SOLUTIONS MANUAL
Fundamentals of Momentum, Heat, and Mass
Transfer 7th Edition By James Welty, Gregory
Rorrer, David Foster | All Chaters 1-31
Part 1: End of Chapter Problem Solutions
Part 2: Instructor Only Problems Solutions
Part 3: Show/Hide Problems Solutions
Part 1: End of Chapter Problem Solutions
Chapter 1 End of Chapter Problem Solutions
1.1
n = 4 x 1020 molecules/in3
=
= 1.32x104 in./s
A= (10-3 in)2
NA = n A=1.04x108 m/s
1.2
Flow Properties: Velocity, Pressure Gradient, Stress
Fluid Properties: Pressure, Temperature, Density, Speed of Sound, Specific Heat
1.3
mass of solid= ρsvs
mass of fluid= ρfvf
mix=
=
=
1.4
Given
=
For P1= 1 atm
7
= 1.01
P= 3001(1.01)7 – 3000 = 217 atm
1.5
At Constant Temperature
= constant
= constant
For 10% increase in
P must also increase by 10 %
1.6
Since density varies as
&
=nM
(M=Molecular wt.)
n250,000= nS.L.
= 4 x 1020
= 2.5 x 1016
1.7
=
+
y
+
y
x
=
+
=
=
Q.E.D.
x
+
1.8
=
=
Q.E.D.
+
=
=
1.9 Transformation from (x,y) to (r, )
=
+
=
+
r2 = x2+y2
=
so:
=
=
=
=
=
=
=
=
=
+
=
1.10
=
+
+
=(
)
=(
Thus:
+
=
+(
)
+
+ (
+
)
+
+
)
+
1.11
P=
+
P(a,b)=
=
2
{[
2
+
+ 2)
+ (
)
(sin1cos1)
1.12
T(x,y)= Toe-1/4 [ (cos cosh )
+
(sin sinh )
T(a,b)= Toe-1/4 [ (cos cosh1)
+
(sin sinh1)
= Toe-1/4[
=
+
[
(1+e-2)
]
+
1.13
In problem 1.12 T(x,y) is dimensionally homogeneous (D.H.)
P(x,y) in Prob 1.11 will be D.H. if
~
LBf s2/ ft4
or using the conversion factor gc
1.14 = 3x2y+4y2
A scalar field is given by the function:
(a) Find
at the point (3,5)
For the value of
at the point (3,5)
(b) Find the component of
that makes a -60 angle with the axis at the point (3,5)
Let the unit vector be represented by
At the point (3,5) this becomes:
1.15
For an ideal gas
P=
From Prob 1.3:
1.16
= Arsin (1- )
a)
=
b)
+
= Asin (1- )
1/2
=A
max is given by d
Requiring
For
= 0 or
=
= 0: -
And for
d =0
) = 0
(1)
=0
(1+ ) +
(1-
=0:
(2)
4a2/r2=0
From Eq. 2:
If a 0, r
dr+
then
Subst. into Eq. 1
for which
= 0,
=0, 1-a2/r2=0
Giving a = r
For =
1+ a2/r2 =0
Thus conditions for
impossible
max are
r = a
(3)
1.17
P=Po+
2
=
2
=+
2
=+
2
P=
2
1.18
Vertical cylinder d=10m, h=6m
V= (10m)2(6m)= 471.2 m3
@ 20
w=998.2 kg/m
3
m= wV= (998.2)(471.2)= 470350 kg
3
@ 80
w=971.8 kg/m
m=(971.8)(471.2)= 457910 kg
12440 kg
1.19
V= 1200 cm3 @ 1.25 MPa
V=1188 cm3 @ 2.5 MPa
Liquid
= -V
T
-V
V= 1194 cm3= 1.194 x 10-3 m3
∆V= -12 cm3= -1.2 x 10-7 m3
= -1.194 x 10-3
= +12440 MPa
1.20
= -V
T
-V
V=0.25 m3
V= -0.005 m3
P= 10 mPa
= -0.25
= 500 MPa
1.21
For H20:
=2.205 GPa
= -0.0075
-V
or ∆P =
∆P= (2.205 GPa)(0.0075) = 0.0165 GPa= 16.5 MPa
1.22
For H20:
P1=100kPa
= -V
=
or
=
P2=120 MPa
2.205
=
= 0.999 x 10-3 = 0.0999 percent
1.23
H20@ 68
(341 K)
0.123 [1-0.00139(341)] = 0.0647 N/m
In a clean tube- =0
h=
=
= 9.37 x 10-3m = 9.37 mm
1.24
Parallel Glass Plates
Gap=1.625 mm
= 0.0735 N/m
For a unit depth:
Surface Tension Force = 2(1) cos
Weight of H20 =
(1)(1.625x10-3)
For clean glass cos =1
Equating Forces:
2(1) =
h=
(1)(1.625x10-3)
= 0.00922m= 9.22 mm
1.25
1.26
H20-Air-Glass Interface @40
Tube Radius= 1 mm
h=
cos
= 0.123[1-0.0139(313)]=0.0695 N/m
h=
= 0.0143m (1.43 cm)
1.26
1.27
Soap Bubble- T=20
d=4mm
= 0.025 N/m (Table 1.2)
Force Balance for Bubble:
r2
r
so
=
= 25 N/m2 = 25 Pa
1.27
1.29
@60 C
= 0.0662 N/m
= 0.44 N/m
Tube Diameter= 0.55 mm
h=
For H20:
For Hg:
= 0.0499m (4.99 cm Rise)
= -0.0157 m (1.57 cm Depression)
1.28
1.30
H20/ Glass Interface
T=30 C
= 0.123[1-0.0139(303)] = 0.0712 N/m
= 996 kg/m3
h
1 mm
h=
r=
=
d = 2r = 2.915 cm
= 0.0146 m (1.457 cm)
1.29
1.31
Bubble Diameter = 0.25 cm = 0.0025 m, and so Radius = 0.00125 m
Capillary Tube: Diameter = 0.2cm = 0.002 m, and so Radius = 0.001 m
Beginning with:
Rearrange and remember the unit conversion Pa=kg/ms2,
Next, we can calculate the height of the fluid in the tube:
1.30
1.32
First, calculate the surface tension of water using the temperature of the water:
Next, using the equation for the height of a fluid in a capillary,
Rearranging and solving for the radius:
Diameter is 2r so D=1.50 mm
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) = 1.86x106 Pa = 1.86 MPa
2.7
r = Measured from Earth’s Surface
R = Radius of Earth
= g= go
P-Patm=
At Center of Earth: r = R
PCtr-Patm=
Since PCtr
PCtr
Patm
=
= 176x109 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(12’’)
P1=P2
P2=P3+ K g(5’’)
P3=P4
P4=PA+ w g(2’’)
P4=P5
Patm+ Hg g(12’’)= PA
’’
PA= Patm+
K g(5’’)
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 – 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(10”)
PII=PB+
wg(4’’)+
Hg g(10”)
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(22.75’)+Patm]A
= (150)g(3x3x0.5)+[62.4g(22.75)+14.7(144)]x(3x3)
= 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) = (369.8x103)(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 = 10.02x103 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
Chapter 4 End of Chapter Problem Solutions
4.1
= 10 +7x
At (2,2) = 10 +14
At -30 from x axis:
2
1
Unit vector:
=
Along this direction the component is : *
* =(
-
)*( 10 +14
=5
-7 = 1.66m/s
4.2
= 10
+ 7x
=
=
10
=7
10y = 7
+c
At (2,1) c = 10 – 14 = - 4
Eqn is: 7
Or x2
10y + C = 0
y-
= 0 (a)
Across the surface connecting points (1,0) and (2,2):
(2,2)
y
(1,0)
x
= 10
+7
=20+ (3)= 20+10.5 =30.5 m3/s (b)
4.3
CV
dA+
dA =
2A2=18
=18
2Avg =
dV=0
2AvgA2 -
)dr
= 72
= 128 ft/s
(1-
) 2 rdr=0
4.4
dA = 0 =
in =
cos30 dA2
out
Ain = 4Aout
out =
= 46.2 ft/s
A = 10( x ) = 1.11 ft.3/s
dA1=0
4.5
Steady Flow:
A1 +
1A1=
2=
A2 +
2A2+
6 (0.2)2=
dA=0
2
dx =
[0.2(0.2+0.5)]
= 1.71 m/s
2A2 +
=
2
+
=
(D+L)
4.6
For Steady, Incompressible Flow:
=A Avg=
i
Ai
From given data set:
Dist. from center
in.
0
3.16
4.45
5.48
6.33
7.07
7.75
8.37
8.94
9.49
10.-
i
Ft./s
7.5
7.10
6.75
6.42
6.15
5.81
5.47
5.10
4.50
3.82
2.4
Ai
in.2
7.844
37.64
31.96
32.10
31.48
29.85
33.22
33.22
31.44
31.57
15.82
Ai
Ft. /s
0.4084
1.856
1.498
1.431
1.344
1.204
1.262
1.176
0.982
0.838
0.264
316.14
18.263
= 316.14 in.2 {exact area= 314.16 in.2} = 2.195 ft.2
Ai = 18.26 ft3/s
=
i
Avg =
/A =
= 8.32 ft/s
3
4.7
(a)
Let M = total mass in tank and S = salt in the tank.
Thus,
Integrating,
t = 1 hour and 40 minutes which equals 100 minutes.
M = 68 lb/ ft3 x 1ft3/7.48 gal x 100 gals = 909.9 lb
(b)
(c)
Begin with the equation developed in part (a),
Rearrange and solve for dt
Integrate,
4.8
For Piston & Cylinder Shown:
At 1: = 1= 2 ft./s
A1 1 = A2 2
2=
a2 = a1
1=
1=
a = a1 = 5ft/s2
)2(2) = 128 ft./s
= 5( )2 = 320 ft./s2
4.9
For steady flow:
dA = 0
or
=constant
=
+
+
=0
Q.E.D.
4.10
dA=
~
(
~
=
)dA=d
dV= M
+
=0
Q.E.D.
4.11
For the C.V. shown:
dA+
Steady flow:
dV=0
dV=0
C.V. moves to right with =
Thus:
2=
w(1-
1A
w+
1/
2)
2A( w-
w
2) = 0
4.12
Avg =
dA =
For z = r/R
Avg = 2
dr
dz = dr/R
Max
For
d = -dz
Avg= -2
Max
Avg=
max= 0.817
max
4.13
dA+
Steady flow=
dV=0
dA= Horizontal=
dV=0
o(6d)-
o(6d)+
Horizontal + 2
o(3d) =
o(3d)
ydy
4.14
+
=0
m= (2L)(b)(1)
=2 L
Where = -v
=2
side=2
Giving: -2
+2
=0
or LV=
(a) For = Avg (A constant)
LV= Avgb
avg = V
(b) =ay+by2
with
=4
(b/2)=
max
LV= 4
max=
max
max
dy
4.15
=
dA = w
(2 -
2000 cm3/m = (10)
(2)
= 150 cm/m= 2.5 cm/s
) dy = w h
4.16
v = wA = wh2cot20
= w cot(20) d h2/dt
=
=
t=( -
t
)[ cot(20)]
4.17
v = wA = wL2tan /2
=
= wL2
~
cos2 =
~
=
=
(tan ) = wL2sec2
4.18
3
Steady Flow:
dA=0
Let the exit from the nozzle be position 3.
1=
A2 + A3
1.3x10-3 = (0.02)2(2.1) + (100) (10-3)2
= 8.15 m/s
4.19
Steady Flow:
=0
A3 - A1 - A2 =0
=
= 5.15 m/s
=
4.20 Volume displaced by plunger
=Ap =
V
Volume of H20 moving past P:
= (A-Ap) = (D2-
)
In steady state operation these must be equal:
V = (D2=V
)
(a)
Relative to plungerR=
+V = V
+1] (b)
4.21
Cons. of Mass - constant
3
out= 6 cm /s - constant
For no leakage
= ApV = (2)2V = 6
v = 1.91 cm/s
For Leakage: = 6+0.6
=
= 2.1 cm/s
4.22
Parallel Plates - Incompressible, Steady Flow
= constant
zo=
=a
(zo-z)dz = a
is max at z=zo/2
= 6 [zo- ]
= 6 Vo/4= 12 cm/s
a
4.23
For steady incompressible flow:
out- in=0
Q+b
(
)dy=
b
~
dy=
~
Q=
b(1-5/8)=
d
= 5/8
4.24 (See sketch for Prob. 4.14)
Plates are circular
+
=0
m= b L2
=
L2 = -
L2v
out=
Lb
=
=
Lv/2b
(a)
exit
L2v
As in Prob. 4.14- Parabolic Exit Profile
2
exit=ay+by
=4
max[
–(
]
~
exit =
max=
2 Ldy =
v
Lb
Max
4.25
Since there are no leaks in the system,
And the equation becomes
4.26
(a) Total mass in the tank after 2 hours
1 liter/min = 0.01667 liters/sec
Total Mass = M
At the 2 hour point, which is 120 minutes,
(b) The amount of sodium sulfate in the tank after 2 hours
Rearrange,
Integrating,
The initial condition given in the problem was that at time = 0, there is 6000 grams of a 10%
solution of sodium sulfate solution.
At the 2 hour point (2 hours = 120 minutes):
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 depthFx + (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
5.6
1
2
C.V. around boat- Steady, Incompressible flow
=
Fx=
dA =
( - )
=
Tension in Rope= Fx/cos30 =215 lbf
= 186 lbf
5.7
Flow is Steady, Incompressible
=
dA = ( - )
=
= 0.8(62.4)(3 ft.3/s) =149.8 lbm/s
= Fx+P1A1-P2A2
{Atmospheric Pressure cancels}
Equating:
Fx + P 1
- P2
=
Fx= (P2
- P1
)+
(
(
P1= 50 Psig P2= 5 Psig
Fx= -5630+392 = -5238 lbf
)
-
)
5.8
Fluid is H20
P1= 60 Psig
P2= 14.7 Psia (units?)
D1= 0.25 ft.
D2 = ft.
= 400 gal/m= 0.892 ft.3/s
Steady, Incompressible Flow
dA =0
=
=
=
= 18.17
Fx-P1A1-P2A2=
= 18.17 ft./s
= 72.7 ft./s
( - )
Fx= ( - )-P1A1+P2A2
=[62.4(0.892)(72.7-18.17)]/32.2-(60+14.7)(144) (0.25)2+(14.7)(144)
=94.3-528.0+25.4
=-408 lbf
2
5.9
H20- Flow is Steady, Incompressible
=
dA
= P 1A 1 - P 2A 2
dA = A2
- (As + Aj )
Equating:
P1-P2= [
-
-
]
By Conservation of Mass:
dA =0
A2 - As - Aj =0
=
P1-P2 =
+
=
(10) +
(90) = 18 ft./s (a)
[(18)2 -
(10)2 -
(90)2] = -1116 lbf/ft.2= -7.75 psi
P2-P1= 7.75 psi
5.10
Flow is Steady, Incompressible, Frictionless
For Frictionless Flow - No Drag on plate
=
Fn=
dA =0
( cos )Aj = -
Fx = (-160)(4/5)= -128 lbf
Fy = (-160)(3/5)= -96 lbf
= -2(100)(4/5)= -160
5.11
Steady, Incompressible, Frictionless Flow
In x-direction:
=
~
Cons. of mass:
dA =
h=
2
b-
2
a-
(a+b)
h=a+b
a= (1- cos
b= (1+ cos
~
=
dA =
Part (b):
=Fy =
Fy =
2
=
2
h sin
dA
(
a)-
(
b)
h(sin
=
-
=
=
(a)
2
h cos =0
5.12
Vj
Vp
Flow is Steady, Incompressible, Frictionless - Atmospheric pressure cancels
C.V. moves to left with velocity, vp
=
dA
Fx= Aj( j+vp)2 =
(5+30)2 = 237.4 lbf (for moving plate)
For vp = 0
Fx =
(30)2 = 174.4 lbf
5.13
Cons. of Mass: For unit cross section
+
=0
M=
2
2x+
+
1y
-
1
Since
=
2( w -
2) -
=-
2 2
2 2= 0
=-
m
m
m= 0
1
(1)
x-momentum:
=
P2-P1 =
From (1):
dA +
2
2
2
2+
2
2 ( m-
2)=
Giving: P2 - P1 =
1
2
2x=
1
2
2
m
m
Q.E.D.
+
2
2
w=
2
2 [ w-
2]
5.14 For situation considered in Prob. 5.13
(a) Air
m= 1130 ft./s
= 0.00238 Slug/ft.3
P2-P1= 1 2 m = (0.00238)(10)(1130) = 26.9 PSF = 0.187 psi
(b) H20
m= 4700 ft./s
= 1.938 Slugs/ft.3
P=(1.938)(10)(4700) = 91,080 PSF= 633 psi
5.15
Cons. of Mass:
Technique is to let
P2 = P1 +
2=
1+
S
S
etc.
By Conservation of Mass- { S= y->0}
( A )=0
By Momentum Theorem, using Cons. of mass result:
dP+ d +gdy=0
-messy-
5.16
D1=0.3m
1=12 m/s
P1=128 kPag
D2=0.38m
P2=145 kPag
2=7.48 m/s
A1 = (0.3m)2 = 0.0707 m2
= A1
A2 = 0.1134 m2
3
1 = (0.0707)(12) = 0.8484 m /s
In x direction:
=
Fx + P1A1 - P2A2cos =
dA
( 2 cos -
1)
Fx=(1000)(0.8484)[7.48(cos30 )-12]-(1000)[(128)(0.0707)+(145)(0.1134) cos30 ] = -505.5 N
In y direction:
Fy – P2A2sin =
=
dA
( 2 sin )
Fy=(1000)(0.8484)(7.48sin30 )+(1000)(145)(0.1134)(sin30 ) = 11395 N=11.395 kN
5.17
Steady incompressible flow:
=
dA
A1= (1.5)2=1.767 ft.2
A2= (1)2=0.785 ft.2
A3= (2)2=3.142 ft.2
For C.V. between (1)&(2) of the figure:
=
dA
Fx+F2+P1A1-P2A2=
Fx
( 2-
1)
+(530-14.7)(144)(0.785)-(990-14.7)(144)(1.767)-F2 = -130,000 lbf –F2
For C.V. between (2)&(3) of the figure:
Fx+P2A2-P3A3= ( 3- 2)
Fx=
(6700-3400)+(26-14.7)(144)(3.14)-(530-14.7)(144)(0.785) = 25777 lbf
Stress at position 2 in the figure:
= =
= 1823 psi Compression
At 1: F1= -130,000+25777 = -104220 lbf
=
= 4915 psi Tension
5.18
Flow is steady & incompressible
No net pressure force
=
Fx=
dA
dA-
Ain = 2
( )2dy+
(3d)-
Momentum out top & bottom
Fx=2
+
Force on cylinder=
(3d-6d) = d
d
(6d)
5.19
Fluid is H20
Vsonic=1433 m/s
This is just like Prob. 5.13 for an observer moving with H20 stream: = 3 m/s
Then P=
w
=(1000)(1433-3)(3) = 4287 kPa
5.20
Static pressure of H20:
On left- P= H
On right- P=
h
=
dA
H2 =
H=[
=(
[ h
2
h)
+
+ h2]1/2
]=
+ h2
5.21
Conservation of Mass:
=0
- hv1+ Lv2 = 0
v2=
(a)
x-momentum:
=
dA
P1A1-P2A2+Fx= ( 2Fx= ( 2-
1)
1)+P2A2- P1A1 =
1h( 2-
1)+
-
=
h(h/L -1)+
(L2-h2)
(b)
5.22
Conservation of Mass:
1h 1= 2h 2
Momentum Thm:
=
dA
P1h1-P2h2= ( 2P 1=
,
1)
P2
-
1h 1( 2-
1)
From Cons. of Mass:
2=
( -
=
)=
1 h1/h2
h1
Factoring & cancelling h1-h2
(h1+h2)=
h1
+ h1h2 -
=0
h 2= [
- 1]
& from continuity
[1+
]
5.23
D1= 8 cm
1= 5 m/s
h= 58 cm
D2= 5 cm
P2= 1 atm
A1= (8 cm)2= 50.3 cm2
A2= (5 cm)2= 19.6 cm2
2= 5 m/s(
)= 12.83 m/s
x-momentum:
=
Fx+P1A1-P2A2=
( 2-
P1-P2=
dA
1)
wgh[13.6-1] = (1000)(9.81)(0.58)(12.6) = 71.69 kPa
Since P2= 1 atm
P1= 71.69 kPa
Fx+P1A1= (1000)(50.3x10-4)(5)(12.83-5)
Fx= 197-71.69(50.3x10-4)(1000) = -163.7 N
5.24
x momentum:
=
+
dV
Fx= - wAj ( cos )+ wAj
= wAj ( - cos )
=1000( )(0.1)2(20)[4.5-20cos45 ]
= -1515 N
Force on car by jet: Fx=1515 N
y momentum:
=
Fy= - sin
+
dV
( Aj)+ 0 = -(20sin45)(1000)(-20)* (0.1)2 = +2220 N
Force exerted by H20:
Fy= -2220 N
Total Force = = 1515 - 2220
N
5.25
Coordinates fixed to cart
~moving to right at
Momentum thm. in x-direction
= Aj (- )- As (- )
Fx= As
- Aj 2
In y-direction
Fy= Aj (0)- As (- s) = As
2
Force of fluid on car is negative of these
5.26
for C.V. shown as a red dotted line in the figure.
+ M=0
M= Ah
= A
=
+
- gAh= - A 2+ A (h
-gh= - 2+ (h
= -g
dV
5.27
Steady in compressible flow
Rotation is about z-axis=
dA
=
out = (rx
-
)
out
At position on x-axis: rx=r2 and ry=0
= 800(
=
tan =
)= 1.783 ft.3/s
)(
= 8.51 ft./s
=
= 8.51 ft./s
Abs. velocity @ r2
= -rw+ tan
= -(8/12)(
)+ t
= -82.38+ 8.51= -73.87 ft./s
Now- in momentum expression:
Mz=(r2 )
=
Power= Mzw = 174(
= 174 ft. lbf
)(
) = 39.1 Hp
5.28 Same configuration of Prob. 5.28
= 1.783 ft.3/s
at inlet- =
=
Vr1
= 25.54 ft./s
rw
r =-r1 =-(2/12) (
) = -20.6 ft./s
=
= 38.9
=
Part (B):
=
Fx=-
dA
(
)A1 =
=
=
= 70.6 lbf
α
5.29
y
x
For C.V. shown as a red dotted line.
=
Mz=
dA
=2 (
)=
= 1.057 ft. lbf
5.30
=
x )z
=
dA
( )
Mz=2()
ry= Rsin
= sin - R
Mz=2 [-Rsin ( sin - r )]
=
+
5.31
=
x )z
dA
=
tdx
=
t xdx= =
M=
t( )
= 64 ft./s
(
) = -5950 ft. lbf
5.32
=
x )z
dA
MB=
+ r2P2A2 =(
=(30 gal/m)(
= 0.0668 ft.3/s
= =
MB =
-
)
+ r 2P 2A 2
=0
= 21.79 ft./s
+ (3)(24)( )(0.75)2 = 8.46+ 31.81= 40.3 ft. lbf
5.33
Linear Momentum: Coordinate System moves with cart
=
Fx= A[ - c)cos ]( - c)- A( - c)2
P=
cFx=
A ( - c)2[cos -1]
For m=
c
P= A[cos -1]
For P=Pmax
3
m(1-m)2
=0
= A[~] 3
(m-2m2+m3)
={ } (1-4m+3m2)=0
m= 1, 1/3
m=1 ~ minimum
m= 1/3~ maximum
m= = 1/3
Part (b) Rotation about z-axis=
x )z
dA =
=r [ - c)cos + c- ]
= r (cos -1)(
c)
P= Mzw= Mz =[ (cos -1)] c( for m=
P=[ ]m 2(1-m)
= [ ] 2(1-2m)=0
or Pmax occurs when
m= = ½
(out)-
c)
(in)
Chapter 6 End of Chapter Problem Solutions
6.1
For V=A+Br
V(ro)=0=A+Bro
V(ri)= = A+Bri
A=-Bro=
B(ro-ri)= B=
A=
V=
- Bri
6.2
Steady Flow:
-
=
-
=
=
=0
dA
[(u2-u1)+
+
=
= 1025(21)= 215.25 kg/s
1=
= 4.278 m/s
2=
= 11.573 m/s
Since T=0
+ g(z2-z1)]
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
6.5
Between A & B:
=
dA
=
=
u=
= 402,000
[
+
+ g(yB-yA)+ u]
=0
={
+ g(yA-yB)+
}
PB=PA+ {
+ g(yA-yB)+
} = 46.5 Psia
6.6
1
= 4 ft.3/s
P1= -6 Psig
=
2
P2= 40 Psig
= 5.10 ft./s
=
= 7.33 ft./s
y2-y1= 5 ft.
Energy Eqn. reduces to:
=
[ +
g(y2-y1)] =
g[ +
+y2-y1] = 27850 ft. LBf/s = 50.6 Hp
6.7
CV is red dotted line. Entrance is 1 and Exit is 2.
For C.V. (1)- Energy Eqn. reduces to
0=
+
+ g(z2-z1)
(z2-z1)=
=0
=
=
= [2 ]1/2= 20 m/s
=A = (0.3)2(20)= 1.417 m3/s
For C.V. (2) Energy Eqn. is:
=
[ u+
+
g(
)=
=(1.22)(1.417)(20)2/2 =346 W
6.8
Steady Flow Energy Eqn:
1(u1+
+ )+
3(u3+
+ )=
2(u2+
Energy Eqn. can be written: A1
1[cvT1+
+ )
Cons. of Mass:
1A1+ 3A3= 2A2 (1)
Momentum:
(P1-P2)A1=
A1-
A1-
+ ]+ A3
3[cvT3+
A3 cos (3)
For u= cvT, T1=T3, P1=P3
& Lots of Algebra
cv(T2-T1)=
[1+
-1] [
]+ [
-2 cos ]
+ ] = A2
2[cvT2+
+
(2)
6.9
=3 ft.3/s
Between A & B- Energy Eqn. is:
)
dA=0
+ uB-uA+
A=
=0
= 3.82 ft./s
B=
= 15.28 ft./s
=
+ 0.45 = 2.15 ft. H20
PA= 2.15+2= 4.15 ft. of H20
6.10
C.V.is red dashed line in the figure. PA= 10 Psig
=
dA
Flow rate must be determined
Energy Eqn. for C.V. shown:
u+
=
A=
B=
+
+ g z=0
u=0
= 743 ft.2/s2
= 2.865
= 3.82
= (2.8652-3.822) = -3.19
g y= 32.2(-5)= -61 ft.2/s2
743-3.19 -61=0
=
= 14.6 ft.3/s
A= 41.9 ft./s
B= 55.9 ft./s
Fy+PAAA- gV= ( A)
PAAA=10( )(8)2= 502 lbf
gV (wt.)=
= 109 lbf
( A)=
= -1185 lbf
Fy= -502+ 109- 1185 = -1578 lbf (Force on lid is 1578 lbf )
6.11
= 6 m3/s
P= 0.10 m Alcohol (S.G.= 0.8) = 0.08 m H20= 785 Pa
A1= (0.6)2= 0.283 m2
1=
=
= 21.2 m/s
Energy Eqn. reduces to:
+
+ g y=0
=
= [
=
= 640 m2/s2
640=
A2= 0.144 m2
D2=0.428m
-
g y=0
]= [
6.12
Energy Eqn. reduces to:
u+
+
+ g y=0
& for u=
1=0
+
+ g(y2-y1)=0
2 = [2
)+g y]1/2
By Conservation of Mass:
ATank(
=Ajet 2
=[2
)+g y]1/2
=
k1=
k2=2g
t=
k1= 344 ft.2/s2
[
[
t= 105 s
(
]1/2= 23.2 ft./s
(yo)]1/2= 25.8 ft./s
6.13
Energy Eqn. reduces to:
+
= +
P1=Patm= 29 “Hg(
)=14.25 Psi
P2=?
=[(85 n /H)(
)]2=(124.7)2
= 1202
P2=14.25+ [(124.7)2-(120)2]
~
=
=
= 0.0770 lbm/ft.2
~
P2= 14.25+1.37= 15.6 Psi
6.14
Energy Eqn:
+g(y2-y1)=0
In x-direction:
ocos
In y-direction:
osin
x=(
ocos
t
y=(
osin
t-gt2/2
=
x=
=
Combining:
y=xtan -
y=0.6 m
x=3.6 m
tan = 0.577
cos =0.866
0.6= 3.6(0.577)= 7.57 ft./s
Total head= 0.6+
= 3.52 m
6.15
= 550 g/m= 1.225 ft.3/s
= =
= 6.35 ft./s
Energy Eqn: 2 is at H20 level outside pipe
+
=
P1=-
+ g(y1-y2)=0
- gy1 =
= =0
- 32.2(6) = -(20.16+193.2) = -213.4 ft.2/s2
= -2.87 Psig
6.16 With reference to Prob. 6.15 between H20 surface & pump inlet energy Eqn. is
+
+ g(y1-y2)= hL
-
=
-(y2-y1)- hL =
2=
P2=Pv
+ y1-y2= hL
2=[2(32.2)(25.35)]
=A
P1=PAtm
2
1/2
= 40.4 ft./s
(40.4)= 7.8 ft.3/s
-4-4 = 25.35 ft.
=0
6.17 From Prob. 5.27
r= 10.22 ft./s
r2= 82.2 ft./s
t= 10.22 ft./s
At r2:
x= 82.2-10.22,
y= 10.22
=( + )1/2= 82.9 ft./s
Head=
= 106.7 ft.
P=
= 62.4(82.9)2/32.2(2) = 6660 LBf/ft.2 = 46.2 Psi
6.18
For the situation shown:
Thrust= F=
Power=
=
~
~ ~
Favorable Choice:
High Volume, low pressure
6.19 From Prob. 5.7:
P1= 50 Psig
D1= 12 in.
= 3 ft.3/s
hL=
P2= 5 Psig
D2= 2.5 in.
S.G.= 0.8
+
~
2
1=3/ (1) = 3.82 ft./s
2=3/
hL=
2
= 88 ft./s
+
= 9.79 ft.
6.20 For a C.V. enclosing the falls:
u+
+
=
=0
+ g y=0
u= g y= 9.81(165)= 1620 m2/s2 =1620 m2/s2(1
For H20: Cp= 4184 J/kg*K
T= =
0.39
)= 1620 J/kg
6.21
Assume Vertical Forces do not include momentum of incoming air:
(P-PAtm)A=mg
{Pressure Force}= Wt.
Energy Eqn. becomes Bernoulli Eqn. between inside & exit=
or
=2
= 4885 m2/s2
=2
= 69.9 m/s
=69.9(24)(0.03) =50.3 m3/s
= 60.6 kg/s
Energy Eqn:
=
With
u=
( u+
g
+
g
)
=0
Energy Equation becomes
=
=
= 148 kW {per person}
6.22 From Prob. 5.22
h2= [
- 1]
For Bernoulli Eqn. to be
vsolid-hL=0
Energy Eqn. for this case is
hL=
+
+ y1-y2
& since P=Patm+
hL=
(h-y)
+ h1-h2
Writing solution to Prob. 5.22 as
h2= [
- 1]
{B=
}
& Note that for h2>h1 B>8
Bernoulli Eqn. applies for B=8
& Obviously hL>0 for B>8
6.23 Energy Eqn. applies in form:
=
=
(Power)
= 1.238x10-4 ft.2/s
=
P=
=
=
= 1.584 Ft. lbf/s = 2.148 W
Per monthP= 2148 W (30)(24) = 1547 Wh= 1.547 kWh
6.24 Bernoulli Eqn. between free stream & a reference point(1) on car
+
=
=
=2
=2
+
6.25 Energy Eqn. is
=
)
= [(u2-u1)+
dA
+
+ g(y2-y1)]
200 kJ/kg
=
= 340 kJ/kg
=0
g y= 9.81(15)= 0.147 kJ/kg
= 200+340+0.15= 540 kJ/kg
6.26
=
2
A
=
B(2
)(1)(
)
2
2
A = 8.22
2
2
B =14.6
A=2.865
B=3.82
For negligible friction- Bernoulli Eqn. applies
+
+ g(y2-y1)=0
= 2
=
=3.32 2
= 743 ft.2/s2
g(yA-yB)= 32.2(-5)= -161
3.32 2= 743-161= 582
= 13.2 ft.2/s
6.27
Bernoulli Eqn. between 1&2
+
=0
1=1/
2
= 5.09 ft./s
2=1/
2
= 11.5 ft./s
=
= -1.65 ft. H20
Manometer reading = -1.457 “ Hg
h[1-1/13.6]= 1.457
h= 1.63 inches
6.28
For a Control volume between 1&2(exit) in mixture region
+
+ g(y2-y1)=0
P1-PAtm=
g y1
=0
(1)
Mass Balance around mixing chamber
Air+ w= m
As given: = /2
m=2
w+2
(2)
Control volume between H20 surface & 1 (H20 only)
+
P1-PAtm=
+ g y2=0
g y2-
(3)
Equating (1) & (3):
g y1=
g y2-
)
= g( y2- y1/2)
w=[2g(0.45m)]
1/2
Substituting Expression for
m=2[g(0.9m)]
Since
>>1
1/2
+
w into (2)
2
2nd term is small
and
1/2
m=2[9.81(0.9)] = 5.94 m/s
6.29 For conditions of Prob. 6.28
Control volume around mixing chbr. (word?)
=
dA
PA= A m[ m m]
~
This neglects momentum of air
~
From Prob. 6.28
Above mixerP=PAtm+
g y1
2
Below mixerP= g(y2-y1)-
P=PAtm+
g y2-
2
2
Equating with Momentum Expression
2
P=
(1- ) = [g(y2-y1)- 2]
2
4
= g(y2-y1)-
For
P=
=2
2
2
= 4.6 m/s
(1-1/2)= ½ (4.6)2(1000/2)
5.3 kPa
6.30
In both gases- Bernoulli Eqn. is:
2
2
2
1
2
+ 2
1
=0
2
2
152
(a) P= 2g=2 9 1 = 11.47 m Air = 1.39 cm H20
2
242 15
(b) P= 2 9 1 = 17.9 m Air = 2.52 cm Oil
6.31
Between liquid surface & exit:
Bernoulli Eqn:
2
=g y
=(2 g y)1/2 = 2(32.2)(10)= 25.35 ft./s
2
1
=4 12 2(25.35)= 1.66 ft.3/s
Between point B & exit:
+ g(yB-yExit)=0
PB=PAtm- g y = 14.7 PsiBy continuity:
Tank= ATank(
1 2
2[10
) =Apipe 2g
2g
dy=
2y1/2 17 =
1/2
2 2 144
2g y1/2
=
7
1
2 4 2 2 14
2g t
1/2
– 7 ]=
1 2
12
2
1
t= 1854 s= 0.515 h
2 22 t
= 8.63 Psi
6.32
Energy Eqn. for this case:
2
2
2
1
2
+ g(y2-y1)+ u2-u1=0
1=0
y2-y1= -10 ft.
2
u2-u1= 3.2 g
2
2
2
+ (-10)+ 3.2 g2 =0
2g
2 2
g
= 10
= 10.03 ft./s
1
=4 12 2(10.03)= 0.0547 ft.3/s
6.33
Between 1 (top of tank) & 2 (at the exit of the nozzle):
2
2
+ g(y2-y1)=0
1/2
2=[2(32.2)(20)] = 35.9 ft./s
2
2
= A =4 12 2(35.9)= 0.783 ft.3/s
I 4” l
= 42 = 8.975 ft./s
2
= 80.55 ft.2/s2
:
Between 1 & A:
1
2
+
2
1
2
+ g(yA-y1)=0
24
55
PA=PAtm+ (- 2 /2)+ g(y1-yA)= PAtm+ 2 2 ( 2 )+ 62.4(23)
= PAtm+ 1356 LB/ft.2= 3475 Psf (24.12 Psi)
A= 8.975 ft./s
Between A & B:
2
+
2
2
+ g(yA-yB)=0
2
2
=0
2
PB=PA+ g(-3)= 3290 Psf (22.83 Psi)
B= 8.975 ft./s
Conditions at D & B are equal
PD= 3290 Psf
D= 8.975 ft./s
Between B & C:
2
2
+
2
2
2
+ g(yB-yC)=0
PC=PB+ g(yB-yC)
= PB+ 62.4(-20)
= 2042 Psf (14.18 Psi)
C= 8.975 ft./s
2
=0
6.34 Between water level (1) & exit (2)
2
2
2
1
2
2=
=4
2
1
+ g(y2-y1)=0
2g
2
[2g ]1/2
H20 in tank:
=4 2
)2 2g y1/2
=
1 2
dy=
2y1/2 24 =
)2 2g t
1
t=
)2 2g
1
2 2 2 42
2 2
12
2
15
22
= 6644 s= 1.846 hours
1 2
6.35
4
1
2
From 1 to 2
1
2
+ 2 2 1 + g(y1-y2)=0
2
+ 22 =0
1
1
2
2
1
2
(1)
From 3 to 4:
2
4
2
4
2
+
2
4
2
2
+ g(y3-y4)=0
4
= 2 +
2
– gL
(2)
2
Note that P4+ 1 gL=P1, giving
2
4
2
1g
1
= 2 +
2
(3)
2
From 2 to 3:
2
1
2
2
+ 22
For 2 &
P2=P3
2
=0
negligible
& from (2)
2
2
= -gL+ 1 gL= gL( 1 - 1)
2
2
3
6.36 From Prob. 6.28:
1
2
2
+ 22 =0
1
= 2 + gL 2
1
2
2
1
2
2
- 2
Cons. of Mass:
= 2 3= 2 /R
1 2
~
H
=R
~
2
2=
2
2
2 2
)(
2
1
= 2
Giving P2-P1=
1
2 2
2
1
2
)
2
2
2
P3-P1= 2 2 + gL( 2 - 1 - 2 2
2
From momentum theorem=
(
)dA
(P2-P3)A= 1 2A( 32
P2-P3= 2 2 1
2
2)
)
1
Bernoulli Eqn:
2
2
1
2
1
) + 21
2
2 2
)
1
Doing the Algebra:
1
2
2g
2=
1
1
1
2
2
1
2
2
2
+ 2 + gL(1
1
2
2
)-
2 2
=0
6.37
Frictionless flow:
From Bernoulli
2
2
1
2
+ 22g 1 + (y2-y1)=0
g
2
2 = 2g(y1-y2)
=[2(9.81)(10)]1/2= 14 m/s
=(1000)(4 (0.04)2(14)= 17.6 kg/s
with nozzle- = 14 m/s {still}
=(1000)(4 (0.01)2(14)= 1.10 kg/s
with u2-u1=3
2
Energy Eqn. reduces to
2
2
2
+ g = 2(y1-y2)
2g
1/2
2=[ 4/7(9.81)10] = 7.49 m/s
Pipe: = 9.42 kg/s
Nozzle: = 0.589 kg/s
6.38 Same tank as in Prob. 6.37 but 2 exit pipesPipe 1: D= 0.04 m
y= 10 m
Pipe 2: D= 0.04 m
y= 20 m
Frictionless Flow:
Pipe 1As in Prob. 6.37
= 2g
= 14 m/s
= 17.6 kg/s
Pipe 2: Also = 2g
=[2(9.81)(20)]1/2 = 19.81 kg/s
=(1000)(4 (0.04)2(19.81) = 24.9 kg/s
6.39
,l ’
First, since the shaft work and heat must be
