Carnot engine working in reverse poblem-

[LandOfStandar]

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?

Homework Equations

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

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.

 

[Andrew Mason]

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?

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

 

[LandOfStandar]

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

the other one was from my text

 

[LandOfStandar]

anyone

 

[Andrew Mason]

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?

\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

 

[LandOfStandar]

I did not have those formulas

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

ok, thank you

 

[LandOfStandar]

<br /> \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

 

[LandOfStandar]

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

 

[Andrew Mason]

<br /> \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

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

 

[LandOfStandar]

what is the coefficient of performance is 3.8

 

[Andrew Mason]

what is the coefficient of performance is 3.8

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

 

[LandOfStandar]

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

 

[Andrew Mason]

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