250+ TOP MCQs on Thermodynamics of Coupled Cycles and Answers

Thermodynamics Multiple Choice Questions on “Thermodynamics of Coupled Cycles”.

1. The efficiency of a binary cycle is given by(here E=efficiency)
a) E = (1-E1)(1-E2)
b) E = 1 – (1-E1)(1-E2)
c) E = 1 / (E1-1)(E2-1)
d) none of the mentioned
Answer: b
Clarification: Here E1 is the efficiency of topping cycle and E2 is the efficiency of bottoming cycle.

2. For n cycles, overall efficiency would be
a) E = (1-E1)(1-E2)(1-E3)……(1-En) – 1
b) E = 1 / (1-E1)(1-E2)(1-E3)……(1-En)
c) E = (1-E1)(1-E2)(1-E3)……(1-En)
d) E = 1 – (1-E1)(1-E2)(1-E3)……(1-En)
Answer: d
Clarification: Hence we can say total loss = product of losses in all cycles.

3. By combining two cycles in series, we can get high combined efficiency even if the individual efficiencies are low.
a) true
b) false
Answer: a
Clarification: Such a high efficiency cannot be achieved by a single cycle.

4. How can we generate required power and required quantity of steam in a single process?
a) by modifying initial steam pressure
b) by modifying exhaust pressure
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: We require modification in both the pressures.

5. In a back pressure turbine,
a) exhaust steam from turbine is used for process heating
b) the process heater replaces the condenser of Rankine cycle
c) exhaust pressure from turbine is the desired saturation pressure
d) all of the mentioned
Answer: d
Clarification: In a back pressure turbine we modify both initial steam pressure and exhaust pressure.

6. A cogeneration plant produces,
a) power
b) process heat
c) both power and process heat
d) none of the mentioned
Answer: c
Clarification: In cogeneration plant, both power and process heat are produced.

7. In a by-product power cycle,
a) the basic need is power produced and process steam is a by-product
b) the basic need is process steam and power produced is a by-product
c) both process steam and power is the basic need
d) both process steam and power is a by-product
Answer: b
Clarification: Here, power produced is a by-product and the basic need is process steam.

8. In a by-product power cycle, condenser losses is ____
a) high
b) low
c) zero
d) infinity
Answer: c
Clarification: In a normal steam plant, this is the biggest loss but in a by-product power cycle it is zero.

9. The fraction of energy utilized in a by-product power cycle is ____
a) very high
b) very low
c) zero
d) infinity
Answer: a
Clarification: The reason being the condenser losses is zero.

10. The power available from back pressure turbine through which the heating steam flows is very less.
a) true
b) false
Answer: a
Clarification: The reason being relatively high back pressure or maybe small heating requirement.

250+ TOP MCQs on Velocity of Pressure Pulse in a Fluid and Stagnation Properties and Answers

Thermodynamics Multiple Choice Questions on “Velocity of Pressure Pulse in a Fluid and Stagnation Properties”.

1. An incompressible fluid is one for which density does not change with change in ____
a) pressure
b) temperature
c) velocity
d) all of the mentioned
Answer: d
Clarification: In an incompressible fluid, either density does not change or changes very little.

2. A compressible fluid is one for which density changes with change in temperature, pressure or velocity.
a) true
b) false
Answer: a
Clarification: This is opposite to an incompressible fluid.

3. Liquids are ____ and gases are ____
a) both are compressible
b) both are incompressible
c) incompressible, compressible
d) compressible, incompressible
Answer: c
Clarification: Liquids are incompressible whereas gases are compressible.

4. The pressure wave velocity c is given by
a) 1 / (dp/d(density))
b) sqrt(dp/d(density))
c) sqrt(d(density)/dp)
d) (dp/d(density))
Answer: b
Clarification: Here dp is the change in pressure.

5. For an ideal gas, velocity of sound is given by
a) 1/ɣ*R*T
b) sqrt(1/ɣ*R*T)
c) sqrt(ɣ/RT)
d) sqrt(ɣ*R*T)
Answer: d
Clarification: Here R is the characteristic gas constant.

