past year exam 5

Question

1. A sharp-edge orifice 52mm in diameter in the side of a large tank, is discharging water under a constant pressure head of 4,5m. The diameter of the vena contracta is 41mm. If the horizontal distance of the water jet is 2,15m and the jet falls 327mm vertically, determine the following: 1.1. the theoretical flow velocity of the water 1.2. the acutual flow velocity of the water 1.3. the theoretical delivery of water 1.4. the actual delivery of the water 1.5. the coefficient of delivery (Cd) 1.6. the coefficient of velocity (Cv) 1.7. the coefficient of contraction (Cc) 2. The centre line of a tapered pipe that is 10,5m long is at an incline of 35degrees to the horizontal. The taper decreases from 350mm at the larger diameter, which is at the upper end of the pipe, to a diameter 300mm at the lower end of the pipe. Water with a density of 1000 kg/m^3 flows from the lower end to the upper end at 130 litres per seconds and the pressure guage at the lower end registers a reading of 125kPa. Calculate the following: 2.1. the pressure at the upper end of the pipe 2.2. the change in the kinetic energy per unit mass over the length of the pipe 3. The centre line of a tapered pipe that is 10,5m long is at an incline of 35 degrees to the horizontal. The taper decreases from 350mm at the larger diameter, which is at the upper end of the pipe, to a diameter of 300mm at the lower end of the pipe. Water with a density of 1000kg/m^3 flows from the lower end to the upper end at 130 litres per second, and the pressure gauge at the lower end registers a reading of 125kPa. Calculate the following: 3.1. the pressure at the upper end of the pipe 3.2. the change in the kinetic energy per unit mass over the length of the pipe 4. The centre of a tapered pipe that is 10,5m long is at 35 degrees to the horizontal. The taper decreases from 350mm at the larger diameter which is at the upper end of the pipe, to a diameter 300mm at the lower end. Water with a density of 1000 kg/m^3 flows from the lower end to the upper end at 130 litres per second, and the pressure gauge at the lower end registers a reading of 125kPa. Calculate the following: 4.1. the pressure at the upper end of the pipe 4.2. the change in kinetic energy per unit mass over the length of pipe. 5. A horizontal tapered pipeline conveying water with a density of 1000kg/m^3 has a diameter of 150mm and 70mm respectively. The pressure gauges at the two different diameters show pressure readings of 750kPa at the large diameter and 700kPa at the small diameter. Determine the flow rate of water in litres per second. 5.1. determine the flow rate of water, in litres per second, flowing through a 200mm diameter pipe equipped with a venture meter having a throat diameter of 130mm. the coefficient of delivery is 0,97 and the mercury manometer shows a reading of 880mm. 5.2. water flows through a pipe with a diameter of 25mm and a length of 50m at a velocity of 2,5m/s. use Darcy’s formula to determine the loss of head due to friction. Assume that the coefficient of friction has a value of 0,005. test the answer using the Chezy formula. Show all calculations

Answer 1

5.2 )

Water density ρ = 1000 kg/m^3

Diameter d = 25 mm

d = 25/1000 = 0.0025 m

Length l = 50 m

Velocity v = 2.5 m/s

Coefficient of friction f = 0.005

Darcy’s formula :

Loss of head due to friction  h = 4flv²/2gd

h = (4(0.005)(50)(2.5)²) / (2(9.8)(0.0025))

h = 127.55 m

Testing the answer using the Chezy formula

v = C√(mi)

head loss due to fiction i = h/l

i = h/50

Hydraulic radius m = A/p

Area A = (¼)πd²

Peremeter p = πd

m = (¼)πd²/(πd)

m = d/4

m = 0.0025/4

m = 0.000625

Constant of Chezy C = √(2g/f)

C = √(2*9.8/0.005)

C = 62.61

Chezy formula : v = C√(mi)

Substitute C = 62.61 , m = 0.000625 v = 2.5 and i = h/50

2.5 = 62.61√(0.000625*(h/50))

(0.000625/50)h = (2.5/62.61)²

(0.000625/50)h = 0.001594

h = ( 0.00159 * 50 ) / 0.000625

h = 127.52 m

Solution:

The loss of head due to friction = 127.52 m

Answer 2

5)

