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

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Homework Help Overview

The discussion revolves around the performance of heat pumps in different climates, specifically comparing northern and southern regions. The original poster presents a problem involving the coefficient of performance of a heat pump operating at varying outside temperatures and the implications for heating identical houses in these climates.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the calculations for the coefficient of performance in different temperature scenarios and question the implications of heat loss through building materials. There is discussion about the assumptions made regarding heat loss rates and whether they can be considered equal for both houses.

Discussion Status

Some participants have validated parts of the original poster's calculations, while others are probing deeper into the assumptions regarding heat loss. There is an ongoing exploration of how heat loss might vary with temperature differences and the mechanisms involved, such as conduction, convection, and radiation.

Contextual Notes

Participants note the lack of specific numerical data regarding heat loss, which complicates the analysis. The discussion acknowledges that the heat loss may not be uniform due to differing external temperatures and the nature of heat transfer mechanisms.

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 coefficient 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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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:
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.
 
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
 
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.
 
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
 
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.
 
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"
 

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