# Need help with Heat Engines

1. Nov 6, 2004

### 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! :)

2. Nov 7, 2004

### andrew410

can anyone help ?

3. Nov 7, 2004

### 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?

4. Nov 7, 2004

### Clausius2

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.

5. Nov 7, 2004

### andrew410

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