Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Otto Cycle”.

1. The Otto cycle is the
a) air standard cycle of CI engine
b) air standard cycle of SI engine
c) vapour power cycle of CI engine
d) vapour power cycle of SI engine
Answer: b
Clarification: The Otto cycle is air standard cycle and is used in SI engine.

2. In a four-stroke internal combustion engine,
a) the piston does four complete strokes within cylinder
b) for each cycle, the crankshaft completes two revolutions
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is the functioning of a four-stroke internal combustion engine.

3. The correct sequence of strokes in a four-stroke SI engine is
a) intake->compression->exhaust->expansion
b) intake->expansion->compression->exhaust
c) intake->exhaust->compression->expansion
d) intake->compression->expansion->exhaust
Answer: d
Clarification: The correct sequence is intake->compression->expansion->exhaust and expansion stroke is also called power stroke.

4. The spark plug fires shortly before the ____ stroke.
a) compression
b) expansion
c) intake
d) exhaust
Answer: b
Clarification: The spark plug fires shortly before the piston reaches TDC and after this ignition the expansion stroke takes place.

5. The pressure in cylinder is ____ the atmospheric value during exhaust stroke and ____ it during intake stroke.
a) above, below
b) below, above
c) equal to, equal to
d) equal to, above
Answer: a
Clarification: This is done to ensure that all the exhaust gases are thrown out of the cylinder and enough amount of intake mixture enters the cylinder.

6. In a two-stroke engine, the four functions performed in SI engine are done in which two strokes?
a) expansion stroke and compression stroke
b) intake stroke and exhaust stroke
c) compression stroke and power stroke
d) compression stroke and expansion stroke
Answer: c
Clarification: In a two-stroke engine, these two strokes perform all the functions.

7. A two-stroke engine is used in motorcycles and scooters.
a) true
b) false
Answer: a
Clarification: The reason is that these vehicles need engines of small size and weight.

8. In a two-stroke engine,
a) the crankcase is sealed
b) the outward motion of piston is used to pressurize the air-fuel mixture
c) the intake and exhaust valves are replaced by opening in lower part of cylinder wall
d) all of the mentioned
Answer: d
Clarification: These are the modifications done in a two-stroke engine.

9. The two-stroke engine is ____ the four-stroke engine.
a) more efficient than
b) less efficient than
c) equally efficient to
d) none of the mentioned
Answer: b
Clarification: The reason being incomplete removal of exhaust gases in a two-stroke engine.

10. The two-stroke engine is
a) simple and expensive
b) high power-to-weight ratio
c) low power-to-volume ratio
d) all of the mentioned
Answer: b
Clarification: The two-stroke engine is inexpensive and has high power-to-volume ratio.

11. The intake and exhaust processes are not considered in the p-V diagram of Otto cycle.
a) true
b) false
Answer: a
Clarification: The reason is these two processes cancel each other.

12. The efficiency of Otto cycle is given by (rk is the compression ratio)
a) 1/(rk)^(ɣ-1)
b) 1 – 1/(rk)^(ɣ)
c) 1 – 1/(rk)^(ɣ-1)
d) 1/(rk)^(ɣ)
Answer: c
Clarification: This is the expression for efficiency of Otto cycle and rk=compression ratio=Vmax/Vmin.

13. Which of the following statement is true?
a) efficiency of air standard cycle is a function of compression ratio and temperature levels
b) higher the compression ratio, higher will be the efficiency
c) efficiency is dependent on the temperature levels at which the cycle operates
d) all of the mentioned
Answer: b
Clarification: The efficiency of air standard cycle = 1 – 1/[(rk)^(ɣ-1)] and it does not depend on the temperature levels.

14. ɣ for air is equal to
a) 1.0
b) 1.2
c) 1.3
d) 1.4
Answer: d
Clarification: In sir standard cycle, air is the working fluid and ɣ for air is 1.4.

15. The Otto cycle consists of
a) two reversible isotherms and two reversible isobars
b) two reversible isochores and two reversible adiabatics
c) two reversible isotherms and two reversible isochores
d) two reversible isobars and two reversible adiabatics
Answer: b
Clarification: This can be shown in a p-V and T-s diagrams.

