
Refrigeration - 36 (ME Bd. Apr. 1985)
e=98"1o
1°C
m., c p (t 2 - t 1)
580A4 = rn., (4.187)(32- 21)
m., = 12.6 kg/sec
=
An am monia compressor operates at an evaporator pressure of 316
Kpa and a condenser pressure of 1514.2 Kpa. The refrigerant is
subcooled 5 degrees and is superheated 8 degrees. A twin cylinder
compressor with a bore to stroke ratio of 0.85 is to be used at 1200
rpm. The mechanical efficiency is 78%. For a load of 87.5 kw,
determine the bore and stroke for 5% clearance.
A. 123A4 mm
B. 109.23 mm
Calculate the bore in mm of a single-acting, 6 cylinder ammonia
compressor running at 1200 rpm to compress 700 kglhr of refrigerant
which vaporized at -15°C, given the following:
a. Bore and stroke = Unity
b. Volumetric Efficiencv = 70%
c. Specific volume of
3 = 8.15 felIb at 5°F
A. 110 mm
C. 114 nun
B. 106 rnm
D. 124 rnm
NU
SOLUTION:
From ammonia charts and table:
P
h, = 1472 KJ/kg
h, = 1715 KJ/kg
h, = h, = hr at 34°C
h, = h, = 361.2 KJ/kg
VI = OAI mJ/kg
V2 = 0.12 m 3/kg
Refrigerating Effect = nuh, - h 4)
87.5 = m (1472 - 361.2)
m = 0.07877 kg/sec
VI = rn VI
VI = 0.07877(OA1)(60)
VI = 1.9378 m'zmin
Since clearance is not given, then assume 5% clearance.
T]v
8.l5fellb x 1m 3/35.31ft3 x 2.2051h/lkg
vi = 0.5089 m 3/kg
VI = m VI
V I = (700/60XO.5089)
=
C. 117.40 mm, 99.79 mm
D. 234.23 mm, 86.79 mm
SOLUTION:
Refrigeration - 35 (ME Bd. Oct. 1988)
VI
409
= 1 + c - C(VI / vz)
I + 0.05 - 0.05(OA I 10.12)
T]v =
T]v =
VD
VD
=
=
0.879
1.9378/0.879
2.204 mJ/min
v
410
Refrigeration
VD~rr/4D2LNc
D/L ~ 0.85
2.204 ~ (11:/4) D 2 (D/0.85)(1200)(2)
D = 0.09979
D = 99.79 mm
L = 99.79/0.85
L = 117.40 mm
In a certain refrigeration system for a low temperature application, a
two stage operation is desirable which employs ammonia system that
serves a 30 ton evaporator at -30°C. The system uses direct contact
cascade condenser, and the condenser temperature is 40°C. Find the
total power required in kw,
A. 23.43 kw
C. 28.34 kw
B. 25.37 kw
D. 45.23 kw
p
SOLUTION:
6
r,
=
~PIP6
Px
=
";'---0-1-9.-9)-(1-55-7-)
I 3i
I
'A~/1:>"2
P x = 432.10 Kpa
From ammonia tables and chart:
h, = 1462 KJ/kg
hi = 1422.86 KJ/kg
h 2 = 1590KJ/kgh6 = 1649KJ/kg
h, = h, = 200.46
h 7 = h g = 390.587
Refrigerating Effect = ml(h 2 - h 3 )
30(3.516) = ml( 1422.6 - 200.46)
. ml = 0.0863 kg/sec
By heat balance in cascade condenser:
m2(h s - hg ) = ml(h 2 - h 3 )
ml1462 - 390.587) = 0.0863( 1590 - 200.46)
m, = 0.112 kg/sec
Total Power required
=
ml(h 2 - hi) + m2(h6 - h s)
Total Power required
=
Total Power required
=
00863( 1590 - ~422.86)
+ 0.112([649 - 1462)
35.37 KW
Refrigeration - 38 (ME Bd. Apr. 1981)
Refrigeration - 37 (ME Bd. Oct. 1984)
From Ammonia table:
PI = 119.9 Kpa
P 6 = 1557 Kpa
411
Refrigeration
A refrigeration system operates on the reversed Carnot cycle. The
minimum and maximum temperatures are -25°C and 72°C,
respectively. If the heat rejected to the condenser is 6000 KJ/min,
draw the T-S diagram and find power input required
A. 20.34 kw
C. 13.45 kw
B. 65.33 kw
D. 28.12 kw
SOLUTION:
T
T[ = -25 + 273
T 1 = 248°K
T 2 = 72 + 273
T 2 = 34SOK
QR = (SI - s4)(T2)
4•
6000 = (s, - 54)(345)
(SI - S4)= 6000/345
Ref. Effect = (SI - s4)T I
Ref. Effect = (6000/345)(248)
Ref. Effect = 4313 Kl/rnin
Power required = 6000 - 4313
Power required = 1687 KJ/min
Power required = 1687/60
Power required = 28.12 KW
2
0
.2
t:J
oC
_2S
•
'""
1
s
Refrigeration - 39 (ME Rd. Oct. 1986)
A refrigerating system operates on the reversed Carnot cycle. The
highest temperature of the refrigerant in the system is 120°F and the
lower temperature is 10°F. The capacity is to be 20 tons. Determine
the heat rejected from the system in Btn/min
A. 4936
C. 3423
B. 5634
D. 7421
412
ReJrigeration
Refrigeration
required plant refrigerating capacity in tons of' refrigeration if the
specific heat of fish is 0.7 above freezing and 0.3 below' freezing point
which is -3°e. The latent heat of freezing is 55.5 Kcallkg.
A. 21.23
C. 28.34
SOLUTION:
T
~.