2
l
5 1
2 2 1
l
77 17
h
l
2 174
l
h
h
h
h
u
1 1 4 5 4
2
:
2
l
Assuming no viscous work and steady state,
Writing in terms of energy efflux,
2
1
Expanding the energy term, and using
2
N
,l ’
2
1
l ul
2
2
1
2
1
2
1
g
2
h
2 2
,
h
u ll
2
h
1
l
5
2
55 7 9
2
2
.
l
2
21 74
l
l
2
5 2
4
4
2
2
2
2
4 12 2
1
2
2 174
2
2
2
l
77 17
2
2
1
h
1
1 1
1
2
2
u
1
2
229 1
2
1 1
2
2
2
g
2
2
22
1
2
44
u
Plug these values back into
2
l
1 1 4 5 4
5 1
2
Note that the same value of
left out since it cancels.
4 11 4 5
2
5
1
l
5
2
l
2
7 9
1
7 9
2
2
2
2
1
1
2 174
215
5
2
l
2
l
l
l
2
55 7 9
2
21 74
2
2
2
5
1
l
2
44
2
was used in both places in the above equation, it could have been
l
1
1
2
2
l
5
1,
l
2
2
7255 9
2
2
5
1
l
6.40
u
2
2
2
2
1
2
2
2
1
12
4
2
2
12
4
2
1
2
1
2
5
1
2
1
1
l
77 17
1 19
91 7
Plug all known values unto equation 1
l
g
2
2
1
1
l
2 174
l
2
29 1
5
2
2
2
71 1
l
2
5
l
22
2
2
29 1 5
2
2
1
2
2
47
2
5
l
l 2
2 174 l
7
1 19
2
2
47
2
91 7
4
l
2
2
1
Chapter 7 End of Chapter Problem Solutions
7.1
(a)
>>
(b)
>>
7.2
For 2-D Flow-in x,y
=0
=0
=0
zx= xz=0
=0
zy= yz=0
7.3
I
II
y
x
I= (x) t
II= (x+ x) t
Axial Strain Rate:
= lim
->0
->0
=
Rate of Volume Change:
=lim
->0
->0
=lim
->0
->0
=
In 3 Dimensions
Both Axial Strain Rate and Volume Change are given by
+
+
7.4
In r-z plane:
= - lim
->0
=- lim [
->0
= lim [
->0
->0
+
rz= zr=
[
+
]
]
In the -z plane
Same procedure
lim [
->0
->0
=
+
= [
+
]
]
& in r- plane
lim [
->0
->0
=
+
= [
+r
]
( )]
7.5
Nitrogen 175 K
= 2.6693x106
T= 175 K
M=28
=91.5
=1.91
11.55x10-6 Pa.s.
=3.681
=1.1942
7.6
Oxygen @ 350 K
Eqn. 7.10
=
=
Yielding
M=32
= 2.6693x106
= 3.097
=1.03
=3.433
= 2.327x10-5 Pa.s.
Table value: = 2.318 Pa.s
7.7
For H20
= 0.76x10-3 LBm/s*ft.
= 0.375x10-3 LBm/s*ft.
Percent change =
= 0.51 or 51%
7.8
Properties of Helium, Glycerin from Appendix
T,F
60
0.00125
80
0.00132
100
0.00141
Plot the data and obtain an intersection very close to 100
,Glycerin
0.0177
0.00762
0.00128
7.9
For H20 ~1/
@ 120 F
w= 0.391x10
@ 32 F
w= 1.2x10
=
-3
3
lbm/s*ft.
lbm/s*ft.
= 3.07
Percent change =
= 3.07-1 = 2.07 or 207 %
7.10
For Air:
@ 140 F
@ 32 F
= 1.34x10-5 LBm/s*ft.
= 1.15x10-5 LBm/s*ft.
For ~1/
=
= 0.852
Percent change =
= 0.852-1= -0.148 = -14.8 %
7.11
oil
Di= 3.175 cm
Do= 3.183 cm
= 0.1 Pa.s
1st Law:
=0
{no flow in or out}
=0
=
Viscous =
i=
=
-at moving boundary
{t= gap width}
( DL)(r ) =
=1700( )= 178 Rad/s
=
= 5.58 W
7.12 Refer to Prob. 7.13
For
=
2=2 1
=4
Percent increase =
= 4-1=3 = 300 %
7.13
Ship 1:
Ship 2:
1= 4 m/s
2= 3.1 m/s
Choose control volume attached to Ship 1
=
dA =
Relative to moving ship
=0
Fx= +
= -0.9 m/s
= 100 kg/s(0.9 m/s) = 90 N
This is force applied to maintain stated conditions
Force exerted by fluid transfer= -90 N
7.14
t= Gap=
F=
=
=
= 0.01 cm
=
=
{Assumes linear profile}
=
F=
=
=
= 1676 N
7.15
Refer to conditions of Prob. 7.14
Load on Ram= 680 kg, L= 244 m
F= mg=
v=
=
= 0.768 m/s
7.16
M=
dF=
dA is on the conical surface = 2 rdL
{dL is along slanted surface}
dL= dr/sin
so:
dF=
=
= rdF =
M=
2 r
7.17
=
[1-(
]=2
[1-(
]
=
=2
[
]
At r=R
= -4
=
~
w= 0.76x10
-3
lbm/s*ft. @ 60
~
=
= -0.0453 lbf/ft.2
7.18 For conditions of Prob. 7.17
=
F= A= DL =
P=
( DL) = (-0.0453) )(0.1/2)(1) = 0.00119 LBf
= 0.00119 lbf ( (0.1/12)2/4) = 21.75 Psf
7.19
Shear work rate=
=
For parabolic profile= max[1-(
2
max [1-(
=
( )=
For
2
max [
( )=0
=
=
]
][
]=
]
2
max [
]
7.20
MOVING AT 3 ft/sec
0.03 inches
FIXED
(a) What is shear stress exerted on the fluid under these conditions?
Solving for shear stress,
(b) What is the force of the upper plate on the fluid?
Chapter 8 End of Chapter Problem Solutions
8.1 Hagen- Poiseulle Eqn.
=
For D=Do
For D1=2Do
=
=
o=[(
)
] Do4
4
1=[ ](2Do)
1=16 o
Percent Change =
=
-1= 15 = 1500 %
8.2
For single pipe: Po=[
]
o(40)
For single- parallel combination
P1= [ ] (22){ single branch}
P2= [ ] (18){parallel branch}
P1+ P2= Po= 3.45x106 Pa
=2
Case 2:
P1+ P2= [ ](22+9)
=
(4000)
o=
= 5161 BBL/Day
8.3
For 1- Reservoir
2- Pipe Entrance
3- Pipe Exit
Between 1 & 2:
=
+
+ g(y1-y2)=0
=(y1-y2)=0
+
Between 2 & 3:
+
+ g(y2-y3)-
=
=0
=(y2-y3)= 0
+
For Inviscid Flow =0
For Laminar Viscous Flow:
=
Inviscid Case:
=
=
~
Assuming fluid is hydraulic fluid @ 60 F – 15.9 K
=849 kg/m3
=0.0165 Pa.s
~
=[
]1/2= 22.08 m/s
=A = (0.006352)(22.08)
7x10-4 m3/s
Viscous Case:
=
2
=
+
+
-2
+
=0
=30.85 m2/s2
=
2
= 487.6 m2/s2
2
+30.85 -487.6=0
=
= 11.51 m/s
= (0.00635)2(11.51) = 3.645x10-4 m/s
=
= 1.92
8.4
From 1 to 2 (Bernoulli)
+
+ g(y2-y1)=0
=
with
=0
=0
with
+ g(y1-y2)
From 2 to 3
+
+ g(y2-y3) +
~
=
=
~
Combining expressions:
=g(y1-y3)~
=
~
=g y
= 1.0605x10-5
= 0.000987x10-5
=(1.0605-0.001)x10-5 = 1.0595x10-5 m2/s
=0
8.5
Using the same development as in Section 8.1:
(r )=r
For an element of length, L
(r )
=r
which becomes
(r )=r
Integrating:
r=
+ c1
=
+ c1/r
For laminar flow, Newtonian
=
so
=
=
+ c1/r
rdr+
Integrating:
= r2+ lnr+c2
Boundary conditions:
(r=R)=0
(r=kR)=0
k<1
Considerable algebra yields
c1=
c2=
R2[1-(1-k2)
]
& with substitution & simplification:
=
[1-
-
ln ]
8.6 This is same configuration as shown in Prob. 8.5
(r )-
r=0
Integrating:
-
=
& for laminar flow, Newtonian fluid:
=
-
=
Integrating:
x-
r2= lnr+c2
Boundary conditions:
(r=D/2)=0
(r=d/2)=v
More algebra:
c1=
[v+
c2=
(D2-d2)]
- ln
Drag force per unit length:
F= A= d)(1) =
( d)
Giving:
F= d [ +
For the case with
F=
= d [
=0
+
]
8.7
In -direction:
= -r
z
+r
z
+
z
component of force on(+ )face
Divide by r
(r
r
)+
z & take limit as
=0
+ 2 =0
+ 2 =0
ln +2lnr=ln(constant)
r2 = constant
= r ( )
r2 = r3 ( )= constant(c)
d( )=
= c1(
=
)+c2
+ rc2
Boundary conditions:
(R)=0
0= + Rc2
(kR)=V
Algebra
=
(
V=
- )
If profile is linear: (word?)
= ar+b
= ( – 1)
Percent error=
= Actual- Linear
+ kRc2
0:
=
(
- )-
=
(
(r– R)
+ )-
=
Vmax occurs at =
=1-
=0.01
Resulting in
=0.99
k= 0.96
[
(
+ )- ] =0
8.8 For flow between 2 horizontal platesGoverning D.E.( yx)- =0
Laminar, Steady Newtonian
yx=
B.C.- At interface (y=0)
(1)
=
(2) yxA= yxB
(3) (-hB)=
hA)=0
8.9
Fully developed, steady, laminar flow; Newtonian fluidyx -
=0
yx=
B.C.
=0 for y= h
=
=
=
y+c1
y2+c1y+c2
Applying Boundary conditions
c1=0
c2=
Giving:
=
(y2-h2)
8.10
Governing D.E. is
yx -
=0
Integrating: yx -
y=c1
Laminar flow, Newtonian fluid:
yx=
-
y=c1
For yx(0)=0 c1=0
-
y=0
-
= c2
@y=h=
o
c2=
o-
Also
@y=0=0
Giving
=
c2=0
8.11
For horizontal pipe flow:
Development in Section 8.1 results in
=(
For =0
+ c2
must=0 for all r
=constant=v
8.12
For an element in liquid film:
y
xy
- xy y - g x y=0
– =0
In limit as x->0
=0
xy xy=
-
=0
- x= c1
x2=c1x+c2
-
Boundary conditions:
(0)= o
(h)=0
c1= - h
c2= o
Giving:
= o - (2hx-x2)
=
o-
[2 -(
]
=
=
[2 -(
ohoh-
(h3-h3/3)
]dy
8.13
Treat fluid layer as a thin linear layer:
In the usual way:
Treat
=0
constant~
=
& =
Giving =
+ c1y+c2
Boundary Conditions:
(0)=R
(h)=0
Giving
R =c2
0=
+ c1h+ R
c1=
= R (1-y/h)
Flow rate=
[( )-(
=
Giving:
)dy =
=
[
Efficiency=
=
~
evaluated at R(y=0)
~
=
+
After doing the algebra:
=
]
8.14
Fluid enters at x=0 & flows equally in +x & -x directions exiting at x=L/2 where P=Patm
Working with the R.H. flow(in +x direction)
The applicable D.E. is
=
& as usual
Giving
=
Integrating: =
y+c1
Boundary conditions: (0)=0
y2+ c2
Again =
Boundary conditions: ( =0
So c2=
)
Velocity expression is:
=
)(
-y2)
=
=
)
=
)
dy
So the expression for
is:
=
&
=
Po-PAtm=
& for the plate of total length,L,
Fy=( Po-PAtm)2L=
c1=0
8.15
Liquid flowing Down the outside of a cylinder:
Governing D.E. is
(r )+ =0
&, as usual
(r )+
r +
=0
= c1
B.C. =0 @ r=R+h
r = [(R+h)2-r2]
And again:
= [(R+h)2 lnr- ]+c2
B.C. (R)=0
Giving
= [(R+h)2 ln +
(1- )
8.16
For result of Prob. 8.15
max=
max occurs where
[2(1+ ) ln(1+ ) -
– ]
=0 which is at r=R+h
8.17
8.18
8.19
Calculate
,
Calculate viscosity,
Chapter 9 End of Chapter Problem Solutions
9.1
~
=
r
zr
=
r
r
zr
+
r
-
r
+
zr
+
-
zr
z
~
Substituting into C.V. relationship & evaluating in limit as r
(r r)+
r
=0
z go to 0
r
9.2
= x +
=
+
* =
x
Note
*
* =
y
+
z
+
(
)+
*
(
y
=1 For j=i
=0 For j i
x
+
y
+
z
*
)+
z
(
*
)
9.3
4
3
4’
3’
1
2
1’
2’
For 2-Dimensional flow:
Volume change= (
)(
)-(
y
x
= x
= x+[
)(
)
= y
x(x+
x,y)-
3 2 = y[ y(x+ x,y+
( )( )= x y
x(x,y)]
)-
y(x+
x,y)]
(
)(
)= x y+[ y(x+ x,y+
[ y(x+ x,y+ )- y(x+ x,y)] 2
x,y)]
+[
Dividing by x y
& evaluating in limit as x y
->0
Volume change =
+
= *
But, from continuity * =0
)-
y(x+
x(x+
x,y)-
x(x,y)]
+
9.4
=
r
+
d =
dr+
d + dt
=
+
+
=
+
+
=
+
+
+
r
r
+
= cos + sin
=- sin
=
=
=0
= - sin
=
In similar fashion
=0
=-
Giving:
=
+
=(
- )
+(
For
to be
then
= +( r
+ r)
=
&
= =
+
- )
r
+
r
+
+
)
9.5 Navier-Stokes Eqn.- Incompressible form:
= -
+
(a) For small- all terms involving (
&
) are small relative to other terms.
(b) For small & large the product cannot be considered small relative to other
terms
9.6
y
x
2L
Incompressible N.S. Eqn. in x-direction
+
With
=
+
x
=gx-
y
=
=
=
=
+
y=0
y+ c1
=
+ c1y+c2
B.C.
c1=0
=0 @ y=
c2=
=
(y2-L2)
L2
9.7
=
=
(r )+
+
~
* =
(
)=
( =0
And continuity is satisfied
9.8
=
+
=-
At y=100,000 ft.
=
=
=
=20,000 ft./s
= 0.0096
s-1
9.9
P= ( - )
=400[(y/L)2 +(x/L)2 ]
=
+
+
= 400(y/L)2800
Evaluated at (L,2L) we get
P=
- [2g+
]
+400(x/L)2800
=
[x2y +xy2 ]
9.10
In x-direction:
= gx= gx-
-
+
)+
(
[
(
(
(
Similarly in y & zA total of 45 terms!
)+ (
+
)+
+
(
)+ (
) [
+
(
)+
+
+
)]
)+
(
)+
(
)
]
9.11
For = o+ r
o- of coordinate origin
r- relative to coordinate origin
=
+
=
N.S. Eqn. reduces to
=
- P and P= ( - )
9.12 Given that
(r )+
a) For
=0
=0
(r )=0
and r ( )=F(
or = F( /r
b) For =0
=f(r)
=0
9.13 N.S. for Incompressible, laminar flow
= -
+
For negligible
a) Vector properties~ and
are independent by themselves but in same
relationship must lie in same plane.
b) If viscous force are negligible
=
is determined by
& is positive in direction of decreasing pressure.
c) In similar fashion, any fluid- either moving or static-will move or tend to move in
direction of decreasing P.
9.14
For 1-D Steady flow:
= (x)
= =0
(
=-
+ [
=-
+3
)=0
2
)+
3
)
]
9.15
Continuity:
+ (
Momentum: (
+
)=0
)=-
9.16
Taking z as positive down with
=
=0 &
=f(r)
Eqn. E.6. yields
z direction
2
(
+
+
+
=
+ gz+ [
(r )+ 2
2
+
2
2
]
Realizing that:
2
=
=
= = =
2
2
=
2
=0
and since gz=-g
= (r )
Proceed as was done in solutions to problems 8.17 & 8.18
9.17
For incompressible, steady flow, with
r direction
= =0 Eqn. (E-4) has the form
2
2
(
+
+
+
=
+ gr+ [
Realizing that the following terms are zero:
2
=
= = =2
2
2
2
2
The initial equation becomes
2
(P+
2
)= gr
2
=
2
=0
)+ 2
2
2
- 2
2
+
2
]
9.18
Governing Eqns. Are
r direction
2
2
(
+
+
+
=
+ gr+ [
)+ 2
2
2 - 2
2
+
2
]
direction
2
(
+
+
+
=
+
+ [
)+ 2
2
2 - 2
2
+
2
z direction
2
(
+
When
+
+
=
=f(r) and with
2
= =
Q.E.D.
(r )+ 2
2
2 +
2
]
= =0 the only non-zero term on the left-hand side of all
2
component eqns. is
+ gz+ [
]
9.19
Eqn. (E-5) is simplified for this case as
direction:
(
+
2
+
+
+
=
Realizing that the following terms are zero:
2
=
=
=
=2
2
=
2=
2
=0
& in the absence of gravity we have
=
[
(r )]
+ [
)+ 2
2
2 - 2
2
+
2
]
9.20
From Prob. 9.19 & steady flow
=0
Giving
(r
=c1
Integrating again:
r =c1lnr+c2
B.C.
(R1)=R1
(R2)=R2
= [R12
+
2
2
2 2
2
2
]
9.21
Fluid flow around center section
RO
RI
r
V0
Need to begin with z-direction Navier-Stokes equation in cylindrical coordinates:
2
2
Applying the assumptions given in the problem statement:
2
2
2
2
Boundary Conditions:
1.
2.
Apply Boundary Condition #1:
2
2
2
2
2
Apply Boundary Condition #2:
2
2
Solve for
,
2
2
2
2
Solve for 2 by plugging back into equation for Boundary Condition #1 (if you plug into
the Boundary Condition #2 equation your result will be slightly different and no less
correct),
2
2
2
2
2
2
2
2
Thus,
2
2
2
2
2
2
9.22
VB
Fluid
1
Fluid
2
L1
L2
VA
(I will show two different solutions, the second solution may be slightly easier. The only
difference is the identification of the boundary conditions that are in red in the figures
below.)
Need y-direction Navier-Stokes Equation
2
2
2
2
2
2
Using the conditions in the problem statement this equation becomes,
2
2
Integrate twice solving for velocity,
2
2
2
2
2
Next, we develop equations for the velocity of Fluid 1 and Fluid 2 using the above
equation.
Fluid 1
VB
VA
Fluid
1
Fluid
2
L1
L2
x=0
x=L1
x=L1+L2
2
2
2
Boundary Conditions:
BC#1:
BC#2:
2
Apply BC#1:
2
2
Apply BC#2:
2
2
Solve for
,
2
2
Thus, plugging in the expression
,
2
2
2
2
Fluid 2
2
2
Boundary Conditions:
BC#2: 2
BC#3: 2
Apply BC#2:
2
2
(this boundary condition applies to both fluids)
2
2
2
2
Apply BC#3:
2
2
2
2
Set
2,
2
2
2
2
,
Solve for
2
2
2
2
2
2
2
Now solve for
2
2
2
2
2
2
2
Insert the equation for
,
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Thus,
2
2
2
2
And, plugging
2
2
2
2
and
2 back into the original equation for
2
2
2
2
2
2
2
2
2
2,
2
2
2
2
2
2
2
2
2
2 2
2
This equation could be simplified, but that is not necessary.
2
2
2
Solution 2
(this may be a slightly easier solution, but both are correct).
Need y-direction Navier-Stokes Equation
2
2
2
Using the conditions in the problem statement this equation becomes,
2
2
Integrate twice solving for velocity,
2
2
2
2
2
2
2
2
Next, we develop equations for the velocity of Fluid 1 and Fluid 2 using the above
equation.
Fluid 1
VB
Fluid
1
Fluid
2
L1
L2
VA
x= -L1
x=0
x= +L2
2
2
2
Boundary Conditions:
BC#1:
BC#2:
2
2
2
2
Apply BC#2:
2
2
2
2
Apply BC#1 (and insert the result of BC#2):
2
2
2
2
2
2
Thus, plugging in the expression for
2
,
2
2
Fluid 2
2
2
2
2
BC#2:
BC#3:
(this boundary condition applies to both fluids)
2
2
2
2
2
2
2
Apply BC#2:
2
2
Apply BC#3, inserting the value for
2
2
2:
2
2
2
Plugging the values of
and
2
2
2
2
2
2
back
into
the
original equation:
2
2
2
2
2
9.23
Thin viscous fluid
with thickness = h
Belt moving
up with a
velocity=Vw
Air
Begin with the y-direction equation in rectangular coordinates:
2
2
2
2
2
2
2
Thus,
2
2
Boundary Conditions:
BC#1: at the moving wall:
BC#2: at the free surface:
Rearrange,
2
2
Integrate,
at x = 0
at x = h
2
2
2
Integrate again,
2
2
From BC#2:
From BC#1:
2
Thus,
2
2
2
9.24
Vz
z-direction Navier-Stokes Equation in Cylindrical Coordinates
2
2
To obtain the desired equation we must make the following assumptions:
2
2
2
2
Rearrange,
Integrate,
2
2
Rearrange,
2
Substitute in Newton's Law of Viscosity
Boundary condition:
2
, so
2
,
2
2
2
2
Chapter 10 End of Chapter Problem Solutions
10.1
Since
and
For reference, see problem 9.4
All remaining (non-zero) terms give:
,
10.2
In the limit: Note that tan(z)→z
10.3
10.4
10.5
since
, continuity can be expressed as
using
Check
or
10.6
In the core: Euler’s Eqn.
Since velocity variation is linear
Outside the central core – Bernoulli Eqn. applies:
v varies inversely with r
Adding (1) and (2)
For P = -10 psf
Pressure will fall from -10 to -38 psf in a distance of 138 ft
At 60 mph = 88 ft/s, Time = 138/88 = 1.57 seconds
Pressure at tornado center = -38 psf
At edge of core
Far from center
Total ΔP = 38 psf
(c)
(b)
10.7
Along the stagnation streamline, θ=π
10.8
From continuity:
10.9
10.10
(a)
Flow configuration is:
When
(b)
(c)
When
10.11
In figure – for ψ=0, or any number – pick 3 –
Choose θ – solve for r – Plot looks like:
10.12
10.13
For source at Origin,
Adding:
, where m=Source Strength. Free stream:
10.14
For Steady, Irrotational Flow,
@ Stagnation point, where v=0,
10.15
Lift Force:
From Bernoulli Equation:
On Hut:
Substitute into the expression for Fy,
10.16
10.17
10.18
For this case -
10.19
when origin is at vortex
10.20
a) Stagnation Point
& at Stagnation Point,
b) Body Height
Stagnation Streamline
So when
c) For X Large – All flow is at
d) Maximum Surface Velocity
10.21
In this case,
In upper half plane they appear as:
Streamlines can be plotted for
10.22
10.23
To be irrotational, the flow must satisfy the equation,
must be equal to zero.
For the equation of the stream function we solve for
function,
And then,
So the condition of irrotational flow is satisfied.
and as a result, Equation 10.1
and
using the definition of the stream
10.24
We defined the stream function as
and
Thus,
We will choose to begin by integrating equation (5) partially with respect to x,
Where
is an arbitrary function of y.
Next, we take the other part of the definition of the stream function, equation (2), and
differentiate equation (6) with respect to y,
Here,
since f is a function of the variable y.
The result is that we now have two equations for
these and solve for
.
Solving for
, equations (4) and (7). We can now equate
,
So
The integration constant C is added to the above equation since f is a function of y only. The
final equation for the stream function is,
The constant C is generally dropped from the equation because the value of a constant in this
equation is of no significance. The final equation for the stream function is,
Which agrees with the result of Example 3.
Chapter 11 End of Chapter Problem Solutions
11.1
Variable
D
ρ
H
g
ω
Q
η
P
Dimensions
L
M/L3
L
L/t2
1/t
L3/t
ML2/t3
11.2
Variable
v
ρ
D
μ
e
Dimensions
L/t
M/L3
L
M/Lt
L
11.3
Variable
ΔP
ρ
ω
D
Q
μ
Dimensions
M/Lt2
M/L3
1/t
L
L3/t
M/Lt
11.4
11.5
Variable
k
D
d
ω
ρ
μ
Dimensions
L/t
L2/t
L
1/t
M/L3
M/Lt
Plot π1 vs. π3 over a range in values of π2
11.5
11.6
Variable
Q
d
σ
ω
ρ
μ
Dimensions
L3/t
L
M/t2
1/t
M/L3
M/Lt
11.6
11.7
Variable
M
d
σ
g
ρ
Dimensions
M
L
M/t2
L/t2
M/L3
11.7
11.8
Variable
n
d
L
T
ρ
Dimensions
1/t
L
L
ML/t2
M/L3
11.8
11.9
Variable
Q
d
P
ω
ρ
μ
Dimensions
L3/t
L
ML2/t3
1/t
M/L3
M/Lt
11.9
11.10
Variable
r
t
E
ρ
Dimensions
L
t
ML2/t2
M/L3
11.10
11.11
Variable
D
d
V
σ
ρ
μ
Dimensions
L
L
L/t
M/t3
M/L3
M/Lt
11.11
11.13
11.14
11.12
We could do all other terms in a like manner, but problem statement only asks for a ratio of
gravity forces to inertial forces. The Gravitational (Buoyancy) Force is
asked for is
Q.E.D.
so the ratio
11.13
11.15
Variable Dimensions Prototype
D
D
6D
v
v
20 kT
ρ
ρ
ρ
μ
μ
μ
F
10 LBf
Fp
2
A
D
(6D)2
Dynamic symmetry requires:
11.14
11.16
Similarity Requires:
Variable Model
v
vm
L
L/10
Prototype
vp
L
11.15
11.17
Model
L=3m
vm
ρA
νA
Fm
Am
Prototype
4L
16 m/s
ρw
νw
Fp
16Am
11.16
11.18
11.17
11.19
Variable
L
a
v
τ
g
Dimensions
L
L
L/t
t
L/t2
11.18
11.20
For Equal Reynolds Numbers:
For Ideal Gas Behavior
11.19
11.21
Variable Model Prototype
L
0.41
2.45
v
2.58
v
Equating Froude numbers:
Equating
:
Thrust Force involved Euler No
Torque = FL – From Euler Equation
11.20
11.22
Variable
Flow rate
Density
Viscosity
Velocity
Surface Tension
Channel Length
Q
3
0
-1
M
L
t
1
-3
0
Symbol Dimensions
Q
L3/t
M/L3
M/Lt
v
L/t
M/t2
L
L
1
-1
-1
v
0
1
-1
1
0
-2
The rank is found to be 3 (rank of a matrix is the number of rows (columns) in the
largest nonzero determinant that can be formed from it). Using Equation 11-4, the
number of dimensionless parameters is
Thus there are 3 dimensionless parameters to be found.
Find
:
Evaluate for M, L and t
M:
L:
t:
Thus,
and
L
0
1
0
Find
:
Evaluate for M, L and t
M:
L:
t:
Thus,
Find
and
:
Evaluate for M, L and t
M:
L:
t:
Thus,
and
Chapter 12 End of Chapter Problem Solutions
12.1
12.2
a)
12.3
12.4
From (1),
:
12.5
Comparison Approximate Exact
12.6
12.7
12.8
v, ft/s
CD
50
100
150
200
250
300
350
400
0.47
0.47
0.46
0.45
0.41
0.35
0.20
0.08
Re
(x104)
4.07
8.14
12.2
16.3
20.3
24.4
28.5
32.5
FD,
LBf
0.0199
0.0797
0.175
0.305
0.434
0.534
0.415
0.217
Plot of Drag vs. Velocity
0.6
Drag (lbf)
0.5
0.4
0.3
0.2
0.1
0
50
100
150
200
250
300
Velocity (feet/sec)
350
400
12.9
12.10
12.11
12.12
12.13
12.14
12.15
12.16
12.17
12.18
Re (x104)
CD
7.5 10
0.48 0.38
15
0.22
Re
(x104)
7.5
10
15
20
25
v
CD
92.2
122.9
184.4
245.8
307.3
0.48
0.48
0.47
0.44
0.10
FD,
LBf
0.069
0.100
0.129
0.125
0.164
20
0.12
25
0.10
0.5
0.45
0.4
Drag (lbf)
0.35
0.3
0.25
Drag for Sphere
0.2
Drag for Golf Ball
0.15
0.1
0.05
0
0
100
200
300
Velocity (ft/s)
400
12.19
12.20
12.21
12.22
12.23
x,m
0
0.1
0.5
1
2
Rex
δL,cm
0
0
5
2x10 0.111
0.249
0.352
0.498
δt,cm
0
0.327
1.186
2.063
3.591
12.24
12.25
12.26
12.27
0
0.1
0.3
0.5
0.7
0.9
1
0
0.156
0.455
0.707
0.89
0.99
1.0
0
0.0244
0.207
0.5
0.795
0.98
1.0
0
0.0038
0.094
0.355
0.708
0.97
1.0
0
0.1
0.3
0.5
0.7
0.9
1
0
0.518
0.709
0.820
0.903
0.970
1.0
0
0.373
0.600
0.743
0.858
0.956
1.0
12.28
12.29
12.30
12.31
12.32
12.33
Chapter 13 End of Chapter Problem Solutions
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
(5.9 in)
13.9
13.10
13.11
Figure 13.1 for a smooth Tube:
13.12
Rectangular duct: 8”x8”x25 ft
Standard Air
Energy equation reduces to:
From Figure 13.1:
13.13
14 inch pipe is the cheapest.
13.14
With
13.15
13.16
13.17
13.18
13.19
13.20
(1) = surface of tank
(2) =pipe exit
13.20
13.21
13.21
13.22
13.22
13.23
13.23
13.24
,
,
,
13.24
13.25
Pipe
1
2
3
Length,
m
125
150
100
Diameter,
cm
8
6
4
Roughness,
mm
0.240
0.120
0.200
13.25
13.26
13.26
13.27
13.27
13.28
Pipe
Length,
m
Diameter,
cm
Roughness,
mm
1
100
8
0.240
2
150
6
0.120
3
80
4
0.200
Chapter 14 End of Chapter Problem Solutions
14.1
14.2
14.3
Equating
14.4
14.5
14.6
14.7
14.8
14.9
14.10
14.11
14.12
14.13
14.14
14.15
14.16
14.17
14.18
14.19
0.1
0.15
0.2
0.25
h
93.48
95.93
100.54
106.5
14.20
(1)
v
0.20
0.25
0.30
0.35
hsyst
7.27
7.71
8.24
8.87
14.21
14.22
Pump Performance:
Capacity, m3/s x
104
0
10
20
30
40
50
Developed Head,
m
36.6
35.9
34.1
31.2
27.5
23.3
Efficiency,
%
0
19.1
32.9
41.6
42.2
39.7
vx
104
20
30
40
50
h,m
20.61
22.63
25.45
29.07
14.23
14.24
Chapter 15 End of Chapter Problem Solutions
15.1
(
)
∫
∫
[
]
[
]
(
)
15.2
15.3
∫
∫
(
(
)
)
[
]
[
]
[
[
(
)]
](
) [
](
)
15.4
(
)( )(
)
15.5
(
(
)(
)
)
15.6
(
(
)(
)
)(
(
)(
(
)
)
)(
)
15.7
|
( )
(
)(
)(
)
15.8
(
)
(
[
(
) ]
)
15.9
(
)(
)
15.10
( )
(
)
(
)
(
)
(
)
( )
15.11
(a)
(b)
(c)
(d)
15.12
(a)
(b)
15.13
(
) [ (
)
(
(
)
) ]
( )
(
(
) (
)(
)
) ( )(
)
15.14
[
(
)
]
( ) [(
)
]
15.15
(
(
)
)(
)(
)
15.16
(
[
(
)
)
[
(
]
)]
15.17
[ (
(
)
)
(
)]
[(
)
]
15.18
(
)
[(
(
( )
)
(
(
)
]
) ( )
)
( )
( )
( )
(
) ( )
15.19
[(
)(
)
(
)(
(
)(
)(
)
)
(
)(
)]
15.20
(
)
(
(
)
)
(
(
)
)
(
)
15.21
(
)(
(
)(
(
(
)(
)(
)
)(
)
)(
)(
)(
)
)
15.22
15.23
( )
( )
( )
15.24
)
)
)
15.25
(
)( )
( )
(
)( )
15.26
15.27
∫
∫
∫
∫
( )
(
)
(
)
Chapter 16 End of Chapter Problem Solutions
16.1
(
)
( )
(
( )
)
( )
(
)
( )
(
)
16.2
(
)
( )
(
(
)
)
(b)
(
) (c)
16.3
( )
( )
(
)
(
(c) ∫
∫
)
( )
( )(
)
16.4
( )
⃑
( )
( )
[
( )
⃑
⃑
( )
( )
( )
⃑⃑⃑
( )
( )
( )]
( )
16.5
̇
̇
̇
̇
̇
( )
̇
( )
(
)
̇
[
(
)]
16.6
̇
( )
[
]
̇
16.7
̇
( )
[
]
̇
16.8
⃑
⃑
̇
̇
⃑
16.9
( )
[
[
( ) ]
]
16.10
(
)
16.11
[
(
(
) ]
(
)
)
(
∫
)
∫
[
(√
)
)](
(
∫ (
∫
[
(√
)][
)(
)
(
)] (
)
)
16.12
̇
̇
(
)
̇
̇
̇
∫
(
̇
∫ (
̇
(
)
)
)
(
)
16.13
̇
̇
(
(
)
( ) )
̇
∫ (
(
̇
(
̇
∫[
]
)
∫
̇
)
( ) )
̇
∫ (
)
16.14
(
̇
)
̇
(
(
)
∫
̇
( ) )
̇
∫ (
)
Chapter 17 End of Chapter Problem Solutions
17.1 Steady State X-directional conduction through a plane wall
qx= -kA
= kA/L (T1-T2)
For T1-T2= 75 K
= 7500 w/m2
q=
dT/dx =
=
-250 K/m
For T1=300 K
dT/dx=
T=
q= -2000 W/m
=
=
= 66.7 K/m
= -20 K
T2= 320 K
For T2=350 K
dT/dx= -300 K/m
q= -(30)(-300)= 9000 W/m2
T= -300 K/m (0.3m)= 90 K
T1= 440 K
For T1=250 K
dT/dx= 200 K/m
q = -(30)(200 K/m)= -6000 W/m2
T= -200(0.3m)= -60 K
T2= 310 K
17.2
T=
2
T=
% Error=
=
x 100
=
x 100
=
x 100
For
=1.5
% error= 1.3 %
=3
=5
% error= 10.0%
% error= 20.7%
T (a)
17.3
q=
Am=
T
(a) =
=2
% Error=
= 1-(1/2)(
)
(b)
= 1.5
% error=8.3 %
=3
=5
% error= 66.6%
% error= 160 %
17.4
q”= -k
= -ko(1+bT)
From 0 to 12:
q”
= -ko
q”=
[T+ T2
= 23(T-300)
Solving: TRH wall= 307.1 K
q”= 163.3 W/m2
From 0 to L:
q”
L=
& Solving: L=0.646 m
17.5
Input Parameters:
Ro= 1.5 cm
(0.015m)
Rm= 1.8 cm
(0.018 m)
h= 80 W/m2 K
To= 70
= 20
L= 6.0 cm
(0.06 m)
q= 15 W
(a) Source: heating rod
Sink: surrounding air
(b) B.C. system= biomaterial r=Ro, T=To; r=Rm, T=Ts
(c) km
Assumptions

