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Assignment-1-solutions

# 00e 02 250e 02 cpp 200e 02 150e 02 100e 02 500e 03

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Unformatted text preview: 2 cp∆p 2.00E-02 1.50E-02 1.00E-02 5.00E-03 0.00E+00 1.80E+07 1.90E+07 2.00E+07 2.10E+07 2.20E+07 Pressure (Pa) The maximum value occurs at the extremes with a value of 0.0273 (unitless) Formula Explanation/Derivation Formula for pore compressibility: But , therefore: Assuming bulk volume is a constant: Rearranging and simplifying to single differentials: This is a first order separable ODE. If does not depend on pressure we find: 6 2.30E+07 2.40E+07 ENGI 9114 – Advanced Reservoir Engineering Assignment # 1 Solutions – Winter 2011 Or: A Taylor series expansion yields: It is normally adequate in solid and liquids to neglect the second order or higher terms since . Therefore: Now we must show the validity of the second formula. Note that: Therefore: May be rewritten as: Now completing a Taylor Series expansion: Again, neglecting second order or higher terms leads to: Therefore showing both formula to be valid. NOTE: THERE IS NO SOLUTION TO PART C – IT WAS ONLY A HINT. 7 ENGI 9114 – Advanced Reservoir Engineering Assignment # 1 Solutions – Winter 2011 Part D Information given: Bulk volume reduced by 0.1% when pore pressure drops 1 bar o when o is measured in bar when is measured in Pa Porosity reduced by 0.3% when pore pressure drops 1 bar o o when is measured in bar when is measured in Pa We have to fine the pore compressibility: But , therefore: Differentiating: Filling in our values: 8 ENGI 9114 – Advanced Reservoir Engineering Assignment # 1 Solutions – Winter 2011 Part E For a real gas: Since we know: Filling in for : Canceling constant terms: Rearranging: Differentiating: This is the formula for a real gas. For an ideal gas, , simplifying to: 9 ENGI 911...
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