31/2 in T-95, 15.5 lb/ft MTC (wall thickness = 0.476 in)...

31/2 in T-95, 15.5 lb/ft MTC (wall thickness = 0.476 in) tubing is landed in a well, which is filled with 9.1 ppg completion fluid (tubing is also filled with the same fluid). A packer, having a bore diameter of 41/2 in, is present at 8,500 ft, which can be assumed as the bottom of the wellbore. The production casing is 65/8 in N-80, 32 lb/ft BTC (wall thickness = 0.475 in). Then, 0.7 sp.gr. oil is produced from this well through the tubing. As the production starts, the measured pressure at the surface inside the tubing is 1375 psi. Frictional losses are negligible. Average temperature change is 30°F. Estimate the length change due to piston effect only. Young's modulus is 30 x $10^6$ psi, and Poisson's ratio is 0.3.

Transcript

00:01 Hi in this question we are given with a boiler drum these dimensions so thickness is 1 .25 mm which is 1 .25 into 10 raised to minus 3 meter and this is 208 gigapascal which is 208 into 10 power 3 megapascal so this is also 1 .53 into 10 power 3 megapascal now the original volume v is given by v original so e original is equal to pi by 4 into d square into l.

00:34 So on substituting the values for d and l we have original volume is 6 .79 meter cube.

00:42 Now by volumetric strain formula so this is first step.

00:46 Volumetric strain formulae is equal to change in volume by original volume so which is equal to pd divided by 40 into e into 5 minus mu where mu is point 3 so on substituting so we know v and all of this value so delta v will be equal to pd divided by 40 into e into 5 minus mu into v therefore delta t on substituting this values delta v1 so this is v1 which is equal to 2 .3817 into 10 power minus 3 meter cube now in third part we have two hemispirical path hemispyric path in cylinder so ev will be equal to delta v2 divided by v so this is for hemispherical path which is equal to 3pd divided by 40 into e into 1 minus mu therefore where this v is the v of sphere so we know don't know the v of sphere therefore it is calculated as v sphere of original is equal to here we have two hemispirical path so 2 into v hemisphere.

02:06 Therefore which is equal to 2 into pi by 12 d cube which is equal to pi by 6 d cube so on substituting the value of c d we have v of spear original will be equal to 0 .9048 meter cube so on substituting for delta v by v which is equal to so the bracket term will be 019385 into 10 power minus 3 or in delta v will be equal to, so taking v that side and substituting this value, delta v2 will be equal to 1 .7539 into 10 power minus 4 meter cube.

02:49 In the next step we have to find total volume due to water pressure.

02:54 So due to water pressure we have total volume, delta v total is equal to delta v1 minus delta v2.

03:05 So on substituting both the values, we have total volume is equal to 2 .2069 into 10 power minus 3 meter cube.

03:17 The original final volume.

03:20 So original final volume is given by original volume of cylinder plus.

03:25 So this is for first case plus original volume of the spherical path.

03:31 So which is equal to on substitution it will be 6 .79 plus.

03:37 2 .385 into 10 power minus 3 plus 0 .9048 minus 1 .7539 into 10.

03:51 Minus 4.

03:52 So this one substitution.

03:54 So this one solving, v final will be equal to 7 .697 meter cube and total volume needed.

04:02 So total volume needed is v3 which is equal to v3 is equal to vf plus p by k into v3.

04:17 So on substitution, so taking to lhs and solving for v3, v3 will be equal to 7 .7377 .74 meter cube.

04:31 Hence, the additional volume, additional volume, v4 will be equal to vf or vf difference v3.

04:45 So v4 will be equal to 1 .0404 meter cube...

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