Water density ρ = 1000 kg/m^3

Large diameter d₁= 150 mm

d₁ = 150/1000 = 0.15 m

Small diameter d₂=  70 mm

d₂=  70/1000 = 0.07 m

Pressure at the small diameter p₁ = 750 kPa

Pressure at the large diameter p₂ = 700 kPa

Pipe  daiameter d = 200 mm

d = 200/1000 = 0.2 m

Venture meter throat diameter d_v =  130mm

d_v =  130/1000 = 0.13 m

Flow rate of water Q = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.15²)

A₁ = 0.01766

A₂  = (¼)πd₂²

A₂  = (¼)(0.07²)

A₂  = 0.00385

Q = A₁V₁ = A₂V₂

A₁V₁ = A₂V₂

Substitute A₁ = 0.01766 and A₂  = 0.00385

0.01766V₁ = 0.00385V₂

V₂ = 4.587 V₁

From Bernoulli Equation : (p₁/ρg)+((v₁)²/2g) + z₁ = (p₂/ρg)+((v₂)²/2g) + z₂

p₁= 750k , ρ =1000 , g = 9.8 ,  z₁ = 0 , p₂= 700k , z₂ = 0

(750000/(1000*9.8)+(v₁²) / (2*9.8) + 0 = (700000/(1000*9.8)+(4.587 V₁)² / (2*9.8) + 0

76.5+(v₁²) / (2*9.8) = 71.4+20.976(V₁)² /( 2*9.8)

(20.976-1)(V₁)² /( 2*9.8) = 76.5 - 71.4

1.019 (V₁)² = 5.1

V₁ = 5.005

Q = A₁V₁

Substitute A₁ = 0.01766 and V₁ = 5.005

Q = 0.01766 * 5.005

Q = 0.884 litres per second

The flow rate of water is 0.884 litres per second

Answer 3

5.1)

Water density ρ = 1000 kg/m³

Pipe  daiameter d₁ = 200 mm

d₁ = 200/1000 = 0.2 m

Venture meter throat diameter d₂ =  130 mm

d₂ =  130/1000 = 0.13 m

The coefficient of delivery = The discharge coefficient C = 0.97

Mercury manometer reading (p_{1 -} p₂) = 880 mmHg

1mmHg = 133.322 Nm^-2g

(p_{1 -} p₂) =133.322*880 = 117323.36 Nm^-2

Flow rate of water Q = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.2²)

A₁ = 0.0314

A₂  = (¼)πd₂²

A₂  = (¼)(0.13²)

A₂  = 0.001725

Flow rate of water

 

Substitute ρ =1000 , A₁ = 0.01766 ,  A₂  = 0.00385 , (p_{1 -} p₂)  = 117323.36 and C = 0.97

Q = 0.05877 litres per second

The flow rate of water is 0.05877 litres per second

Answer 4

1.1)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The Theoretical flow velocity

v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

The Theoretical flow velocity is 9.39 m/s

Answer 5

1.2)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The theoretical flow velocity v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

Orifice area A  = d_o²/4

A_o= (0.052)²/4

A_o = 0.000676 m²

Area at vena contracta A_v = d_v²/4

A_v = 0.041²/4

A_v = 0.00042 m²

The coefficient of velocity C_v = A_v / A_o

C_v = 0.00042/ 0.000676

C_v = 0.621

The coefficient of contraction C_c = x / √(4yH) = v_a / v_t

C_c = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

C_c = 2.15 / √(4*0.327*4.5)

C_c = 0.8862

Theoretical delivery of water Q_t = A_ov_t

Substitute A_o = 0.000676 and v_t = 9.39

Q_t = (0.000676)(9.39)

Q_t = 0.00635

The coefficient of delivery C_d = C_v C_c = Q_a / Q_t

C_d = C_v C_c

C_d = 0.621*0.8862

C_d = 0.621*0.8862

C_d = 0.55

The coefficient of delivery C_d =  Q_a / Q_t

The actual delivery of the water Q_a = C_d * Q_t

Q_a = 0.55 * 0.0635

Q_a = 0.035

The coefficient of contraction C_c = v_a / v_t

Actual flow velocity v_a = C_c * v_t

v_a =0.8862 * 9.39

v_a = 8.32 m/s

Actual flow velocity is 8.32 m/s

Answer 6

1.3)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The theoretical flow velocity v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

