Power and Industrial Plant Engineering
1. Find the
work possess for a Helium gas at 200C
*A. 609
KJ/kg B. 168 KJ/kg C. 229 KJ/kg D. 339 KJ/kg
Solution:
W = m R T
= m (8.314 / M) T
For helium, M = 4
W/m
= (8.314/4)(20 + 273) = 609 KJ/kg
2. Two
kilogram of gas is confined in a 1 m3 tank at 200 kpa and 880C.
What type of gas is in the tank?
A.
Helium *B. Ethane C. Methane D. Ethene
Solution:
Therefore:
the gas is Ethane (C2 H8)
3. Find the enthalpy of Helium if its internal energy is 200 KJ/kg
- 144 KJ/kg B.
223.42 KJ/kg *C. 333.42 KJ/kg D. 168 KJ/kg
Solution:
4. Compute the
mass of a 2 m3 propane at 280 kpa and 40˚C.
A. 6.47 kg B. 5.1 kg C. 10.20 kg *D.
9.47 kg
Solution:
Propane is C3 H3--------------M
= 12(3) + 8(1) = 44
PV = m R T
280(2) = m (8.314/44)(40 + 273)
m = 9.47 kg
5. Compute the
air flow in ft3/min of mechanical ventilation required to exhaust an
accumulation of refrigerant due to leaks of the system capable of revolving air
from the machinery room for a mass of 4 lbs refrigerant.
*A. 200 B.
210 C. 220 D. 230
Solution:
Q = 100 x G0.5 ft3/min
Q = 100 x (4)0.5 = 200 ft3/min
6. Compute the
free-aperture cross section in m2 for the ventilation of a machinery
room if the mass of refrigerant is 9 kg.
- 0.314 *B.
0.414 C. 0.514 D. 0.614
Solution:
F = 0.138 G0.5 m2
F = 0.138 (9)0.5
= 0.414 m2
7. A 29.53” x 39.37” pressure vessel contains ammonia with f =
0.041. Compute the minimum required discharge capacity of the relief device in
kg/hr.
A. 106.71 kg/hr B. 108.71 kg/hr *C. 110.71 kg/hr D.
112.71 kg/hr
Solution:
C = f D L, kg/s
C = 0.041(29.53/39.37)(39.37/39.37) = 0.03075 kg/s
(3600) = 110.71 kg/hr
8. Compute the
maximum length of the discharge pipe installed on the outlet of a
pressure-relief device in feet for internal pipe diameter of 0.5 inch and rated
discharge capacity is 8 lb/min of air. The rated pressure of relief valve is 16
psig.
*A. 0.286 ft B.
0.386 ft C. 0.486 ft D. 0.586 ft
Solution:
P = Pg + Patm = 16 x 1.1 + 14.7
= 32.3 psia
L = 9P2d5/16Cr2
= 9(32.3)2(0.5)5/16(8)2 = 0.286 ft
9. A thermal
power plant has a heat rate of 11,363 Btu/kw-hr. Find the thermal efficiency of
the plant.
- 28% *B. 30% C. 34% D.
40%
Solution:
eth = 3412/Heat rate = 3412/11,363 = 30%
10. What is
the hydraulic gradient of a 1 mile, 17 inches inside diameter pipe when 3300
gal/min of water flow with f = 0.03.
*A. 0.00714 B.
0.00614 C. 0.00234 D. 0.0018
Solution:
v = (3300/7.481)/(π/4)(17/12)2(60) = 4.66
ft/s
L = 1 mile = 5280 ft
hL = fLv2/2_D =
0.03(5280)(4.66)2/2(32.2)(17/12) = 37.7 ft
Hydraulic gradient = 37.7/5280 = 0.007.14
11. Find the
loss of head in the pipe entrance if speed of flow is 10 m/s.
- 5.10 m B. 10.2 m C. 17.4 m *D. 2.55 m
Solution:
Loss at entrance = 0.5 (v2/2g) = 0.5 [102
/ 2(9.81)] = 2.55 m
12. Wet
material, containing 220% moisture (dry basis) is to be dried at the rate of
1.5 kg/s in a continuous dryer to give a product containing 10% (dry basis) .
Find the moisture removed, kg/hr
*A. 3543.75 kg/hr B.
3513.75 kg/hr C. 3563.75 kg/hr D. 3593.75 kg/hr
Solution:
Solid in wet feed = solid in dried product
[1/(1 + 2.2)](1.5) = [1/(1 + 0.1)](x)
x = 0.5156 kg/s (total dried product)
Moisture removed = 1.5 – 0.5156 = 0.984 kg/s =
3543.75 kg/hr
13. Copra
enters a dryer containing 70% moisture and leaves at 7% moisture. Find the
moisture removed on each pound on solid in final product.
- 6.258 lb B. 1.258 lb C. 4.258 lb *D.
2.258 lb
Solution:
Solid in wet feed = solid in dried product
0.3x = 1
x = 3.333 lbs
1 = 0.93y
y = 1.07527 lb
Moisture removed = x – y = 3.333 – 1.07527 = 2.258
lb
14. A 1 m x
1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at
the bottom of the tank in dynes.
- 106.33 x 103
dynes B. 106.33 x 104
dynes C. 106.33 x 105 dynes *D. 106.33 x 106 dynes
Solution:
P = w h = (0.92 x 9.81) (1.5) = 13.5378 kpa
F = PA = 13.5378(π/4 x 12) = 10.632 KN =
10,632.56 N x 10,000 dynes/N
F = 106.33 x 106 dynes
15. Find the
pressure at the 100 fathom depth of water in kpag.
*A. 1,793.96 kpag B.
1,893.96 kpag C. 1,993.96 kpag D.
1,693.96 kpag
Solution:
H = 100 fathom x 6 = 600 ft
P = w h = (600/3.281)(9.81) = 1,793.96 kpag
16. Find the
depth in furlong of the ocean (SG = 1.03) if the pressure at the sea bed is
2,032.56 kpag.
*A. 1 B.
2 C. 3 D. 4
Solution:
P = w h
2,032.56 = (1.03 x 9.81) h
h = 201.158 m
x 3.281 ft/m x 1
yd/3ft x
1 furlong/220yd = 1 furlong
17. Find the
mass of 10 quartz of water.
- 10.46 kg *B. 9.46 kg C. 11.46 kg D. 8.46 kg
Solution:
V = 10 quartz
x 1gal/4quartz x
3.785li/1gal x 1m3/1000li
V = 0.0094625
x 10-3m3
w = m/V
1000 = m/0.0094625 x 10-3
m = 9.46 kg
18. Find the
mass of carbon dioxide having a pressure of 20 psia at 200˚F with 10 ft3
volume.
- 1.04 lbs B. 1.14 lbs *C. 1.24 lbs D. 1.34 lbs
Solution:
PV = m R T
(20 x 144)(10) = m (1545/44)(200 + 460)
m = 1.24 lbs
19. Find the
heat needed to raise the temperature of water from 30˚C to 100˚C with 60%
quality. Consider an atmospheric pressure of 101.325 kpa. Use the approximate
enthalpy formula of liquid.
- 293.09 KJ/kg B. 1,772.90 KJ/kg C. 1,547.90 KJ/kg
*D. 1,647.29 KJ/kg
Solution:
At 100°C
hf = Cp t = 4.187(100) 418.7 KJ/kg
hfg = 2257 KJ.kg
h2 = hf + xhfg = 418.7 +
0.60(2257) = 1,772.9 KJ/kg
Q = 1(4.187) (100-30) + 1(1772.9 – 418.7)
= 1647.29 KJ/kg
20. Find the enthalpy of water at 212˚F and 14.7 psi if the
dryness factor is 30%. Use the approximate enthalpy formula of liquid.
- 461 Btu/lb *B. 471 Btu/lb C. 481 Btu/lb D. 491 Btu/lb
Solution:
ht = (˚F – 32) = (212 – 32) = 180 Btu/lb
htg = 970 Btu/lb
h = ht
+ x htg
h = 180
+ 0.3(970) = 471 Btu/lb
21. An air
compressor consumed 1200 kw-hr per day of energy. The electric motor driving
the compressor has an efficiency of 80%. If indicated power of the compressor
is 34 kw, find the mechanical efficiency of the compressor.
- 117.65 % B. 75 % *C. 85 % D. 90 %
Solution:
P/m = 1200kw-hr/24 hrs = 50 kw
BP = 50(0.80) = 40 kw
em = 34/40 = 85 %
22. A
refrigeration system consumed 28,000 kw-hr per month of energy. There are 20 %
of energy is lost due to cooling system of compressor and motor efficiency is
90 %. If COP of the system is 6, find the tons of refrigeration of the system.
