ISRO Mechanical Engineering/Refrigeration and Air Conditioning Thermal Engineering Questions, Answers and Explanation
ISRO Mechanical Engineering Thermal Engineering Questions, Answers and Explanation
System Properties
2007.2. The universal gas constant of a gas is the product of molecular weight of the gas and
a) Gas constant
b) Specific heat at constant pressure
c) Specific heat at constant volume
d) None of the above
Answer
a) Gas constant
Explanation
The value of universal gas constant is 8.314 J mol-1K-1. This when divided by the molecular weight of gas we get specific gas constant
Molecular mass of air = 28.97 g/mol
Rair = 8.314x1000/28.97 J kg-1K-1 = 287 J kg-1K-1.
2007.3. Van der Waal's equation of state of a gas is
a) pV = nRT
b) (p+a/V2,/sup>)(v+b) = RT
c) (p+a/V2,/sup>)(v-b) = RT
d) (p-a/V2,/sup>)(v-b) = RT
Answer
c) (p+a/V2,/sup>)(v-b) = RT
Explanation
a) Gas constant
b) Specific heat at constant pressure
c) Specific heat at constant volume
d) None of the above
Answer
a) Gas constant
Explanation
The value of universal gas constant is 8.314 J mol-1K-1. This when divided by the molecular weight of gas we get specific gas constant
Molecular mass of air = 28.97 g/mol
Rair = 8.314x1000/28.97 J kg-1K-1 = 287 J kg-1K-1.
2007.3. Van der Waal's equation of state of a gas is
a) pV = nRT
b) (p+a/V2,/sup>)(v+b) = RT
c) (p+a/V2,/sup>)(v-b) = RT
d) (p-a/V2,/sup>)(v-b) = RT
Answer
c) (p+a/V2,/sup>)(v-b) = RT
Explanation
The constant 'a' provides correction for inter molecular forces and b for finite molecular size. The effect of the finite volume of the particles is to decrease the available void space in which the particles are free to move. We must replace V by V − b, where b is called the excluded volume or "co-volume".
The pressure of a gas is due to the hits of the molecules on the walls of the containing vessel. The attractive force between the molecules comes into play when the molecules are brought close together by compressing the gas. A molecule in the body of the gas is attracted in all the directions when forces acting in opposite directions cancel out, but a molecule, the boundary of the gas is subjected to an inward pull due to unbalanced molecular attraction. In this way some of the energy of the molecule about to strike the wall of the vessel is used up in overcoming this inward pull. Therefore, it will not strike the opposite wall with the same force. The observed pressure consequently will be less than the ideal pressure. Therefore the ideal pressure, will be equal to observed pressure P plus a pressure correction, depending upon the attractive forces.
Answer
c) P/T = Const
Explanation
For an isochoric process volume remains constant, therfore PV = nRT reduces to P/T = constant
a) nR/VP
b) nRT/VP2
c) nR/VP2
d) nRT/VP
Answer
b) nRT/VP2
Explanation
Compressibility is the inverse of bulk modulus
Bulk modulus is the ratio change in pressure to strain = -dP/(dV/V)
Therefore, compressibility = -dV/(VdP)
For an ideal gas PV = nRT
Since the process is isothermal dT = 0, only differentiable variables are P and V
V = nRT/P
Differentiating dV = -nRTdP/P2
Compressibility = -dV/(VdP) = nRT/VP2
2017.2.74. Isochoric changes in ideal gas are characterized by
a) V/T = Const
b) PV = Const
c) P/T = Const
d) None of these
Answer
c) P/T = Const
Explanation
For an isochoric process volume remains constant, therfore PV = nRT reduces to P/T = constant
2017.2.39. Isothermal compressibility of an ideal gas is
b) nRT/VP2
c) nR/VP2
d) nRT/VP
Answer
b) nRT/VP2
Explanation
Compressibility is the inverse of bulk modulus
Bulk modulus is the ratio change in pressure to strain = -dP/(dV/V)
Therefore, compressibility = -dV/(VdP)
For an ideal gas PV = nRT
Since the process is isothermal dT = 0, only differentiable variables are P and V
V = nRT/P
Differentiating dV = -nRTdP/P2
Compressibility = -dV/(VdP) = nRT/VP2
First Law of Thermodynamics
2006. 44. For a closed system, difference between the heat added to the system and work done by the gas, is equal to the change ina) Internal Energy
b) Entropy
c) Enthalpy
d) Temperature
Answer
a) Internal Energy
Explanation
Closed system is one in which only energy transfer takes place. Open system is one in which both energy and mass transfer takes place.
