Chapter 05 Solution Manual Fundamentals of Momentum Heat and Mass Transfer 7th Edition James Welty Gregory Rorrer

Chapter 5 End of Chapter Problem Solutions

5.1




Cons. of Mass: dA=0

1A1=[2



4 =2 =


= = 26.7 /s

,5.2 System shown in Prob. 5.1

= dA

-Assuming unit depth-
Fx + (P1-P2)4=2 [

=2 [ + - ]

=2 [ +1)]-4

From 5.1 =

Fx+(P1-P2)4=

Fx= -800N/m= 52.8 lbf/ft.

P1-P2 = = 157 lbf/ft2 7500 Pa = 7.5 kPa

,5.3 Same general configuration except exit velocity distribution is = 2(1-cos )
As in 5.1 the expression to be used is:

A1 = 2


4 =2


= = = 55 ft/s

, 5.4




Steady Flow

Fx= dA

= 2 A 2- 1 A1

= -

= (1.02 - )

= 1A1 = (0.0805 LBm/ft.3)(10.8 ft.2)(300 ft./s) =260.8 LBm/s

Fx = (260.8 lbm/s)[1.02(900 ft./s)-300 ft./s] = 5005 lbf