- 43.15 TR B. 46.15 TR *C. 49.15 TR D. 41.15 TR
Solution:
P/m = 28,800/(24 x 30)
= 40 kw
BP = 40(0.90) = 36 kw
Wc = 36(1 – 0.20) = 28.80 kw
COP = RE/Wc
6 = RE/28.80
RE = 172.8/3.516 = 49.15 TR
23. A 23 tons
refrigeration system has a heat rejected of 100 kw. Find the energy efficiency
ratio of the system.
- 13.42 *B. 14.42 C. 15.42 D.
16.42
Solution:
QR = RE
+ Wc
100 = 23(3.516)
+ Wc
Wc = 19.132 kw
COP = RE/Wc = (23 x
3.516) / 19.132 = 4.32
EER = 3.412 COP = 3.412(4.23) = 14.42
24. A 200
mm x
250 mm, 8-cylinder, 4-stroke diesel engine has a brake power of 150 kw.
The mechanical efficiency is 80%. If two of the cylinders were accidentally cut
off, what will be the new friction power?
- 31.50 kw B. 33.50 kw C. 35.50 kw *D. 37.50 kw
Solution:
em = BP/IP
0.8 = 150/IP
IP = 187.5 kw
FP1 = IP – BP = 187.5 – 150 = 37.50 kw
FP1 = FP2 = 37.50 kw
25. If the
energy efficiency ratio of the refrigeration system is 12.6, what is the COP of
the system?
*A. 3.69 B.
4.23 C. 5.92 D. 6.83
Solution:
EER = 3.412 COP
12.6 = 3.412 COP
COP = 3.69
26. An air
compressor has a power of 40 kw at 4 % clearance. If clearance will increase to
70 %, what is the new power?
- 70 kw *B. 40 kw C. 53kw D.
60 kw
Solution:
The power of compressor will not be affected with
the changes in clearance. Therefore the power will still be 40 kw.
27. What is
the approximate value of temperature of water having enthalpy of 208 Btu/lb?
- 138.67 ˚C *B. 115.55 ˚C C. 258.67 ˚C D. 68.67 ˚C
Solution:
h = ˚F – 32
208
= F – 32
˚F = 240 ˚F = 115.55 ˚C
28.
Convert 750˚R to ˚K
A. 390.33 ˚K B. 395.33 ˚K C. 410.33 ˚K *D.
416.33 ˚K
Solution:
˚R = 1.8 ˚K
750 = 1.8 ˚K
˚K = 416.667
29. An otto cycle has a
compression ratio of 8. Find the pressure ratio during compression.
*A. 18.38 B.
16.38 C. 14.38 D. 12.38
Solution:
P1V1k = P2V22
(V1/V2)k = (P2/P1)
rkk = rp
rp = (8)1.4 = 18.38
30. A diesel
cycle has a cut off ratio of 2.5 and expansion ratio of 4. Find the clearance
of the cycle.
A.
9.11
% B. 5.55 % *C. 11.11 % D.
15.15 %
Solution:
rk
= rc re
rk = 2.5(4) = 10
rk = (1 + c)/c
10 = (1
+ c)/c
c =
11.11 %
31. A dual cycle has an
initial temperature of 30 ˚C. The compression ratio is 6 and the heat addition
at constant volume process is 600 KJ/kg. If cut-off ratio is 2.5, find the
maximum temperature of the cycle.
A.
3638.50
˚C *B. 3365.50 ˚C C. 3565.50 ˚C D. 3965.50 ˚C
Solution:
T2 = T1 rkk-1
= (30 +273)(6)1.4-1 = 620.44 ˚K
QAV = m cv (T3
– T2)
600 = 1(0.7186)(T3 – 620.44)
T3 = 1455.396 ˚K
rc
= T4/T3
2.5 = T4/1455.396
T4 = 3638.49 ˚K = 3365.50 ˚C
32. A three stages air
compressor compresses air from 100 kpa to 1000 kpa. Find the intercooler
pressure between the first and second stage.
A.
505.44
kpa B. 108.44 kpa C. 316.23 kpa *D. 215.44 kpa
Solution:
Px = (P12P2)1/3
Px = [(100)2(1000)]1/3
= 215.44 kpa
33. A 10-stages air
compressor compresses air from 100 kpa to 800 kpa. Find the intercooler
pressure between 1st and 2nd stage.
A. 282.84 kpa B. 113.21 kpa *C. 123.11 kpa D.
333.51 kpa
Solution:
Px = (P1s-1P2)1/s
Px = [(100)10-1(1000)]1/10
= 123.11 kpa
34. A 3-stages air
compressor compresses air from 100 kpa to 700 kpa. Find the intercooler
pressure between the 2nd and 3rd stage.
*A. 365.88 kpa B.
375.88 kpa C. 385.88 kpa D. 395.88 kpa
Solution:
Px = (P12P2)1/3
Px = [(100)2(700)]1/3
= 191.28 kpa
Px/P1 = Py/Px
Py = Px2/P1
= (191.28)2/100 = 365.88 kpa
35. Carnot
cycle A, B and C are connected in series so that the heat rejected from A will
be the heat added to B and heat rejected from B will be added to C, each cycle
operates between 30 ˚C and 400 ˚C. If heat added to A is 1000 kw, find the work
output of C.
*A. 111.44 kw B.
549.78 kw C. 247.53 kw D. 141.89 kw
Solution:
e1 = e2 = e3
= (400 – 30)/(400 + 273) = 54.98 %
e1 = W1/QA1
= (QA1 – QR1)/QA1
0.5498 = (1000 – QR1)/1000
QR1 = 450.22 = QA2
0.5498 = (450.22 – QR2)/450.22
By heat balance:
Qgain
= Qloss
mw
cp (tb- ta) = mg cpg (t2-t1)
(0.30)(4.187)(
tb – 15) = (0.5)(1.0717)(150 – 80)
tb
= 44.86 oC
36. A 350 mm X 450 mm steam
engine running at 280 rpm has an entrance steam condition of 2 Mpa and 230 oC and exit at 0.1 Mpa. The steam
consumption is 2,000 kg/hr and mechanical efficiency is 85%. If indicated mean effective pressure is 600 Kpa,
determine brake thermal efficiency.
At 2 Mpa
and 230 oC (Table 3): h1
= 2849.6 s1 = 6. 4423
At 0.1 Mpa:
sf = 1.3026 hf
= 417.46 sfg = 6.0568 hfg = 2258 hf2 = 417.46 KJ/kg
A. 23.34% *B.
15.25% C. 14.16% D. 27.34%
SOLUTION:
VD =
2[3.1416/4 (0.35) 2
(0.45)(280/60)]= 0.4041 m3/sec
Indicated
Power = Pmi x VD = 600 x 0.4041 =242.45 KW
Brake
Power = IP (em) = 242.45
(0.85) = 206.08KW
Brake
Power 206.08 =
15.25%
etb
= ms (h1-hf2) = (2000/3600)(2849.6 –
417.46)
37. A steam turbine receives
5,000 kg/hr of steam at 5 Mpa and 4000oc and velocity of 30 m/sec. It leaves
the turbine at 0.06 Mpa and 85% quality and velocity of 15 m/sec. Radiation
loss is 10,000 KJ/hr. Find he KW developed.
At 5 Mpa and 400oC: h1 = 3195.7 KJ/kg s1 =6.6459
At 0.006 Mpa:
hf = 151.53 hfg
= 2415.9
A.
1273.29 B. 2173.29 *C. 1373.60 D. 7231.29
SOLUTION:
h2 = hf + xhfg
= 151.53 + 0.85(2415.9) = 2205.045 KJ/
kg
KE1
= ½ m v2 = ½ (5,000/3600)(30)2 = 625 W = 0.625 KW
KE2 = ½ m
v2 = ½ (5,000/3600)(15)2
= 156.25 W = 0.15625 KW
By energy
balance:
KE1
+ mh1 = KE2 + mh2 + Q + W
W = (KE1
– KE2) + m(h1-h2) – Q
5000 10,000
W =
(0.625 – 0.156) + (3600)(3195.7 – 2205.045) – 3600 = 1373.60 KW
38. A steam turbine with 85%
stage efficiency receives steam at 7 Mpa and 550oC and exhausts as 20 Kpa. Determine
the turbine work.
