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.
