1. An oil well has been dug to a depth of 333 m and oil is gushing out. At this

Question

1. An oil well has been dug to a depth of 333 m and oil is gushing out. At this depth the estimated pressure on the top of the oil reservoir is 81.6 atmospheres. The density of oil is 873 kg/m &1 atmosphere- 1.013x10 N/m2. (a) What is the pressure of the oil as it shoots out of the ground? (b) What is the speed at which the oil shoots out of the well? (Hint: Far away from the pipe opening the reservoir oil speed can be taken to be zero.) (c) What is the pressure of the oil just inside the mouth of the well tube in the reservoir in atmospheres?

Explanation / Answer

a) Pressure = 1 atm as the oil exits the ground

b) Using Bernouli eqn,

Change in pressure = 1/2 * density * v2 + density *g*h

81.6 atm - 1 atm = 0.5*873*v2 + 873*9.80665*333

v = 111 m/s

c)At the ground level, P= Pbottom - d*g*h

81.6 atm = 8266.08 kPa

Thus, P = 8266.08 kPa - (0.873)*9.80665*333

P=5415 kPa = 53.45 atm --- Just before exiting

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