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Temperature and heat

  1. May 20, 2006 #1
    the question is that:
    A steam pipe with a radius of 2.00cm, carrying steam at 140 degree celcius, is surrounded by a cylindrical jacket with inner and outer radii 2cm and 4cm and made of a type of cork with thermal conductivity 4x10^(-2) W/m K. This in turn is surrounded by a cylindrical jacket made of a brand of Styrofoam with thermal conductivity 1x10^(-2) W/m K and having inner and outer radii 4 cm and 6 cm. The outer suface of the Styrofoam is in contact with air at 15 degree celcius. Assume that this outer surface has a timperature of 15 degree celcius.
    a) What is the timperature at a radius of 4 cm, where the two isulating layers meet?
    b) What is the total rate of transfer of heat out of a 2 m length of pipe?

    thx of attention
  2. jcsd
  3. May 20, 2006 #2


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    Thoughts? Work?
  4. May 20, 2006 #3
    It is necessary to apply the formula H=kA(T2-T1)/L
    (T2 is higher temperature, T1 is lower temperature)

    but the temperature at 2cm and 4cm is different ,
    i can't apply this formula directly.......
    what formula should be used to work out the temperature
    of the same material but different thickness.
  5. May 20, 2006 #4


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    The formula is correct, but you must be aware that you cannot assume a linear temperature gradient here because the area through which heat flows varies with the radius.

    Do you know how to solve a first order differential equation?

    In a cylindrical shell (annulus) of length [tex]L[/tex], thermal conductivity [tex]\kappa[/tex], inside radius [tex]r_i[/tex] and outside radius [tex]r_o[/tex], with the inside held at temperature [tex]T_i[/tex] and the outside at temperature [tex]T_o[/tex], the constant rate of heat flow [tex]\dot{Q}[/tex] is given by

    [tex]\dot{Q} = 2\pi\kappa rL\frac{dT}{dr}[/tex]

    You need to solve that diff equation by separation of variables and integration with the given bounds to get an expression for [tex]\dot{Q}[/tex] in terms of the temperature difference and the inner and outer radii. There will be a natural logarithm in the expression.

    After doing that, let the temperature at the 4cm interface be T. The temperature gradient between the steam pipe and the outside of the cork jacket will be (140-T). That between the outside of the cork jacket and the outside of the styrofoam will be (T-15). You know the heat flux through the whole thickness is the same constant, and you've worked out a formula for it. From that, form a simple linear equation in T and solve it.

    That will give you the first answer. For the second, just use the formula you've derived for the heat flow and the given length to work out the heat flux in watts.
    Last edited: May 20, 2006
  6. May 21, 2006 #5
    I have got the answer , thx so much
  7. May 21, 2006 #6


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    Sure. :smile:
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