Maximum efficiency of an engine taking heat from two hot reservoirs

In summary, the maximum efficiency of a heat engine that takes identical amounts of heat from two hot reservoirs at temperatures ##T_{H1}## and ##T_{H2}## and then transfers heat to a cold reservoir ##T_C## is given by the equation ##\eta = 1 - \frac{T_C(T_{H1}+T_{H2})}{2 T_{H1}T_{H_2}}##, where the temperatures are taken to be for a Carnot engine and the process is assumed to be reversible. This equation can be derived by applying the first and second laws of thermodynamics and solving for the efficiency, showing that it is dependent on the temperatures of the hot and cold reservoirs.
  • #1
Gregg
459
0

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?
 
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  • #2
Gregg said:

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.
 
  • #3
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}} ##
 
  • #4
Gregg said:
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".
 
  • #5


Your approach is correct. The maximum efficiency of a heat engine is given by the Carnot efficiency equation, which is based on a reversible process. Therefore, by considering a new process that is reversible and has the same inputs and outputs as the original heat engine, you can use the Carnot efficiency equation to calculate the maximum efficiency for the original engine. This approach is commonly used in thermodynamics to find the maximum efficiency of a heat engine or any other system.
 

1. What is the maximum efficiency of an engine taking heat from two hot reservoirs?

The maximum efficiency of an engine taking heat from two hot reservoirs is known as the Carnot efficiency, which is given by the formula η = 1 - Tc/Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

2. How does the temperature of the hot reservoir affect the maximum efficiency of the engine?

The temperature of the hot reservoir directly affects the maximum efficiency of the engine. The higher the temperature of the hot reservoir, the higher the maximum efficiency of the engine. This is because the Carnot efficiency formula is inversely proportional to the temperature of the hot reservoir.

3. Can the maximum efficiency of the engine ever reach 100%?

No, the maximum efficiency of the engine can never reach 100%. This is due to the second law of thermodynamics, which states that it is impossible to convert all heat energy into work. There will always be some energy loss in the form of heat, resulting in an efficiency less than 100%.

4. How does the use of two hot reservoirs improve the efficiency of the engine?

The use of two hot reservoirs in an engine allows for a higher maximum efficiency compared to an engine with only one hot reservoir. This is because the temperature difference between the hot and cold reservoirs is greater, resulting in a larger Carnot efficiency.

5. Are there any real-world engines that operate at maximum efficiency?

No, there are no real-world engines that operate at maximum efficiency. This is due to various factors such as energy loss through friction and other inefficiencies in the engine's design. However, engineers continue to strive for higher efficiencies in engine designs.

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