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2.1 )

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Tapered pipe line length l = 10.5 m

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Inclination angle θ = 35^o

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Water density ρ = 1000 kg/m³

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Large diameter d₁= 350 mm

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d₁ = 350/1000 = 0.35 m

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Small diameter d₂=  300 mm

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d₂=  300/1000 = 0.3 m

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Flow rate of water Q = 130 litres per second

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Liter per second = 0.0010 m³/sec

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Q = 130*0.001 = 0.13 m³/sec

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Pressure at the small diameter  p₂ = 125 kPa

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Pressure at the large diameter  p₁ = ?

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A₁ = (¼)πd₁²

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A₁ = (¼)π(0.35²)

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A₁ = 0.0962

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A₂  = (¼)πd₂²

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A₂  = (¼)π(0.3²)

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A₂  = 0.0707

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Q = A₁V₁ = A₂V₂

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V₁ = Q / A₁ = 0.13 / 0.0962

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V₁ = 1.351 m/s

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V₂ = Q / A₂ = 130 / 0.0707

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V₂ =  1.839 m/s

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θ = 35^o

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l = 10.5 m

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Consider z₁ = 0 m

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Then from figure z₂ = 10.5 sin35^o

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z₂ = 6.02 m

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

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

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

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

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(p₁/980) = 18.8545

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p_{1= 184.77 kPa}

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Pressure at the large diameter  p₁ is 184.77 kPa

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