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Energy use, conductive-convective layers and air films

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    The door of a refrigerator is 1.5 m high, 0.80 m wide and 6.0 cm thick. Its thermal conductivity is 0.21Wm-1 degrees Celsius-1.
    a) what is the heat loss per hour through the door neglecting convection effects?
    b) Air is usually still inside the refrigerator so there will be a conductive convective layer on the inside but in a busy kitchen such a layer will be absent on the outside. What is the heat loss per hour in this case and what is the temperature of the inside surface of the door?
    C) at night the kitchen is also still; what is the heat loss per hour now, and what are the temperatures of the inside and outside surfaces?

    A=1.2 m^2
    Δx=.06 m
    k=.21Wm-1°C-1
    ΔT=27°C
    t=3600 seconds

    2. Relevant equations
    For a) Q=kA(ΔT/Δx)t
    For b) part one, Q=(Tf-Ta)xhxAxt, where Tf is Temp in fridge and Ta is temp of air outside fridge.
    For b) part two, I have no idea.
    For c) I have no idea.

    3. The attempt at a solution
    I was able to solve a) through the equation above. I know that k=.21Wm-1°C-1, A=1.2m^2, ΔT=27, Δx=.06, and t=3600 seconds.

    For b) part one, I used the equation above to solve for the first part. I'm not even sure if I did it right because the book rounds the answer up kind of by a lot. Using that equation, I got 1.7 * 10^5 Joules. The book says 2.0*10^5 Joules. I realized though that for this part, I thought "h" meant height, but it's actually the conduction-convection parameter which is 1.8(ΔT)^1/4. In this case, ΔT is 27°C, but the equation already incorporates ΔT into the equation so I don't understand why we use ΔT twice in this equation. If I solve for "h" using 27°C, I get h=4.1 and plugging that back in to the equation, I get 4.8*10^5 Joules which is definitely wrong.

    For b) part two, is it right to assume that the temperature at the surface inside the door is right between the temperature in the fridge and outside of the fridge? Which means it would be 13.5°C. Or is there a way to actually solve for this? The book says it's 13.5°C so it seems like we just assume it's in the middle.
     
    Last edited: Feb 26, 2017
  2. jcsd
  3. Feb 27, 2017 #2
    In part b, the ##\Delta T## refers to the temperature difference between the air inside the refrigerator and the inner surface of the door.
    In part c, the ##\Delta T## refers separately to the temperature difference between the outer surface of the door and the room temperature.

    And, yes, there is a way of solving all this.
     
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