I am currently attempting to calculate the heat transfer when compressed air is flowing isothermally through a pipe with frictional losses. I realise this might seem like an odd question, but I am aiming to demonstrate the difference between assuming isothermal flow and isentropic on the calculated pressure drop and wish to calculate the entropy generation.

Note that I am assuming the pipe has a constant cross sectional area.

I have been following the book "Fundamentals of pipe flow" by Benedict. Which writes the modified darcy-weisbach equation (differential form):

$$ \delta F=f_d\frac{dx}{D}\frac{V^2}{2} $$

and defines the compressible loss coefficient as: $$dK=f_d\frac{dx}{D} $$

For the isothermal case: $$K_{1,2}=\frac{A^2}{\dot{m}^2RT}(p_1^2-p_2^2) +2ln(\frac{p_2}{p_1}) $$

I am struggling to understand how to calculate the heat transfer from the Darcy-Weisbach equation. The form that the Dracy-Weisbach equation is written is suggests integrating it as:

$$\Delta F= K_{1,2}\frac{V^2}{2} $$

However, I realise that velocity obviously increases along the pipe due to pressure drop requiring velocity to increase to preserve the mass flow rate. So would it be true to simply write

$$F_{1,2}= K_{1,2}\frac{V_2^2-V_1^2}{2} $$

I personally thought when integrating you should taking into account the fact that $V=V(x)$ as:

$$\Delta F= \frac{f_D}{D}\frac{V_2^3-V_1^3}{3} $$ But this would result in dimensional inconsistency (dimension on right is not equal to J/kg).

Any help on this would be much appreciated!

  • $\begingroup$ I don't think you mean isentropic flow in your first paragraph. I think you mean adiabatic flow, correct. This flow with viscous friction is certainly not isentropic. You seem to be asking two questions: How do they obtain their equation for the pressure drop, and how do I estimate the average rate of entropy generation (or the overall entropy change) in the pipe? Is this correct? $\endgroup$ Jul 5, 2018 at 11:05
  • $\begingroup$ Sorry, yes I do mean adiabatic (I'm attempting to see when isentropic would be a better approximation for the actual flow and when isothermal would be a better approximation). Yes my main question is how do I estimate the rate of entropy generation? The way I thought of doing it was to calculate the heat transfer and then isothermal flow would provide an upper bound for the entropy generation (I think). Secondly, I understand where the compressible loss coefficient comes from, I just don't understand how to estimate the heat transfer from those equations given that velocity changes. Thank you $\endgroup$
    – user401751
    Jul 5, 2018 at 12:22
  • $\begingroup$ Suppose the pipe is adiabatic. And, if you know where the pressure change comes from, then you know that the equation assumes you have an ideal gas. From the open system (control volume) version of the first law of thermodynamics, what is the change in enthalpy per unit mass of the gas passing through the pipe under steady state flow? $\endgroup$ Jul 5, 2018 at 12:49
  • $\begingroup$ Oh yes of course! From the 1st law, since it is adiabatic and no work output then change in enthalpy is zero. Then from $Tds=dh-vdp$ with ideal gas law you get $\Delta S=-Rln(p2/p1) $. For the isothermal case since $ dh=cpdT $ the change in enthalpy is also zero and the change in entropy is also given by $\Delta S=-Rln(p_2/p_1) $. Just wanted to double check that is correct? $\endgroup$
    – user401751
    Jul 5, 2018 at 15:07
  • $\begingroup$ Actually I'm still confused as to how I would calculate the heat transfer required to maintain isothermal conditions. $\endgroup$
    – user401751
    Jul 5, 2018 at 15:30

1 Answer 1


Yes. This is an interesting situation. As you said, the change in enthalpy per unit mass of the gas passing through the pipe is zero, and so, for an ideal gas, this means that the temperature change is also zero. But the paradox here is that, even though there is viscous heating, the temperature change is zero. So what is the solution to the paradox? Well mechanistically, there are two things happening: (1) the gas is experiencing viscous heating and (2) the gas is experiencing expansion cooling, caused by the work which each expanding parcel of gas does on its neighbors. For an ideal gas, these two effects exactly cancel one another, so that, even though the flow is adiabatic, it is also isothermal.


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