# Calculating pressure at pressure relief valve outlet

I have a tank which has a Pressure Relief Valve (PRV) connected to the atmospheric vent (Carbon steel with friction coefficient of 0.14). The PRV is opened at Set pressure of 30 PSIG and has a capacity of 60,000 SATSTM (I assumed it to be in lb/hr). I am trying to calculate the discharge pressure at the vent exit (opening).

The steps,

1. From steam table, specific volume of dry saturated steam at 30 PSIG = 9.5 cu.ft./lb
2. Calculating the volume flow rate at the exit = 60,000 x 9.5 = 570,000 cu.ft./hr = 158.33 cu.ft./sec
3. Calculating the nozzle outlet cross sectional area = 0.55 sq.ft. (Pipe nozzle diameter = 10 in = 0.83 ft)
4. Velocity at the exit = 158.33 x 0.55 = 290 ft/sec
5. Calculating pressure drop = (Friction coefficient/2) x (Pipe Length/Pipe Diameter) x (Velocity squared/gravity) = 0.14 x (10 ft /0.83 ft) x (290.35x290.35/32.17) = 2201 ft = 953 psi (Impossible!)

here is the excel file http://www.filedropper.com/prv_1

If I include the density in the above pressure drop calculation, the unit turns out to be in PSI than ft. But then I found no such equation that includes the density in calculating the pressure drop. What am I missing? Thanks.

Last edited:

Nidum
Gold Member
Pressure at the outlet of the vent pipe will be equal to atmospheric pressure - 0 psig .

If you are working this as an exercise then Nidum can be considered close enough. If this is a real relief valve application you are better (and much safer) relying on a reliable supplier to do the calculation for you. Pressure at the outlet nozzle can be well above atmospheric pressure because a pressurized area will form at the outlet under actual relief.

I used a large Friction coefficient (3.5E-06 is the right one).
The pressure turned out to be 0.15 PSIG. So Nidum is right.