I'm looking at viscous dissipation of an incompressible fluid in fully developed pipe flow.

I'd like to generally derive the bulk temperature variation as a function of pipe length due to viscosity, and understand how viscous dissipation requires more work to make a fluid flow.

The general bulk flow energy equation for a pipe is:

$$ \dot{m} c_p \frac{dT_b}{dx} = q_w\pi D+q'_d$$

where $T_b$ is the bulk temperature, $q_w$ is the wall heat flux into the fluid, and $q'_d$ is the viscous dissipation per unit length.

Now, by definition, viscous dissipation is mechanical energy that irreversibly increases the entropy of the flow, and this is given by the shaft work into the fluid minus the change in kinetic energy, minus the reversible flow work (per unit length, denoted by the primes $'$):

$$ q'_d = \dot{W'} - \dot{m}\frac{du_b^2}{dx} - \dot{V}\frac{dP}{dx}$$

where I'll ignore shaft work and kinetic energy changes, therefore

$$q'_d = \dot{V}\frac{\Delta P}{L} $$

This basically says that a higher viscous losses require a larger pressure drop to drive the flow. Makes sense, right?

Now, for Poiseuille flow in a tube, $\dot{V} = \frac{\pi R^4 \Delta P}{8 \mu L}$ therefore I can substitute $\dot{V}\frac{\Delta P}{L} = \frac{8\dot{V}^2\mu}{\pi R^4}$ into my expression for $q'_d$:

$$ q'_d = \frac{8\dot{V}^2\mu}{\pi R^4} = 8 \pi \mu u_b^2 $$

This is a nice result, basically our viscous losses increase with viscosity, and they increase even more so with bulk velocity.

Now, I can substitute this into the bulk flow energy equation:

$$ \dot{m} c_p \frac{dT_b}{dx} = q_w\pi D+8 \pi \mu u_b^2$$

Now I can easily obtain the bulk temperature as a function of pipe length due to viscous dissipation and added heat. Is this analysis correct?

  • $\begingroup$ Do you mean $\dot {m}c_pdT_b/dx=…$ ? $\endgroup$ Jan 24, 2019 at 23:13
  • $\begingroup$ @AlexTrounev yes, I'm trying to get that in terms of viscosity and bulk velocity. $\endgroup$ Jan 24, 2019 at 23:27
  • $\begingroup$ Then correct the typo. $\endgroup$ Jan 24, 2019 at 23:44
  • $\begingroup$ @AlexTrounev what typo? I changed the title to be more specific, if that's what you're referring to. $\endgroup$ Jan 25, 2019 at 1:42
  • $\begingroup$ No, just use $\dot {m}c_p\frac {dT_b}{dx}$ instead of $\dot {m}c_p\frac {T_b}{dx}$. $\endgroup$ Jan 25, 2019 at 12:11

2 Answers 2


Your analysis is correct, but a much easier approach is to apply the open system version of the first law directly: $$\dot{Q}-\dot{m}(h_{out}-h_{in})=0$$with $$h_{out}-h_{in}=C_p\Delta T+\frac{\Delta P}{\rho}$$assuming constant density. Plus, $$\dot{m}\frac{\Delta P}{\rho}=\dot{V}\Delta P$$where $\dot{V}$ is the volumetric flow rate.

Note, however, that this all assumes that the effect of temperature on viscosity can be neglected.


Heat equation in viscous incompressible flow

$\rho c_p(\frac {\partial T}{\partial t}+\vec {v}.\nabla T)=\lambda \nabla ^2 T+\frac {\mu}{2}(\frac {\partial v_i}{\partial x_k}+\frac {\partial v_k}{\partial x_i})^2$

In the case of a Poiseuille flow, the terms can be calculated and averaged over the cross section of the pipe.

  • $\begingroup$ This would just lead to $\dot{V}\Delta P$ when averaged out. $\endgroup$ Jan 25, 2019 at 0:14
  • $\begingroup$ Why use questionable hypotheses if there is an exact equation? $\endgroup$ Jan 25, 2019 at 0:26
  • 1
    $\begingroup$ Would you call the first law of thermodynamics a questionable hypothesis? $\endgroup$ Jan 25, 2019 at 0:40
  • $\begingroup$ @ChesterMiller We are discussing here not the first law of thermodynamics, but how viscous dissipation affects the temperature of the liquid in the pipe. $\endgroup$ Jan 25, 2019 at 12:19
  • $\begingroup$ Are you really admitting that you don't see the connection? The equation you wrote is just the differential form of the open-system version of the first law of thermodynamics, combined with the mechanical energy balance equation (i.e., the equation of motion dotted with the velocity vector) and the mechanical constitutive equation for a viscous Newtonian fluid. I submit that, if I integrate this equation over the volume of the pipe, I will get the same result that @Drew and I obtained without actually solving for the temperature distribution. $\endgroup$ Jan 25, 2019 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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