ENTROPY REPORT 2

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(1) Determine the entropy change for an irreversible process between states 1 and 2, should the integral ∮12 dQ/T be performed along the actual process path or an imaginary reversible path? Explain.- The integral should be performed along a reversible path to determine the entropy change.

(2) Is an isothermal process necessarily internally reversible? Explain your answer with an example - No, not necessarily. It can be irreversible. Example: A system that involves paddle wheel work while losing an equivalent amount of heat.

(3) How do the values of the integral ∮12 dQ/T compare for a reversible and - - irreversible process between the same end states ?- The value of this integral is always larger for reversible processes. (4) The entropy of a hot baked potato decreases as it cools. Is this a violation of the increase of entropy principle? Explain.- No, because the entropy of the surrounding air increases even further during the process, making the total entropy change positive.

(5) Is it possible to create entropy? Is it possible to destroy it?- It is possible to create entropy, but not possible to destroy it.

(6) A piston–cylinder device contains helium gas. During a reversible, isothermal process, the entropy of the helium will (never, sometimes, always) increase.- Sometimes.

(7) A piston–cylinder device contains nitrogen gas. During a reversible, adiabatic process, the entropy

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of the nitrogen will (never, sometimes, always) increase.- Never

(8) A piston–cylinder device contains superheated steam. During an actual adiabatic process, the entropy of the steam will (never, sometimes, always) increase.- Always

(9) The entropy of steam will (increase, decrease, remain the same) as it flows through an actual adiabatic turbine.- Increase

(10)The entropy of the working fluid of the ideal Carnot cycle (increases, decreases, remains the same) during the isothermal heat addition process.- Increase

(11) The entropy of the working fluid of the ideal Carnot cycle (increases, decreases, remains the same) during the isothermal heat rejection process.- Decreases

(12) During a heat transfer process, the entropy of a system (always, sometimes, never) increases.- Sometimes

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(13)Is it possible for the entropy change of a closed system to be zero during an irreversible process? Explain.- Yes. This will happen when the system is losing heat and the decrease in entropy as a result of this heat loss is equal to the increase in entropy as a result of irreversibilities.

(18.) In large compressors, the gas is frequently cooled while being compressed to reduce the power consumed by the compressor. Explain how cooling the gas during a compression process reduces the power consumption.- The work associated with steady-flow devices is proportional to the specific volume of the gas. Cooling a gas during compression will reduce its specific volume, and thus the power consumed by the compressor.

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(19) The turbines in steam power plants operate essentially under adiabatic conditions. A plant engineer suggests to end this practice. She proposes to run cooling water through the outer surface of the casing to cool the steam as it flows through the turbine. This way, she reasons, the entropy of the steam will decrease, the performance of the turbine will improve, and as a result the work output of the turbine will increase. How would you evaluate this proposal ?- Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the work output of the turbine. Therefore, this is not a good proposal.

(20) It is well known that the power consumed by a compressor can be reduced by cooling the gas during compression. Inspired by this, somebody proposes to cool the liquid as it flows through a pump, in order to reduce the power consumption of the pump. Would you support this proposal? Explain.- We would not support this proposal since the steady-flow work input to the pump is proportional to the specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.

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(21) On a T-s diagram, does the actual exit state (state 2) of an adiabatic turbine have to be on the right-hand side of the isentropic exit state (state 2) ? Why ?- Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a result of irreversibilities. Therefore, the actual exit state has to be on the right hand side of the isentropic exit state.

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