First law of thermodynamics states that the change in internal energy equals the difference between heat supplied and work done.
2018.RAC.8. A mixture of gases expand from 0.03m3 to 0.06m3 at constant pressure of 1MPa and absorb 84kJ of heat during the process. The change in internal energy of the mixture is
a) 54kJ
b) 30kJ
c) 84kJ
d) 110kJ
Answer
a) 54kJ
Explanation
Work done during a non-flow constant pressure process is PdV = 1000kPa x (0.06-0.03)m3
= 30kJ
From the first law of thermodynamics change in internal energy is the difference of heat absorbed by the system and work done by the system.
Change in internal energy = 84kJ - 30kJ = 54kJ
Entropy
2017.1.20. The entropy of a hot baked potato decreases as it coolsa) The above statement is correct
b) The above statement is incorrect as it violates the increase in entropy principle
c) Too less information to comment
d) The above statement is incorrect as it violates second law of thermodynamics
Answer
a) The above statement is correct
Explanation
Magnitude of increase of entropy of the surroundings is more than the magnitude of the decrease of entropy of the hot baked potato. As the overall entropy increases the process is not reversible and practical.
Answer
b) Reversible Adiabatic Expansion
Explanation
2017.2.72. Isentropic process is related to
a) Adiabatic Expansion
b) Reversible Adiabatic Expansion
c) Isothermal Expansion
d) Reversible Isothermal Expansion
Answer
b) Reversible Adiabatic Expansion
Explanation
Irreversible free expansion of a gas in adiabatic condition is not isentropic. There is an increase of entropy. The equation of dS=dQ/T is not an accurate equation, the actual equation should be dS=dQreversible/T. For dQ of irreversible the equation should be changed into the clausius inequality form which is dS> dQ irreversibl/T. Under this clausius inequality equation, even dQ is zero in the adiabatic condition, and the right hand side equation is equal to zero, the left side dS is greater than zero which means that the dS entropy is increasing.
An isentropic process is an idealized thermodynamic process that is both adiabatic and reversible. The work transfers of the system are frictionless, and there is no transfer of heat or matter.
In reversible isothermal process the entropy change of the universe is zero. But as temperature of the system remains constant, there is net heat flow into the system thereby increasing the entropy of the system.
In an isothermal expansion there is net increase in entropy as explained before because entropy is a state property
2018.42. A simple compressible substance inside a cylinder undergoes a change of state quasistatically, isothermally and at a constant pressure of 1.5MPa. If the enthalpy change for the system is 4000 kJ at 150oC, what is the corresponding change in its entropy?
a) 9.45 kJ/K
b) 26.67 kJ/K
c) 14.17 kJ/K
d) 0
Answer
a) 9.45 kJ/K
Explanation
dH = TdS + Vdp
At constant pressure, dH = T dS
Here dH = 4000 kJ, T = 150oC = 423 K
dS = dH/T = 4000/423 = 9.45 kJ/K
2018.42. A simple compressible substance inside a cylinder undergoes a change of state quasistatically, isothermally and at a constant pressure of 1.5MPa. If the enthalpy change for the system is 4000 kJ at 150oC, what is the corresponding change in its entropy?
a) 9.45 kJ/K
b) 26.67 kJ/K
c) 14.17 kJ/K
d) 0
Answer
a) 9.45 kJ/K
Explanation
dH = TdS + Vdp
At constant pressure, dH = T dS
Here dH = 4000 kJ, T = 150oC = 423 K
dS = dH/T = 4000/423 = 9.45 kJ/K
Air Standard Cycles
a) True
b) False
c) Can't say
d) Insufficient data
Answer
a) True
Explanation
Thermal efficiency = 1 - T2/T1 = 1 - 267/300 = 33/300 = 11%. As his claim is less than the theoretical efficiency the claim is true.
