qz10sol_3706Hs11

# qz10sol_3706Hs11 - Δ p s = 150 kN/m 2 a For the conditions...

This preview shows page 1. Sign up to view the full content.

TA: Tomoyuki Nakayama Tuesday, April 12, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 10 (Homework Set #12) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ Fresh water flows horizontally from pipe section 1 of cross-sectional area A 1 into pipe section 2 of cross-sectional area A 2 . The figure below gives a plot of the pressure difference p 2 - p 1 versus the inverse area squared A 1 -2 that would be expected for a volume flow rate of a certain value if the water flow were laminar under all circumstances. The scale on the vertical axis is set by
This is the end of the preview. Sign up to access the rest of the document.

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 2-2 = 16 A 2 = 0.25 m 2 b) What is the value of the volume flow rate? When A 1-2 = 32 m-4 , (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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online