Heat Pump Thermodynamics Question

In summary, the heat pump needs W.in to operate, and without it, the minimum work required is determined by the temperatures between which it is operating.
  • #1
mcomputing
2
0
I have a general question, it's not homework or anything, we are studying heat pumps and thermal efficiencies and COP. There's an example in my book that has a heat pump that is heating a house and it asks to find the minimum theoretical cost per day. Of course the solution and detailed steps are in my textbook.

However I was thinking what if I wanted to raise the temperature from any given T1 to T2 using a heat pump that gives me the COP and the heat loss or Q_loss and the mass of the house. I want to find how long it would take to raise the temperature from let's say 10 degrees Celsius to 22 degrees Celsius. I think I have an idea of how to go about solving such a problem.

Is the the process:

I have the energy balance equation which gives me sum(Q) = mC_v*delta(T)
I have a sum to be the Qin - Qloss = mC_v*delta(T). If I am given the Qloss and the mC_v and the Qin I can obviously solve for delta(T) and use that to find Tf if I am given a Ti.

Then I can use the COP equation of COP = Q.in/W.in to find Q.in. and Q.in would be equal to Qin/delta(t_sec).

Now my question is what if I wasn't given W.in for the heat pump, how would I be able to solve for it with my given information. Please advise. Thank you very much.
 
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  • #2
Then you will have some sort of differential equation. How good is your math?
 
  • #3
mcomputing said:
Now my question is what if I wasn't given W.in for the heat pump, how would I be able to solve for it with my given information.
The minimum work required by the heat pump is a function of the temperatures between which it is operating. For a Carnot heat pump:

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

W= Qh(Th-Tc)/Th = Qh(1-Tc/Th)

AM
 
  • #4
Thanks for all your help. I was able to figure this out by asking my professor. He basically told me that not being given W.in is beyond the scope of the class I am taking. I think I was over thinking the question and imagining all possibilities.
 
  • #5
mcomputing said:
Thanks for all your help. I was able to figure this out by asking my professor. He basically told me that not being given W.in is beyond the scope of the class I am taking. I think I was over thinking the question and imagining all possibilities.

I hate it when teachers say this.
 
  • #6
khemist said:
I hate it when teachers say this.



This is a real-world problem (which I thought of and solved myself a while back) but he's too lazy to explain it to the student.That's the human mentality:

"If it don't make my paycheck bigger, I don't give a flying f***."
 

FAQ: Heat Pump Thermodynamics Question

1. What is a heat pump and how does it work?

A heat pump is a device that uses thermodynamics to transfer heat from one location to another. It uses a compressor, evaporator, condenser, and expansion valve to circulate a refrigerant and absorb heat from the outside environment. This heat is then compressed and released into the inside of a building.

2. What is the coefficient of performance (COP) of a heat pump?

The COP is a measure of the efficiency of a heat pump. It is calculated by dividing the amount of heat output by the amount of energy input. The higher the COP, the more efficient the heat pump is at transferring heat.

3. How does the outside temperature affect the performance of a heat pump?

The efficiency of a heat pump is affected by the outside temperature. As the outside temperature drops, the heat pump has to work harder to extract heat from the environment. This can decrease its COP and make it less efficient. However, modern heat pumps have been designed to work well in colder temperatures.

4. Is a heat pump a cost-effective heating and cooling option?

In general, heat pumps are considered to be a cost-effective heating and cooling option, especially compared to traditional heating systems like furnaces. They are more energy-efficient and can save homeowners money on their utility bills in the long run. However, the initial cost of installing a heat pump may be higher.

5. Can a heat pump be used in any climate?

Heat pumps can be used in most climates, but they are most efficient in moderate climates where the temperature does not regularly drop below freezing. In colder climates, a supplemental heating source may be needed to keep the home warm during extreme cold temperatures. In hotter climates, a heat pump can also be used for cooling purposes.

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