Optimizing Heat Engines: Finding the Temperature of the Hot Reservoir

[andrew410]

A 20% efficient real engine is used to speed up a train from rest to 5 m/s. It is known that an ideal (Carnot) engine using the same cold and hot reservoirs would accelerate the same train from rest to a speed of 6.5 m/s using the same amount of fuel. The engines use air at 300 K as a cold reservoir. Find the temperature of the steam serving as the hot reservoir.

I don't know how the velocity comes into play in order to get the temperature of the steam serving as the hot reservoir. I know that the formula for efficiency is
e = \frac {W_{eng}} {|Q_{h}|}.
Any help would be great! thx! :)

 

[andrew410]

can anyone help ?

 

[andrew410]

I think I got it maybe...
Since efficiency of a Carnot engine is
e = 1 - \frac {T_{c}} {T_{h}}
and e = .2.
So, we can reorder the formula to get,
T_{h} = \frac {T_{c}} {1-e}
What I am confused about is then why do they give us the initial and final velocity of each engine?

 

[Clausius2]

I think I got it maybe...
Since efficiency of a Carnot engine is
e = 1 - \frac {T_{c}} {T_{h}}
and e = .2.
So, we can reorder the formula to get,
T_{h} = \frac {T_{c}} {1-e}
What I am confused about is then why do they give us the initial and final velocity of each engine?

You're wrong and have a misconception. The efficiency of a Carnot engine is not 0.2. Your problem states:

"A 20% efficient real engine is used to speed up a train from .."

A real engine is not a Carnot engine. That phrase also not mean the efficiency of the engine is the 20% of the carnot efficiency.

Let' see:

v_1=5m/s
v_2=6.5m/s
m=mass of the train
m_f=mass of the fuel employed in combustion.
L_i= Caloric power or combustion enthalpy.

From this phrase: "It is known that an ideal (Carnot) engine using the same cold and hot reservoirs would accelerate the same train from rest to a speed of 6.5 m/s using the same amount of fuel. The engines use air at 300 K as a cold reservoir." one can write:

\eta_{carnot}=\frac{W}{Q_h}=1-\frac{T_c}{T_h}=\frac{1/2 m v_2^2}{m_f L_i} because you need to produce a work equal to the kinetic energy needed, and the combustion heat is the fuel mass multiplied by the combustion enthalpy. So that we have the first equation:

1-\frac{T_c}{T_h}=\frac{1/2 m v_2^2}{m_f L_i} (1);

From this phrase: "A 20% efficient real engine is used to speed up a train from rest to 5 m/s." one can write again:

\eta_{real engine}=0.2=\frac{W}{Q_h}=\frac{1/2 m v_1^2}{m_f L_i} (2)

Dividing (1) and (2) we have:

\frac{1-\frac{T_c}{T_h}}{\eta_{real engine}}=\Big(\frac{v_2}{v_1}\Big)^2

and solve for T_h.

You should know that going straightforward to the solution is not always the best way for solving a problem. They might be more difficult than we could think.

 

[andrew410]

thx for the help!
I did think about the problem first, but that's all I could think of. yea...I need to study more or something...I'm not fully understanding the material...
anyways, thanks for the help again! :)