Answer:
a. Pressure head: 33.03,
Velocity Head: 6.53
b. Pressure Head: -1.97,
Velocity Head: 6.53
Explanation:
a.
Given
Diameter = 0.15-m, radius = 0.075
rate = 0.2 m3/s
Pressure =275 kPa
elevation =25 m.
We'll consider 3 points as the water flow through the pipe
1. At the entrance
2. Inside the pipe
3. At the exit
At (1), the velocity can be found using continuity equation.
V1 = ∆V/A
Where A = Area = πr² = π(0.075)² = 0.017678571428571m²
V1 = 0.2/0.017678571428571
V1 = 11.32 m/s
The value of pressure at point 1, is given by Bernoulli equation between point 1 and 2:
P1/yH20 + V1²/2g + z1 = P2/yH20 + V2²/2g + z2
Substitute in the values
P1/yH20 + 20 = (275 * 10³Pa)/yH20 + 25
P1/yH20 = (275 * 10³Pa)/yH20 + 25 - 25
=> P1/yH20 = (275/9.81 + 5)
P1/yH20 = 33.03
The velocity head at point one is then given by
V2²/2g = 11.32²/2 * 9.8
V2²/2g = 6.53
b.
The value of pressure at point 1, is given by Bernoulli equation between point 1 and 3:
P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3
Substitute in the values
33.03 + 20 = P3/yH20 + 55
P3/yH20 = 33.03 + 20 - 55
=> P1/yH20 = -1.97
The velocity head at point three is then given by
V2²/2g = V3²/2g = 6.53
