Heat engine and Heat pump in combination

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SUMMARY

The discussion focuses on the analysis of a heat engine and heat pump system, specifically calculating the efficiency of the heat engine and the coefficient of performance (COP) of the heat pump. The efficiency of the heat engine is determined to be 0.93808 using the formula QL/Qw1 = TL/Twaste. For the heat pump, the COP is calculated as 4.23 using the formula COP = TH/(TH-TL). The conversation emphasizes the importance of applying the first and second laws of thermodynamics to derive the necessary equations for solving for heat transfer rates, specifically dQh/dt.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first and second laws of thermodynamics.
  • Familiarity with heat engine efficiency calculations and formulas.
  • Knowledge of heat pump performance metrics, including coefficient of performance (COP).
  • Ability to manipulate equations involving energy transfer and state functions.
NEXT STEPS
  • Study the derivation of the first and second laws of thermodynamics in the context of heat engines and heat pumps.
  • Learn how to calculate heat transfer rates in thermodynamic cycles.
  • Explore advanced topics in thermodynamics, such as entropy changes in reversible and irreversible processes.
  • Investigate the relationship between work output and heat input in thermodynamic systems.
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Students and professionals in mechanical engineering, thermodynamics researchers, and anyone involved in the design and analysis of heat engines and heat pumps.

canadiansmith
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Question is attached

I have spent a great deal of time on this problem and I am hoping someone can help me out.
My attempt at problem.

first for the heat engine
QL/Qw1 = TL/Twaste
so efficiency = 1- Tl/Twaste
efficiency = 0.93808
and I know that efficiency = Wout/ Qw1


This is as far as i got for the heat engine...

For Heat pump
i found cop = TH/(TH-TL) = 4.23
 

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Never mind efficiency. The problem doesn't ask for efficiency. It only wants you to compute dQh/dt.

ANYWAY ...

write the equations for the 1st and 2nd laws. That's 4 equations.

Then write the relationship between Qw1, Qw2 and 5 MW.

5 equations, 5 unknowns. Away you go. Solve for Qh.

(Note - the 1st and 2nd laws are usually written in terms of energy rather than power. Don't let that slow you down. Just pretend it's energy for 1 second, which of course is 1 J/s = 1 W.
 
rude man said:
Never mind efficiency. The problem doesn't ask for efficiency. It only wants you to compute dQh/dt.

ANYWAY ...

write the equations for the 1st and 2nd laws. That's 4 equations.

Then write the relationship between Qw1, Qw2 and 5 MW.

5 equations, 5 unknowns. Away you go. Solve for Qh.

(Note - the 1st and 2nd laws are usually written in terms of energy rather than power. Don't let that slow you down. Just pretend it's energy for 1 second, which of course is 1 J/s = 1 W.

I am still a little confused. There are 2 equations for both first and second law?
i only have delta U = Q-W
and Q = integral of TodeltaS
 
canadiansmith said:
I am still a little confused. There are 2 equations for both first and second law?
i only have delta U = Q-W
and Q = integral of TodeltaS

Yes, but you have two mechanisms, the heat engine and the heat pump. Each has its own
1st & 2nd law equations.

BTW dQ = TdS is not an expression of the second law. It's just a formula to calculate entropy. What you need is an expression reflective of the fact that entropy is unchanged in a reversible cycle.

As for the 1st law, you are right, but what can you say about Qw1, QL, Qw2, QH and W? Remember that U is a state function so over a cycle it doesn't change either, reversible or not.
 
Last edited:

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