Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Vapor Compression Refrigeration Cycle-2”.

1. Which of the following statement is true for a condenser?
a) it can be air-cooled or water-cooled
b) small self-contained units use water-cooled condenser
c) large installations use air-cooled condenser
d) all of the mentioned
Answer: a
Clarification: We use air-cooled condenser for small self-contained units and water-cooled condenser for large installations.

2. For an expansion device, which of the following is true?
a) it increases the pressure of refrigerant
b) it regulates the flow of refrigerant to evaporator
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: The expansion device reduces the pressure of refrigerant.

3. Which of the following is a type of expansion device?
a) capillary tubes
b) throttle valves
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These are the two types of expansion devices.

4. Throttle valves are used in ____
a) small units
b) larger units
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: They regulate the flow of refrigerant according to load on evaporator.

5. Capillary tubes are used in ____
a) small units
b) larger units
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: In their case, id size and length are fixed, the evaporator pressure also gets fixed.

6. Types of compressor include
a) reciprocating
b) centrifugal
c) rotary
d) all of the mentioned
Answer: d
Clarification: These are the three types of compressor.

7. When volume flow rate of refrigerant is large, which compressor is used?
a) reciprocating
b) centrifugal
c) rotary
d) all of the mentioned
Answer: b
Clarification: For plants with higher capacities, centrifugal compressors are used.

8. Which of the following statement is true?
a) rotary compressors are mostly used for small units
b) reciprocating compressors are employed in plants with capacity up to 100 tonnes
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is the difference in reciprocating and rotary compressors.

9. In reciprocating compressors, actual volume of gas drawn in cylinder is ____ the volume displaced by piston.
a) less than
b) more than
c) equal to
d) none of the mentioned
Answer: a
Clarification: The reason being, leakage, clearance and throttling effects.

10. The clearance volumetric efficiency is equal to
a) 1 + C + C(p2/p1)^(1/n)
b) 1 – C – C(p2/p1)^(1/n)
c) 1 – C + C(p2/p1)^(1/n)
d) 1 + C – C(p2/p1)^(1/n)
Answer: d
Clarification: Here C is the clearance and p1,p2 are pressures.

11. Plate evaporator is a common type of evaporator.
a) true
b) false
Answer: a
Clarification: In a plate evaporator, a coil is brazed on to a plate.

12. Why is multistage compression with intercooling adopted?
a) using a single stage with high pressure ratio decreases volumetric efficiency
b) high pressure ratio with dry compression gives high compressor discharge temperature
c) the refrigerant is damaged
d) all of the mentioned
Answer: d
Clarification: Because of these reasons we have to use multistage compression with intercooling.

13. The intercooler pressure is given by
a) p1*p2
b) sqrt(p1*p2)
c) (p1*p2)/(p1+p2)
d) (p1+p2)/2
Answer: b
Clarification: Here p1 is the evaporator pressure and p2 is the condenser pressure.

14. The most widely used refrigerants are
a) freon
b) genetron
c) arcton
d) all of the mentioned
Answer: d
Clarification: These are a group of halogenated hydrocarbons.

15. Why is ammonia used in food refrigeration?
a) high COP
b) low cost
c) lower energy cost
d) all of the mentioned
Answer: d
Clarification: Also ammonia can be detected easily in case of a leak.

Thermodynamics Questions and Answers for freshers on “Specific Heat at Constant Volume and Pressure and Control Volume”.

1. The specific heat of a substance at constant volume is defined as the rate of change of ___ with respect to ___
a) specific internal energy, temperature
b) work, pressure
c) specific internal energy, pressure
d) heat, temperature
Answer: a
Clarification: cv=∂u/∂T at constant volume.

2. Heat transferred at constant _____ increases the _____ of a system.
a) pressure, increases
b) volume, increases
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: At constant pressure, (dQ)=dh and at constant volume, Q=Δu.

3. Specific heat of a substance at constant volume is a property of the system.
a) true
b) false
Answer: a
Clarification: Since T,v and u are the properties of the system, specific heat at a constant volume is a property of the system.

4. The specific heat of a substance at constant pressure is defined as the rate of change of ___ with respect to ___
a) work, pressure
b) enthalpy, temperature
c) enthalpy, pressure
d) heat, temperature
Answer: b
Clarification: cp=∂h/∂T at constant pressure.

