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Given the temperature $T_1$ and volume flow rate $\dot{V}$ at the start of the tube, which has a temperature of $T_0$ and an inner surface of $A_0$, what is the resulting temperature of the stream going out of the tube? ($T_2$).

I assumed that the temperature $T_0$ is fixed, and the material is aluminium and the flow is laminar. Maybe i have to assume more?

I think it boils down to finding the heat transfer coefficient between the copper and the air stream - but this depends on the temperature difference, which changes as the air is heated up.

Here a small sketch of my problem:

sketch of the problem

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As long as you have enough data about the inlet conditions you don't need to assume whether the flow is laminar or turbulent (this is a very dangerous assumption), you can calculate Reynolds number to find that out: $$ Re = \frac{\rho v D}{\mu} $$ where $\rho$ is the flow density, $v$ is the velocity, $D$ is the hydraulic diameter of the tube and $\mu$ is the dynamic viscosity.

Flow inside tubes is laminar for $Re < 2300$ and turbulent for $Re > 10,000$ (and transition in between), Assuming that you have a constant surface temperature tube $ T_s = \text{constant}$, making energy balance we have: $$ Q^o = hA_s \theta_{lmtd} = m^o C_p \triangle T $$

Where $A_s$ is the surface area $\pi D L$ and $\theta_{lmtd}$ is logarithmic mean temperature difference.

For laminar flow, Nusselt number can be calculated as (for fully developed flow and moderate temperature difference): $$ Nu = \frac{hD}{k} = 3.66$$

For turbulent flow, you can use Gnielinski correlation, Dittus-Boelter equation or Sieder-Tate correlation. (This Wikipedia page is very helpful).

After calculating $h$ from Nusselt number correlations (You might need to assume an exit temperature for the first iteration) substituting in energy balance equation, after some iterations the solution will converge and you'll finally obtain the exit temperature from the tube.

(You can refer to the forced convection heat transfer chapter in this book for more details and solved examples.)

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  • $\begingroup$ The logarithmic mean temperature difference is defined for two streams - in my case, I assume, I just insert the temperature difference between the stream and the tube for both ends, right? $\endgroup$ – John H. K. Sep 24 '15 at 11:28
  • $\begingroup$ @JohnH.K. This is one way, but the logarithmic mean temperature diff. is better. In your case, $\triangle T_A = T_s - T_i$ and $\triangle T_B = T_s - T_o$ where A and B are the subscripts in the Wikipedia page and i and o are the inlet and outlet tempertures. $\endgroup$ – Algo Sep 24 '15 at 11:38
  • $\begingroup$ Uhh, I was trying to say exactly that, my bad. Thank you! $\endgroup$ – John H. K. Sep 24 '15 at 11:55
  • $\begingroup$ @JohnH.K. You're welcome :) $\endgroup$ – Algo Sep 24 '15 at 12:04
  • $\begingroup$ Is $m^0$ the mass of the fluid in the tube? $\endgroup$ – John H. K. Sep 24 '15 at 12:21

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