Heat Engine: Maximizing Work Output

[so_gr_lo]

Homework Statement

Heat engine operates between a tank of 1x10^3 m^3 of water and a river at constant temperature of 10°C . If the temperature of the tank is initially 100°C what is the maximum work the heat engine can perform?

Not sure if my attempt is correct -am I supposed to be considering entropy ?

Relevant Equations

dQ = mcdT
Q1/Q2 = T1/T2
W = Q1 - Q2

 

[kuruman]

I would consider a Carnot cycle operating between the tank and the river and calculate the work $dW$ performed by the engine when heat is moved from tank to river. I would take into account that the efficiency of the engine depends on the temperature of the water in the tank and is continuously changing.

 

[Chestermiller]

Suppose we regard this combination of a finite reservoir (the water in the tank), the heat engine, and an infinite reservoir (the river) as an adiabatic system. The work done will be maximum if this combined system is operated reversibly. That would mean that no entropy would be generated and the overall entropy change for this overall adiabatic combination would be zero.

What is the final temperature of the water in the tank? What is its entropy change and how much heat has flowed out of the tank?

 

[so_gr_lo]

The final temperature would be 10°C as it thermalises with the river

would this be the entropy change of the water in the tank ?

 

[Chestermiller]

The final temperature would be 10°C as it thermalises with the river

would this be the entropy change of the water in the tank ?
View attachment 300689

You left out the heat capacity of the water and you lost the minus sign.

Your original relevant equation Q1/T1=Q2/T2 was incorrect for this situation. The correct equation should be $$\Delta S_{tank\ water}+\Delta S_{river\ water}=0$$What equation would you write for the entropy change of the river water (in terms of the heat removed by the river water and in terms of its temperature)? Based on this and the above equation, how much heat is removed by the river water?

 

[so_gr_lo]

 

[Chestermiller]

View attachment 300723

Yes, if $T_{tank\ water}$ is the initial value of the tank water temperature. How much heat does the tank water lose to the engine? What is the heat lost by the tank water to the engine minus the heat rejected by the engine to the river water?

 

[so_gr_lo]

 

[Chestermiller]

So what value do you get for the maximum work?

 

[so_gr_lo]

seems kind of high

 

[kuruman]

View attachment 301151
seems kind of high

It's the same number I got through a different method that I outlined in post #2. Here is what I did. Maximum work is achieved when the heat engine has maximum efficiency which means operating a Carnot cycle between the variable tank temperature $T$ and the constant river temperature $T_R$. When heat $dQ$ is transferred from tank to river, the work done by the Carnot engine is $$dW=\eta~dQ=\eta~ m~c~dT=m~c\left(1-\dfrac{T_R}{T}\right)dT.$$ Then, $$W=m~c\int_{T_0}^{T_R}\left(1-\dfrac{T_R}{T}\right)dT=m~c\left[ (T_0-T_R)+T_R\ln\left(\frac{T_R}{T_0}\right) \right].$$