Heat pumps and coeffiecent of performance

In summary, the user is asking for help with a thermodynamics problem involving a heat pump and coefficient of performance. They are given specific temperatures and need to find the rate of work needed to run the pump. The user mentions the equation for coefficient of performance and is confused about what to do with the temperatures given. They also mention converting heat to SI units in order to get the result in watts.
  • #1
rmorelan
8
0
Hi everyone, I was hoping to ask a thermodynamics problem, specificially one about heat pumps and coeffiecent of performance. I am wondering if I am missing something.

I am given that heat pump is heat a building. Inside temperature is to be maintained at 22 degree celcius, outside is -5.0 celcius. The coefficent of performance is 3.8, and the the pump delivers 7.6MJ of heat each hour to the building. I need to find out at what rate work needs to be applied to run the pump.

Homework Equations



Now I believe the equation of the the coeffiecient of performance (CP) is equal to:

cp = |Q|/W

Now this is where I am confused, as I don't understand what (if anything) I am supposed to do with the temperatures given. I know cp, and I know Q, so I can solve for W needed to run the pump can't I? I would need to convert the heat produced to be in SI in order to get Watts as a result of course.

thanks for your help!
rob
 
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  • #2
If we treat the heat pump as a Carnot heat pump we can say that;

[tex]\frac{Q_2}{Q_1} = \frac{T_2}{T_2}[/tex]

Hence;

[tex]\eta^{HP}_{C} = \frac{Q_1}{W} = \frac{Q_1}{Q_1-Q_2} = \frac{T_1}{T_1-T_2}[/tex]

Where T1 and T2 are the temperatures of the hot and cold reservoirs respectively.
 
  • #3


Hello Rob,

Yes, you are on the right track. The coefficient of performance (COP) of a heat pump is defined as the ratio of heat output (Q) to the work input (W). So, using the equation you provided, we can rearrange it to solve for W:

W = Q/CP

In this case, Q is given to be 7.6 MJ per hour. However, we need to convert this to SI units, which is Joules per second (J/s) or Watts (W). To do this, we divide 7.6 MJ by 3600 seconds (1 hour) to get 2111.11 J/s or 2111.11 W.

Next, we need to calculate the COP. This is given to be 3.8. So, we can plug this into the equation for COP and solve for W:

3.8 = 2111.11/W

W = 2111.11/3.8

W = 555.55 W

Therefore, the rate of work needed to run the heat pump is 555.55 Watts.

Hope this helps! Keep up the good work in thermodynamics.
 

1. What is a heat pump?

A heat pump is a device that transfers heat from a colder area to a warmer area, using mechanical energy. It can be used for heating or cooling purposes depending on the direction of heat transfer.

2. How does a heat pump work?

A heat pump works by using a refrigerant to absorb heat from a cold source (such as outdoor air, water or the ground) and then transferring it to a warmer area (such as a building interior). This process is achieved by compressing and expanding the refrigerant, which changes its temperature and pressure.

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

The coefficient of performance (COP) is a measure of a heat pump's efficiency. It is the ratio of the heat output to the energy input required to operate the heat pump. Generally, the higher the COP, the more efficient the heat pump is.

4. How is the COP of a heat pump calculated?

The COP of a heat pump is calculated by dividing the heat output by the energy input. The heat output is typically measured in BTUs or kilowatts, while the energy input is measured in watts or kilowatts.

5. What factors affect the COP of a heat pump?

The COP of a heat pump can be affected by several factors, such as the outdoor temperature, the size and efficiency of the heat pump, the quality of insulation in the building, and the desired temperature inside the building. Higher outdoor temperatures and smaller, more efficient heat pumps generally result in a higher COP.

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