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So I have this exercise: A hot water pipe consists of a copper tube of length $L = 1$ m, of thermal conductivity $\lambda_{1} = 380$, has an internal radius of $R_{1} = 6\cdot10^{-3}$ m and an external radius of $R_{2} = 7\cdot10^{-3}$ m. Using an insulating material of thermal conductivity $\lambda_{2} = 0.1$, a coaxial sheath with an internal radius $R_{2}$ and an external radius $R_{3} = 8\cdot10^{-3}$ is produced. The temperature of the internal wall is $T_{1} = 80ºC$ and the temperature of the ambient air is $T_{2} = 20ºC$. $h = 10$ is the heat transfer coefficient by convection at the outer surface of the insulator (or of the copper tube in the absence of insulation)."

I had to calculate the heat loss per meter of a pipe. First of the uninsulated pipe, and then of the pipe with a coaxial sheath of an insulating material.

The resistance of the copper can be neglected so we get this expression for the total resistance:

$R_{T} = \frac{ln(\frac{R_{3}}{R_{2}})}{2\pi\lambda_{2}L} + \frac{1}{2\pi hR_{3}L} = 2.2$

When there's no insulator we just have the second term (due to convection):

$R_{T} = \frac{1}{2\pi hR_{2}L} = 2.27$

Then to calculate the heat loss we just simply do:

$\phi = \frac{(80 - 20)}{R_{T}} = 27.3$ for the insulated pipe

$\phi = \frac{(80 - 20)}{R_{T}} = 26.4$ for the uninsulated pipe

How is it possible that I get a greater heat loss when the pipe is insulated than when it isn't?

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  • $\begingroup$ Let's see your calculation. Your result doesn't make sense mathematically. $\endgroup$
    – Chet Miller
    Commented Mar 5, 2021 at 19:33
  • $\begingroup$ Why not? @ChetMiller $\endgroup$
    – schrodingal
    Commented Mar 5, 2021 at 19:43
  • $\begingroup$ Because the sum of the two terms is going to be larger than the 2nd term alone. $\endgroup$
    – Chet Miller
    Commented Mar 5, 2021 at 22:01
  • $\begingroup$ I looks like there is a worst value for the insulation thickness. This is for R3 = 0.01. By making R3 larger, the 1st term gets larger and the 2nd term gets smaller. $\endgroup$
    – Chet Miller
    Commented Mar 6, 2021 at 14:49

3 Answers 3

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This is a very careless error in the calculation as pointed out by others. The sum of two positive quantities should be greater than either or them taken individually. Please do check your calculations.

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As already pointed out this is an error in the calculation, given that the formulas you have used is correct(I haven't checked the formula because the error was very clear).

The value for $$\frac{ln(\frac{R_{3}}{R_{2}})}{2\pi\lambda_{2}L} = \frac{0.133}{0.628}=0.2125$$ My point is its positive and cannot be negative since both the numerator $ln(R_3/R_2)$ and the denominator is positive. This added to the second term will always be greater than the second term alone.

Please redo the calculation, since I am not allowed to provide Homework solutions in this platform.

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It looks like there is a worst value for the insulation thickness. This is for R3 = 0.01, for which R = 2.16. By making R3 larger, the 1st term gets larger and the 2nd term gets smaller. For values of R3>0.1, R increases monotonically with R3.

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