How Does Heat Pump Performance Vary Between Northern and Southern Climates?

In summary, a heat pump is designed for southern climates to extract heat from the outside air and deliver air at 20C to the inside of houses. The average coefficient of performance for heat pumps in southern climates is 19.54, while in northern climates it is 9.77. When considering heat loss through walls, windows, and roofs, the ratio of electrical power required to heat and maintain the interior temperature of two identical houses (one in the north and one in the south) is 0.5. This assumes that heat loss is proportional to the temperature difference between the inside and outside of the house.
  • #1
sapiental
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1. North vs. South. A heat pump is designed for southern climates extracts heat from the outside air, and delivers air at 20C to the inside of the houses. What is the average coefficent of performance of the heat pump: (Consider this to be a Carnot device)

a) In the south where the average outside temperature is 5C?
b) In the north where the average outside temperature is -10C?

c)Two Identical houses, one in the north and one in the south, are heated by this pimp, and maintain indoor temperatures of 20C. Considering heat loss through the walls, windows, and roof, what is the ratio of the electrical powers required to heat the houses and to maintain the interiors at 20C. Express your result as P_n_/P_s_

a) K = T_h_/(T_c_-T_h_)

T_c_ = 5C + 273.15 = 278.15K
T_h_ = 20C + 273.15 = 293.15K

K = 293.15/(278.15-293.15)
= 19.54

b) K = T_h_/(T_c_-T_h_)

T_c_ = -10C + 273.15 = 263.15K
T_h_ = 20C + 273.15 = 293.15K

K = 293.15/(263.15K-293.15)
= 9.77

c) solving for P_n_

(9.77kWh)(3.6x10^6J)= 3.51x10^7 J/s

solving for P_s_

(19.54kWh)(3.6x10^6J)=7.03x10^7 J/s

P_n_/P_s_ = (3.51x10^7 W)/(7.03x10^7 W)

= .5


I would really appreciate it if someone reviewed my answeres. What confuses me is the line "Considering heat loss through the walls, windows, and roof". Did I miss plugging in a number somehwere, how do I calculate it? Or is it just theoretical?

Thanks a lot in advance!
 
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  • #2
sapiental said:
I would really appreciate it if someone reviewed my answeres. What confuses me is the line "Considering heat loss through the walls, windows, and roof". Did I miss plugging in a number somehwere, how do I calculate it? Or is it just theoretical?

Thanks a lot in advance!
a) and b) appear to be correct. c) is simply stating that the heat pump is continually working to replenish heat lost. So the ratio of power consumed is a ratio of input work per unit time of one house compared to the other. Since:

[tex]\eta = \frac{Q_{out}}{W_{in}}[/tex]

What is:

[tex]\frac{W_{in-n}}{W_{in-s}}[/tex]?

I think you have to assume that Qout per unit time is the same for both (ie the heat is lost - and therefore replaced by the heat pump - at the same rate for both houses). You do not have enough information otherwise.

AM
 
Last edited:
  • #3
is it

(3.6x10^6)/(2x3.6x10^6)

damn, again .5??

Correct me if I'm wrong but 9.77 = 9.77kWh (Q_out_) per 1 kwh (W_in_)

Thanks again.
 
  • #4
sapiental said:
is it

(3.6x10^6)/(2x3.6x10^6)

damn, again .5??

Correct me if I'm wrong but 9.77 = 9.77kWh (Q_out_) per 1 kwh (W_in_)

Thanks again.
It is simply:

[tex]P_n/P_s = W_{in-n}/W_{in-s} = \frac{Q/\eta_n}{Q/\eta_s} = \frac{\eta_s}{\eta_n} = 2[/tex]

AM
 
  • #5
Andrew Mason said:
I think you have to assume that Qout per unit time is the same for both (ie the heat is lost - and therefore replaced by the heat pump - at the same rate for both houses). You do not have enough information otherwise.

AM
It occurs to me that the rate of heat loss for identical houses should be proportional to the temperature difference between the interior and exterior.
 
  • #6
OlderDan said:
It occurs to me that the rate of heat loss for identical houses should be proportional to the temperature difference between the interior and exterior.
I agree that they won't have the same rate of heat loss, due to the difference in exterior temperatures. But I would think that the rate of heat loss might depend on the way the heat is lost: eg: radiation, convection or conduction. Would heat loss necessarily vary linearly with temperature difference?

AM
 
  • #7
Andrew Mason said:
I agree that they won't have the same rate of heat loss, due to the difference in exterior temperatures. But I would think that the rate of heat loss might depend on the way the heat is lost: eg: radiation, convection or conduction. Would heat loss necessarily vary linearly with temperature difference?

AM
It probably would not be strictly linear, but I believe it is the usual assumption for defining R-Factor in construction. For example

http://esa21.kennesaw.edu/activities/rfactor/rfactor.pdf

The model assumes conduction with rate proportional to temperature difference.
 
  • #8
I think they want you to assume the heat loss Q is proportional to the temp difference between the inside and outside of the house as the house has the same u value.ie heat loss per degree C. as it has the same resistance to heat loss as it is the "same house"
 

What is a heat pump and how does it work?

A heat pump is a device that uses a small amount of energy to transfer heat from one location to another. It works by using a refrigerant to absorb heat from a low-temperature source (such as outdoor air or underground) and then transferring it to a high-temperature source (such as a building or hot water tank).

What factors affect the performance of a heat pump?

The performance of a heat pump can be affected by several factors, including the outdoor temperature, the size and efficiency of the heat pump, the condition of the heat pump and its components, and the design and insulation of the building it is heating or cooling.

What is the coefficient of performance (COP) and how is it related to heat pump performance?

The coefficient of performance (COP) is a measure of a heat pump's efficiency, calculated by dividing the heat output by the energy input. A higher COP indicates a more efficient heat pump, as it is able to produce more heat with less energy input.

How can the performance of a heat pump be improved?

The performance of a heat pump can be improved by ensuring proper maintenance and regular servicing, installing an appropriately sized and efficient heat pump for the building's needs, and optimizing the design and insulation of the building to maximize the heat transfer from the heat pump.

What is the difference between heating and cooling performance of a heat pump?

The heating and cooling performance of a heat pump may differ due to the different temperature ranges required for each function. In general, heat pumps are more efficient at heating than cooling, as they are able to extract heat from the colder outdoor air and transfer it into the warmer indoor air. However, with proper sizing and maintenance, the difference in performance can be minimized.

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