1.

Water is flowing smoothly through a closed pipesystem. At one point A, the speed of the water is 3.0 m s^(-1)while at another point B, 1.0 m higher, the speed is 4.0 m s^(-1) . The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops. Then

Answer»

the pressure at B when water is flowing is 6.7 KPA.
the pressure at B when water is flowing is 8.2 kPa.
the pressure at B when water STOPS flowing is 10.2 kPa.
the pressure at B when water stops flowing is 8.2 kPa.

Solution :Let `P_1 , h_1 , v_1` and `P_2 , h_2 , v_2` represent the pressures , heights and VELOCITIES of flow at the two points A and B respectively . According to the Bernoulli.s principle
`P_1 + rho gh_1 + 1/2 rho v_1^2 = P_2 + rhogh_2 + 1/2 rho v_2^2`….(i)
Putting `v_1=30 m s^(-1) , v_2=4.0 m s^(-1) , (h_2-h_1) = 1M , P_1`= 20 kPa
we get ,
`P_2=20+ [ 10^3 xx 9.8 (-1)+ 10^3/2 [ 9-16]]`
=20-9.8-3.5 =6.7 kPa
Also when the flow stops , `v_1=v_2=0` and then from (i) ,
`P._2`=18-9.8 = 8.2 kPa


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