Entropy Generation Homework: 100kPa-500kPa, 300K-500K, 600K

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of work and entropy generation during the reversible adiabatic compression and internally reversible isobaric heating of air. The specific conditions include an initial state of 100 kPa and 300 K, and a final state of 500 kPa and 500 K, with heat exchange occurring with a reservoir at 600 K.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The first participant outlines the problem and attempts to calculate work and entropy generation for both the adiabatic compression and isobaric heating processes.
  • Another participant requests clarification on the entropy change during the adiabatic compression and asks for the work to be shown for the heating step.
  • One participant provides specific calculations for the work done and entropy generation during both processes, noting that the entropy generation appears negative, suggesting a potential error in their approach.
  • There is a repeated emphasis on the final temperatures at the end of each step and the calculation of heat removed from the reservoir, with a focus on the relationship between heat, work, and enthalpy.
  • Participants discuss the application of the first law of thermodynamics and the need to correctly account for boundary work in their calculations.
  • A later reply indicates a misunderstanding of the first law equation, prompting a correction and suggesting that the participant should re-evaluate their calculations to achieve the correct entropy increase.

Areas of Agreement / Disagreement

Participants express differing views on the calculations of entropy generation and the application of the first law of thermodynamics. There is no consensus reached on the correct approach or final values, as participants continue to challenge and refine each other's calculations.

Contextual Notes

Some calculations depend on assumptions about specific heat capacities and the ideal gas behavior of air. The discussion includes unresolved mathematical steps and varying interpretations of the first law of thermodynamics.

Who May Find This Useful

This discussion may be useful for students studying thermodynamics, particularly those interested in entropy generation and the application of the first law in thermodynamic processes.

albertov123
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Homework Statement


Air at 100 kPa, 300 K is to be delivered to a pipeline at 500 kPa, 500 K. The scheme involves reversible adiabatic compression of the air and then internally reversible isobaric heating. Assume that, heat is exchanged with a reservoir at 600 K.

Determine the work and entropy generation.

Homework Equations



v = RT/P

The Attempt at a Solution


[/B]
  • For the reversible adiabatic compression part. (Between 1-2)
I have found both the work and the entropy generation.
  • For the internally reversible isobaric heating (Between 2-3)
I have found boundary work from PdV but I'm not clear on how to find the entropy generation for this heating process.
 
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I'm interested in what you got for the entropy change in the adiabatic reversible compression. Do you know the general equation for the entropy change of an ideal gas between two thermodynamic equilibrium states? Please show your work for step 2 also.

Chet
 

  • For the reversible adiabatic compression part. (Between 1-2)
v1 = 0.861 m3/kgFrom T2/T1 = (P2/P1)(k-1)/k

T2 = 475.15 K- W = U = cv (T2-T1) W = - 126 kJReversible adiabatic means isentropic. Sgen = 0

  • For the internally reversible isobaric heating (Between 2-3)
W = PdV = 500 kPa . (v3-v2)From v=RT/P

v2 = 0.273 m3/kg

v3 = 0.287 m3/kgW = 7 kJ

q - w = ΔH = cp (T3-T2)

q = 32 kJ/kg

Sgen = cpln(T3/T2) - Q/T

This leads me to negative entropy generation. So, I must be making some wrong turns on the way.
 
The final temperature at the end of step 1 (according to your calculations) is 475 K. The final temperature at the end of step 2 is 500K. So the entropy change of the system is Cp ln(500/475). What is the amount of heat removed from the reservoir is step 2? I get Cp (500-475).
 
Chestermiller said:
The final temperature at the end of step 1 (according to your calculations) is 475 K. The final temperature at the end of step 2 is 500K. So the entropy change of the system is Cp ln(500/475). What is the amount of heat removed from the reservoir is step 2? I get Cp (500-475).

From q - w = ΔH = cp (T3-T2)
I get q = 32 kJ/kg

sgen = Δs - q/T = 0,052 - 32 kJ / 600 K = 0.052 - 0.053 = - 0.01

Edit: Why didn't you involve work in your first law equation? Isn't there some boundary work PdV also. That is why I get q = w + cp(500-475) = 7 + 25 = 32 kJ/kg
 
Last edited:
Albertov123 said:
From q - w = ΔH = cp (T3-T2)
I get q = 32 kJ/kg

sgen = Δs - q/T = 0,052 - 32 kJ / 600 K = 0.052 - 0.053 = - 0.01

Edit: Why didn't you involve work in your first law equation? Isn't there some boundary work PdV also. That is why I get q = w + cp(500-475) = 7 + 25 = 32 kJ/kg
It's q-w = ΔU, not ΔH. The constant pressure work w is included in ΔH=q=ΔU+w.

Try again. You almost have it. With this change, you will get your desired increase in entropy.
 
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Oh, I see. That helped a lot. Thank you so much.
 

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