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Unformatted text preview: NAME: MAE 101B – Advanced Fluid Mechanics  Spring 2010 Midterm # 2 May 3, 2010 50 minutes, open book, open notes, calculator allowed, no cell phones. Write your answers directly on the exam. 30 points total + 5 extra credit points (optional). The exam is 4 pages long. (1) (10 points) Water flows in a horizontal constantarea pipe. The pipe diameter is 50 mm and the average flow speed is 1.5 m/s. At the inlet the gage pressure is 588 kPa, and the outlet is at atmospheric pressure. We ignore any minor losses. (a) (2 points) Using the energy equation, determine the head loss in the pipe. Solution : We take (1) to be inlet and (2) the outlet. The governing equation between inlet (1) and exit (2) is the energy balance p 1 ρ + 1 2 α 1 ¯ V 2 1 + gz 1 p 2 ρ + 1 2 α 2 ¯ V 2 2 + gz 2 = h lT . (1) Along the pipe, we will always have α 2 = α 1 and ¯ V 1 = ¯ V 2 so these terms cancel out. For question (a) we have z 1 = z 2 therefore the head loss is given by h lT = p 1 p 2 ρ = 588 10 3 = 0 . 588 J/g. (b) (3 points) If the pipe is now aligned so that the outlet is 25 m above the inlet, what will the inlet pressure need to be to maintain the same flow rate?...
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This note was uploaded on 05/05/2010 for the course MAE 101b taught by Professor Staff during the Spring '08 term at UCSD.
 Spring '08
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