Thermodynamics Interview Questions and Answers on “Other types of Work Transfer-2”.

1. Find the rate of conduction heat transfer through a 1.5 cm thick board(k = 0.16 W/m K), with a temperature difference of 20°C between the two sides.
a) 113 W/m²
b) 413 W/m²
c) 313 W/m²
d) 213 W/m²
Answer: c
Clarification: The rate of conduction heat transfer = k(∆T/∆x)
= (0.16 *20)/0.015 = 213 W/m².

2. A window having area of 2m^2 has a surface temperature of 15°C and the air is blowing at 2°C across it with convection heat transfer coefficient of h = 125 W/m2K. Find the total heat transfer loss?
a) 3250 W
b) 2250 W
c) 4250 W
d) 5250 W
Answer: a
Clarification: Total heat transfer loss = h A ∆T
= 125*2*(15-2) = 3250 W.

3. A radiant heating lamp has a temperature of 1000 K with ε = 0.8. What should be the surface area to provide 250 W of radiation heat transfer?
a) 0.0035 m²
b) 0.0055 m²
c) 0.0075 m²
d) 0.0095 m²
Answer: b
Clarification: Radiation heat transfer = εσA(T⁴)
A = 250/[0.8 × 5.67 × 10^-8× 1000⁴]
= 0.0055 m².

4. A piston of mass 2 kg is lowered by 0.5 m. Find the work involved in the process.
a) 7.805 J
b) 8.805 J
c) 9.805 J
d) 10.805 J
Answer: c
Clarification: F = ma = 2 kg × 9.80665 m/s² = 19.61 N
W = ∫ F dx = F ∫ dx = F ∆x = 19.61 N × 0.5 m = 9.805 J.

5. An escalator raises a 100 kg bucket of water 10 m in 60 seconds. Determine the amount of work done during the process.
a) 9807 J
b) 9307 J
c) 9507 J
d) 9107 J
Answer: a
Clarification: F = mg
W = ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s² × 10 m
= 9807 J.

6. A hydraulic cylinder has a piston(cross sectional area 25cm2) and a fluid pressure of 2 MPa. If piston moves by 0.25m, how much work is done?
a) 0.25 kJ
b) 1.25 kJ
c) 2.25 kJ
d) 3.25 kJ
Answer: b
Clarification: W = ∫ F dx = ∫ PA dx = PA ∆x
= 2000 kPa × 25 × 10^(-4) m2 × 0.25 m = 1.25 kJ.

7. In a thermally insulated kitchen, a refrigerator with 2 kW motor for running the compressor provides 6000 kJ of cooling to the refrigerated space during 30 min operation. If the condenser coil placed behind the refrigerator rejects 8000 kJ of heat to the kitchen during the same period, calculate the change in internal energy of the kitchen.
a) 3600 kJ
b) 2400 kJ
c) 4800 kJ
d) none of the mentioned
Answer: a
Clarification: QKitchen = 0 (Insulated!),
W(Electrical) = – P*∆τ = – 2 kW*30*60 sec = – 3600 kJ
(It is negative because work is done on to the system)
Change in internal energy of the kitchen (∆UKitchen) = QKitchen – WElectrical
= 0 – (–3600) = 3600 kJ.

8. An escalator raises a 100 kg bucket of sand 10 m in a minute. Determine the rate of work done during the process.
a) 143 W
b) 153 W
c) 163 W
d) 173 W
Answer: c
Clarification: The work is force with a displacement and force is F = mg, which is constant
W = ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s² × 10 m = 9807 J
The rate of work is work per unit time = W/∆t = 9807 J / 60 s
= 163 W.

9. A crane lifts a bucket of cement with a mass of 450 kg vertically up with a constant velocity of 2 m/s. Find the rate of work.
a) 8.83 kW
b) 8.33 kW
c) 8.53 kW
d) 8.63 kW
Answer: a
Clarification: Rate of doing work = FV = mg × V = 450 kg × 9.807 ms^(−2) × 2 ms^(−1)
= 8826 J/s = 8.83 kW.