2017.2.53. Which of the following is most appropriate for various power cycles?
a) In Carnot cycle, expansion of vapour occurs without change in entropy
b) In Rankine cycle, transformation of liquid water in steam generator occurs at constant temperature
c) In Carnot cycle, heating of vapor occurs at constant volume
d) In Rankine cycle, entropy of vapor increases during its expansion in steam turbine
Answer
a) In Carnot cycle, expansion of vapour occurs without change in entropy
Explanation
Carnot cycle
1. Isothermal compression
2. Isentropic compression
3. Isothermal expansion
4. Isentropic expansion
1 and 2 rules out choice c)
Rankine cycle
1. Isentropic compression
2. Constant pressure heat addition (choice b) ruled out)
3. Isentropic expansion (choice d) ruled out)
4. Constant pressure heat rejection Answer
a) In Carnot cycle, expansion of vapour occurs without change in entropy
Explanation
Carnot cycle
1. Isothermal compression
2. Isentropic compression
3. Isothermal expansion
4. Isentropic expansion
1 and 2 rules out choice c)
Rankine cycle
1. Isentropic compression
2. Constant pressure heat addition (choice b) ruled out)
3. Isentropic expansion (choice d) ruled out)
2018.RAC.1. Carnot cycle consists of
a) Two constant pressure and two isentropic processes
b) One constant isenthalpic, one constant volume and two constant pressure processes
c) Two isothermal and two isentropic processes
d) One constant pressure, one constant volume and two isothermal processes.
Answer
c) Two isothermal and two isentropic processes
Explanation
The most efficient heat engine cycle is the Carnot Engine cycle and it is represented as a rectangle in T-S diagram
The most efficient heat engine cycle is the Carnot Engine cycle and it is represented as a rectangle in T-S diagram
Otto Cycle - 2 constant volume and 2 isentropic
Diesel Cycle - 1 constant pressure, 1 constant volume and 2 isentropic
2018.RAC.2. One reversible heat engine operates between 1600K and T2 K and another reversible heat engine operates between T2 K and 400K. If both engines have the same heat input and output, then the temperature T2 is equal to
a) 800K
b) 1600K
c) 1200K
d) 6400K
Answer
a) 800K
Explanation
Reversible engine is a theoretical engine operating in according to Carnot cycle. If both engines have the same input and output their efficiencies must be the same as efficiency = work done/ heat input = (heat input - heat output)/heat input. The efficiency of the Carnot cycle operating between cold reservoir temperature Tc and hot reservoir temperature Th is given by 1 - Tc/Th.
1 - T2/1600 = 1 - 400/T2
T2/1600 = 400/T2
(T2)2 = 20x20x40x40
T2 = 800K
2018.RAC.2. One reversible heat engine operates between 1600K and T2 K and another reversible heat engine operates between T2 K and 400K. If both engines have the same heat input and output, then the temperature T2 is equal to
a) 800K
b) 1600K
c) 1200K
d) 6400K
Answer
a) 800K
Explanation
Reversible engine is a theoretical engine operating in according to Carnot cycle. If both engines have the same input and output their efficiencies must be the same as efficiency = work done/ heat input = (heat input - heat output)/heat input. The efficiency of the Carnot cycle operating between cold reservoir temperature Tc and hot reservoir temperature Th is given by 1 - Tc/Th.
1 - T2/1600 = 1 - 400/T2
T2/1600 = 400/T2
(T2)2 = 20x20x40x40
T2 = 800K
Combustion
2017.2.66. The temperature of the products of combustion when the maximum amount of chemical energy is converted to thermal energy is
a) Higher than adiabatic flame temperature
b) Lower than adiabatic flame temperature
c) Equal to adiabatic flame temperature
d) Independent of adiabatic flame temperature
Answer
c) Equal to adiabatic flame temperature
Explanation
Maximum amount of chemical energy is converted to thermal energy when the air fuel ratio is stoichiometric. At that ratio, the adiabatic flame temperature is equal to the temperature of the products of combustion.