At 7 Mpa
and 550oC: h1
= 3530.9 Kj/kg s1 =
6.9486
At 20 Kpa
(0.020 Mpa): sf = 0.8320 hf
= 251.4 sfg = 7.0766 hfg = 2358.3
A. 1,117 KJ/kg B. 1,132 KJ/ kg C. 1,123.34 KJ/ kg *D.
1,054.95 KJ/kg
SOLUTION:
s1
= s2 = sf + xsfg
6.9486 =
0.8320 + x(7.0766)
x =
0.8643
h2
= 251.40 + 0.8643(2358.3) = 2289.78 KJ/kg
h1 - h2a
eST = h1
– h2
0.85 =3530.9
– h2a
3530.9 – 2289.78
h2a =
2475.95 KJ/kg
WT
= h1 = h2a = 3530.9 – 2475.95 = 1,054.95 KJ/kg
39. A steam turbine with 80%
stage efficiency receives steam at 7 Mpa and 550oC and exhaust as 20 Kpa.
Determine the quality at exhaust.
At 7 Mpa
and 550oC: h1
= 3530.9 Kj/kg s1 =
6.9486
At 20 Kpa
(0.020 Mpa): sf = 0.8320 hf
= 251.4
*A.
96.96% B. 76.34% C. 82.34% D. 91.69%
SOLUTION:
sfg =
7.0766 hfg
= 2358.3
s1 = s2 = sf + sfg
6.9486 =
0.8320 + x(7.0766)
x =
0.8643
h2 =
251.40 + 0.8643(2358.3) = 2289.78 KJ/kg
nST
= h1 – h2a
h1 – h2
0.80 = 3530.9
– h2a
3530.9 – 2289.78
h2a =
2538.004 KJ/kg
h2a = hf + x hfg
2538.004
= 251.40 + x (2358.3)
x =
96.96%
40. A 16,000KW
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 400, 000 kg/hr, determine the number of wells are required to
produce if the charge of enthalpy if the change of enthalpy at entrance and
exit of turbine is 500KJ/kg.
A. 4 wells *B.
2 wells C.
6 wells D.
8 wells
SOLUTION:
WT = ms(h3 – h4)
16,000
= ms (500)
0.9(0.8)
ms =
44.44 kg/sec
ms =
160,000 kg/hr
160,000 =
0.20 mg
mg =
800,000 kg/hr
No. of
wells = 800,000/400,000 = 2 wells
41. A liquid
dominated geothermal plant with a single flash separator receives water at 204oC.
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 60 MW, what is the mass flow rate of the well-water in kg/s?
At 204oC: hf = 870.51 KJ/kg
At 1.04
Mpa: hf = 770.38 hfg = 2009.2 hg = 2779.6 sg =
6.5729
At 0.034
MPa: hf = 301.40 hfg = 2328.8 sf = 0.9793 sfg =
6.7463
*A. 2,933
B. 2,100 C. 1,860 D. 2,444
SOLUTION:
h3 = hg
at 1.04 MPa = 2779.6 KJ/kg
Solving
for h4:
s3 = s4 =
sf + xsfg
6.5729 =
0.9793 + x4(6.7463)
x4 =
0.829
h4 =
301.4 + 0.829(2328.8) = 2232.3 KJ/kg
WT = ms
(h3 – h4)
60,000 =
ms (2779.6 – 2232.3) 0.75
ms =
146.17 kg/sec
Solving for x2: (h1 = h2)
h1 = h2 =
hf + xhfg
870.51 =
770.38 + x2(2009.2)
x2 =
0..049836
ms = x mg
146.17 =
0.049836 mg
mg =
2,933.06 kg/sec
42. An engine-generator
rated 9000 KVA at 80% power factor, 3 phase, 4160 V has an efficiency of 90%.
If overall plant efficiency is 28%, what is the heat generated by the fuel.
A. 18,800 KW B.
28,800 KW C. 7500 KW *D.
25,714 KW
SOLUTION:
Gen.
Output = pf x KVA = 0.8 x 9000 = 7200 KW
eoverall=
Gen. Output
Qg
0.28 =
7200/Qg
Qg =
25,714.28 KW
43. The indicated thermal
efficiency of a two stroke diesel engine is 60%. If friction power is 15% of
heat generated, determine the brake thermal efficiency of the engine.
A. 43% *B. 45 % C.
36% D. 37%
SOLUTION:
ne = IP/
Qg
0.60 =
IP/Qg
IP = 0.60
Qg
BP = IP-
FP = 0.60Qg – 0.15Qg = 0.45Qg
etb =
BP/Qg = 0.45Qg/Qg = 45%
44. 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 0.56 kg/KW-hr with a heating value of 43,912
KJ/kg. Calculate brake thermal efficiency
A.
10.53% B.
27.45% *C. 14.64% D. 18.23%
SOLUTION:
mf =
0.56 kg/KW-hr x 150 KW = 84 kg/hr = 0.0233 kg/sec
Brake thermal efficiency =
45. A waste heat
recovery boiler produces 4.8 Mpa(dry saturated) steam from 104°C feedwater. The boiler
receives energy from 7 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 per second? Note: At 4.80 Mpa dry saturated, h = 2796.
A.
1.30 B.
0.92 *C. 1.81 D. 3.43
SOLUTION:
hf = approximate enthalpy of feedwater
hf =
Cpt
hf =
4.187(104)
hf =
435.45 KJ/kg
Heat loss
= Heat gain
m gc
p(t 1 - t 2) = m s(h - h
f)
7(1.0)(954
– 343) = ms(2796.0 – 436.45)
m s =
1.81 kg/sec
46. A diesel
electric plant supplies energy for Meralco. During a 24-hour period, the plant
consumed 240 gallons of fuel at 28°C and produced 3930 KW-hr. Industrial fuel used is 28°API and was
purchased at P30 per liter at 15.6°C. What is the cost of the fuel be to
produce one KW-hr?
*A. P6.87 B.
P1.10 C. P41.07 D. P5.00
SOLUTION:
SG
15.6C = 141.5/(131.5 + 28) = 0.887
Density
at 15.6°C = 0.887(1kg/li) = 0.887 kg/li
SG 28C = 0.887[1-.0007(1 – 15.6)] = .879
Density
at 28°C = 0.879(1 kg/li) = 0.879 kg/li
V28C / V15.6C = SG15.6C / SG28C
240 / V15.6C =
0.887 / 0.879
V15.6C = 237.835
gallons x 3.785 li/gal = 900.21 li
Cost = [(30)(900.21)] / 3930
= P6.87/KW-hr
47. In a gas
turbine unit, air enters the combustion chamber at 550 kpa, 277°C and 43 m/s. The products of
combustion leave the combustor at 511 kpa, 1004°C and 180 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?
A.
70.38% B. 79.385% *C.
75.38% D. 82.38%
SOLUTION:
Heat supplied by
fuel = mfQh =
0.0229(43,000) = 984.7 KJ/kg air
Q = heat absorbed by
fuel
Q/m = Cp(T2 –
T1)
+ ½(V22 –
V12)
Q/m = (1.0)(1004 – 277) + ½[(180) 2 –(43) 2]/1000
=742.28 KJ/kg air
Combustor Efficiency = = 75.38%
48. The specific speed of turbine is 85 rpm and running at 450 rpm. If
the head is 20 m and generator efficiency is 90%, what is the maximum power
delivered by the generator.
A.
450.51 KW B. 354.52 KW C. 650.53 KW *D.
835.57 KW
SOLUTION:
NS = (N√HP)/h5/4
85 = (450√HP)/(20 x 3.281) 5/4
Hp = 1244.52
Generator Output = (1244.52 x 0.746)(0.9)
= 835.57 KW
49. In Francis
turbine, the pressure gage leading to the turbine casing reads 380 Kpa. The
velocity of water entering the turbine is 8 m/sec, if net head of the turbine
is 45 m, find the distance from center of spiral casing to the tailrace.
*A. 3.0 m B. 3.5 m C.
4.0 m D. 4.5m
SOLUTION :
h = V2/2g
45 =
(380/9.81) + z + [82/(2 x 9.81)]
z = 3 m
50. A turbine
has a mechanical efficiency of 93%, volumetric efficiency of 95% and total
efficiency of 82%. If effective head is 40 m, find the total head.
A.
48.72 m B.
40.72 m *C. 36.22 m D. 34.72 m
SOLUTION:
eT
= emehev
0.8 = 0.93(eh)(.95)
ηh
= 0.9055
Total head = h eh =
(40)(0.9055) = 36.22 m
51. A Pelton type turbine has 25 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. 1,355.73 mm B. 3,476.12 mm C.
6771.23 mm D. 1686.73 mm
SOLUTION:
h
=25- 4.5 = 20.5
v = √(2gh) = [(2 x 9.81 x 20.5)1/2]
= 20.55 m/sec
hL = (2fLv2)/gD
4.5 = (2)(0.00093)(80)(20.055)2 / 9.81D
D =
1,355,730 m = 1,355.73 mm
52. In an
9,000 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?
A.
58,960 KW-hrs *B.
