14.15 Figs. 14.27A, 14.27B, and 14.27C show, respectively, the dual induction/La
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Question
14.15 Figs. 14.27A, 14.27B, and 14.27C show, respectively, the dual induction/LaterologSM, FDC/CNL, and borehole- compensated logs obtained from a section of a well drilled in the southeastern U.S a. Give a general description of the lithology present in the interval shown b. For Zone S, give a detailed interpretation in terms of li- thology, lithology fractions, porosity, porosity type hydrocarbon type, and saturation. Well heading informa- tion is as follows R,-0.534 .rn at 100°F. Rny' 0.348 ·m at 79°F Bottomhole-276°F at 15,760 ft Bit size-5%-in Mud weight 10.7 lbm/galExplanation / Answer
14.15
a) Above 15,600 m, from the log data SP-log gave a positive high approximately 16mV value, gamma ray log gave high 75 API units, resistivity log gave 20 ohm.m; neutron porosity log gave almost very few porosity, and sonic log gave 6 s/ft. Thus, the bed should be a high compacted bed is having less porosity, which means it should be the hard rock.
In between 15600 to 15700 m the rock is having a negative SP value of 17 mv, resistivity readings of 200 ohm.m for LLs and 1000 ohm.m for LLd reading, the neutron porosity of that rock is 15%, and gamma ray gave less value. It should be an oil zone because for gas zone we may observe cross over of neutron and density logs. Here, there is no cross over in between the zones. Hence, the layer in between 15,600 to 15,700 is oil zone of constituent reservoir rock.
b).
The layer in between 15,600 to 15,700 is a reservoir rock layer. It may be sandstone, limestone, or dolomite. The porosity of the formation is 15-20% directly; we will get directly from the neutron log.
The saturation of the formation we can get by calculating of true resistivity and resistivity of the water by using the long data.
From the log data we can easily say, it should be an oil zone because in that data there is no cross over of neutron and density logs, so it should be an oil zone.
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