# Thermodynamics - efficiency question

1. May 14, 2006

### N_L_

The burning of gasoline in a car releases about 3.0 x 10^4 kcal/gal. If a car averages 41 km/gal when driving 90 km/h, which requires 25 hp, what is the efficiency of the engine under those conditions?

I think I have an answer, but I'm not sure if it's correct.

1 horsepower = 746 watts

25 hp = 18650 watts

1 kcal = 4186 joules (watts / sec)

e = W / Qh = 1 - (Ql / Qh)

1 gal / 41 km at 91 km / h means that that 2.26 gallons will be used in one hour.

Since no temperatues are given I don't think that the 1 - Ql / Qh is needed...

Converting 30000 kcal / gal to watts I get 34883 joules/sec. Since 2.26 gal are used, I multiplied the above by 2.26 to get 78836.

Dividing 18650 watts (from hp) by 78836 I got an efficiency of 23.46%.

2. May 14, 2006

### Andrew Mason

You don't need to find Qc but it is just:$$Qc = Qh - W$$. You don't need temperatures to find this.

$$\eta = W/Q_h = (dW/dt)/(dQ_h/dt)$$

Your answer is almost right. I get 2.195 gal/hr (90/41) not 2.26.

$$dQ_h/dt = 2.195 * gal/hr = 3 x 10^4 * 2.195 * 4.186 KJ/hr = 2.76 x 10^8 J/hr = 7.66 x 10^4 J/sec$$

$$dW/dt = Power = 25 * 746 J/sec = 1.87 x 10^4 J/sec$$

So:

$\eta$ = 1.87/7.66 = 24.4%

AM

Last edited: May 14, 2006
3. May 14, 2006

### Hootenanny

Staff Emeritus
I have not rigorously checked your answer, however the efficency appears reasonable for a car, just beware of rounding too early in your calculations, this could induce significant errors. For example you obtained 78836 watts for the input power of the engine, however I obtained $78798\frac{2}{3}$.

Just a small point that I sould point out is that here you said;
I'm sure this is just a typo, but this should be joules = watts * sec. Power is work done (energy) divided by time, therefore it follows that energy is the product of power and time.

~H

Sorry AM, didn't see your post. I sould learn to type faster

4. May 14, 2006

Thank you.