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A $5\,\mathrm{m}$-long heated pipe is used heat up water from $15\sideset{^{\circ}}{}{\mathrm{C}}$ to $65\sideset{^{\circ}}{}{\mathrm{C}}$. Water flow through the pipe is $10\frac{\mathrm{L}}{\mathrm{min}}$. The heating gives a constant heat flow in all parts of the pipes surface. The inner and outer diameter of the pipe is $30$ and $50\,\mathrm{mm}$, respectively. Calculate the heat transferred to the water and the inner surface temperature of the pipe at the end (point where water leaves the pipe at temperature of $65\sideset{^{\circ}}{}{\mathrm{C}}$), the inner heat transfer coefficient is $h_{\text{i}}=1548\frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}}$ at the pipe exit.

The heat transferred to the water, $m$, is mass flow:$$ Q_1~~=~~m \cdot C_{\text{p}} \, \left(65\sideset{^{\circ}}{}{\mathrm{C}} \, -\ \, 15\sideset{^{\circ}}{}{\mathrm{C}}\right) ~~=~~34.8\, \mathrm{kW} $$

To calculate the inner surface temperature, I want to use$$ Q_2~~=~~h \cdot A \, \left(T_{\text{wall}} \, - \, 65\sideset{^{\circ}}{}{\mathrm{C}}\right) \,,$$where $A$ is the inner surface area of the pipe.

Questions:

  1. Under what conditions can we say that $Q_1 = Q_2$?

  2. Assuming that $Q_1$ can equal $Q_2$, is this possibly only because the problem assumes that the heat flow is constant in all parts of the pipe?

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  • $\begingroup$ I am wondering under what conditions we can say that Q1 = Q2. That is what is done in the solution to this problem to find Twall. $\endgroup$ – Jmei May 22 '18 at 9:01
  • $\begingroup$ In the problem statement it is given that the "heating gives a constant heat flow in all parts of the pipe". I am wondering if that is why we can, in this case, say that Q1 = Q2 and solve for Twall. $\endgroup$ – Jmei May 22 '18 at 9:05
  • $\begingroup$ Anyone able to give me an answer here? $\endgroup$ – Jmei May 22 '18 at 13:11
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If the pipe heats the water in a steady state conditions, no heat is spent on heating the pipe itself (it is already done) so the pipe serves as a heat conductur with the effective surface of exchange determined with the inner diameter. No longitudial heat flow along the pipe is taken into accound, it is negligible despite a temperature gradient in the pipe towards the end. In order to have the same heat flux, the inner wall temerature must grow from the inlet to the ourlet of the pipe to keep the same $\Delta T$ between the liquid and the surface.

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  • $\begingroup$ And to have the same heat flux across the pipe the heat flow over the length of the pipe must be constant, I assume? $\endgroup$ – Jmei May 22 '18 at 8:41
  • $\begingroup$ Heat flux is a local thing, the exchanged power in the pipe is an integral thing over the length. Locally the temperature difference between the liquid and the surface must be the same, but as the liquid gets hotter to the end, the pipe surface temperature must be higher. $\endgroup$ – Vladimir Kalitvianski May 22 '18 at 8:53
  • $\begingroup$ $Q_1$ is the total power and it is an integral of local heat flux with local temperature difference over the pipe length. $\endgroup$ – Vladimir Kalitvianski May 22 '18 at 9:01
  • $\begingroup$ The question has now been edited, hopefully it´s more clear now! $\endgroup$ – Jmei May 22 '18 at 9:13

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