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For a project, I am looking to calculate the drop in temperature across a length, $L$ of PVC pipe of diameter, $d$. Water, initially at temperature $T_1$ enters the pipe with a constant flow rate of $Q$. The ambient temperature outside of the pipe is $T_A$.

From what I understand,

$$\frac{dQ}{dt} = \frac{\lambda \times A \times (T_A - T)}{s}$$

Where $\frac{dQ}{dt}$ is the rate at which heat is leaving the pipe, $s$ is the thickness of the pipe, $\lambda$ is the thermal conductivity of PVC (0.19 W/mK) and $T$ is the temperature of the fluid inside of the pipe. Can this equation be applied to determine an expression for $T_2$ assuming turbulent flow? If so, how?

Otherwise, how else would you determine an expression for the temperature of $T_2$?

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  • $\begingroup$ If $Q$ is volumetric flow rate, then $dQ/dt$ has units of volumetric acceleration? You might review your fundamental units on the equations and fix the confusion. $\endgroup$ Commented Oct 5, 2020 at 16:00
  • $\begingroup$ @JeffreyJWeimer Thanks for pointing that out! I should have specified that dQ/dt is the rate of heat transfer out of the pipe. I'll fix that up now. $\endgroup$ Commented Oct 6, 2020 at 0:13

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The resistance to heat transfer across the pipe wall is not the only resistance to heat flow involved. There is also series resistance between the inner pipe wall and the bulk flow, through the turbulent internal thermal boundary layer, and external series resistance between the outer pipe wall and the surroundings at TA. And the temperature difference between the ambient surroundings and the flowing fluid is not constant, because the latter is changing along the pipe. The overall heat transfer equation is given by $$Q\rho C\frac{dT}{dx}=UP(T_A-T)$$where x is distance along the pipe, T is the fluid temperature, C is the fluid heat capacity, $\rho$ is the fluid density, P is the wetted perimeter, and U is the overall heat transfer coefficient. U is the inverse of the overall series resistance. To get the individual contribution to U from the inner resistance, one uses the so-called Dittus-Boelter equation, relating the internal heat transfer coefficient to the turbulent flow Reynolds number and Prantdl number of the fluid. The heat transfer resistance to the surroundings is determined in a similar approach considering external natural convection and forced convection. For details on all this, see Transport Phenomena by Bird, et al, Chapter 14.

I might mention that the conductive resistance of the wall is typically not the dominant resistance to the heat flow.

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  • $\begingroup$ This is basically an abridged version of a fuller solution with an equation for the profile as posted at this link. What is also lacking here is the analytical equation for the temperature profile and units on the terms (e.g. to avoid ambiguity that $Q$ volumetric flow not heat). $\endgroup$ Commented Oct 5, 2020 at 15:58
  • $\begingroup$ Good point regarding units. $\endgroup$ Commented Oct 5, 2020 at 17:41
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The temperature difference between the two end is constant. The situation is very much similar to electric water heater. The equation you should apply for the heat loss through the fluid should be - $$\frac{dQ}{dt} = \frac{dm}{dt}s|∆T|$$

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  • $\begingroup$ Thanks. Just a point of clarification, if your $s$ is the thickness of the pipe, how would that equation be valid? Since when checking the units, $\frac{J}{s} = \frac{kg}{s} \times m \times K$. Isn't that problematic since $\frac{J}{s} \neq \frac{kg m K}{s}$ ? $\endgroup$ Commented Oct 5, 2020 at 5:43
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    $\begingroup$ @Elliot Fekete $s$ represents the specific heat of water. $\endgroup$
    – Ankit
    Commented Oct 5, 2020 at 5:45
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    $\begingroup$ @Ankit Okay, that clears things up. Thank you! $\endgroup$ Commented Oct 5, 2020 at 5:47

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