10. A battery is well insulated while being charged by 12.3 V at a current of 6 A. Take the battery as a control mass and find the instantaneous rate of work.
a) 63.8 W
b) 73.8 W
c) 83.8 W
d) 93.8 W
Answer: b
Clarification: Battery thermally insulated ⇒ Q = 0
For constant voltage E and current i, Power = E i = 12.3 × 6 = 73.8 W.

11. A current of 10 amp runs through a resistor with resistance of 15 ohms. Find the rate of work that heats the resistor up.
a) 1200 W
b) 1300 W
c) 1400 W
d) 1500 W
Answer: d
Clarification: Power = E i = R i^(2) = 15 × 10 × 10 = 1500 W.

12. A pressure of 650 kPa pushes a piston of radius 0.125 m with V = 5 m/s. What is the transmitted power?
a) 139.5 kW
b) 149.5 kW
c) 159.5 kW
d) 169.5 kW
Answer: c
Clarification: A = π/4(D)² = 0.049087 m²;
Volume flow rate = AV = 0049087 m^2 × 5 m/s = 0.2454 m³/s
Power = F V = P(Volume flow rate) = 650 kPa × 0.2454 m³/s = 159.5 kW.

13. Air at a constant pressure in a piston-cylinder is at 300 K, 300 kPa and V=0.1 m^3. It is heated to 600 K in 30 seconds in a process with constant piston velocity. Find the power delivered to the piston.
a) 1 kW
b) 2 kW
c) 3 kW
d) 4 kW
Answer: a
Clarification: Process: P = constant : dW = P dV
V2 = V1× (T2/T1) = 0.1 × (600/300) = 0.2
Rate = P (∆V / ∆t) = 300 × (0.2-0.1)/30 = 1 kW.

14. A torque of 650 Nm rotates a shaft of radius 0.125 m with ω = 50 rad/s. What is the transmitted power?
a) 22.5 kW
b) 32.5 kW
c) 42.5 kW
d) 52.5 kW
Answer: b
Clarification: V = ωr = 50 × 0.125 = 6.25 m/s
Power = Tω = 650 × 50 Nm/s = 32 500 W = 32.5 kW.

Thermodynamics for Interviews,

Thermodynamics Multiple Choice Questions on “Reversibility, Irreversibilty and causes of Irreversibilty”.

1. A reversible process is performed in such a way that
a) at the conclusion of process, both system and surroundings can be restored to their initial states without producing any change
b) it should not leave any trace to show that the process had ever occurred
c) it is carried out infinitely slowly
d) all of the mentioned
Answer: d
Clarification: These are some basic concepts of a reversible process.

2. A reversible process coincides with a quasi-static process.
a) true
b) false
Answer: a
Clarification: A reversible process is carried out very slowly and every state it passes through is an equilibrium state.

3. Irreversibility of a process may be due to
a) lack of equilibrium during the process
b) involvement of dissipative effects
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These two are the major causes of irreversibility.

4. A heat transfer process approaches reversibility as the temperature difference between two bodies approaches
a) infinity
b) zero
c) -1
d) 1
Answer: b
Clarification: For heat transfer to be reversible, heat must be transferred through an infinitesimal temperature difference.

5. All actual heat transfer processes are
a) irreversible
b) take place through a finite temperature difference
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: An infinitesimal temperature difference is not easy to attain.

6. Free expansion is irreversible.
a) true
b) false
Answer: a
Clarification: It can be demonstrated by the second law.

7. Which of the following can be a cause of irreversibility?
a) friction, viscosity
b) inelasticity
c) electrical resistance, magnetic hysteresis
d) all of the mentioned
Answer: d
Clarification: These effects are known as dissipative effects.

8. The continual motion of a movable device in the complete absence of friction is known as
a) PMM2
b) PMM3
c) PMM1
d) PMM0
Answer: b
Clarification: This is not possible since lubrication cannot be completely eliminated.

9. The friction present in moving devices makes a process reversible.
a) true
b) false
Answer: b
Clarification: Friction lakes the process irreversible.

10. Which of the following is irreversible?
a) stirring work
b) friction work in moving devices
c) current flowing through a wire
d) all of the mentioned
Answer: d
Clarification: All these processes includes a particular cause of irreversibility.

