Chapter 06 Solution Manual Fundamentals of Momentum Heat and Mass Transfer 7th Edition James Welty Gregory Rorrer

Chapter 6 End of Chapter Problem Solutions


6.1

For V=A+Br

V(ro)=0=A+Bro

V(ri)= = A+Bri

A=-Bro= - Bri

B(ro-ri)= -

B=


A=


V=

,6.2
Steady Flow: = =0

- = dA


- = [(u2-u1)+ + + g(z2-z1)]

= = 1025(21)= 215.25 kg/s

1= = 4.278 m/s


2= = 11.573 m/s

Since T=0 u2-u1=0

z2-z1= 1.8 m

P2= 175 kPa

P1= -0.15 m Hg = -19.9 kPa

= = 190 m2/s2


= = -57.7 m2/s2

g(z2-z1)=9.81 (1.8)= 17.7 m2/s2

- =(190-57.7+17.7)(215.3) = 32,295 w= 32.3 kw

,6.3

= dA +


=0


- [h1+ + gz1]+ [mu]sys=0 gz1=negligible


vcv = A ( )


= = = 21.8 F/s

, 6.4 Energy equation reduces to

dA =0


u2-u1+ + + g(y2-y1)=0


= g(y2-y1)=0

u= c T=


T= = = 0.0297 F