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Poiseuille's Law Answered
Aedan Gonzales 2022-05-08

Steady isothermal flow of an ideal gas
So I have a steady isothermal flow of an ideal gas through a smooth duct (no frictional losses) and need to compute the mass flow rate (per unit area) as a function of pressures at any two different arbitrary points, say 1 and 2. I have the following momentum equation in differential form:
ρ v d v + d P = 0
where v is the gas the flow velocity and P is static pressure. The mass flow rate per unit cross section G, can be calculated by integrating this equation between points 1 and 2. This is where it gets confusing. I do the integration by two ways:
1) Use the ideal gas equation P = ρ R T right away and restructure the momentum equation:
v d v + R T d P P = 0
, integrate it between points 1 and 2 and arrive at:
v 1 2 v 2 2 + 2 R T l n P 1 P 2 = 0
Since the flow is steady, I can write G = ρ 1 v 1 = ρ 2 v 2 , again use the ideal gas law to write density in terms of gas pressure and finally arrive at the mass flow rate expression:
G 2 = 2 l n P 2 P 1 R T ( 1 P 1 2 1 P 2 2 )
2) In another way of integrating (which is mathematically correct), I start by multiplying the original momentum equation by ρ to get
ρ 2 v d v + 1 R T P d P = 0
write ρ 2 v = G 2 / v, integrate between points 1 and 2 to arrive at
G 2 l n v 2 v 1 = P 1 2 P 2 2 2 R T
Using the ideal gas law the velocity ratio can be written as the pressure ratio to finally arrive at the mass flow rate equation
G 2 = P 1 2 P 2 2 2 R T l n P 1 P 2
Both the expressions are dimensionaly sound and I know that the second expression is the correct one. My question is, whats wrong with first expression.

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