Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Absorption Refrigeration Cycle”.

1. In absorption refrigeration cycle, which of the following is used?
a) refrigerant
b) absorbent
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Both refrigerant and absorbent are used in absorption refrigeration cycle.

2. In absorption system, compressor in vapour compression cycle is replaced by absorber-generator assembly.
a) true
b) false
Answer: a
Clarification: The absorber-generator assembly involves less mechanical work.

3. In the aqua-ammonia absorption system,
a) water is the refrigerant and ammonia is the absorbent
b) ammonia is the refrigerant and water is the absorbent
c) both ammonia and water can be used as refrigerant or absorbent
d) none of the mentioned
Answer: b
Clarification: This is the basic absorption refrigeration cycle.

4. Which of the following statement is true?
a) ammonia vapour is absorbed in water
b) boiling point of ammonia is more than that of water
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: The boiling point of ammonia is less than that of water.

5. why is an analyser-rectifier combination is used in absorption refrigeration cycle?
a) to increase the amount of water vapour in ammonia vapour
b) to decrease the amount of water vapour in ammonia vapour
c) to eliminate the water vapour from ammonia vapour
d) all of the mentioned
Answer: c
Clarification: This is done to prevent the blocking of expansion valve because of frozen ice.

6. Which of the following condenses first?
a) ammonia vapour
b) water vapour
c) both condense at same temperature
d) none of the mentioned
Answer: b
Clarification: The saturation temperature of water is higher than ammonia at any given pressure.

7. The vapour going to condenser is ____ in temperature and ____ in ammonia.
a) higher, less
b) higher, richer
c) lower, less
d) lower, richer
Answer: d
Clarification: When passing through analyser, the vapour is cooled and enriched by ammonia.

8. Lithium bromide-water vapour is another absorption refrigeration system.
a) true
b) false
Answer: a
Clarification: In this, water is the refrigerant and solution of lithium bromide in water is the absorbent.

9. Water is used as a ____ in air conditioning units.
a) absorbent
b) refrigerant
c) absorbent and refrigerant
d) none of the mentioned
Answer: b
Clarification: The reason is that water cannot be cooled below 0 degree Celsius.

10. The COP of absorption refrigeration system is
a) low
b) high
c) equal to that of vapour compression refrigeration system
d) none of the mentioned
Answer: a
Clarification: The Cop of absorption refrigeration system is low.

Thermodynamics Multiple Choice Questions on “Examples of Steady Flow Processes”.

1. What does a nozzle do?
a) decreases the velocity of a fluid at the cost of its pressure gain
b) increases the velocity of a fluid at the cost of its pressure drop
c) increases the velocity of a fluid and also its pressure
d) none of the mentioned.
Answer: b
Clarification: A nozzle increases KE of fluid and reduces its pressure.

2. What does a diffuser do?
a) increases the pressure of the fluid at the expense of its KE
b) decreases the pressure of the fluid and also increases its KE
c) increases the pressure of the fluid and also its KE
d) decreases the pressure of the fluid and also its KE
Answer: a
Clarification: A diffuser increases the pressure at the expense of its KE.

3. For an insulated nozzle, SFEE of the control surface gives ( considering change in PE is zero and inlet velocity is small compared to exit velocity)
a) V2=sqrt(4*Δh)
b) V2=sqrt(Δh)
c) V2=sqrt(Δh/2)
d) V2=sqrt(2*Δh)
Answer: d
Clarification: dQ/dm=0, dW/dm=0, Δh=h1-h2.

4. Fluid flow through which of the following throttles the flow?
a) partially opened valve
b) orifice
c) porous plug
d) all of the mentioned
Answer: d
Clarification: In all of the given cases, there is an appreciable drop in pressure and hence the flow is throttled.

5. In a throttling device, what do we get as SFEE when changes in PE and KE are taken zero?
a) dQ/dm≠0
b) dW/dm≠0
c) h1=h2
d) none of the mentioned
Answer: c
Clarification: Enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling.

6. Turbines and engines ____ positive power output, and compressors and pumps ____ power input.
a) require, give
b) give, require
c) give, give
d) require, require
Answer: b
Clarification: This is the basic information about turbines, engines, compressors and pumps.

7. For a turbine, it is seen that work is done by the fluid at the expense of its enthalpy.
a) true
b) false
Answer: a
Clarification: For a turbine, W/m=h1-h2.

8. For an adiabatic compressor or pump,
a) the enthalpy of fluid remains constant with the amount of work input
b) the enthalpy of fluid decreases by the amount of work input
c) the enthalpy of fluid increases by the amount of work input
d) none of the mentioned
Answer: c
Clarification: For an adiabatic pump or compressor, W/m=h2-h1.