~1
~,
-
s
T1
T,
T2
T2
=
=
=
=
10 + 460
470 0 R
120 + 460
580 0 R
B. 24.38
D. 32.12
_10°C
200 e
SOLUTION:
Ref. Effect = (s, - S4)Tl
20 x 200 = (s, - S4 )(470)
(s, - S4) = 4000/470
Heat Rejected = (s, - s4)T2
Heat Rejected = (4000/470)(580)
Heat Rejected = 4,936.17 BtuJmin
'11000k 9\
fish
I
n~
~
Ref. capacity = m[ c, (t, - tr) + L + c2Ctr - t 2) ]
t,::::3°e
11,000
Ref. capacity =
[0.7(20-(-3»+55.5+0.3{-3-(-10)}]
11(3600)
,
Ref. capacity = 20.47 Kcal/s x 4.187
Ref. capacity = 85.72 KW
Refrigeration - 40{ME Bd, Oct. 1989)
An Ice plant produces 20 tons of ice per day at -15°C from water at
25°e. If miscellaneous losses are 12% of the freezing and chilling load,
calculate the refrigeration capacity of the plant in tons of
refrigeration.
A. 21.35
C. 31.5
B.43.12
Tons of Refrigeration
Tons of Refrigeration
85.72/3.516
24.38 tons ref.
Refrigeration - 42 (ME Bd. Apr. 1986)
D. 36.3
25°e
SOLUTION:
.....1 _1
-15°e
n
n.. ''. r:::l
~lJL:J
~
2OtOM.
Product Load = mjc, (t. - t r) + L + C2 (t r- t 2 ) ]
!,aooe
20(907)
Product load =
[4.187(25 - 0) + 335 + 2.09{0-(-15)}]
24(3600)
Product load = 98.904 kw
Considering the 12% freezing and chilling load:
Ref. capacity = 98.904(1.12)
Ref. capacity = 110.773 kw
Tons of Refrigeration = 110.773/3.516
Tons of Refrigeration = 31.5 tons ref.
Compute tm; neat to be removed from 110 kg of lean beef if it were to
be cooled from 20°C to 4°C, after which it is to be frozen and cooled to
-18°e. Specific heat above freezing is given as 3.23 KJ/kgOC and below
freezing as 1.68 KJ/kg_°C. freezing point is -2.2°C and latent heat of
fusion is 233 KJ/kg.
A. 32,455 KJ
C. 36,437.5 KJ
B. 23,455 KJ
D. 54,223.2 KJ
200e
SOLUTION:
Q =m[cj(t,-tr}+L+c2(tr- t2)]
Q = 110[3.23(20 - 4) + 3.23(4 - (-2.2»
Q = 36,437.5 KJ
§-{][J------§
11...
.
11_
......
_18°e
• • ,.
. a.r • •
+ 233 + 1.68(-2.2 - (-18»]
Refrigeration - 41 (ME Bd. Apr. 1989)
Refrigeration - 43 (ME Bd. Apr. 1992)
Fish weighing 11,000 kg with a temperature of 20°C is brought to a
cold storage and which shall be cooled to -10°C in 11 hours. Find the
Magnolia Dairy products plant must cool 4000 gallons of fresh milk
received from the farm per day from an initial temperature of 80°F to
j
l
414
Refrigeration
415
Refrigeration
a temperature of 38°C in 5 hours. If the density of milk is 8.6
Ibs/gaIlon, specific gravity is 1.03, and specific heat is 0.935 how much
brine must be circulated if the change in temperature is 15°F, specific
gravity is 1.182, and specific heat is 0.729?
A. 41.77 gpm
C. 34.11 gpm
B. 54.22 gpm
D. 65.23 gpm
Heat of fusion of ice ------------- 144 Btu/lb
A. 34.56 Ibs
C. 43.23 Ibs
B. 74.23 Ibs
D. 64.8 Ibs
ITJ
SOLUTION:
8S0F
SOF
Vegetable
2S0lb
Ice
BO°F
Ref. capacity = m c p ~t
4000x8.6
Brine
Cooler
3BOF
m
m
=
m
Cooling Load = m, cp ~t(U)
SOLUTION.
5
6880 Ib/hr
.
Coolmg Load
250(0.80)(85 -- 45)(1.30)
=
--------
Cooling Load
=
24
433.33 Btu/hr
Heat gained by vegetable
=
(m/24)[ 0.463(32 - 25)+ 144 + 1.01(45 - 32)]
m = 64.81bs
Q = m c p ~t
270,177.6 = m(0.729)(IS)
m = 24,707.6 Ib/hr
m = 411.793 Ib/min
Refrigeration - 45 (ME Bd. Oct. 1990)
v
411.793
1.182 x 62.4
V = 5.583 fe/min
V = 5.583 x 7.481
V = 41.766 gal/min
I
i
Ice
Ref. Capacity = 6880(0.935)(80 - 38)
Ref. Capacity = 270,177.6 Btu/hr
=
+
Vegetable
433.33
A 10 tons ice plant using ammonia refrigerant operates between
evaporator and condenser temperature of -20°C and 35°C
respectively. The ice plant is to produce ice at -12°C from water at
30°C in 24 hours. Assuming losses to be 18% of the heat absorbed
from water, determine the power required by the compressor
A. 21.23 kw
C. 56.32 kw
B. 16.79 kw
D. 43.23 kw
300 e
Refrigeration - 44 (ME Bd. Apr. 1982)
A mass of ice at 25°F is needed to cool 250 pounds of vegetables in a
bunker, for 24 hours. The initial temperature of the vegetables is
assumed to be 85°F. It is also assumed that the average temperature
inside the bunker is 45°F, within the 24-hour period. If the heat gained
per hour in the bunker is 30% of the heat removed to cool the
vegetables from 85°F to 45°F, what would be the required mass of ice?