80,080 KW-hrs C. 65,960 KW-hrs D. 70,960 KW-hrs
SOLUTION:
Plant
Capacity = 9,000(0.88)(24) = 190,080 KW-hrs
Secondary
Power = 190,080 – 110,000 = 80,080 KW-hrs
53. A Pelton
type turbine was installed 30 m below the gate of the penstock. The head loss
due to friction is 12 percent of the given elevation. The length of penstock is
100 m and coefficient of friction is 0.00093. Determine the power output in KW.
( Use Moorse equation)
A.
22,273 B.
23,234 C. 32,345 *D. 34,452
SOLUTION:
hL =
0.12(30) = 3.6 m
h = 30 –
3.6 = 26.40 m
v = (2gh)1/2
= [(2)(9.81)(26.4)]1/2 = 22.759 m/sec
hL=
(2fLv2)/gD
3.6 = (2
x .00093 x 100 x 22.759) / (9.81D)
D = 2.728
m
Q = A x v
= [2](22.759) = 133.03 m3/sec
Power = w Q h =
9.81(133.03)(26.4) = 34,452 KW
54. Water flows
steadily with a velocity of 3.05 m/s in a horizontal pipe having a diameter of
25.4 cm. At one section of the pipe, the temperature and pressure of the water
are 21C and 689.3 Kpa, respectively. At a distance of 304.8 m downstream
55. A hydro
electric plant having 30 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 13.5 x 106 kwh. The
overall generation efficiency is 75%. Find the fall in the height of water in
the reservoir after the 5-hour period.
A.
5.13 m B.
1.32 m C. 3.21 *D. 2.20 m
SOLUTION
Energy
Output = Power x time = (w Q h) x time
13.5 x 106 = 9.81(Q)(100)(0.75)(5)
Q = 3669.725 m3/s
Volume
after 5 hrs = 3669.725(5 x 3600) = 66,055,050 m3
Volume =
A x height
66,055,050
= (30 x 106) h
H =2.202
m
56. The gas
density of chimney is 0.75 kg/m3 and air density of 1.15 kg/m3.
Find the driving pressure if the height of chimney is 63.71 m.
A.
0.15 kpa *B.
0.25 kpa C. 0.35 kpa D. 0.45 kpa
SOLUTION:
hw = H(da – dg) =
63.71(1.15 – 0.75) (0.00981) = 0.25 kpa
57. The actual
velocity of gas entering in a chimney is 8 m/sec. The gas temperature is 25C
with a gas constant of 0.287 KJ/kg-K. Determine the gas pressure for a mass of
gas is 50,000 kg/hr and chimney diameter of 1.39m.
A.
95 kpa *B. 98
kpa C. 101 kpa D. 92 kpa
SOLUTION:
Vg
= A x v = / 4 (1.39)2(8)
= 12.139 m3/s
PgVg
= mgRgTg
P(12.139)
= (50,000/3600)(.287)(25 +273)
P = 97.85
kpa
58. A steam
generator with economizer and air heater has an overall draft loss of 25.78 cm
of water. If the stack gases are at 177C and if the atmosphere is at 101.3 Kpa
and 26C, what theoretical height of stack in meters is needed when no draft fan
are used? Assume that the gas constant for the flue gases is the same as that
for air.
A 611.10 B. 631.10 *C.651.10 D.671.10
SOLUTION:
w = P/RT
da = (101.325)/[(.287)(26 + 273)] = 1.180
kg/m3
dg
= (101.3)/[(0.287)(177 +273)] = 0.784 kg/m3
Draft = (0.2578)(1000) =
257.80 kg/m3
Draft =
H(da – dw)
257.80 = H(1.18 – 0.784)
H = 651.10 m
59. A
foundation measures 12 ft x 14 ft x16 ft. Find the number of sacks of cement
needed for 1:2:4 mixture.
A.
302 B.
404 C. 356 *D. 598
SOLUTION:
V = 12 X
14 X 16 = 2,688 ft3 (1 yd3 / 33 ft3)
= 99.55 yd3 of concrete
60. For every
1 yd3 concrete, it needs 6 sacks of cement
Therefore:
No. of sacks = 6(99.55) = 597.33 sacks or 598 sacks
61. A
rectangular foundation cross-section has a bed plate dimension of 8 ft x 10 ft.
The uniform clearance on each side is 1 ft. The height of foundation is 4.5 ft.
If the weight of the steel bar reinforcements needed is 1/2% of weight of
foundation, find the weight of steel bars. Use concrete density of 2400 kg/m3
.
A.
173.47 kg *B.
183.47 kg C. 163.47 kg D. 153.47 kg
SOLUTION:
A = (8 +
2) (10 + 2) = 120 m2
V = Ah = 120(4.5) = 540 ft3 = 15.29 m3
W = wV = (2400)(15.29) = 36,693.25 kg
Weight of steel bars = (1/2%) Wf = 0.005(36,693.25)
= 183.47 kg
62. A steam
pipe having a surface temperature of 250C passes through a room where the
temperature is 27 C. The outside diameter of pipe is 100 mm and emissivity
factor is 0.8. Calculate the radial heat loss for 3 m pipe length.
A.
1434.7 W B.
37.46 W *C. 2651.82 W D. 3545.45 W
SOLUTION:
A = DL =
= 0.425m2
Solving for heat due
to radiation:
Tg = 250
+273 = 523K
T2 = 27
+273 = 300K
Qa =
20,408.4 x 104 AF(T14 – T24),
J/hr = 20,408.4 x 104(0.8)(0.7539)[(523)4 – (300)4]
Qr =
10,266,539.06 j/hr x 1hr/3600sec = 2851.82 W
63. Brine enters a
circulating brine cooler at the rate of 60 m3/hr at -*C and leaves
at -18C. Specific heat of brine is 1.072 KJ/kg-K and specific gravity of 1.12.
Determine the tons of refrigeration.
A.
53.5 TR B. 65.3 TR C.33.5 TR *D.
56.9 TR
SOLUTION:
Density of brine =
1.12(1000 kg/m3) = 1120 kg/m3
m = (1120)(60)/3600 =
18.67 kg/sec
Q = mcp = 18.67(1.072)(-8 + 18) = 200.11 KW
TR = 200.11/3.516 =
56.91 Tons of refrigeration
64. A turbo-charged, 16
cylinder, Vee-type diesel engine has an air consumption of 3,000 kg/hr per
cylinder at rate load and speed. This air is drawn in through a filter by a
centrifugal compressor directly connected to the exhaust gas turine. The
temperature of the air from the compressor is 135C and a counter flow air
cooler reduces the air temperature to 45C before it goes to the engine suction
heater. Cooling water enters air cooler at 30C and leaves at 40C. Calculate the
log mean temperature difference.
A.
47.23C B. 87.82C *C. 43.34C D.
65.24C
SOLUTION:
a = 45-30 = 15C
b = 135 – 40 = 95C
mean = [
a -
b] / [ln(
a
b)] = [95-15] / ln(95/15) = 43.34C
65. Water is flowing in
a pipe with radius of 30 cm at a velocity of 5 m/s at the temperature in 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. 2647 B. 96.2 C.
3100 D. 1140
SOLUTION:
n = Dvg /
v
Where:
D = 2(0.30) = 0.60 m
vg = 5
M/SEC
v = 1.131/997.9 =
0.0011334 m2 / sec
Nm =
0.60(5)/0.0011334 = 2,647
66. Compute the amount
of condensate form during 10 minutes warm-up of 180 meter pipe conveys the
saturated steam with enthalpy vaporization hfg = 1,947.8 LJ/kg. The
minimum external temperature of pipe is 2C. The final temperature of pipe is
195C. The specific heat of pipe material is 0.6 KJ/kg-C. The specific weight is
28 kg/m.
A.
249.69 kg B.
982.45 kg *C. 299.64 kg D. 423.45 kg
SOLUTION:
mp = mass
of pipe = 28(180) = 5,040 kg
Heat Loss by steam =
Heat loss from pipe
m(hg - hf)
= mpcp (t2 – t1)
m(1947.8) =
(5040)(0.6)(195-2)
m = 299.64 kg
67. The
discharge pressure of an air compressor is 5 times the suction pressure. If
volume flow at suction is 0.1 m3/sec, what is the suction pressure
if compressor work is 19.57 kw? (use n=1.35)
A.
97 kpa *B.98
kpa C. 99 kpa D.100 kpa
SOLUTION:
W = [(P2/P1)n-1/n – 1]
19.57 = 1.35(P1)(0.1)/(1.35-1)[(5)1.35-1/1.35
– 1]
P1
= 98 KPa
68. The
initial condition of air in an air compressor is 98 KPa and 27C and discharge
air at 450 KPa. The bpre and stroke are 355 mm and 381 mm, respectively with
percent cleared of 8% running at 300 rpm. Find the volume of air at suction.
