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 Joules