Unformatted text preview: Δ p s = 150 kN/m 2 . a) For the conditions of the figure, what is the value of A 2 (in square meters)? Applying Bernoulli’s equation at section 1 and section 2, we have p 1 + (1/2) ρ v 1 2 = p 2 + (1/2) ρ v 2 2 ⇒ ∆ p = (1/2) ρ (v 1 2 – v 2 2 ) The continuity equation yields A 1 v 1 = A 2 v 2 v 1 = (A 2 /A 1 )v 2 Combining the two equation leads to the mathematical expression for the graph: ∆ p = (1/2) ρ [(A 2 /A 1 ) 2 – 1]v 2 2 According to this expression, Δ p = 0 when A 1 = A 2 . Therefore, A 22 = 16 A 2 = 0.25 m 2 b) What is the value of the volume flow rate? When A 12 = 32 m4 , (A 2 /A 1 ) 2 = (32/16) = 2 and ∆ p = 150 kN/m 2 . The expression obtained in part a) yields ∆ p = (1/2) ρ [2 – 1]v 2 2 v 2 = √ (2 ∆ p/ ρ ) = 17.3 m/s The volume flow rate is A 1 v 1 = 4.33 m 3 /s...
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 Spring '08
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 Fluid Dynamics, Expression, Mathematical Expression

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