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

2.1 )

Tapered pipe line length l = 10.5 m

Inclination angle θ = 35^o

Water density ρ = 1000 kg/m³

Large diameter d₁= 350 mm

d₁ = 350/1000 = 0.35 m

Small diameter d₂=  300 mm

d₂=  300/1000 = 0.3 m

Flow rate of water Q = 130 litres per second

Liter per second = 0.0010 m³/sec

Q = 130*0.001 = 0.13 m³/sec

Pressure at the small diameter  p₂ = 125 kPa

Pressure at the large diameter  p₁ = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.35²)

A₁ = 0.0962

A₂  = (¼)πd₂²

A₂  = (¼)π(0.3²)

A₂  = 0.0707

Q = A₁V₁ = A₂V₂

V₁ = Q / A₁ = 0.13 / 0.0962

V₁ = 1.351 m/s

V₂ = Q / A₂ = 130 / 0.0707

V₂ =  1.839 m/s

θ = 35^o

l = 10.5 m

Consider z₁ = 0 m

Then from figure z₂ = 10.5 sin35^o

z₂ = 6.02 m

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

p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , V₁ = 1.351 , V₂ =  1.839 , z₂ = 6.02

(p₁/(1000)(9.8))+((1.351)²/2*9.8) + 0 = (125000/(1000)(9.8))+((1.839)²/2*9.8) + 6.02

(p₁/9800) + 0.093 = 12.755 + 0.1725 + 6.02

(p₁/980) = 18.8545

p_{1= 184.77 kPa}

Pressure at the large diameter  p₁ is 184.77 kPa

Answer 2

2.2 )

Tapered pipe line length l = 10.5 m

Inclination angle θ = 35^o

Water density ρ = 1000 kg/m³

Large diameter d₁= 350 mm

d₁ = 350/1000 = 0.35 m

Small diameter d₂=  300 mm

d₂=  300/1000 = 0.3 m

Flow rate of water Q = 130 litres per second

Liter per second = 0.0010 m³/sec

Q = 130*0.001 = 0.13 m³/sec

Pressure at the small diameter  p₂ = 125 kPa

The change in kinetic energy per KE = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.35²)

A₁ = 0.0962

A₂  = (¼)πd₂²

A₂  = (¼)π(0.3²)

A₂  = 0.0707

Q = A₁V₁ = A₂V₂

V₁ = Q / A₁ = 0.13 / 0.0962

V₁ = 1.351 m/s

V₂ = Q / A₂ = 130 / 0.0707

V₂ =  1.839 m/s

θ = 35^o

l = 10.5 m

Consider z₁ = 0 m ( Large diameter side is taken as datum )

Then from figure z₂ = 10.5 sin35^o

z₂ = 6.02 m

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

Substitute p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , V₁ = 1.351 , V₂ =  1.839 , z₂ = 6.02

(p₁/(1000)(9.8))+((1.351)²/2*9.8) + 0 = (125000/(1000)(9.8))+((1.839)²/2*9.8) + 6.02

(p₁/9800) + 0.093 = 12.755 + 0.1725 + 6.02

(p₁/980) = 18.8545

p₁ = 184.77 kPa

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

Substitute KE₁ = (v₁)²/2g and KE₂ = (v₂)²/2g

(p₁/ρg)+ KE₁ + z₁ = (p₂/ρg)+ KE₂ + z₂

Substitute p₁ = 184.77k _, p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , z₂ = 6.02

(184770/(1000)(9.8))+ KE₁ + 0 = (125000/(1000)(9.8))+ KE₂ + 6.02

KE₂ - KE₁ = (184770 - 125000/(1000)(9.8)

KE₂ - KE₁ = 6.098979

The change in kinetic energy KE = | KE₂ - KE₁ |

= | 6.1 |

= 6.1 J

The change in kinetic energy is 6.1 Jouls

Answer 3

3.1 )

