Fomula estimating pressure increase vs pump rate in fixed volume

[mhldrillpipe]

Gentlemen and laydies

Looking for a formula to predict simple pressure increase within a cylinder as a function of fluid pumped into the cylinder (pump rate).

suspect pressure to be dependent on size of fixed volume, fluid compressibility (oil in this case), and amount of fluid pumped.....(?)

hope you can help

many thanks

 

[BigInch]

I'm saying you don't need to consider it in the problem at hand, unless you are specifically asked "HOW LONG WILL IT TAKE TO FILL..."

Ahhhh. But considering this further, it can stand still.

But... I have better things to do than consider relativistic effects on pipelines.

http://virtualpipeline.spaces.msn.com

 

[mhldrillpipe]

deltaV = Cp*deltaP*V ------> YES, love it!

This is the formula, as i mentioned earlier as well. Casing is pre filled with oil based mud, casing is WAY thin (compared to ID) and cemented to formation wall and delta V is a fuction of cement pumping unit that is used for pressure testing.

Plotting P against volume pumped or strokes of pump and then record the actual pumping pressure gives a quick way of recognizing a deviation, i.e. a leak in hte casing or whatever.....


thanks again

 

[sailoday28]

BigInch (Petroleum)

Bulk modulus K=-v(dp/dv) at T=constant

dp =-K(dv/v)

Approximation of constant K and isothermal process
p2-p1=-K*ln(v2-v1)

v2-v1= e^[(p1-p2)/K]

using your nomenclature
if v1-v2= dVwater
then
change in volume= -e^[(p1-p2)/K]

Please review your past posts which included:


Its dependent on the bulk modulus (dP/dV) of the liquid.

Bulk Modulus [PSI] of Water at 15 psia

change in water volume
dVwater = (P2-P1)/ BulkModulus

 

[BigInch]

http://cannoli.mps.ohio-state.edu/phy847/set3.pdf

http://virtualpipeline.spaces.msn.com

 

[sailoday28]

BlgInch-thanks for nothing.

To others
Please note my correction on the integration of the isothermal bulk modulus where K is constant.
WRONG integration
Approximation of constant K and isothermal process
p2-p1=-K*ln(v2-v1)

Correct integration
p2-p1=-K*ln(v2/v1)

Note for small changes where v2/v1=1+epsilon and epsilon is small
ln(1+epsilon)=epsilon

Therefore p2-p1 = -K(v2/v1 -1)
p2-p1 = K(v1-v2)/v1

or
dv water = v1*(p2-p1)/bulk modulus

BigInch check your units of
dVwater = (P2-P1)/ BulkModulus Your left hand side is change in volume, Your right hand side has no dimensions.

 

[BigInch]

you're welcome see thread378-188060

http://virtualpipeline.spaces.msn.com

 

[sailoday28]

BigInch-Don't you understand that units have to be consistent?