HI guys,now this is a problem trouble me long time ,i am a new guy in the reservior,so i am dout the state:
Co=-1/Bo*dBo/dP+Bg/Bo*dRs/dP when the reservior is saturate ? any effort you maked will be great appreciate!
thanks in advance!
regards
arek
HI guys,now this is a problem trouble me long time ,i am a new guy in the reservior,so i am dout the state:
Co=-1/Bo*dBo/dP+Bg/Bo*dRs/dP when the reservior is saturate ? any effort you maked will be great appreciate!
thanks in advance!
regards
arek
Arek,
I'm not sure what your question is, but the equation your wrote is right, that's the general equation for oil compressibility and it works for saturated and undersaturated reservoirs. I'll elaborate a little bit on it:
As you wrote: Co = -1/Bo * (dBo/dP - (dRs/dP)*Bg); derivatives are taken under constant temperature which is the expected condition for hydrocarbon reservoirs due to the large thermal capacity of the rock matrix and the large area for heat transfer.
For reservoir pressures above Bubble Point Pressure, there is no liberation of gas, so dRs/dP (at constant temperature) is zero; so for undersaturated oil reservoirs oil compressibility equation reduces to the form:
Co = -1/Bo * (dBo/dP)
Once below the Bubble Point Pressure, gas compressibility (in its free state) should be taken into account, so the general form of the oil compressibility equation must be considered.
You can get more information by looking at McCain's book "The Properties of Petroleum Fluids" (page 235)
Thank you so much DAH7542,your answer was so appreciated and useful!
Here:I still confused Once below the Bubble Point Pressure,the gas compressibility equation is(dRs/dP)*Bg)?
cheers
regards
arek
Dear Arek, summarizing:
For reservoir pressures above Bubble Point Pressure:
Co = -1/Bo * (dBo/dP)
For reservoir pressures below Bubble Point Pressure:
Co = -1/Bo * (dBo/dP - (dRs/dP)*Bg)
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