Question
Let's say that for a given power plant the top of the water carrying tube is connected to the bottom of a dam of a large lake. This tube has a radius of 40.0 cm and the tube is located 60.0 m below the surface of the lake. The bottom end of the tube, where the hydroelectric plant is located, is another 250.0 m below the top of the tube (and 310.0 m below the surface of the lake). The outlet nozzle of the tube, which feeds the turbine, has a radius of 20.0 cm.
If water wasn't flowing, you could easily calculate the pressure at the inlet to the pipe (60.0 m below the lake surface). How much greater or less than this pressure is the pressure at the inlet when water is flowing? (if the pressure is less when water is flowing, your answer should be negative)
Explanation / Answer
Given Radius of tube
r1 = 40 cm = ( 40 cm ) ( 0.01 m / 1 cm ) = 0.4 m Radius of turbine
r2 = 20 cm = ( 20 cm ) ( 0.01 m / 1 cm ) = 0.2 m Area of the tube
A1 =
r12 = ( 0.4 m ) 2 = 0.5026 m2 Area of turbine
A2 =
r22 = ( 0.2 m ) 2 = 0.1256 m 2 From equation of continuity
A1
v1 =
A2
v2
v1 =
A2
v2 /
A 1 = 0.125
v2 / 0.5026 = 0.2499
v 2 ...... (1) According to bernoullis equation
p2 + ( 1 /2 )
v22 +
gh2 =
p1 +( 1 /2 )
v12 +
gh1
p2 -
p1 = (1/2)
(
v12 -
v 22 ) +
g (
h 1 -
h2 ) From equation (1)
p2 -
p1 = (1/2)
( 0.623
v22 -
v 22 ) +
g (
h 1 -
h2 ) ...... (2) Here water was not flowing so
v 2 = 0 Therefore
p2 -
p1 = ( 1000 kg /m3 ) ( 9.8 m/s2 ) ( 60 m ) = 5.88 * 105 Pa ________________________________________________________________________ If water is flowing we required the velocity of water at the inlet or outlet. So plug the valkues in equation (2) and get the answer ________________________________________________________________________ If water is flowing we required the velocity of water at the inlet or outlet. So plug the valkues in equation (2) and get the answer