# Solution Manual for Fundamentals of Momentum, Heat and Mass Transfer, 6th Edition

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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)
From (1) & (2): h-y=1.275 ft.
(2)
(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: yor at 44.6 ft. Below Surface
2.4
=
=
o
=
= 1ln(1-
) = 300,000 ln(1-0.0462) = 14190 Psi
Density Ratio:
=
= 1.0484
so P= 1.0484
2.5
Buoyant Force:
FB= V=
For constant volume: F varies inversely with T
2.6
Sea H20: S.G.=1.025
At Depth y=185m
Pg= 1.025 wgy = 1.025(1000)(9.81)(185) = 1.86×106 Pa = 1.86 MPa
2.7
r = Measured from Earthโs Surface
R = Radius of Earth
= g= go
P-Patm=
At Center of Earth: r = R
PCtr-Patm=
Since PCtr
PCtr
Patm
=
= 176×109 Pa = 176 MPa
2.8
=
g
=
g
P-Patm= g(+h) = (1050)(9.81)(11034) = 113.7 MPa
1122 Atmospheres
2.9
As in Previous Problem
P-Patm= gh
For P-Patm= 101.33 kPa
h=101.33/ g
for H20: h =
Sea H20: h =
Hg: h =
= 10.33m
= 10.08m
= 0.80m
2.10
5
4
3
2
1
P1=Patm+ Hg g(12โโ)
P1=P2
P2=P3+ K g(5โโ)
P3=P4
P4=PA+ w g(2โโ)
P4=P5
Patm+ Hg g(12โโ)= PA
โโ
PA= Patm+
K g(5โโ)
g[(13.6)(12)-2-0.75(5)]=Patm+5.81
PA = 5.81 PSIG
2.11
Force Balance on Liquid Column: A=Area of Tube
-3A + 14.7A โ ghA = 0
h=
= 26.6 in.
2.12
A
B
D
C
PA=PB- o g(10 ft.)
PC=PB+ w g(5 ft.)
PD=PC- Hg g(1 ft.)
PA-PD=
PA-PAtm=
Hg g(1)-
w g(5)-
o g(10)
w g(13.6 x1-5-0.8 x 10 x 1) = 37.4 LBf/ft
2
2.13
P3=PA-d1g
w = PB+(
Hg g)x(d3+d4sin45)
PA-PB =
= 245 LBf/ft2= 1.70 Psi
3
2.14
3
P3 = PA –
wgd1
P3= PB –
wg(d1+d2+d3) +
Hg gd2
Equating:
PA-PB= Hg gd2- wg(d2+d3) =
2
wg[(13.6)(1/12)-7.3/12] = 32.8 LBf/ft
= 0.227 Psi
2.15
I
PI=PA+ wg(10โ)
PII=PB+
wg(4โโ)+
Hg g(10โ)
PI=PII
PA-PB =
wg[-6+13.6(10)] = 56.3 psi
II
2.16
Pressure Gradient is in direction of – & isobars are perpendicular to ( – )
( – )
String will assume the ( – ) direction & Balloon will move forward.
2.17
At Rest: P=
o
Accelerating: P=
Equating: ya=
Level goes down.
= (g+a)ya
which
yo
2.18
F = P6.6A – PatmA =
h( r2)
h=2m
r=0.3m
F=5546N
yC.P.= + Ibb/A
For a circle: Ibb= r4/4
yC.P.= 2m +
= 2.011m
2.19
Height of H20 column above differential
element= h-4+y
For (a) – Rectangular gate- dA = 4dy
dFw=[ wg(h-4+y)+Patm]dA
dFA=[Patm+(6Psig)(144)]dA
=
=0
[ g(h-4+y)-864](4dy)=0
h=15.18 ft.
For (b): dA=(4-y)dy
[ g(h-4+y)-864](4-y)dy=0
h=15.85 ft.
