siriusengineer
Mechanical
- Nov 27, 2002
- 2
Hi,
I am a new engineer working on a derivation for maximum flow of a gas in a nozzle.
API RP 520, Pt. I, Para. 4.3.2.1 gives the equation for max flow to be
W=C*A*P*sqrt(T/M)
I've left out the correction factors for back pressure, non-ideal flow, and compressibility of a non-ideal gas.
Now I've been able to start with a tank of initial pressure and temperature, Po and To, and get to the following equation on my own.
W=Po*A*[(M/Ru/To)(k(2/(k+1))^((k+1)/(k-1))]^1/2
M being molecular weight, and Ru the universal gas constant.
Now the term C in the first equation is given as
C=520*[(k(2/(k+1))^((k+1)/(k-1))]^1/2
which makes the first equation almost match up with mine except for the 520 term, and the universal gas constant in my equation.
My question is this. Where did this C term come from and what are its units?
After putting in 1545.35 ft*lbf/lbmol/R for Ru and converting all the units to make sure W came out in lbm/s my term for C instead of being 520 would be 0.1443
A big thank you for any help on this.
I am a new engineer working on a derivation for maximum flow of a gas in a nozzle.
API RP 520, Pt. I, Para. 4.3.2.1 gives the equation for max flow to be
W=C*A*P*sqrt(T/M)
I've left out the correction factors for back pressure, non-ideal flow, and compressibility of a non-ideal gas.
Now I've been able to start with a tank of initial pressure and temperature, Po and To, and get to the following equation on my own.
W=Po*A*[(M/Ru/To)(k(2/(k+1))^((k+1)/(k-1))]^1/2
M being molecular weight, and Ru the universal gas constant.
Now the term C in the first equation is given as
C=520*[(k(2/(k+1))^((k+1)/(k-1))]^1/2
which makes the first equation almost match up with mine except for the 520 term, and the universal gas constant in my equation.
My question is this. Where did this C term come from and what are its units?
After putting in 1545.35 ft*lbf/lbmol/R for Ru and converting all the units to make sure W came out in lbm/s my term for C instead of being 520 would be 0.1443
A big thank you for any help on this.