A.
541.62 m3/hr B. 551.62 m3/hr C.
561.62 m3/hr *D. 571.62
m3/hr
SOLUTION:
ev
= 1 + c – c(P2/P1)1/n = 1 + 0.08 -
0.08(450/98)1/1.4 = 0.842
VD
= D2 LN =
(0.355)2
(0.381)(300/60) = 0.1885 m3/sec
V1 =
0.1885(0.842) = 0.15878 m3/sec = 571.62 m3/hr
69. An air
compressor has a suction volume of 0.35 m3/sec t 97 KPa and
discharges to 650 KPa. How much power saved by the compressor of there are two
stages?
A.
18.27 KW B.
16.54 KW C. 13.86 KW *D.
11.58 KW
SOLUTION:
W = [(P2/P1)n-1/n – 1] = (1.4 x 97
x 0.35)/(1.4 -1) [(650/97)1.4-1/1.4 – 1] = 85.79 KW
For two stages :
Px = (P1P2)1/2
= (97 x 650)1/2 = 251.097 KPa
W = [(Px/P1)n-1/n – 1] =
2(1.4)(97)(0.35)/(1.4 – 1) [(251.0.97/97)1.4-1/1.4 – 1] = 74.208 KW
POWER SAVED = 85.79 –
74.208 = 11.582 KW
70. A two stage air
compressor has an intercooler pressure of 4 kg/cm2. What is the
discharge pressure if suction pressure is 1 kg/cm2?
A.
3 kg/cm2 B.
9 kg/cm2 C. 12 kg/cm2
*D. 16 kg/cm2
SOLUTION:
Px
= (P1P2)1/2
Px2
= P1(P2)
42
= 16 kg/cm2
71. A two
stage air compressor compresses air at 100 KPa and 22C discharges to 750 KPa.
If intercooler intake is 105C. Determine the value of n.
A.
1.400 *B.
1.325 C. 1.345 D. 1.288
SOLUTION:
Px = = 273.86 Kpa
1.281 =
n = 1.326
72. A single
acting compressor has a volumetric efficiency of 89%, operates at 500 rpm. It
takes in air at 900 KPa and 30C and discharges it at 600 KPa. The air handled
is 8 m3/min measured at discharge condition. If compression is
isentropic, find mean effective pressure in KPa.
*A.
233.34 B. 973.17 C. 198.34 D. 204.82
SOLUTION:
P1V1K
= P2V2K
100(V11.4)
= 600(6)1.4
V1 =
28.768 m3/min
VD
= 28.768/0.89 = 32.32 m3/min
W = n P1V1/n-1
x [(P2 / P1)n-1/n
– 1] = [(1.4 x 100 x 32.32)/(1.4 – 1)] x [(600/100)1.4-1/1.4 – 1]
W =
7562.19 KJ/min
W = Pm
x Vd
7562.19 = Pm x 32.32
Pm = 233.34 KPa
73. A
water-jacketed air compressed handles 0.343 m3/s of air entering at
96.5 KPa and 21C and leaving at 460 KPa and 132C; 10.9 kg/h of cooling water
enters the jacket at 15C and leaves at 21C. Determine the compressor brake
power.
A.
26.163 KW *B.
62.650 KW C. 34.44 KW D. 19.33 KW
SOLUTION:
T2/T1
= (P2/P1) n-1/n
(132+273) / (21+273) =
(480/96.5)n-1/n
n = 1.249
W =
(1.249 x 96.5 x 0.343) / (1.249-1) [(480 / 96.5)1.249-1/1.249 – 1]
W = 62.57
KW
Q = heat
loss = mcp(t2 – t1) = (10.9/3600)(4.187)(21 –
15) 0.075 KW
Brake
power = W + Q = 62.57 + 0.076 = 62.65 KW
74. A double
suction centrifugal pumps delivers 20 ft3/sec of water at a head of
12 m and running at 650 rpm. What is the specific speed of the pump?
A.
5014.12 rpm B.
6453.12 rpm *C. 2770.73 rpm D. 9966.73 rpm
SOLUTION:
N = N(Q)1/2
/ h3/4
Q = 20/2
ft3/sec x 7.481 gal/ft3 x 60 sec/1min = 4,488.6 gal/min
h = 12 x
3.281 = 39.37 ft
N = (650
x (4,488.6)1/2)/(39.37)3/4
N = 2,770.73 rpm
75. 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 30
ft.
A.
6 B. 4 *C. 5 D.
7
SOLUTION:
Wp
= w Q h
15 x
0.746 = 9.81(400 gal/min x 0.00785m3/gal x 1/60)h
h = 45.20
m x 3.281 ft/m = 148.317 ft
Number of
stages = 148.317/40 = 4.94 stages = 5 stages
76. The
suction pressure of a pump reads 3 in. of mercury vacuum and discharge pressure
reads 140 psi is use to deliver 120 gpm of water with specific volume of 0.0163
ft3/lb. Determine the pump work.
A.
4.6 KW B. 5.7
KW *C. 7.4 KW D. 8.4 KW
SOLUTION:
P1
= -3 in Hg x 101.325/29.92 = -10.16 KPa
P2 =
140 psi x 101.325/14.7 = 965 KPa
w = 1/v =
1/0.163 = 61.35 lb/ft3 x 9.81/62.3 = 9.645 KN/m3
h = (P2
– P1)/w = (965 +10.16)/9.645 = 101.105 m
Q = 120
gal/min x 3.785/1gal x 1m3/1000li x 1/60 = 0.00757 m3/sec
P = w Q h
= 9.645(0.00757)(101.105) = 7.38 KW
77. A
submersible pump delivers 350 gpm of water to a height of 5 ft from the ground.
The pump were installed 150 ft below the ground level and draw down of 8 ft
during the operation. If water level is 25 ft above the pump, determine the
pump power.
A.
7.13 KW B.
4.86 KW C. 7.24 KW *D. 9.27 KW
SOLUTION:
h = 5 +
150 – (25 – 8) = 138/3.281 = 42.06 m
Q = 350
gal/min x 0.003785 m3/gal x 1 min/60sec = 0.02246 m3/sec
Wp =
w Q h = 9.81(0.02246)(42.06) = 9.27 KW
78. A vacuum
pump is used to drain a flooded mine shaft of 20 water. The pump pressure of water at this
temperature is 2.34 KPa. The pump is incapable of lifting the water higher than
16 m. What is the atmospheric pressure?
*A.
159.30 B. 32.33 C. 196.22 D. 171.9
SOLUTION:
Using
Bernoulli’s Theorem:
P1/w
+ V12/2g + z1 = P2/w + V2/2g
+ z2
P1/w = P2/w
+ (V22 - V12)/2g + (z2 -
z1)
P1/9.81
= 2.34/9.81 + 0 + 16
P1
= 159.30 KPa
79. 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 3000
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 B.
5.33 *C. 3.71 D. 6.34
SOLUTION:
V = D N
V =
D (3000/60) =
(2(32.2)(38))1/2
D = 0.315 ft = 3.708
inches
80. A fan pressure of
2.54 cm of water t 1.42 m3 per second of air at static pressure of
2.54 cm of water through a duct 300 mm diameter and discharges it through a
duct 275 mm diameter. Determine the static fan efficiency if total fan
mechanical is 75% and air measured at 25 and 60 mm Hg.
A.
50.11% *B.
53.69% C. 65.67% D. 45.34%
SOLUTION:
wA
= P/RT = 101.325/(0.287)(25 + 273) = 1.18 kg/m3
hA
= hwww/wA = (0.0254)(1000)/1.18 = 21.52 m
vA =
1.42/(/4)(0.3)2 = 20.09 m/s
Vd = 1.42/(/4)(0.275)2 = 23.9 m/s
hv = (23.9)2 –
(20.09)2 / 2(9.81) = 8.54 m
h = ha
+ hv = 21.52 + 8.54 = 30.06 m
eT
= wa Q h/BP
0.75 =
(1.18 x 0.00981)(1.42)(30.06) / BP
BP =
0.6588 KW
ep =
wa Q hs/BP = (1.18 x 0.00981)(1.42)(21.52) / 0.6588 =
53.69%
81. A water cooler uses 50 lb/hr of melting ice to cool running
water from 80 to 42
. Based on te inside coil area, U1 =
110 Btu/hr-ft2-
. Find the gpm of water
cooled.