Constant Source & Sink
steady state

Sealed ends
1-D heat flux along “r”

Constant km, Ro, Rm, L, q
q= 2 RmLh(TsTs=
+
Ts= 20
+
)
(1
Ts= 47.6
For a hollow cylinder
km =
=
km= 0.324 W/m*K
q=
17.6
Input Parameters:
h= 50 W/m2 K
ks= 16 W/m-K (stainless steel)
L= 1.6 cm
Assumptions:

Steady state

1-D heat flux along x by conduction

convection of surface (x=L)

ks & h not a function of T
(a) q=?
Topic 17.1
=
(T2-T1)
= h(T1=
= T2-T1
at surface
=
= T1= 10,952
(b) T1= ?
T1=
+
=
+ 20
= 239
(c) Most of drop in temperature is across gas film
17.7
Q=
RGL=
Rair=
RConv=
= 0.06633 K/W
q= 37/0.06633= 585 W
Ti=27 -
22.94 C
17.8
Brick Size= 9”x 4.5”x 3”
Brick #1
k=0.44
Tmax=1500 F
Brick # 2
k=0.94
Tmax=2200 F
Most economical arrangement is to use as much of #1 as possible (low k). Use #2 next to high
temp such that its cooler surface has T
.
q”=
= 28.5 in
= 2.35 Ft.
= 28.2 in.
(9x2+4.5+2x3)
=
= 31.6 in.
Most economical:
17.9
Ri=1/115= 8.696x10-5 K/W
R1= 0.25/1.13= 0.221 K/W
R2= 0.20/1.45= 0.138 K/W
R3= 0.10/0.66= 0.152 K/W
Ro= 1/23 = 0.0435 K/W
q=
=
q= 23(To-300)
= 1900 W/m2 = 176.5 W/ft2
To= 383 K
17.10
To= 325 K
Ri=1/115= 8.696x10-5 K/W
R1= 0.25/1.13= 0.221 K/W
R2= L/1.45
R3= 0.10/0.166= 0.152 K/W
= 0.416 + L/1.45=
0.416+ L/1.45=
L= 2.03 m
q=23(325-300)
= 575 W/m2
17.11
LSt= 1/8”, k= 10 BTU/Hr ft. F
LASB= 1/8”, k= 0.15 BTU/Hr ft. F
q=
=
1351 W/ft2 (a)
2305= h
T= 768 F
=3
TSurf = 838 F (b)
17.12
R1 =
= 0.0694
R2 =
0.00104
R3 =
0.154
RConduction, Equiv =
=
= 0.0483
per side = 0.0483+ 1/3= 0.3817
New Ht. Flux =
Increase =
2437 BTU/Hr-ft2
= 0.057= 5.7 %
17.13
Ti= 325 K
h1= 12 W/m2 K
h2= “
k1= 2.42 W/ m K
k2= 229 W/m K
a) Applied to Plastic:
q = 12(T1- 295) +
(T1-325)
(325-T2) + 12(T2-295)
T2= 324.9 T1= 328.7
q = 359+404= 763 W
b) Applied to Al:
q = 12(T2- 295) +
(T2-325) +
(T2-325)
(325-T1)
T2= 325 K T1= 322 K
q = 320+361= 681 W
17.14
Bolts in a square array with 4 equiv. Bolts/ ft2
R1(bolts) =
=
= 5.7 Arf/ BTU – steel
= 0.475 Arf/ BTU - alum.
R2 SS=
= 0.0021
R3 CB=
= 10
R4 PL=
= 0.0278
= 10.03
Per ft2 of x-section
REquiv=
=
= 3.63 Hr F/BTU (a)
= 0.454 Hr F/BTU (b)
17.15
For a section 36 cm wide & 1 m deep:
R1 =
K/W
R2 =
= 0.0683 K/W
R3 =
= 16.67 K/W
R2 =
= 14.28 K/W
R5 =
K/W
R Stud Wall Equiv =
q=
= 4.28 W
q3=
q4=
=
q3=
= 7.691 K/W
= 32.92 K
= 1.975 W
q4= 2.305 W
17.16
q=
= 292.7- 70= 227.7 F
Room Temp (Assumed)
For 2-in sched 40
RST =
ID= 2.067 in
OD=2.375 in
= 1.475x10-5
RINS =
= 0.0529
ROuts = 1/hAo= 0.00537 w/o Insul
= 0.00237 w/o Insul
= 0.0553
= 0.00537
= 37470 BTU/hr
Cost = 37470(
= $0.255/hr
Time =
= 177 Hours
q= 4030 BTU/hr
q= 41,500 BTU/hr
17.17
R1 =
R2 =
R3 =
= 0.0015
= 6.19x10-5
=
= 0.0725
= 0.072 Hr F/BTU
q=
= 2530 BTU/Hr (a)
R1+R2= 0.00156 Hr F/BTU
R3 =
= 0.461 Hr F/BTU
R4 =
= 0.0221
= 0.485 q=
= 386 BTU/Hr (b)
For Bare Pipe:
=
= 2.71 LBm/Hr (c)
17.18
= 30 = 303 K
h= 100 W/m2-K
L1= 0.5 cm
k1= 1.0 W/m-K
L2= 0.3 cm
k2= 230 W/m-K
W= 4.0 cm
V= 10.0 W
Aside: not required for your calculations. Verify T2=T3
For System II
x=L2
x=L1
T2=c2
T(x) =T2=T3
c1=0
=0
T(x) =c1x+c2
(a) x=0
T1
q= AL1=Ah(T1-
=
1.25 W/cm3 * 0.5 cm = 100
T1= 365.5 K
)
System I
= 1.25 W/cm3
(T1-303 K)
(b) T2 For system I
Assumptions

Steady state

Const

1-D flux along x

X=L1,
(no heat flux)
From Topic 17.2 & assumptions given
T(x) = T1+
(L1x – )
x=L1
T=T2= T1 +
B.C. x=0, T=T1
T2= 365.5 K +
= 381.1 K
x=L1,
17.19
Input Parameters
kIC= 1.0 W/m-K
= 10.0 W
h= 100 W/m2 K
L= 0.5 cm
A= (4 cm) 2= 16cm2
Vchip= A*L1
Assumptions:

Steady state (constant source & sink)

Constant power

1-D heat flux aligned along Source and Sink
Vchip
=
= 1.25 x 106 W/m3
(1.25 W/cm3)
(a) T1 surface temp (x=0)
From Topic 17.2
A*L1= h A (T1 -
)
T1=
+
T1= 365.5 K
(92.5
)
(b) T2 at base of IC chip
From Topic 17.2 heat balance is
T(x) =
+ xc1 + c2
kIC
=
+ c1
Boundary conditions
x=0, T=T1
x=L,
T1=0+0+c2
0=
=0
+ c1
c2= T1
T(x)=
+x
+ T1
at x=L T=T2
T2=
+
+ T1=
+ T1
T2=
+ 365.5 K
T2= 381.1 K
Tav=
=
chip failure= 22%
(108
= 100
)
c1=
17.20
(Ti-To) =
I2R= 3.43 W
R=
= 2.95 x10-4
I2 =
I = 107.9 Amp
(120-70) = 11.72 BTU/Hr-Ft= 3.43 W
17.21
=
I = 106.1 Amp
11.34 = 4[
T = 71.3 F
17.22
Same as Problem 17.21 except wire is aluminumR= 4.85x10-4
I= 83.9 Amp
17.23
Assume thin-walled inner vessel is 77 K throughout
17.24
For q = ½ of value in 17.23
Insulation thickness= 0.0555m
Added thickness: 0.0055 m or 5.5 mm
17.25
L=0.06 m
= qo[1-x/L]
Poisson Eqn. applies: (Energy Balance)
=0
=
+ c1
17.26
In waste material:
For Stainless steel:
Equating with equation (1):
Putting in values
@ center of waste material:
17.27
17.28
Input Variables
Ro= 0.5 m
R1=0.6 m
q= 2 x 105 W/m3
h= 1000 W/m2-K
krw= 20 W/m-K
ksteel= 15 W/m-K
=25
(298 K)
(a) Model for T(r) System I
Assumptions: System I

Steady state

1-D heat flux along r (sealed ends)

Const. q

k is not a function of T
Energy Balance (System I) differential volume element (fix in Lect notes)
IN-OUT+GEN=ACC ACC=0
2
2
*
+ *
=0
=volume of vol. element
Divide by
k,
, rearrange, r -> 0
+
(r )=
Ordinary diff. equation (O.D.E.) is perfect differential
B.C.
r=Ro
r=0
Qr=0
T=To
=0
based on O.D.E., integrate twice
r =
+ c1
=
+
T(r)=
+ c1ln(r) + c2
Apply B.C.
At r=0
=0
0*0=0+c1
At r=Ro
T=To
To=
+ 0*ln(Ro)+c2
c2= To+
T(r) =
(
-r2) +To
c1=0
(b) To & T1
Assumptions: System II

Steady state

1-D flux along r

=0
B.C.
r=Ro
T=To
r=R1
T=T1
From Topic 17.1
qr=
At surface by convection
qr= 2 LR1h(T1-
If =constant
= *
(To-T1)
=qr
= energy generation/volume
= vol. of radwaste
=2
R1h(T1-
)
T1=
T1=
298 K= 339.7 K
=
(To-T1)
To=T1+
To=339.7 K +
(c) max T system I
= 643.5 K
(r=0)
)
Tmax(r=0)=
To
=
+ 643.5 K
= 1268.5 K (995.5
-ouch!)
17.29
Input Parameters:
k=ktissue= 0.6 W/m-K
L=0.5 cm
Ro2= 0.5 mmol O2/cm3 tissue-hr
R, O2= 468 J/mol O2
Ts= 30
Assumptions:

steady state

1-D flux along x

Insulated boundary

k(T)= constant
From Topic 17.2
Heat balance is (on diff vol. element)
-kA
kA
Divide by k, A, rearrange,
=
A
=0
->0
B.C.
x=0 T=Ts, x=L
=RO2*
T(x) = Ts+ (L*x – x2/2)
=
At x=L TL=Ts+
=6.5 x 10-2 W/cm2
TL= 30
(1
TL= 31.4
+
=1 K)
(tissue is reasonably at constant temp…)
*
(1 W= 1 J/sec)
17.30
Assumptions

steady state

1-D flux along x

kA & kB constant with T
L1= 0.5 cm
L2= 0.7 cm
kA=50 W/m-K
kB= 20 W/m-K
(a)
At steady state
= *L1=
(b) T1=?
=
T1=
* 0.5 cm = 10 W/cm2
At x=L1
(T1-Ts)
T1=
+
+ Ts
(c) h=?
= h(Ts-
)
h=
=
h= 0.20 W/cm2-K= 200 W/m2-K
(d) To=?
at x=0 (see aside for Model Development)
T(x) = To + (x2/2 –L1*x)
T1=To-
To=T1+
at x=L1, T=T1
= 90
+
17.31
Input Parameters:
A= 2.0 m2
L1= 0.02 m (inert layer)
Lc= 10 cm (compost layer)
System I= inert layer
k1= 0.4 W/m-K
System II= compost layer
kc= 0.1 W/m-K
(a) q= A*Lc= 600 W/m3* 2.0 m2* 0.1 m= 120 W
(b) Tc=?
=
Tc=
q=
=
+q
System I
1 source-> sink
= 0.25 m2K/W
= 288 K + 120 W *
(c) TL
at base (x=Lc) dT/dx=0
TL= Tc +
System II
303 K (30 )
insulated boundary
TL= 303 K +
*
= 333 K (60 )
(d) As Lc increases, q increases
Tc increases and TL increases and Ts increases
17.32
Input Parameters:
km= 0.419 W/m-K
Ro= 1.5 cm
= 15 k W/ m3 (15,000 W/m3)
T1= 37
Assumptions:

Steady state

1-D flux along “r” only (neglect ends)

constant q

km not a function of “T”

Ro constant along z
Fourier’s Law:
(a) Model for T(r)
Heat Balance on Differential Volume Element
IN-OUT+GEN=ACC
2 Lr (
2 Lr= flux area
= vol. of volume
element
,
, rearrange (watch signs!)
+
=0
Or
ordinary differential equation (ODE)
Boundary conditions (make the O.D.E. unit)
r= Ro, T=To
r=0, Qr=0=
by symmetry
Based on “type” of O.D.E. (perfect differential) integrate twice
1) r
2) T
Apply B.C. at r=0,
for 1) 0*0=0+c1
At r=Ro
To
T(r) =
+ To
c1=0
c2
At r=0.75 cm (0.0075 m)
T=
+ 37
T= 38.5
(b) Where and what is Tmax? Find the maxima of curve
=
Tmax=
=0
maximum T at r=0
+ To =
(1.5 cm *
)2 + 37
Tmax= 39
(c)
=
= 112 W/m2
=
(d) Revised Physical System
Figure
R1= 1.8 cm
For System I tissue T(r)=
+ To
For System II sheath cylindrical geometry, =0
Topic 17.1
qr=
(To-T1)
Eliminate To
Realize qr= *
To=
+T1
=
+T1
Ro=1.5 cm
17.33
Per foot of bare tube:
For longitudinal fins:
For circular fins:
Per fin:
Per foot:
17.34
Longitudinal case:
Circular case
Increase:
Long:
Circle:
17.35
Solution for
Is in form
Long fin approximation:
17.36
Airside:
Waterside:
For
Without fins:
With fins:
Waterside:
Airside:
Fins added to airside only:
To waterside:
Both sides:
17.37
One dimensional conduction with internal heat generation
Or:
At L/2:
Mid-point temperature:
17.38
17.39
Energy balance:
(d^2 )/(dx^2 )-m^2 =-W/kA {W in Btu/ft}
=C_1 e^mx+C_2 e^(-mx)+W/hP
(0)=0
C_1=-(W/hP (1-e^(-mx) ))/(e^mx+e^(-mx) )
(L)=0
C_2=-(W/hP (e^mx-1))/(e^mx-e^(-mx) )
Putting in values:
m=2.28 C_1=-0.0017 W/hP
C_2=-0.999
= _max @ 1.5 ft
W_max=I_max^2 R
R= L/A=(1.72x〖10〗^(-6) )(3)/( /4 (1/48)^2 (30.48) )=4.97x〖10〗^(-4)
θ_max=90[-0.00107e^2.28(1.5) -0.999e^(-2.28(1.5) )+1] W/hP
W/hP=96.3 W=96.3(6)( )(0.25/12)=37.8Btu/(hr ft)=11.08Btu/ft
I_max^2=11.08/(4.97x〖10〗^(-4) )=2.23x〖10〗^4 A^2
I_max=150 A
/ft
17.40
Substituting:
17.41
Substituting:
17.42
For aluminum I-beam: Same procedure as previous problem
17.43
17.45
Without fins:
For longitudinal fins:
Increase = 2068 Btu/hr % increase = 290%
For varying values of h:
2
0.731
0.84 412
5
1.156
0.71 346
8
1.462
0.60 290
15
2.00
0.48 229
50
3.65
0.27 122
100 5.17
0.19 81
i.e. fins are most effective when h is small
17.44
17.46
Same as previous problem except material is aluminum
Without fins:
h
% incr.
2
488
5
477
15
448
50
370
100
303
17.45
17.47
For known end temps:
Substituting:
17.46
17.48
Figure
For IC Chip: *Vchip= 15 W= qx
(a) T1 & chip failure rate without “extended surface”
Assumptions:

Steady state

1-D flux along x

is constant
Energy Balance
At x=0
T1=
T1= 413 K= 140
qx= *Vchip = Achip*h*(T1=
293 K
chip failure rate > 80 % (off the chart)
(b) T1 with extended heat transfer surfaces
Af= 2WL= 2(5.0 cm)(1.0 cm)= 20 cm2
Nf= 5
For rectangular fin with ho=h, heat transfer rate is
q= (Ao+ NfAf )h(T1with
m=
“f in efficiency”
=
if t<<w
for Al from Appendix H W3Rk=229 W/m-K (293-373K) for Aluminum
= 24.12 m-1
m=
mL= 24.12 m-1 * 1.0 cm (
0.241
=
= 0.981
Ao= WW- Nf(t*w)= 5.0 cm*5.0 cm -5(0.03 cm)(5.0 cm)= 24.25 cm2
Back to heat transfer rate
15 W= (24.25 cm2+ 5* 10 cm2*0.981)
T1= 395 K= 122
New chip failure rate
(T1-293 K)
50%
Chapter 18 End of Chapter Problem Solutions
18.1
 One can use lumped parameters
Combination of Newton’s law of cooling and Fourier’s 1st Law:
Where
If:
The solution is:
18.2
Clearly a lumped parameter case
Energy Balance:
Letting:
18.3
Aluminum wire:
This shows the lumped parameter solution can be used
Steady state case per mole:
Transient case:
18.4
A lumped parameter problem
18.5
Lumped parameter solution applies if
or if
Therefore
ut
must be
se istributed arameter solution
18.6
Use lumped parameter solution
Case
a
b
c
d
e
L(in)
3
6
12
24
60
V/A
0.5
0.6
0.67
0.706
0.732
T(min)
19.7
23.6
26.3
27.9
28.9
18.7
a) Cu – Bi<0.1 Lumped
b) Al – Bi<0.1Lumped
c) Zn - Bi<0.1Lumped
d) Steel - Bi<0.1Lumped
18.8
Water is well-stirred, therefore lumped solution applicable, thus T=T(t) only
t (hr.) 0
T ( ) 40
0.1
81
0.2 0.4 0.6 0.8 1.0
115 169 210 237 253
18.9
Lumped parameter case. Temperature may be considered uniform at any time.
18.10
Lumped parameter
18.11
A distributed parameter problem
By trial and error
18.12
Must use distributed parameter solution fig. f.8
18.13
18.14
Using chart for Cyl.
Interpolating @ 0.33
18.15
Distributed parameter solution
From chart
18.16
Using chart solution
By trial and error
18.17
Distributed parameter problem
By trial and error
18.18
Distributed Parameter problem
From chart
18.19
For a finite cylinder:
For both r & x directions:
Trial and error using charts:
18.20
Same cycle as in Prob. 18.15 but H varies:
As
With ends considered
For plane
H
1.3
0.65
0.26
0.13
10
5
2
1
t,s
3114
3114
~3000
~1600
~600
18.21
Assumptions:

Unsteady state

1- flux along “x”

=0

Constant

Small penetration depth ( Error function)
For “meat” (steak)
= 1500 kg/m3
k= 0.6 W/m-K
Cp= 4000 J/kg-K
What is time required for T(x, t) = 100
Based on Physical System and Assumptions,
Consider
=
Appendix L
= erf (
) =erf ( )
= 0.60= erf ( )
at erf ( ) =0.6 -> =0.59
at x= 2 mm?
=
=> t=
=
= 1.0 x10-7 m2/sec
t=
28.7 sec
Now consider Ts= 150
=
From Appendix L,
t=
= 0.400= erf ( )
,
=0.37
73 sec
(by linear interpolation)
18.22
L= 10 cm
a=L/2= 5 cm
= 120
To=20
h=60 W/m2K
2cm=0.02 m
Thermo physical Parameters
= 1500 kg/m3
k= 0.6 W/m-K
Cp= 4000 J/kg-K
= 1.0 x10-7 m2/sec
=
Assumptions

USS conduction

2-D flux r & z (cylindrical coordinates)

Constant h,

Neglect contribution of metal container to heat transfer resistance
What is T(r,z,t) at t=2500 sec, r= 0 cm, z= 0 cm?
z-direction:
x a=
0.1
R=
na=
0
ma=
0.2
Fig F.7
Ya=0.95 (does accounting for z direction matter in this calculation?)
r-direction:
Xcyl=
0.625
ncyl=
0
mcyl=
Fig F.8
Ycyl= 0.27
Y=YaYcyl= 0.95*0.27= 0.257=
T(r,z,t)= 94
0.5
18.23
Input Parameters
ksi=22 W/m-K
si=2300 kg/m
Cp,si=1000 J/kg-K
3
h=147 W/m2-K
=303 K (
R=
)
0.075 m
To=1600 K
What is T(r,t) at t= 583 sec, r=R=0.075 m
Assumptions
=

Unsteady state conduction

Finite volume system

1-D flux along r (neglect end effects)

Cylindrical geometry
=
= 9.57 x10-6 m2/sec
For cylinder, use Temp Time charts (Fig F.8 or Fig F.2
n=
m=
= 2.0
X=
1.0
Fig F.8
(Fig F.2 also works)
Y= 0.375=
Ts= 789 K (still too hot to handle!)
)
18.24
Input Parameters:
To= 25
=0
kliq= 0.3 W/m-K
=1.0 g/cm3= 1000 kg/m3
Cp,liq= 2000 J/kg-K
h= 0.86 W/m2-K
Tc= 2.0
precipitation temperature of drug
D=0.70 cm
R=0.35 cm
Assumptions:

Unsteady state

1-D flux along r

23>10 (neglect conduction)

Thermo hysical ro erties constant with “T”

Neglect conduction through syringe wall
Heat Transfer Model:
=
r=0
0
r=R
T(0,t)=Ts or
= h(Ts -
)
T=0, T(r,0)=To
se the analytical solution of the heat transfer model as contained in the “charts” to hel solve
the problem.
Two Approaches:

Get “t” at center (r=0) where T(0,t) =2.0

Get T(0,t) at center for t= 2.0 hr
Time-Temperature Charts (core calculation)
n=
m=
0
= 99.7 convection resistance is significant!
Call “m”= 100 to facilitate use of gra h
0.080
Fig F.2 for cylinder is “off the chart”. Try Fig. F.5 with r=0
X=130=
t=
= 1.5x10-7 m2/sec
t=
= 10,616 sec (2.9 hr)
Problem 18.25
Cannot use “Charts”, as no thickness of water is s ecified
Assumptions:

