SAMPLE PROBLEMS
1. An air-standard Diesel Cycle operates with a compression ratio of 15. The maximum
cycle temperature is 1,700ºC. At the beginning of compression the air is at 100 KPa abs
and 40ºC. What is the cut-off ratio?
T2 = T1 rk K-I = (313 915) 0.4 = 924.657 ºK
rc = V3 /V2 = T3 /T2 =197.3/924.657 = 2.13
A) 2.13
B) 2.31
C) 1.23
D) 2.87
2. A spark- ignition engine produces brake power of 224 KW while using 0.0169 kg/s of
fuel. The fuel has a higher heating value of 44,186 KJ/kg, and the engine has a
compression ratio of 8. The frictional power lost is found to be 22.4 KW. Determine the
indicated thermal efficiency.
A) 11
B) 22
C) 33
D) 44
ei = WKi / QA = WK i / mf x H V = (224 + 22.4) / (0.0169 x 44,186) = 33%
3. The volume in the clearance space of a 152.40 mm x 254.0 mm Otto gas engine is 1.70
liters. Find the ideal thermal efficiency of the engine on the standard air basis, if the
exponent of the expansion and compression lines is 1.35. Express in percent.
A) 38.55
B) 36.89
C) 26.98
D) 35.95
2
3
VD =[π
] (0.1524) (0.254) = 0.004633m ;
C = V2 / VD = 0.0017/0.004633 = 36.69%
rk = (1+ C)/C = (1 +0.3669)/0.3669 = 3.726
e = 1 – (1 / rk k-1 ) = 1 – [1/3.726] 0..35 = 36.89%
4. Calculate from the following data, the kg of air per kg of fuel used by an automobile
engine: air temperature is 21.50ºC, barometer reads 760 mm Hg, air entering 1.75
m3 /min., measured gasoline 14.25 liters per hr, specific gravity of gasoline 0.735.
A)12.02
B) 15.45
C) 14.45
D) 21.45
ma =PV/RT = (101.325 x 1.75) / 0.287)(294.5) = 2.098 Kg/min
mf = ρ x v =0.735 (1,000) x (0.01425)/(60) = 0.1746 kg/min
ra/f = 2.098 / 0.1746 = 12.02 kga /kgf
5. An engine has an efficiency of 26%. It uses 2 gallons of gasoline per hour. Gasoline has a
heating value of 20,500 Btu/lbf and a specific gravity of 0.8. What is the power output of
the engine?
A. 20.8KW
B. 29KW
C. 19.8KW
D. 18.9KW
Solution:
mf 
2 gal
ft 3
0.862.4lb
lbs


 13.346
3
hr
7.481gal
ft
hr
Power Output  m f Q f  eff
Power Output 
13.346lb
hr
20500 Btu 1.055kj



 0.26
hr
3600 sec
lb
Btu
Power Output  20.8Kw
6. A supercharged six-cylinder four stroke cycle diesel engine of 10.48 cm bore and 12.7
cm stroke has a compression ratio of 15. When it is tested on a dynamometer with a
53.34 cm arm at 2,500 rpm, the scale reads 81.65 kg, 2.86 kg of fuel of 45,822.20 KJ/kg
heating value are burned during a 6 min test, and air metered to the c ylinders at the rate
of 0.182 kg/s. Find the brake thermal efficiency.
A. 0.327
B. 0.367
C. 0.307
D. 0.357
Solution:
T = load x lever arm = [81.65kg x 9.91N/kg](0.5334m) = 427.25N- m = 0.42725 KN-m
Brake Power = 2nT = 2rad/rev[(2,500/60)rev/sec](0.42725 KN-m) = 111.854 KW
Mass of fuel, mf = 2.86kg / 6(60)sec = 0.00794 kgf /sec
Brake thermal efficiency = brake power/ mf x HVf
= 111.854 KW / [0.00794 kg/sec x 45,822.20 KJ/kg]
= 0.307 = 30.7%
7. A Carnot engine receives 130 BTU of heat form a hot reservoir at 700°F and rejects 49
BTU of heat. Calculate the temperature of the cold reservoir. (April ’97)
A. -21.9ºF
B. -24.2
C. -20.8
D. -22.7
Solution:
TH = 700 + 460 = 1160°R
Q A  QR
T  TL
130  49 1160  TL
= H
=
=
QA
TH
1160
130
TL = 437.23°R
TL = -22.77°F
Carnot Engine efficiency
=
8. A Carnot engine requires 35 KW from the
power and the temperature of the sink is
source? (April ’97)
A. 245.6ºC
B. 210.1
Solution:
TC = 26 + 273 = 299°K
T  TL
W
Efficiency =
= H
TH
QA
TH = 523.25°K
hot source. The engine produces 15 KW of
26°C. What is the temperature of the hot
C. 250.2
15
35
TH = 523.25 – 273
=
D. 260.7
TH  299
TH
TH = 250.25°C
9. The compression ratio of an ideal Otto cycle is 6:1. Initial conditions are 101.325 KPa
and 20°C. Find the temperature at the end of the adiabatic compression. (October ’94)
A. 60ºC
B. 600ºK
C. 60ºK
D. 600ºC
Solution:
k 1
T2  V1 


