# Carnot engine working in reverse poblem- need help

1. Mar 25, 2008

### LandOfStandar

1. The problem statement, all variables and given/known data

A heat pump is used to heat a building. The outside temp is -5.0 Celsius and the temp inside the building is to be maintained at 22 Celsius. The pump's coefficient of performance is 3.8, and the heat pump delivers 7.54MJ as heat to the building each hour. If the heat pump is Carnot engine working in reverse, at what rate must work be done to run it?

2. Relevant equations

Kc = QL /(QH - QL) = TL/(TH-TH)

3. The attempt at a solution

I am lost, help, Am I using the right equation, what do I do first.

I have tried this many times and I just need help understanding what I am looking for and how to start to get it.

2. Mar 25, 2008

### Andrew Mason

Your definition of Kc (COP) is incorrect. What is the definition of COP for a heat pump in terms of heat out, work in?

For a Carnot cycle, how is that efficiency related to temperature?

AM

3. Mar 25, 2008

### LandOfStandar

k =Qin/W = Qin/(Qin - Qout) = Tin/(Tin - Tout)

the other one was from my text

4. Mar 26, 2008

### LandOfStandar

anyone

5. Mar 26, 2008

### Andrew Mason

So how is the overall COP related to the COPs of each part?

$$\kappa_{total} = Q_{total}/W_{total} = (Q_{h1} + Q_{h2})/(W_1 + W_2)$$

Now we know W1:

$$W_1 = Q_{h1}/\kappa_1$$

We also know W2:

$$W_2 = Q_{h2}/\kappa_2$$

We know $\kappa_1$ and $\kappa_2$: eg:

$$\kappa_1 = T_1/(T_1-T_2)$$

Work out the expression $\kappa_{total}$ in terms of these known variables.

AM

Last edited: Mar 26, 2008
6. Mar 26, 2008

### LandOfStandar

I did not have those formulas

I wish I could find these in my text book Fundamentals of Physics by Halliday and Resnick

ok, thank you

7. Mar 26, 2008

### LandOfStandar

$$\kappa_1$$ is that the pump's coefficient of performance

with one being the given 3.8 and finding the other as 9.926

are one of the Q 7.54x10^6J what is the other, or is that Q total?

k total = 3.8 + 9.9 = 13.7

Q total = 7.54x10^6J

what am I trying to get?

rate must work be done
W net / 3600sec (one hour) to get Watts or power

Last edited: Mar 26, 2008
8. Mar 26, 2008

### LandOfStandar

Please someone help I need to understand this the exam is coming up and I need an understand of this

9. Mar 27, 2008

### Andrew Mason

I confused you on my last answer. I was thinking of the other problem you had posted on the efficiency of two Carnot engines. Sorry about that. Just ignore my last answer.

You are trying to get the amount of work done (per unit time).

You have correctly stated the relationship between COP and temperature:

COP = Qh/W = Th/(Th-Tc)

You are given Qh (actually dQh/dt) = 7.54MJ (per hour)

You are given Th and Tc. You just need to determine W (per hour).

AM

10. Apr 2, 2008

### LandOfStandar

what is the coefficient of performance is 3.8

11. Apr 2, 2008

### Andrew Mason

The question is not clear. If it is a Carnot heat pump, the COP is:

$$COP = Q_h/W = Q_h/(Q_h - Q_c) = T_h/(T_h - T_c)$$

For these temperatures, the COP is 10.9

However, the question says that the COP is 3.8. So it is not a Carnot heat pump. Work out the rate of work for both COPs and tell your prof that the question is not clear which one you are supposed to use.

AM

12. Apr 7, 2008

### LandOfStandar

the answer is 440Watt and I can not get it either way

13. Apr 7, 2008

### Andrew Mason

440 watts is not correct. 440 watts = 440 x 3600 = 1,584,000 Joules/hour. At a COP of 3.8, this means it is pumping 6,019,000 Joules of heat per hour. At a COP of 10.9 (295/27), it would pump 30.1 MJ per hour.

AM