Chapter 01 Solution Manual Fundamentals of Momentum Heat and Mass Transfer 7th Edition James Welty Rorrer Foster

Chapter 2 End of Chapter Problem Solutions


2.1
Assume Ideal Gas Behavior

= =

For T = a + by

= 530 - 24 y/n

=


=


ln = ln

With P = 10.6 PSIA, Po = 30.1 in Hg

h = 9192 ft.

,2.2




(a) =0 on tank

(1)

At H20 Level in Tank: P=Patm+ wg(h-y) (2)

From (1) & (2): h-y=1.275 ft. (3)

For Isothermal Compression of Air
PatmVtank=P(Vair)

P= Patm (4)

Combing (1) & (4): y=0.12 ft. and h=1.395 ft.

(b) For Top of Tank Flush with H20 Level

=0
P=Patm+

At H20 Level in Tank: P=Patm+ wg(3-y)


Combining Equations:
F= 196(3-y)- 250

For Isothermal Compression of Air:
(As in Part (a))
3-y = 2.8 ft.

F = 196(2.8) - 250 = 293.6 LBf

, 2.3

When New Force on Tank = 0

Wt. = Buoyant Force = 250 Lbf

Vw Displaced = 250/ wg = 4.01 ft3

Assuming Isothermal Compression
PatmA(3ft.)= P(4.01 ft3) = (Patm+ gy)(4.01)
y=45.88 ft.

Top is at Level: y-

or at 44.6 ft. Below Surface