HW5s - formation. Assume a=1. F = a -1.57 ; = (0.39)-1.57 =...

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PGE 424 Home Work 5 Due: April 15 1. Chemical analysis of an oil reservoir brine yielded the following results: 25,000 ppm Na + , 30,000 ppm Cl - , 7500 ppm Mg ++ , 6,000 ppm SO 4 -2 , 4000 ppm HCO 3 - . Calculate the equivalent salinity in ppm sodium chloride, and the resistivity of brine at 100, 175, and 250 ° F. Discuss the effect of temperature on the resistivity of water. Total Dissolved Solid = 25,000+30,000+7500+6,000+4000 = 72,500 ppm Mg multiplier = 0.75 SO4 multiplier = 0.3 HCO3 multiplier = 0.2 Equiv NaCl conc. = 25,000+30,000+(0.75x7500)+(0.3x6000)+(0.2x4000)=63225 ppm From Fig. 8.1, at 100F Rw = 0.081 ohm-m 175F Rw = 0.05 ohm-m 250F Rw = 0.036 ohm-m As T increases, Rw decreases. 2. A normally pressured well has a producing zone that is completely water-wet. The apparent porosity of the shale zone is approximately 0.39. The cementation exponent for shales in this field is 1.57. The true resistivity of shale zone is 0.9 ohm-m. Calculate water resistivity of the
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Unformatted text preview: formation. Assume a=1. F = a -1.57 ; = (0.39)-1.57 = 4.386 = Ro/Rw; Rw = Ro/F = 0.9/4.386 = 0.205 ohm-m 3. The results of lab measurements (porosity, Ro) made on water-wet clean sandstone samples and well log analysis (Rt) are shown in the table below. The resistivity of the formation brine is 0.056 ohm-m at 25 C. The formation temperature is 89 C. a. Calculate the formation factor for each sample F = Ro/Rw = Ro/0.056 b. Estimate the cementation factor Assume, m=2, a= F/ -2 c. Determine the water saturation in each sample Assume n=2. Rw89/Rw25 = Ro89/Ro25 = 0.024/0.056= 0.429; Sw = sqrt (0.429Ro/Rt) Sample Porosity Ro, ohm-m Rt, ohm-m F a Sw 1 0.204 0.665 30 11.875 0.49 0.097 2 0.178 0.83 24 14.82 0.47 0.122 3 0.163 0.96 22 17.14 0.455 0.137 4 0.201 0.68 21 12.14 0.49 0.118 5 0.143 1.19 20 21.25 0.43 0.16...
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This note was uploaded on 10/11/2009 for the course PGE 19460 taught by Professor Sharma during the Spring '09 term at University of Texas at Austin.

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