# Steam Heat Engine Question

GreenPrint

## Homework Statement

18. (II) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate ehat input of the second. Th eoperating temeprautres of the first are 680 degrees C and 430 degrees C, and of the second 415 degrees C and 280 degrees C. If the heat of combuston of coal is 2.8 E8 J/kg, at what rate msut coal be burned if the plant is to put out 900 MW of power. Assume the efficiency of the engines is 65 percent of the ideal (Carnot) efficiency.

## Homework Equations

e = |W|/|QH|

were the H is the subscript

e ideal = 1-TL/TH

were L and H are subscripts

## The Attempt at a Solution

Ok I calculated the efficiency of the first engine to be about 17.05 % efficient and the second one to be about 14.64 % efficient.

I know that efficiency is

e = |W|/|QH|

were the H is the subscript

I know that 900 MW is a value of power which is not work

I know that |QH| is 2.8 E8 J/kg but I do not like the units for this, J/kg, as I thought Q was measured in joules only...

How do I find the work done?

## Answers and Replies

Homework Helper
Consider one second of time. 900 MJ of energy is produced, so |W|=900 MJ. e is the total efficiency of the entire system, first and second engine combined. Once you get |QH|, remember that this is the energy used per second.

GreenPrint
but it's the power done not the work
900 MW of power
hmmmm ok hold up

Homework Helper
I know, but you can always consider the energy expended in a single second of time. After all, that's what power is: energy expended in a single second of time!

If you want, you can rearrange e = |W|/|QH| into |QH| * e = |W|. Take the time derivative of both sides and you get e*d(|QH|)/dt = d|W|/dt: in other words, the rate of heat usage times e is equal to the power generated. If you don't know what a time derivative is or what I'm talking about, don't worry about it; you aren't required to know this.

GreenPrint
ok so I got for |QH| 2816 E 7 J were do I go from here
hmmm hold up

GreenPrint
Ok so I'm doing this wrong lets see here

GreenPrint
e ideal = 1 - TL/TH = 1 - 703.16 K/953.16 K = .2623 [first engine]
e = .65 (e ideal) = .65(.2623) = .1705 [first engine]

e ideal = 1 - TL/TH = 1 - 533.16 K/688.16 K = .2252 [second engine]
e = .65 (e ideal) = .65 (.2522) = .1639 [second engine]

GreenPrint
e = .1705 + .1639 = .3344 [system]
P = W/t = 900 MW = W/1 s
then
W = 900 MW (1 s)
seconds cancle out I'm getting
900 MJ for |W|

Homework Helper
e ideal = 1 - TL/TH = 1 - 533.16 K/688.16 K = .2252 [second engine]

533 K corresponds to 260 C. Is it 260 or 280?

GreenPrint
e = |W|/|QH|
then
|QH| = |W|/e = (900 MJ)/.3344 = 2691 J

Homework Helper
e = .1705 + .1639 = .3344 [system]

That's not right. Suppose an amount of energy, Q, goes into the first engine. Q*e_1 gets spat out as work and goes into the second engine. Then, Q*e_1*e_2 gets spat out as work. So the total efficiency is e1*e2, not e1+e2.

GreenPrint
It was 280 in the problem

GreenPrint
That's not right. Suppose an amount of energy, Q, goes into the first engine. Q*e_1 gets spat out as work and goes into the second engine. Then, Q*e_1*e_2 gets spat out as work. So the total efficiency is e1*e2, not e1+e2.

ok intersection and union...
makes sense lets see now then...

Homework Helper
It was 280 in the problem

OK, so it's 553 K and not 533 K.

GreenPrint
e = e_1 intersection e_2 = (.1705) .1639 = .0279 [system]
that does not seem correct :O

Homework Helper
Well, you have to repeat the calculation for e2 because, as I was saying, you used 533 K for the temperature when it should be 553 K. Other than that, you calculation is correct. You're right that this efficiency is unrealistically low; Google tells me that coal-fired plants typically have an efficiency of 30-50%.

GreenPrint
e = |W|/|Q_H|
then |Q_H| = |W|/e = 500 MJ/.0279 = 1.792 E 10 J

GreenPrint
Well, you have to repeat the calculation for e2 because, as I was saying, you used 533 K for the temperature when it should be 553 K. Other than that, you calculation is correct. You're right that this efficiency is unrealistically low; Google tells me that coal-fired plants typically have an efficiency of 30-50%.

I'm lost...

ok the T_L for the second engine was 280 degrees Celceius

Kelvin = degrees Celceius + 273.16
= 280 degrees Celceius + 273.16
= 533.16 K
What am I doing wrong here?

GreenPrint
1.792 E 10 J /(2.8 E 7 J/Kg)

I'm getting 640 Kg of coal every second?

GreenPrint
Chapter 20: 18

18.First, find the efficiencies of each engine. That’ll make things easier later and help you fill in more information on your diagram. You should find an efficiency of 17.1% for the first engine and an efficiency of 12.8% for the second. Worth noting is the fact that the output of the first engine is the input of the first (or, QL1 = QH2). That means that the work done by the first engine is 17.1% of the energy fed into the first engine, and the work done by the second engine is 12.8% of the wasted heat from the first engine. If your diagram doesn’t make sense at this point, eMail me.