Specific heat of ice --------------- 0.463
Specific heat of vegetables ----- 0.80
Specific heat of water ---------- 1.01
SOLUTION:
10 tons I
Water
-12°e
I
n
.r:::l
LI1.:J
t,=ooe
RE =
10(907)
I
,I
4S0F
Heat loss from ice
!
!
[4.187(30-0)+335+2.0935{0-(-12)}](1.l8)
24(3600)
RE = 60.17kw
From Ammonia chart and table:
hi = 1437.23 KJ/kg
h z = 1736 KJlkg
h, = !4 = 366.072 KJlkg
416
Relllgnalmg Effect = rruh, - h.)
60.17
111(1437.23 - 366.072)
m
00562 kg/sec
Compressor Power
m(h 2 - h)
Compressor Power
0.0562(1736 - 1437.23)
Compressor Power
16.79 kw
SOLUTION
Ref. Effect = rruh, - h 4 )
5(3.516) = m(353.6 - 238.5)
m = 0.153 kg/sec
Heat gained by water
m., cp (t) - t 2
Refrigeration - 46 (ME Bd. Oct. 1994)
An ideal vapor compression refrigeration cycle requires 2.5 KW to
power the compressor. You have found the (ollowing data for the
cycle: the enthalpy at condenser entrance = 203 KJ/kg, exit = 55;
evaporator entrance = 55 KJ/kg, exit = 178. If mass flow rate of the
refrigerant is 0.10 kg/sec, then the coefficient of performance of this
refrigeration cycle is most nearly:
A. 592
C. 5.92
B. 59.2
D.4.92
SOLUTION:
hi ~ h,
COP
h, - hi
178 - 55
COP
417
Refrigeration
Refrigeration
Vw
Vw
=
=
Heat loss by refrigerant
m(h 2 - h 3 )
~w(4.187)(7) = 0.153(377 - 238.5)
m., = 0.723 kg/sec
(0.723 kg/sec)(1 lilkg)(60sec/min)(1 gal/3 .785Ii)
11.46 gpm
=
)
=
Refrigeration - 48 (ME Bd. Apr. 1996)
A refrigeration system operates on an ideal vapor-compression using
Refrigerant-12 with an evaporator temperature of -30°C and a
condenser exit temperature of 49.3°C and requires a 74.6 KW motor
to drive the compressor. What is the capacity of the refrigerator in
tons of refrigeration? Enthalpy of condenser entrance = 382 KJ/kg,
exit = 248.15; at evaporator entrance = 248.15, exit = 338.14.
A. 43.1
C. 21.3
B. 34.5
D. 18.2
SOLUTION:
203 -178
COP
=
4.92
Refrigeration - 47 (ME Bd. Oct. 1994)
A Freon -12 waste water system operating at a 5°C suction
temperature and a 40°C condensing temperature has an evaporator
load of 5 tons. If the condenser is selected for a 7°c water temperature
rise, how many gpm must be circulated through the condenser?
The following enthalpies have been found: condenser entrance = 377
KJ/kg, exit = 238.5; evaporator entrance = 238.5 KJ/kg, exit = 353.6
A. 11.46gpm
C. 13.45gpm
B. 15.23 gpm
D. 23.22 gpm
W = m(h 2 - hi)
74.6 = m(382 - 338.14)
m = 1.7 kg/sec
QA
QA
QA
m/h, - h 4 )
1.7(338.14 - 248.15)
153 kw
Ref. Capacity
Ref. Capacity
153/3.516
43.5 tons ref.
Refrigeration - 49 (ME Bd. Apr. 1995)
Liquid ammonia at a temperature of 26°C is available at the expansion
valve. The temperature of the vaporizing ammonia in the evaporator
418
Refrigeration
Refrigeration
is 2°C. Find the percentage of liquid vaporized while flowing through
the expansion valve.
Temperature
Pressure
(0C)
(Kpa)
hf
hfg
hg
2°C
462.49
190.4
1255.2 1445.6
26°C
1033.97
303.6
1162.0 1465.6
A. 9.02
C. 91.08
B. 90.98
D. 8.92
SOLUTION:
From 350 kpa to 1300 kpa:
RE = m (h, - h 4 )
40 = m(189.023 - 87.796)
m = 0.39515 kg/s
From 350 kpa to 1400 kpa
RE = m'(h, - h;')
40 = m'(189.023-91.355)
m' = 0.40955 kg/s
SOLUTION:
h3
h,
=
=
303.6
14
=
303.6 KJlkg
Tlv
hf + xh fg
=
419
Tlv
mV
190.4 + x(1255.2)
m'v I
I
(n/4)(D)2 LN - (n/4)(D)2 LN,
x = 9.02%
Refrigeration - 50 (ME Bd. Apr. 1998)
A 140 mm x 140 mm single effect, twin-cylinder, single actmg Freon12 compressor with a refrigeration capacity of 40 kw operates between
a discharge pressure of 1300 kpa and a suction pressure Of 350 kpa.
The speed of the compressor is 600 rpm. If the discharge pressure
shall be raised to 1400 kpa, at what speed (rpm) should the
compressor be run to produce the same refrigeration capacity and
assuming the volumetric efficiency to remain the same?
Freon 12 Properties:
At 350 kpa
p
~j
-;r- \.
h = 189.023 KJ/~
J
y = 0.04923 m /kg
At 1300 kpa
h = 211.314 KJ/kg
h, = 87.796 KJ/kg
At 1400 kpa
h = 213.692 KJ/kg
h, = 91.355 KJ/kg I
•
A. 610
C. 620
v
B. 615
D. 630
m
m'
N
N'
0.39515
0.40955
600
N'
N'
=
621.865 rpm
Refrigeration - 51 (ME Bd, Oct. 1997)
lJ)
A refrigeration system using R-22 has a capacity of 320 kw of
refrigeration. The evaporating temperature is minus 10 degrees C and
the condensing temperature is 40"C. Calculate the fraction of vapor in
the mixture before the evaporator.