Tapered pipe line length l = 10.5 m

Inclination angle θ = 35^o

Water density ρ = 1000 kg/m³

Large diameter d₁= 350 mm

d₁ = 350/1000 = 0.35 m

Small diameter d₂=  300 mm

d₂=  300/1000 = 0.3 m

Flow rate of water Q = 130 litres per second

Liter per second = 0.0010 m³/sec

Q = 130*0.001 = 0.13 m³/sec

Pressure at the small diameter  p₂ = 125 kPa

Pressure at the large diameter  p₁ = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.35²)

A₁ = 0.0962

A₂  = (¼)πd₂²

A₂  = (¼)π(0.3²)

A₂  = 0.0707

Q = A₁V₁ = A₂V₂

V₁ = Q / A₁ = 0.13 / 0.0962

V₁ = 1.351 m/s

V₂ = Q / A₂ = 130 / 0.0707

V₂ =  1.839 m/s

θ = 35^o

l = 10.5 m

Consider z₁ = 0 m ( Large diameter side is taken as datum )

Then from figure z₂ = 10.5 sin35^o

z₂ = 6.02 m

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

p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , V₁ = 1.351 , V₂ =  1.839 , z₂ = 6.02

(p₁/(1000)(9.8))+((1.351)²/2*9.8) + 0 = (125000/(1000)(9.8))+((1.839)²/2*9.8) + 6.02

(p₁/9800) + 0.093 = 12.755 + 0.1725 + 6.02

(p₁/980) = 18.8545

p_{1= 184.77 kPa}

Pressure at the large diameter  p₁ is 184.77 kPa

 

Answer 4

4.1 )

Tapered pipe line length l = 10.5 m

Inclination angle θ = 35^o

Water density ρ = 1000 kg/m³

Large diameter d₁= 350 mm

d₁ = 350/1000 = 0.35 m

Small diameter d₂=  300 mm

d₂=  300/1000 = 0.3 m

Flow rate of water Q = 130 litres per second

Liter per second = 0.0010 m³/sec

Q = 130*0.001 = 0.13 m³/sec

Pressure at the small diameter  p₂ = 125 kPa

Pressure at the large diameter  p₁ = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.35²)

A₁ = 0.0962

A₂  = (¼)πd₂²

A₂  = (¼)π(0.3²)

A₂  = 0.0707

Q = A₁V₁ = A₂V₂

V₁ = Q / A₁ = 0.13 / 0.0962

V₁ = 1.351 m/s

V₂ = Q / A₂ = 130 / 0.0707

V₂ =  1.839 m/s

θ = 35^o

l = 10.5 m

Consider z₁ = 0 m ( Large diameter side is taken as datum )

Then from figure z₂ = 10.5 sin35^o

z₂ = 6.02 m

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

p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , V₁ = 1.351 , V₂ =  1.839 , z₂ = 6.02

(p₁/(1000)(9.8))+((1.351)²/2*9.8) + 0 = (125000/(1000)(9.8))+((1.839)²/2*9.8) + 6.02

(p₁/9800) + 0.093 = 12.755 + 0.1725 + 6.02

(p₁/980) = 18.8545

p_{1= 184.77 kPa}

Pressure at the large diameter  p₁ is 184.77 kPa

Answer 5

3.2 )

Tapered pipe line length l = 10.5 m

Inclination angle θ = 35^o

Water density ρ = 1000 kg/m³

Large diameter d₁= 350 mm

d₁ = 350/1000 = 0.35 m

Small diameter d₂=  300 mm

d₂=  300/1000 = 0.3 m

Flow rate of water Q = 130 litres per second

Liter per second = 0.0010 m³/sec

Q = 130*0.001 = 0.13 m³/sec

Pressure at the small diameter  p₂ = 125 kPa

The change in kinetic energy per KE = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.35²)

A₁ = 0.0962

A₂  = (¼)πd₂²

A₂  = (¼)π(0.3²)

