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Conduction problem!

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose the insulating qualities of the wall of a house come mainly from a 4.0-in. layer of brick and an R-19 layer of insulation, as shown in the figure. What is the total rate of heat loss through such a wall, if its total area is 190ft^2 and the temperature difference across it is 10F?


    2. Relevant equations

    dQ/dt = -kA dT/dx

    3. The attempt at a solution

    uh...so i don't know the thermal conductivity of "r-19" (w/e that is) and it says the temperature difference across both the brick and the r-19 layer is 10F which baffles me on how to solve this problem. :(
     
  2. jcsd
  3. Sep 17, 2008 #2

    LowlyPion

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    This might help:
    http://en.wikipedia.org/wiki/R-value_(insulation)

    But it looks like you need the R-value of the brick as well.
     
  4. Sep 17, 2008 #3
    I have the R-value for the brick wall but I can't find the r-value for r-19 through that wikipedia link...:(
     
  5. Sep 17, 2008 #4

    LowlyPion

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    Don't the R values add?

    Aren't its units (K·m²/W)?

    If the R-19 is US units, then the conversion factor to SI units is 1 ft²·°F·h/Btu ≈ 0.1761 K·m²/W isn't it?
     
  6. Sep 17, 2008 #5
    um....huh? all i have is that the thermal conductivity of brick is .84J/ s * m * C

    what exactly is r-19 anyway? arrgh so confusing~
     
  7. Sep 17, 2008 #6

    LowlyPion

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    J/s are Watts so that becomes immediately .84 W/(m2*C)
    For temperature note Degrees C are Degrees K - 273. But we can treat them the same since ΔC will = ΔK. (That is we are going to be looking at changes in temperature, and the differences will be the same.)

    The value they give is apparently U which is 1/R.

    And U is in SI units.

    So now the trick is to determine R-19 in SI units for the insulation and match up the U value of the brick to get the total insulating property.
     
  8. Sep 17, 2008 #7

    LowlyPion

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    To convert US R-19 to proper units you use 19*(0.1761 K·m²/W) = 3.35 K·m²/W

    And using the R value of the brick = 1/.84 = 1.19 K·m²/W

    Now you can add the two 1.19 + 3.35 = 4.54 K·m²/W
    Now all that's left is to figure the rate of total heat loss across the area given in the problem with the heat difference between inside and outside of 10 degrees F - Note Fahrenheit not C. And area is in Sq - Ft not Sq - m.

    Any idea what units they want the answer in? Watts right?
     
    Last edited: Sep 17, 2008
  9. Sep 17, 2008 #8
    The units for thermal conductivity should be W/m-K and in English units Btu-in/hr-ft^2-F. If you want to work in ft then Btu/hr-ft^2-F. The unit for heat loss will be in W/m^2 or Btu/hr-ft^2.
     
  10. Sep 17, 2008 #9

    LowlyPion

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    Yes the units can go either British or SI. And British is probably the easier way to go, meaning only his value for brick need to be converted.

    I should have taken the final question into account at the beginning and saved the extra conversions.

    But that said they are looking for the Total Watts over the 190 Sq-feet given.

    So long as everything is converted to a consistent set of units it should arrive at the same answer.
     
  11. Sep 17, 2008 #10
    Ohh, I didn't see that a total area was given. Yes then you are correct.
     
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