Properties of R-22 are:
At -10 nC
h g = 401.60 KJ/kg
h r = 188.426 KJ/kg
At 40°C
h r = 249.686 KJ/kg
C. 0.245
A. 0.287
D.
0.227
B. 0.315
420
Refrigeration
Refrigeration
SOLUTION:
p
h,
=
"
h, = h, + x (h g - he)
249.686
=
-- - .. \
,,0
SOLUTION:
Condinser
188.426 + x(401.60 - 188.426
Expansion
Valve
x = 0.287
v
Refrigeration - 52 (ME Bd. Oct. 1999)
What is the coefficient of a vapor compression refrigeration system
having the following data: Enthalpy entering the compression is
181.79; enthalpy after compression work is 207.3 KJ/kg.
After
condensation the enthalpy is 58.2 and the throttled from 0.19 Mpa to
0.18 Mpa.
A. 5.8
C. 4.75
B. 3.2
D. 5.6
SOLUTION:
COP = hI - h,
h, - h I
COP
181.79-58.2
207.3 -181.79
@
h,=181.79KJ
Kg
COP = 4.84
Refrigeration - 53 (ME Bd. Oct. 1999)
A refrigeration system having a 22 kw capacity needs 7.8 hp
compressor. Find the COP of the system.
C. 3.78
A. 312
B. 4.62
D. 6.34
COP = RE/W c
COP == 22/(7.8 x 0.746)
COP = 3.78
421
422
Air-conditioning
Air-conditioning
AIR-CONDITIONING
42')
Air Conditioning - 3
The change of enthalpy of air in a cooling tower is 35 Btu/lb and the
mass flow of air is 453.17 Ib/min. Water enters the tower at the rate of
50 gpm and 115°F. Determine the exit temperature.
A. 45°F
C. 65°F
D. 77°F
B. 55aF
Air Conditioning - 1
When 100 kg/min of outside air at 32°C dry bulb and 200 kg/min recirculated air at 22°C dry bulb are mixed with an air conditioning
system, the resulting dry bulb temperature will be:
C. 4633°C
A 2533°C
B. 35.44°C
D. 26.88°C
SOLUTION:
=
Heat
Heat
Heat
Heat
carried by
carried by
carried by
carried by
air
air
air
air
=
=
m(h l - h 2)
453.17(35)
15860.95 Btu/min
Heat remove from water
Heat carried by air = m cp (t 2 - t.)
m; = 50 gal/min x 62.4/7 .481
m., = 417.056Ib/min
rna to + m, t, = m., t m
m., = rna + rn,
rn., = 100 + 200
m., = 300 kg/min
100(32) + 200(22)
tm = 25.33°C
SOLUTION:
300
tm
Air Conditioning - 2
15860.95 = 417.056(1)(115 - t2 )
t 2 = 76.97°F
Air Conditioning - 4
The amount water carried by air in a cooling tower is 15 Ib/min. The
change in humidity ratio in outlet and inlet is 0.025 Ib/lb. Determine
the volume flow of air needed if specific volume is 13 fe/lb.
A. 6000 ft3/min
C. 7500 ft3/min
B. 7800 ft 3imin
D. 5000 ft3/tnin
SOLUTION:
The change of enthalpy in an air conditioning unit is 10 Btu/lb. The
mass of supply air is 150,000 Ib/hr. What is the conditioner capacity?
A. 125 TR
C. 150 TR
B. 100 TR
D. 200 TR
SOLUTION:
m., = tn. (W2 - WI)
15 = m, (0.025)
rna = 600 Ib/min
Va = rna V
Va = 600(13)
V, = 7800 ftl/min
Conditioner capacity
m, (h I - h4 )
Conditioner capacity = 150,000 (10)
Conditioner capacity = 1,500,000 Btulhr x 1.055/3600
Conditioner capacity = 439.58 KW
Cond itioner capacity
439.58/3.516
Conditioner capacity = 125.02 ton ref
424
Air-conditioning
Air-conditioning
425
SOLUTION:
Air Conditioning' - 5
o
SHF
Re-circulated air of 8 kg/sec with 53 KJ/kg enthalpy and outside air of
2 kg/sec with 90 KJ/kg enthalpy enters the conditioning unit.
Determine the air conditioning capacity if supply enthalpy to
conditioned space is 42 KJ/kg.
C. 174 KW
A 154 KW
D. 184 KW
B 164 KW
OL + Os
120
SHF
120+47
Air Conditioning - 8
SOLUTION:
rna h, + m, h, = m, h,
2(90) + 53(8) = (2 -t- 8)(h 4 )
h, = 60A KJ/kg
A ir conditioning capacity
Air conditioning capacity
Air conditioning capacity
ills
(h, - hi)
(2 + 8)(60A - 42)
184 KW
The total heat load and latent heat load of theater is 150 KW and 60
KW, respectively. Supply air is at 15°C and has a mass of 9 kg/sec.
Determine the temperature to be maintained in the theater.
A. 20 0 e
c. 30 0 e
B. 25°e
D. 35°e
SOLUTION:
Air Conditioning - 6
Os
=
Os + OL
Os + 60
90 KW
Os
=
Ins c p (t2 - t j )
90
=
9(1 )(tz- 15)
25°e
OT
=
150
Outside air of an ail' conditioning system is 25% of re-circulated air.
Determine the mass of outside air if mass of supply air is 15 kg/sec.
