Maximum efficiency of an engine taking heat from two hot reservoirs

[Gregg]

Homework Statement

A heat engine is taking identical amounts of heat from two hot reservoirs at temperatures $T_{H1}, T_{H2}$ doing work and then heat to a cold reservoir $T_C$

What is the maximum efficiency of this heat engine?

Homework Equations

For a Carnot cyle it is $\eta = 1 - T_C/T_H$

The Attempt at a Solution

First of all, the heat engine takes $Q_1$ from $T_{H1}$ and $Q_2$ from $T_{H2}$ it then does work $W$ and heats $T_C$

I thought that since the maximum efficiency was for a reversible process that does the same thing then I could make a new process that takes heat $Q_1+Q_2$ from a reservoir of temperature $T_H = \frac{T_{H1}+T_{H2}}{2}$

Making the efficiency (max) $\eta = 1 - T_C/T_H = 1 - \frac{2 T_C}{T_{H1}+T_{H2}}$

Am I able to do this?

 

[rude man]

Homework Statement

A heat engine is taking identical amounts of heat from two hot reservoirs at temperatures $T_{H1}, T_{H2}$ doing work and then heat to a cold reservoir $T_C$

What is the maximum efficiency of this heat engine?

Homework Equations

For a Carnot cyle it is $\eta = 1 - T_C/T_H$

The Attempt at a Solution

First of all, the heat engine takes $Q_1$ from $T_{H1}$ and $Q_2$ from $T_{H2}$ it then does work $W$ and heats $T_C$

I thought that since the maximum efficiency was for a reversible process that does the same thing then I could make a new process that takes heat $Q_1+Q_2$ from a reservoir of temperature $T_H = \frac{T_{H1}+T_{H2}}{2}$

Making the efficiency (max) $\eta = 1 - T_C/T_H = 1 - \frac{2 T_C}{T_{H1}+T_{H2}}$

Am I able to do this?

'Fraid not.

Write down the equation for the 1st and 2nd laws and solve for η = W/2Q.

 

[Gregg]

Take it to be a reversible process

$\eta = |\frac{2Q-Q_C}{2Q}|$

Now the problem is to relate that to the temperatures $T_{H1}$ and $T_{H2}$. Going to say that they are Carnot engines. I'm not sure this right but my attempt:

$\eta = 1 - \frac{Q_C}{2Q}$

For an ireversible process we have

$\oint \frac{\delta q}{T} = 0$

$\frac{Q}{T_{H1}} + \frac{Q}{T_{H2}} - \frac{Q_C}{T_C} = 0$

$\frac{Q(T_{H1}+T_{H2})}{{T_{H1}}{T_{H2}}} = \frac{Q_C}{T_C}$

$2Q = \frac{2 T_{H1}T_{H_2}Q_C}{T_C(T_{H1}+T_{H2})}$

$\eta = 1 - \frac{T_C(T_{H1}+T_{H2})}{2 T_{H1}T_{H_2}}$

 

[rude man]

Take it to be a reversible process

$\eta = |\frac{2Q-Q_C}{2Q}|$

Now the problem is to relate that to the temperatures $T_{H1}$ and $T_{H2}$. Going to say that they are Carnot engines. I'm not sure this right but my attempt:

$\eta = 1 - \frac{Q_C}{2Q}$

For an ireversible process we have

$\oint \frac{\delta q}{T} = 0$

$\frac{Q}{T_{H1}} + \frac{Q}{T_{H2}} - \frac{Q_C}{T_C} = 0$

$\frac{Q(T_{H1}+T_{H2})}{{T_{H1}}{T_{H2}}} = \frac{Q_C}{T_C}$

$2Q = \frac{2 T_{H1}T_{H_2}Q_C}{T_C(T_{H1}+T_{H2})}$

$\eta = 1 - \frac{T_C(T_{H1}+T_{H2})}{2 T_{H1}T_{H_2}}$

You get an A+! Good shot!

It's always best to go back to fundamentals instead of relying on formulas that may or may not apply. You did it the right way.

PS - you said "irreversible process". You meant "reversible".