Aerodynamics Questions and Answers – A Brief Review of Thermodynamics – 2

This set of Advanced Aerodynamics Questions and Answers focuses on “A Brief Review of Thermodynamics – 2”.

1. The equation of the second law of thermodynamics says that____
a) Total change in entropy is always zero
b) We cannot find entropy for an irreversible process
c) Entropy change is defined as the heat added in an irreversible process
d) Entropy is a state process
View Answer

Answer: d
Explanation: According to the second law, entropy is a state process. We can find the entropy change by the heat added in a reversible process. Using the total head added, we can find the heat added in the irreversible process also (since entropy is a state process).

2. Which is the correct result according to the second law of thermodynamics?
a) Total change in entropy is always zero
b) Total change in entropy in a reversible process is always zero
c) Total change in entropy in an irreversible process is always zero
d) Total change in entropy for an irreversible process is negative
View Answer

Answer: b
Explanation: In an irreversible process, entropy is generated. This means the change in entropy for an irreversible process is positive. Also, for a reversible process, change in entropy is zero since the entropy is conserved in the universe. Thus, the change in total entropy for an irreversible process is greater or equal to zero while that for the reversible process is zero.

3. For a block of ice on a plate, second law of thermodynamics says that____
a) The plate will get hotter
b) Ice will get cooler
c) Cannot give conclusive direction
d) Ice will melt
View Answer

Answer: d
Explanation: The second law of thermodynamics states which direction the process will take place, unlike the first law which just tells whether the process is possible or not. From entropy considerations, ice will melt in this case. This is visible from real life observation as well. Total entropy has to increase or stay the same.
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4. First law when expressed in terms of entropy takes the form_____
a) T ds = de – p dv
b) T ds = dh – p dv
c) T ds = de + p dv
d) T ds = dh + p dv
View Answer

Answer: c
Explanation: The first law of thermodynamics when expressed in terms of entropy gives us two alternate equations. These equations are T ds = de + p dv and T ds = dh – v dp. We have used the relation of heat and entropy in the first law equation to get these equations.

5. The incorrect formula for entropy for a calorically perfect gas is____
a) ds=0
b) s=s(p,T)
c) s2-s1=cvln\(\frac {T_2}{T_1}\)+Rln\(\frac {v_2}{v_1}\)
d) s2-s1=cpln\(\frac {T_2}{T_1}\)-ln\(\frac {p_2}{p_1}\)
View Answer

Answer: a
Explanation: The change in entropy is zero only for an adiabatic process of a gas. Calorically perfect gas has constant specific heats, which makes the integration between the initial and final state easier. The final form of the equation comes out as s2-s1=cvln\(\frac {T_2}{T_1}\)+Rln\(\frac {v_2}{v_1}\) and s2-s1=cpln\(\frac {T_2}{T_1}\)-ln\(\frac {p_2}{p_1}\). As visible from these equations s=s(p, T).
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6. For a calorically perfect gas, entropy is not________
a) State function
b) Conserved in a reversible process
c) A function of two thermodynamic variables
d) Zero always
View Answer

Answer: d
Explanation: The enthalpy is a state function. This is true for all type of gases. It is conserved for a reversible process (since no generation of entropy takes place due to dissipation). From the formula, it is visible that entropy always is a function of two thermodynamic variables (p, T) or (v, T), etc. It is not zero always.

7. The isentropic relation between pressure and temperature of a calorically perfect gas is_____
a) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)
b) \(\frac {p_1}{p_2} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)
c) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma -1}{\gamma }}\)
d) \(\frac {p_1}{p_2} =(\frac {T_2}{T_1} )^{\frac {\gamma -1}{\gamma }}\)
View Answer

Answer: a
Explanation: The isentropic relation means no heat exchange and no dissipative forces. Also, for the calorically perfect gas, specific heats are constant. When we put these conditions in the entropy equation, we obtain the relation between pressure and temperature which is \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\).
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8. An isentropic process is restrictive and is not feasible in real life.
a) False
b) True
View Answer

Answer: a
Explanation: Isentropic process seems restrictive in a practical case, i.e. a process which is both reversible and adiabatic is not feasible. This is what we think generally. Actually, a large number of practical compressible flow cases are good assumptions of isentropic processes only.

9. Flow over a rocket outside the boundary layer is isentropic.
a) False
b) True
View Answer

Answer: b
Explanation: The boundary layer regime over flow has dissipative effects and viscosity. This is not isentropic. But the flow outside has negligible viscosity, dissipative effects. Thus, it is a reversible process to a good approximation. Moreover, there is no heat exchange taking place to or from the fluid element, making this an adiabatic flow. So, outside the boundary layer, the flow is isentropic.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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