A. 1 kg/sec
C. 3 kg/sec
B. 2 kg/sec
D. 4 kg/sec
t2
SOLUTION-
rna + 01, = m,
0.25 rn,
rna
ill,
4 Ina
m, + -lm, = 15
rn,
3 kg/sec
Air Conditioning - 7
The sensible heat load and latent heat load in an air conditioning
svstem is 120 KW and 47 KW, respectively. What is the sensible heat
factor?
A. 65.34%
C. 76.54%
R. 29A5%
D 71.86%
.
=
=
Air Conditioning - 9
A cooling tower has an efficiency of 65%. Water enters the tower at
55°C. The wet bulb temperature of surrounding air is 27°C. What is
the temperature of water leaving the tower?
A. 36.8°e
c. 46.9°e
B. 44.5°e
D. 30Aoe
SOLUTION:
"'
Efficiency
ta - tb
t a - t wb
A it-conditioning
426
o. 65
__
_ ')') - I h
Air-conditioning
427
S()UJ!]()N
5') - 27
~
0.65(28)
tb
Approach = t b - t wb
10 = 37 - t Wb
twb = 27°C
.
ta - t b
55 - t b
= 36.8°C
Efficiency
=
---
t. - t wb
0.65
Air Conditioning - 10
t. - 27
t,
The change of temperature entering the cooling tower and wet bulb
temperature of surrounding air is 25°C and efficiency is 65%. If mass
of water leaving the tower is 10 kg/sec, determine the heat carried by
air.
C. 680 KW
A. 720 KW
D. 700 KW
B. 540 KW
SOLUTION:
Efficiency -
t. - t h
t. - t Wb
0.65 =
Heat
Heat
Heat
Heat
t a - 37
ta carried
carried
carried
carried
t. -
tb =
by air
by air
by air
by air
= 52°C
Air Conditioning - 12
A cooling tower is used to cool a jacket water loss from the engine. The
heat generated by fuel is 2500 KW and cooling loss is 30%. If
temperature range of the tower is 15°C. Determine the mass flow of
water entering the tower.
A. 12 kg/sec
C. 16 kg/sec
B. 14 kg/sec
D. 18 kg/sec
SOLUTION:
lb
25
16.25
= Heat loss by water
= me, (t a - tb)
= 10(4.187)(16.25)
= 680.39 KW
Cooling loss
Cooling loss
0.3(2500)
750KW
«, -
Cooling loss = m., c p
lb)
750 = mw (4 . 187)( 15)
m., = 1 1.94 kg/sec
Air Conditioning - 11
Air Conditioning - 13
The approach and efficiency of cooling tower is 10°C and 60%,
respectively. If temperature of water leaving the tower is 37°C, what is
the temperature of water entering?
A. 46°C
C. 68°C
B S2°C
D. 48°
A dryer is to deliver 1000 kg/hr of cassava with 2% moisture and 20%
moisture in the feed. Determine the mass of air required if change in
humidity ratio is 0.0165.
A. 3.57 kg/sec
C. 3.79 kg/sec
B. 4.67 kg/sec
D. 5.36 kg/sec
428
Air-conditioning
429
Air-conditioning
SOLUTIUN:
SOLUTION:
Beat supplied
Let x = total amount of feed materials
Amount of solid in product = 1000(0.98) = 980 kg/hr
Amount of solid in the product = Amount of solid in the feed
980 = 0.8(x)
x = 1225 kglhr
Moisture removed = 1225 - 1000
Moisture removed = 225 kg/hr
'225 = rna (0.165)
rna = 1363636 kg/hr
rna = 3.7878 kg/sec
Air Conditioning The moisture remove from a material is 250 lb/hr and change of
humidity ratio in dryer is 0.0175. Determine the fan capacity if specific
volume of air entering is 35 fellb.
A. 400,000 ft3/hrc. 500,000 ft3/hr
B. 450,000 fe /hr D. 550,000 ft3/hr
Heat supplied
Heat supplied
- WI)
Va = rnava
Va = 14285.7(35)
Va = 500,000 ft3lhr
Air Conditioning The change of enthalpy in a heating chamber of dryer is 25 Btu and
the mass of air supplied is 30,000 lb/hr, What is the heat supplied by
heater?
A. 560 KW
C. 350 KW
B. 450 KW
o 220 KW
30,000
(--)(25 x 1.055)
3600
219.79kw
Water at 55°C is cooled in a cooling tower which nas an efficiency of
65%. The temperature of the surrounding air is 32°C 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.
C. 6.34
A. 8.55
D. 9.23
B. 7.34
®
SOLUTION:
ta - t
ta
Moisture removed = rna (W2
250 = rna (0.0175)
m••~ 14285.71b/hr
ma(h 2 - h.)
Air Conditioning - 16 (ME Bd. Apr. 1986)
Efficiency
SOLUTION:
=
-
0
t Wb
From psychrometric chart:
At32°C and 70% RH:
twb = 27.35°C
55 - t b
0.65
55 ~ 27.35
tb = 37.03°C
By heat balance in the condenser:
m cp (t a - tb ) = 2,300,00013600
m(4.187)(55 - 37.03) = 2,300,000/3600
m = 8.49 kg/sec
From steam table(Table 1), At 37.03°C ,
Vf = 1.0067 li/kg
Pump capacity = 8.49( 1.0067)
Pump capacity = 8.547 Ii/sec
430
~ '
4 .J
I
Air-conditioning
Air-conditioninp
SOIlJTION:
Air Conditioning - 17
Fifty gallons per minute of water enters a cooling tower at U5°F.
Atmospheric air at 60°F and 55% relative humidity enters the tower
at 6,009 cfm and leaves at 90°F saturated. Determine the volume of
water that leaves the tower in gpm.