Unsteady state

1-D flux

Semi infinite medium
Heat transfer model:
@ x=0, t= 330 sec
= exp [
Input Parameters:
= 1.716x10-5 m2/sec
=
=
erf(
= 0.05
)= 0.0564
Appendix L
Ts= (1020 -20 ) exp
Ts= 966
q=hA(Ts q= 334 W
(1-0.0564)
(1239 K)
= 20 W/m2K * 176.7 cm2 *
(966-20)
18.26
Use Semi-infinite wall solution.
@ surface: x=0
Approximation: Use 1st term in series expansion.
At this time:
Solving for T:
18.27
Use semi-infinite solution:
Given:
For y=0
18.28
Use semi-infinite wall solution
At x=0 this reduces to
Where
t
Substituting:
By trial and error
18.29
where
So
is max when
is max
18.30
This is a semi infinite case
18.31
Solution is amenable to either numerical or analytical approach
Trial and error at each t
T
T( )
1 hr
580
6 hr
396
24 hr
275
18.32
18.36
at
18.33
18.37
Input Parameters:
=2.5 x10-7 m2/sec
Ts= 250
To=37
t=2.0 sec
Assumptions:




Constant source, transient sink
1-D flux along x
=0
semi infinite medium
unsteady state
Model:
= erf(
(a)
)
=
for living skin
= 9.165x10-8 m2/sec
=
Compare to = 2.8 x10-4 cm2/sec *
(b) Get x at t=2.0 sec for
2
= 2.8 x10-8 m2/sec
=2.5x10-7 m2/sec and T(x,t)= 62.5
for dry cadaverous epidermis
=
=0.8803=erf( )
5th =1.1
From Appendix L
=
=> x=2
= 1.56 x10-3m= 1.56 mm
x= 2* 1.1
Deep dermal burn
(c) T(x,t) at x=0.5 mm, t= 2.0 sec
=
=
= 0.354
By Linear interpolation Appendix L
erf ( ) = 0.3814=
T(x, t) = 169
(ouch)
=
18.34
Problem 18.38
Input Parameters:
For tissue
k= 0.35 W/m-K
Cp=3.8 kJ/kg K
= 1005 kg/m3
Ts=55.5
To=37
x= 3 mm (0.003 m)
Assumptions:

Constant source, transient sink

1-D flux along x

Semi-infinite medium

Skin tissue and tumor tissue has some thermo physical properties
(a) Thermal diffusivity
unsteady state
= 9.165 x10-8 m2/sec
=
(b) Active treatment time tact
ttotal=tact+ t
final tmin for T(x, t) = 45 , x= 3.0 mm
Model:
=erf ( ) = erf (
=
= 0.568= erf ( )
= 0.556
Appendix L
t=
= 79.5 sec
tactual= ttotal-t= 120 sec- 79.5 sec= 40.5 sec
5th by linear interpolation
Chapter 19 End of Chapter Problem Solutions
19.1
For a plane wall:
Variables
T
x
L
Α
k
t
h
Dimension
T
T
T
L
L
t
If temps are grouped as
19.2
Air
1.9e-5
k
Re
Pr
Nu
St
1.008e3
0.0293
2.3e5
0.699
348
2.16e-3
H2O
0.474e5
1.0
Benz.
0.473e-5
0.383
1.02e7
2.72
15.4
5.55e-7
0.0762
1.02e7
5.21
77.3
1.45e-6
0.45
Hg
1.06e6
0.033
Glyc.
0.128e2
0.598
5.03
4.57e7
0.021
1.17
1.22e6
0.165
37800
13.1
35.7
7.21e-5
19.3 ~Plots~
19.4
As per development in text:
In limit as
0
1
2
3
4
250
1310
1750
1310
250
0
27
80.3
134
161
19.5
For a single 4 ft long plate:
For stack of plates:
19.6
Energy balance
Standard procedure, resulting expression is:
For
x
0
0.4
0.8
1.2
1.22
900
3040
3110
1030
900
T
100
171
285
360
361
116
226
340
378
377
19.7
19.8
Laminar flow for all values of
T,F
30
50
80
79.5
79
18.2
6.61e-2
1.52e-2
0.13e-2
TCl
TFl
Tf
30
50
80
80
100
130
55
75
105
T
30
50
80
hL
8.74
11.4
16
437
570
800
Re
303
1320
15400
fd
37.6
17.9
5.2
0.0419 21.9 7.25
0.0101 44.6 4.64
0.0011 134.8 2.16
19.9
x
0
0.5
1
1.5
2
10.92
15.46
18.92
21.85
8.74
6.19
5.05
4.37
22.2
31.45
38.5
44.5
11.37
8.06
6.57
5.70
67.5
95.2
117
135
16.06
11.32
9.29
8.03
x
0
0.5
1
1.5
2
x
0
0.5
1
1.5
2
19.10
N2 @
k
Pr
a)
b)
c)
d)
e)
f)
g)
h)
100
0.069
1.77e-3
0.0154
0.71
200
0.0583
0.236e-3
0.0174
0.71
150
0.209e-3
0.0164
0.71
19.11
For air @
1.008
a)
b)
c)
19.12
19.13
For conditions specified:
Probably turbulent boundary layer
Use CouBurn analogy:
From Chapter 13 for turbulent boundary layer
{Assuming all surface exposed to turbulent B.L.}
19.14
For O2 gas at 300 K
= 1.3 kg/m3
Cp=920 J/kg-K
= 2.06 x10-5 kg/m-sec
k= 0.027 W/m-K
Assumptions:


Steady state
Constant
(a) Pr=
=
= 0.702
(b) h required for Ts= 310 K
Energy balance (Topic 17.2)
*l*A= hA (Ts-
)
h=
= qO2*
h=
=
= 32.5 W/m2-K
(c) Nu= =
= 60.2
(d) Consider laminar flow with h=
Nux= 0.332 Rex1/2 Pr1/3
= 0.332
Or h=
=
h=
2L1/2
=
Or NuL= = 0.664 ReL1/2Pr1/3
ReL=
(e) What is ?
Nu= 0.664 ReL1/2Pr1/3
ReL=
Re=
=
= 10,401
=> =
= 3.3 m/sec
High flow rate! How could this system be redesigned? How is T s affected by O2 flow rate?
19.15
Input Parameters:
Kpoly= 0.12 W/m-K
l= 1.5 mm
L= 2.0 m
W= 1.5 m
(a) Tav=
= 67 = 340 K ( film average temperature)
1.9553 x10-5 m2/sec
From Appendix I
Prair= 0.699
kair= 0.629282 W/m-K
(b) q= A*h*(
ReL=
L*W= 2.0 m* 1.5 m= 3.0 m2
)
= 51,143 (<2x105
=
laminar)
Nu=0.664 ReL1/2Pr1/3= 0.664 (51,143)1/2(0.699)1/3
Nu= 133.3=
h=
h= 1.95 W/m2-K
q= 2*W*L*h (
)
= 2* (3.0 m2) (1.95 W/m2-K) (107 -27 ) (1 = 1K)
= 936 W
19.16
Input Parameters:
Ts= 60
=20
Tav=
=40
For fluid (water at Tav= 40 )
= 992.2 kg/m3
= 658x10-6 kg/m-sec
Cp= 4175 J/kg-K
k= 0.663 W/m-K
=
= 0.10 m/sec
(a) Re & Pr
ReL=
=
ReL= 120,632 < 2x105
Pr=
Pr= 4.15
laminar flow
(b) hx at x= 0.5 m from leading edge
Nux= 0.332 Rex0.5Pr1/3= 0.332 (75,395)1/2(4.15)1/3= 146.5
Rex=
= 75,395
Nux=
hx=
q/A=hx (
hx= 194 W/m2-K
=
= 7770 W/m2
)=
(c) total “q” from plate
realize h(x)!
Let q= A* (
)
with A=W*L
=
h= 0.664
ReL=
(120,632)1/2(4.15)1/3
h=0.664
= 307 W/m2-K
q= 0.2 m *0.8 m * 307 W/m2-K (60
) = 1966 W
(d) Boundary Layer thickness at x=L
Hydrodynamic
=>
=
= 1.15 cm < 12 cm liq depth
H=
Pr-1/3 = 1.15 cm (4.15)-1/3= 0.717 cm << 12 cm
19.17
Momentum theorem in x direction:
At low velocity:
& Steady state
LHS:
Buoyant force (B.F.):
{per unit volume}
Viscous Force (V.F.):
RHS:
Equating: (LHS)=(RHS) & dividing by
In limit as
Energy equation: Same for both natural & forced convection
19.18
Into energy equation:
Equating:
Into momentum equation:
Equating:
Letting:
Previous two equations become:
Exponents on x must agree
Giving:
So equations for A&B becomes
So we have
& finally upon substitution
19.19
Air@ 310 K
x=
15 cm
30 cm
1.5 m
Gr=
4.31e7
3.45e8
4.31e10
=
0.985 cm
1.17 cm
1.75 cm
Nux=
30.5
51.2
171.3
Hx=
5.48
4.61
3.08
19.20
19.22
B.C.
B.C
Into momentum equation:
Equating & solving:
Energy equation:
Substituting & solving:
Giving:
If X=0
19.21
19.24
Assuming
B.C.
Temperature profile becomes:
Similarly for velocity:
Into the integral expression:
Left hand side (LHS):
(RHS)
Substitute equations (1)&(2):
Giving
Equating (LHS)=(RHS)
& letting
Substitution & some algebra give:
Separating variables:
Nusselt #:
19.22
19.25
B.C.
Into energy equation
{Presume
for integration}
19.23
19.27
Reynolds analogy: Assume
Close enough, doing over with
Colburn Analogy: Assume
Summary
Reynolds
Colburn
174
141
43500
35300
will yield
as a result.
19.24
19.28
For air @ 180 :
Reynolds:
Colburn:
19.25
19.29
From Colburn analogy:
Re=
{Laminar}
19.26
19.30
19.27
19.31
Reynolds analogy:
Colburn analogy:
Chapter 20 End of Chapter Problem Solutions
20.1
Ends are neglected
For vertical orientation
By trial and error
HTR surface temp
Horizontal orientation
Trial and error:
HTR surface temp
20.2
Bismuth
As in problem 20.1
Vertical – same formula as above
By trial and error
HTR surface temp
Horizontal – same formula as above
By trial and error
HTR surface temp
Continued hydraulic fluid
Same formulas and procedures
Vertical
{Properties used at 200
Horizontal
- highest temp in tables}
20.3
16 cm (0.525 ft) – vertical height
(20.3 continued)
By trial and error:
For 10 cm (0.328 ft) – height
Trial and error:
English units used – tables easier to use
20.4
For
Use lumped parameter
Since lumped parameter solution is valid answers to parts (a) & (b) are the same when:
20.5
Air @ 355 K:
Horizontal Cylinder:
For
Remainder of input goes to electrical & conduction losses & to illumination
20.6
For a horizontal cylinder
Trial and error:
20.7
For Cu cylinder with:
Height = 20.3 cm, Diameter= 2.54 cm
For
For an h value < 6340
Bi<0.1 ∴Lumped parameter
(20.7 continued)
20.8
For a sphere:
D, cm
h
7.5
615
0.0104
5
710
0.0135
1.5
1180
0.0271
L.E. surface resistance is very small ∴
~Time 0
& Falls to this value almost instantaneously
20.9
For
to reach 320 K – use values calculated in problem 20.8
D, cm
h
t
7.5
615
0.16
1.19 hr
5
710
0.16
31.7 min
1.5
1180
0.16
2.86 min
at all times
20.10
a) Horizontal
b) Vertical
20.11
Same conditions as prob. 20.10 except fluid is H2O @ 60
Horizontal:
Vertical:
20.12
Spherical tank
For sphere:
Properties @ 260 K
20.13
Assuming each plate is independent
20.14
Fraction of total =
With 1ftx1ft ridges
Same as in part (a) fraction = 0.54
20.15
For forced convection
Flow is
Plate is mostly in transition regime
Assume laminar boundary layer
Fraction due to F.C.=0.34
If B.L. is turbulent
Fraction = 1.49
For this case there is more capacity to transfer heat than there is solar energy supplied. Surface
temperature will be <150
20.16
By trial and error
Resistance of insulation
20.17
Trial and error
{see problem 20.43}
(20.19 continued)
20.18
Assume
T in R
Trial and error T=1147 R = 687
20.19
Forced convection outside
{B,n functions of Re}
Assume T 650 =1110R Tf=360
Table 20.3 B=0.027 n=0.805
@ This temp
{Pretty good}
20.20
Insulation on outside w/natural convection on surface
With
Trial and error
20.21
Equations to be solved are:
Trial and error:
Insulation thickness =
20.22
For natural convection case plane upward – facing hot surface
if 2x107<Re<1010
Assume top surface is square ~A=L2
Now for same Heat loss and L=0.982 m
{forced convection}
Assume
Transition regime
Assume laminar B.L.
20.23
Process Input Parameters:
W= 20 cm (0.2 m)
L= 60 cm (0.6 m)
ks= 16 W/m-K
km= 2.0 W/m-K
Air flow:
= 3.0 /sec
Tav=
=400 K
air= 2.5909x10
-5
m2/sec
Prair= 0.689
kair=0.033651 W/m-K
(a) & (b)
From heat balance *
qT=
Vm+ qloss
qloss= h*A (
Vm= 600 W
)
A= W*L
Get h, for flow over a flat plate
Re=
=
= 69,474 (<2x105) ∴ laminar
Use Nu= 0.664 Re1/2Pr1/3= 0.664 (69,474)1/2(0.689)1/3=154.6
Nu=
h= 8.67 W/m2-K
qloss= 8.67 W/m2-K * 0.2 m x 0.6 m (480 K-320 K) = 166.5 W
qT= 600 W+ 166.5 W= 766.5 W
(d) T2=?
q=
(T1-T2)
T2=
T1
T2=
T2= 22.5
+ 217
(490 K)
(498 K)
*Energy Balance on stainless steel slab + load material
Assumptions:
Steady state energy balance
IN-OUT+GEN=ACC
qT-qloss- m m=0
qT= qloss* m m
(c) T1=?
Show T1=Ts+
Q=q/A
m= -1.0 x10
T1= 217
6
W/m2
-qmVm (since measured)
20.24
20.25
a) Flow parallel to tube
If
B.L is laminar
If
B.L. is in transition
If laminar over total length
b) Crossflow case
20.26
Water:
a) Flow parallel to tube
{Turbulent}
b) Crossflow case
20.27
a) Flow parallel to tube
{turbulent}
b) Crossflow
20.28
{From problem 20.18}
Insulation resistance = 0.842 {Prob. 20.43}
20.29
Sphere
20.30
(a)
Input Parameters:
Gas=Air
540 K (247 )
kair= 0.0417 W/m-K
= 5.94 x10-5 m2/sec
4.04 x10-5 m2/sec
Solid = popcorn kernel
D= 0.5 cm (0.005 m)
k= 0.2 W/m-K
Cp= 2000 J/kg-K
= 1300 kg/m3
(b) h
For fluid flow around a sphere in a Fluidized Bed
NuD=
ReD=
= 2.0 + 0.6 ReD1/2Pr1/3
= 371.3
Pr=
= 0.6801
NuD= 2.0 + 0.6 (371.3)1/2 (0.6801)1/3= 12.16
= 101.5 W/m2-K
h=
(c) For the sphere
Bi=
Bi= 0.422 ∴ cannot neglect convection resistance!
Bi>10 to neglect
(d) Time “t” required for popcorn kernel to pop
T (0, t) = 175 = 448 K
USS conduction in a sphere with finite “m”
m=
= 0.78
call it m=0.8
n=
0 (center)
=
= 0.32
Fig F.9
X= 0.5=
=
t=
= 7.7 x10-8 m2/sec
= 40.6 sec
20.31
Input Parameters:
For liq H20 @ 30 =
k= 0.633 W/m-K
Pr=4.33
= 6.63 x10-7 m2/sec
@ Ts= 100
=278 x10-6 kg/m-sec for H20
kpoly= 0.08 W/m-K for bead
Heat Transfer Model:
q=
)
(A=
for sphere)
for external fluid flow around a sphere falling in a fluid
Nu=
2.0+0.6 Re1/2Pr1/3 for the fluid
Re=
Nu= 2.0 + (452.5)1/2(4.33)1/3= 22.8
= 452.5
h=
= 1443 W/m2-K
∴ q= (0.01 m) 2 1443 W/m2-K (100
If Whitaker correlation is used
) = 31.7 W
Nu= 2.0 +[0.4Re1/2+0.06Re2/3]Pr0.4
For ( ) = 1, h=1497 W/m2-K, q=32.9 W
For ( ) = 2.97, h=1925 W/m2-K, q=42.3 W
20.32
Input Parameters:
Ts=200
(383 K)
d= 4.0
(4.0x10-6 m)
L=1.2 mm (1.2x10-3 m)
P=I2R=q= 13 mW
For N2 @ Tav= 110
= 0.8948 kg/m3
= 2.13 x10-5 kg/m-sec
Cp= 1047.3 J/kg-K
k= 0.032 W/m-K
(a) Pr=
(b) Nu=?
=
= 0.697
q= dLh(Ts-
)
h=
h = 4789.4 W/m2-K
Nu=
= 0.599
Small “d”  large h, small Nu
(c)
=?
Nu= 0.30 +
[1+
for flow around a cylinder
] 4/5
[1+
] 4/5 1 for small Re
Back out Re
0.599= 0.30 +
*1
Re= 0.601 =
= 3.6 m/sec
Hot wire anemometer in tube with flowing N2 gas
20.33
Input Parameters:
D= 30 cm (0.3m), L=3.0 m
ksi= 30 W/m-K
Ts= 860 K
= 300 K
= 0.50 m/sec
(a), (b) Heat transfer rate, q
q=
For flow normal to a cylinder
NuD=
For air at Tav= (Ts+
[1+
)/2= (860 K+300 K)/2= 580 K
= 4.8512 x10-5 m2/sec, Prair= 0.68, kair= 0.45407 W/m-K
ReD=
NuD= 0.30+
= 3092
[1+
)
NuD= 28.1
h=
= 4.25 W/m2-K
=
q= (0.3 m) (3.0 m) 4.25 W/m2-K (860 K-300 K) = 6734 W
(c) Bi=
=
=
=0.011
∴ lumped parameter analysis can be used (Bi<0.1)
Asides:
Glycerin at 38
= 1253 kg/m3
= 78.2
Cp=0.598
=
k=
*
J/kg-K
= 0.15 kg/m-sec
= 0.285 W/m-K
20.34
From Figure 20.13
{extrapolation}
20.35
20.36
Figures 20.12&20.13 apply strictly for liquids flowing through tube banks but they will be used
for lack of other resources.
Using Fig 20.13
At this Re: j 0.01
For a bank of 10 tubes, 10 rows deep
20.37
For same conditions as Prob. 20.51 except staggered tube arrangement
& both arrangements give same value for j
∴
20.38
20.44
{Laminar}
Assume
20.39
20.45
Assume
{Laminar}
20.40
20.46
Putting everything together
By trial and error:
20.47
20.41
(20.44 continued)
Assume T0 is outside tube temperature
20.42
20.48
Outside of tube insulated ∴ all heat generated goes into H2O
At constant flux
Chapter 21 End of Chapter Problem Solutions
21.1
Plate is assumed to be copper
For H2O @ 323 K
Natural convection
Nucleate boiling:
L=0.565 ft
Equating (1)=(2)
Part b): Plot q/A from (1), q/A from (2), and their sum
21.2
In English units:
For Ni & Brass
For Cu & Pt
Ts(K)
A
B
390
17
31
0.033 5.04
0.364
420
47
85
0.09
4.67
0.36
450
77
139
0.148 3.79
0.355
Ni, Brass
Cu, Pt
390
168
2e5
165
.533e5
420
3516
8.5e5
346
4.07e5
450
16340
24e5
1610
1610e5
Ts
21.3
21.4
Boiling H2O @ 1 atm: Burnout point is
Film boiling part:
(21.4 continued)
Substituting into formula:
Nucleate boiling part:
As it cools the cylinder is in:
With
21.5
Using English units:
Assume nucleate boiling:
Surface temp = 221
21.6
Per plate:
It appears that nucleate boiling on one plate can achieve this:
{Ok one plate will do it}
21.7
21.8
{Film boiling}
21.9
{at a given depth y}
t, hours
ft x102
inches
0
0
0
0.2
0.586
0.0681
0.4
0.804
0.0964
0.6
1.0
0.12
0.8
1.14
0.137
1.0
1.27
0.152
21.10
{for thickness y}
if pan is horizontal
For pan inclined:
Per unit detla:
Length of surface exposed:
Assume accumulation of condensate due principally to condensation on exposed surface
21.11
21.12
Vertical tube:
Horizontal:
21.13
Horizontal cylinder:
(21.16 continued)
21.14
Horizontal tube case {see prob. 21.16}
21.15
Single horizontal tube:
Chapter 22 End of Chapter Problem Solutions
22.1
Using dittus-boelter correlation
As diameter increases the required area increases as
22.2
Oil:
22.3
(22.4 continued)
Assuming turbulent flow:
Solving for
22.4
Water
Oil:
For
max
will be associated with minimum
If T0 out max=160
For H2O as minimum fluid:
{Fig 22.12 a
}
Problem 22.5
Biodiesel (30-60 )
= 880 kg/m3
Cp,B= 2400 J/kg-K
(a) qshell=
= 2.5 m3/hr * 880 kg/m3 * 2400 J/kg-K (35
= -44,000 W
qtube= =
= - qshell
at Tav=
= 293 K
Cp, H20= 4182 J/kg-K
H20=
(b) A=?
= 27.4
=
= 0.526 kg/sec
A=
* 1 hr/3600 sec
= 0.462 m/sec
Re=
= 17,698 is turbulent (>10,000)
NU= 0.023 Re0.8 Pr0.4 = 0.023 (17,698)0.8(6.96)0.4
Nu= 125.06 =
=>
hi= 1959.6 W/m2-K
U=
= 289.7 W/m2-K
A=
5.53 m2
(c) New “A” for co-current flow
Figure
recalculate
= 19.5 K ( )
A=
=
7.77 m2 (larger!)
Problem 22.6
Input Parameters:
Tav=
38
= 1253 kg/m3
Cp= 2501 J/kg-K
= 0.15 kg/m-sec
k= 0.295 W/m-K
Di= 0.0135 m (0.532 in)
Do= 0.01905 m (0.75 in)
(a) Required steam flow,
s
qtube=
qshell= -qtube=
550,220 W
=
=
= 0.244 kg/sec
(b)
(c)
= 59.3
tw=
= 2.969 x 10-3 m
Tube-side convective heat transfer
Re=
= 203 (<2100)
Pr=
laminar
= 1316
For laminar flow inside a tube
Nu= 1.86 (Re Pr)1/3( )1/3 = 1.86 (203*1316)1/3 (
)1/3= 16.68
= 352 W/m2-K
Nu=
U= 322 W/m2-K
(d) A
q= UA
Heat Exchanger Design Equation
= 28.8 m2
A=
n=
= 96 tubes
Note: the next level of analysis could be to adjust the desired
so that A by the Heat
Exchanger Design Equation matches the actual tube velocity in “n” tubes.
Aside to part (c)
Rigorously, for a “thick walled” tube
+
+
Uo= 233 W/m2-K
A= 39.8 m2
22.7
𝑤 ,𝑖𝑛 = 20
𝑤 ,𝑜𝑢𝑡
𝑠 = 120
(a) Change
to
𝑠𝑎𝑡 = 120
in problem statement
(b) Change desired Re in the problem to 40,000.
(c)
Since water side resistance is >> condensate side,
The heat transfer surface area is,
22.8
a)
b) Water in shell; oil ~ 2 passes
can’t be done
c)
Fig (22.12) ~
22.9
To find
:
Fig 22.9 a:
F 1
22.10
22.11
(22.11 Continued)
To find F:
Fig. 22.9a F 1
22.11
22.14
Counter flow:
Cross flow – air mixed:
Cross flow – both mixed:
Cross flow configuration with both mixed is most compact
22.12
22.15
Tubes:
{fig 22.12c}
2 Tubes passes will work
37 tubes per pass
L=1.64 m per pass
22.13
22.16
Steam condensation rate:
22.14
Problem 22.17
Hot fluid: ethanol, shell side (B)
Cold fluid: water, tube side (A)
= 250 W/m2-K
based on outer tube area
(a) What is
a (cooling water mass flowrate)
Energy balance
shell=
B(
B,1 –
out
B,2) = =
B,1 (sat liq)
B,2 (sat vap)
in
qshell= 2.0 kg EtOH/sec * (-838 kJ/kg)= -1676 kJ/sec
qtube= -qshell=
ACp,A,liq (TC,2- TC,1)
A=
A= 13.3 kg/sec
(b)
=
= 41.2
(c)
A=
A= 162.7 m2 (based on outer tube area ) stop
(d) ntubes=? For 16 ft, 1.5 in schedule 40 pipes Appendix , W3R 5th; Do= 1.90 inches= 0.048 m
n=
= 220 tubes
What would be the diameter of the tube bundle of the
pitch was 3 inches?
22.15
Problem 22.18
(a) TC,2
qshell=
if
H20= 0.5 kg/sec,
CO2 Cp,CO2(TH,1-TH,2)
= 325 K
Appendix I, W3R
Cp,CO2= 875.8 J/kg-K at 325 K
CO2= 𝑛CO2 * MW,CO2= 360 kgmol/hr * 44 kg/kgmol * 1 hr/3600 sec = 4.4 kg/sec
qshell= 4.4 kg/sec * 875.8 J/kg-K (300 K-350 K)= -1.93 x 105 J/sec (193 kW)
qtube=
H20CP,H20(TC,2-TC,1)= -qshell
TC,2= TC,1+
=
+ 293 K= 385 K
TC,2 > TH,2 – ouch 2ND Law Thermo violated
Recalc with TH,2-TC,2=17
(or 17 K)
= 350-17= 333 K
=
(a) U=?
1.15 kg/sec
ok
ho= 300 W/m2-K
tw= 0.109 in= 4.291 x 10-4 m
0.5 inch sched 40 tube
Di= 0.546 in= 0.00215 m
Do=0.622 in= 0.00245 m
Tav=
313 K tube side
0.663 x 10-6 m2/sec
Pr= 4.333
Appendix I W3R
k= 0.633 W/m-K
= 992.2 kg/m3
Re= 10,000=
= 3.08 m/sec
For Cu
kw= 386 W/m-K at 293 K ( Appendix H W3R)
For tube
Nu= 0.023 Re0.8Pr0.4 (fluid is heated)
0.023 (10,000)0.8(4.33)0.4= 40.86
Nu=
h=
= 1.203 x 10-4 W/m2-K
=
U= 292.6 W/m2-K
ho
shell side is obviously limiting
(d) A=?
Design Equation
q= UA
=
A=
(e) n=?
=
= 11.3 K (11.3
= 58.4 m2
by continuity
)
𝑛
=
or n=
also A=
= 104 tubes
𝑛
L=
73 m
wow
Go multi-pass and limit Lp= 5 m
# passes=
14 passes
Will this configuration work?
Calculate heat-exchanger correction factor “F”
Y=
=
Off chart
won’t work
22.16
Problem 22.19
Appendix I, M W3R
kw= 386 W/m-K
Schedule 40 1” pipe
Di= 1.049 in= 2.67 x 10-2 m
Do= 1.315 in= 3.34 x 10-2 m
tw= 0.133 in= 3.378 x 10-3 m
Input Parameters:
Tube side CO2
Tav=
Cp,CO2= 875.8 J/kg-K
kCO2= 0.01851 W/m-K
325 K
1.67 kg/m3
PrCO2= 0.763
9.76 x 10-6 m2/sec
(Appendix H, W3R 5th)
(a)
qtube=
and q= qtube= - qshell
CO2 on tube side
=𝑛
𝑠
= 0.44 kg/sec
qtube= 0.44 kg/sec * 875.4 J/kg-K ( 350 K-300 K)= 1.927 x 104 J/sec (W)
For water on shell side @ Tav= 40 , Cp,H20= 4184 J/kg-K
qshell=
= -qtube
𝑠
=
(b) as
(414.5 kg/hr)
, TC,1
(c)
Overall Heat Transfer Coeff. “U”
For tube side, Re= 20,000 is desired
Nu= 0.023 Re0.8Pr0.3= 0.023 (20,000)0.8(0.763)0.3
Nu= 58.52=
= 40.57 W/m2-K
U= 35.73 W/m2-K (why lower than hi?)
(d) A
q=UA
=
A=
=
= 11.3 K
= 47.85 m2
(e) ntube and Ltube
n=
Re=
CO2 flows in one tube
= 7.31 m/sec
n=
(call it 65)
A= (
𝑛
L= 7.0 m for each tube
If 10 ft tubes were available, what are your options?
1) multipass
2) work backwards: change Re, n etc. until L=10 ft.
(one tube)
22.17 22.20
Problem
Thermo physical properties of CO2: (Tav 425 K)
= 1.268 kg/m3
Cp= 960.9 J/kg-K
= 2.0325 x 10-5 kg/m-sec
k= 0.02678 W/m-K
Tube Properties
Do= 1.05 in= 0.0267 m
Di= 0.824 in= 0.0209 m
tw= 0.113 in= 0.0029 m
kw= 42.9 W/m-K
(a) Heat Exchanger Area “A”
q=U*A*
note TH>Ts!
=> U= 49.6 W/m2-K
=
= 47.7
q=qshell=
= 1,128,500 W
m2
A=
(b)
for CO2 flow in tube
-qshell=qtube=
*Cp,CO2(TH,2-TH,1)
=
(c)
= 23.5 kg CO2/sec
for single tube, fluid= CO2 @ Tav= 425 K
For turbulent flow, Nu=
Nu=
Pr=
= 0.729
Re=
12,279
Re=
=
(d) For A->
= 9.4 m/sec
TH,2 ->Ts=100
22.18
22.21
22.19
22.22
22.20
22.23
To find F:
Fig. 22.10 a F 0.96
22.21
22.24
Input data – see problem 22.3
Crossflow:
Shell and tube:
Same data
22.22
22.27
For cool fluid in tubes:
Hot fluid in tubes:
Both are off the charts. Neither is possible
can’t use this configuration.
Chapter 23 End of Chapter Problem Solutions
23.1
Radiant emission from sun
All passes through a spherical surface of radius, L
At the earth:
Flux at earth
23.2
Area subtended by earth
Incident solar energy =
{from problem 23.1}
Energy balance:
23.3
23.4
Energy balance for collector:
)
Equating:
800=
By trial and error T=322 K
23.5
a)
b) With window
23.6
From Wien’s displacement law:
Fraction in visible band:
{table 23.1}
or
23.7
T (K)
( m)
Sun
5790
1.998
L. bulb
2910
1.004
surface
1550
0.535
Skin
308
0.1063
23.8
Filament at 2910 K
q=100 W
Visible range:
Fraction in V.R. = 0.1008
23.9
For
For
23.10
For solar irradiation:
Plain glass:
Tinted glass:
In the visible range:
{For tinted glass
Plain glass:
Tinted glass:
}
23.11
T (K)
500
-0
0.1613
0.1613
2000
0.0197
0.9142
0.8945
3000
0.1402
0.9689
0.8287
4500
0.4036
0.9896
0.586
23.12
For:
Fraction in visible range = 0.3696
In U.V. range
Fraction in U.V. range = 0.12
In I.R. range
Fraction in I.R. range = 0.88
Wein’s law
23.13
For surroundings at 0 K:
In visible range 0.4< <0.7 m
Fraction = 0.0078
In U.V. range 0< <0.4
Fraction = 0
In I.R. range 0.4< <100
Fraction = 0.992
23.14
T=1500 K Peephole D=10 cm
for 0< <3.2 m
for 3.2< <
For:
Total heat loss
23.15
23.16
through hole with D=0.0025 m2
23.17
With no intervening plate:
With intervening plate present:
Per unit area:
(
{Emissivity of intervening plate has no effect}
22.18
23.19
Entire hole interior is surf. 2
Opening (surroundings) is surf. 1
23.20
1 is inner cylinder
2 is outer cylinder
With radiation shield
23.21
23.22
With no reflector:
23.22
23.23
per ft {radiant loss}
Convection:
For Tf=137
23.23
23.24
23.24
23.25
Diameter of hole = 5 cm
Gray case:
23.25
23.28
Opening diameter= 5 mm
Equivalent surface (1) sees interior as a single surface
Analogous circuit:
All interior may be considered a single surface
23.26
23.39
Problem statement asks for radiant energy reaching tank bottom, i.e. the irradiation
23.27
23.30
Surroundings are considered and equivalent surface (3) at 0 K
(23.30 continued)
To surroundings
23.28
23.32
Solving:
23.29
23.33
From Figs 23.14 & 23.15
For black disks:
For gray bodies:
For loops shown:
Solving simultaneously:
23.30
23.36
(23.36 continued)
23.31
23.37
{surface 3 is surroundings}
{Fig 23.14 F12 0.37} → F13=0.63
Loops equation:
Solving:
23.37 (Alternate solution)
Using equations 23.37 & 23.38
Applying them to each node:
Solving these equations simultaneously gives same result as above
23.32
23.38
Test specimen is 1
Tube is 2
Viewing port W
becomes:
From specimen
Loss through window
23.33
23.41
In limit as
:
By graphical integration:
1900
3.2
2.93
1700
4.0
3.0
1500
5.4
3.15
1300
8.0
3.33
1100
14.6
3.65
Radiant fraction
For v doubled:
Integrate graphically until x=35.2
At this location T=1265
24.1
Assume: 1) Ideal gas mixture, 2) constant 𝜌𝜌 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑), 3) constant c (total molar concentration).
Given
𝐍𝐍𝐴𝐴 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑦𝑦𝐴𝐴 + 𝑦𝑦𝐴𝐴 (𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 )
With
𝑤𝑤𝐴𝐴 =
And
𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴
𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴 +𝑦𝑦𝐵𝐵 𝑀𝑀𝐵𝐵
𝐧𝐧𝐴𝐴 = 𝜌𝜌𝐴𝐴 𝐯𝐯𝐴𝐴 = 𝜌𝜌𝐴𝐴
with M = 𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴 + 𝑦𝑦𝐵𝐵 𝑀𝑀𝐵𝐵
𝐍𝐍𝐴𝐴
𝜌𝜌𝐴𝐴
=
𝐍𝐍 = 𝑀𝑀𝐴𝐴 𝐍𝐍𝐴𝐴
𝑐𝑐𝐴𝐴
𝑐𝑐𝐴𝐴 𝐴𝐴
𝐍𝐍𝐴𝐴 𝑀𝑀𝐴𝐴 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴 ∇𝑦𝑦𝐴𝐴 + 𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴 (𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 )
𝐧𝐧𝐴𝐴 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴
∇𝑐𝑐𝐴𝐴
𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵
+ 𝑤𝑤𝐴𝐴 𝑀𝑀( + )
𝑐𝑐
𝑀𝑀 𝑀𝑀
𝐧𝐧𝐀𝐀 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴 ∇𝑐𝑐𝐴𝐴 + 𝑤𝑤𝐴𝐴 (𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 )
𝐧𝐧𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝜌𝜌𝐴𝐴 + 𝑤𝑤𝐴𝐴 (𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 )
24-1
24.2
a. Prove 𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 = 𝑐𝑐𝐕𝐕
𝐍𝐍𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑐𝑐𝐴𝐴 +
𝐍𝐍𝐴𝐴 = 𝐉𝐉𝐴𝐴 +
𝑐𝑐𝐴𝐴
(𝐍𝐍 + 𝐍𝐍𝐵𝐵 )
𝑐𝑐 𝐴𝐴
𝑐𝑐𝐴𝐴
(𝐍𝐍 + 𝐍𝐍𝐵𝐵 )
𝑐𝑐 𝐴𝐴
𝐉𝐉𝐴𝐴 + 𝑐𝑐𝐴𝐴 𝐕𝐕 = 𝐉𝐉𝐴𝐴 +
Thus,
𝑐𝑐𝐴𝐴
(𝐍𝐍 + 𝐍𝐍𝐵𝐵 )
𝑐𝑐 𝐴𝐴
𝑐𝑐𝐴𝐴
(𝐍𝐍 + 𝐍𝐍𝐵𝐵 ) = 𝑐𝑐𝐴𝐴 𝐕𝐕
𝑐𝑐 𝐴𝐴
𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 = 𝑐𝑐𝐕𝐕
b. Prove 𝐧𝐧𝑨𝑨 + 𝐧𝐧𝑩𝑩 = 𝜌𝜌𝐕𝐕
𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 = 𝑐𝑐𝐕𝐕
𝐧𝐧𝐴𝐴 = 𝑀𝑀𝐴𝐴 𝐍𝐍𝐴𝐴
Then
𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵
+
= 𝑐𝑐𝐕𝐕
𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵
1
𝐕𝐕 = (𝑐𝑐𝐴𝐴 𝐯𝐯𝐴𝐴 + 𝑐𝑐𝐵𝐵 𝐯𝐯𝐵𝐵 )
𝑐𝑐
𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵
1
+
= 𝑐𝑐 ∙ (𝑐𝑐𝐴𝐴 𝐯𝐯𝐴𝐴 + 𝑐𝑐𝐵𝐵 𝐯𝐯𝐵𝐵 )
𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵
𝑐𝑐
𝜌𝜌𝐴𝐴 = 𝑐𝑐𝐴𝐴 𝑀𝑀𝐴𝐴
𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵
𝜌𝜌𝐴𝐴 𝐯𝐯𝐴𝐴 𝜌𝜌𝐵𝐵 𝐯𝐯𝐵𝐵
+
=(
+
)
𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵
𝑀𝑀𝐴𝐴
𝑀𝑀𝐵𝐵
Thus
𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 = (𝜌𝜌𝐴𝐴 𝐯𝐯𝐴𝐴 + 𝜌𝜌𝐵𝐵 𝐯𝐯𝐵𝐵 )
24-2
𝐯𝐯 =
1
(𝜌𝜌 𝐯𝐯 + 𝜌𝜌𝐵𝐵 𝐯𝐯𝐵𝐵 )
𝜌𝜌 𝐴𝐴 𝐴𝐴
𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 = 𝜌𝜌𝐯𝐯
c. Prove 𝐣𝐣𝐴𝐴 + 𝐣𝐣𝐵𝐵 = 0
𝐣𝐣𝐴𝐴 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑤𝑤𝐴𝐴
So
𝐣𝐣𝐴𝐴 + 𝐣𝐣𝐵𝐵 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑤𝑤𝐴𝐴 − 𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑤𝑤𝐵𝐵 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 (∇𝑤𝑤𝐴𝐴 + ∇𝑤𝑤𝐵𝐵 )
∇𝑤𝑤𝐴𝐴 + ∇𝑤𝑤𝐵𝐵 = ∇𝑤𝑤𝐴𝐴 + ∇(1 − 𝑤𝑤𝐴𝐴 ) = 0
Thus
𝐣𝐣𝐴𝐴 + 𝐣𝐣𝐵𝐵 = (−𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0) + (−𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0) = 0
24-3
24.3
Given
𝑛𝑛
𝑦𝑦𝑖𝑖 𝑦𝑦𝑗𝑗
𝑑𝑑𝑦𝑦𝑖𝑖
= �
(𝑣𝑣 − 𝑣𝑣𝑖𝑖 )
𝑑𝑑𝑑𝑑
𝐷𝐷𝑖𝑖𝑖𝑖 𝑗𝑗
𝑗𝑗=1,𝑗𝑗≠𝑖𝑖
i =A, j = B, and n = 2 (species A and B)
𝑑𝑑𝑦𝑦𝐴𝐴 𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵
=
�𝑣𝑣 − 𝑣𝑣𝐴𝐴,𝑧𝑧 �
𝑑𝑑𝑑𝑑
𝐷𝐷𝐴𝐴𝐴𝐴 𝐵𝐵,𝑧𝑧
=
=
=
So
=
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
Thus
𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵 𝑣𝑣𝐵𝐵,𝑧𝑧 − 𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵 𝑣𝑣𝐴𝐴,𝑧𝑧
𝐷𝐷𝐴𝐴𝐴𝐴
𝑦𝑦𝐴𝐴 (𝑐𝑐𝐵𝐵 ⁄𝑐𝑐 )𝑣𝑣𝐵𝐵,𝑧𝑧 − (𝑐𝑐𝐴𝐴 ⁄𝑐𝑐 )𝑦𝑦𝐵𝐵 𝑣𝑣𝐴𝐴,𝑧𝑧
𝐷𝐷𝐴𝐴𝐴𝐴
𝑦𝑦𝐴𝐴 (𝑐𝑐𝐵𝐵 𝑣𝑣𝐵𝐵,𝑧𝑧 ) − 𝑦𝑦𝐵𝐵 (𝑐𝑐𝐴𝐴 𝑣𝑣𝐴𝐴,𝑧𝑧 )
𝑐𝑐𝑐𝑐𝐴𝐴𝐴𝐴
𝑦𝑦𝐴𝐴 𝑁𝑁𝐵𝐵,𝑧𝑧 − (1 − 𝑦𝑦𝐴𝐴 )𝑁𝑁𝐴𝐴,𝑧𝑧
𝑐𝑐𝑐𝑐𝐴𝐴𝐴𝐴
𝑑𝑑𝑦𝑦𝐴𝐴
= 𝑦𝑦𝐴𝐴 𝑁𝑁𝐵𝐵,𝑧𝑧 − 𝑁𝑁𝐴𝐴,𝑧𝑧 +𝑦𝑦𝐴𝐴 𝑁𝑁𝐴𝐴,𝑧𝑧
𝑑𝑑𝑑𝑑
= 𝑦𝑦𝐴𝐴 �𝑁𝑁𝐴𝐴,𝑧𝑧 + 𝑁𝑁𝐵𝐵,𝑧𝑧 � − 𝑁𝑁𝐴𝐴,𝑧𝑧
𝑁𝑁𝐴𝐴,𝑧𝑧 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑦𝑦𝐴𝐴
+ 𝑦𝑦𝐴𝐴 �𝑁𝑁𝐴𝐴,𝑧𝑧 + 𝑁𝑁𝐵𝐵,𝑧𝑧 �
𝑑𝑑𝑑𝑑
24-4
24.4
NA = 5.0 × 10-5 kgmole A/m2∙s
NB = 0 kgmole B/m2∙s
𝑐𝑐𝐴𝐴 = 0.005 kgmole/m3
𝑐𝑐𝐵𝐵 = 0.036 kgmole/m3
Find 𝑣𝑣𝐴𝐴 , 𝑣𝑣𝐵𝐵 , and V
NA = 𝑐𝑐𝐴𝐴 𝑣𝑣𝐴𝐴
𝑁𝑁𝐴𝐴 5.0 × 10−5 kgmole A/m2 ∙ s
𝑣𝑣𝐴𝐴 =
=
= 0.010 m/s
𝑐𝑐𝐴𝐴
0.005 kgmole/m3
NB = 𝑐𝑐𝐵𝐵 𝑣𝑣𝐵𝐵 = 0
Thus vB = 0 m/s
𝑁𝑁𝐴𝐴
1
1
𝑁𝑁𝐴𝐴
=
𝑉𝑉 = (𝑐𝑐𝐴𝐴 𝑣𝑣𝐴𝐴 + 𝑐𝑐𝐵𝐵 𝑣𝑣𝐵𝐵 ) = (𝑁𝑁𝐴𝐴 + 𝑁𝑁𝐵𝐵 ) =
𝑐𝑐
𝑐𝑐
𝑐𝑐
𝑐𝑐𝐴𝐴 + 𝑐𝑐𝐵𝐵
=
5.0 × 10−5 kgmole A/(m2 ∙ s)
= 0.0012 m/s
0.005 kgmole/m3 + 0.036 kgmole/m3
24-5
24.5
Given:
P = 6.1 × 10−3 bar, T = 210 K
yCO2 = 0.9532, yN2 = 0.027, yAr = 0.016, yO2 = 0.0013, yCO = 0.0008
a. Partial pressure 𝑝𝑝𝐴𝐴 for CO2
A = CO2
 105 Pa 
−3
×
0.9532
6.1
10
bar
(
)