5. The heat capacity at constant pressure Cp
a) m/cp
b) cp/m
c) mcp
d) none of the mentioned
Answer: c
Clarification: Cp=(mass*specific heat at constant pressure).

6. Specific heat of a substance at constant pressure is a property of the system.
a) true
b) false
Answer: a
Clarification: cp is a property of a substance just like cv.

7. When there is mass transfer across the system boundary, the system is called
a) isolated system
b) closed system
c) open system
d) none of the mentioned
Answer: c
Clarification: Basic definition of an open system.

8. If a certain mass of steam is considered as the thermodynamic system, then the energy equation becomes
a) Q=ΔKE + ΔPE – ΔU + W
b) Q=ΔKE + ΔPE – ΔU – W
c) Q=-ΔKE – ΔPE + ΔU + W
d) Q=ΔKE + ΔPE + ΔU + W
Answer: d
Clarification: Q=ΔE + W and E=KE + PE + U.

9. The surface of the control volume is known as the control surface.
a) true
b) false
Answer: a
Clarification: This is same as the system boundary of the open system.

10. Steady flow means that the rates of flow of mass and energy across the control surface
a) varies
b) remains constant
c) depends on the control surface
d) none of the mentioned
Answer: b
Clarification: In a steady flow rate of flow remains constant.

Thermodynamics for Freshers,

Thermodynamics Interview Questions and Answers for experienced on “Entropy Principle and its Applications-2”

1. For the flow of electric current through a resistor,
a) at steady state, internal energy of resistor is constant
b) at steady state, temperature of resistor is constant
c) W=Q
d) all of the mentioned
Answer: d
Clarification: Internal energy is dependent on temperature and by first law Q=ΔE+W.

2. When stirring work is supplied to a viscous thermally insulated liquid, temperature of the liquid
a) remains constant
b) increases
c) decreases
d) none of the mentioned
Answer: d

3. A car uses power of 25 hp for a one hour in a round trip. A thermal efficiency of 35% can be assumed? Find the change in entropy if we assume ambient at 20°C?
a) 554.1 kJ/K
b) 654.1 kJ/K
c) 754.1 kJ/K
d) 854.1 kJ/K
Answer: b
Clarification: E = ⌠ W dt = 25 hp × 0.7457 (kW/hp) × 3600 s = 67 113 kJ = η Q
Q = E / η = 67 113 / 0.35 = 191 751 kJ
∆S = Q / T = 191 751 / 293.15 = 654.1 kJ/K.

4. In a Carnot engine working on ammonia, the high temperature is 60°C and as QH is received, the ammonia changes from saturated liquid to saturated vapor. The ammonia pressure at low temperature is 190 kPa. Find the entropy. thermodynamics-interview-questions-answers-experienced-q4″ alt=”thermodynamics-interview-questions-answers-experienced-q4″ width=”135″ height=”124″ class=”alignnone size-full wp-image-166423″>
a) 4.6577 kJ/kg K
b) 5.6577 kJ/kg K
c) 6.6577 kJ/kg K
d) 7.6577 kJ/kg K
Answer: a
Clarification: qH = ∫ Tds = T (s2 – s1) = T s(fg) = h2 – h1 = h(fg) = 997.0 kJ/kg
TL = T3 = T4 = Tsat(P) = –20°C
η(cycle) = 1 – (Tl/Th) = 1 – (253.2/333.2) = 0.24
s3 = s2 = sg(60°C) = 4.6577 kJ/kg K.

5. A slab of concrete, 5 × 8 × 0.3 m, is used as a thermal storage mass in a house. The slab cools overnight from 23°C to 18°C in an 18°C house, find the net entropy change associated with this process?
a) 0.4 kJ/K
b) 1.4 kJ/K
c) 2.4 kJ/K
d) 3.4 kJ/K
Answer: d
Clarification: V = 5 × 8 × 0.3 = 12 m^3; m = ρV = 2200 × 12 = 26400 kg
V = constant so 1W2 = 0; 1Q2 = mC∆T = 26400 × 0.88(-5) = -116160 kJ
∆S(SYST) = m(s2 – s1) = mC ln(T2/T1) = 26400 × 0.88 ln (291.2/296.2) = -395.5 kJ/K
∆S(SURR) = -1Q2/T0 = +116160/291.2 = +398.9 kJ/K
∆S(NET) = -395.5 + 398.9 = +3.4 kJ/K.