A. 32.34
C. 34.23
B. 48.62
D. 65.33
SOLUTION:
5''lPm
From psychrometric chart:
hi ~ 21 Btullb
~m
h 2 = 56 Btu/lb
~
WI = 0.0061 lb/lb
W2 = 0.03121b/lb
VI = 13.24 ftl/lb
From steam table, at 115°F, Vf = 0.01618 ftllib
Heat lost by water = heat
leal galIll:U, ,oy. air ~
a
50
= (6000/13.24)(56-21)
)(1.0KI15-t)
(
7.481(0.01618)
e
t, = 76.61°F
~
From psychrometric chart:
hi = 21 Btullb
h2 = 56 Btulb
WI = 0.0061 lb/lb
_____ t.
~
Q
W2 = 0.03I21b/lb
3/1b
Vi = 13.24 ft
m., C~ Amount of water carried by air
mw = maC w 2 - WI)
m., = (V, I V.)(W2 - WI)
m.; = (6000/13.24)(0.0312 - 0.0061)
rn., = 11.37 Ib/min
Air Conditioning - 19 (ME Bd. Apr. 1987)
GO°F
55%RH
From steam table, at 115°F,
3/1b
\if = 001618ft
Q = Volume of water that leaves the tower
Q = 50·· 1l.3'";'(001618)(7.48)
Q = 48.624 gpm
Air Conditioning - 18
An atmospheric cooling tower is to provide cooling for the jacket
water of a four stroke, 800 KW diesel generator with useful output of
34% and cooling loss of 30% . The cooling tower efficiency is 60% at a
temperature of approach of 10°e. If ambient air has a relative
humidity of 70% and dry bulb temperature of 32°C, determine the
cooling water supplied to the diesel engine in IiIhr. Generator
efficiency is 97%.
C. 43,345
A. 41,713
D. 47,234
B. 45,232
SOLUTION:
From psychrometric chart, at 32°C and 70% RH,
t wb = 27.50°C
Approach = t b - twb
10 = t b - 27.5
t b = 37.5°C
Efficiency
Fifty gallons per minute of water enters a cooling tower at 115°F.
Atmospheric air at 60°F and 55% relative humidity enters the tower
at 6,000 cfm and leaves at 90°F saturated. Determine the exit
temperature of water, of
A. 56.23
C. 76.61
B. 65.33
D. 45.34
=
ta
o
- tb
t a - t wh
0.60 =
t a - 37.5
ta
-
27.5
t, = 52.5°C
Brake power = 800/0.97
Brake Power = 824.742 KW
e=60%
+--air
432
Air-conditioning
0.34
=
Air-conditioning
824.742
rn., = 5,555 kg/hr
Percentage Make-up water
Percentage Make-up water
- - -.._ - -
Heat sup plied
Heat supplied - 2425.71 KW
Cooling loss ~ 0.30(2425.71)
Cooling loss = 727.713
Cooling loss = mCp(t 2 - t j )
727.713 = m(4.187)(52.5 - 37.5)
m = 11.587 kg/sec x 3600
m = 41,713 kg/hr
m = 41,713 kh!hr(I li/kg)
m = 41,713li/hr
'~'j?
..2.:
5,555/250,000
2.22%
An auditorium is to be maintained at a temperature of 26°C drybulb
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 heat
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 the
refrigeration capacity in tons.
A 43.45
C. 63.28
B. 54.23
D. 76.34
250,000 kg/hr of water at 35°C enters a cooling tower where it is to be
cooled to 17.5°C. The energy is to be exchanged with atmospheric air
entering the unit at 15°C and leaving the unit at 30°C. 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. 3.44%
C. 3.94%
B. 8.34%
D. 2,22%
From psychrometric chart
At 15°C and 30% RH:
h, = 23.0 KJ/kg
w, = 0.0032 kg/kg
At 30°C and 85% RH:
h 2 = 89.0 KJ/kg
W2 = 0.0232 kg/kg
Heat loss by water = Heat gain by air •
I
m w cp i1t = ma(h 2 - hI)
250,000(4.187)(35-17.5) = m.(89-23)
rna = 277,746 kg/hr
rn., = amount of make-up water
rn., = rna (W2 - WI)
m., = 277,746(0.0232 - 0.0037)
=
Air Conditioning - 21 (ME Bd. Oct. 1995)
Air Conditioning - 20 (ME Bd. Oct 1991)
SOLUTION:
=
433
®S5%RH
30 0 e
SOLUTION:
Qs = m, cp (t, - t])
110 = ms(1.0)(26 -15)
m, = 10 kg/sec
From psychrometric chart,
h, = 90.5 KJ/kg
h, = 53 KJ/kg
QT = Qs + QL
QT = m s(h 4 - h])
110+37.5 = 10(53-h])
h] = 38.25 KJ/kg
01
_
m - O.25 m.
o
m,
Air
I
By mass balance:
rna + rn, = m,
0.25m, + m, = 10
~
rn, = 8 kg/sec
By heat balance:
0
m.h 1 + m.h, = m.h,
(0.25 x 8)(90.5) + 8(53) = IOh2
h 2 = 60.5 KJ/kg
Refrigerating capacity = m s(h 2 - h])
Refrigerating capacity = 10(60.5 - 38.25)
Refrigerating capacity = 222.5/3.516
Refrigerating capacity = 63.28 tons ref
Q,
37.5 kW
434
Air-conditio 11illg
Air Conditioning - 22 (ME Bd. Apr. 1988)
----
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 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 conditioner's refrigeration capacity
in tons refrigeration and its ventilation load in, KW.
A. 83.22, 132
C. ~6.23, 45.32
B. 75.34, 412
D. 54.23, 83.23
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 dry bulb and nOF wet bulb
temperature is cooled to 63°F dry bulb and 59°F wet bulb temperature
by chilled water cooling coils and delivered as supply air to the
theater. Calculate the tons of refrigeration required.
