HW4_solution

HW4_solution - Problem 8.86 Given: Flow through sudden...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 8.86 [2] Given: Flow through sudden expansion Find: Inlet speed; Volume flow rate Solution: Basic equations p 1 ρ α V 1 2 2 + gz 1 + p 2 ρ α V 2 2 2 + 2 + h lm = h lm K V 1 2 2 = QV A = Δ p ρ H2O g Δ h = Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal Hence the energy equation becomes p 1 ρ V 1 2 2 + p 2 ρ V 2 2 2 + K V 1 2 2 = From continuity V 2 V 1 A 1 A 2 = V 1 AR = Hence p 1 ρ V 1 2 2 + p 2 ρ V 1 2 AR 2 2 + K V 1 2 2 = Solving for V 1 V 1 2p 2 p 1 () ρ 1A R 2 K = AR D 1 D 2 2 = 75 225 2 = 0.111 = so from Fig. 8.14 K 0.8 = Also p 2 p 1 ρ H2O g Δ h = 1000 kg m 3 9.81 × m s 2 5 1000 × m Ns 2 kg m × = 49.1 Pa = Hence V 1 2 49.1 × N m 2 m 3 1.23 kg × 1 1 0.111 2 0.8 × kg m 2 × = V 1 20.6
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/03/2010 for the course MAE 101b taught by Professor Staff during the Spring '08 term at UCSD.

Page1 / 7

HW4_solution - Problem 8.86 Given: Flow through sudden...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online