How to calculate flow rate of ammonia escaping ruptured line 1

[ProjectEng]

Assume ammonia vapor at 100F, 100psi, in a 3" SS pipe. Assume pipeline is very long and is connected to a bulk tank (also at 100psi, 100F) with an unlimited quantity of ammonia.

Now assume this pipeline breaks. So you have a 3" line that is vented to atmospheric. How can I calculate steady-state flow through this rupture?

I don't need a very precise answer, a good estimate is what I need.

Thanks.

 

[bchoate]

Bill Choate
A rough estimate can be made with a few assumptions:
1) assume the size of the hole; 1/4" D hole, 1/2"D hole. Don't worry about the shape of the hole; just assume it is circular.
2) assume that the upstream pressure does not drop with time as the ammonia escapes.

Estimate the volumetric flow rate using an orifice calculation. Assume that the upstream pipe diameter is the ID of the 3" pipe for d/D. Calculate the volumetric flow rate. Convert to a mass flow rate using the conditions of the gas in the pipe. This can be done at a few pressures since this is a long pipeline and the pressure may not equal the source pressure. Similarly bracket the gas temperature.

This will give you a crude estimate of the gas loss. With the volumetric flow rate, the cumulative loss over time can be estimated.

This could also be done in something like ChemCad.
Bill Choate

 

[mbeychok]

ProjectEng:

At 100 psia, liquified ammonia would boil at 56 deg F .. so to have ammonia vapor at 100 psia and 100 deg F, the ammonia vapor must have been superheated somehow. Then transporting that in a "long" pipeline, depending on the surrounding environmental temperature, the vapor might cool down to 56 deg F or lower and begin to condense into liquid (for example, in Wisconsin or Montana on a cold winter night). You really should explain your problem in more detail:

(1) How "long" is the pipeline?
(2) What is the surrounding environmental temperature of the pipeline?
(3) What do you mean when you say the bulk storage tank has an "unlimited" amount of ammonia? Is ammonia vapor constantly being fed into the tank to keep it full of vapor at 100 psia and 100 deg F?
(4) What do you mean when you say the pipeline "breaks"? Do you mean it completely ruptures leaving in effect a 3-inch opening to the atmosphere? Or does it have a leaking hole caused by corrosion and, if so, what size is the hole?

In any event, if we assume that the vapor does not condense in the pipeline and that the pipeline frictional pressure drop is not large enough to reduce the pressure below about 28 psia, then the ratio of the absolute upstream pressure to the absolute downstream pressure at the pipeline break is more than 28/14.7 = 1.84 and the flow through the hole will be under "choked" conditions. The equation for the choked flow of a gas through a restriction orifice is:

Q = C A P [ g k M / ( R T ) ]^1/2 [ 2 / ( k + 1 ) ] ^{(k + 1) / (2k - 2)}

where:
Q = mass flow rate of the gas, lb/s
C = discharge coefficient
A = discharge hole area, ft²
P = absolute upstream pressure, lb/ft²
g = gravitational acceleration of 32.17 ft/s²
k = Cp / Cv of the gas
M = gas molecular weight
R = Universal Gas Law constant = 1545.3 ( ft-lb ) / ( lbmol ) ( deg R )
T = gas temperature, deg R

Now, you will have calculate or assume what the upstream pressure will be within the pipeline at the point of breakage in lb/ft² absolute; what the gas temperature will be within the pipeline at the point of breakage in degrees Rankine; and assume the hole is circular and what its diameter is so that you can calculate the hole area. The molecular weight is 17 and Cp/Cv is 1.31 for ammonia vapor. As for the discharge coefficient, you can use about 0.7 to 0.8 as an approximation. Then plug all of that into the above equation and you will obtain the initial instantaneous flow rate through the pipeline break.

I don't believe your "unlimited" amount of vapor in the bulk storage tank. Thus, the upstream pressure will decrease with time as the leak continues and the flow rate will diminish. So the initial instantaneous flow rate calculated with the above equation is a maximum or "worst case" flow.


Milton Beychok
mbeychok@xxx.net (replace xxx with cox)
(Visit me at www.air-dispersion.com)