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I would like to know if there is a way to determine using calculation the inner wall temperature of a pipe which has a steady flow of water through it knowing the temperature measured on the surface of the pipe (outside)?

As the measurement takes place on a very small points, one can consider there will not have any variation of the temperature on the length of the pipe at this point. Moreover if we consider the pipe to be perfectly cylindrical it will not have vairation of the temperature by rotation on the pipe. Therefore, in steady state, the heat equation boils down to :

$$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)=0 \tag 1$$

Which give a solution as, $$T(r)=C_1 \ln (r) + C_2 \tag 2$$

Now I am only having a few difficulties to apply the rights limit conditions. If I say, at the limits $T(r_0) = T_0$ (wall temperature of the pipe inside the pipe) and $T(r_1) = T_1$ (wall temperature of the pipe outside the pipe) then it gives the solution of $T(r)$ as a function of $T_0, T_1, r_0$ and $r_1$. But here the point is that $T_0$ is the temperature I want to determine...

I would like to know the difference of temperature so that I can determine $Q$ the power lost as, $$Q = \frac{T_1 - T_0}{2 \pi \lambda L}$$ and I have absolutely no possibility to measure the temperature inside directly.

I think there is probably something wrong on my way to approach the problem, but I haven't found anything concluant on the web to do such a thing.

I am stuck there and I need some help..

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    $\begingroup$ What is keeping the inside temperature different from the outside temperature different? If the answer is "nothing", then they're probably the same. $\endgroup$ Commented Mar 29, 2019 at 23:55
  • $\begingroup$ As Peter Shor correctly mentioned, if the pipe is not losing heat to the rest of the universe, then $T_1=T_0$. But if there's convection loss, then you need to use that as a boundary condition. So which is it? $\endgroup$
    – Gert
    Commented Mar 30, 2019 at 0:14
  • $\begingroup$ And if the thickness of the wall is small compared to the diameter of the pipe, you can treat the pipe wall as a straight, flat wall. $\endgroup$
    – Gert
    Commented Mar 30, 2019 at 0:35

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The equation you've been trying to derive is $$\dot{Q}=2\pi k\frac{(T_1-T_0)}{\ln{(r_0/r_1)}}$$where $\dot{Q}$ is the rate of heat loss per unit length of pipe and k is the thermal conductivity of the pipe. Note that there are two unknowns ($\dot{Q}$ and $T_1$) but only one equation. To provide closure on this, as @Gert has indicated, you need to characterize the rate of heat loss from the pipe to the surrounding air in the room: $$\dot{Q}=2\pi r_0 h(T_0-T_{surr})$$where h is the convective heat transfer coefficient on the outside of the pipe. You can then get both $\dot{Q}$ and $T_1$ by combining these equations (using an estimate of h).

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    $\begingroup$ Still, quite a bizarre way to measure a temperature! +1 $\endgroup$
    – Gert
    Commented Mar 30, 2019 at 0:38
  • $\begingroup$ @Gert As a chemist, you certainly seem to have some natural talent for engineering. $\endgroup$ Commented Mar 30, 2019 at 0:45
  • $\begingroup$ That calculation for h on both sides of the pipe, as well as an estimate of the rate of heat conduction through the pipe wall given that some fouling will be present, is definitely non-trivial. $\endgroup$ Commented Mar 30, 2019 at 1:04
  • $\begingroup$ @David White If all that is wanted is the temperature at the inside pipe wall, the h on the inside of the pipe is not needed. $\endgroup$ Commented Mar 30, 2019 at 1:18
  • $\begingroup$ @ChetMiller: that's because I'm a chemical engineer. $\endgroup$
    – Gert
    Commented Mar 30, 2019 at 13:50
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Here is a cheap way to get a reading:

For a distance of ~10 pipe diameters, wrap the pipe with a thick layer of insulation after having mounted a thermocouple tightly to the outside of the pipe at the midpoint of the insulated section.

Where the pipe is insulated, heat loss is minimized and the pipe's outside temperature will therefore rise to approximately the inside temperature.

Note that the outside temperature will transiently lag behind the inside temperature in an amount determined by the thickness of the pipe. The thicker the pipe, the slower the response. For thin copper pipe the lag time will be of order ~seconds. You will have to take this time lag into account in your measurements.

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