Efficiency and Performance of a Reversible Heat Engine

[asdf1]

for the following question:

a reversible heat engine, opearting in a cycle, withdraws heat from a high-temperature reservoir(the temperature of which consequentyly decreases), performs work w, and rejects heat into a low-temperature reservoir(the tempertaure of which consequentyly increases). The two reswervoirs are, initially, aht the temperatures T1 and T2 and have constant heat capacities
C1 and C2, respectively. Calculate the final temperature of the system and the maximum amount of work which can be obtained from the engine.


how do you solvethat kind of question?

 

[Andrew Mason]

for the following question:
a reversible heat engine, opearting in a cycle, withdraws heat from a high-temperature reservoir(the temperature of which consequentyly decreases), performs work w, and rejects heat into a low-temperature reservoir(the tempertaure of which consequentyly increases). The two reswervoirs are, initially, aht the temperatures T1 and T2 and have constant heat capacities
C1 and C2, respectively. Calculate the final temperature of the system and the maximum amount of work which can be obtained from the engine.
how do you solvethat kind of question?

What is the expression for efficiency (dW/dQ_h) in terms of T1 and T2. How do T1 and T2 change with the heat flow? As these temperatures change, how does the efficiency change? What is the condition for flow to cease?

AM

 

[asdf1]

expression for efficiency (dW/dQ_h) ? what's that?

 

[Andrew Mason]

expression for efficiency (dW/dQ_h) ? what's that?

What is the efficiency of the Carnot engine (expressed as Work output/heat input)?

AM

 

[asdf1]

[q2-q1]/q2 = [T2-T1]/T2?

 

[Andrew Mason]

[q2-q1]/q2 = [T2-T1]/T2?

Right idea. But I think T1 = hot reservoir and T2 = cold reservoir. So, for the Carnot (reversible) cycle:

[tex]\Eta = \frac{dW}{dQ_H} = \frac{Q_H - Q_C}{Q_H} = \frac{T_1 - T_2}{T_1}[/tex]

So:
[tex]dW = \left(\frac{T_1 - T_2}{T_1}\right)dQ_H[/tex]

and:

[tex]W = \int dW = \int \left(\frac{T_1 - T_2}{T_1}\right)dQ_H[/tex]

Now, T1 and T2 change with heat flow. How are dQ_H and temperatures T1 and T2 related to the change in temperature, dT1 and the heat capacities of the two reservoirs? Find the expression for T1 and T2 with change in temperature dT1 and express the integral in terms of temperature and heat capacities, then integrate. You will have to work out the limits of integration (heat flow stops when T1 = T2).

AM

 

[asdf1]

but this isn't a carnot cycle, right? so does the same thing apply?
@@

 

[Andrew Mason]

but this isn't a carnot cycle, right? so does the same thing apply?
@@

'Carnot cycle' and 'reversible heat engine cycle' are synonymous.

AM

 

[Dr.Brain]

The heat being radiated by the hotter reservoir is m(C1)(T1) INITIALLY , and heat being gained by colder reservoir after the work is done by the engine is m(C2)(T2) initially , this will go on until an equilibrium position is established when no automatic heat flows from hotter to colder temperature.

Let the total amount of work done be W

Therefore at the end of it all:

m(C1)(T1-T) = W + m(C2)(T-T2)

As quoted by Andrew:

(heat flow stops when T1 = T2).

which will not be the case since T1 starts decreasing and T2 starts increasing as soon as the process starts. There will be some equilibrium T , where heat flow stops.

 

[asdf1]

thank you very much!

 

[Andrew Mason]

The heat being radiated by the hotter reservoir is m(C1)(T1) INITIALLY , and heat being gained by colder reservoir after the work is done by the engine is m(C2)(T2) initially , this will go on until an equilibrium position is established when no automatic heat flows from hotter to colder temperature.
Let the total amount of work done be W
Therefore at the end of it all:
m(C1)(T1-T) = W + m(C2)(T-T2)
which will not be the case since T1 starts decreasing and T2 starts increasing as soon as the process starts. There will be some equilibrium T , where heat flow stops.

I agree. The question is: what is that temperature T where equilibrium is reached? It depends on how much work is done by the system. If no work is done, it is simple algebra. If work is being done (ie. heat flowing out of the system as work) the final temperature T will be lower. In order to find W you have to use the Carnot efficiency.

AM

 

[asdf1]

ok, thanks again!