11 June, 13:42

# Determine the flow velocities at the inlet and exit sections of aninclined tapering pipe using fluid flow theory and given pressurereadings and flow rates.There is a sloping pipeline that has one end 1.35 m higher than theother. The pipe section tapers from 0.95 m diameter at the top end to0.44m diameter at the lower end. The difference in pressure betweenthe two sections is 12.35kPa, with pressure being greater at higherlevel.Your task is to determine the inlet and exit velocities and thevolumeflow rate of the inclined pipe.

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1. 11 June, 14:20
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The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

Explanation:

Using Bernoulli's equation

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

ΔP + ρgΔh = 1/2ρ (v₂² - v₁²) (1)

where ΔP = pressure difference = 12.35 kPa = 12350 Pa

Δh = height difference = 1.35 m

From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.

Substituting v₁ into (1), we have

ΔP + ρgΔh = 1/2ρ (v₂² - (d₂²v₂/d₁²) ²)

ΔP + ρgΔh = 1/2ρ (v₂² - (d₂/d₁) ⁴v₂²)

v₂ = √[2 (ΔP + ρgΔh) / ρ (1 - (d₂/d₁) ⁴) }

substituting the values of the variables, we have

v₂ = √[2 (12350 Pa + 1000 kg/m³ * 9.8 m/s² * 1.35 m) / (1000 kg/m³ (1 - (0.44 m/0.95 m) ⁴)) }

= √[2 (12350 Pa + 13230 Pa) / (1000 kg/m³ * 0.954) ]

= √[2 (25580 Pa) / 954 kg/m³]

= √[51160 Pa/954 kg/m³]

= √53.627

= 7.32 m/s

v₁ = d₂²v₂/d₁²

= (0.44 m/0.95 m) ² * 7.32 m/s

= (0.954) ² * 7.32 m/s

= 6.66 m/s

The volume flow rate Q = A₁v₁

= πd₁²v₁/4

= π (0.95 m) ² * 6.66 m/s : 4

= 18.883 m³/s : 4

= 4.72 m³/s

So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s