# How does viscous dissipation influence fluid temperature in pipe flow?

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?

• Do you mean $\dot {m}c_pdT_b/dx=…$ ? – Alex Trounev Jan 24 '19 at 23:13
• @AlexTrounev yes, I'm trying to get that in terms of viscosity and bulk velocity. – Drew Jan 24 '19 at 23:27
• Then correct the typo. – Alex Trounev Jan 24 '19 at 23:44
• @AlexTrounev what typo? I changed the title to be more specific, if that's what you're referring to. – Drew Jan 25 '19 at 1:42
• No, just use $\dot {m}c_p\frac {dT_b}{dx}$ instead of $\dot {m}c_p\frac {T_b}{dx}$. – Alex Trounev Jan 25 '19 at 12:11

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.

• This would just lead to $\dot{V}\Delta P$ when averaged out. – Chet Miller Jan 25 '19 at 0:14
• Why use questionable hypotheses if there is an exact equation? – Alex Trounev Jan 25 '19 at 0:26
• Would you call the first law of thermodynamics a questionable hypothesis? – Chet Miller Jan 25 '19 at 0:40
• @ChesterMiller We are discussing here not the first law of thermodynamics, but how viscous dissipation affects the temperature of the liquid in the pipe. – Alex Trounev Jan 25 '19 at 12:19
• 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. – Chet Miller Jan 25 '19 at 13:30