Orifice area A  = d_o²/4

A_o= (0.052)²/4

A_o = 0.000676 m²

Area at vena contracta A_v = d_v²/4

A_v = 0.041²/4

A_v = 0.00042 m²

The coefficient of velocity C_v = A_v / A_o

C_v = 0.00042/ 0.000676

C_v = 0.621

The coefficient of contraction C_c = x / √(4yH) = v_a / v_t

C_c = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

C_c = 2.15 / √(4*0.327*4.5)

C_c = 0.8862

Theoretical delivery of water Q_t = A_ov_t

Substitute A_o = 0.000676 and v_t = 9.39

Q_t = (0.000676)(9.39)

Q_t = 0.00635 m³/s

Theoretical delivery of water is 0.00635 m³/s

Answer 7

1.4)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The theoretical flow velocity v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

Orifice area A  = d_o²/4

A_o= (0.052)²/4

A_o = 0.000676 m²

Area at vena contracta A_v = d_v²/4

A_v = 0.041²/4

A_v = 0.00042 m²

The coefficient of velocity C_v = A_v / A_o

C_v = 0.00042/ 0.000676

C_v = 0.621

The coefficient of contraction C_c = x / √(4yH) = v_a / v_t

C_c = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

C_c = 2.15 / √(4*0.327*4.5)

C_c = 0.8862

Theoretical delivery of water Q_t = A_ov_t

Substitute A_o = 0.000676 and v_t = 9.39

Q_t = (0.000676)(9.39)

Q_t = 0.00635

The coefficient of delivery C_d = C_v C_c = Q_a / Q_t

C_d = C_v C_c

C_d = 0.621*0.8862

C_d = 0.621*0.8862

C_d = 0.55

The coefficient of delivery C_d =  Q_a / Q_t

The actual delivery of the water Q_a = C_d * Q_t

Q_a = 0.55 * 0.0635

Q_a = 0.035

The actual delivery of the water 0.035 m³/s

Answer 8

1.5)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The theoretical flow velocity v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

Orifice area A  = d_o²/4

A_o= (0.052)²/4

A_o = 0.000676 m²

Area at vena contracta A_v = d_v²/4

A_v = 0.041²/4

A_v = 0.00042 m²

The coefficient of velocity C_v = A_v / A_o

C_v = 0.00042/ 0.000676

C_v = 0.621

The coefficient of contraction C_c = x / √(4yH) = v_a / v_t

C_c = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

C_c = 2.15 / √(4*0.327*4.5)

C_c = 0.8862

Theoretical delivery of water Q_t = A_ov_t

Substitute A_o = 0.000676 and v_t = 9.39

Q_t = (0.000676)(9.39)

Q_t = 0.00635

The coefficient of delivery C_d = C_v C_c = Q_a / Q_t

C_d = C_v C_c

C_d = 0.621*0.8862

C_d = 0.621*0.8862

C_d = 0.55

The coefficient of delivery is 0.55

Answer 9

1.6)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The theoretical flow velocity v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

Orifice area A  = d_o²/4

A_o= (0.052)²/4

A_o = 0.000676 m²

Area at vena contracta A_v = d_v²/4

A_v = 0.041²/4

A_v = 0.00042 m²

The coefficient of velocity C_v = A_v / A_o

C_v = 0.00042/ 0.000676

C_v = 0.621

The coefficient of velocity is 0.621

Answer 10

1.7)

Given Data :

Orifice diameter d = 52 mm

d_o = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta d_v = 41 mm

d_{v =} 41/1000 = 0.041 m

The horizontal distance of the water jet x = 2.15 m

The vertical distance of the water jet y = 327 mm

y = 327/1000 = 0.327 m

The theoretical flow velocity v_t = √(2gH)

Substitute H = 4.5 m and g = 9.8

v_t = √(2*9.8*4.5)

v_t = 9.39 m/s

Orifice area A  = d_o²/4

A_o= (0.052)²/4

A_o = 0.000676 m²

Area at vena contracta A_v = d_v²/4

A_v = 0.041²/4

A_v = 0.00042 m²

The coefficient of velocity C_v = A_v / A_o

C_v = 0.00042/ 0.000676

C_v = 0.621

The coefficient of contraction C_c = x / √(4yH) = v_a / v_t

C_c = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

C_c = 2.15 / √(4*0.327*4.5)

C_c = 0.8862

The coefficient of contraction is 0.8862