• Support PF! Buy your school textbooks, materials and every day products Here!

Carnot engine working in reverse poblem- need help

1. Homework Statement

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. Homework 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.
 

Answers and Replies

Andrew Mason
Science Advisor
Homework Helper
7,546
318
1. Homework Statement

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. Homework Equations

Kc = QL /(QH - QL) = TL/(TH-TH)
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
 
k =Qin/W = Qin/(Qin - Qout) = Tin/(Tin - Tout)

the other one was from my text
 
anyone
 
Andrew Mason
Science Advisor
Homework Helper
7,546
318
k =Qin/W = Qin/(Qin - Qout) = Tin/(Tin - Tout)

the other one was from my text
So how is the overall COP related to the COPs of each part?

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

Now we know W1:

[tex]W_1 = Q_{h1}/\kappa_1 [/tex]

We also know W2:

[tex]W_2 = Q_{h2}/\kappa_2 [/tex]

We know [itex]\kappa_1[/itex] and [itex]\kappa_2[/itex]: eg:

[tex]\kappa_1 = T_1/(T_1-T_2)[/tex]


Work out the expression [itex]\kappa_{total}[/itex] in terms of these known variables.

AM
 
Last edited:
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
 
[tex]
\kappa_1[/tex] 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:
Please someone help I need to understand this the exam is coming up and I need an understand of this
 
Andrew Mason
Science Advisor
Homework Helper
7,546
318
[tex]
\kappa_1[/tex] 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
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
 
what is the coefficient of performance is 3.8
 
Andrew Mason
Science Advisor
Homework Helper
7,546
318
what is the coefficient of performance is 3.8
The question is not clear. If it is a Carnot heat pump, the COP is:

[tex]COP = Q_h/W = Q_h/(Q_h - Q_c) = T_h/(T_h - T_c)[/tex]

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
 
the answer is 440Watt and I can not get it either way
 
Andrew Mason
Science Advisor
Homework Helper
7,546
318
the answer is 440Watt and I can not get it either way
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
 

Related Threads for: Carnot engine working in reverse poblem- need help

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
13
Views
3K
Replies
14
Views
5K
Replies
2
Views
928
  • Last Post
Replies
6
Views
2K
Top