A. 123
C. 124
D. 128
B. 125
Qs = ills c p (t 2 - t 1)
ISO = fisC 1.0)(26 - 15)
rn, = 13.636 kg/sec
From psychrometric chart:
h 2 = 53 Kl/kg
h) = 86.5 KJ/kg
SOLUTION:
01 m60%RH
o
n
m.
11\
Assembly Han
j
r
I
Q
0
1 m,
0.80(13.636)
= 10.909kg/sec
rn , = 0.20(13.636)
rn, = 2.727 kg/sec
By heat balance:
m, h) + fir h2 = m, h,
2.727(86.5) + 10.909(53) = 13.636 h,
h, = 59.7 Kl/kg
Capacity = m,(h 4 - hi)
Capacity = 13.636(59.7 - 38.24)
Capacity = 292.629 KW / 3.516
Capacity = 83.23 tons ref
Ventilation Load = maCh) - hi)
Ventilation Load = 2.727(86.5 - 38.24)
Ventilation Load = 131.60 KW
=
01
34°C db
QT = Qs + QL
150+51.3 = 13.636(53-h l )
hi = 38.24 Kl/kg
fir
fir
---
Air Conditioning - 23 (ME Bd. Oct. 1981)
SOLUTION:
k
435
Air-conditioning
150 kW
Q,
51.3 kW
From psychrometric chart: '1" 184°F db
hi = 35.82 Btu/lb \..:.J m'72°F wb
h2 = 25.78 Btu/lb
h) = 31.3 5 Btu/Ib
Q.=620000
~U/h
Q,=210000
Btuth
Total heat load = rruh, - h z)
620,000 + 210,000 = m(31.35 - 25.78)
m = 149,013 lb/hr
Conditioner capacity = fi(h, - h 2 )
149,013(35.82 - 25.78)
Conditioner Capacity
=
Conditioner Capacity
=
12,000
125 tons of ref
Air Conditioning - 24 (ME Bd. Oct. 1996)
A room being air conditioned is being held at 25°C dry bulb and 50%
relative humidity. A flow rate of 5 ml/s of supply air at 15°C dry bulb
and 80% RH is being delivered to the room to maintain that steady
condition at 100 Kpa, What is the sensihte heat absorbed from the
room air in KW?
C. 40.5
A. 50.8
D. 70.9
B. 60.8
436
Air-conditioning
A it-conditio II illl{
S()[
SOLliTION
UTION:
PY = m R T
100(5) = m(O.287)(15 +273)
III = 6.049 kg/sec
Os
Os
Os
Os
=
=
=
=
437
sensible heat
mCp(t z - t.)
6.049(1003)(25 - 15)
60.80 KW
Let x
=
weight of original ")roduct per Ib of wet feed
Solid in wet feed
0.95 x = OAO( I)
x = OA21 Ib
Weight
Weight
Weight
Weight
Weight
=
Solid in dried product
of water removed = 1 - OA21
of water removed = 0.579 Ib/lb of orig. product
of water removed per Ib of final product
removed = 0.579/0A21
removed = 1.3751b/lb
Air Conditioning - 25 (ME Bd. Oct. 1992)
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. 0.34 Ib
C. 0.86 lb
B. 0.63
D. 0.58 lb
SOLUTION:
Consider 1 Ib of wet feed:
Let x = weight of original product per Ib of wet feed
Solid in wet feed = Solid in dried product
0.95 x = OAO(I)
x = OA21 Ib
Weight of water removed = I - OA21
Weight of water removed = O.5791b/lb oforig. product
Air Conditioning - 26 (ME Bd. Oct. 1992)
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 final product
A. 1.3751b
C. 1.872 Ib
B. 119 Ib
D. 2.345 lb
Ii
j
Air Conditioning - 27 (ME Bd. Apr. 1983)
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 1 metric ton per hour. The initial moisture content is 3
kilograms moisture per kilograms dry solid and will be dried to a
moisture content of 0.10 kg moisture per kg solid. Air enters the dryer
with a humidity ratio of 0.016 kg moisture per kg dry air and leaves
with a relative humidity of 100% while the temperature remains
constant at 60 a C. If the total pressure of air is 101.325 Kpa, determine
capacity of forced draft fan to handle this air in m 'zmin.
A. 80
C. 82
B. 84
D. 86
SOLUTION:
Let x= mass in kg of dry solid
3x + x = 1000
x = 250 kglhr
Since the given value is out ofrange, therefore:
From steam table at 60 aC, P sat = 19.94 Kpa
19.94
w = 0.622(
)
J 01.3 - 19.94
w = 0.152 kg/kg
Moisture removed = mtw, - WI)
438
Air-conditioning
Air-conditioning
3(~5())-0.1(250) = m(0152-0.016)
m
5,331 kglhr
Using high temperature psychrometric chart:
At 60°C(140°F) and 0.016 humidity ratio,
v = 15.5 ttl/lb
v .~ 0.968 rrr'zkg
Fan capacity = (5331/60)(0.968)
Fan capacity = 86 m 3/min
4j'j
0.029 kg/kg
amount of air
m a ( W 2 - WI) = amount of moisture removed
m a(0.029 - 0.0087145) = 0.998747
rna ~ 49.234 kg/sec
m, = 177,242.4 kg/hr
W2
lct
Ina
Air Conditioning - 29 (ME Rd. Oct. 1985)
Air Conditioning - 28 (ME Rd. Oct. 1990)
Wet material, containing 215% moisture(dry basis) is to be dried at
the rate of 1.5 kg/sec 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 0 K and 70%
saturated. calculate how much air will be required to remove the
moisture.
A. 213,233 kg/hr C. 177,242 kglhr
B. 177,142kg/hrD. 198,234kg/hr
A Dryer is to deliver 1000 kg/hr 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 degrees centigrade 'wet bulb. The
dryer is' maintained 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 heat supplied by heater in kw.
