Ahorizontal pipe has an abrupt expansion from d1 = 5 cm to d2 = 10 cm. the water velocity in the smaller section is 8 m/s and the flow is turbulent. the pressure in the smaller section is p1 = 380 kpa. taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure p2, and estimate the error that would have occurred if bernoulli’s equation had been used. take the density of water to be rho = 1000 kg/m3.

P₂ = 392720.38 Pa = 392.72 kPa

Explanation:

Given

D₁ = 5 cm = 0.05 m

D₂ = 10 cm = 0.10 m

v₁ = 8 m/s

P₁ = 380 kPa = 380000 Pa

α = 1.06

ρ = 1000 kg/m³

g = 9.8 m/s²

We can use the following formula

(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL

knowing that z₁ = z₂ we have

(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL   (I)

Where

V₂ can be obtained as follows

V₁*A₁ = V₂*A₂      ⇒     V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)

⇒    V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)  

⇒    V₂ = 2 m/s

and

hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)

⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)

⇒ hL = 1.9469 m

Finally we get P₂ using the equation (I)

⇒  P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)

⇒  P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)

⇒  P₂ = 392720.38 Pa = 392.72 kPa

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