6. The lower the molecular weight of fluid, ____ the value of ɣ and ____ is the sonic velocity at same temperature.
a) lower, lower
b) higher, higher
c) lower, higher
d) higher. lower
Answer: b
Clarification: This comes from the expression of velocity of sound in an ideal gas.

7. The Mach number is given by
a) (c/V)2
b) (V/c)2
c) V/c
d) c/V
Answer: c
Clarification: Here V is the actual velocity and c is the sonic velocity.

8. For the isentropic stagnation state,
a) it is a reference state
b) designated with subscript zero
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: It is a reference state in compressible fluid flow.

9. The reference temperature To and normal temperature T are related by
a) (To/T) = 1 + (V2)/(2*cp*T)
b) (To/T) = 1 – (V2)/(2*cp*T)
c) (To/T) = 1 + (V2)/(cp*T)
d) (To/T) = 1 – (V2)/(cp*T)
Answer: a
Clarification: Here cp is the specific heat at constant pressure and V is the actual velocity.

10. The reference temperature To and normal temperature T can be related in terms of M by
a) (To/T) = 1 + (ɣ-1)*(M2)
b) (To/T) = 1 + (ɣ-1)*(M2)/2
c) (To/T) = 1 – (ɣ-1)*(M2)/2
d) (To/T) = 1 – (ɣ-1)*(M2)
Answer: b
Clarification: Here M is the Mach number.

250+ TOP MCQs on Temperature Basics and Answers

Thermodynamics Multiple Choice Questions on “Temperature Basics”.

1. When a body A is in thermal equilibrium with a body B, and also separately with a body C, then B and C will be in thermal equilibrium with each other.
a) True
b) False
Answer: a
Clarification: Zeroth law of thermodynamics.

2. Which of the following were used as fixed points before 1954?
a) The ice point
b) The steam point
c) All of the mentioned
d) None of the mentioned
Answer: c
Clarification: Both of these were used.

3. What is the standard fixed point of thermometry?
a) The ice point
b) The steam point
c) The triple point of water
d) None of the mentioned
Answer: c
Clarification: After 1954, only one fixed point has been used.

4. All gases and vapours approach ideal gas behaviour at?
a) High pressure and high density
b) Low pressure and low density
c) High pressure and low density
d) Low pressure and high density
Answer: b
Clarification: Comes from ideal gas equation of state.

5. The value of ratio of the steam point temperature to the ice point temperature is?
a) 1.466
b) 1.266
c) 1.166
d) 1.366
Answer: d
Clarification: This value is a universal constant.

6. Celsius temperature of the triple point of water is ( in degree Celsius)?
a) -0.00
b) 0.00
c) 0.01
d) None of the mentioned
Answer: c
Clarification: Zero point of degree Celsius is shifted.

7. Which of the following is chosen as the standard thermometric substance?
a) Gas
b) Liquid
c) Solid
d) All of the mentioned
Answer: a
Clarification: Smallest variation is observed among different gas thermometers.

8. A real gas behaves as an ideal gas when?
a) Temperature approaches zero
b) Pressure approaches zero
c) Both temperature and pressure approaches zero
d) None of the mentioned
Answer: b
Clarification: It is a property of gas.

9. The temperature interval from the oxygen point to the gold point is divided into how many parts?
a) 2
b) 3
c) 4
d) 1
Answer: b
Clarification: Taken as international temperature scale.

10. Optical method is adopted for measuring temperatures higher than the gold point?
a) True
b) False
Answer: a
Clarification: Temperature is determined with the help of Planck’s law of thermal radiation.

250+ TOP MCQs on Second Law of Thermodynamics and Answers

Thermodynamics Multiple Choice Questions on “Second Law of Thermodynamics”.

1. Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output for this heat engine.thermodynamics-questions-answers-second-law-thermodynamics-q1″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q1″ class=”alignnone size-full wp-image-166407″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q1 213w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q1-183×300 183w” sizes=”(max-width: 213px) 100vw, 213px”>
a) 30 MW
b) 40 MW
c) 50 MW
d) 60 MW
Answer: a
Clarification: Net power output = 80 – 50 MW = 30 MW.

2. Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the thermal efficiency for this heat engine.thermodynamics-questions-answers-second-law-thermodynamics-q1″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q1″ class=”alignnone size-full wp-image-166407″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q1 213w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q1-183×300 183w” sizes=”(max-width: 213px) 100vw, 213px”>
a) 47.5 %
b) 27.5 %
c) 37.5 %
d) none of the mentioned
Answer: c
Clarification: The thermal efficiency of heat engine = net work output / heat input
= 30/80 = 0.375 = 37.5 %.

3. A car engine with a power output of 50 kW has a thermal efficiency of 24 percent. Determine the fuel consumption rate of this car if the fuel has a heating value of 44,000 kJ/kg . thermodynamics-questions-answers-second-law-thermodynamics-q3″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q3″ width=”237″ class=”alignnone size-full wp-image-166408″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q3 237w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q3-201×300 201w” sizes=”(max-width: 237px) 100vw, 237px”>
a) 0.00273 kg/s
b) 0.00373 kg/s
c) 0.00473 kg/s
d) 0.00573 kg/s
Answer: c
Clarification: Q = 50/0.24 = 208.3 kW,
hence fuel consumption rate = 208.3 kW / 44000 kJ/kg = 0.00473 kg/s.

4. The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2kW, determine the coefficient of performance of the refrigerator.
thermodynamics-questions-answers-second-law-thermodynamics-q4″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q4″ class=”alignnone size-full wp-image-166409″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q4 265w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q4-223×300 223w” sizes=”(max-width: 265px) 100vw, 265px”>
a) 4
b) 3
c) 2
d) 1
Answer: b
Clarification: COP = (360/2)(1/60) = 3.

5. The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2kW, determine the rate of heat rejection to the room that houses the refrigerator.thermodynamics-questions-answers-second-law-thermodynamics-q4″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q4″ class=”alignnone size-full wp-image-166409″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q4 265w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q4-223×300 223w” sizes=”(max-width: 265px) 100vw, 265px”>
a) 450 kJ/min
b) 460 kJ/min
c) 470 kJ/min
d) 480 kJ/min
Answer: d
Clarification: Q = 360 + (2)(60/1) = 480 kJ/min.

6. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine the power consumed by the heat pump.thermodynamics-questions-answers-second-law-thermodynamics-q6″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q6″ class=”alignnone size-full wp-image-166410″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q6 295w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q6-250×300 250w” sizes=”(max-width: 295px) 100vw, 295px”>
a) 32000 kJ/h
b) 33000 kJ/h
c) 34000 kJ/h
d) 35000 kJ/h
Answer: a
Clarification: W = Q/COP = 80000 kJ/h / 2.5 = 32000 kJ/h.

7. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine the rate at which heat is absorbed from the cold outdoor air.thermodynamics-questions-answers-second-law-thermodynamics-q6″ alt=”thermodynamics-questions-answers-second-law-thermodynamics-q6″ class=”alignnone size-full wp-image-166410″ srcset=”/2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q6 295w, /2017/06/thermodynamics-questions-answers-second-law-thermodynamics-q6-250×300 250w” sizes=”(max-width: 295px) 100vw, 295px”>
a) 32000 kJ/h
b) 48000 kJ/h
c) 54000 kJ/h
d) 72000 kJ/h
Answer: b
Clarification: The rate at which heat is absorbed = 80000 – 32000 = 48000 kJ/h.

8. An air-conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric air at 35°C. Estimate the amount of power needed to operate the air-conditioner.
a) 1.09 kW
b) 1.19 kW
c) 1.29 kW
d) 1.39 kW
Answer: d
Clarification: Q = m*cp*(temperature change) = 20.08 kW
COP = (15+273)/(35-15) = 14.4
hence power needed = 20/14.4 = 1.39 kW.

