I'm trying to solve the following problem:

We want to know what the outflow temperature of the water, $\rm T_{water, out}$ is as a function of time.


  • Water is flowing through a thick pipe in a turbulent way
  • The initial temperature of the thick pipe is known, and uniform, $\rm T_{0,Ext}$.
  • We know the internal radius of the pipe, $\rm r_{Int}$, and the external radius, $\rm r_{Ext}$.
  • We know the inflow temperature into the pipe, $\rm T_{water,in}$. This value remains constant through time.
  • The pipe's external wall is perfectly insulated (i.e. no heat can flow through it and the initial amount of heat held in the pipe is the only heat available to heat the water).

enter image description here

Current Approach :

I feel like the above should be a textbook example, but I can't find the solution to that problem anywhere and my maths are not good enough to be able to solve that problem from scratch. I've tried using papers describing similar problems but I just can't re-arrange the equations to fit my perfectly insulated external wall assumption.

My current approach has been to equate the heat gained by the water (1) to the heat flow transferred from the pipe internal wall to the fluid (2), to the heat transmitted through the pipe itself (3). But with a time dependent $T_{Ext}$ who will gradually cool I seem to always have an additional unknown I can't solve for as soon as I move away from $t_{0}$.

(1) $\rm Q = density_{water} × Flow\ Rate_{water} × Specific\ Heat_{water}(T_{water, out}-T_{water, in})$

(2) $\rm Q = 2× \pi × r_{Int}× Length_{pipe} × \left[ \frac {Thermal\ Conductivity_{water}× \nu} {2× r_{Int}}\right]×\left(T_{r_{Int}}-\frac {T_{water, out}-T_{water, in}}{2}\right)$

with Nu the Nusslet Number.

(3) $\rm Q = 2× \pi× Length_{pipe}× \frac {Thermal\ Conductivity_{pipe}}{(r_{Ext}/r_{Int})}×(T_{Ext}-T_{r_{Int}})$

Any help would be much appreciated.


  • $\begingroup$ You have neglected the thermal inertial of the wall. Probably, the internal heat transfer resistance of the fluid is going to be much higher than the radial conductive resistance of the pipe, so the pipe can, at least tentatively, be assume to be of constant temperature radially. The wall temperature and the fluid temperature are going to be functions of both axial position and time. This has to be taken into account. $\endgroup$ Mar 25 at 13:06
  • $\begingroup$ @ChetMiller thanks for the comment. Not sure if that helps but the ID of the pipe is about 4 m and its OD is between 8 and 20 m. That's why I thought the temperature wouldn't be constant radially. I agree with you that the wall temperature and water are a function of both axial position and time. $\endgroup$
    – Sorade
    Mar 25 at 13:31
  • $\begingroup$ Then you need to be solving the transient heat conduction PDE for the wall at every axial location as a function of r and t, neglecting axial conduction. $\endgroup$ Mar 25 at 18:11

1 Answer 1


The easiest way to initially proceed is to assume everything has come to steady-state conditions and there are no thermal gradients anywhere i.e., the system is equilibrated.

Then you equate the total friction losses (work performed on the parcels of water traversing the pipe length) to the increase in internal energy of the water flowing through the pipe at a certain rate and obtain the temperature rise in the exiting water as a function of flow rate.

This yields the steady-state solution. Then you add in the time-dependent terms one at a time, solve, and check the result against the steady-state solution.

  • $\begingroup$ Is it fair to say that is taking the problem from the end? So you mean, assume the water anywhere in the pipe and the entire pipe have reached the constant inflow temperature? And work backwards from there, by gradually increasing the temperature of the water and the pipe based on friction? I've got no idea how I would go about doing that however. $\endgroup$
    – Sorade
    Mar 25 at 17:47
  • $\begingroup$ Yes. assume the pressure drop x flow rate (units = power) is the input and assume all of the input power goes into warming up the water. Then solve for the delta T of the exiting water. $\endgroup$ Mar 26 at 1:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.