When the air fuel ratio is less or more the temperature of products of combustion is always less than the adiabatic flame temperature.
When the air fuel ratio is less or more the temperature of products of combustion is always less than the adiabatic flame temperature.
Refrigeration
2017.2.18. What is the maximum rate that a heat pump which uses 1kW of electric power can supply heat to a house at 27C when the outside temperature is 12C
a) 50J/s
b) 25kJ/s
c) 20kJ/s
d) 30J/s
Answer
c) 20kJ/s
Explanation
At maximum efficiency, Q2/Q1 = T2/T1
Q2/Q1 = T2/T1=300/285, Q1=285Q2/300
Q1+W=Q2
W = 15/300 Q2
Q2 = 20*1kJ/s = 20kJ/s
2018.RAC.4. Reversed Joule cycle is known as
a) Rankine cycle
b) Carnot cycle
c) Bell Coleman cycle
d) Otto cycle
Answer
c) Bell Coleman cycle
Explanation
Joule cycle is also called as Brayton cycle. Brayton cycle consists of
- Isentropic compression
- Constant pressure heat addition (mostly combustion)
- Isentropic expansion
- Constant pressure heat rejection
Bell Coleman cycle consists of
- Isentropic compression
- Constant pressure heat rejection
- Isentropic expansion
- Constant pressure heat absorption (refrigeration effect)
Steam Turbines
2017.2.55. The reheat factor Rf for a steam turbine is usually in the range of
a) 0 to 1
b) 1 to 1.065
c) 2 to 2.065
d) 5 to 10
Answer
b) 1 to 1.065
Explanation
Reheat factor is the ratio of cumulative heat drop to direct isentropic or Rankine heat drop. It is always greater than 1.
2017.2.51. The critical pressure ratio for maximum discharge in steam nozzle is
a) 2⁄γ+1
b) (2⁄(γ+1))γ
c) (2/(γ+1))γ-1
d) (2/(γ+1))γ/(γ-1)
Answer
d) (2/(γ+1))γ/(γ-1)
Explanation
Derivation of critical pressure ratio for maximum discharge
Answer
b) 1 to 1.065
Explanation
Reheat factor is the ratio of cumulative heat drop to direct isentropic or Rankine heat drop. It is always greater than 1.
2017.2.51. The critical pressure ratio for maximum discharge in steam nozzle is
a) 2⁄γ+1
b) (2⁄(γ+1))γ
c) (2/(γ+1))γ-1
d) (2/(γ+1))γ/(γ-1)
Answer
d) (2/(γ+1))γ/(γ-1)
Explanation
Derivation of critical pressure ratio for maximum discharge
IC Engines
2018.RAC.21. The purpose of thermostat in an engine cooling system is to
a) Indicate coolant temperature
b) Prevent coolant from boiling
c) Allow the engine to warm up quickly
d) Pressurize coolant for effective cooling
Answer
c) Allow the engine to warm up quickly
Explanation [Thermal Engineering]
The thermostat is like a valve that opens and closes as a function of its temperature. The thermostat isolates the engine from the radiator until it has reached a certain minimum temperature. Without a thermostat, the engine would always lose heat to the radiator and take longer to warm up. Once the engine has reached the desired operating temperature, the thermostat adjusts flow to the radiator to maintain a stable temperature.
a) Indicate coolant temperature
b) Prevent coolant from boiling
c) Allow the engine to warm up quickly
d) Pressurize coolant for effective cooling
Answer
c) Allow the engine to warm up quickly
Explanation [Thermal Engineering]
The thermostat is like a valve that opens and closes as a function of its temperature. The thermostat isolates the engine from the radiator until it has reached a certain minimum temperature. Without a thermostat, the engine would always lose heat to the radiator and take longer to warm up. Once the engine has reached the desired operating temperature, the thermostat adjusts flow to the radiator to maintain a stable temperature.