 = 581 Pa
 1 bar 
b. Molar concentration 𝑐𝑐𝐴𝐴 for CO2
(
p A y=
=
AP
cA =
)
pA
581 Pa
gmole
=
= 0.333
3
RT
m3
8.314 m ⋅ Pa/gmole ⋅ K ( 210 K )
(
)
c. Total molar concentration c for the Martian atmosphere
=
c
P
=
RT
(
610 Pa
gmole
= 0.349
m3
8.314 m ⋅ Pa/gmole ⋅ K ( 210 K )
3
)
d. Total mass concentration ρ for the Martian atmosphere
M avg= yCO2 M CO2 + yN2 M N2 + yAr M Ar + yO2 M O2 + yCO M CO



g 
g 
g 
M avg = ( 0.9532 )  44.01
 + ( 0.027 )  28.00
 + ( 0.0163)  9.95
+
gmole 
gmole 
gmole 





g 
g 
( 0.0013)  32.00
 + ( 0.0008 )  28.01

gmole 
gmole 


M avg = 43.41
g
gmole

g 
gmole 
g
c  43.41
=
ρ M avg ⋅=
  0.349
 =15.2 3
3
gmole  
m 
m

24-6
24.6
A = benzene solute (C6H6) in B = liquid ethanol solvent (C2H5OH)
𝑤𝑤𝐴𝐴 = 0.0050
T = 293 K
𝜌𝜌𝐵𝐵 = 789 kg/m3
𝑀𝑀𝐴𝐴 = 78.11
Estimate 𝑐𝑐𝐴𝐴
g
, 𝑀𝑀𝐵𝐵 = 46.07
gmole
g
gmole
= 46.07
kg
kgmole
The solution is dilute with respect to A
𝜌𝜌𝐵𝐵
𝑐𝑐 ≈
=
𝑀𝑀𝐵𝐵
kg
kmole
m3
= 17.1
kg
m3
46.07
kgmole
789
𝑤𝑤𝐴𝐴 + 𝑤𝑤𝐵𝐵 = 1
𝑤𝑤𝐵𝐵 = 1 − 𝑤𝑤𝐴𝐴 = 1‒ 0.0050 = 0.9950
0.0050
g
78.11
gmole
𝑥𝑥𝐴𝐴 = 𝑤𝑤
= 0.0030
𝑤𝑤 =
𝐴𝐴
0.0050
0.9950
+ 𝐵𝐵
𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵
g +
g
78.11
46.07
gmole
gmole
𝑤𝑤𝐴𝐴
𝑀𝑀𝐴𝐴
𝑐𝑐𝐴𝐴 = 𝑥𝑥𝐴𝐴 𝑐𝑐 = (0.0030)(17.1 kgmole⁄m3 ) = 0.051 kgmole⁄m3 = 51 gmole⁄m3
24-7
24.7
a. Estimate 𝐷𝐷𝐴𝐴𝐴𝐴 for 𝐶𝐶𝐶𝐶2 in air
A = 𝐶𝐶𝐶𝐶2 and B = air
𝑀𝑀𝐴𝐴 = 44.01
T = 310 K
g
, 𝑀𝑀𝐵𝐵 = 28.97
gmole
𝑃𝑃 = (1.5 × 105 Pa) �
g
gmole
1atm
� = 1.49 atm
1.01 × 105 Pa
Hirschfelder Equation
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
1 1⁄2
0.001858 𝑇𝑇 3⁄2 �𝑀𝑀 + 𝑀𝑀 �
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴
2Ω
𝐴𝐴
𝐷𝐷
𝐵𝐵
From Appendix K, Table K.2: 𝜎𝜎𝐴𝐴 = 3.996 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å,
𝜎𝜎𝐴𝐴𝐴𝐴 =
𝜀𝜀𝐴𝐴
κ
𝜀𝜀
= 190 K, 𝐵𝐵 = 97 K
κ
𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 3.996 Å + 3.617 Å
=
= 3.807 Å
2
2
𝜅𝜅𝜅𝜅
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 310 K = 2.28
𝜀𝜀𝐴𝐴𝐴𝐴
𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵
190 K 97 K
From Table K.1, interpolate to find ΩD = 1.029
1/2
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
1
0.001858 ∙ 310 K 3/2 �
g +
g �
44.01
28.97
gmole
gmole
2
(1.49 atm)�3.807 � (1.029)
= 0.109 cm2 ⁄s
b. Estimate DAB for ethanol in air
A = ethanol and B = air
𝑀𝑀𝐴𝐴 = 46.07
g
, 𝑀𝑀𝐵𝐵 = 28.97
gmole
g
gmole
24-8
T = 325 K
𝑃𝑃 = (2.0 × 105 Pa) �
1atm
� = 1.98 atm
1.01 × 105 Pa
𝜀𝜀
𝜀𝜀
From Appendix K, Table K.2: 𝜎𝜎𝐴𝐴 = 4.455 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å, 𝐴𝐴 = 391 𝐾𝐾, 𝐵𝐵 = 97 𝐾𝐾
𝜎𝜎𝐴𝐴𝐴𝐴 =
κ
𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 4.455 Å + 3.617 Å
=
= 4.036 Å
2
2
κ
𝜅𝜅𝜅𝜅
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 325 K = 1.67
𝜀𝜀𝐴𝐴𝐴𝐴
𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵
391K 97K
From Table K.1, interpolate to find ΩD = 1.148
𝐷𝐷𝐴𝐴𝐴𝐴 =
0.001858 ∙ 325 𝐾𝐾 3/2 �
1
1/2
1
g +
g �
46.07
28.97
gmole
gmole
2
(1.98 atm)�4.036 � (1.148)
= 0.0697 cm2 ⁄s
c. Estimate DAB for CCl4 in air
A = CCl4 and B = air
𝑀𝑀𝐴𝐴 = 153.82
T = 298 K
g
, 𝑀𝑀𝐵𝐵 = 28.97
gmole
𝑃𝑃 = (1.913 × 105 Pa) �
g
gmole
1atm
� = 1.894 atm
1.01 × 105 Pa
𝜀𝜀
𝜀𝜀
From Appendix K, Table K.2, 𝜎𝜎𝐴𝐴 = 5.881 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å, 𝐴𝐴 = 327 K, 𝐵𝐵 = 97 K
𝜎𝜎𝐴𝐴𝐴𝐴 =
𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 5.881 Å + 3.617 Å
=
= 4.749 Å
2
2
κ
κ
𝜅𝜅𝜅𝜅
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 298 K = 1.67
𝜀𝜀𝐴𝐴𝐴𝐴
𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵
327K 97K
From Table K.1, ΩD = 1.148
24-9
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
0.001858 ∙ 298 K 3/2 �
153.82
1/2
1
g �
g +
28.97
gmole
gmole
2
(1.894 atm)�4.749 � (1.148)
= 0.0395 cm2 ⁄s
24-10
24.8
Estimate 𝐷𝐷𝐴𝐴𝐴𝐴 for NH3 in air
A = NH3, B = air
𝑀𝑀𝐴𝐴 = 17.03
g
, 𝑀𝑀𝐵𝐵 = 28.97
gmole
g
gmole
P = 1.0 atm, T = 373 K, (𝜇𝜇𝑝𝑝 )𝐴𝐴 = 1.46 debye, (𝑉𝑉𝑏𝑏 )𝐴𝐴 = 25.8
Brokaw equation
𝑐𝑐𝑐𝑐3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
, (𝑇𝑇𝑏𝑏 )𝐴𝐴 = 239.7 K
1.94 × 103 (𝜇𝜇𝑝𝑝 )𝐴𝐴 2 1.94 × 103 (1.46 debye)2
𝛿𝛿𝐴𝐴 =
=
= 0.669
𝑐𝑐𝑐𝑐3
(𝑉𝑉𝑏𝑏 )𝐴𝐴 (𝑇𝑇𝑏𝑏 )𝐴𝐴
(25.8
)(239.7 K)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝜀𝜀𝐴𝐴
= 1.18�1 + 1.3𝛿𝛿𝐴𝐴 2 �(𝑇𝑇𝑏𝑏 )𝐴𝐴 = 1.18(1 + 1.3(0.669)2 )(239.7 K) = 447 K
𝜅𝜅
𝜎𝜎𝐴𝐴 = �
1.585(𝑉𝑉𝑏𝑏 )𝐴𝐴
1 + 1.3𝛿𝛿𝐴𝐴 2
1/3
�
1/3
𝑐𝑐𝑚𝑚3
)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
=�
�
1 + 1.3(0.669)2
1.585(25.8
= 2.957 Å
Since the air has no dipole, δB = 0, δAB = 0
From Appendix K, Table K.2,
𝜀𝜀𝐵𝐵
κ
= 97 K
From Table 24.4 of Ch.24, (𝑉𝑉𝑏𝑏 )𝐵𝐵 = 29.9
𝜎𝜎𝐵𝐵 = �
1.585(𝑉𝑉𝑏𝑏 )𝐵𝐵
1 + 1.3𝛿𝛿𝐵𝐵 2
1/3
�
𝑐𝑐𝑐𝑐3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1/3
𝑐𝑐𝑐𝑐3
1.585(29.9
)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
=�
�
1 + 1.3(0)2
= 3.619 Å
𝜎𝜎𝐴𝐴𝐴𝐴 = (𝜎𝜎𝐴𝐴 𝜎𝜎𝐵𝐵 )1/2 = (2.957 Å ∙ 3.619 Å)1/2 = 3.271 Å
𝑇𝑇 ∗ =
𝜅𝜅𝜅𝜅
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 373 K = 1.79
𝜀𝜀𝐴𝐴𝐴𝐴
𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵
447K 97K
Ω𝐷𝐷 = Ω𝐷𝐷,0 =
𝐴𝐴
𝐶𝐶
𝐸𝐸
𝐺𝐺
+
+
+
∗
𝐵𝐵
∗
∗
(𝑇𝑇 )
exp(𝐷𝐷𝑇𝑇 ) exp(𝐹𝐹𝑇𝑇 ) exp(𝐻𝐻𝑇𝑇 ∗ )
24-11
=
1.06036
0.19300
1.03587
1.76474
+
+
+
= 1.12
0.15610
(1.79)
exp(0.47635 ∙ 1.79) exp(1.52996 ∙ 1.79) exp(3.89411 ∙ 1.79)
Hirschfelder Equation
𝐷𝐷𝐴𝐴𝐴𝐴 =
=
1
1 1⁄2
0.001858 𝑇𝑇 3⁄2 �𝑀𝑀 + 𝑀𝑀 �
𝐴𝐴
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 2 Ω𝐷𝐷
𝐵𝐵
0.001858 ∙ 373 K 3/2 �
17.03
1/2
1
g �
g +
28.97
gmole
gmole
1
= 0.341 cm2 ⁄s
2
(1.0 atm)�3.271 � (1.12)
For comparison, from Appendix J, Table J.1, DABP = 0.198
𝑐𝑐𝑐𝑐2 𝑎𝑎𝑎𝑎𝑎𝑎
s
𝑐𝑐𝑐𝑐2 𝑎𝑎𝑎𝑎𝑎𝑎
𝐷𝐷𝐴𝐴𝐴𝐴 𝑃𝑃 0.198
𝑐𝑐𝑐𝑐2
s
𝐷𝐷𝐴𝐴𝐴𝐴,1 =
=
= 0.198
1.0 atm
s
𝑃𝑃
at T1 = 273 K
𝜅𝜅𝜅𝜅
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
𝑇𝑇 =
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 273 K = 1.31
𝜀𝜀𝐴𝐴𝐴𝐴
𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵
447K 97K
∗
𝐴𝐴
𝐶𝐶
𝐸𝐸
𝐺𝐺
+
+
+
(𝑇𝑇 ∗ )𝐵𝐵 exp(𝐷𝐷𝑇𝑇 ∗ ) exp(𝐹𝐹𝑇𝑇 ∗ ) exp(𝐻𝐻𝑇𝑇 ∗ )
1.06036
0.19300
1.03587
1.76474
=
+
+
+
= 1.27
0.15610
(1.31)
exp(0.47635 ∙ 1.31) exp(1.52996 ∙ 1.31) exp(3.89411 ∙ 1.31)
Ω𝐷𝐷,2 =
Table J.1: DAB = 0.198
𝑐𝑐𝑐𝑐2
s
at T = 273 K and P = 1.0 atm
𝑃𝑃1 𝑇𝑇2 3/2 Ω𝐷𝐷,1
𝑐𝑐𝑐𝑐2 1 atm 373 K 3⁄2 1.12
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴,2 = 𝐷𝐷𝐴𝐴𝐴𝐴,1 � � � �
= �0.198
��
= 0.279
��
�
𝑃𝑃2 𝑇𝑇1
Ω𝐷𝐷,2
s
1 atm 273 K
1.27
s
Brokaw equation, scaled to 373K and 1.0 atm, DAB = 0.279
𝑐𝑐𝑐𝑐2
s
24-12
24.9
T = 1073 K
𝑃𝑃 = (1.5 × 105 Pa) �
1atm
� = 1.49 atm
1.01 × 105 Pa
a. Estimate DAB for SiCl4 in H2
A = SiCl4 and B = H2
𝑀𝑀𝐴𝐴 = 169.9
𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
, 𝑀𝑀𝐵𝐵 = 2.02
𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝜀𝜀
𝜀𝜀
From Appendix K, Table K.2, 𝜎𝜎𝐴𝐴 = 5.08 Å, 𝜎𝜎𝐵𝐵 = 2.968 Å, 𝐴𝐴 = 358 K, 𝐵𝐵 = 33.3 K
𝜎𝜎𝐴𝐴𝐴𝐴 =
𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 5.08 Å + 2.968 Å
=
= 4.02 Å
2
2
κ
κ
𝜅𝜅𝜅𝜅
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 1073 K = 9.83
𝜀𝜀𝐴𝐴𝐴𝐴
𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵
358K 33.3K
From Table K.1, interpolate to find ΩD = 0.7448
𝐷𝐷𝐴𝐴𝐴𝐴 =
=
1
1 1⁄2
0.001858 𝑇𝑇 3⁄2 �𝑀𝑀 + 𝑀𝑀 �
𝐴𝐴
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 2 Ω𝐷𝐷
𝐵𝐵
0.001858 ∙ 1073 K 3/2 �
169.9
1
1/2
1
𝑔𝑔 +
𝑔𝑔 �
2.02
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
2
(1.49 atm)�4.02 � (0.7448)
= 2.58
𝑐𝑐𝑐𝑐2
s
b. Estimate DA-m for SiCl4 in mixture gas
Find DAC for SiCl4 in HCl
A = SiCl4 , B = H2, C = HCl
𝑀𝑀𝐴𝐴 = 169.9
𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
, 𝑀𝑀𝐶𝐶 = 36.46
𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝜀𝜀
𝜀𝜀
From Appendix K, Table K.2, 𝜎𝜎𝐴𝐴 = 5.08 Å, 𝜎𝜎𝐶𝐶 = 3.305 Å, 𝐴𝐴 = 358 K, 𝐶𝐶 = 360 K
κ
κ
24-13
𝜎𝜎𝐴𝐴𝐴𝐴 =
𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐶𝐶 5.08 Å + 3.305 Å
=
= 4.19 Å
2
2
𝜅𝜅 𝜅𝜅 1/2
1
1 1⁄2
𝜅𝜅𝜅𝜅
= � ∙ � 𝑇𝑇 = �
∙
� ∙ 1073 K = 2.99
𝜀𝜀𝐴𝐴 𝜀𝜀𝐶𝐶
358K 360K
𝜀𝜀𝐴𝐴𝐴𝐴
From Table K.1, interpolate to find ΩD = 0.9499
𝐷𝐷𝐴𝐴𝐴𝐴 =
=
1
1 1⁄2
+
�
𝑀𝑀𝐴𝐴 𝑀𝑀𝐶𝐶
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 2 Ω𝐷𝐷
0.001858 𝑇𝑇 3⁄2 �
0.001858(1073K)3/2 �
1
1
𝑔𝑔 +
𝑔𝑔 �
169.90
36.46
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
2
(1.49 atm)�4.19 � (0.9499)
𝑦𝑦𝐴𝐴 = 0.4, 𝑦𝑦𝐵𝐵 = 0.4, 𝑦𝑦𝐶𝐶 = 0.2
𝐷𝐷𝐴𝐴−𝑀𝑀 =
𝑦𝑦 ′ 𝐵𝐵 =
1/2
1
𝑦𝑦′𝐵𝐵 𝑦𝑦′𝐶𝐶
+
𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴
0.4
1−0.4
𝐷𝐷𝐴𝐴−𝑀𝑀 =
,
𝑦𝑦′𝐵𝐵 =
= 0.667 , 𝑦𝑦 ′ 𝐶𝐶 =
1
𝑦𝑦𝐵𝐵
,
1 − 𝑦𝑦𝐴𝐴
0.2
1−0.4
0.667
0.333
+
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
2.58
0.480
s
s
𝑦𝑦′𝐶𝐶 =
= 0.480
𝑐𝑐𝑐𝑐2
s
𝑦𝑦𝐶𝐶
1 − 𝑦𝑦𝐴𝐴
𝑦𝑦𝐵𝐵
𝑦𝑦𝐶𝐶
,
=
+
1 − 𝑦𝑦𝐴𝐴
𝐷𝐷𝐴𝐴−𝑚𝑚
𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴
= 0.333
= 1.05
𝑐𝑐𝑐𝑐2
s
24-14
24.10
A = CO, B = H2
yA = 0.01, yB = 0.99
dpore = 15 × 10−7 cm, ε = 0.30, T = 673 K, P = 5.0 atm
MA = 28.01
a. Estimate cA
cA = yA c =
g
, MB = 2.02
gmole
yA P
=
RT
g
gmole
0.01 ∙ 5.0 atm
mol
= 9.054 × 10−7
3
cm ∙ atm
cm3
∙ 673 K
82.06
mol ∙ K
b. Estimate molecular diffusion coefficient, DAB
Hirschfelder Equation
1
3
2
1
1
0.001858 ∙ T 2 ∙ �M + M �
A
B
DAB =
2
P ∙ σAB ΩD
εA
= 110 K,
κ
εB
= 33.3 K,
κ
εAB
εA εB
= � ∙ = 60.52 K,
κ
κ κ
Use Appendix K. 1 and interpolate for ΩD = 0.7336
σAB =
σA + σB
3.590 Å + 2.968 Å
=
= 3.279 Å
2
2
3
DAB =
1
2
1
g +
g �
2.02
cm2
gmole
gmole
= 0.599
s
5.0 atm ∙ 3.279 Å2 ∙ 0.7336
0.001858 ∙ 673 K 2 ∙ �
28.01
1
κT
= 11.12
εAB
c. Estimate Knudsen diffusion and effective diffusion coefficient
T
673
cm2
−7
DKA = 4850 ∙ dpore ∙ �
= 4850 ∙ 15 × 10 ∙ �
= 0.0357
MA
28.01
s
24-15
1
1
1
1
1
= �
+
+
�=�
�
2
cm
cm2
DAe
DAB DKA
. 0357
0.599
s
s
DAe = 0.0337
cm2
s
′
𝐷𝐷𝐴𝐴𝐴𝐴
= 𝜀𝜀 2 DAe = (0.3)2 �0.0337
cm2
s
� = 3.03 x 10−3
cm2
s
Knudsen diffusion is important at current conditions. DAe ≈ DKA
1
When DKA = DAB , P =?
2
1
3
2
1
1
2
1 0.001858 ∙ T ∙ �MA + MB �
DKA = ∙
2
P ∙ σ2AB ΩD
1
2
3
1
1
0.001858 ∙ 673 K 2 ∙ �
g +
g �
28.01
2.02
cm2
1
mol
mol
= 0.0357
= ∙
2
s
P ∙ 3.279 Å2∙ 0.7336
∴ P = 42.0 atm
24-16
24.11
Given:
A = H2S, B = CH4
MA = 34.08 g/gmole, MB = 16.05 g/gmole
yA = 0.010,
yB = 0.99
T = 303 K, P = 15.0 atm
ε = 0.50, dpore = 20 nm = 20 ×10−7 nm
a. Molar concentration of H2S, cA
Ideal Gas Law
=
c A y=
Ac
( 0.010 )(15.0 atm ) = 6.033 × 10−6 gmole
yA P
=
RT 
cm3
cm3 ⋅ atm 
 82.06
 ( 303 K )
gmole ⋅ K 

b. Molecular diffusion coefficient of H2S−CH4, DAB
Hirschfelder Equation
σA = 3.623 Å, σB = 3.822 Å, Appendix K.2
σ AB =
σ A +σ B
2
=
3.623 Å+3.822 Å
=3.723 Å
2
εA
ε
ε
=301.1 K, B =136.5 K, AB = 301.1 K ⋅136.5 K = 202.7 K
κ
κ
κ
κT
303 K
=
=1.495 , Appendix K.1, ΩD = 1.198
ε AB 202.7 K
1/2
1/2
 1
1 
1
1

