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
Answer:
Pressure head at point 1 (P1/√w) = 28.04m
velocity head at point 1
V1²/2g = 6.51m
Pressure head at point 3 (P3/√w) = -6.96m
velocity head at point 3 (V3²/2g) = 6.51m
Explanation:
Given details
pipe diameter = 0.15-m
volume flow rate = 0.2 m3/s
pressure of flow = 275 kPa
Heads = 25m, 20m, 55m
√ = specific density
V = Velocity
A= Area
When water flows through vertical pipe, it will consists of three point which are;
Point 1: Point of entry of water
Point 2: Point above the entry point inside the pipe
Point 3: Point at which the water exit the pipe.
Point 1
Using continuity equation
Velocity at point 1 V1 = ∆V/A
But A = πd²/4 = π*0.15²/4 = 0.0177m³
V1 = 0.2/0.0177 = 11.30m/s
Using Bernoulli's equation between point 1 and 2 to get pressure head at point point 1
(P1/√w) + (V1²/2g) + Z1 = (P2/√w) + (V2²/2g) + Z2
Note that the Velocity head at point 1 and 2 are equal and thus the Velocity components are nullified.
(P1/√w) + 20 = (275*10³/2*9.81) + 25
(P1/√w) = (275*10³/2*9.81) + (25 - 20)
(P1/√w) = (275*10³/2*9.81) + 5
(P1/√w) = 28038mm = 28.04m
Pressure head at point 1 (P1/√w) = 28.04m
For velocity head at point 1
V1²/2g = 11.30²/2*9.18 = 6.51m
Point 2
Using Bernoulli's equation between point 1 and 3 to get pressure head at point point 1
(P1/√w) + (V1²/2g) + Z1 = (P3/√w) + (V3²/2g) + Z3
Velocity components are nullified and thus;
28.04. + 20 = (P3/√w) + 55
(P3/√w) = 28.04 + 20 - 55
P3/√w) = -6.96m
Pressure head at point 3 (P3/√w) = -6.96m
And the velocity components of this can be find by using continuity equation
(V1²/2g) = (V3²/2g)
(V3²/2g) = 6.51m
Thus, it means that Velocity head at point of entry is the same as that of exist.