11. A process will be reversible if it has
a) no dissipative effects
b) dissipative effects
c) depends on the given conditions
d) none of the mentioned
Answer: a
Clarification: Without any dissipative effects, a process can perform in a reversible manner.

12. Irreversibility can be distinguished in how many types?
a) 0
b) 1
c) 2
d) 3
Answer: c
Clarification: Tow types of irreversibility are internal and external irreversibility.

13. Internal irreversibility is caused by
a) internal dissipative effects
b) friction, turbulence
c) electrical resistance, magnetic hysteresis
d) all of the mentioned
Answer: d
Clarification: Internal dissipative effects are the major cause of internal irreversibility.

14. The external irreversibility occurs at the system boundary.
a) true
b) false
Answer: a
Clarification: This mainly includes heat interaction with the surroundings due to a finite temperature gradient.

15. Which of the following is true?
a) mechanical irreversibility is due to finite pressure gradient
b) thermal irreversibility is due to finite temperature gradient
c) chemical irreversibility is due to finite concentration gradient
d) all of the mentioned
Answer: d
Clarification: These are some other distinctions of irreversibility.

Thermodynamics Multiple Choice Questions on “Boundary Work in a Multistep Process”.

1. A cylinder contains 1kg of ammonia. Initially the ammonia is at 180°C, 2 MPa and is now cooled to saturated vapour at 40°C, and then further cooled to 20°C, at which point the quality is 50%. Find the total work for the process, assuming a linear variation of P versus V.
a) -19.4 kJ
b) -29.4 kJ
c) -39.4 kJ
d) -49.4 kJ
Answer: d
Clarification: State 1: (T, P) v1 = 0.10571 m3/kg; State 2: (T, x) sat. vap. P2 = 1555 kPa, v2 = 0.08313 m³/kg
State 3: (T, x) P3 = 857 kPa, v3 = (0.001638 + 0.14922)/2 = 0.07543 m³/kg
Work = ⌠PdV = (P1 + P2)m(v2 – v1)/2 + (P2 + P3)m(v3 – v2)/2
= (2000 + 1555)1(0.08313 – 0.10571)/2 + (1555 + 857)1(0.07543 – 0.08313)/2
= -49.4 kJ.

2. A piston cylinder has 1.5 kg of air at 300 K, 150 kPa. It is now heated up in a two step process. First constant volume to 1000 K (state 2) then followed by a constant pressure process to 1500 K, state 3. Find the work in the process.
a) 205.3 kJ
b) 215.3 kJ
c) 225.3 kJ
d) 235.3 kJ
Answer: b
Clarification: 1 -> 2: Constant volume V2 = V1 and 2 -> 3: Constant pressure P3 = P2
State 1: T, P => V1 = mRT1/P1 = 1.5×0.287×300/150 = 0.861 m³
State 2: V2 = V1 => P2 = P1 (T2/T1) = 150×1000/300 = 500 kPa
State 3: P3 = P2 => V3 = V2 (T3/T2) = 0.861×1500/1000 = 1.2915 m³
Total work = P3(V3 – V2) = 500(1.2915 – 0.861) = 215.3 kJ.

3. A piston-cylinder assembly has 1kg of R-134a at state 1 with 600 kPa, 110°C, and is then brought to saturated vapour, state 2, by cooling. The cooling continues to state 3 where the R-134a is saturated liquid. Find the work in each of the two steps, 1 to 2 and 2 to 3.
a) 0, -20.22 kJ
b) -20.22 kJ,0
c) 0, 0
d) -20.22 kJ, -20.22 kJ
Answer: a
Clarification: State 1: (T,P) => v = 0.04943 m3/kg; State 2: v2 = v1 and x2 = 1.0
v2 = v1 = vg = 0.04943 m3/kg => T = 10°C
State 3 reached at constant P (F = constant) v3 = vf = 0.000794 m³/kg
Since no volume change from 1 to 2 => 1W2 = 0
From 2 to 3: ∫P dV = P(V3 – V2) = mP(v3 – v2) = 415.8 (0.000794 – 0.04943)(1)
= -20.22 kJ.