3/2 
+
0.001858T 
0.001858 ( 303) 
+


MA MB 
34.08 16.05 


=
2
ΩD
Pσ AB
(15.0 ) ( 3.723) 2 (1.198)
3/2
DAB
DAB = 0.0119
cm 2
s
24-17
Fuller Schettler Giddings Equation
Table 24.3
2 ⋅1.98 + 17.0 =
20.96
( Σv ) A =
( Σv )B= 16.5 + 4 ⋅1.98= 24.42
1/2
 1
1 
+
0.001T 

MA MB 

=
1/3
1/3 2
P ( Σv ) A + ( Σv ) B
1.75
DAB
(
)
0.001 ⋅ ( 303 )
1.75
1/2
1 
 1
+


 34.08 16.05 
(15.0 ) ( ( 20.96 ) + ( 24.42 ) )
1/3
1/3 2
cm 2
DAB = 0.0138
s
b. Effective diffusion coefficient of H2S in porous material, DAe
Knudsen diffusion coefficient
T
303
cm 2
= 4850 20 ×10−7
= 0.0289
34.08
s
MA
Effective diffusion coefficient






1
1
1
1
1
= 
+
+ =
2
2 
cm
cm 
DAe  DAB DKA  
0.0289
 0.0138

s
s 

cm 2
DAe = 9.37 ×10−3
s
Effective diffusion coefficient corrected for porosity
2s
cm 2 
2
'
−5 cm
=
×
DAe
ε 2 DAe =
=
2.34
10
( 0.050 )  9.37 ×10−3

s 
s

DKA
4850 d pore
(
)
24-18
24.12
A = SiH4, B = He
yA = 0.01, yB = 0.99
dpore = 10 × 10−4 cm, T = 900 K
MA = 32.12
g
, MB = 4.00
gmole
κ = 1.38 × 10−16 ergs/K
g
gmole
a. Determine wA
g
0.01 ∙ 32.12
yA ∙ MA
gmole
wA =
=
= 0.750
yA ∙ MA + yB ∙ MB 0.01 ∙ 32.12 g + 0.99 ∙ 4.00 g
gmole
gmole
b. Estimate DAB at P = 1.0 atm and P = 100 Pa, T = 900 K
P = 1.0 atm
εA
κ
= 207.6 K,
εB
κ
= 10.22 K ,
εAB
κ
ε
=� A∙
κ
εB
κ
= 46.06 K,
Using Appendix K. 1, interpolate to get ΩD = 0.6676
σA = 4.08 Å,
σB = 2.576 Å, σAB =
σA +σB
2
= 3.328 Å
κT
εAB
= 19.54
1
2
3
1
1
1
0.001858 ∙ 900 K 2 ∙ �
3
2
g +
g �
1
1
32.12
4.00
0.001858 ∙ T 2 ∙ �M + M �
gmole
gmole
A
B
DAB =
=
2
P ∙ σAB ΩD
1.0 atm ∙ 3.328 Å2 ∙ 0.6676
= 3.60
cm2
s
𝑃𝑃 = 100 Pa ∙
1 atm
= 9.87 × 10−4 atm
101325 Pa
24-19
1
2
3
1
1
1
0.001858 ∙ 900 K 2 ∙ �
3
2
g +
g �
1
1
32.12
4.00
0.001858 ∙ T 2 ∙ �
+
�
gmole
gmole
MA MB
DAB =
=
2
P ∙ σAB ΩD
9.87 × 10−4 atm ∙ 3.328 Å2 ∙ 0.6676
= 3645
cm2
s
c. Assess importance of Knudsen Diffusion
Pressure = 1.0 atm
J
∙ 900 K
K
λ=
=
= 1.66 × 10−7 cm
−10
2
2
√2πσA P √2π ∙ (4.08 × 10 m) 101325 Pa
1.38 × 10−23
κT
Kn =
λ
dpore
=
6.76×10−7 cm
10×10−4 cm
Pressure = 100 Pa
= 1.66 × 10−4 < 1, Knudsen diffusion is not important.
J
∙ 900 K
K
λ=
=
= 1.68 × 10−4 cm
√2πσA 2 P √2π ∙ (4.08 × 10−10 m)2 100 Pa
κT
Kn =
λ
dpore
=
1.38 × 10−23
1.68 × 10−4 cm
= 0.168,
10 × 10−4 cm
Knudsen diffusion plays moderate role
d. Gas velocity for Pe = 5.0 x 10-4
T
900 K
cm2
DKA = 4850 ∙ dpore ∙ �
= 4850 ∙ 10 × 10−4 cm ∙ �
=
25.67
g
MA
s
32.13
gmole
P = 1.0 atm
1
1
1
1
1
= �
+
+
�=�
�
2
cm2
cm
DAe
DAB DKA
3.60
25.67
s
s
DAe = 3.157
cm2
s
24-20
Pe =
v∞∙dpore
DAe
v∞ = 1.58
V̇ = v∞
1
= �
Pe =
cm
s
cm
πd2
cm3
= 1.58
∙ π ∙ (5 × 10−4 cm)2 = 1.24 × 10−6
s
4
s
P = 100 Pa
DAe
= 5 × 10−4
1
DAB
+
v∞∙dpore
DAe
1
DKA
= 5 × 10−4
∴ v∞ = 12.75
V̇ = 12.75
� , ∴ DAe = 25.49 cm2 /s
cm
s
cm π
cm3
−4
2
−5
∙ ∙ (10 × 10 cm) = 1.00 × 10
s 4
s
24-21
24.13
A = CH4, B = H2O
T = 673 K, dpore = 5 × 10−3 cm, v∞ = 4.0 cm/s
MA = 16.04
Pe =
g
, MB = 18.02
gmole
v∞ dpore
=2
DAe
g
gmole
a. Knudsen diffusion coefficient for CH4
T
673 K
2
DKA = 4850 ∙ dpore ∙ �
= 4850 ∙ 5 × 10−3 cm ∙ �
g = 157 cm /s
MA
16.04
gmole
DAB will dominate DAe since DKA is large
b. Total system pressure for Pe = 2.0
𝑃𝑃𝑃𝑃 =
𝑣𝑣∞ 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝐷𝐷𝐴𝐴𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 =
DAe =
1
=
DAe
𝑣𝑣∞ 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑃𝑃𝑃𝑃
cm
4.0 s ∙ 5 × 10−3 cm
2
cm2
= 0.01
s
1
1
1
1
1
= �
+
+
�=�
�
2
cm
DAB DKA
DAB 157 cm2 /s
0.01
s
DAB = 0.01
cm2
s
Fuller, Schettler, and Giddings Equation:
DAB =
1
1
0.001 ∙ T1.75 ∙ (M + M )1/2
A
B
1
1
P ∙ ((Σv)3A + (Σv)3B )2
24-22
(Σvi )A = 16.5 + 4 ∙ 1.98 = 24.42
(Σvi )B = 12.7
DAB =
P=
1
1/2
g +
g )
16.05
18.02
gmole
gmole
0.001 ∙ 673 K1.75 ∙ (
1
3
P((24.42)
0.001 ∙ 673 K1.75 ∙ (
1
1
1
+ (12.7)3 )2
1
1/2
g +
g )
16.04
18.02
gmole
gmole
1
1
cm2
0.01
∙ ((24.42)3 + (12.7)3 )2
s
= 0.01
cm2
s
= 111.6 atm
24-23
24.14
A = albumin, B = H2O
DAB = 5.94 × 10−7 cm2 /s, T = 293K, κ = 1.38 × 10−16 ergs/K
1 Pa ∙ s = 1
kg
m∙s
1 erg = 1 g·cm2/s2
μB (293 K) = 993 × 10−6 Pa ∙ s ∙ �
Stokes Einstein Equation:
DAB =
κT
6πrA μB
1000 g
1m
g
�∙�
� = 993 × 10−5
1kg
100 cm
cm ∙ s
κT
diameterA = rA ∙ 2 = 2 ∙
=2∙
6πDAB μB
= 7.27 nm
cm2
1.38 × 10−16 g ∙ 2
∙ 293K
s ∙K
g
cm2
6 ∙ π ∙ 5.94 × 10−7
∙ 993 × 10−5
s
cm ∙ s
This answer is close to the known value of 7.22 nm
24-24
24.15
Given:
A = O2 (dissolved solute), B = H2O (solvent)
T = 310 K
Solute: VA = 25.6 cm3/gmole (Table 24.4)
 1000 cP 
Solvent: ΦB = 2.6, MB = 18.02 g/gmole, µ B =7.083 ×10−4 Pa ⋅ s 
 =0.7083 cP
 1 Pa ⋅ s 
Wilke Chang Equation
DAB
7.4 ×10−8 (Φ B M B )1/2 T
=
VA0.6
µB
7.4 ×10−8 (2.6 ⋅18.02 )1/2
( 25.6 )
0.6
310
0.7083
D=
3.17 ×10−5 cm 2 / s
AB
Hayduk and Laudie Equation (for nonelectrolytes in liquid H2O)
DAB =
13.26 ⋅10−5 µ B−1.14VA−0.589 =
13.26 ×10−5 ( 0.7083)
−1.14
( 25.6 )
−0.589
DAB = 2.91×10−5 cm 2 /s
24-25
24.16
A = Benzene (C6H6, liquid), B = Ethanol (C2H6O, liquid)
MA = 78.11 g/gmole, MB = 46.07 g/gmole
Interpolate and convert Pa∙s to cP from Appendix I to find viscosity of liquid benzene at 288K
(59 F)
lbm 0.45359 kg
1 ft
μA (59℉) = 44.5 × 10−5
∙�
��
� = 0.662 × 10−3 Pa ∙ s = 0.662 cP
ft ∙ s
lbm
0.3048 m
From Perry’s Chemical Engineering Handbook, interpolate and convert units to cP:
μB (288 K) = 1.34 cP
Using Table 24.4 and 24.5
VA = 6 ∙ 14.8 + 6 ∙ 3.7 − 15 (ring) = 96.0
VB = 2 ∙ 14.8 + 6 ∙ 3.7 + 7.4 = 59.2
ΦA = 1.0, ΦB = 1.5
cm3
gmole
cm3
gmole
Wilke Chang correlation (liquid benzene in ethanol)
7.4 × 10−8 (ΦB MB )1/2 T
DAB =
∙
μB
VA 0.6
DAB =
7.4 × 10−8 (1.5 ∙ 46.07
96
3 0.6
cm
gmole
g 1/2
)
gmole
∙
288 K
= 8.55 × 10−6 cm2 /s
1.34 cP
Schiebel correlation (liquid benzene-solute in ethanol-solvent)
K
2
T
3VB 3
DAB = 1/3 ∙ , K = 8.2 × 10−8 ∙ (1 + �
� )
μB
VA
V
A
2
cm3 3
3
∙
59.2
K
288 K
⎛
gmole ⎞
−8
DAB =
∙ ⎜1 + �
� ⎟
1 ∙ 1.34 cP , K = 8.2 × 10
3
cm
3
96.0
cm 3
gmole
�96.0
�
gmole
⎝
⎠
DAB = 9.65 × 10−6 cm2 /s
24-26
Wilke Chang correlation (liquid ethanol in benzene)
DBA =
7.4 × 10−8 (ΦA MA )1/2 T
∙
μA
VB 0.6
7.4 × 10−8 (1.0 ∙ 78.11 g/mole)1/2 288 K
cm2
−5
DBA =
∙
= 2.46 × 10
59.2 cm3 /gmole0.6
0.662 cP
s
Schiebel correlation (liquid ethanol-solute in benzene-solvent)
Note VB < 2VA, benzene is the solvent
DBA =
DBA =
K
1/3
VB
∙
T
, K = 18.9 × 10−8
μA
18.9 × 10−8
59.2
288 K
1 ∙ 0.662 cP,
3
cm 3
gmole
DBA = 2.11 × 10−5 cm2 /s
DAB ≠ DBA
24-27
24.17
A = CuCl2 (an ionic solute), B = H2O
T = 298 K, ℱ = 96,500 C/gmole
R = 8.316
J
gmole ∙ K
λ°+ (Cu2+ ) = 108
λ°− (Cl− ) = 76.3
A ∙ cm2
V ∙ gmole
A ∙ cm2
V ∙ gmole
n+ (valence of cation) = 2
n− (valence of anion) = 1
Nernst-Haskell Equation
1
1
( + + − )RT
n
n
DAB =
1
1
( ° + ° )(ℱ)2
λ+ λ−
1 1
J
( + )(8.316
) ∙ 298K
cm2
2 1
gmole ∙ K
= 1.78 × 10−5
DAB =
1
1
s
(
+
)(96,500 C/gmole)2
A ∙ cm2
A ∙ cm2
108
76.3
V ∙ gmole
V ∙ gmole
24-28
24.18
For liquids
DAB µ B
=constant
T
 T  µ (T ) 
DAB (T2 ) = DAB (T1 )  2  B 1 
 T1  µ B (T2 ) 
2
cm 2  303K   1.14 cP 
−5 cm
=
×
DAB (T2 ) =
1.28 × 10−5
1.93
10



s  288 K   0.797 cP 
s
24-29
24.19
a. DAB for glucose in water
A = C6H12O6, B = H2O
Stokes-Einstein equation
DAB =
𝑟𝑟𝐴𝐴 =
κT
6π rA µ B
1
1
𝑑𝑑𝐴𝐴 = (0.86 𝑛𝑛𝑛𝑛) = 4.3 𝑥𝑥10−8 𝑐𝑐𝑐𝑐
2
2
Appendix I, µΒ = 0.00826 g cm ⋅ s at 303 K
)
(1.38 ×10 g ⋅ cm s ⋅ K ) ( 303 K=
D=
6.25 ×10
6π ( 4.3 ×10 cm ) ( 0.00826 g cm ⋅ s )
−16
AB
2
2
−8
−6
cm 2 s
b. DAe of glucose through the membrane
DAe = D o AB F1 (ϕ ) F2 (ϕ )
=
ϕ
ds
0.86 nm
=
= 0.287
d pore
3.0 nm
F1 (ϕ ) =
(1 − ϕ ) 2 =
(1 − 0.287 ) =0.508
2
F2 (ϕ ) =
1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5 =
1 − 2.104 ( 0.287 ) + 2.09 ( 0.287 ) − 0.95 ( 0.287 ) =
0.444
3
5
DAe =
1.41×10−6 cm 2 s
( 6.25 ×10−6 cm2 s ) ( 0.508)( 0.444 ) =
24-30
24.20
Given:
o
= 3.46 × 10−7 cm2/s
DAB
ds = 12.38 nm, dpore = 100 nm
Effective diffusion coefficient
Solute diffusion through solvent-filled pore
o
DAe = DAB
F1 (ϕ ) F2 (ϕ )
Reduced pore diameter
=
ϕ
ds
12.38 nm
=
= 0.124
100 nm
d pore
Correction factors F1(φ), F2(φ)
F1 (ϕ ) =
(1 − ϕ ) 2 =
(1 − 0.124 ) =0.767
2
F2 (ϕ ) =
1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5 =
1 − 2.104 ( 0.124 ) + 2.09 ( 0.124 ) − 0.95 ( 0.124 ) =
0.743
3
5
DAe = ( 3.46 × 10−7 cm 2 s ) ( 0.767 )( 0.743) = 1.97 × 10−7 cm 2 s
24-31
24.21
Find dpore when DAe = 5.0 x 10-7 cm2/s
DAe = D o AB F1 (ϕ ) F2 (ϕ )
DoAB =DAB = 1.19 x 10-6 cm2/s (dilute solution)
F1 (ϕ ) F2 =
(ϕ )
DAe
5.0 ×10−7 cm 2 s
=
= 0.420
DAB 1.19 ×10−6 cm 2 s
0.420 =
(1 − ϕ ) (1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5 )
2
Solve for ϕ
ϕ = 0.183
ϕ=
ds
d pore
=
d pore
3.6 nm
= 19.7 nm
0.183
24-32
24.22
a. Determine molecular diffusion coefficients of solute A and solute B in water at 293 K
Wilke-Chang Equation
DA− H 2O µ H 2O 7.4 ×10−8 (φ M H 2O )1/2
=
T
VA0.6
From Appendix I, µH2O = 9.93 x 10-4 Pa ⋅ s = 0.993 cP at 293 K
MH2O = 18.02 g/gmole
For solute A
DA − H 2 O =
7.4 x 10-8 (2.6×18.02 g/gmole)1/2
( 233.7 cm gmole )
3
0.6
( 293 K ) = 5.67 x 10-6 cm 2 s
( 0.993 cP )
DA− H2O = 5.67 x 10-6 cm2/s
For solute B
DB − H 2O =
7.4 x 10-8 (2.6×18.02 g/gmole)1/2
( 347.4 cm gmole )
3
0.6
( 293 K ) = 4.47 x 10-6 cm 2 s
( 0.993 cP )
DB − H2O = 4.47 x 10-6 cm2/s
b. Determine dpore to achieve a separation factor of α = 3
ϕ=
ds
d pore
DAe = D o A− H 2O F1 (ϕ ) F2 (ϕ )
F1 (ϕ )= (1 − ϕ ) 2
=
α
F2 (ϕ ) =
1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5
DAe DA− H2O F1 (ϕ A ) F2 (ϕ A )
=
DBe DB − H2O F1 (ϕ B ) F2 (ϕ B )
24-33
ϕA =
2.0 nm
d pore
ϕB =
3.0 nm
d pore
DA− H2O (1 − ϕ A ) (1 − 2.104ϕ A + 2.09ϕ A3 − 0.95ϕ A5 )
2
α=
α=
DB − H2O (1 − ϕ B ) (1 − 2.104ϕ B + 2.09ϕ B 3 − 0.95ϕ B 5 )
2
2
3
5
 2.0 nm  
 2.0 nm 
 2.0 nm 
 2.0 nm  
DA − H 2 O  1 −
 1 − 2.104 
 + 2.09 
 − 0.95 
 

d
d
d
d

pore 

 pore 
 pore 
 pore  

2
3
5
 3.0 nm  
 3.0 nm 
 3.0 nm 
 3.0 nm  
DB − H2O 1 −
 1 − 2.104 
 + 2.09 
 − 0.95 
 

d
d
d
d

pore
pore
pore
pore

 





 
Trial-and-error: guess dpore and solve for α
Guess dpore = 7.5 nm
( 5.67 ×10
α=
−6
2
3
5
 2.0 nm  
 2.0 nm 
 2.0 nm 
 2.0 nm  
cm s ) 1 −
 1 − 2.104 
 + 2.09 
 − 0.95 
 
 7.5 nm  
 7.5 nm 
 7.5 nm 
 7.5 nm  
2
2
3
5
 3.0 nm  
 3.0 nm 
 3.0 nm 
 3.0 nm  
2
−6
×
−
−
+
4.47
10
cm
s
1
1
2.104
2.09
−
0.95
(
)  7.5 nm  





 
 7.5 nm 
 7.5 nm 
 7.5 nm  

α = 3.20
Guess dpore = 7.88 nm
( 5.67 ×10
α=
−6
2
3
5
2.0 nm  

 2.0 nm 
 2.0 nm 
 2.0 nm  
cm s ) 1 −
 1 − 2.104 
 + 2.09 
 − 0.95 
 
 7.88 nm  
 7.88 nm 
 7.88 nm 
 7.88 nm  
2
2
3
5
3.0 nm  

 3.0 nm 
 3.0 nm 
 3.0 nm  
−6
2
×
−
−
+
4.47
10
cm
s
1
1
2.104
2.09
−
0.95
(
)  7.88 nm  





 
 7.88 nm 
 7.88 nm 
 7.88 nm  

α = 3.00
To achieve a separation factor (α) of 3, dpore = 7.88 nm
24-34
24.23
a. CA
=
C A y=
yA
AC
C A = ( 0.05 )
P
RT
(1.5 atm )

-5 m ⋅ atm 
 8.206×10
 ( 353K )
gmole ⋅ K 

3
=2.58
gmole A
m3
'
b. DAe
at T = 80 oC (353 K) and 1.5 atm
 Pref 
cm 2  1.0 atm 
cm 2
=
DAB (T , P) D=
=
 0.10
AB (T , Pref ) 

 0.0.067
s  1.5 atm 
s
 P 
DKA =
4850 d pore
T
353
cm 2
−5
4850(1.5 × 10 )
0.202
=
=
MA
46
s
1
1
1
=
+
=
DAe DKA DAB
1
0.202
cm
s
2
+
1
0.067
cm 2
s
cm 2
DAe = 0.050
s

cm 2 
cm 2
'
2
DAe
DAe (0.5) 2  0.05 =
=
ε=
0.0125

s 
s

24-35
24.24
A = solute, B = solvent
Find solvent viscosity to achieve a D'Ae of 1.0 x 10-7 cm2/s
DAB = 2.04 x 10-7 cm2/s
D ' Ae = ε 2 DAe
D ' Ae 1.0 × 10 −7 cm 2 s
=
=
= 2.04 × 10 −7 cm 2 s
D
Ae
2
2
ε
( 0.70 )
For hindered diffusion of a solute in a solvent-filled pore
DAe = D o AB F1 (ϕ ) F2 (ϕ )
Hindered diffusion of the solute within the pore can be neglected so DAe ≅ DAB
Stokes-Einstein Equation
DAB =
κT
6π rA µ B
rA = 10 nm = 1 x 10-6 cm
Rearrange to solve for µΒ
K)
(1.38 ×10 g ⋅ cm s ⋅ K ) ( 303
=
=
= 0.0109 g=
cm ⋅ s 0.00109 Pa ⋅ s
µ
6π D r
6π (1×10 cm )( 2.04 ×10 cm s )
κT
−16
−6
B
2
2
−7
2
AB A
24-36
24.25
a. CB inside pipe
=
CB y=
yB
BC
CA
P
RT
( 3.0 atm )
gmole B
0.20 )
25.2
(=
3
m3

−5 m ⋅ atm 
K
×
8.206
10
290
)

(
gmole ⋅ K 

b. DAB inside pipe
Fuller Schettler Giddings Equation
Table 24.3
+ 24.92
( Σv=
) A 16.5 + 4 ⋅ 1.98=
( Σv )B = 2 ⋅ 16.5 + 6 ⋅ 1.98 = 44.88
1/ 2
 1
1 
+
0.001T 

MA MB 

=
1/ 3
1/ 3 2
P ( Σv ) A + ( Σv ) B
1.75
DAB
(
DAB = 0.0505
)
1/2
0.001 ⋅ ( 290 )
1.75
1 1 
 + 
 16 30 
( 3.0 ) ( ( 24.42 ) + ( 44.88) )
1/3
1/3 2
cm 2
s
 Pref 
cm 2  1.0 atm 
cm 2
DAB (T , P) D=
=
=
 0.10
AB (T , Pref ) 

 0.0.067
s  1.5 atm 
s
 P 
'
c. DAe
in porous layer
 Pref 
cm 2  3.0 atm 
cm 2
=
DAB (T , P ) D=
T
P
=
(
,
)
0.505
0.151

AB
ref 


s  1.0 atm 
s
 P 
T
290
cm 2
=
DKA 4850
=
d pore
4850(0.0010)
= 20.6
MA
16
s
24-37
1
1
1
=
+
=
DAe DKA DAB
DAe = 0.150
1
1
+
2
cm
cm 2
20.6
0.151
s
s
cm 2
s

cm 2 
cm 2
D
DAe (0.3)  0.150 =
=
ε=
 0.0135
s 
s

'
Ae
2
2
24-38
24.26
A = carbon, B = solid iron
DAB for carbon diffusing into fcc iron and bcc iron at 1000 K
Arrhenius equation
 Q 
=
DAB Do exp  −

 RT 
DAB for carbon diffusing into fcc iron at 1000 K
From Table 24.8, Do = 2.5 mm2/s and Q = 144.2 kJ/gmole = 144200 J/gmole



mm 2 
144200 J/gmole
7.34 ×10−8 mm 2 / s =
7.34 ×10−10 cm 2 / s
 =
 exp  −
s 
 ( 8.314 J/gmole ⋅ K )(1000 K ) 
DAB =
 2.5

DAB for carbon diffusing into bcc iron at 1000 K
From Table 24.8, Do = 2.0 mm2/s and Q = 84.1 kJ/gmole = 84100 J/gmole



mm 2 
84100 J/gmole
8.09 ×10−5 mm 2 / s =
8.09 ×10−7 cm 2 / s
DAB =
 =
 2.0
 exp  −
s 

 ( 8.314 J/gmole ⋅ K )(1000 K ) 
24-39
24.27
a. Comparison of diffusion coefficients
For As-Si, Do = 0.658 cm2/sec, Q = 348.1 kJ/gmole (348,100 J/gmole)
For P-Si, Do = 11.1 cm2/sec, Q = 356.2 kJ/gmole (356,200 J/gmole)
DAB Do exp(−Q / RT )
=
At T = 600 oC = 873 K
For As(A)-Si(B),
DAB = ( 0.658cm 2 /sec ) (exp(-348,100 J/gmole)/((8.314J/gmole-K)(873 K))) =9.76 x10-22cm 2 /sec
For P(A)-Si(B),
DAB = (11.1 cm 2 /sec ) (exp(-356,200 J/gmole)/((8.314J/gmole-K)(873 K))) =5.39 x 10-21cm 2 /sec
At T = 1000 oC = 1273 K
For As(A)-Si(B),
DAB = ( 0.658cm 2 /sec ) (exp(-348,100 J/gmole)/((8.314J/gmole-K)(1273 K))) =3.42 x 10-15cm 2 /sec
For P(A)-Si(B),
DAB = (11.1 cm 2 /sec ) (exp(-356,200 J/gmole)/((8.314J/gmole-K)(1273 K))) =2.69 x 10-14cm 2 /sec
At 600 oC, the diffusion coefficients are ~106 lower relative to 1000 oC
b. At what temperature might DAs-Si = 0.12 DP-Si?
Let
Do , As − Si exp(−QAs − Si / RT=
) 0.12 Do ,P − Si exp(−QP − Si / RT )
 0.12 Do ,P − Si 
QP − Si − QAs − Si
= ln 

 D
RT
o , As − Si


T
( 356,200-348,100 ) J/gmole = 1382 K
QP − Si − QAs − Si
=
 0.12 Do ,P − Si 
8.314 J  0.12×11.1 cm 2 /sec 
ln 
R ln 


2
 D
o , As − Si

 gmole-K  0.658 cm /sec 
24-40
25.1

∂ρ
∇ ⋅ n A + A − rA = 0
∂t
If ρ and DAB are assumed constant,


nA =
− DAB ∇ρ A + ρ A v


∇ n A = − DAB ∇ 2 ρ A + ∇ ⋅ ρ A v
Substitution yields
 ∂ρ
− DAB ∇ 2 ρ A + ∇ ⋅ ρ A v + A = rA
∂t
25-1
25.2
impermeable barrier for A
y=H
y
x
Control Volume
still mixture of A and B
(initially uniform conc. at cA = cAo = 0)
cAs
y=0
x=0
CAs
x=L
a. Assumptions
1.
2.
3.
4.
5.
Unsteady state process - constant source for A, but control volume is sink for A
No reaction of A in control volume (RA = 0)
2-D flux in x, y directions
Dilute with respect to A
No bulk flow in control volume
b. Differential Equation for Mass Transfer
Rectangular Coordinate System
Flux Equation
By Assumption 4
∂c
∂c
c
− DAB A + A ( N A, x + N B , x )  − DAB A
N A, x =
∂x
∂x
c
∂c
∂c
c
− DAB A + A ( N A, y + N B , y )  − DAB A
N A, y =
∂y
∂y
c
∂c
∇N A + RA =A
∂t
Assumptions 1-3
∂N A, x
∂x
+
∂N A, y
∂c A
=
∂y
∂t
Assumption 4 CA(x,y)
25-2
∂ 
∂c A  ∂ 
∂c A  ∂c A
=
 − DAB
 +  − DAB
∂x 
∂x  ∂y 
∂y  ∂t
 ∂ 2c
∂ 2c  ∂c A
DAB  2A + 2A  =
∂y  ∂t
 ∂x
c. Boundary Conditions
y= 0, 0 ≤ x ≤ L, c A ( x,0)= c As
y H , 0 ≤ x < L,
=
∂c A ( x, H )
= 0
∂y
∂c A ( 0, H )
= 0
∂x
x= L, 0 ≤ y < H , c A ( L, y=
) cAs
x= 0, 0 ≤ y ≤ H ,
25-3
25.3
a. Diagram
liquid MEK
vx(y)
x=0
thin polymer film
x=L
exiting MEK
+ dissolved polymer
b. Assumptions
1.
2.
3.
4.
5.
Steady-state process
No homogenous reaction, RA = 0
Dilute with respect to solute A
No fluid motion in y-direction (vy = 0)
Bulk flow dominates in the x-direction
c. Flux Equation
− DAB
N A, y =
∂C A
∂C
+ v yC A =
− DAB A (vy = 0)
∂y
∂y
  y 2 
∂C A
N A, x =
− DAB
+ v x (y) CA with =
v x ( y ) v max 1 −   
 H 
∂x


d. Differential Equation for Mass Transfer
Rectangular Coordinate System
∂N A, y ∂N A, y 
 ∂N
∂C A
−  A, x +
+
 + RA =
∂y
∂y 
∂t
 ∂x
By assumptions 1 and 2
∂N A, x
∂x
+
∂N A, y
∂y
=
0
25-4
∂ 
∂C A
∂C A 
 ∂ 
0
+ v x (y)  +  − DAB
=
 − DAB
∂x 
∂x
∂y 
 ∂y 
 ∂ 2C A ∂ 2C A 
∂ CA
DAB 
+
- v x (y)
=
0
2
2 
∂y 
∂x
 ∂x
  y 2  ∂ CA
 ∂ 2C A ∂ 2C A 
DAB 
0
+
=
 - v max 1 −   
2
∂y 2 
 ∂x
  H   ∂x
If bulk flow dominates in the x-direction, the final model is
  y 2  ∂ CA
∂ 2C A
DAB
- v max 1 −   
0
=
  H   ∂x
∂y 2


e. Boundary Conditions
x = 0, 0 < y < H, CA(0,y) = 0
y = 0, 0 < x < L, CA(x,0) = CA*
y = H, 0 < x < L,
∂C A (x,0)
=0
∂y
25-5
25.4
a. General differential equation for O2 transfer in liquid film in terms of the fluxes
Assumptions:
1) Steady-state
2) The process is dilute
3) Falling liquid film has a flat velocity profile with velocity vmax
4) Gas space always contains 100% oxygen,
5) W >> L
6) No homogenous reaction of O2 (RA = 0)
7) δ is constant
Mass transfer only in x-direction and z-direction (rectangular coordinates)
∂c A  ∂N A, x ∂N A, y ∂N A, z 
=
+
+
RA

=
∂y
∂t
∂z 
 ∂x
∂N A, x
+
∂N A, z
0
=
∂x
∂z
General flux equation for O2 in liquid film
∂c
∂c
N A, x =
− DAB A + vx∗c A = − DAB A
(x-direction)
∂x
∂x
∂c
∂c
N A, z =
− DAB A + vz∗c A =
− DAB A + vmax c A (z-direction)
∂z
∂z
𝜕𝜕𝑐𝑐𝐴𝐴
If 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 ≫ −𝐷𝐷𝐴𝐴𝐴𝐴
, then N A, z = vmax c A
𝜕𝜕𝜕𝜕
b. Differential equation in terms of the oxygen concentration cA
Insert flux equations into differential equation for mass transfer
∂c 
∂c



