Question

Exercise 5.Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m3/s and a pressureof 275 kPa at an elevation of 25 m. Determine the velocity head and pressure head at elevations of20 and 55 m.

Answer 1

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.

Answer 2

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