9. A heat exchanger is a device in which heat is transferred from one fluid to another.
a) true
b) false
Answer: a
Clarification: Basic fact about heat exchanger.

10. For an inviscid frictionless fluid flowing through a pipe, Euler equation is given by
a) Vdp+VdV+gdZ=0
b) Vdp-VdV+gdZ=0
c) Vdp-VdV-gdZ=0
d) none of the mentioned
Answer: a
Clarification: Euler equation is derived from steady flow energy equation.

11. The Bernoulli equation is restricted to _____ fluids but the SFEE is valid for _____ fluids as well.
a) viscous compressible, frictionless incompressible
b) frictionless incompressible, viscous compressible
c) viscous incompressible, frictionless compressible
d) none of the mentioned
Answer: b
Clarification: This statement tells us that the Bernoulli equation is a limiting case of SFEE.

Thermodynamics Multiple Choice Questions & Answers (MCQs) on “Entropy Transfer Mechanisms”.

1. Entropy can be transferred to or from a system in which of the following forms?
a) heat transfer
b) mass flow
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Entropy is transferred by these two forms while energy id transferred by work also.

2. Entropy transfer for an adiabatic transfer is zero.
a) true
b) false
Answer: a
Clarification: The only form of entropy interaction associated with a fixed mass or closed system is heat transfer.

3. If heat Q flows reversibly from the system to the surroundings at To,
a) entropy increase of the surroundings is Q/To
b) entropy of the system is reduced by Q/To
c) system has lost entropy to the surroundings
d) all of the mentioned
Answer: d
Clarification: We can say that there is entropy transfer from the system to the surroundings along with heat flow.

4. The sign of entropy transfer is opposite to the sign of heat transfer.
a) true
b) false
Answer: b
Clarification: The sign of entropy transfer is same as the sign of heat transfer: positive, if into the system, and negative, if out of the system.

5. ____ is exchanged during work interaction, whereas both ____ and ____ are exchanged during heat transfer.
a) energy, energy and entropy
b) entropy, energy and entropy
c) mass, energy and entropy
d) none of the mentioned
Answer: a
Clarification: This is the distinction between heat transfer and work which is brought about by the second law.

6. Mass contains
a) entropy
b) energy
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is a basic fact and the entropy and energy of a system are proportional to the mass.

7. The entropy of a system ____ by ____ when the mass of amount m enters it.
a) decreases, ms
b) increases, ms
c) decreases, s/m
d) increases, s/m
Answer: b
Clarification: When mass m enters a system, an entropy of amount ms, s being the specific entropy, accompanies it.

8. What happens when heat is added to the system?
a) dQ is positive
b) dS=dQ/T
c) entropy of the system increases
d) all of the mentioned
Answer: d
Clarification: dS=dQ/T and when heat is added, dQ=positive and thus dS=positive.

9. The first law of thermodynamics makes no distinction between heat transfer and work.
a) true
b) false
Answer: a
Clarification: The first law of thermodynamics considers both work and heat transfer equal.

10. Which of the following explains that there is no entropy transfer associated with work.
a) working of flywheel
b) compression of spring
c) raising of weight by a certain height
d) all of the mentioned
Answer: d
Clarification: In all these examples, there is work done but there is no entropy transfer.

Thermodynamics Questions & Answers for entrance exams on “Availability or Energy Balance”.

1. Which of the following clearly defines availability or exergy?
a) it is the maximum useful work obtainable from a system as it reaches the dead state
b) it is the minimum work required to bring the closed system from the dead state to the given state
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: The given statements clearly explain exergy.

2. The maximum work or exergy cannot be negative.
a) true
b) false
Answer: a
Clarification: This is because any change in state of the closed system to the dead state can be accomplished with zero work.

3. Energy is ____ conserved and exergy is ____ conserved.
a) always, generally
b) always, not generally
c) not always, always
d) always, always
Answer: b
Clarification: Exergy is destroyed due to irreversibilities.

4. When the closed system is allowed to undergo a spontaneous change from a given state to a dead state, its exergy is ____ destroyed ____ producing useful work.
a) not completely, though
b) not completely, without
c) completely, though
d) completely, without
Answer: d
Clarification: The potential to develop work which was originally present is completely wasted in such a spontaneous process.

5. The difference in exergy entering a system and that leaving out is the exergy which is destroyed.
a) true
b) false
Answer: a
Clarification: This is because exergy is not conserved.

6. The exergy of an isolated system can ____
a) increase
b) decrease
c) never increase
d) never decrease
Answer: c
Clarification: It is the counterpart of the entropy principle which states that the entropy of an isolated system can never decrease.

