Increase Carnot Efficiency: Lower Low-Temp Res?

[endeavor]

Suppose you had the choice of either raising the high-temperature reservoir a certain number of kelvins, or lowering the low-temperature reservoir the same number of kelvins. Which would you choose (assuming you wanted to increase the efficiency, of course)?

After doing experimenting with some random numbers, I thought the answer to be: lower the low-temperature reservoir.

For example,
1 - 100/300 = 66.7%
1 - 95/300 = 68.3%
1 - 100/305 = 67.2%
both of the last two are greater than the first, but the second has the greatest efficiency.
However, my book says to raise the high-temperature reservoir. Is my book wrong?

 

[Chi Meson]

Could you tell us what the question asks, exactly? I'm thinking there's something missing. Carnot efficiency approaches 100% as the cold resevoir approaches zero kelvins; for a constant temperature difference, the greatest efficiency will be found for the coldest cold resevoir. This is what you have noticed. But it would take more energy to remove 5 K from the cold than it would to add 5 K to the hot.

 

[endeavor]

The exact question reads:
"The Carnot efficiency shows that the greater the temperature difference of the reservoirs of a heat engine, the greater the efficiency. Suppose that you had the choice of either raising the high-temperature reservoir a certain number of kelvins, or lowering the low-temperature reservoir the same number of kelvins. Which would you choose (assuming you wanted to increase the efficiency, of course)?"

 

[Cyrus]

The carnot heat engine efficiency is given to us as:

\eta_{th, rev} = 1 - \frac{T_L}{T_H}

What can you conclude?

In other words, what happens as the denominator, T_H approaches infinity? similarly, what happens as the numerator, T_L approaches zero?

Edit: I think you and Chi are right and your book is wrong.

 

[Chi Meson]

I've been over this several ways, but if you look at the other version of the carnot efficiency formula, it seals it:
e=(Thot-Tcold)/ Thot

this says that if the difference in temperature is the same, the greater that T-hot is, the lesser the efficiency is.

 

[Cyrus]

As I am sure you are already aware, your formula is the same thing Chi Meson. The formula\eta_{th, rev} = 1 - \frac{T_L}{T_H} is derived from the reduction of your formula. (Just simplify your fraction)

 

[Chi Meson]

As I am sure you are already aware, your formula is the same thing Chi Meson. The formula\eta_{th, rev} = 1 - \frac{T_L}{T_H} is derived from the reduction of your formula. (Just simplify your fraction)

Yeah. It's just that looking at the other version (of the same thing) it was more obvious to me that the book is incorrect. You can simply say that efficiency is inversely proportional to T hot (if delta T is constant).

 

[endeavor]

I thought my book was wrong.
Thanks, all.