T1  V2 
T2
 (6)1.41
20  273
T2 = 599.96°K
10. The maximum thermal efficiency possible for a power cycle operating between 1200°F
and 225°F is: (April ’97)
A. 58%
B. 58.73
Solution:
TH = 1200 + 460 = 1660°R
Efficiency =
C. 57.54
D. 57.40
TL = 225 + 460 = 685°R
TH  TL
1660  685
=
= 58.73%
TH
1660
11. A heat engine is operated between temperature limits of 1370 degree C and 260 degree C.
Engine supplied with 14,142 per KW-hr. Find the Carnot cycle efficiency in percent.
(October ’97)
A. 67.56
B. 70.10
C. 65.05
D. 69.32
Solution:
T1 = 1370 + 273 = 1643°K
Efficiency = 1
T2 = 260 + 273 = 533°K
T2
533
= 1
= 67.56%
T1
1643
12. An Otto engine has clearance volume of 7%. It produce s 300 KW power. What is the
amount of heat rejected in KW? (October ’97)
A. 152 KW
B. 95.32
C. 92.16
D. 90.72
Solution:
r = compression ratio =
e = 1
e=
1
r
k 1
= 1
W
QA
1 c 1 0.07
=
c
0.07
1
= 0.664
(15.286)1.41
QA = 452 KW
W = QA – QR
300 = 452 – Q R
QR = 152 kW
13. Calculate the energy transfer rate across 6 in wall of firebricks with a temperature
difference across the wall of 50°C. The thermal conductivity of the fireb rick is 0.65
Btu/hr- ft-°F at the temperature interest. (October ’94)
A. 285 W/m2
B. 369
C. 112
D. 429
Solution:
x = 6 in = 0.50 ft
9
(ta – tb ) = 50 ×
= 90°F
5
kA( t a  t b ) 0.65(1)(90)

QC =
x
0.50
Btu
1055 J
hr
(3.28)2 ft 2



QC = 117
= 369 W/m2
Btu
3600 sec
hr  ft 2
m2
14. The sun generates 1 kW/m2 when used as a source for solar collectors. A collector with
an area of 1 m2 heats water. The flow rate is 3.0 liters/min. What is the temperature rise
in the water? The specific heat of water is 4,200 J/kg -°C. (April ’97)
A. 4.8ºC
B. 0.48
C. 0.56
D. 0.84
Solution:
1kW
Q
 (1m 2 )  1kW  1000 W
2
m
m
3 li 1kg
min