4. R-22 is contained in a piston-cylinder, where the volume is 11 L when the piston hits the stops. The initial state is 150 kPa, −30°C with V=10 L. This system warms up to 15°C. Find the work done by R-22 during this process.
a) 0.35 kJ
b) 0.25 kJ
c) 0.15 kJ
d) 0.05 kJ
Answer: c
Clarification: Initially piston floats, V 3/kg; m = V/v = 0.010/0.1487 = 0.06725 kg
now, P1a = 150 kPa, v = V(stop)/m and v1a = V/m = 0.011/0.06725 = 0.16357 m3/kg
=> T1a = -9°C & T2 = 15°C
Since T2 > T1a then it follows that P2 > P1 and the piston is against stop.
Work = ∫ P(ext) dV = P9ext)(V2 – V1) = 150(0.011 – 0.010) = 0.15 kJ.

5. A piston-cylinder contains 50 kg of water at 200 kPa with V=0.1 m³. Stops in the cylinder restricts the enclosed volume to 0.5 m³. The water is now heated to 200°C. Find the work done by the water.
a) 50 kJ
b) 60 kJ
c) 70 kJ
d) 80 kJ
Answer: d
Clarification: Initially the piston floats so the equilibrium lift pressure is 200 kPa
1: 200 kPa, v1= 0.1/50 = 0.002 m³/kg, and 2: 200°C
v(stop) = 0.5/50 = 0.01 m³/kg;
State 2 two phase => P2 = Psat(T2) = 1.554 MPa, V2 = V(stop) = 0.5 m³
1W2 = 1W(stop) = 200 (0.5 – 0.1) = 80 kJ.

6. Ammonia in a piston/cylinder arrangement is at 80°C, 700 kPa. It is now cooled at constant pressure to saturated vapour (state 2) at which point the piston is locked with a pin. The cooling continues to −10°C (state 3). Find the work.
a) -28.64 kJ/kg
b) -38.64 kJ/kg
c) -48.64 kJ/kg
d) -58.64 kJ/kg
Answer: b
Clarification: 1W3 = 1W2 + 2W3 = ⌠PdV = P1(V2 – V1) = mP1(v2 – v1)
Since constant volume from 2 to 3; v1 = 0.2367 m³/kg, P1 = 700 kPa,
v2 = vg = 0.1815 m³/kg, 1w3 = P1(v2- v1) = 700 × (0.1815 – 0.2367)
= -38.64 kJ/kg.

7. A piston-cylinder contains 1 kg of liquid water at 300 kPa, 20°C. Initially the piston floats, with a maximum enclosed volume of 0.002 m³ if the piston touches the stops. Now heat is added so that the final pressure is 600 kPa. Find the work in the process.
a) 0.30 kJ
b) 0.40 kJ
c) 0.50 kJ
d) 0.60 kJ
Answer: a
Clarification: State 1: Compressed liquid v = vf(20) = 0.001002 m³/kg
v(stop) = 0.002 m3/kg , 300 kPa
v2 = v(stop) = 0.002 m³/kg and V = 0.002 m³
Work is done while piston moves at P(lift) = constant = 300 kPa
1W2 = ∫ P dV = m*P(lift)*(v2 -v1) = 1 × 300(0.002 − 0.001002) = 0.30 kJ.

8. 10 kg of water in a piston-cylinder exists as saturated liquid/vapour at 100 kPa, with a quality of 50%. It is now heated till the volume triples. The mass of the piston is such that a cylinder pressure of 200kPa will float it. Find the work given out by the water.
a) 3090 kJ
b) 3190 kJ
c) 3290 kJ
d) 3390 kJ
Answer: d
Clarification: Process: v = constant until P = P(lift) then P is constant.
State 1: v1 = vf + x vfg = 0.001043 + 0.5 × 1.69296 = 0.8475 m³/kg
State 2: v2, P2 ≤ P(lift) => v2 = 3 × 0.8475 = 2.5425 m³/kg;
T2 = 829°C ; V2 = m*v2 = 25.425 m³
Work = ∫ P dV = P(lift)×(V2 – V1) = 200 kPa × 10 kg × (2.5425 – 0.8475) m³/kg
= 3390 kJ.