7. Since irreversibility > 0, the only processes allowed by the second law are those for which the exergy of the isolated system
a) increases
b) decreases
c) remains constant
d) none of the mentioned
Answer: b
Clarification: The above statement also means that the exergy of an isolated system can never increase.

8. For a closed system, availability or exergy transfer occurs through
a) heat interactions
b) work interactions
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These are the two ways in which exergy transfer can take place.

9. For an isolated system, the exergy balance gives
a) ΔA=-I
b) ΔA=I
c) ΔA=0
d) none of the mentioned
Answer: a
Clarification: Change in availability = – irreversibility.

10. Which law is used for exergy balance?
a) first law
b) second law
c) first law and second law
d) third law
Answer: c
Clarification: Both first and second laws are used to balance exergy.

11. Evaluate the steady state exergy flux due to a heat transfer of 250 W through a wall with 400 K on one side and 600 K on the other side. Also find the exergy destruction in the wall. thermodynamics-questions-answers-entrance-exams-q11″ alt=”thermodynamics-questions-answers-entrance-exams-q11″ width=”206″ height=”211″ class=”alignnone size-full wp-image-166461″>
a) 52 W
b) 62 W
c) 72 W
d) 82 W
Answer: b
Clarification: Φ(Q) = [1 – (T0/T)]Q
Φ1(Q) = [1 – (T0/T1)]Q = [1-(298/600)](250) = 125.8 W
Φ2(Q) = [1 – (T0/T2)]Q = [1-(298/400)](250) = 63.8 W
Φ(destruction) = Φ1 – Φ2 = 125.8 – 63.8 = 62 W.

12. A constant pressure piston/cylinder contains 2 kg of water at 5 MPa and 100°C. Heat is added from a reservoir at 700°C to the water until it reaches 700°C. Find the total irreversibility in the process.
a) 1572 kJ
b) 1672 kJ
c) 1772 kJ
d) 1872 kJ
Answer: a
Clarification: This process is : P = C hence 1W2 = P(V2 – V1)
1Q2 = m(u2 – u1) + 1W2 = m(h2 – h1)
= 2(3900.13 – 422.71) = 6954.8 kJ
1S2 gen = m(s2 – s1) – 1Q2/T
= 2(7.5122 – 1.303) – 6954.8273 + 700 = 5.2717 kJ/K
irreversibility = m 1i2 = T0 1S2 gen
= 298.15 K × 5.2717 kJ/K
= 1572 kJ.

13. A flow of air at 1000 kPa, 300 K is throttled to 500 kPa. What is the irreversibility?
a) 47.63 kJ/kg
b) 57.63 kJ/kg
c) 67.63 kJ/kg
d) 77.63 kJ/kg
Answer: b
Clarification: A throttle process is a constant enthalpy process
Process: he = hi so ideal gas => Te = Ti
Entropy Eq.: se – si = sgen
sgen = – 0.287 ln (500/1000) = 0.2 kJ/kg K
i = To*s = 288.15 X 0.2 = 57.63 kJ/kg.

Thermodynamics Multiple Choice Questions on “Law of Corresponding States”.

1. For a gas, the compressibility factor Z depends on
a) pressure and volume
b) pressure and temperature
c) volume and temperature
d) pressure, volume and temperature
Answer: b
Clarification: For a particular gas, Z depends on pressure and temperature.

2. We can use Z instead of directly plotting v.
a) true
b) false
Answer: a
Clarification: The main advantage of using Z instead of v is a smaller range of values in plotting.

3. How could we use one compressibility factor chart for all the substances?
a) it would be more convenient
b) the general shapes of the vapour dome and of the constant temperature lines on the p-v plane can be similar for all substances
c) their similarity can be exploited using dimensionless properties
d) all of the mentioned
Answer: d
Clarification: These dimensionless properties called reduced properties can be used for all the substances.

4. Which of the following property is used as the dimensionless property?
a) reduced pressure
b) reduced volume
c) reduced temperature
d) all of the mentioned
Answer: d
Clarification: The reduced pressure is the ratio of existing pressure to the critical pressure of the substance and like this all other properties are written.

5. Which of the following statement is true?
a) the specific volumes of different gases at same pressure and temperature are different
b) the reduced volumes of different gases at same reduced pressure and reduced temperature are same
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is found from the experimental data and because of this reduced properties are used.

6. Value of critical compressibility factor Zc is taken to be constant.
a) true
b) false
Answer: a
Clarification: The experimental values of Zc fall within a narrow range of 0.20-0.30, hence it is taken constant.

7. The generalized compressibility chart is a plot in
a) when reduced pressure is plotted as a function of reduced temperature and Z
b) when reduced temperature is plotted as a function of reduced pressure and Z
c) when Z is plotted as a function of reduced pressure and reduced temperature
d) none of the mentioned
Answer: b
Clarification: The generalized compressibility chart is found to be satisfactory for a great variety of substances.