 0.05 kg/sec
2
li
60 sec
m
Q = m × cp × Δt
1000 = 0.05 × (4200) × Δt
Δt = 4.76°C
15. An electric current is passed through a wire 1mm diameter and 10 cm long. The wire is
submerged in liquid water at atmospheric pressure, and the current is increased until the
water boils. For this situation h = 5000W/m2 -K, and the water temperature will be 100 0 C.
How much electric power must be supplied to the wire to maintain the wire surface at
1140 C?
A. 11.99
B.31.99
C. 41.99
D. 21.99
Solution: convection heat transfer because there is a flow of elec tricity
Heat Transfer = electric power supplied
Q = hAt = h (πdL) t = 5000 [π(1x10-3 )(1x10-2 )](114 - 100)
Q = 21.99 W
16. What is the temperature in degree C of 2 liters of water at 30 degree C after 500 calories
of heat have been added to? (April ’98)
A. 35.70ºC
B. 30.25
C. 38.00
D. 39.75
Solution:
Q = m × cp × (t2 – t1 )
0.500 Kcal (4.187 KJ/Kcal) = 2 liters (1kg/liter)(4.187 KJ/kg-°C)(t2 – 30°C)
t2 = 30.25°C
17. At an average temperature of 100°C, hot air flows through a 2.5 m long tube with an
inside diameter of 50 mm. The temperature of the tube is 20°C along its entire length.
Convective film coefficient is 20.1 W/m2 -°K. Determine the convective heat transfer
form air to the tube. (April ‘ 97)
A. 900W
B. 909
C. 624
D. 632
Solution:
A = heat transfer area = π × D × L
= π × (0.050) × (2.5) = 0.3927 m2
QC = hC × A × (t2 – t1 )
QC = 20.1 × (0.3927 × (100 – 20)
QC = 631.5 W
18. One kg of water initially at 15ºC is to be freeze at –3ºC. Calculate the total amount of
heat to be removed during the cooling process? Cpice = 0.5(Cpw)
A) 39.831KJ
B) 403
Q 1 = mc∆T = 1 (4.187)15 = 62.8
Q 3 = 1(0.5)(4.187)3 = 6.28
C) 3983.1
Q2 = 334
QT = 403KJ
D) 39831
19. Calculate the enthalpy in KJ of 1.50 kg of fluid that occupy a volume of 0.565 m3 if the
internal energy is 555.60 Kcal per kg and the pressure is 2 atm.
A. 2,603.7 KJ
B. 3,603.7
C. 4,603.7
D. 5,603.7
h = u + pv = [555.6 kcal x (1.055/0.252)] + 2(101.325)(0.565/1.5) = 2402.35 KJ/kg
H = mh = 1.5(2402.35) = 3,603.5KJ
20. Steam flows into a turbine at the rate of 10 kg/s and 10 KW of heat are lost from the
turbine. Ignoring elevation and kinetic energy effects, calculate the power output from
the turbine. (October ’94)
Given:
hin = 2739.0 KJ/kg;
hout = 2300.5 KJ/kg
A. 4605KW
B. 4973
C. 4375
D. 4000
Solution:
H1 + PE1 + KE1 = H2 + PE2 + KE2 + QR + W T
W T = H1 – H2 + PE1 – PE2 + KE1 – KE2 – QR
= m(h 1 – h 2 ) – QR
= [10 kg/s (2739 – 2300.5)KJ/kg] – 10KW
= 4385 – 10
= 4375 KW
21. The enthalpy of air is increased by 139.586 KJ/kg in a compressor. The rate 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? (April ’97)
A. 10KW
B. 90.95
C. 100
D. 995
Solution:
H1 + PE1 + KE1 + W C = H2 + PE2 + KE2 + QR
QR = H1 – H2 + PE1 – PE2 + KE1 – KE2 – W C
= m(–∆h) + W C
= – (16.42/ 60)kg/s (139.586KJ/kg ) + 48.2 KW
= –38.2 KW + 48.2 KW
= 10 KW
22. A volume of 400 cc of air is measured at a pressure of 740 mmHg abs and a temperature
of 18°C. What will be the volume at 760 mmHg and 0°C?
A. 376 cc
B. 326
C. 356
D. 365
Solution:
760 V2
P1V1
PV
740( 400)
= 2 2
=
V2 = 365.4 cc
T2
0  273
T1
18  273
23. When the pressure of an ideal gas is doubled while the absolute temperature is halved, the
volume is:
(A) quadrupled
(B) doubled
(C) halved
(D) quarte red.
P1 V1 / T1 = P2 V2 / T2
V2 / V1 = P1 T2 / P2 T1 = P1 (0.5T1 ) / 2P1 T1
= (0.5)(0.5) = 0.25
24. An ideal gas at 45 psig and 80°F is heated in a closed container to 130°F. What is the
final pressure? (April ’97)
A. 54psia
B. 65
C. 75
D. 43
Solution:
P1 = 45 + 14.7 = 59.7 psia
P1 P2