∂  − DAB A  ∂  − DAB A + vmax c A 
∂x 
∂z

=
+ 
0
∂x
∂z
 ∂ 2cA ∂ 2cA 
∂c
0
DAB  2 + 2  − vmax A =
∂z 
∂z
 ∂x
c. Boundary conditions for the oxygen mass-transfer process
2 boundary conditions on x-direction, 1 boundary condition on z-direction
25-6
At z= 0, 0 ≤ x ≤ δ , c A ( x, 0)= 0
At x = 0, 0 ≤ z ≤ L, c A (0, z ) = c As = c∗A = p A / H
At x = δ , 0 ≤ z ≤ L, N A (δ , z ) = 0 ∴
∂c A (δ , z )
= 0
∂x
25-7
25.5
bulk gas phase
z = L = 2.0 cm
0.01 cm diameter
channels
catalytic surface
A (g) → B (g)
z=0
inert support
a. Assumptions and simplification of General Differential Equation for Mass Transfer
Physical system: gas space inside pore
Source for A: bulk gas phase
Sink for A: reaction of A to B at catalyst surface on cylindrical walls of pore
Assumptions:
1. Constant source and sink – steady-state process
2. No homogeneous reaction of A within control volume
3. Rapid heterogeneous surface reaction so that flux of A to catalyst surface is diffusion limited
with cAs = 0 at r = R for all z
4. Two-dimensional flux in r and z direction, given orientation of source and sink for A
5. Binary mixture of A and B, not dilute
yA∞ = 0.60
yB∞ = 0.40
z = L = 2.0 cm
cAs = 0
CA(r,z)
impermeable
barrier
z=0
r = R = 0.005 cm
r=0
25-8
General Differential Equation for Mass Transfer in terms of NA
∂c
∇N A + RA =A
∂t
∂c A
By assumptions 1 and 2, =
RA 0
0,=
∂t
∂N A, z
1 ∂
0
rN A,r ) +
=
(
r ∂r
∂z
b. General Flux Equations and Differential Equation for Mass Transfer in terms of cA
∂c
c
− DAB A + A ( N A,r + N B ,r )
N A, r =
∂r
c
Note N A,r = − N B ,r
∂c A
∂r
∂c
N A,z = − DAB A
∂z
N A,r = − DAB
Insert flux equations into differential equation for mass transfer
1 ∂ 
∂c A  ∂ 
∂c A  
0
−
 −rDAB
 +  − DAB
 +0=
∂r  ∂z 
∂z  
 r ∂r 
 ∂ 2 c A 1 ∂c A ∂ 2 c A 
0
DAB  2 +
+ 2 =
r ∂r
∂z 
 ∂r
c. Boundary conditions for gas phase concentration cA
z= L, 0 ≤ r ≤ R, c A ( r , L=
) c A, ∞
bulk gas
∂c A ( r , 0 )
= 0
impermeable barrier
∂z
∂c A ( 0, z )
r= 0, 0 ≤ z ≤ L,
= 0
symmetry condition
∂r
r = R, 0 ≤ z ≤ L, c A ( R, z ) = c As = 0 rapid surface reaction
z= 0, 0 ≤ r ≤ R
25-9
25.6
a. General Differential Equation for Mass Transfer
Coordinate System: x, y, z
Assumptions:
1) Rapid reaction, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0
2) Gas velocity profile is flat, 𝑣𝑣𝑥𝑥 (𝑦𝑦) = 𝐯𝐯
3) Gas velocity (𝑣𝑣) is slow (doesn’t dominate over diffusion flux in “x”)
𝜕𝜕𝑐𝑐
4) Steady state process, constant source and sink for CO, 𝐴𝐴 = 0
𝜕𝜕𝜕𝜕
5) RA = 0 (no homogeneous reaction
6) 2-D flux in x and y
7) Dilute with respect to A
Source: Inlet gas
Sink: Catalyst surface
−∇𝑁𝑁𝐴𝐴 + 𝑅𝑅𝐴𝐴 =
𝑁𝑁𝐴𝐴,𝑥𝑥 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑁𝑁𝐴𝐴,𝑦𝑦 = −𝐷𝐷𝐴𝐴𝐴𝐴
∴
𝜕𝜕𝑐𝑐𝐴𝐴
,
𝜕𝜕𝜕𝜕
∴ −�
𝜕𝜕𝑐𝑐𝐴𝐴
+ 𝑐𝑐𝐴𝐴 ∙ 𝑣𝑣
𝜕𝜕𝜕𝜕
𝜕𝜕𝑁𝑁𝐴𝐴,𝑥𝑥 𝜕𝜕𝑁𝑁𝐴𝐴,𝑦𝑦
+
�=0
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝑐𝑐𝐴𝐴
𝜕𝜕𝜕𝜕
𝜕𝜕
𝜕𝜕𝑐𝑐𝐴𝐴
𝜕𝜕
𝜕𝜕𝑐𝑐𝐴𝐴
�−𝐷𝐷𝐴𝐴𝐴𝐴
+ 𝑐𝑐𝐴𝐴 ∙ 𝑣𝑣� +
�−𝐷𝐷𝐴𝐴𝐴𝐴
�=0
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝐷𝐷𝐴𝐴𝐴𝐴 �
𝜕𝜕 2 𝑐𝑐𝐴𝐴 𝜕𝜕 2 𝑐𝑐𝐴𝐴
𝜕𝜕𝑐𝑐𝐴𝐴
+
∙ 𝑣𝑣 = 0
�−
2
2
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
b. Boundary Conditions
x = 0, 𝑐𝑐𝐴𝐴 (0, 𝑦𝑦) = 𝑐𝑐𝐴𝐴0
x = L, 𝑐𝑐𝐴𝐴 (L,y) = 𝑐𝑐𝐴𝐴𝐴𝐴
y=0,
y=
𝐻𝐻
2
,
𝑐𝑐𝐴𝐴 (𝑥𝑥, 0) ≅ 0
𝐻𝐻
𝜕𝜕𝜕𝜕𝐴𝐴 (𝑥𝑥, 2 )
=0
𝜕𝜕𝜕𝜕
25-10
25.7
𝑘𝑘1
𝐴𝐴 → 𝐵𝐵, 𝑅𝑅𝐴𝐴 = −𝑘𝑘1 𝑐𝑐𝐴𝐴 ,
𝑘𝑘𝑠𝑠
𝐴𝐴 → 2𝐶𝐶, 𝑟𝑟𝐴𝐴,𝑠𝑠 = −𝑘𝑘𝑠𝑠 𝑐𝑐𝐴𝐴𝐴𝐴 ,
1
𝑘𝑘1 [ ]
𝑠𝑠
𝑘𝑘𝑠𝑠 [
𝑐𝑐𝑐𝑐
]
𝑠𝑠
4 species: A=1, B=2, C=3, D=4 (inert)
( 𝑅𝑅1 = −𝑘𝑘1 𝑐𝑐1 , 𝑟𝑟1,𝑠𝑠 = −𝑘𝑘𝑠𝑠 𝑐𝑐1𝑠𝑠 )
a. Assumptions, Source and Sink for species 1 (A)
Assumptions:
1) Dilute process with respect to species 1
2) Constant source and sink for species 1 (steady state)
3) Right side and top side are impermeable barriers
4) 2-D flux of species 1 for catalyst II (source and sink antiparallel)
Source: Flowing fluid containing reactant 1, of constant concentration (𝑐𝑐1,∞ )
Sinks: First-order homogeneous reaction of 1 in porous layer (catalyst I)
First-order heterogeneous surface reaction of 1 at nonporous boundary
surface (catalyst II)
b. Differential model for C1(x,y) (Shell balance on differential volume element for species 1)
IN – OUT + GEN = ACC = 0
𝑛𝑛1,𝑥𝑥 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑖𝑖𝑖𝑖 𝑥𝑥 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑤𝑤 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑖𝑖𝑖𝑖 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
[𝑛𝑛1,𝑥𝑥 ∆𝑦𝑦𝑦𝑦|𝑥𝑥,𝑦𝑦� + 𝑛𝑛1,𝑦𝑦 ∆𝑥𝑥𝑥𝑥|𝑥𝑥̅ ,𝑦𝑦 ] − [𝑛𝑛1,𝑥𝑥 ∆𝑦𝑦𝑦𝑦|𝑥𝑥+∆𝑥𝑥,𝑦𝑦� + 𝑛𝑛1,𝑦𝑦 ∆𝑥𝑥𝑥𝑥|𝑥𝑥̅ ,𝑦𝑦+∆𝑦𝑦 ] + 𝑟𝑟1 ∆𝑥𝑥∆𝑦𝑦𝑦𝑦 = 0
÷ ∆𝑥𝑥, ∆𝑦𝑦; 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 lim ∆𝑥𝑥 → 0, 𝑡𝑡𝑎𝑎𝑘𝑘𝑘𝑘 lim ∆𝑦𝑦 → 0 ; ÷ 𝑤𝑤
25-11
−
𝜕𝜕𝑛𝑛1𝑦𝑦
𝜕𝜕𝑛𝑛1,𝑥𝑥
+−
+ 𝑟𝑟1 = 0
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
General Flux Equation
x-direction: 𝑛𝑛1,𝑥𝑥 = −𝐷𝐷1
y-direction: 𝑛𝑛1,𝑦𝑦 = −𝐷𝐷1
𝜕𝜕𝐶𝐶1
𝜕𝜕𝜕𝜕
𝜕𝜕𝐶𝐶1
𝜕𝜕𝜕𝜕
(𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ≈ 0, 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)
(𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ≈ 0, 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)
Homogeneous reaction: 𝑟𝑟1 = −𝑘𝑘1 𝑐𝑐1
Combine: 𝐷𝐷1 �
𝜕𝜕2 𝐶𝐶1
𝜕𝜕𝑥𝑥
2 +
𝜕𝜕2 𝐶𝐶1
𝜕𝜕𝑦𝑦 2
� − 𝑘𝑘1 𝑐𝑐1 = 0, 𝑐𝑐1 (𝑥𝑥, 𝑦𝑦)
c. Boundary Conditions, 4 required: (2 for x, 2 for y)
𝑦𝑦 = 0, 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿,
𝑐𝑐1 (𝑥𝑥, 0) = 𝑐𝑐1,∞
𝑦𝑦 = 𝐻𝐻, 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿,
𝑛𝑛1 (𝑥𝑥, 𝐻𝐻) = 0,
𝑥𝑥 = 0, 0 < 𝑦𝑦 < 𝐻𝐻,
𝑟𝑟1,𝑠𝑠 = 𝑛𝑛1,𝑠𝑠 ,
𝑥𝑥 = 𝐿𝐿, 0 < 𝑦𝑦 < 𝐻𝐻,
𝑛𝑛1 (𝐿𝐿, 𝑦𝑦) = 0
∴
𝜕𝜕𝑐𝑐1 (𝑥𝑥, 𝐻𝐻)
=0
𝜕𝜕𝜕𝜕
−𝑘𝑘𝑠𝑠 𝑐𝑐1,𝑠𝑠 = −𝐷𝐷1
∴
𝜕𝜕𝑐𝑐1 (𝐿𝐿,𝑦𝑦)
𝜕𝜕𝜕𝜕
𝜕𝜕𝑐𝑐1 (0, 𝑦𝑦)
,
𝜕𝜕𝜕𝜕
=0
∴
𝜕𝜕𝑐𝑐1 (0, 𝑦𝑦) 𝑘𝑘𝑠𝑠
=
𝑐𝑐 (0, 𝑦𝑦)
𝜕𝜕𝜕𝜕
𝐷𝐷1 1
25-12
25.8
well-mixed flowing bulk fluid (cAo)
A
dead
tissue (B)
A
clump of
unhealthy
cells
inert surface
D
A→D
r = R1
r = R2
a. Define physical system and assumptions
Physical system: dead tissue between bulk fluid and unhealthy cells
Source for A: bulk liquid phase
Sink for A: reaction of A inside dead cells, flux of A into unhealthy cells
1.
2.
3.
4.
5.
6.
Constant source and sink – steady-state process
First-order homogeneous reaction of species A within dead tissue (control volume)
Dilute process with respect to species A and D
One-dimensional flux in r direction, given orientation of source and sink for species A
R1 and R2 constant
No reaction of A at boundary surface (r = R1)
b. Differential Forms of General Flux Equation for species A
Fick’s Flux Equation
N A , r = DA − m
dc A c A
+ ( N A, r + N B , r + N D , r )
dr
c
By assumption 3, dilute process with c A / c → 0 , also NB,r = 0
dc
N A,r = DAB A
dr
c. General Differential Equation for Mass Transfer in terms of NA and cA
25-13
−∇N A + RA =
dc A
dt
By assumptions 1 and 2,
−
(
∂c A
= 0, RA = −k c A
∂t
)
d 2
r N A, r − k c A =
0 dr
Insert simplified form of General Flux Equation
dc A 
1 d  2
− 2
0
 −r DAB
 − k cA =
r dr 
dr 
 d 2 c A 2 dc A 
DAB  2 +
0
 − k cA =
r dr 
 dr
d. Boundary conditions for species A and D
=
r R=
c Ao
2 , cA
bulk fluid
=
r R=
c As
1 , cA
active concentration for drug A at interface to unheathy tissue
=
r R=
0
2 , cD
bulk fluid
dcD
0
=
r R=
1,
dr
impermeable barrier to D
25-14
25.9
a. System for Analysis and Assumptions
System: Tissue in quadrant IV (A = dissolved O2, B = tissue)
Assumptions:
1) Steady state O2 transport, constant source and sink
2) 2-D flux along x and y, via orientation of source and sink
3) Homogeneous first-order reaction 𝑅𝑅𝐴𝐴 = −𝑘𝑘1 𝑐𝑐𝐴𝐴
4) Dilute solution, tissue acts as a homogeneous medium approximating water
5) Constant dimensions (𝐿𝐿1 , 𝐿𝐿2 , 𝐻𝐻1 , 𝐻𝐻2 , 𝑑𝑑), constant 𝐷𝐷1
Sources for O2: O2 gas in duct, liquid surrounding monolith
Sink for O2: homogeneous O2 consumption within tissue
Differential Model (System IV, rectangular geometry, mass concentration equation)
−�
−�
𝜕𝜕𝑁𝑁𝐴𝐴,𝑥𝑥 𝜕𝜕𝑁𝑁𝐴𝐴,𝑦𝑦 𝜕𝜕𝑁𝑁𝐴𝐴,𝑧𝑧
𝜕𝜕𝑐𝑐𝐴𝐴
+
+
� + 𝑅𝑅𝐴𝐴 =
=0
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝑁𝑁𝐴𝐴,𝑥𝑥 𝜕𝜕𝑁𝑁𝐴𝐴,𝑦𝑦
+
� − 𝑘𝑘1 𝑐𝑐𝐴𝐴 = 0
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
General Flux Equation, Species A
𝑁𝑁𝐴𝐴,𝑥𝑥 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑁𝑁𝐴𝐴,𝑦𝑦 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝜕𝜕𝑐𝑐𝐴𝐴
,
𝜕𝜕𝜕𝜕
𝑑𝑑𝑑𝑑𝐴𝐴
,
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠:
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠:
𝑐𝑐𝐴𝐴
≅0
𝑐𝑐
𝑐𝑐𝐴𝐴
≅0
𝑐𝑐
Combine mass conservation equation and general flux equation
𝜕𝜕 2 𝑐𝑐𝐴𝐴 𝜕𝜕 2 𝑐𝑐𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 � 2 +
� − 𝑘𝑘 𝑐𝑐𝐴𝐴 = 0,
𝜕𝜕𝑥𝑥
𝜕𝜕𝜕𝜕 2
b. Boundary Conditions
(𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 𝐼𝐼 − 𝐼𝐼𝐼𝐼)
Symmetry between quadrants I and II & III and IV
System IV
𝑥𝑥 = 0,
0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,
𝑁𝑁𝐴𝐴 (0, 𝑦𝑦) = 0,
∴
𝜕𝜕𝜕𝜕𝐴𝐴 (0, 𝑦𝑦)
=0
𝜕𝜕𝜕𝜕
25-15
𝑥𝑥 = 𝐿𝐿,
0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,
𝑦𝑦 = 𝐻𝐻1 ,
0 ≤ 𝑥𝑥 ≤ 𝐿𝐿1 ,
𝑦𝑦 = 0,
0 ≤ 𝑥𝑥 ≤ 𝐿𝐿1 ,
𝑐𝑐𝐴𝐴 (𝐿𝐿, 𝑦𝑦) =
𝑃𝑃𝐴𝐴
= 𝑐𝑐𝐴𝐴∗
𝐻𝐻
𝑐𝑐𝐴𝐴 (𝑥𝑥, 0) = 𝑐𝑐𝐴𝐴,∞
𝑃𝑃𝐴𝐴
𝑐𝑐𝐴𝐴 (𝑥𝑥, 𝐻𝐻1 ) =
= 𝑐𝑐𝐴𝐴∗
𝐻𝐻
System III
𝑥𝑥 = 𝐿𝐿1 + 𝑑𝑑,
𝑥𝑥 = 𝐿𝐿2 ,
𝑦𝑦 = 0,
𝑦𝑦 = 𝐻𝐻1 ,
0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,
0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,
𝐿𝐿1 + 𝑑𝑑 ≤ 𝑥𝑥 ≤ 𝐿𝐿2 ,
𝑐𝑐𝐴𝐴 (𝐿𝐿1 + 𝑑𝑑, 𝑦𝑦) =
𝜕𝜕𝜕𝜕𝐴𝐴 (𝐿𝐿2 , 𝑦𝑦)
=0
𝜕𝜕𝜕𝜕
𝐿𝐿1 + 𝑑𝑑 ≤ 𝑥𝑥 ≤ 𝐿𝐿2 ,
𝑃𝑃𝐴𝐴
= 𝑐𝑐𝐴𝐴∗
𝐻𝐻
𝑐𝑐𝐴𝐴 (𝑥𝑥, 0) = 𝑐𝑐𝐴𝐴,∞
𝑐𝑐𝐴𝐴 (𝑥𝑥, 𝐻𝐻1 ) =
𝑃𝑃𝐴𝐴
= 𝑐𝑐𝐴𝐴∗
𝐻𝐻
25-16
25.10
A = C6H6, B = H2O (liq)
T = 298 K, 𝑐𝑐𝐴𝐴,∞ 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
a. Differential model for cA(r,t)
Assumptions:
1) Unsteady state (control volume is also the sink)
2) No homogeneous reaction (𝑅𝑅𝐴𝐴 = 0)
3) Dilute with respect to A
4) 1-D flux along r (unimolecular diffusion)
General Differential Equation
′
𝑁𝑁𝐴𝐴,𝑟𝑟 ≅ −𝐷𝐷𝐴𝐴𝐴𝐴
𝜕𝜕𝑐𝑐𝐴𝐴
,
𝜕𝜕𝜕𝜕
(𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
Mass Conservation Equation
−∇𝑁𝑁𝐴𝐴 =
𝜕𝜕𝑐𝑐𝐴𝐴
, (𝑅𝑅𝐴𝐴 = 0),
𝜕𝜕𝜕𝜕
Combine Equations
𝜕𝜕𝑐𝑐𝐴𝐴
1 ′ 𝜕𝜕 2 𝜕𝜕𝑐𝑐𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 �𝑟𝑟
,
�=
2
𝑟𝑟
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
∴ ∇𝑁𝑁𝐴𝐴 =
1 𝜕𝜕 2
�𝑟𝑟 𝑁𝑁𝐴𝐴,𝑟𝑟 �,
𝑟𝑟 2 𝜕𝜕𝜕𝜕
′
𝑜𝑜𝑜𝑜 𝐷𝐷𝐴𝐴𝐴𝐴
�
𝑐𝑐𝐴𝐴 (𝑟𝑟, 𝑡𝑡)
𝜕𝜕 2 𝑐𝑐𝐴𝐴 2 𝜕𝜕𝜕𝜕𝐴𝐴
𝜕𝜕𝑐𝑐𝐴𝐴
+
�=
2
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
b. Boundary and Initial Conditions
IC:
𝑡𝑡 = 0,
BC:
𝑟𝑟 = 0,
𝑐𝑐𝐴𝐴 (𝑟𝑟, 0) = 𝑐𝑐𝐴𝐴0 = 0
𝜕𝜕𝑐𝑐𝐴𝐴 (0, 𝑡𝑡)
=0
𝜕𝜕𝜕𝜕
𝑟𝑟 = 𝑅𝑅, 𝑐𝑐𝐴𝐴 (𝑅𝑅, 𝑡𝑡) = 𝑐𝑐𝐴𝐴𝐴𝐴
25-17
25.11
a. System
A = phenol, B = H2O (A = species 1 in analysis below)
Adsorbent Surface = carbon
Ong, S. (1984). Simplified Approach for Designing Carbon Adsorption Column.
J. Environ. Eng., 110(6), 1184–1188.
b. Differential model for cA(z,t)
Assumptions:
1) Unsteady state
2) 1-D flux along z
3) Dilute with respect to species 1 (A)
4) No homogeneous reaction (𝑅𝑅𝐴𝐴 = 0)
5) Linear adsorption isotherm is valid
6) Fast adsorption
General Flux Equation
𝑁𝑁𝐴𝐴,𝑥𝑥 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝜕𝜕𝑐𝑐𝐴𝐴
𝜕𝜕𝜕𝜕
Differential volume element:
25-18
Mass Conservation Equation
Shell balance on Species 1 (A)
IN – OUT + GEN = ACC
𝑁𝑁𝐴𝐴
𝜋𝜋𝑑𝑑2
𝜋𝜋𝑑𝑑 2
1
𝜋𝜋𝑑𝑑2
1
|𝑧𝑧,𝑡𝑡̅ − 𝑛𝑛1
|𝑧𝑧+∆𝑧𝑧,𝑡𝑡̅ + 0 = �𝑞𝑞𝐴𝐴 |𝑧𝑧̅,𝑡𝑡+∆𝑡𝑡 − 𝑞𝑞𝐴𝐴 |𝑧𝑧̅,𝑡𝑡 �𝜋𝜋𝜋𝜋∆𝑧𝑧 + [𝑐𝑐𝐴𝐴 |𝑡𝑡+∆𝑡𝑡,𝑧𝑧̅ − 𝑐𝑐𝐴𝐴 |𝑡𝑡,𝑧𝑧̅ ]
∆𝑧𝑧
4
4
4
∆𝑡𝑡
∆𝑡𝑡
÷ 𝜋𝜋,
𝑑𝑑2
, ∆𝑧𝑧
4
−
�𝑁𝑁𝐴𝐴 |𝑧𝑧+∆𝑧𝑧,𝑡𝑡̅− 𝑁𝑁𝐴𝐴 |𝑧𝑧,𝑡𝑡̅ � [𝑞𝑞𝐴𝐴 |𝑧𝑧̅,𝑡𝑡+∆𝑡𝑡 − 𝑞𝑞𝐴𝐴 |𝑧𝑧̅,𝑡𝑡 ] 4 [𝑐𝑐𝐴𝐴 |𝑧𝑧,� 𝑡𝑡+∆𝑡𝑡 − 𝑐𝑐𝐴𝐴 |𝑧𝑧̅,𝑡𝑡 ]
=
+
∆𝑡𝑡
𝑑𝑑
∆𝑡𝑡
∆𝑧𝑧
−
𝜕𝜕𝑁𝑁𝐴𝐴 (𝑧𝑧, 𝑡𝑡) 4 𝜕𝜕𝑞𝑞𝐴𝐴 𝜕𝜕𝑐𝑐𝐴𝐴
=
+
𝜕𝜕𝜕𝜕
𝑑𝑑 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
lim∆𝑡𝑡 → 0
𝑞𝑞𝐴𝐴 =
𝑞𝑞𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚
𝑐𝑐𝐴𝐴 ,
𝐾𝐾
∴
𝜕𝜕𝑞𝑞𝐴𝐴
𝑞𝑞𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚 𝜕𝜕𝑐𝑐𝐴𝐴
=
𝜕𝜕𝜕𝜕
𝐾𝐾 𝜕𝜕𝜕𝜕
𝜕𝜕 2 𝑐𝑐𝐴𝐴
4 𝑞𝑞𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚
𝜕𝜕𝜕𝜕𝐴𝐴 (𝑧𝑧, 𝑡𝑡)
=
�
+
1�
𝐷𝐷𝐴𝐴𝐴𝐴
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 2
𝑑𝑑 𝐾𝐾
𝐷𝐷𝐴𝐴𝐴𝐴
𝜕𝜕 2 𝑐𝑐𝐴𝐴 (𝑧𝑧, 𝑡𝑡)
𝜕𝜕𝜕𝜕𝐴𝐴 (𝑧𝑧, 𝑡𝑡)
� 𝑞𝑞
�
=
2
4 𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+1
𝑑𝑑 𝐾𝐾
Boundary and Initial Conditions
𝑧𝑧 = 0,
𝑐𝑐1 (0, 𝑡𝑡) = 𝑐𝑐1,𝑠𝑠
25-19
𝑧𝑧 = 𝐿𝐿,
𝑡𝑡 = 0,
𝜕𝜕𝜕𝜕1 (𝐿𝐿, 𝑡𝑡)
=0
𝜕𝜕𝜕𝜕
𝑐𝑐1 (𝑧𝑧, 0) = 𝑐𝑐1,0 = 0
25-20
25.12
Sealed Ends
Catalyst
Surface
Porous
Layer
Bulk Gas (cA = cA,∞)
A
B
L = 5.0 cm
C
r=0
r = Ro
(0.5 cm)
1.5 atm, 150 oC
4 mole% C2H4 (A)
94 mole% H2 (B)
2 mole% C2H6 (C)
r = R1
a. Species A, B, and C are undergoing mass transfer, as each has a source and sink
b. Assumptions
1.
2.
3.
4.
5.
Steady-state, constant source and sink
No homogeneous reaction within the control volume, RA = 0
1-D flux along coordinate r
Dilute process with respect to species A
Rapid reaction of A at catalyst surface
c. General Differential Equation for Mass Transfer (cylindrical system)
Cylindrical Coordinate System
By assumptions 1 and 2
∂N A, z 
1 ∂
∂c A
rN A, r ) +
−
(
 + RA =
∂z 
∂t
 r ∂r
1 d
( r ⋅ N A, r ) = 0
r dr
d
( r ⋅ N A, r ) = 0
dr
By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r
Flux Equation, 1-D flux along r
25-21
dc
c
dc
c
N A, r =
− DAB A + A ( N A, r + N B , r + N C , r ) =
− DAB A + A ( N A, r − N A, r + 0 + N A, r )
dr C
dr C
− DAB dc A
N A, r =
1 − y A dr
N A, r = − DAB
dc A
(if dilute with respect to A)
dr
d. Boundary Conditions
r = Ro, cA = 0 (rapid reaction at surface)
r = R1, cA = cA,∞ (bulk fluid)
25-22
25.13
flow A+B
cAo
cAo
z=0
A →B
z=L
porous disk with
sides and bottom
sealed
impermeable
barrier
r=0 r=R
a. Assumptions
1. Steady state process – constant source and sink for A and B
2. Homogeneous, first-order reaction of A within control volume
3. 1-D flux along z direction (sealed circumference and base)
b. General Flux Equation
dc
c
dc
c
− DAB A + A ( N A, z + N B , z ) =
− DAB A + A ( N A, z − N A, z )
N A, z =
dz C
dz C
dc
N A, z = − DAB A (equilimolar counter diffusion with respect to A and B)
dz
c. Differential Equation for Mass Transfer
∂N A, z 
1 ∂
∂c A
rN A, r ) +
−
(
 + RA =
∂z 
∂t
 r ∂r
∂N
− A , z + RA = 0
∂z
Homogeneous, first-order reaction, RA = −kc A
DAB
d 2 cA
− k cA =
0
dz 2
d. Boundary Conditions
z = 0, cA = cAo
z = L,
dc A
=0
dz
25-23
26.1
A = SiH4 vapor, B = H2 gas
𝑇𝑇 = 900 𝐾𝐾, 𝑃𝑃 = 70 𝑃𝑃𝑃𝑃 (6.91 𝑥𝑥 10−4 𝑎𝑎𝑎𝑎𝑎𝑎), 𝛿𝛿 = 5.0 𝑐𝑐𝑐𝑐, 𝑦𝑦𝐴𝐴𝐴𝐴 = 0.20, 𝜎𝜎𝐴𝐴 = 4.08 Å,
𝜀𝜀𝐴𝐴
= 207.6 𝐾𝐾, 𝜎𝜎𝐵𝐵 = 2.968 Å, 𝜀𝜀𝐵𝐵 /𝜅𝜅 = 33.3 𝐾𝐾
𝜅𝜅
𝑔𝑔
𝜌𝜌𝑆𝑆𝑆𝑆 = 2.32 3
𝑐𝑐𝑐𝑐
a. Estimate the rate of Si film formation
𝑦𝑦𝐴𝐴𝐴𝐴 ≈ 0
𝑃𝑃
=
𝑐𝑐 =
𝑅𝑅𝑅𝑅
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
70 𝑃𝑃𝑃𝑃
−3
=
9.355
×
10
𝑚𝑚3 ∙ 𝑃𝑃𝑃𝑃
𝑚𝑚3
8.314
∙ 900 𝐾𝐾
𝐾𝐾 ∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
Determine DAB
1
1 2
0.001858 ∙ 𝑇𝑇 3/2 �𝑀𝑀 + 𝑀𝑀 �
𝐴𝐴
𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 =
2
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴
ΩD
σAB = (2.968 + 4.08)/2 = 3.524 Å
𝜀𝜀𝐴𝐴𝐴𝐴 = √207.6 𝐾𝐾 ∙ 33.3 𝐾𝐾 = 83.145
900 𝐾𝐾
𝜅𝜅𝜅𝜅
=
= 10.82, ∴ Ω𝐷𝐷 = 0.73597
𝜀𝜀𝐴𝐴𝐴𝐴 83.145 𝐾𝐾
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
1
0.001858 ∙ 900 𝐾𝐾 3/2 [2.02 + 32.12]1/2
6.91 × 10−4 ∙ 3.524 2 ∙ 0.73597
Material balance on Si
= 5672
𝑐𝑐𝑐𝑐2
𝑠𝑠
IN – OUT + GEN = ACC
dm
0 − N A ⋅ S + 0 = Si
dt
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
𝑁𝑁𝐴𝐴 =
ln(1 + 𝑦𝑦𝐴𝐴𝐴𝐴 )
𝛿𝛿
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
𝜌𝜌𝑆𝑆𝑆𝑆 𝑑𝑑ℓ
ln(1 + 𝑦𝑦𝐴𝐴𝐴𝐴 ) =
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝛿𝛿 ≫ ℓ
𝛿𝛿
𝑀𝑀𝑆𝑆𝑆𝑆 𝑑𝑑𝑑𝑑
𝑔𝑔
𝑐𝑐𝑐𝑐2
−9 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
9.355
×
10
∙
5672
∙ 28.09
∙ ln(1 + 0.2)
3
𝑑𝑑ℓ 𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝑆𝑆𝑆𝑆
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐
=
ln(1 + 𝑦𝑦𝐴𝐴𝐴𝐴 ) =
𝑔𝑔
𝑑𝑑𝑑𝑑
𝛿𝛿𝜌𝜌𝑆𝑆𝑆𝑆
5.0 𝑐𝑐𝑐𝑐 ∙ 2.32 3
𝑐𝑐𝑐𝑐
𝑐𝑐𝑐𝑐
𝜇𝜇𝜇𝜇
−5
= 2.34 𝑥𝑥 10
= 1406
𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
26-1
b. Estimate the rate of Si film formation with ks
𝑘𝑘𝑠𝑠 = 1.25
𝑐𝑐𝑐𝑐
𝑠𝑠
ln(1 + 𝑦𝑦) ≅ 𝑦𝑦
𝑁𝑁𝐴𝐴 =
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 (𝑦𝑦𝐴𝐴𝐴𝐴 − 𝑦𝑦𝐴𝐴𝐴𝐴 )
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
1 + 𝑦𝑦𝐴𝐴𝐴𝐴
ln �
�=
𝛿𝛿
1 + 𝑦𝑦𝐴𝐴𝐴𝐴
𝛿𝛿
𝑦𝑦𝐴𝐴𝐴𝐴 =
𝑁𝑁𝐴𝐴
𝑘𝑘𝑠𝑠 𝑐𝑐
𝑁𝑁𝐴𝐴 = 𝑘𝑘𝑠𝑠 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝑘𝑘𝑠𝑠 𝑐𝑐𝑐𝑐𝐴𝐴𝐴𝐴
𝑁𝑁𝐴𝐴 =
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴𝐴𝐴
𝐷𝐷
𝛿𝛿 �1 + 𝐴𝐴𝐴𝐴 �
𝛿𝛿𝑘𝑘𝑠𝑠
Material balance on Si
IN – OUT + GEN = ACC
𝑁𝑁𝐴𝐴 𝑆𝑆 − 0 + 0 =
𝑑𝑑𝑑𝑑𝑆𝑆𝑆𝑆
𝑑𝑑𝑑𝑑
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴𝐴𝐴
𝜌𝜌𝑆𝑆𝑆𝑆 𝑑𝑑ℓ
𝑆𝑆 =
𝑆𝑆
𝐷𝐷𝐴𝐴𝐴𝐴
𝑀𝑀
𝑑𝑑𝑑𝑑
𝑆𝑆𝑆𝑆
𝛿𝛿 �1 +
�
𝛿𝛿𝑘𝑘𝑠𝑠
𝑑𝑑ℓ
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴𝐴𝐴 𝑀𝑀𝑆𝑆𝑖𝑖
=
𝑑𝑑𝑑𝑑 𝛿𝛿 �1 + 𝐷𝐷𝐴𝐴𝐴𝐴 � 𝜌𝜌
𝛿𝛿𝑘𝑘𝑠𝑠 𝑆𝑆𝑆𝑆
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔
𝑐𝑐𝑐𝑐2
∙
5672
∙ 0.2 ∙ 28.09
3
𝑑𝑑ℓ
𝑐𝑐𝑐𝑐
𝜇𝜇𝜇𝜇
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐
−8
=
=
2.83
𝑥𝑥
10
=
1.7
𝑐𝑐𝑐𝑐2
𝑑𝑑𝑑𝑑
𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
5672 𝑠𝑠
𝑔𝑔
5.0 𝑐𝑐𝑐𝑐 ∙ �1 +
𝑐𝑐𝑐𝑐� 2.32 𝑐𝑐𝑐𝑐3
5.0 𝑐𝑐𝑐𝑐 ∙ 1.25
𝑠𝑠
In this case, surface reaction controls the overall rate of Si thin film formation
9.355 × 10−9
26-2
26.2
A = drug, B = gel diffusion barrier
𝑐𝑐𝑐𝑐2
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝐷𝐷𝐴𝐴𝐴𝐴 = 1.5 × 10−5
, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0.01
𝑠𝑠
𝑐𝑐𝑐𝑐3
a. Develop model equation for WA
Assume: 1) Steady-state (PSS), 2) 1-D flux transfer along r, 3) No reaction, 4) Dilute system,
5) UMD, 6) Saturation concentration A ( cAs = cA*)
Flux equation
𝑑𝑑𝑑𝑑𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
General Differential Equation for Mass Transfer
1 d 2
d
0 ⇒ ( r 2 N A, r ) =
0
r N A, r ) =
(
2
r dr
dr
∴WA = N A,R i ⋅ 4π Ri 2 = N A,R o ⋅ 4π Ro 2 = N A,r ⋅ 4π r 2 = const
𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴,𝑟𝑟 ∙ 4𝜋𝜋𝑟𝑟 2 �−𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑𝐴𝐴
�
𝑑𝑑𝑑𝑑
𝑅𝑅𝑜𝑜
𝑐𝑐𝐴𝐴𝐴𝐴
𝑅𝑅𝑖𝑖
𝑐𝑐𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
𝑊𝑊𝐴𝐴 � 2 = −4𝜋𝜋𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑑𝑑𝐴𝐴
𝑟𝑟
𝑊𝑊𝐴𝐴 =
4𝜋𝜋𝐷𝐷𝐴𝐴𝐴𝐴 (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 )
1
1
−
𝑅𝑅𝑖𝑖 𝑅𝑅𝑜𝑜
b. Determine WA
At maximum transfer rate,𝑐𝑐𝐴𝐴𝐴𝐴 ≈ 0
𝑊𝑊𝐴𝐴 =
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐2
∙ 0.01
𝑠𝑠
𝑐𝑐𝑐𝑐3 = 3.17 × 10−3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1
1
ℎ𝑟𝑟
−
0.2 𝑐𝑐𝑐𝑐 0.35 𝑐𝑐𝑐𝑐
4𝜋𝜋 ∙ 1.5 × 10−5
26-3
26.3
Sealed
SealedEnds
Ends
DAe = 1.2 x 10-5 cm2/s
/sec
CAA** == 0.05
c
0.05 mmole/cm
mmole/cm33
gel
layer
DAe
solid
solid
solute
solute AA
(0.2 mmol)
L =L 1.5
= 1.5
cmc
CAA**
c
CAA∞∞ == 0.01
c
0.01 mmole/cm
mmole/cm33
CAA∞∞
c
r = Ro
(0.75 cm)
r = 0 r = Ri
(0.25 cm)
a. Boundary Conditions
The system is the gel layer surrounding the source reservoir for solute A
r = Ri, CA = CA*
r = Ro, CA = CA∞
b. Assumptions
1.
2.
3.
4.
5.
Steady state - constant source and sink for A as long as solid A is not completely dissolved
No homogeneous reaction of A within gel layer
1-D flux along position r - ealed top and bottom ends
Constant CA* - source reservoir is well mixed, solid A is not yet completely dissolved
Dilute with respect to solute A
c. Total transfer rate, WA
Flux Equation based on assumption 3
N A, r = − DAe
dC A
dr
General Differential Equation for Mass Transfer based on assumptions 1-3
26-4
d
( r N A, r ) = 0
dr
By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r
=
WA 2=
π rL N A, r 2π R=
Ro WA is constant along r
o L N A, r r
C A∞
dr
= − DAe ∫ dC A
Ri r
CA *
2π LDAe ( C A* − C A∞ )
WA =
R 
ln  o 
 Ri 
WA ∫
Ro