A₂  = 0.0707

Q = A₁V₁ = A₂V₂

V₁ = Q / A₁ = 0.13 / 0.0962

V₁ = 1.351 m/s

V₂ = Q / A₂ = 130 / 0.0707

V₂ =  1.839 m/s

θ = 35^o

l = 10.5 m

Consider z₁ = 0 m ( Large diameter side is taken as datum )

Then from figure z₂ = 10.5 sin35^o

z₂ = 6.02 m

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

Substitute p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , V₁ = 1.351 , V₂ =  1.839 , z₂ = 6.02

(p₁/(1000)(9.8))+((1.351)²/2*9.8) + 0 = (125000/(1000)(9.8))+((1.839)²/2*9.8) + 6.02

(p₁/9800) + 0.093 = 12.755 + 0.1725 + 6.02

(p₁/980) = 18.8545

p₁ = 184.77 kPa

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

Substitute KE₁ = (v₁)²/2g and KE₂ = (v₂)²/2g

(p₁/ρg)+ KE₁ + z₁ = (p₂/ρg)+ KE₂ + z₂

Substitute p₁ = 184.77k _, p₂= 125k , ρ =1000 , g = 9.8 , z₁ = 0 , z₂ = 6.02

(184770/(1000)(9.8))+ KE₁ + 0 = (125000/(1000)(9.8))+ KE₂ + 6.02

KE₂ - KE₁ = (184770 - 125000/(1000)(9.8)

KE₂ - KE₁ = 6.098979

The change in kinetic energy KE = | KE₂ - KE₁ |

= | 6.1 |

= 6.1 J

The change in kinetic energy is 6.1 Jouls

Answer 6

4.2 )

Tapered pipe line length l = 10.5 m

Inclination angle θ = 35^o

Water density ρ = 1000 kg/m³

Large diameter d₁= 350 mm

d₁ = 350/1000 = 0.35 m

Small diameter d₂=  300 mm

d₂=  300/1000 = 0.3 m

Flow rate of water Q = 130 litres per second

Liter per second = 0.0010 m³/sec

Q = 130*0.001 = 0.13 m³/sec

Pressure at the small diameter  p₂ = 125 kPa

The change in kinetic energy per KE = ?

A₁ = (¼)πd₁²

A₁ = (¼)π(0.35²)

A₁ = 0.0962

A₂  = (¼)πd₂²

A₂  = (¼)π(0.3²)

A₂  = 0.0707

Q = A₁V₁ = A₂V₂

V₁ = Q / A₁ = 0.13 / 0.0962

V₁ = 1.351 m/s

V₂ = Q / A₂ = 130 / 0.0707

V₂ =  1.839 m/s

θ = 35^o

l = 10.5 m

Consider z₁ = 0 m ( Large diameter side is taken as datum )

Then from figure z₂ = 10.5 sin35^o

z₂ = 6.02 m

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

Substitute p₂= 125k , ρ =1000 , g = 9.8 ,  z₁ = 0 , V₁ = 1.351 , V₂ =  1.839 , z₂ = 6.02

(p₁/(1000)(9.8))+((1.351)²/2*9.8) + 0 = (125000/(1000)(9.8))+((1.839)²/2*9.8) + 6.02

(p₁/9800) + 0.093 = 12.755 + 0.1725 + 6.02

(p₁/980) = 18.8545

p₁ = 184.77 kPa

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

Substitute KE₁ = (v₁)²/2g and KE₂ = (v₂)²/2g

(p₁/ρg)+ KE₁ + z₁ = (p₂/ρg)+ KE₂ + z₂

Substitute p₁ = 184.77k _, p₂= 125k , ρ =1000 , g = 9.8 , z₁ = 0 , z₂ = 6.02

(184770/(1000)(9.8))+ KE₁ + 0 = (125000/(1000)(9.8))+ KE₂ + 6.02

KE₂ - KE₁ = (184770 - 125000/(1000)(9.8)

KE₂ - KE₁ = 6.098979

The change in kinetic energy KE = | KE₂ - KE₁ |

= | 6.1 |

= 6.1 J

The change in kinetic energy is 6.1 Jouls