T1 T1
T1 = 80 + 460 = 540°R
59.7 P2

540 590
T2 = 130 + 460 = 590°R
P2 = 65.23 psia
25. A closed vessel contains air at a pressure of 160 KPa gauge and temperature of 30 degree
C. The air is heated at constant volume to 60 degree C with the atmospheric pressure as
759 mmHg. What is the final gauge pressure? (April ’98)
A. 186KPag
B. 169
C. 167
D. 172
Solution:
101.325 KPa
Patm = 759 mmHg 
= 101.2 KPa
760 mmHg
P1 P2
P2
(160  101.2)


T1 T2
(30  273)
(60  273)
P2 = 287 KPa abs
P2 = 287 – 101.325 = 185.675 KPag
26. Three resistors with 4, 5 and 6 ohms respectively are connected in parallel. Calculate the
total resistance in the circuit.
A. 6.122
B. 1.622
C. 2.661
D. 1.226
1/ RT = 1/R1 + 1/R2 + 1/R3 = ¼ + 1/5 + 1/6 = 37/60
RT = 60/37 = 1.622 ohms
27. What current would flow for three hours through a resistance of 50 ohms to produce
enough heat to raise the temperature of ten kg water from the freezing point to boiling
point?
A) 7.76amp
B) 77.6
C) 0.776
D) 776
Q = mc∆T = 10(4.187)100 = 4,187 KJ in 3 hrs
Heat rate = 4,187KJ / 3hrs = 1396 KJ/hr = 0.388 KW = 388 W
Therefore;
I = √ P/R = √388/50 = 7.76 amp
28. How many 100-W lamps connected in parallel can be placed in a 230 volts circuit, which
is protected by a 30-amp fuse?
A. 69
B. 70
C. 71
D. 72
P = E x I = 230volts x 30amp = 6900 watts
NL = 6900 / 100 = 69
29. A 12-V automotive storage battery is charged with a constant current of 10 amp for 3
hours. Determine the amount of energy supplied to the battery in kw- hr.
A. 0.33
B. 0.36
C. 0.39
D. 0.42
E = 12V x 10amp x 3hrs = 360watt-hrs = 0.36 KW-hr
30. Electrical resistance of 7 ohms and 11 ohms are connected in parallel and the
combination of which is then connected in series with a resistance of 15 ohms and a
source of 220 volts. Calculate the total current flowing in the circuit.
A) 0.57amp
B) 11.41
RT = 15 + 1 / (1/7 + 1/11) = 19.28 ohms
C) 10.41
D) 0.057
I = E / R = 220 / 19.28 = 11.41 amp
31. What retarding force is required to stop a 0.45 caliber bullet of mass 20 grams and speed
of 300m/sec as it penetrates a wooden block to a depth of 2 inches?
A. 17,716 N
B. 19,645
C. 15,500
D. 12,500
x = 2 in = 0.0508 m
Energy -- Work relation:
Work = KE
F(x) = ½ mV²
F(0.0508m) = ½(0.02kg)(300m/s)²
F = 17,716.53 N
32. A 50 mm dia. pipe is 15 m long extends vertically downward from the bottom of an
elevated tank and discharges into air. When the water in the tank is 3 m from the entrance
to the pipe, what is the discharge in liters/sec?
A. 36.88L/s ___
B. 38.86
C. 68.38
D. 63.88
Vel = √2gh = √2(9.81)18 = 18.783 m / s
Q = V x A = (18.783) x 0.7854 x (0.05)2 = 0.03688 m3 / s = 36.88 liters / s
33. An opened topped cylindrical tank with oil (S.G. = 0.8) has a circular base 10 m in
diameter when filled to a height of 8 m. Calculate the force exerted in its base.
A. 4296kN
B. 4926
C. 4629
D. 4962
2
F = pressure x area = hγA = 8 x 0.8(1000)9.81 x 0.7854(10) = 4,926,028.8 N
34. What pressure is a column of water 100 cm high equivalent to? (October ’94)
A. 9807 D/cm2
B. 9807 N/m2
C. 0.1 bar
D. 98,100 Pa
Solution:
1m
N
Pressure = γ × h = 9807 3 × 100 cm ×
= 9807 N/m2
100 cm
m
35. A pressure of 1 millibar is equivalent to: (April ‘95)
A. 1000 D/cm2
B. 1000 cm Hg
C. 1000psi
D. 1000kg/cm2
Solution:
1 millibar = 0.001 bar = 0.1 kPa = 0.1
kN
m2
kN 1000 N 100,000 dynes
m2



