Pressure Increase in Liquid Full Pipework

[Green999]

I have been asked to calculate the increase in pressure when adding a certain mass of LPG to an already liquid full LPG line. The pipework volume is fixed, there is no increase in temperature, only increase in LPG mass, so the pressure increase will be almost instantaneous due to the very low compressibility, no doubt raising above the flange rating, but now sure how I calculate the final pressure to confirm this.

 

[zdas04]

The bulk modulus of elasticity of a 50-50 propane/butane mix is around 70,000 psi. That means that it would take 70,000 psi to reduce a volume of LPG 1%. I read that to mean that if I add 0.0001% of the pipe volume, the pressure would increase 7 psi (50 kPa). There is no way to answer "if I add a mL of LPG, how much would the pressure increase" unless you know the fluid volume of the system.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. ùGalileo Galilei, Italian Physicist

 

[Green999]

Thanks for the info. I agree, I think this is the only way I can do this. However, my understanding is that if bulk elasticity is approx 70,000 psi then it will take 700 psi to reduce volume by 1%, not 70,000 psi.

 

[zdas04]

The bulk modulus is the amount of pressure required to reduce a volume by 1%.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. ùGalileo Galilei, Italian Physicist

 

[Green999]

No, the Bulk Modulus is the ratio of change in pressure against change in volume, see formula below, where V = initial volume.

Bulk Modulus = -V dp/dv

If changing volume by 1%, dv = 0.01V

Therefore, dp = - Bulk Modulus x 0.01V/V
= 0.01 x Bulk Modulus

 

[plontaj]

But in that system are two phases: liquid and gas PG, they have got different Bulk Modulus?

 

[zdas04]

Plontaj
I took "liquid full" to mean that it was full of liquid (i.e. the pressure and temperature put the fluid in the "sub-cooled liquid" region). I'm not sure why you think it is two phase, the OP never mentioned the actual pressure or temperature of the line.

Green999
I think you are confusing the constant of integration in that equation with what has become known as Bulk Modulus. So:

dp = - Constant x 0.01V/V

Bulk Modulus = Constant * 0.01 so

dP = -Bulk Modulus

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. ùGalileo Galilei, Italian Physicist

 

[Green999]

The system is liquid full, so only one phase.

I have looked at several examples on sites now including manufacturers and they all seem to agree with what I wrote above.

 

[25362]

As stated by Green999, the system volume and temperature don't change, but the pressure p, and density [ρ] do. The expression for the bulk modulus E_T is, then,

E_T = [ρ] (dp/d[ρ])_T​

Using [Δ] instead of the derivative function:

E [≅] [ρ] [Δ]p/[Δ][ρ]​

A 0.01 mass addition is, in fact, a 0.01 increase in density.

Therefore, E [≅] [Δ]p/0.01 and [Δ]p [≅] 0.01 E.

Errors and omissions accepted.

 

[25362]

Should I say excepted?