A. 0.10 GPM B. 0.21 GPM*C. 0.38 GPM D. 0.45 GPM
SOLUTION:
Q
= mf L = mwcpw(t1 – t2)
50
(144) = mW(1)(80-42)
mw
= 189.474 lb/hr
V
= (189.474/62.4) (7.48/60) = 0.38 GPM
82. The charge in a Diesel engine consists of 18.34 grams of
fuel, with lower heating value of 42,571 KJ/kg, and 409 grams of fuel and
products of combustion. At the beginning of compression, t1 = 60. Let rk =
14. For constant cP = 1.11 KJ/kg-C, what should be the cut-off ratio
in the corresponding ideal cycle?
A.
2.05 B. 2.34 C. 5.34 *D.
2.97
SOLUTION:
QA
= mfQh = 0.01283(42,571) = 780,752 KJ
T2/T1
= rkk-1
T2
= (60 + 273)1.4-1 = 956.964K
mt
+ mg = 409
mt
+ ma + mf = 409
ma
= 409 – 2(18.34) = 372.32 grams
QA
= macp(t3 – t2)
780.752
= 0.37232(1.11)(T3 – 956.964)
T3
= 2846,146
rC = T3/T2
= 2846.146/956.964 = 2.97
83. The gain of entropy during isothermal nonflow process of 5
lb of air at 60 is 0.462 Btu/R. Find the V1/V2.
A.
3.85 *B. 0.259 C. 1.0 D.
0.296
SOLUTION:
s = m R T ln(V2/V1)
0.462
= 5 (53.33/778) ln (V2/V1)
V2/V1
= 3.85
V1/V2
= 1/3.85 = 0.259
84. An auditorium seating 1500 people is to be maintained at 80 dry bulb and 85
wet bulb temperature when outdoor air is at 91
dry bulb and 75
wet bulb. Solar heat load is 110,000 Btu/hr
and supply air at 60
determine the amount of supply air.
*A. 93,229.17 lb/hr B. 83,229.17 lb/hr C. 73,229.17 D. 63,229.17 lb/hr
SOLUTION:
Sensible
heat per person = 225 Btu/hr
Qa
= 225(1500) + 110,000 = 447,500 Btu/hr
Qa
= m cp(t1 – t2)
447,500
= ma(0.24)(80 – 60)
ma
= 93,229.17 lb/hr
85. In a Brayton cycle that operates between temperature limits
of 300K and 1773K wit k = 1.4, determine the temperature at the end of the
compression (isentropic) for maximum work of the cycle.
A.
700K B. 590.5K *C. 730K D. 350K
SOLUTION:
For
maximum work: T2 = (T1T3)1/2
T2
= (300 x 1773)1/2 = 730K
86. At 35% solution leaves the absorber and 30% solution enters
the absorber. The heat removed from the absorber by cooling water is 547.6 Btu
and ammonia is superheated by 10. Find the pound per
pound of ammonia gas from the evaporating coils.
A.
11 B. 12 *C. 13 D.
14
SOLUTION:
n
= lb/lb of ammonia gas from the coils
n
= (1 - x2) / (x1 - x2) – 1
n
= (1- 0.3) / (0.35 – 0.3) – 1 = 13
87. A Carnot refrigeration system operates at Tmax /
Tmin = 1.5. Find the KW per tons of refrigeration.
A.
1.91 B. 2.15 *C. 1.76 D. 1.55
SOLUTION:
Wo
/ TR = 3.516 / COP = 3.516 / (Tmin / (Tmax – Tmin)
Wo
/ TR = [3.516(Tmax – Tmin)] / Tmin
Wo
/ TR = 3.516[(Tmax / Tmin) – 1] = 3.516(1.5-1) =
1.758 KW/TR
88. Assume 8 ft3 of air at 100 psi, 100 are compressed isothermally to a volume of 2
ft3. For each of end states of the process, find the bulk modulus.
*A. 400 and 100 psi B. 400 and 110 psi C.
400 and 120 psi D. 400 and 130
89. An empty
open can is 30 cm high with a 15 cm diameter. The can, with the open end and
down, is pushed under water with a density of 1000kg/m3. Find the water
level in the can when the top of the can is 50 cm below the surface.
A.
17.20 cm *B.
2.12 cm C. 4.20 cm D. 5.87cm
SOLUTION:
Consider
the water pressure
Pw
= w h + 1010.325 + (0.8-x)(9.81) + 101.325 = 109.173 – 9.81x
Consider
the air pressure
P1V1 = P2V2
101.325(Ax0.3) = P2[A(0.3-x)]
P2 =
Pw
= P2
109.173
– 9.81x =
9.81x2
– 112.116x + 2.3705 = 0
By
quadratic formula:
X
= 0.02118 m = 2.12 cm
90. A cylindrical pope with water flowing downward at 0.03 m3/s
having top diameter of 0.08, bottom diameter of 0.04 m and a height of 1.5m.
Find the pressure between the pipe
A. 154.63 Kpa B. 197.93 Kpa *C. 252.44
Kpa D. 243.92 Kpa
SOLUTION:
+ Z1 =
+ Z2
Z1
– Z2 = 1.5 m
Z2
– Z1 = -1.5 m
V1
= = 5.968 m/s
V2
= = 23.87 m/s
P1
– P2 = 252.44 Kpa
91. Determine the size of pipe which will deliver 8 liters of
medium oil (v= 6.10 x 10-6 m2/s) assuming laminar flow
conditions:
A. 622 mm B.
754 mm C. 950 mm *D. 835 mm
SOLUTION:
V
=
Re
=
For
laminar flow, Re = 200
2000
=
d
= 0.835 m = 835 mm
92. The type of flow occupying in a 1 cm diameter pipe which
water flows at a velocity of 2.50 m/s. Use v = 1.13 x 10-6 m2/s
for water.
*A.
turbulent B. constant C. laminar D. none of the above
SOLUTION:
Re
=
Re
=
Re
= 22,124
Since
it is greater than 2000 then it is turbulent flow
93. What is the force is exerted by water jet 60 mm diameter if
it strikes a wall at the rate of 15 m/s?
*A.
636.17 N B.442.62 N C. 764.23 N D.
563.34 N
SOLUTION:
F
= w Q v
Q
= A v = = 0.0424 m3/s
F
= (1000)(0.0424)(15) = 636.17 N
94. A 300 mm diameter pipe discharges water at the rate of 200
li/s. Point 1 on the pipe has a pressure of 260 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. 4.29 m B.
2.59 m C. 6.32 m *D. 1.87 m
SOLUTION:
hL
hL
=
95. Water flowing at the rate of 10 m/s from an orifice at the
bottom of a reservoir. Find the pressure at the bottom of the reservoir.
A. 30 kpag B.
40 kpag *C. 50 kpag D. 60 kpag
SOLUTION:
h
= V2/ 2g = 102/ 2(9.81) = 5.0968 m
P
= w h = 9.81(5.0968) = 50 kpag
96. Steam flows through a nozzle at 400oC and 1 Mpa
(h = 3263.9 KJ/kg) with velocity of 300 m/s. Find the stagnation enthalpy.
A. 3300 KJ/kg B. 3290 KJ/kg *C. 3320
KJ/kg *D. 3309 KJ/kg
SOLUTION:
ho
= h + v2/2000 = 3263.9 + 3002/2000 = 3309 KJ/kg
97. Air flows through a nozzle at a speed of 350 m/s. Find the
stagnation temperature if entrance temperature is 200oC.
A. 241.25oC B. 251.25oC *C. 261.25oC D.
271.25oC
SOLUTION:
To
= T1 + v2/2000Cp = (2000 + 273) + 3502/2000(1)
To
= 534.25oK = 261.25
98. Carbon dioxide flows through a nozzle with a speed of 400
m/s. Compute the dynamic temperature.
A. 92.56oK *B. 94.56oK C. 96.56oK D. 98.56oK
SOLUTION:
For
CO2: Cp = 0.846 KJ/kg-K
Dynamic
temperature = v2/2000Cp = 4002/2000(0.846) = 94.56oK
99. Carbon dioxide flows through a nozzle with a speed of 380
m/s. The entrance condition of nozzle is 250oC and 1200 kpa. Find
the stagnation pressure.
*A.
2,136.34 kpa B. 2,146.34 kpa C. 2,156.34 kpa D. 2,166.34 kpa
SOLUTION:
T1
= 250 + 273 = 523oK
To
= T1 + v2/2000 = 523 = 3802/2000 = 595.2oK
P1
= 1200 kpa
T1/To
= (P1/Po)k-1/k
For
CO2: k = 1.289
523/595.2
= (1200/Po)1.289-1/1.289
P0
= 2,136.34 kpa
100. Air enters a diffuser with a velocity of 200 m/s. Determine
the velocity of sound if air temperature is 30oC.
*A.
349 m/s B. 359 m/s C. 369 m/s D.
379 m/s
SOLUTION:
C
=
101. Air flows through a nozzle with temperature of entrance of
420oK stagnation temperature of 468oK. Find the mach
number.