kN
N
m2
10,000 cm 2
dynes
1 millibar  1000
cm2
0.1
36. If the velocity in a 300 mm pipe is 0.50 m/s, determine the discharge in a 75 mm
diameter jet attached to the pipe.
A. 33.5L/s
B. 35.3
C. 53.3
D. 55.3
Q = Vel1 x Area1 = Vel2 x Area2
= 0.50 m /s x 0.7854 x 0.3m2 = 0.03534 m3 / sec = 35.3 L/s
37. A wheel starts from rest and acquires a speed of 500 rad/sec in 15 seconds. Calculate the
approximate number of turns made by the wheel.
A. 119.4
B. 1194
C. 11940
D. 119400
NT = 500rad/s x 15 sec x 1rev / 2πrad = 1,194 rev
38. A ball dropped from the roof of a building 40 m high will hit the ground with a velocity
of:
A. 20 m/s
B. 28
C. 38
D. 40
V = √2gh = √2(9.81)(40) = 28
39. A water temperature rise of 18°F in the water cooled condenser is equivalent in °C to:
(October ’94)
A. 7.78ºC
B. 10
C. 263.6
D. -9.44
Solution:
5
5
Temperature change, Δ°C
= F  (18)  10°C
9
9
40.
One kg-mole of oxygen gas has a molecular mass of:
A. 16kg
B. 32kg
C. 16g
D. 32g
1 kg- mole = 32kg
1lb- mole = 32lb
1g- mole = 32gram
41. Ten liters of 25% salt solution and 15 liters of 35% salt solution are poured into a drum
originally containing 30 liters of 10% salt solution. What is the percent concentration of
salt in the mixture?
A. 19.55%
B. 22.15%
C. 27.05%
D. 25.72%
Solution:
25%
10
+
35%
+
15
10%
30
=
x%
55
0.25(10) + 0.35(15) + 0.10(30) = x(55)
x = 19.55%
42. As2 O 3 + 3 C →3 CO + 2As
Atomic weights may be taken as 75 for arsenic, 16 for oxygen, and 12 for ca rbon.
According to the equation above, the reaction of 1 gram- mole of As2 O 3 with carbon will
result in the formation of
(A) 1 gram- mole of As.
(C) 150 grams of As
(B) 28 grams of CO.
(D) a greater amount of weight of CO than of As.
43. Ethane gas burns according to the equation, 2C 2 H6 + 7 O2 → 4CO2 + 6H2 O.
What volume of CO 2 , measured at standard temperature and pressure, is formed for each
gram- mole of C2 H6 burned?
(A) 22.4 liters
(B) 44.8 liters
(C) 88.0 liters
(D) 89.6 liters
44. Express X = 0.351351351…. into fractional number
A.) 351/999
B.) 351/100
C.) 99/351
D.) 351/98
45. Express X = 0.353535 into fractional number
A.) 35/99
B.) 35/100
C.) 99/35
D.) 35/98
46. Express X = 0.77777777 into fractional number : X = _______
A.) X = 6/9
B.) X = 7/ 10
C.) X = 7/9
D.) X = 6/10
47.
Evaluate: x  2  2  2  .......
A.) X= 1
B.) X = 2
C.) X= 3
D.) X= 4