1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat Pump performance

  1. Oct 13, 2006 #1
    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 alot in advance!
     
  2. jcsd
  3. Oct 13, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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: Oct 13, 2006
  4. Oct 13, 2006 #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.
     
  5. Oct 13, 2006 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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
     
  6. Oct 13, 2006 #5

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Oct 13, 2006 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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
     
  8. Oct 13, 2006 #7

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  9. Aug 15, 2008 #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"
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Heat Pump performance
  1. Heat Pump (Replies: 3)

  2. Heat Pumps (Replies: 3)

  3. Heat Pump (Replies: 6)

  4. Heat Pump (Replies: 3)

Loading...