9. Ammonia at 10°C with a mass of 10 kg is in a piston-cylinder arrangement with an initial volume of 1 m³. The piston initially resting on the stops has a mass such that a pressure of 900 kPa will float it. The ammonia is now slowly heated to 50°C. Find the work in the process.
a) 483.2 kJ
b) 583.2 kJ
c) 683.2 kJ
d) 783.2 kJ
Answer: b
Clarification: Process: V = constant unless P = P(float)
State 1: T = 10°C, v1 = V/m = 1/10 = 0.1 m³/kg;
also v(f) T1a then v2 > v1a
State 2: 50°C => 900 kPa which is superheated vapor hence
v2 = 0.1648 m³/kg, V2 = mv2 = 1.648 m³
Work = ∫ P dV = P(float) (V2 – V1) = 900 (1.648 – 1.0) = 583.2 kJ.

10. A piston-cylinder contains 0.1 kg saturated liquid and vapour water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 500 kPa will float it. The water is heated to 300°C. Find the work.thermodynamics-questions-answers-boundary-work-multistep-process-q10″ alt=”thermodynamics-questions-answers-boundary-work-multistep-process-q10″ width=”151″ height=”144″ class=”alignnone size-full wp-image-166447″ srcset=”/2017/06/thermodynamics-questions-answers-boundary-work-multistep-process-q10 151w, /2017/06/thermodynamics-questions-answers-boundary-work-multistep-process-q10-150×144 150w” sizes=”(max-width: 151px) 100vw, 151px”>
a) 2.91 kJ
b) 3.91 kJ
c) 4.91 kJ
d) 5.91 kJ
Answer: c
Clarification: Process: v = constant until P = P(lift)
To locate state 1: v1 = 0.001043 + 0.25×1.69296 = 0.42428 m³/kg
1a: v1a = v1 = 0.42428 m3/kg > vg at 500 kPa, so state 1a is Sup.Vapor T1a = 200°C
State 2 is 300°C so heating continues after state 1a to 2 at constant P
=> 2: T2, P2 = P(lift) => v2 =0.52256 m3/kg ; V2 = mv2 = 0.05226 m³
1W2 = P(lift)*(V2 – V1) = 500(0.05226 – 0.04243) = 4.91 kJ.

11. A constant pressure piston cylinder contains 0.2 kg water in the form of saturated vapour at 400 kPa. It is now cooled to occupy half the original volume. Find the work in the process.
a) -12.5 kJ
b) -24.5 kJ
c) -8.5 kJ
d) -18.5 kJ
Answer: d
Clarification: v1= 0.4625 m³/kg, V1 = mv1 = 0.0925 m³
v2 = v1/ 2 = 0.23125 m³/kg, V2 = V1 / 2 = 0.04625 m³
Process: P = C
W = ∫ PdV = P(V2-V1) = 400 kPa × (0.04625 – 0.0925) m³ = -18.5 kJ.

12. A piston cylinder contains air at 600 kPa, 290 K and volume of 0.01 m³. A constant pressure process gives out 54 kJ of work. Find the final temperature of the air.
a) 2700 K
b) 2800 K
c) 2900 K
d) 3000 K
Answer: c
Clarification: W = ∫ P dV = P∆V hence ∆V = W/P = 54/600 = 0.09m³
V2 = V1 + ∆V = 0.01 + 0.09 = 0.1 m³
Assuming ideal gas, PV = mRT,
T2 = P2*V2/(m*R) = [(P2*V2)/(P1*V1)]*T1 = T1*(V2/V1) = (0.1*290)/0.01
= 2900 K.

13. A piston/cylinder has 5m of liquid 20°C water on top of piston with cross-sectional area of 0.1 m². Air is let in under the piston that rises and pushes the water out. Find the necessary work to push all the water out.
a) 62.88 kJ
b) 52.88 kJ
c) 92.88 kJ
d) 42.88 kJ
Answer: a
Clarification: P1 = Po + ρgH = 101.32 + 997 × 9.807 × 5 / 1000 = 150.2 kPa
∆V = H × A = 5 × 0.1 = 0.5 m³
Work = Area = ∫ P dV = ½ (P1 + Po )(Vmax -V1)
= ½ (150.2 + 101.32) kPa × 0.5 m³
= 62.88 kJ.