9. A cyclic machine, as shown below, receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200kJ of work as output. Is this cycle reversible, irreversible, or impossible?
a) reversible
b) irreversible
c) impossible
d) none of the mentioned
Answer: c
Clarification: The Carnot efficiency = 1 – (400/1000) = 0.6 and real efficiency = (300/325) = 0.615 which is greater than the Carnot efficiency hence cycle is impossible.

10. In a cryogenic experiment you need to keep a container at -125°C although it gains 100 W due to heat transfer. What is the smallest motor you would need for a heat pump absorbing heat from the container and rejecting heat to the room at 20°C?
a) 97.84 kW
b) 98.84 kW
c) 99.84 kW
d) 95.84 kW
Answer: a
Clarification: COP = 1.022 and thus power required = 100/1.022 = 97.84 kW.

11. A car engine operates with a thermal efficiency of 35%. Assume the air-conditioner has a coefficient of performance of 3 working as a refrigerator cooling the inside using engine shaft work to drive it. How much fuel energy should be spend extra to remove 1 kJ from the inside?
a) 0.752 kJ
b) 0.952 kJ
c) 0.852 kJ
d) none of the mentioned
Answer: b
Clarification: W = thermal efficiency * Q(fuel) thus Q(fuel) = 1/(0.35*3) = 0.952 kJ.

250+ TOP MCQs on Reversible Adiabatic Work in a Steady Flow System and Entropy and Disorder and Answers

Thermodynamics Quiz on “Reversible Adiabatic Work in a Steady Flow System and Entropy and Disorder”.

1. The equation W=∫vdp holds good for
a) work-producing machine like an engine or turbine
b) work-absorbing machine like a pump or a compressor
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: The equation given here is used for steady flow process and also when the fluid undergoes reversible adiabatic expansion or compression.

2. Only those processes are possible in nature which would give an entropy ____ for the system and the surroundings together.
a) decrease
b) increase
c) remains same
d) none of the mentioned
Answer: b
Clarification: The entropy of an isolated system can never decrease.

3. A process always occurs in such a direction as to cause an increase in the entropy of the universe.
a) true
b) false
Answer: a
Clarification: This comes from the second law which indicates the direction in which a process takes place.

4. When the potential gradient is ____, the entropy change of the universe is ____
a) large, zero
b) infinitesimal, zero
c) infinitesimal, negative
d) none of the mentioned
Answer: b
Clarification: This makes the process reversible.

5. At equilibrium, the isolated system exists at the peak of the entropy-hill and
a) dS=-1
b) dS=1
c) dS=infinity
d) dS=0
Answer: d
Clarification: At equilibrium, the entropy becomes maximum and hence change in entropy is zero.

6. Which of the following is true?
a) the KE of a gas is due to the coordinated motion of of all the molecules with same average velocity in same direction
b) the PE is due to the displacement of molecules from their normal positions
c) heat energy is due to the random thermal motion of molecules in a disorderly fashion
d) all of the mentioned
Answer: d
Clarification: These are the main causes of Kinetic energy, Potential energy and thermal energy of gas molecules.

7. Orderly energy can be easily converted into disorderly energy.
a) true
b) false
Answer: a
Clarification: An example can be, converting mechanical and electrical energy into internal energy by friction.

8. When work is dissipated into internal energy, what is the change in the disorderly motion of molecules.
a) decreases
b) increases
c) remains same
d) none of the mentioned
Answer: b
Clarification: We know that increase in internal energy causes more random motion.

9. When heat is imparted to a system,
a) the disorderly motion of molecules increases
b) the entropy of the system increases
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: As heat is given to a system, its internal energy increases, thus increasing the entropy of the system.

10. Which of the following relation is correct?
a) S=lnK/W
b) S=K/lnW
c) S=lnK*W
d) S=K*lnW
Answer: d
Clarification: S=K*lnW where S is the entropy, W is the thermodynamic probability, and K is the Boltzmann constant.