2.20
Per unit depth:
Fy|up =
wg
=0
r2/2 {buoyancy}
Fy|down = g r2+
2
wg(r –
r2/4)
Equating:
=
g r2+
= w
/ =
w
2
wgr (1-
/4)
= 0.432 w = 432 kg/m3
2.21
a) To lift block from bottom
F = {wt. of concrete}+{wt. of H20}
=
cgV+[
wg(22.75โ)+Patm]A
= (150)g(3x3x0.5)+[62.4g(22.75)+14.7(144)]x(3×3)
= 675+31828 = 32503 lbf
b) To maintain block in free position
F = {wt. of concrete}-{Buoyant force of H20)}
= 675-
wgV = 675-[62.4g(3x3x0.5)] = 675 – 281 = 394 lbf
2.22
Distance z measured along gate surface from
bottom
=500(15)-
g(h-zsin60)dz=0
g
=7500
g
=7500
(62.4)g
=7500
h3 =
h=8.15 ft.
= 541
2.23
Using spherical coordinates for a ring
At y=constant
dA=2 r2sin d
P= g[h-rcos +rcos ]
dFy=dFcos
Fy =
(h-rcos +rcos )(2 r2sin
=2 gr2
(sin
Let: c=2
= c[
d +
=c
d ]
+r
)(1-
d )
gr2
sin
= c[(h-
d )
]
)- (0-
)]
Now for Fy=0
sin =
=[1-
0=(h-
)(
]1/2
& r=d/2
)+ (
) = h-
+
Giving h=
=
=
1/2
For d=0.6m
h=
=
1/2
=
1/2
2.24
0
y
12 ft
H2O,
A
10 ft
Mud,
B
PA-Patm= wg(12)=24g
PB-Patm=24g+40g=64g
Between 0 & A: P-Patm= wgy
Between A & B: P= wg(12)+
mg(y-12)
Per unit depth:
F=
=
dy
(192)+
(50)
=18790 lbf
Force Location:
Fxy=
=
=
dy+
(576+2040)+
=288,400 ft. LBf
=
= 15.35 ft.
(2973-2040)
)dy
2.25
Force on gate=
A = (1000)(9.81)(12) (2)2 = 369.8 kN
yC.P.=
= 0.0208m (Below axis B)
=
=0
P(1) = (369.8×103)(0.0208) = 7.70 kN
2.26
Fc
Fw
10 m
Fw =
A= (1000)(9.81)(4)(10)(1) = 392 kN
ycp is 2/3 distance form water line to A
~ 6.66m down form H20 line
~ 3.33m up from A
= Fc(9)=392(3.33)
Fc = 145.2 kN
2.27
Width = 100m
H20@27
=997 kg/m3
F=
A = (997)(9.81)(64)x(160)(100) = 10.016 x 109 N = 10.02×103 MN
For a free H20 surface
ycp= (128m) = 85.3m {below H20 surface} =106.7 m {measured along dam surface}
2.28 Spherical Float
Upward forces
F + FBuoyant
Downward forces
WT
W= gV=
R3 )
Fb=
g
gVz=
R3)z
z= fraction submerged
F=
z=
R3)-
g
R3 )
2.29
=
G is center of mass of solid
= 2[1/2(L/2)(0.1L)(L)
(
)-(0.9L)(L)(L)
(0.05L
)]+M
{Part of original submerged volume is now out of H20}
(Part that was originally out is now submerged}
M=
L4
[
+ 0.045] =
L4
(0.02833) = 0.00556
L4
2.30
Air
Butyl Alcohol
Benzene
Water
17 feet
19 feet
34 feet
25 feet
(a) the pressure at the butyl alcohol/benzene interface
(b) the pressure at the benzene/water interface
(c) the pressure at the bottom of the tank
2.31
For Blood:
For Mercury:
So,
Solve for the height of the blood,
Eliminate the gravity terms,
We take 120 mm Hg as the height here, and 120 mm = 0.12 m,
and the density of mercury is 845
=
:
2.32
Freon
12 ft
2 ft
Nitrogen
4 ft
7 ft
Benzene
2.5 ft
4 ft
7.5 ft
10 ft
5 ft
8 ft
11 ft
5 ft
10 ft
7 ft
7.5 ft
Water
5.8 ft
Mercury
Glycerin

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