FinalSolutions

# FinalSolutions - MAE 101A Summer Session I 2008 Final...

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Unformatted text preview: MAE 101A Summer Session I 2008 Final Solution July 30, 2008 1. (a) Using the usual convention of acceleration towards the center of the earth as positive and away as negative, then a z =- 4 . 5 m/s 2 and g = 9 . 81 m/s 2 . Therefore applying equation (2.38) and (2.40) we get P = ( g- a ) = dP dz = ( g- a z ) . Hence, P = P + ( g- a z ) 4 z. Then with 4 z = 1 . 8 m we get that the net gage pressure on the bottom of the tank is P gage = ( g- a z ) 4 z = 1260 kg m 3 (9 . 81 m s 2 + 4 . 5 m s 2 )(1 . 8 m ) = 32 , 455 Pa. (b) For P gage = 0 = ( g- a z ) 4 z then a z = g = 9 . 81 m s 2 i.e. the elevator is free falling with the tank of glycerin inside. The elevator must accelerate towards the center of the earth (i.e. in the direction of g ). 2. Applying the Navier-Stokes x-compponent equation we get g x- p x + 2 u x 2 + 2 u y 2 + 2 u z 2 = du dt with g x = 0 , Since g is the direction of y, where y is the vertical direction....
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## This note was uploaded on 03/30/2011 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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FinalSolutions - MAE 101A Summer Session I 2008 Final...

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