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past year exam 5

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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
asked Oct 28, 2014 in PHYSICS by anonymous

16 Answers

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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

answered Oct 29, 2014 by lilly Expert
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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 dv =  130mm

dv =  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

answered Oct 29, 2014 by lilly Expert
0 votes

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 (p1 - p2) = 880 mmHg

1mmHg = 133.322 Nm-2g

(p1 - 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 image

 

Substitute ρ =1000 , A1 = 0.01766 ,  A₂  = 0.00385 , (p1 - p2)  = 117323.36 and C = 0.97

image

image

image

Q = 0.05877 litres per second

The flow rate of water is 0.05877 litres per second

answered Oct 29, 2014 by lilly Expert
0 votes

1.1)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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

vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

The Theoretical flow velocity is 9.39 m/s

answered Oct 29, 2014 by lilly Expert
0 votes

1.2)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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 vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

Orifice area A  = do²/4

Ao= (0.052)²/4

Ao = 0.000676 m²

Area at vena contracta Av = dv²/4

Av = 0.041²/4

Av = 0.00042 m²

The coefficient of velocity Cv = Av / Ao

Cv = 0.00042/ 0.000676

Cv = 0.621

The coefficient of contraction Cc = x / √(4yH) = va / vt

Cc = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

Cc = 2.15 / √(4*0.327*4.5)

Cc = 0.8862

Theoretical delivery of water Qt = Aovt

Substitute Ao = 0.000676 and vt = 9.39

Qt = (0.000676)(9.39)

Qt = 0.00635

The coefficient of delivery Cd = Cv Cc = Qa / Qt

Cd = Cv Cc

Cd = 0.621*0.8862

Cd = 0.621*0.8862

Cd = 0.55

The coefficient of delivery Cd =  Qa / Qt

The actual delivery of the water Qa = Cd * Qt

Qa = 0.55 * 0.0635

Qa = 0.035

The coefficient of contraction Cc = va / vt

Actual flow velocity va = Cc * vt

va =0.8862 * 9.39

va = 8.32 m/s

Actual flow velocity is 8.32 m/s

answered Oct 29, 2014 by lilly Expert
reshown Oct 29, 2014 by bradely
0 votes

1.3)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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 vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

Orifice area A  = do²/4

Ao= (0.052)²/4

Ao = 0.000676 m²

Area at vena contracta Av = dv²/4

Av = 0.041²/4

Av = 0.00042 m²

The coefficient of velocity Cv = Av / Ao

Cv = 0.00042/ 0.000676

Cv = 0.621

The coefficient of contraction Cc = x / √(4yH) = va / vt

Cc = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

Cc = 2.15 / √(4*0.327*4.5)

Cc = 0.8862

Theoretical delivery of water Qt = Aovt

Substitute Ao = 0.000676 and vt = 9.39

Qt = (0.000676)(9.39)

Qt = 0.00635 m³/s

Theoretical delivery of water is 0.00635 m³/s

answered Oct 29, 2014 by lilly Expert
0 votes

1.4)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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 vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

Orifice area A  = do²/4

Ao= (0.052)²/4

Ao = 0.000676 m²

Area at vena contracta Av = dv²/4

Av = 0.041²/4

Av = 0.00042 m²

The coefficient of velocity Cv = Av / Ao

Cv = 0.00042/ 0.000676

Cv = 0.621

The coefficient of contraction Cc = x / √(4yH) = va / vt

Cc = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

Cc = 2.15 / √(4*0.327*4.5)

Cc = 0.8862

Theoretical delivery of water Qt = Aovt

Substitute Ao = 0.000676 and vt = 9.39

Qt = (0.000676)(9.39)

Qt = 0.00635

The coefficient of delivery Cd = Cv Cc = Qa / Qt

Cd = Cv Cc

Cd = 0.621*0.8862

Cd = 0.621*0.8862

Cd = 0.55

The coefficient of delivery Cd =  Qa / Qt

The actual delivery of the water Qa = Cd * Qt

Qa = 0.55 * 0.0635

Qa = 0.035

The actual delivery of the water 0.035 m³/s

answered Oct 29, 2014 by lilly Expert
0 votes

1.5)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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 vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

Orifice area A  = do²/4

Ao= (0.052)²/4

Ao = 0.000676 m²

Area at vena contracta Av = dv²/4

Av = 0.041²/4

Av = 0.00042 m²

The coefficient of velocity Cv = Av / Ao

Cv = 0.00042/ 0.000676

Cv = 0.621

The coefficient of contraction Cc = x / √(4yH) = va / vt

Cc = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

Cc = 2.15 / √(4*0.327*4.5)

Cc = 0.8862

Theoretical delivery of water Qt = Aovt

Substitute Ao = 0.000676 and vt = 9.39

Qt = (0.000676)(9.39)

Qt = 0.00635

The coefficient of delivery Cd = Cv Cc = Qa / Qt

Cd = Cv Cc

Cd = 0.621*0.8862

Cd = 0.621*0.8862

Cd = 0.55

The coefficient of delivery is 0.55

answered Oct 29, 2014 by lilly Expert
0 votes

1.6)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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 vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

Orifice area A  = do²/4

Ao= (0.052)²/4

Ao = 0.000676 m²

Area at vena contracta Av = dv²/4

Av = 0.041²/4

Av = 0.00042 m²

The coefficient of velocity Cv = Av / Ao

Cv = 0.00042/ 0.000676

Cv = 0.621

The coefficient of velocity is 0.621

answered Oct 29, 2014 by lilly Expert
0 votes

1.7)

Given Data :

Orifice diameter d = 52 mm

do = 52/1000 = 0.052 m

Pressure head H = 4.5 m

Diameter of the vena contracta dv = 41 mm

dv = 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 vt = √(2gH)

Substitute H = 4.5 m and g = 9.8

vt = √(2*9.8*4.5)

vt = 9.39 m/s

Orifice area A  = do²/4

Ao= (0.052)²/4

Ao = 0.000676 m²

Area at vena contracta Av = dv²/4

Av = 0.041²/4

Av = 0.00042 m²

The coefficient of velocity Cv = Av / Ao

Cv = 0.00042/ 0.000676

Cv = 0.621

The coefficient of contraction Cc = x / √(4yH) = va / vt

Cc = x / √(4yH)

Substitute  x = 2.15 and y = 0.327 m

Cc = 2.15 / √(4*0.327*4.5)

Cc = 0.8862

The coefficient of contraction is 0.8862

answered Oct 29, 2014 by lilly Expert

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