This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 NavierStokes xcompponent 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....
View
Full
Document
This note was uploaded on 03/30/2011 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

Click to edit the document details