- #1

physicsss

- 319

- 0

I know that the QL of the first engine is the QH of the second engine, but how do I get the actual values with only the efficiency and the ratio of the temperatures?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter physicsss
- Start date

- #1

physicsss

- 319

- 0

I know that the QL of the first engine is the QH of the second engine, but how do I get the actual values with only the efficiency and the ratio of the temperatures?

- #2

physicsss

- 319

- 0

Anyone have an idea?

- #3

SpaceTiger

Staff Emeritus

Science Advisor

Gold Member

- 2,957

- 4

physicsss said:I know that the QL of the first engine is the QH of the second engine, but how do I get the actual values with only the efficiency and the ratio of the temperatures?

You have more than just the ratio of temperatures, you have their actual values. Does this formula look familiar?

[tex]\epsilon=\frac{T_H-T_C}{T_H}[/tex]

That's the efficiency of a perfect Carnot cycle. Using this and what they gave you in the text, you should be able to figure out the efficiency of the machines themselves.

Once you have that, it's just a matter of using the definition of efficiency to get your answer. Here's another formula that might come in handy

[tex]\epsilon=\frac{|W|}{|Q_H|}[/tex]

Remember, efficiency is just the ratio of the energy you get out to what you put in.

- #4

LeonhardEuler

Gold Member

- 860

- 1

[tex]\eta_{th,C}=1-\frac{T_c}{T_h}[/tex]

You know the hot and cold sink temperatures, so you can find the actual thermal efficiency of the engines by multiplying the Carnot efficiency by .6. You also know the thermal efficiency to be defined as:

[tex]\eta_{th}=\frac{w}{q_h}[/tex]

You know the power output you want to be 1000MW, which is 1000MJ in one second, or 8.64*10^7J per day. This is w. All you need to do is find an expression for the amount of work that comes from a given amount of heat input. To do this, you need to know how much energy is rejected by the first engine and put into the second. This comes from conservation of energy. It is just the difference between the heat supplied and the work performed.

- #5

Astronuc

Staff Emeritus

Science Advisor

- 20,995

- 5,053

W

W

see also - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html#c1

- #6

Astronuc

Staff Emeritus

Science Advisor

- 20,995

- 5,053

Correction on this.Astronuc said:

W_{total}= W_{1}+ W_{2}, and

W_{i}= Q_{H,i}- Q_{L,i}

see also - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html#c1

W

if eff. < 100%, the W

Share:

- Last Post

- Replies
- 1

- Views
- 375

- Last Post

- Replies
- 10

- Views
- 542

- Last Post

- Replies
- 1

- Views
- 351

- Replies
- 5

- Views
- 337

- Replies
- 30

- Views
- 920

- Last Post

- Replies
- 14

- Views
- 526

- Replies
- 4

- Views
- 412

- Last Post

- Replies
- 4

- Views
- 460

- Replies
- 1

- Views
- 467

- Last Post

- Replies
- 4

- Views
- 541