Exergy value at double temperature

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Discussion Overview

The discussion centers on the concept of exergy in relation to heat at different temperatures, specifically comparing the exergy value of heat at 400°C to that at 200°C. Participants explore the mathematical relationships and implications of exergy calculations, raising questions about the formulas used and the assumptions behind them.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • HS2 questions whether heat at 400°C carries double the exergy value of heat at 200°C, presenting a formula for lost work and deriving a relationship between the exergy values.
  • Another participant expresses uncertainty about the lost work formula, suggesting that if the surroundings were at absolute zero, the lost work might equal zero instead of being negative.
  • There is a proposal that exergy values could be expressed as proportional to heat capacities at the respective temperatures, but it is noted that this would not result in a doubling of exergy values.
  • Further discussion raises the ambiguity of whether to use specific heat at constant pressure (Cp) or constant volume (Cv), indicating that the choice may depend on the system and that heat capacity could vary with temperature.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating exergy or the validity of the formulas used. Multiple competing views remain regarding the interpretation of lost work and the appropriate heat capacity to use.

Contextual Notes

There are unresolved assumptions regarding the conditions under which exergy is calculated, particularly concerning the temperature of the surroundings and the behavior of heat capacities across a wide temperature range.

gfd43tg
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HS2: Does heat at 400°C carry two times exergy value than heat a 200 °C? If not, how much should it be?

Exergy is the maximum useful work that can be achieved. The lost work of a process is

##W_{lost} = T_{\sigma}\Delta S - Q##

Where is the surrounding absolute temperature.

If the Exergy is ##W_{ideal} - W_{lost}##, then

##Exergy_{400} = W_{ideal} - (673 K)\Delta S + Q##
##Exergy_{200} = W_{ideal} - (473 K)\Delta S + Q##

Subtracting these equations,

##Exergy_{400} - Exergy_{200} = -200 \Delta S##
##Exergy_{400} = -200 \Delta S + Exergy_{200}##

Which is clearly not the same as a doubling in value. Is this the right way to do it?
 
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This is just my shot at it, look for others to respond also:

I don't know about your Wlost formula. Never seen it. But if the surroundings are at zero K, is the work lost = -Q? I would think zero instead.

Since the low temperature T2 can be zero K, all the heat extracted at T = T1 can go into work. So W = Q1. But Q1 = C*T1. So assuming C constant between 200C and 400C, what W can be extracted at each temperature?
 
So then you would say for 400C, then ##Exergy_{400} = C_{p}(673K)##, and ##Exergy_{200} = C_{p}(473 K)##, which still would not be double?
 
Maylis said:
So then you would say for 400C, then ##Exergy_{400} = C_{p}(673K)##, and ##Exergy_{200} = C_{p}(473 K)##, which still would not be double?

Yes, except I'm not sure if it would be ##C_{p}## or ##C_{V}## or something else. I guess it would depend on the system. I would just call it C. In reality C would be a function of temperature anyway, for such a large range of temperatures all the way down to zero K.
 

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