Heat Engines and efficiency

  • Thread starter Epictetus
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Homework Statement



Suppose that two heat engines are connected in a series, such that the heat exhaust of the first engine is used as the heat input of the second (attached diagram below). The efficiencies of the engines are e1 and e2, respectively. Show that the net efficiency of the combination is given by:

e(net)= e1 + (1-e1)e2

Homework Equations



e(max)= 1 - Tc/Th
e= w/Qh = Qh-Qc/Qh = 1 - Qc/Qh
Qc/Qh = Tc/Th


The Attempt at a Solution



I broke up the diagram into two free-body diagrams allowing me to for solve e1 and e2:

e1 = 1-Th/Tm (Am I allowed to apply Qc/Qh = Tc/Th into 1 - Qc/Qh ?[/B])

and

e2= 1-Tm/Tc

Applying e1 and e2 in the given equation:

e(net)= e1 + (1-e1)e2
= (1-Th/Tm) + [1-(1-Th/Tm)](1-Tm/Tc)

which leaves
=1-Tm/Tc (The total efficiency of the engine equals to only the second engine because all the heat input eventually ends up there? )
 

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Answers and Replies

  • #2
OlderDan
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I think you want the starting point for e(net) to be the ratio of total work (w_t) to Qh. Instead of starting with the relationship you are trying to prove, start with the fundamental definition of efficiency

e(net) = w_t/Qh = (w1 + w2)/Qh
 
  • #3
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some of your assumptions appear really bad to me. You are not working with Carnot engines are you?
 
  • #4
andrevdh
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The thermal efficiency of a heat engine is a measure of converting the "input heat" to the engine to useful mechanical work

[tex]\varepsilon = \frac{W}{Q_H}[/tex]

if the engines are in series the [tex]Q_{C1}[/tex] will become the [tex]Q_{H2}[/tex].
Note that the thermal efficiency of a heat engine would be 100% if [tex]Q_C = 0[/tex].
 

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  • #5
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The thermal efficiency of a heat engine is a measure of converting the "input heat" to the engine to useful mechanical work

[tex]\varepsilon = \frac{W}{Q_H}[/tex]

if the engines are in series the [tex]Q_{C1}[/tex] will become the [tex]Q_{H2}[/tex].
Note that the thermal efficiency of a heat engine would be 100% if [tex]Q_C = 0[/tex].

And figure out work by substituting W= Qh-Qc?
Does Qm play a role in this or does it not effect the work of the heat engine?
 
  • #6
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some of your assumptions appear really bad to me. You are not working with Carnot engines are you?
No, I guess I'm working with heat pumps. I see now that I should've not used e(max) since that would be used when figuring out the effieciency of a Carnot engine...
 
  • #7
OlderDan
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And figure out work by substituting W= Qh-Qc?
Does Qm play a role in this or does it not effect the work of the heat engine?
Qm does play a role. Qm is the heat output of the first engine and the heat input of the second engine. Write the two work contributions in terms of the three heat values, and write the individual efficiencies in terms of the respective work to heat-input ratios.
 
  • #8
andrevdh
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And figure out work by substituting W= Qh-Qc?
Does Qm play a role in this or does it not effect the work of the heat engine?
Yes you can calculate the work done by a heat engine this way. It will apply for both engines. What it means is that the engine converts the input heat (energy) to usefull work and some left-over heat. A heat engine will always need to shed some left-over heat since it can only function between a hot and cold reservior. That is it can only function if heat can flow between a hot and cold region.

In the case of this problem the heat output of engine one becomes the heat input of engine 2:

[tex]Q_m = Q_{C1} = Q_{H2}[/tex]

We are therefore assuming that no heat escapes from the mid reservior. It just acts to transfer the heat from the first to the second engine.
 
Last edited:
  • #9
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I think you should use
enet=(WA+WB)QH1
 

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