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Thermodynamics

61.

Adequate insulation is not that important for low temperature fluids.

Answer: A

This is because the loss of available energy from such fluids would be low.

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62.

A hot gas flowing through a pipeline can be considered as a

Answer: B

The process given here is irreversible.

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63.

The second law affixes a quality to energy of a system at any state.

Answer: A

For example we can say that the quality of energy of a gas at say 1000K is superior to that at say 300K.

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Answer: D

No answer description available for this question.

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65.

Work done in all reversible processes is equal

Answer: A

Reversible processes between the same end states must coincide and and produce equal amounts of work.

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66.

Argon in a light bulb is at 90 kPa and heated from 20°C to 60°C with electrical power. Find the total entropy generation per unit mass of argon.

Answer: D

1s2(gen) = s2 – s1 = Cp ln (T2/T1) – R ln (P2/ P1)
= Cp ln (T2/T1) – R ln (T2/ T1) = Cv ln(T2/T1)
= 0.312 ln [ (60 + 273)/(20 + 273) ] = 0.04 kJ/kg K.

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67.

It is better to state that “the change of state is irreversible, rather than say it is an irreversible process”.

Answer: A

This is because no irreversible path or process can be plotted on thermodynamic coordinates.

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Answer: D

The decrease in exergy is given by the product of lowest feasible temperature of heat rejection and the additional entropy change in the system.

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69.

Orderly energy can be easily converted into disorderly energy.

Answer: A

An example can be, converting mechanical and electrical energy into internal energy by friction.

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70.

A pan contains 5 L of engine oil at 20°C, 100 kPa. Now 2 L of hot 100°C oil is mixed into the pan. Find the entropy generation.

Answer: A

ρ = 885 kg/m3; From energy equation,
T2 = (mA/m2)TA + (mB/m2)TB = (5/7)20 + (2/7)100 = 42.868°C = 316.02 K
S2 – S1 = m2s2 − mAsA – mBsB = mA(s2 – sA) + mB(s2 – sB)
= 0.005 × 885 × 1.9 ln (316.02/293.15) + 0.002 × 885 × 1.9 ln (316.02/373.15)
= 0.6316 – 0.5588 = + 0.0728 kJ/K.

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