The next trick is an algebraic one. You already have the work from the first engine in terms of the input heat (it’s the input heat times the efficiency of course). Now you need to express the work done by the second engine in terms of the input heat of the first engine. You can do this by first expressing the heat input of the second engine in terms of the input heat of the first (so relate the exhaust of the first engine to its input; the first exhaust equals the second intake).

Now you should have two expressions for the output work, both in terms of the input heat of the first engine. That’s good, because you know the output work (per second) is 600MW. From this, you should be able to calculate the input heat. Your answer for this should be in the neighborhood of 3.25 gigajoules per second.

Knowing the energy density of the fuel, you can easily find out how many kilograms of fuel are needed to supply the appropriate energy per second. It’s a lot, so don’t panic. If you followed all this through, you should have an answer to the tune of 116 kilograms per second.

GreenPrint
Chapter 20: 18

18.First, find the efficiencies of each engine. That’ll make things easier later and help you fill in more information on your diagram. You should find an efficiency of 17.1% for the first engine and an efficiency of 12.8% for the second. Worth noting is the fact that the output of the first engine is the input of the first (or, QL1 = QH2). That means that the work done by the first engine is 17.1% of the energy fed into the first engine, and the work done by the second engine is 12.8% of the wasted heat from the first engine. If your diagram doesn’t make sense at this point, eMail me.

The next trick is an algebraic one. You already have the work from the first engine in terms of the input heat (it’s the input heat times the efficiency of course). Now you need to express the work done by the second engine in terms of the input heat of the first engine. You can do this by first expressing the heat input of the second engine in terms of the input heat of the first (so relate the exhaust of the first engine to its input; the first exhaust equals the second intake).

Now you should have two expressions for the output work, both in terms of the input heat of the first engine. That’s good, because you know the output work (per second) is 600MW. From this, you should be able to calculate the input heat. Your answer for this should be in the neighborhood of 3.25 gigajoules per second.

Knowing the energy density of the fuel, you can easily find out how many kilograms of fuel are needed to supply the appropriate energy per second. It’s a lot, so don’t panic. If you followed all this through, you should have an answer to the tune of 116 kilograms per second.

Hmmmmmmmmmmmmm help

Homework Helper
I'm lost...

ok the T_L for the second engine was 280 degrees Celceius

Kelvin = degrees Celceius + 273.16
= 280 degrees Celceius + 273.16
= 533.16 K
What am I doing wrong here?

Do the calculation again. 280 + 273.16 does not equal 533.16K.

GreenPrint
It does on my calculator :O
weird

oh wow thanks
lets see here

Homework Helper
I'm getting tired of pointing out careless mistakes.

In the first post you said the power plant puts out 900 MW of power. Alright, great. In post #17 you used 500 MW. In the answer in post #20, it's 600 MW. I mean, what the heck?

Also, I'm disappointed at the answer in post #20. It's wrong, period. They took into consideration the fact that the second engine is 0.65 times the Carnot efficiency, but didn't do the same for the first. I mean, what the heck?

GreenPrint
for the second efficiency I'm not getting 12.8

e = 1 - TL/TH = 1 - 553.16/608.16

mulitiplied by .65 gave me a number less than .128

GreenPrint
I'm getting tired of pointing out careless mistakes.

In the first post you said the power plant puts out 900 MW of power. Alright, great. In post #17 you used 500 MW. In the answer in post #20, it's 600 MW. I mean, what the heck?

Also, I'm disappointed at the answer in post #20. It's wrong, period. They took into consideration the fact that the second engine is 0.65 times the Carnot efficiency, but didn't do the same for the first. I mean, what the heck?

hmm do you mean they took into consideration that the engine is .65 times the Carnot efficiency in the first and not the second becasue for some reason I'm not getting .128

GreenPrint
for the second engine I'm getting about .0588

here's what I did

.65 (1 - 553.16/608.16)

Homework Helper
First, can you please tell me whether it's 900 MW, 500 MW, or 600 MW? Second, e = 1 - TL/TH = 1 - 553.16/608.16 is not right because the bottom should NOT be 608.16 K. Recalculate, and make sure you get 0.128 before making another post.

GreenPrint
oh sorry 688 gee I'm really tired sorry

GreenPrint
in book it's 900 MW

GreenPrint
ok i get 12.8

what were you saying about the first one not taking into account that it was .65 Carnot efficiency because I don't see how this is so

GreenPrint
I'm getting tired of pointing out careless mistakes.

In the first post you said the power plant puts out 900 MW of power. Alright, great. In post #17 you used 500 MW. In the answer in post #20, it's 600 MW. I mean, what the heck?

Also, I'm disappointed at the answer in post #20. It's wrong, period. They took into consideration the fact that the second engine is 0.65 times the Carnot efficiency, but didn't do the same for the first. I mean, what the heck?

i don't see how

e (Carnot) = 1 - 703.16 K/953.16 K = .2623
e = .65 e Carnot = .65(.2633) = .1705

what am I doing wrong

Homework Helper
Sorry, my mistake. Ignore that comment; the answer's correct.

GreenPrint
Wait so then...

I got e_system to be .0217
because I did

e_system = e_first intersection e_second = .1705(.1275) = .0217

now

e_system = |W|/|Q_H|

Therefore

|Q_H|= |W|/e_system
= (900 MJ)/.0217
= 4.147 E 10 J
what am I doing wrong now

GreenPrint
So I got the wrong input heat and don't see why...