PNGE333-hw3

# PNGE333-hw3 - P roblem 3 T he plugs from the core sample in...

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Problem 3 The plugs from the core sample in Homework 2 were analyzed through conventional analysis. In Homework 2, you determined the porosity and permeability of each core plug. The following steps have been also taken during the experiments: The core plug was weighed as received (W AR ). The water and oil ( = 0.85 g/cc) were extracted from the core plug and volume of the ρ water ( V w ) was recorded. The core plugs were dried and were saturated with water. Each saturated core plug was placed in a closed container whose bottom consists of a porous plate. A constant air pressure was applied and when equilibrium was attained, the core plug was re- weighed. The air pressure was gradually increased until the sample weight reached a minimum. The minimum sample weight ( W Min ) was recorded at the conclusion of each experiment. The results of experimental measurements are given below. You are asked to determine saturations as received and irreducible water saturation for each core plug and plot them against the depth. Samp le Depth Range, ft W AR V w W Min Vs Vb W dry # Beginning En d g cc g cc cc g 1 4000 400 1 80.4 0.0 80.4 30.00 30.00 80.4 2 4001 400 2 76.8 0.1 76.8 28.61 28.75 76.7 3 4002 400 3 77.2 0.3 77.2 28.71 29.00 76.9 4 4003 400 4 80.2 0.5 80.1 29.75 30.20 79.7 5 4004 400 5 78.3 0.6 78.2 29.01 29.60 77.7 6 4005 400 6 79.1 0.8 79.0 29.25 30.00 78.4 7 4006 400 79.4 0.9 79.2 29.29 30.20 78.5

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7 8 4007 400 8 79.3 0.6 79.4 29.34 30.40 78.6 9 4008 400 9 80.6 0.7 80.6 29.76 31.00 79.8 10 4009 401 0 78.2 0.8 78.1 28.84 30.20 77.3 11 4010 401 1 78.9 0.8 78.3 29.07 30.60 77.9 12 4011 401 2 75.8 0.7 75.2 27.93 29.40 74.9 13 4012 401 3 76.3 0.7 75.7 28.12 29.60 75.4 14 4013 401 4 76.2 0.9 75.5 28.01 29.80 75.1 15 4014 401 5 76.7 0.9 76.0 28.20 30.00 75.6 16 4015 401 6 76.1 1.1 75.3 27.90 30.00 74.8 17 4016 401 7 75.6 1.0 74.8 27.71 29.80 74.3 18 4017 401 8 75.5 1.2 74.6 27.60 30.00 74.0 19 4018 4019 78.5 1.2 77.6 28.70 31.20 76.9 20 4019 4020 77.5 1.2 76.6 28.34 30.80 75.9 21 4020 4021 77.0 1.2 76.1 28.15 30.60 75.4 22 4021 4022 76.4 1.4 75.3 27.85 30.60 74.6 23 4022 4023 75.9 1.4 74.8 27.66 30.40 74.1 24 4023 4024 76.9 1.4 75.8 28.03 30.80 75.1 25 4024 4025 77.4 1.4 76.3 28.21 31.00 75.6 26 4025 4026 74.9 1.4 73.8 27.30 30.00 73.2 27 4026 4027 72.8 1.5 71.6 26.46 29.40 70.9 28 4027 4028 73.6 1.7 72.4 26.70 30.00 71.6 29 4028 4029 74.5 1.8 73.1 26.93 30.60 72.2 30 4029 4030 73.0 1.8 71.7 26.40 30.00 70.8 31 4030 4031 74.5 1.8 73.1 26.93 30.60 72.2 32 4031 4032 73.6 1.7 72.4 26.70 30.00 71.6 33 4032 4033 72.6 1.6 71.4 26.34 29.60 70.6 34 4033 4034 72.8 1.5 71.6 26.46 29.40 70.9 35 4034 4035 73.7 1.5 72.6 26.82 29.80 71.9 36 4035 4036 70.5 1.4 69.5 25.65 28.50 68.7 37 4036 4037 71.1 1.3 70.1 25.94 28.50 69.5 38 4037 4038 71.7 1.1 70.8 26.22 28.50 70.3 39 4038 4039 72.9 0.9 72.2 26.79 28.50 71.8 40 4039 4040 73.5 0.7 38.5 14.25 28.50 38.2 41 4040 4041 78.2 0.8 78.1 28.84 30.20 77.3 42 4041 4042 80.6 0.7 80.6 29.76 31.00 79.8
43 4042 4043 79.3 0.6 79.4 29.34 30.40 78.6 44 4043 4044 79.4 0.9 79.2 29.29 30.20 78.5 45 4044 4045 78.1 0.7 77.9 28.86 29.60 77.3 46 4045 4046 79.9 0.6 79.8 29.60 30.20 79.3 47 4046 4047 77.0 0.4 76.9 28.57 29.00 76.6 48 4047 4048 76.6 0.3 76.5 28.46 28.75 76.3 49 4048 4049 76.1 0.1 76.1 28.36 28.50 76.0 50 4049 4050 76.4 0.0 76.4 28.50 28.50 76.4 Solution Method To find the saturation: = - Vp Vb Vs cc # Vs and Vb are giuen from the table of HW 2 = = - + Vo Woρo WAR Wd ρVwρo cc While W AR , V w and W d are given = ρ 1gcc = . ρo 0 85 gcc = * , % So VoVp 100 saturation oil = * , % Sw VwVp 100 saturation water = - + , % Sg 1 So Sw saturation gas Result: S a m p l e Depth Range, ft

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