14. A piston/cylinder contains 1 kg water at 20°C with volume 0.1 m³. While the water is heated to saturated vapour, the piston is not allowed to move. Find the final temperature.
a) 201.7°C
b) 211.7°C
c) 215.7°C
d) 221.7°C
Answer: b
Clarification: V2 = V1 = 0.1 m³
T2 = Tsat = 210 + 5[(0.1 – 0.10324)/(0.09361 – 0.10324)]
= 211.7°C.

15. A piston cylinder contains 3 kg of air at 20°C and 300 kPa. It is now heated at a constant pressure to 600 K. Find the work in the process.
a) 244.2 kJ
b) 254.2 kJ
c) 264.2 kJ
d) 274.2 kJ
Answer: c
Clarification: V1 = mR T1 / P1 = 3 × 0.287 × 293.15/300 = 0.8413 m³
V2 = mR T2 / P2 = 3 × 0.287 × 600/300 = 1.722 m³
work = ⌠ PdV = P (V2 – V1) = 300 (1.722 – 0.8413) = 264.2 kJ.

Thermodynamics Multiple Choice Questions on “Steam Tables “.

1. The properties of water are arranged in the steam tables as functions of
a) pressure
b) temperature
c) pressure and temperature
d) none of the mentioned
Answer: c
Clarification: The properties of water are arranged in steam tables as functions of both pressure and temperature.

2. The internal energy of saturated water at the triple point is
a) 1
b) 0
c) -1
d) infinity
Answer: b
Clarification: This value is arbitrarily chosen.

3. The entropy of saturated water is chosen to be zero at triple point.
a) true
b) false
Answer: a
Clarification: This is arbitrarily chosen and form the basic assumptions for steam tables.

4. When a liquid and its vapour are in equilibrium at a certain pressure and temperature, then which of the following is required to identify the saturation state.
a) pressure
b) temperature
c) both pressure and temperature
d) pressure or temperature
Answer: d
Clarification: If one of the quantity is given, then other gets fixed.

5. Saturated liquid or the saturated vapour has how many independent variables?
a) one
b) two
c) three
d) none of the mentioned
Answer: a
Clarification: Only one property is required to be known to fix up the state.

6. If data are required for intermediate temperatures or pressures, linear interpolation is normally accurate.
a) true
b) false
Answer: a
Clarification: To reduce the amount of interpolation required, two tables are provided.

7. For a liquid-vapour mixture, which of the following can give us all the properties of the mixture?
a) p or t and the quality of the mixture are given
b) p or t and any one of the property is given
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: In first case, properties can be directly evaluated and in second case we can find the quality first and then evaluate all other properties.

8. When does a vapour become superheated?
a) when the temperature of vapour is less than the saturation temperature at given pressure
b) when the temperature of vapour is more than the saturation temperature at given pressure
c) when the temperature of vapour is equal to the saturation temperature at given pressure
d) none of the mentioned
Answer: b
Clarification: For a superheated vapour, temperature of vapour must be greater than the saturation temperature.

9. The superheat or degree of superheat is given by
a) difference between the temperature of saturated liquid and saturation temperature
b) difference between the temperature of superheated vapour and saturation temperature
c) sum of the temperature of superheated vapour and saturation temperature
d) none of the mentioned
Answer: b
Clarification: Superheat= T1(temperature of superheated vapour) – T(saturated).

10. When the temperature of a liquid is less than the saturation temperature at the given pressure, the liquid is called compressed liquid.
a) true
b) false
Answer: a
Clarification: For a compressed liquid, temperature of liquid must be less than the saturation temperature.

11. The properties of liquid _____ with pressure.
a) do not vary
b) vary largely
c) vary little
d) none of the mentioned
Answer: c
Clarification: This is the reason why properties are taken from the saturation tables at the temperature of the compressed liquid.

12. Which of the following statement is true?
a) a subcooled liquid is one which is cooled below its saturation temperature at a certain pressure
b) subcooling is the difference between the saturation temperature and the actual liquid temperature
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is what a subcooled liquid means.