6. A foundry form box with 25 kg of 200°C hot sand is dropped into a bucket with 50 L water at 15°C. Assuming there is no heat transfer with the surroundings and no boiling away of water, calculate the net entropy change for the process.
a) 2.37 kJ/K
b) 2.47 kJ/K
c) 2.57 kJ/K
d) 2.67 kJ/K
Answer: c
Clarification: C.V. Sand and water, constant pressure process
m(sand)∆h(sand) + m(H2O)∆h(H2O) = 0
m(sand)C∆T(sand) + m(H2O)C(H2O)∆T(H2O) = 0
25 × 0.8×(T2 – 200) + (50×10^(-3)/0.001001) × 4.184 × (T2 – 15) = 0
hence T2 = 31.2°C
∆S = 25 × 0.8 ln(304.3/473.15) + 49.95 × 4.184 ln(304.3/288.15)
= 2.57 kJ/K.

7. Calculate the change in entropy if 1 kg of saturated liquid at 30°C is converted into superheated steam at 1 bar and 200°C .
a) 5.3973 kJ/K
b) 6.3973 kJ/K
c) 7.3973 kJ/K
d) none of the mentioned
Answer: c
Clarification: si= sf @30 C = 0.4369 kJ/kg.K,
se = sg @1 bar and 200 C = 7.8342 kJ/kg.K
Change in entropy (∆S) = m*( se – si) = 1*(7.8342 – 0.4369)
= 7.3973 kJ/K.

8. Two kilograms of water at 120°C with a quality of 25% has its temperature raised by 20°C in a constant volume process. What is the new specific entropy?
a) 3.01517 kJ/kg.K
b) 4.01517 kJ/kg.K
c) 5.01517 kJ/kg.K
d) 7.01517 kJ/kg.K
Answer: b
Clarification: v1 = vf @120 C + x1*vfg @120 C = 0.00106 + 0.25*0.8908 = 0.22376 m3/kg
v2 = v1 = vf @145 C + x2*vfg @145 C = 0.00108 + x2*0.50777 ∴ x2 = 0.4385
New specific entropy (s2) = sf @145 C + x2*sfg @145 C
= 1.739 +0.4385*5.1908 = 4.01517 kJ/kg.K.

9. A thermal reservoir at 538°C is brought into thermal communication with another thermal reservoir at 260°C, and as a result 1055 kJ of heat is transferred only from the higher to lower temperature reservoir. Determine the change in entropy of the universe due to the exchange of heat between these two thermal reservoirs.
a) 0.378182 kJ/K
b) 0.478182 kJ/K
c) 0.578182 kJ/K
d) 0.678182 kJ/K
Answer: d
Clarification: (∆S)System = ∫δQ/T = –1055/(538 + 273.15) + 1055/(260 + 273.15)= 0.678182 kJ/K
(∆S)Surroundings = ∫δQ/T = 0
Change in entropy of the universe ((∆S)Universe)
= (∆S)System + (∆S)Surroundings = 0.678182 kJ/K.

10. A glass jar is filled with saturated water at 500 kPa of quality 25%, and a tight lid is put on. Now it is cooled to -10°C. What is the mass fraction of solid at this temperature?
a) 99.98%
b) 98.98%
c) 93.98%
d) 95.98%
Answer: a
Clarification: Constant volume v1=v2=V/m
from steam table, Psat = 500 kPa and hence Tsat = 151.8°C
v1 = 0.001093 + 0.25*0.3738 = 0.094543
v2 = 0.0010891 + x2*466.756 = v1 = 0.094543
x2 = 0.002 mass fraction vapour
x(solid) = 1- x2 = 0.9998 or 99.98%.

Thermodynamics for interviews,

Thermodynamics question bank on “Irreversibility and Gouy-Stondola Theorem”.

1. The actual work done by a system is always ____ than the reversible work, and the difference between the two is called ____ of the process.
a) more, irreversibility
b) less, irreversibility
c) more, reversibility
d) less, reversible
Answer: b
Clarification: Irreversibility=Maximum work – Actual work.