11. In the reversible adiabatic expansion of a gas the increase in disorder due to an increase in volume is compensated by the decrease in disorder due to a decrease in temperature.
a) true
b) false
Answer: a
Clarification: This ensures that the disorder number or entropy remains constant.

12. When does the entropy of a system become zero?
a) W=0
b) W=1
c) W=-1
d) none of the mentioned
Answer: b
Clarification: When thermodynamic probablity W=1, we get S=0 from S=K*lnW and this happens only at T=0K.

13. According to the Boltzmann,
a) he introduced the thermodynamic probability with each state
b) increase in entropy implies that the system proceeds by itself towards a state of higher thermodynamic probability
c) an irreversible process goes on happening until the most probable is achieved
d) all of the mentioned
Answer: d
Clarification: This is how Boltzmann introduced statistical concepts to define disorder.

14. When W=1, we get S=0 which can occur only at T=0K. This is the Nernst-Simon statement of third law of thermodynamics.
a) true
b) false
Answer: a
Clarification: But the state of T=0K cannot be reached in a finite number of operations.

Thermodynamics for quizzes,

250+ TOP MCQs on P-V diagram for Pure Substance and Answers

Thermodynamics Multiple Choice Questions on “P-V diagram for Pure Substance”.

1. Which of the following is a property of a pure substance?
a) it has constant chemical composition throughout its mass
b) it is a one-component system
c) it may exist in one or more phases
d) all of the mentioned
Answer: d
Clarification: These are some of the properties of a pure substance.

2. For water, as temperature increases, volume always increases?
a) true
b) false
Answer: b
Clarification: From 0 degree Celsius to 4 degree Celsius as temperature increases, volume of water decreases which is a peculiarity of water.

3. A saturation state is a state from which a change of phase may occur
a) without a change of pressure or temperature
b) with a change of pressure or temperature
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: For example, water at 0 degree Celsius and at 100 degree Celsius.

4. In which of the following state does water exist?
a) saturated solid state
b) saturated liquid state
c) saturated vapour state
d) all of the mentioned
Answer: d
Clarification: Water exists in these states at 0 degree Celsius and at 100 degree Celsius.

5. Which of the following exists in a p-V diagram for water?
a) saturated solid line
b) saturated liquid lines
c) saturated vapour line
d) all of the mentioned
Answer: d
Clarification: The p-V diagram for water has all these three lines.

6. The triple point is a line on the p-V diagram, where all the three phases, solid, liquid and gas exist.
a) true
b) false
Answer: a
Clarification: At triple point, all these three phases exists in equilibrium.

7. At a pressure below the triple point line,
a) the substance cannot exist in the liquid phase
b) the substance when heated transforms from solid to vapour
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This phenomenon is known as sublimation and takes place by absorbing the latent heat of sublimation from the surroundings.

8. Which of the following statement is true?
a) to the left of saturated solid line is the solid region
b) between saturated solid line and saturated liquid line with respect to solidification there exists the solid-liquid mixture region
c) between two saturated liquid lines is the compressed liquid region
d) all of the mentioned
Answer: d
Clarification: These statements come from the p-V diagram for a pure substance.

9. The isotherm passing through the critical point is called the critical isotherm.
a) true
b) false
Answer: a
Clarification: At critical point, all the quantities like pressure, temperature and volume attain their critical values.

10. The greater the temperature, the ____ is the vapour pressure.
a) lower
b) higher
c) depends on the substance
d) none of the mentioned
Answer: b
Clarification: The vapour pressure mainly depends on the temperature.

11. Phase change occurs at
a) constant pressure
b) constant temperature
c) constant pressure and temperature
d) none of the mentioned
Answer: c
Clarification: For phase change, pressure and temperature must be constant like water at 0 degree Celsius and at 100 degree Celsius.

12. Which of the following statement is true?
a) saturation temperature is a function of pressure
b) saturation pressure is a function of temperature
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: At saturation temperature, a pure liquid transforms into vapour and at saturation pressure, the liquid boils.

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