# Heat loss of an insulated pipe vs an uninsulated pipe

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?

• Let's see your calculation. Your result doesn't make sense mathematically. Commented Mar 5, 2021 at 19:33
• Why not? @ChetMiller Commented Mar 5, 2021 at 19:43
• Because the sum of the two terms is going to be larger than the 2nd term alone. Commented Mar 5, 2021 at 22:01
• 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. Commented Mar 6, 2021 at 14:49

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.