A. 0.744 *B.
0.754 C. 0.764 D. 0.774
SOLUTION:
To
= T1 + v2/2000Cp
468
= 420 + v2/2000
v = 309.838 m/s
C
=
M
= v/C = 309.838/410.8 = 0.754
102. Air at 300oK and 200 kpa is heated at constant
pressure to 600oK. Determine the change of internal energy.
A. 245.58 KJ/kg B. 235.58 KJ/kg C. 225.58
KJ/kg *D. 215.58 KJ/kg
SOLUTION:
ΔU
= mCv (T2 – T1) = 1(0.7186)(600 -300) = 215.58
KJ/kg
103. An insulated rigid tank initially contains 1.5 lb of helium
at 80oF and 50 psia. A paddle wheel with power rating of 0.02 hp is
operated within the tank for 30 min. Determine the final temperature.
A. 159.22oF B. 169.22oF *C. 179.22oF D.
189.22 oF
SOLUTION:
W
= ΔU = m Cv (T2 – T1)
0.02
hp (0.50hr)(2545Btu/hr/hp) = 1.5(0.171)(t2 – 80)
t2
= 179.22oF
104. A 4m2 asphalt pavement with emissivity of 0.85
has a surface temperature of 50oC. Find the maximum rate of
radiation that can be emitted from the surface.
A. 2,068.32 watts B. 2,078.32 watts C.
2,088.32 watts *D. 2.098.32 watts
SOLUTION:
Qr
= e kev A Ts4
Kev
= 5.67 x 10-8 ( Stefan Boltzman constant)
Qr
= 0.85(5.67 x 10-8)(4)(50 +273)4 = 2,098.32 watts
105. Air at 10oC and 90 kpa enters a diffuser of a
jet engine steadily with a velocity of 200 m/s. The inlet area diffuser is 0.40
m2. Determine the mass flow rate of air.
A. 72.79 kg/s B.
74.79 kg/s C. 76.79 kg/s *D. 78.79 kg/s
SOLUTION:
W
= P/RT = 80/0.287(10 + 273) = 0.985 kg/m3
m = w v
A = 0.985(200)(0.40) = 78.79 kg/s
106. Consider a refrigeration whose 40 watts light bulb remains
on continuously as a result of a malfunction of the switch. If the refrigerator
has a COP of 1.3 and the cost of electricity is 8 cents per kw-hr., determine
the increase in the energy consumption of the refrigerator and its cost per
year if the switch is not fixed.
*A.
P49.59 B. P47.59 C. P45.59 D. P43.59
SOLUTION:
COP
= RE/Wref
1.3
= 40/Wref
Wref
= 30.769 watts
W
= Wb + Wref = 40 + 30.769 = 70.77 watts
W
= 0.07077 Kw
Cost
= 0.07077(8760)(P0.08) = P49.59
107. A 75 hp motor that has an efficiency of 91% is worn out and
is replaced by a high-efficiency motor that has an efficiency of 95.4%.
Determine the reduction in heat gain of the room due to higher efficiency under
full-load conditions.
A. 2.24 KW *B. 2.44 KW C. 2.64 KW D. 2.84 KW
SOLUTION:
P01
= (75 x 0.746)(0.91) = 50.91 KW
P02
= (75 x 0.746)(0.954) = 53.376 KW
Qreduced
= 53.376 – 50.91 = 2.44 KW
108. A household refrigerator that has a power input of 450
watts and a COP of 2.5 is to cool five large watermelons, 10 kg each, to 8oC.
If the watermelons are initially at 20oC, determine how long will
take for the refrigerator cool them. The watermelons can be treated as a water
whose specific heat is 4.2 KJ/kg-oK.
A. 2220 seconds B. 2230 seconds *C.2240
seconds D. 2250 seconds
SOLUTION:
COP
= RE/Wc
2.5
= RE/450
RE = 1,125 watts
RE
= m cp (t2 – t1)
450
t = (10 x 5)(4.2)(20-8)
t
= 2240 seconds
109. When a man returns to his wall-sealed house on a summer
day, he finds that the house is at 32oC. He returns on the air
conditioner which cools the entire house to 20oC in 15 minutes, if
COP is 2.5, determine the power drawn by the airconditioner. Assume the entire
mass within the house is 800 kg of air for which cv = 0.72 KJ/kg-K,
cp = 1.0KJ/kg-K.
A. 1.072 KW B.
2.072 KW *C. 3.072 KW D. 4.072 KW
SOLUTION:
RE
= m cv (T2 –T1) = (800/15x60)(0.72)(32-20)
RE
= 7.66 KW
Wc
= 7.68/2.5 = 3.072 KW
110. A heat source at 8000K losses 2000 KJ of heat to
a sink at 500oK. Determine the entropy generated during this
process.
*A.
1.5 KJ/K B. 2.5 KJ/K C. -2.5 KJ/K D. 4 KJ/K
SOLUTION:
ΔSsource
= -2000/800 = -2.5
ΔSsink
= 2000/500 = 4
ΔSgen.
= -2.5 + 4 = 1.5 KJ/K
111. Helium gas is compressed in an adiabatic compressor from an
initial state of 14 psia and 50oF to a final temperature of 320oF
in a reversible manner. Determine the exit pressure of Helium.
A. 38.5 psia *B. 40.5 psia C.
42.5 psia D. 44.5 psia
SOLUTION:
T2/T1
= (P2/P1)n-1/n
(320 + 460)/(50 +460) = (P2/14)1.587-1/1.587
P2
= 40.5 psia
112. Air pass thru a
nozzle with efficiency of 90%. The velocity of air at the exit is 600 m/s. Find
the actual velocity at the exit.
A. 382 m/s B.
540 m/s C. 458 m/s *D. 568 m/s
SOLUTION:
e
= (v2/v3)2
0.9
= (v2/600)2
v2
= 568.21 m/s
113. A 50 kg block of iron casting at 500K is thrown into a
large lake that is at a temperature of 258oK. The iron block
eventually reaches thermal equilibrium with the lake water. Assuming average
specific hear of 0.45 KJ/kg-K for the iron, determine the entropy generated
during this process.
*A.
-12.65 KJ/k B. 16.97KJ/K C. 4.32 KJ/K D. 6.32 KJ/K
SOLUTION:
ΔSiron
= m c ln (T2/T1) = 50(0.45)ln(285/500) = -12.65 KJ/K
ΔSlake
= Q/T = [50(0.45)(500-285)]/285 = 16.97 KJ/K
ΔSgen.
= -12.65 + 16.97 = 4.32 KJ/K
114. A windmill with a 12 m diameter rotor is to be installed at
a location where the wind is blowing at an average velocity of 10 /s. Using
standard conditions of air (1 atm, 25oC), determine the maximum that
can be generated by the windmill.
A. 68 KW *B.
70 KW C. 72 KW D. 74 KW
SOLUTION:
w
= P/RT = 101.325/(0.28)(25+ 273) = 1.1847 kg/m3
m
= w A v = 1.1847(π/4 x 122)(10) = 1,1339.895 kg/s
KE
= v2/2000 = 102/2000 = 0.05 KJ/kg
Power
= m KE = 1,339.895(0.05) = 70 KW
115. Consider a large furnace that can supply heat at a
temperature of 2000oR at a steady rate of 3000Btu/s. Determine the
energy of this energy. Assume an environment temperature of 77oF.
A. 2305.19 KW *B. 2315.19 KW C. 2325.19 KW D. 2335.19 KW
SOLUTION:
e
= = 0.7315
W
= e Q = 0.7315(3000) = 2194.5 Btu/s = 2315.19 KW
116. A heat engine receives hat from a source at 1200oK
at a rate of 5000KJ/s and rejects the waste heat to a medium at 3000oK.
The power output of the heat engine is 180 KW. Determine the irreversible rate
for this process.
A. 190 KW *B.
195 KW C. 200 KW D. 205 KW
SOLUTION:
e
= (1200 – 300) / 1200 = 0.75
W
= 0.75(500) = 375 KW
Irreversibilities
= 375 – 180 195 KW
117. A dealer advertises that he has just received a shipment of
electric resistance heaters for residential buildings that have an efficiency
of 100 percent. Assuming an indoor temperature of 21oC and outdoor
temperature of 10oC, determine the second law efficiency of these
heaters.
A. 8.74% B.
6.74% *C. 3.74% D. 4.74%
SOLUTION:
COP1
= 100% efficient = 1
COP2
= (21 + 273) / (21 – 10) = 26.72
e
= COP1 /COP2 = 1/
26.72 = 3.74 %
118. A thermal power plant has a heat rate of 11,363 Btu/KW-hr.
Find the thermal efficiency of the plant.
