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!
Best Answer
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.