8. The law of corresponding states is a relation among
a) reduced pressure and reduced temperature
b) reduced volume and reduced temperature
c) reduced volume and reduced pressure
d) reduced pressure and reduced temperature and reduced volume
Answer: d
Clarification: This relation can be derived from the equations of state.

9. The corrected gas equation is a cubic in v, which of the following statement is true about its roots?
a) at low temperature, three positive real roots exists
b) at critical temperature, the three real roots become equal
c) at high temperature, only one real root exists
d) all of the mentioned
Answer: d
Clarification: The equation has three roots which are different for different temperature and pressure.

10. The value of compressibility factor at the critical state for a van der Waals gas is
a) 0.325
b) 0.350
c) 0.375
d) 0.400
Answer: c
Clarification: This value comes from the relation, R=(8/3)(p*v/T) where p,v,T are critical values.

11. At very low pressures Z approaches
a) zero
b) unity
c) 0.50
d) infinity
Answer: b
Clarification: At very low pressures, a real gas approaches the ideal gas behaviour.

12. Which of the following statement is true about the reduced equation of state?
a) the individual coefficients a,b,R for a particular gas have disappeared
b) it reduces the properties of all gases to one formula
c) it tells us that to what extent a real gas obeys van der Waals equation
d) all of the mentioned
Answer: d
Clarification: These are the properties of the reduced equation of state.

13. The Boyle temperature is given by
a) a/(b*R)
b) b/(a*R)
c) R/(a*b)
d) 1/(a*b*R)
Answer: a
Clarification: It is obtained when the value of B=b-a/(RT) is zero, i.e., B=0.

Thermodynamics Multiple Choice Questions on “Reheat-Regenerative Cycle and Feedwater Heaters”.

1. The reheating of steam is used when the vaporization pressure is ____.
a) low
b) high
c) both when low or high
d) always
Answer: b
Clarification: When vaporization pressure is high, the reheating of steam is adopted.

2. Why both reheating and regeneration is used together?
a) the effect of reheat alone on efficiency is very small
b) regeneration has a marked effect on efficiency
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Thus a modern steam power plant has both reheating and regeneration.

3. How many types of feedwater heaters are present?
a) one
b) two
c) three
d) four
Answer: b
Clarification: The two types are open heaters and closed heaters.

4. Which of the following statement is true?
a) open heater is also known as contact-type heater
b) in an open type heater the extracted or bled steam is allowed to mix with the feedwater
c) in a closed heater, the fluids are not allowed to mix together
d) all of the mentioned
Answer: d
Clarification: These are the details of open and closed type heater.

5. The temperature of feedwater leaving a heater is ____ the saturation temperature at steam extraction pressure.
a) less than
b) equal to
c) more than
d) none of the mentioned
Answer: a
Clarification: Their difference is known as the terminal temperature difference of heater.

6. Which of the following is true for an open heater?
a) it is simple, has low cost and low heat transfer capacity
b) a pump is required at each heater
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: The open heater has high heat transfer capacity.

7. Deaerator is a type of open heater.
a) true
b) false
Answer: a
Clarification: In steam power plants, closed heaters are favoured but one open heater is used for the purpose of feedwater deaeration.

8. Which of the following is true for a closed heater?
a) it requires a single pump regardless of the number of heaters
b) it is costly
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Closed heaters may not give as high feedwater temperature as do open heaters.

9. The higher the number of heaters used, the ____ will be the cycle efficiency.
a) lower
b) higher
c) efficiency does not depend on number of heaters
d) none of the mentioned
Answer: b
Clarification: The cycle efficiency varies according to the number of heaters.

10. If n heaters are used, the greatest gain in efficiency occurs when overall temperature rise is ____ times the difference between condenser and boiler saturation temperatures.
a) (n-1) / n
b) (n+1) / n
c) n / (n-1)
d) n / (n+1)
Answer: d
Clarification: This gives us the greatest gain in efficiency.

11. The efficiency gain follows the law of diminishing return with the increase in the number of heaters.
a) true
b) false
Answer: a
Clarification: This is because the cycle efficiency is proportional to the temperature rise of feedwater.

12. Which of the following statement is true?
a) in some cases, an increase in feedwater temperature may reduce the boiler efficiency
b) number of heaters are optimized
c) most often, five points of extraction are used
d) all of the mentioned
Answer: d
Clarification: The number of heaters get fixed by the exergy balance of the whole plant.

13. The thermal irreversibility should be ____ to improve the performance.
a) reduced
b) increased
c) kept constant
d) none of the mentioned
Answer: a
Clarification: The major exergy destruction due to irreversibility takes place in the steam generation.