.
C 4.23
A. 323
B. 5.46
D 6.23
SOLUTION:
Let m
SOLUTION:
Let x = rate of flow of dried product
Solid in wet feed = solid in dried product
I
---(1.5) = 0.95x
1 + 2.15
x = 0.501253 kg/sec
Amount of moisture removed
Amount of moisture removed
Solving for WJ :
(
W
Pv
0.622lp-p v
j
"'
1.4
)
101.325 + 1.4
W = 0.0087145 kg/kg
From Psychrometric chart:
W
=
0.622 (
1.5 - 0.501253
0.998747 kg/sec
amount ofpalay in wet feed
Solid in wet feed = solid in product
0.85(m) = 0.90(1000)
m = 1,058.832 kg/hr
From psychrometric chart:
hi = 60.5 KJ/kg
h2 = 74 KJ/kg
h, = 196 KJ/kg
WI = W2 = 0.0111 kg/kg
W3 = 0.0515 kg/kg
3/kg
V2 = 0.915 m
Amount of moisture removed = J058.823 - 1000
Amount of moisture removed = 58.823 kg/hr
=
ma ( W 3 - W2) = 58.823
m a(0.05l5 - 001l!) = 58.823
rna = 1456.015 kglhr
Heat supplied by heater = m, (h 2 - hJ)
Heat supplied by heater = (1456.015/3600)(74 - 60.5)
Heat supplied by heater = 5.46 kw
440
Air-conditioning
Air-conditioning
Air Conditioning - 30 (ME Bd. Oct. 1985)
A Dryer is to deliver 1000 kg/hr 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 degrees centigrade wet bulb. The
dryer is maintained 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 dryer in m 3/hr.
A. 1332.25
C. 1234.23
B. 1532.34
D. 1982.34
=
SOLUTION:
Amount of moisture removed
1058.823 - 1000
Amount of moisture removed
58.823 kg/hr
m a(w3 - W2) = 58.823
m.(0.0515 - 00111) = 58.823
rna = 1456.015 kg/hr
v, 1456.015(0.915)
Va = 1332.25 m 3/hr
P,
Psat
Pv
RH
0.7 =
P,
=
w
=
4.29722
2.9722 kpa
Pv
622 - P - P,
w
~,
101.325 - 2.9722
001879 kg/kg
Air Conditioning - 32
The humidity ratio of air is 0.045. If barometric pressure is 101 kpa,
find the partial pressure of water vapor.
C. 6.81 kpa
A. 4.23 kpa
D. 5.23 kpa
B. 7.34 kpa
SOLUTION
w
At 30°C, air-vapor mixture has a relative humidity of 70%. Find the
humidity ratio if barometric pressure is 100"C. At 30"C, P sn t = 4.246
kpa
"
'I
P
622l-v
P-Pv
(
0.045
Air Conditioning - 31
\j
2.9722
w = 0.622 (
amount of palay in wet feed
Solid in wet feed = solid in product
0.85(m) = 0.90(1000)
m = 1,058.832 kg/hr
From psychrometric chart:
hi = 60.5 Kl/kg
h 2 = 74 KJ/kg
h, = 196 KJ/kg
WI = W2 = 0.0111 kg/kg
W3 = 0.0515 kg/kg
V2 = 0.915 m 3/kg
C 0.054
D. 0.019
A. 0.123
B. 0.986
SOLUTION
Let m
441
=
J
P
101 - P, = 13.82P,
= 6.82 kpa
P,
'1
--J
v
0.622I\.IOI-P v
442
.·1 ir-cotulitioning
443
A ir-conditioning
Air Conditioning - 33
p\
RH
=
Psa t
Air at 36°C and pressure of 101.2 kpa has a density of 1.08 kg/m'.
Find the humidity ratio of air.
A. 0.0352
C 00635
B. 0.6350
D. 00173
060
P,
w
SOLUTION:
=
P,
5.62~
3.3768 kpa
3.3768
= 0.622 (
)
101.2- 3.3768
w = 0.0213
PY
h = 1(35) + 0.0213(2565.3)
h = 89.63 KJ/kg
mRT
=
m
P"
Y
RT
108
=
---"--
0287(36 + 273)
P, = 95.777 kpa
P = P" + P"
1012 = 95.777+P,
P, = 5.422 kpa
p
Air Conditioning - 35
Air-vapor mixture has an enthalpy of 75 KJ/kg at 30°C. Find the
partial pressure of water vapor. At 30°C: h g = 2556.3 KJ/kg
A. 6.34 kpa
C 4.231 kpa
B 1.34 kpa
D. 2.791 kpa
w = 0622 -----'-,P-P v
5.422
0622 ( - - - - - )
1012 - 5.422
w = 0.0352 kg/kg
w
=
SOLUTION:
h = c p t + W hg
75 = 1(30) + w(2556.3)
w = 0.0176
Air Conditioning - 34
What is the enthalpy of the air-vapor mixture at 60% RH and 35°C
when the barometric pressure is 102 kpa? At 35"C: P sat = 5.628, h g =
2565.3
A. 89.63 KJ/kg
C 67.34 KJ/kg
B. 7423 KJ/kg
D. 53.34 KJ/kg
SOLUTION:
0.0176 = 0.622(
101.325 - P, =
P" = 2.79 kpa
P,
)
101.2 - P,
35.33 P,
Air Conditioning - 36
Air in an air-conditioner enters at 60% RH with w = 0.021 and leaves
at 25"C dry bulb and 16°C wet bulb. If mass of air is 10 kg/s, find the
refrigeration capacity in to ns of refrigeration.'
3/kg
At 60% RH, W = 0.021: It = 87 KJ/kg, v = 0.903 m
At 25°C db, 16°C wb: h = 45 KJ/kg;