A. 34% B.
24% C. 26% *D. 30%
SOLUTION:
e
= 3412 / Heat rate = 3412 / 11363 = 30 %
119. A rigid tank contains 2 kmol of N2 and 6 kmol of
CO2 gasses at 300oK and 115 Mpa. Find the tank volume us
ideal gas equation.
A. 7.33 m3 B. 5.33 m3 C. 3.33 m3 *D. 1.33 m3
SOLUTION:
PmVm
= Nm R Tm
15,000
Vm = (6 + 2)(8.314)(300)
Vm
= 1.33 m3
120. A spherical balloon with a diameter of 6 m is filled with
helium at 20oC and 200 kpa. Determine the mole number.
*A.
9.28 Kmol B. 10.28 Kmol C. 11.28 Kmol D. 13.28 Kmol
SOLUTION:
P
V = N R T
(200)[(4/3)(π)(6/2)3]
= N (8.314)(20 + 273)
N
= 9.28 Kmol
121. The air in an automobile tire with a volume of 0.53 ft3
is at 90oF and 20 psig. Determine the amount of air that must be
added to raise the pressure to the recommended value of 30 psig. Assume the
atmospheric to be 14.7 psia and the temperature and the volume to remain
constant.
*A.
0.026 lb B. 0.046 lb C. 0.066 lb D. 0.086 lb
SOLUTION:
P
V = m R T
(20
+ 14.7)(144)(0.53) = m1 (53.3)(90 + 460)
m1
= 0.09034 lb
(30
+ 14.7)(144)(0.53) = m2(53.3)(90 + 460)
m2
= 0.11634 lb
madded
= m2 – m1 = 0.11634 – 0.09034 = 0.026 lb
122. A rigid tank contains 20 lbm of air at 20 psia and 70oF.
More air is added to the tank until the pressure and temperature rise to 35
psia and 90 oF, respectively. Determine the amount of air added to
the tank.
A. 11.73 lb *B.
13.73 lb C. 15.73 lb D. 17.73 lb
SOLUTION:
P1V1
= m1 R1T1
(20
x 144)(V1) = 20 (53.3)(70 + 460)
V
= 196.17 ft3
P2V2
= m2R2T2
(35
x 144)(196.17) = m2 (53.3)(90 + 460)
m2
= 33.73 lbs
madded
= m2 –m1 = 33.73 – 20 = 13.73 lb
123. A rigid tank contains 5 kg of an ideal gas at 4 atm and 40oC.
Now a valve is opened, and half of mass of the gas is allowed to escape. If the
final pressure in the tank is 1,5 atm, the final temperature in the tank is:
*A.
-38oC B. -30oC
C. 40oC D. 53oC
SOLUTION:
PV
= m R T
(4
x 9.81)(V) = 5(0.287)(40 + 273)
V
= 11.446 m3
PV
= mRT
(1.5
x 9.81)(11.446) = (5/2)(0.287)(T)
T
= 234.74oK = -38.26oC
124. The pressure of an automobile tire is measured to be 200
kpa(gage) before the trip and 220 kpa(gage) after the trip at a location where
the atmospheric pressure is 90 kpa. If the temperature of the air in the tire
before the trip is 25oC, the air temperature after the trip is:
*A.
45.6oC B. 54.6oC C. 27.5oC D. 26.7oC
SOLUTION:
T2
/ T1 = P2 / P1
T2
/ (25+ 273) = (220 +90) / (200 + 90)
T2
= 318.55 K
t2
= 45.55oC
125. Water is boiling at 1 atm pressure in a stainless steel pan
on an electric range. It is observed that 2 kg of liquid ater evaporates in 30
mins. The rate of heat transfer to the water is:
A. 2.97 KW B.
0.47 KW *C. 2.51 KW D. 3.12 KW
SOLUTION:
Q
= mL = (2257) = 2.51 KW
126. Consider a person standing in a breezy room at 20oC.
Determine the total rate of heat transfer from this person if the exposed
surface area and the average outer surface temperature of the person are 1.6 m2
and 29oC, respectively, and the convection heat transfer coefficient is 6 W/m2
with emissivity factor of 0.95.
A. 86.40 watts B. 61.70 watts C. 198.1
watts *D. 168.1 watts
SOLUTION:
Qc
= h A (t2 – t1) = (6)(1.6)(29.20) = 86.40 watts
Qr
= (0.95)(5.67 x 10-6)[(1.6)(29 + 273)4 – (20 + 273)4]
= 81.7 watts
Q
= Qc + Qr = 86.40 + 81.7 = 168.1 watts
127. Water is boiler in a pan on a stove at sea level. During 10
minutes of boiling, it is observed that 200 grams of water has evaporated. Then
the rate of heat transfer to the water is:
A. 0.84 KJ/min *B. 45.1 KJ/min C. 41.8
KJ/min D. 53.5 KJ/min
SOLUTION:
Q
= m L = (0.2 / 10) (2257) = 45.1 KJ/min
128. An aluminum pan whose thermal conductivity is 237 W/m-C has
a flat bottom whose diameter is 20 cm and thickness 0.4 cm. Heat is transferred
steadily to boiling water in the pan through its bottom at a rate of 500 watts.
If the inner surface of the bottom of the pan is 105oC, determine
the temperature of the surface of the bottom of the pan.
A. 95.27 oC *B. 105.27oC C. 115.27oC D. 125.27oC
SOLUTION:
A
= π / 4 ( 0.20)2 = 0.0314 m2
Q
=
500
=
T2 = 105.27oC
129. For heat transfer purposes, a standing man can be modeled
as a 30 cm diameter, 170 cm long vertical cylinder with both the top and bottom
surfaces insulated and with the side surface at an average temperature of 34oC.
For a convection heat transfer coefficient of 15 W/m2- oC,
determine the rate of heat loss from this man by convection in an environment
at 20oC.
A. 316.46 watts B. 326.46 watts *C. 336.46
watts D. 346.46 watts
SOLUTION:
Qc
= k A (t2 – t1) = 15 (π x 0.30 x 1.7) (34 – 20) = 336.46
watts
130. A 5cm diameter spherical ball whose surface is maintained
at a temperature of 70oC is suspended in the middle of a room at 20oC.
If the convection heat transfer coefficient is 15 W/m2 – C and the
emissivity of the surface is 0.8, determine the total heat transfer from the
ball.
A. 23.56 watts *B. 32.77 watts C. 9.22
watts D. 43.45 watts
SOLUTION:
A
= 4 π r2 = 4 π (0.05)2 = 0.0314 m2
Qc
= h A (t2 – t1) = 15 (0.0314)(70 – 20) = 23.56 watts
Qr
= (0.80)(5.67 x 10-6)(0.0314)[(70 + 273)4 – (50 + 273)4]
= 9.22 watts
Q
= Qr + Qc = 23.56 + 9.22 = 32.77 watts
131. A frictionless piston-cylinder device and rigid tank
contain 1.2 kmol of ideal gas at the same temperature, pressure, and volume.
Now heat is transferred, and the temperature of both system is raised by 15oC.
The amount of extra heat that must be supplied to the gas in the cylinder that
is maintained at constant pressure.
SOLUTION:
A. 0 B.
50 KJ C. 100 KJ *D. 150 KJ
Q
= m cp (t2 – t1) = (1.2 x 8.314)(1)(15) = 150
KJ
132. A supply of 50 kg of chicken needs at 6oC
contained in a box is to be frozen to -18oC in a freezer. Determine
the amount of heat that needs to be removed. The latent heat of chicken is 247
KJ/kg, and its specific heat is 3.32 KJ/kg-oC above freezing and
1.77 KJ/kg-oC below freezing. The container box is 1.5 kg, and the
specific heat of the box material is 1.4 Kj/kg-oC. Also the freezing
temperature of chicken is -2.8oC.
*A.
15,206.4 KJ B. 50.4 KJ C. 15,156 KJ D.
1,863 KJ
SOLUTION:
Qchicken
= 50 [3.32(6 + 2.8) = 247 1.77(-2.8 + 18)] = 15,156 KJ
Qbox
= 1.5(1.4)(6 + 8) = 50.4 KJ
Q
= 15,156 + 50.4 = 15, 206.4 KJ
133. Water is being heated in a closed pan on top of a range
while being stirred by a paddle wheel. During the process, 30 KJ of heat is
transferred to the water, and 5 KJ of heat is lost to the surrounding air. The
paddle wheel work amounts to 500 N-m. Determine the final energy of the system if
its initial energy is 10 KJ.
*A.
35.5 KJ B. 45.5 KJ C. 25.5 KJ D. 14.5 KJ
SOLUTION:
Final energy = Qa + ΔU – Qloss + W = 30 + 10 – 5 + 0.50 =
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