2. Irreversibility(I) is also called
a) degradation
b) dissipation
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These are other names of irreversibility.

3. For a non-flow process between equilibrium states, when the system exchanges heat only with the environment
a) I=0
b) I>0
c) I<0
d) I>=0
Answer: d
Clarification: I>0 for all processes bit for a reversible process, I=0.

4. For irreversibility, same expression applies to both flow and non-flow processes.
a) true
b) false
Answer: a
Clarification: I=To*(sum of change in entropy of the system and surroundings).

5. The quantity [To*(ΔSsystem + ΔSsurroundings)] represents an increase in
a) available energy
b) unavailable energy
c) exergy
d) all of the mentioned
Answer: b
Clarification: The quantity [To*(ΔSsystem + ΔSsurroundings)] represents an increase in unavailable energy.

6. Which of the following is true?
a) rate of loss of exergy does not depend on the rate of entropy generation
b) rate of loss of exergy is inversely proportional to the rate of entropy generation
c) rate of loss of exergy is directly proportional to the rate of entropy generation
d) none of the mentioned
Answer: c
Clarification: This comes from the Gouy-Stondola theorem.

7. A thermodynamically efficient process would involve ____ exergy loss with ____ rate of entropy generation.
a) minimum, minimum
b) maximum, maximum
c) minimum, maximum
d) maximum, minimum
Answer: a
Clarification: This is because rate of loss of exergy is directly proportional to the rate of entropy generation.

8. Heat transfer through a finite temperature difference is equivalent to the destruction of its exergy.
a) true
b) false
Answer: a
Clarification: When heat transfers through a final temperature difference, all of its exergy is lost.

9. The decrease in availability or lost work is proportional to
a) pressure drop
b) mass flow rate
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Lost work = (mass flow rate)*R*To*Δp/p1.

10. Entropy generation number can be given by
a) (rate of entropy generation)*(mass flow rate/specific heat)
b) (rate of entropy generation)/(mass flow rate/specific heat)
c) (rate of entropy generation)*(mass flow rate*specific heat)
d) (rate of entropy generation)/(mass flow rate*specific heat)
Answer: d
Clarification: Entropy generation number is a dimensionless quantity and given by above formula.

11. If two streams with equal temperature are mixing, then the entropy generation number becomes
a) 0
b) 1
c) -1
d) infinity
Answer: a
Clarification: Putting T1=T2 in the relation for entropy generation number, we get the value as zero.

11. A flow of air at 1000 kPa, 300 K is throttled to 500 kPa. What is the irreversibility?
a) 39.6 kJ/kg
b) 49.6 kJ/kg
c) 59.6 kJ/kg
d) 69.6 kJ/kg
Answer: c
Clarification: A throttle process is constant enthalpy if we neglect kinetic energies.
Process: he = hi, so ideal gas => Te = Ti
se – si = s(gen), s(gen) = 0 – R*ln(Pe/Pi)
s(gen) = – 0.287 ln (500 / 1000) = 0.2 kJ/kg K
i = (T0)*s(gen) = 298*0.2 = 59.6 kJ/kg.

12. A heat exchanger increases the availability of 3 kg/s water by 1650 kJ/kg by using 10 kg/s air which comes in at 1400 K and leaves with 600 kJ/kg less availability. What is the irreversibility?
a) 1020 kW
b) 1030 kW
c) 1040 kW
d) 1050 kW
Answer: d
Clarification: The irreversibility is the destruction of exergy (availability) so
I = Φ(destruction) = Φ(in) – Φ(out) = 10 × 600 – 3 × 1650 = 1050 kW.

13. A 2-kg piece of iron is heated from temperature 25°C to 400°C by a heat source which is at 600°C. What is the irreversibility in the process?
a) 96.4 kJ
b) 86.4 kJ
c) 76.4 kJ
d) 66.4 kJ
Answer: a
Clarification: 1Q2 = m(h2 – h1) = mC(T2 – T1) = 2 × 0.42 × (400 – 25) = 315 kJ
S(gen) = m(s2 – s1) – 1Q2/[email protected] = mC ln (T2/T1) – 1Q2/[email protected]
= 2 × 0.42 × ln (673.15/298.15) – (315/873.15) = 0.3233 kJ/K
I = To (S gen ) = 298.15 × 0.3233 = 96.4 kJ.