A cylindrical tank is full of water and at the top of a hill. A pump pulls water

Question

A cylindrical tank is full of water and at the top of a hill. A pump pulls
water out of this tank at a pressure of 8 atm and a velocity of 50 meters per
second and sends it down a circular pipe of circumference 30 centimeters.
The pipe goes down a hill of height 40 meters and comes out. The pipe at
this point is a square of 9 centimeters to a side.

What is the pressure and velocity of the water as it comes out? What is the
pressure and velocity of the fluid coming out if all variables remain the same
except that the fluid is now liquid mercury? The units for your answers must
be in atm or meters per second.

Explanation / Answer

Given: pressure P1 = 8 atm velocity V1 = 50 m/sec circumference C = 30*10-2 m radius of pipe r = 4.77*10-2 m Area of the pipe A1 = (3.14)(4.77*10-2 m )2                                = 71.65*10-4 m2 Area of squared shaped pipe A2 = (9*10-2)2                                                   = 81*10-4 m 2 According to equation of continuity A1V1 = A2V2   ....... (1) V2 = A1V1/A2      = (71.65*10-4 m2 )(50 m/sec) / (81*10-4 m 2)      = 44.23 m/sec now Accoding to Bernoulis Equation P1 + 1/2v12+gh1 = P2 + 1/2v22+gh2 .........    (2) in this case h2 = 0 P2 = P1 + 1/2v12+gh1 - 1/2v22   ........      (3)     = (8atm)+1/2(1000 kg/m3)(50m/sec)2+(1000 kg/m3)(9.8 m/sec2)(40 m ) - 1/2(1000 kg/m3)(44.23)2     = 663861 atm     = 6.6*105 atm for mercury: replace density of fluid in equation 3 as mercury . 13534 kg/m3

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