cm 2 
mmol A
2π(1.5 cm) 1.2×10-5
 ( 0.05-0 )
s 
cm3

WA =
 0.75 cm 
ln 

 0.25 cm 
WA = 4.12×10-6 mmol A/s
d. Time t
Mass balance on solute A in reservoir
IN − OUT =
ACCUMULATION
dm
0 − WA =A
dt
mAo − mA =
WAt
When mA = 0
0.2 mmol A
mAo
 3600 s 
t =
=

 =13.5 h
WA 4.12×10-6 mmol A  1 h 
s
26-5
26.4
A = drug, B = gel diffusion barrier
ℓ = 2.0 mm, 𝑆𝑆 = 9.0 𝑐𝑐𝑐𝑐2 , 𝑇𝑇 = 293 𝐾𝐾, 𝑐𝑐𝐴𝐴∗ = 0.50
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝑐𝑐𝑐𝑐3
Assume: 1) Steady state, 2) 1-D flux along z, 3) dilute system, 4) UMD process
a. Estimate diffusion coefficient of drug (A) in gel diffusion barrier (B)
d
( N A,z ) = 0 , constant flux along z
dz
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝐿𝐿
𝑑𝑑𝑑𝑑𝐴𝐴
𝑑𝑑𝑑𝑑
0
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴 ,
∗
𝑐𝑐𝐴𝐴
0
𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 𝑆𝑆 = 𝐷𝐷𝐴𝐴𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 =
𝑊𝑊𝐴𝐴 ∙ 𝐿𝐿
𝑆𝑆 ∙ 𝑐𝑐𝐴𝐴∗
𝑐𝑐𝐴𝐴∗
𝑆𝑆
𝐿𝐿
𝑐𝑐𝐴𝐴∗
𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴
𝐿𝐿
Linear drug release profile, from data
𝑊𝑊𝐴𝐴 =
∆𝑚𝑚𝐴𝐴
0.16 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
=
= 0.008
Δ𝑡𝑡
20 ℎ𝑟𝑟
ℎ𝑟𝑟
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
∙ 0.2 𝑐𝑐𝑐𝑐
𝑐𝑐𝑐𝑐2
ℎ𝑟𝑟
−8
𝐷𝐷𝐴𝐴𝐴𝐴 =
= 9.88 × 10
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝑠𝑠
9.0 𝑐𝑐𝑐𝑐2 ∙ 0.50
𝑐𝑐𝑐𝑐3
0.008
b. Determine new WA at 35oC
DAB (T2 )
T µ (T )
WA (T2 ) W=
WA (T1 ) 2 B 2
=
A (T1 )
DAB (T1 )
T1 µ B (T1 )
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 308 𝐾𝐾 993 × 10−6 𝑃𝑃𝑃𝑃 ∙ 𝑠𝑠
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝑊𝑊𝐴𝐴 = 0.008
∙�
= 0.0113
= 0.270
�
−6
ℎ𝑟𝑟
293 𝐾𝐾 742 × 10 𝑃𝑃𝑃𝑃 ∙ 𝑠𝑠
ℎ𝑟𝑟
𝑑𝑑𝑑𝑑𝑑𝑑
26-6
26.5
A = H2O vapor, B = air (inside pore)
𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 10 𝜇𝜇𝜇𝜇, 𝐿𝐿 = 0.05 𝑐𝑐𝑐𝑐, 𝑛𝑛𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 20,000
𝑇𝑇 = 303 𝐾𝐾, 𝑃𝑃𝐴𝐴 = 0.044 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑝𝑝𝐴𝐴∞ = 0.022 𝑎𝑎𝑎𝑎𝑎𝑎
a. Determine the effective diffusion coefficient, DAe
A = H2O, B = air
𝑐𝑐𝑐𝑐2 303 𝐾𝐾 1.5
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴 = 0.260
∙�
� = 0.267
298 𝐾𝐾
𝑠𝑠
𝑠𝑠
𝑇𝑇
303
𝑐𝑐𝑐𝑐2
= 4850 ∙ 10 × 10−4 �
= 19.89
𝑠𝑠
𝑀𝑀𝐴𝐴
18.02
𝐷𝐷𝐾𝐾𝐾𝐾 = 4850𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 �
1
1
1
=
+
=
𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐾𝐾𝐾𝐾
1
0.267
𝑐𝑐𝑐𝑐2
𝑠𝑠
+
1
19.89
𝑐𝑐𝑐𝑐2
𝑠𝑠
𝑐𝑐𝑐𝑐2
∴ 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.263
𝑠𝑠
Knudsen Diffusion is not important in this case.
b. Determine WA for single egg
Assume: 1) Steady state, 2) 1-D flux in pore, 3) Dilute system, 4) Constant P and T
𝑑𝑑𝑑𝑑𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
d
( N A,z ) = 0 , constant flux along z
dz
𝐿𝐿
𝐶𝐶𝐴𝐴∞
0
∗
𝑐𝑐𝐴𝐴
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴 ,
𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴
(𝑐𝑐𝐴𝐴∗ − 𝑐𝑐𝐴𝐴∞ )
(𝑃𝑃𝐴𝐴 − 𝑝𝑝𝐴𝐴∞ )
= 𝐷𝐷𝐴𝐴𝐴𝐴
𝐿𝐿
𝐿𝐿 ∙ 𝑅𝑅𝑅𝑅
𝑐𝑐𝑐𝑐2
(0.044 − 0.022) 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑠𝑠 ∙
𝑁𝑁𝐴𝐴 =
= 4.654 𝑥𝑥 10−6
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠
0.05 𝑐𝑐𝑐𝑐 ∙ 82.06
∙ 303 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
0.263
2
𝜋𝜋𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑛𝑛𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 =
4
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝜋𝜋(10−3 𝑐𝑐𝑐𝑐)2
18.02 𝑔𝑔 86400 𝑠𝑠
= 4.654 × 10−6
∙ 20,000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙
∙
2
𝑐𝑐𝑐𝑐 ∙ 𝑠𝑠
4
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑑𝑑𝑑𝑑𝑑𝑑
𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 𝑆𝑆 = 𝑁𝑁𝐴𝐴
= 0.114 𝑔𝑔/𝑑𝑑𝑑𝑑𝑑𝑑
c. To reduce water loss by 50%, Increase 𝑝𝑝𝐴𝐴∞ to 0.033 atm or double temperature.
26-7
26.6
A = drug, B = water-swollen polymer
𝑆𝑆 = 16.0 𝑐𝑐𝑐𝑐2 , 𝑊𝑊𝐴𝐴 =
𝑐𝑐𝐴𝐴∗ = 0.50
0.02 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
= 5.56 × 10−6
ℎ𝑟𝑟
𝑠𝑠
𝑐𝑐𝑐𝑐2
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
−7
, 𝐷𝐷𝐴𝐴𝐴𝐴 = 2.08 × 10
𝑐𝑐𝑐𝑐3
𝑠𝑠
Estimate drug patch thickness, L
Assumptions: 1) Steady state, 2) 1-D flux along z, 3) dilute system, 4) UMD process
d
( N A,z ) = 0 , constant flux along z
dz
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝐿𝐿
𝑑𝑑𝑑𝑑𝐴𝐴
𝑑𝑑𝑑𝑑
0
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴
0
𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴
𝑐𝑐𝐴𝐴∗
𝐿𝐿
∗
𝑐𝑐𝐴𝐴
𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 ∙ 𝑆𝑆 = 𝐷𝐷𝐴𝐴𝐴𝐴
𝐿𝐿 = 𝐷𝐷𝐴𝐴𝐴𝐴
𝑐𝑐𝐴𝐴∗
∙ 𝑆𝑆
𝑊𝑊𝐴𝐴
𝑐𝑐𝐴𝐴∗
∙ 𝑆𝑆
𝐿𝐿
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
2
0.50
𝑐𝑐𝑐𝑐
𝑐𝑐𝑐𝑐3
𝐿𝐿 = 2.08 × 10−7
∙ 16.0 𝑐𝑐𝑐𝑐2 = 0.30 𝑐𝑐𝑐𝑐
𝑠𝑠 5.56 × 10−6 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
𝑠𝑠
26-8
26.7
A = solute, D = reaction product, B = gel diffusion barrier
𝑐𝑐𝑐𝑐2
𝑆𝑆 = 2.0 𝑐𝑐𝑐𝑐2 , 𝐿𝐿 = 0.30 𝑐𝑐𝑐𝑐, 𝐷𝐷𝐴𝐴𝐴𝐴 = 4.0 × 10−7
, 𝑇𝑇 = 293 𝐾𝐾
𝑠𝑠
𝑐𝑐𝑐𝑐3 𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝐴𝐴′ = 𝐾𝐾𝑐𝑐𝐴𝐴 , 𝐾𝐾 = 0.8 3
surrounding liquid
𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙𝑙𝑙
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐷𝐷
𝑐𝑐𝐷𝐷𝐷𝐷 ≈ 0, 𝑊𝑊𝐷𝐷 = 1.0 × 10−8
𝑠𝑠
Reaction at boundary surface: 𝐴𝐴 → 2𝐷𝐷
Mass Transfer Model
Assume: 1) Steady state, 2) 1-D flux transfer along z, 3) No homogenous reaction, 4) Diffusion
limited reaction at the boundary surface z = L, 5) Dilute process, gel
d
( N A,z ) = 0 , constant flux along z
dz
𝑑𝑑𝑑𝑑𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
𝐿𝐿
′
𝑐𝑐𝐴𝐴𝐴𝐴
𝑐𝑐𝐴𝐴𝐴𝐴 = 𝐾𝐾
0
0
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴
�
𝑑𝑑𝑐𝑐𝐴𝐴 ,
𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴
Relate 𝑊𝑊𝐴𝐴 to 𝑊𝑊𝐷𝐷 by moles (stoichiometry).
′
𝑐𝑐𝐴𝐴𝐴𝐴
𝐿𝐿𝐿𝐿
′
1
𝑐𝑐𝐴𝐴𝐴𝐴
𝑊𝑊𝐴𝐴 = 𝑊𝑊𝐷𝐷 = 𝑁𝑁𝐴𝐴 ∙ 𝑆𝑆 = 𝐷𝐷𝐴𝐴𝐴𝐴
𝑆𝑆
𝐿𝐿𝐿𝐿
2
′
𝑐𝑐𝐴𝐴𝐴𝐴
=
𝑊𝑊𝐷𝐷 𝐿𝐿𝐿𝐿
=
2 ∙ 𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 𝑆𝑆
= 1.5 × 10−3
1.0 × 10−8
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑐𝑐𝑐𝑐3
𝑐𝑐𝑐𝑐3 𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐷𝐷
∙ 1.0𝑐𝑐𝑐𝑐2 ∙ 0.30 𝑐𝑐𝑐𝑐 ∙ 0.8
𝑠𝑠
𝑐𝑐𝑐𝑐3 𝑙𝑙𝑙𝑙𝑙𝑙
2 ∙ 4.0 × 10−7 𝑐𝑐𝑐𝑐2 ∙ 2.0 𝑐𝑐𝑐𝑐2
26-9
26.8
well-mixed bulk fluid (+ drug A)
CAO
NA
healthy
tissue (B)
CAs
clump of
cancer
cells
inert surface
r=0
R1 = 0.05 cm R2= 0.1 cm
Given:
A = drug, B = tissue
R1 = 0.05 cm, R2 = 0.10 cm, DAB = 2 × 10−7 cm2/s
NA =
−6.914 × 10 –4
mmole A day
mmole
= − 8.00 × 10−9
2
cm day 86,400 s
cm 2s
a. System, source and sink for drug, assumptions
Physical System: healthy tissue between bulk fluid and unhealthy cells
Source for A: well mixed bulk fluid containing “A”
Sink for A: clump of cancer cells
Assumptions
1. Steady state
2. 1-D flux along r
3. No homogenous reaction of drug within tissue
4. Dilute system with respect to drug (A)
b. Boundary conditions
=
r R1 , c=
c As ≈ 0
A
=
r R=
c Ao
2 , cA
surface of unhealthy tissue
bulk fluid
26-10
c. Integral form of flux equation NA
General Differential Equation for Mass Transfer
∂c A
∂c A
1 d 2
− 2
R=
r N A,r ) + R=
0, =
0
(
A
A
∂t
∂t
r dr
1 d 2
∴ 2
0
( r N A, r ) =
r dr
d 2
( r N A, r ) = 0
dr
N A,r ⋅ r 2 is constant along r
General Flux Equation
dc
N A = − DAB A
dr
Note N A,R1 ⋅ R12 = N A,R 2 ⋅ R2 2 = N A,r ⋅ r 2 (constant along r)
dc 

2
N A, R1 R=
r 2  − DAB A 
1
dr 

Separate variables cA and r, pull N A, R1 R12 outside of integral, then integrate with respect to limits
shown below
c
R2
Ao
dr
=
−
D
AB ∫ dc A
2
r
R1
0
N A, R1 R12 ∫
N A, R1 =
DAB c Ao
 1
1 
R12  − 
 R2 R1 
 1
1 
R12  −  N A, R1
 R2 R1 
c Ao =
DAB
d. cAo at r = R2
( 0.05 cm ) 
2
c Ao =
1
1

−9 mmole 
−
  −8.00 × 10

mmole
cm 2s 
 0.10 cm 0.05 cm  
=1.0 × 10−3
2
cm
cm3
2.0 × 10−7
s
26-11
26.9
A = H2O vapor, B = O2 gas
𝑇𝑇 = 310 𝐾𝐾, 𝑃𝑃 = 1.2 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 50 × 10−7 𝑐𝑐𝑐𝑐
𝜀𝜀 = 0.40, 𝑃𝑃𝐴𝐴 = 0.0618 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑦𝑦𝐴𝐴∗ = 0.0515, 𝐻𝐻 = 800
a. Determine effective diffusion coefficient, DAe
𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
Fuller-Schettler-Giddings Correlation for DAB
1
1 1/2
1
1 1/2
1.75
0.001𝑇𝑇 1.75 � + �
0.001
∙
310
�
+
�
𝑐𝑐𝑐𝑐2
𝑀𝑀𝐴𝐴 𝑀𝑀𝐴𝐴
18.02
32.0
=
= 0.236
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
1
1/3 2
1/3
𝑠𝑠
𝑃𝑃�𝑉𝑉𝐴𝐴 + 𝑉𝑉𝐵𝐵 �
1.2 ∙ [12.73 + 16.63 ]2
Knudsen Diffusion Coefficient
𝑇𝑇
310
𝑐𝑐𝑐𝑐2
= 4850 ∙ 50 × 10−7 �
= 0.1006
𝑀𝑀𝐴𝐴
18.02
𝑠𝑠
𝐷𝐷𝐾𝐾𝐾𝐾 = 4850𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 �
Effective Diffusion Coefficient
1
1
1
1
𝑐𝑐𝑐𝑐2
1
=
+
=
+
∴
𝐷𝐷
=
0.0705
𝐴𝐴𝐴𝐴
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐾𝐾𝐾𝐾
𝑠𝑠
0.236
0.1006
𝑠𝑠
𝑠𝑠
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
′
𝐷𝐷𝐴𝐴𝐴𝐴
= 𝜀𝜀 2 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.42 ∙ 0.0705
= 0.0113
𝑠𝑠
𝑠𝑠
b. Determine L
Assume: 1) Steady state, 2) no homogenous reaction, 3) 1-D flux along z, 4) constant T and P
𝑐𝑐 =
𝑃𝑃
=
𝑅𝑅𝑅𝑅
1.2 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 4.72 × 10−5
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐3
82.06
∙ 310𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
d
( N A,z ) = 0 , constant flux along z
dz
𝑑𝑑𝑑𝑑𝐴𝐴 𝑐𝑐𝐴𝐴
𝑑𝑑𝑑𝑑𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
+ �𝑁𝑁𝐴𝐴,𝑧𝑧 � = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐
+ 𝑦𝑦𝐴𝐴 �𝑁𝑁𝐴𝐴,𝑧𝑧 �
𝑑𝑑𝑑𝑑
𝑐𝑐
𝑑𝑑𝑑𝑑
𝑁𝑁𝐴𝐴 = −
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐 𝑑𝑑𝑑𝑑𝐴𝐴
1 − 𝑦𝑦𝐴𝐴 𝑑𝑑𝑑𝑑
Boundary Conditions:
𝑧𝑧 = 0, 𝑦𝑦𝐴𝐴𝐴𝐴 = 𝑦𝑦𝐴𝐴∗
26-12
𝑧𝑧 = 𝐿𝐿, 𝑦𝑦𝐴𝐴∞ = 0.2𝑦𝑦𝐴𝐴∗
𝐿𝐿
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 �
0
𝑁𝑁𝐴𝐴 =
𝐿𝐿 =
𝐿𝐿 =
𝑦𝑦𝐴𝐴∞
𝑦𝑦𝐴𝐴𝐴𝐴
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
1 − 𝑦𝑦𝐴𝐴∞
𝑙𝑙𝑙𝑙 �
�
𝐿𝐿
1 − 𝑦𝑦𝐴𝐴𝐴𝐴
𝑑𝑑𝑦𝑦𝐴𝐴
1 − 𝑦𝑦𝐴𝐴
𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴
1 − 𝑦𝑦𝐴𝐴∞
𝑙𝑙𝑙𝑙 �
�
𝑁𝑁𝐴𝐴
1 − 𝑦𝑦𝐴𝐴𝐴𝐴
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐2
∙ 4.72 × 10−5
𝑠𝑠
𝑐𝑐𝑐𝑐3 ln �1 − 0.2 ∙ 0.0515� = 0.20 𝑐𝑐𝑐𝑐
𝑚𝑚𝑚𝑚𝑚𝑚
1 − 0.0515
1.15 × 10−7
𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠
0.0113
∗
c. Determine 𝑐𝑐𝐵𝐵𝐵𝐵
∗
𝑐𝑐𝐵𝐵𝐵𝐵
=
𝑝𝑝𝐵𝐵 𝑃𝑃 − 𝑝𝑝𝐴𝐴 (1.2 − 0.062) 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
=
=
= 1.42 × 10−3
𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝐻𝐻
𝐻𝐻
𝐿𝐿
800
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
26-13
26.10
A = O2, B = silicone polymer layer
𝑐𝑐𝐴𝐴′
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
, 𝑇𝑇 = 298 𝐾𝐾
𝑙𝑙 = 0.10 𝑐𝑐𝑐𝑐, 𝐴𝐴 = 50 𝑐𝑐𝑐𝑐2 , 𝑝𝑝𝐴𝐴 = , 𝑆𝑆 = 3.15 × 10−3
𝑆𝑆
𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐2
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝐷𝐷𝐴𝐴𝐴𝐴 = 1 × 10−7
, 𝑊𝑊𝐴𝐴 = 1.42 × 10−5
= 2.37 × 10−7
𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
𝑠𝑠
Find 𝑝𝑝𝐴𝐴 given 𝑊𝑊𝐴𝐴
Assume: 1) Steady state, 2) 1-D flux along z, 3) Dilute UMD process, 4) No homogenous
reaction, 5) Rapid consumption of O2
d
( N A,z ) = 0 , constant flux along z
dz
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝐿𝐿
𝑑𝑑𝑑𝑑𝐴𝐴
𝑑𝑑𝑑𝑑
0
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴
0
𝑁𝑁𝐴𝐴 =
′
𝑐𝑐𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴′ 𝐷𝐷𝐴𝐴𝐴𝐴 𝑝𝑝𝐴𝐴 𝑆𝑆
=
𝐿𝐿
𝐿𝐿
𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 ∙ 𝐴𝐴 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑝𝑝𝐴𝐴 𝑆𝑆
∙ 𝐴𝐴
𝐿𝐿
𝑊𝑊𝐴𝐴 ∙ 𝐿𝐿
𝑝𝑝𝐴𝐴 =
=
𝐴𝐴 ∙ 𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 𝑆𝑆
2.37 × 10−7
𝑐𝑐𝑐𝑐
50 𝑐𝑐𝑐𝑐2 ∙ 1 × 10−7
𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
∙ 0.1 𝑐𝑐𝑐𝑐
𝑠𝑠
2
∙ 3.15 × 10−3
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
= 1.5 𝑎𝑎𝑎𝑎𝑎𝑎
26-14
26.11
A = benzene, B = water, C = air
𝑇𝑇 = 298 𝐾𝐾, 𝐷𝐷𝐴𝐴𝐴𝐴 = 1.1 × 10−5
𝑐𝑐𝑐𝑐2
𝑚𝑚3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐2
, 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.093
, 𝐻𝐻𝐴𝐴 = 4.84 × 10−3
𝑠𝑠
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑃𝑃𝐴𝐴 = 0.13 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑃𝑃𝐵𝐵 = 0.031 𝑎𝑎𝑎𝑎𝑎𝑎, 𝜌𝜌𝐵𝐵 = 1000
a. System Information
𝑘𝑘𝑘𝑘
𝑔𝑔
𝑔𝑔
, 𝑀𝑀𝐴𝐴 = 78
, 𝑀𝑀𝐵𝐵 = 18
3
𝑚𝑚
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
Assume: 1) Steady state, 2) 1-D flux along z, 3) Dilute UMD system, 4) No homogenous
reaction
Benzene in air
Source: Liquid benzene
Sink: Air
Boundary: Top of liquid to top of well
Benzene in water
Source: Liquid benzene
Sink: Water vapor
Boundary: Bottom of liquid to top of liquid
Water in air
Source: Liquid water
Sink: Air
Boundary: Top of liquid to top of well
b. Boundary Conditions
Benzene in air
𝑧𝑧 = 𝐿𝐿, 𝑐𝑐𝐴𝐴 = 𝑐𝑐𝐴𝐴∞ = 0
𝑧𝑧 = 0, 𝑐𝑐𝐴𝐴 = 𝑐𝑐𝐴𝐴∗
𝑃𝑃𝐴𝐴∗ = 𝐻𝐻𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝐻𝐻𝐴𝐴
𝑐𝑐𝐴𝐴∗ =
𝑃𝑃𝐴𝐴∗
=
𝑅𝑅𝑅𝑅
𝜌𝜌𝐴𝐴
𝑚𝑚3 𝑎𝑎𝑎𝑎𝑎𝑎 156 𝑔𝑔 𝐴𝐴 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 4.83 𝑥𝑥 10−3
= 9.68 𝑥𝑥 10−3 𝑎𝑎𝑎𝑎𝑎𝑎
𝑀𝑀𝐴𝐴
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚3
78 𝑔𝑔
9.68 𝑥𝑥 10−3 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
−7
=
3.95
×
10
𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐3
82.06
∙ 298 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
26-15
Water in air
𝑧𝑧 = 𝐿𝐿, 𝑐𝑐𝐵𝐵 = 𝑐𝑐𝐵𝐵∞
𝑧𝑧 = 0, 𝑐𝑐𝐵𝐵 = 𝑐𝑐𝐵𝐵∗
𝑝𝑝𝐵𝐵,∞ = 0.4 ∙ 0.031 𝑎𝑎𝑎𝑎𝑎𝑎 = 0.0124 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝐵𝐵∞ =
𝑐𝑐𝐵𝐵∗ =
𝑝𝑝𝐵𝐵∞
=
𝑅𝑅𝑅𝑅
𝑃𝑃𝐵𝐵∗
=
𝑅𝑅𝑅𝑅
0.0124 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 5.07 × 10−7
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐3
82.06
∙ 298 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
0.031 𝑎𝑎𝑎𝑎𝑎𝑎
= 1.27 × 10−6
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐3
82.06
∙ 298 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
c. Determine WA, WB, and mA (t = 30 days)
Benzene emitted to air
𝑐𝑐𝑐𝑐2
−7 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴∗ 𝜋𝜋𝑑𝑑 2 0.093 𝑠𝑠 ∙ 3.95 × 10
𝜋𝜋(10 𝑐𝑐𝑐𝑐)2
3
𝑐𝑐𝑐𝑐
𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴 =
∙
=
∙
𝐿𝐿
4
300 𝑐𝑐𝑐𝑐
4
= 9.62 × 10−9
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 8.31 × 10−4
𝑑𝑑𝑑𝑑𝑑𝑑
𝑠𝑠
Total Benzene in 30 days
𝑚𝑚𝐴𝐴 = 𝑊𝑊𝐴𝐴 𝑀𝑀𝐴𝐴 𝑡𝑡 = �8.31 × 10−4
Water evaporation to air
78 𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
��
� (30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) = 1.95 𝑔𝑔 𝐴𝐴
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑑𝑑𝑑𝑑𝑑𝑑
𝐷𝐷𝐵𝐵𝐵𝐵 (𝑐𝑐𝐵𝐵∗ − 𝑐𝑐𝐵𝐵∞ ) 𝜋𝜋𝑑𝑑 2
𝑊𝑊𝐵𝐵 = 𝑁𝑁𝐵𝐵 ∙ 𝐴𝐴 =
∙
𝐿𝐿
4
=
0.260
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑒𝑒
𝑐𝑐𝑐𝑐
2
∙ (1.27 × 10−6 − 5.07 × 10−7 )
𝑠𝑠
𝑐𝑐𝑐𝑐3 ∙ 𝜋𝜋(10 𝑐𝑐𝑐𝑐)
300 𝑐𝑐𝑐𝑐
4
= 5.19 × 10−8
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 4.49 × 10−3
𝑠𝑠
𝑑𝑑𝑑𝑑𝑑𝑑
26-16
26.12
A = O2, B = silicone rubber
𝑅𝑅𝑖𝑖 = 0.635 𝑐𝑐𝑐𝑐, 𝑅𝑅𝑜𝑜 = 0.955 𝑐𝑐𝑐𝑐
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
, 𝑆𝑆 = 3.15
PA = 2.0 atm, 𝐻𝐻 = 0.78
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ′,∗
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝐴𝐴∞ = 0.005
,
𝑐𝑐
=
6.30
𝑎𝑎𝑎𝑎 𝑃𝑃𝐴𝐴 = 2.0 𝑎𝑎𝑎𝑎𝑎𝑎
𝐴𝐴
𝑚𝑚3
𝑚𝑚3
a. Develop model equation for NA from r = Ri to r = Ro
Assume: 1) PSS, 2) 1-D mass transfer along r, 3) Dilute UMD process, 4) No homogenous
reaction, 5) Convective resistances neglected.
Flux Equation
𝑑𝑑𝑑𝑑𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
General Differential Equation for Mass Transfer (cylindrical system)
d
1 d
rN A,r ) =
0 ⇒ ( rN A,r ) =
0
(
r dr
dr
N A,R o ⋅ R=
N A,r ⋅ r= const
∴ N A,R i ⋅ R=
i
o
𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 ∙ 𝑅𝑅𝑜𝑜 = 𝑟𝑟 �−𝐷𝐷𝐴𝐴𝐴𝐴
𝑅𝑅𝑜𝑜
𝑑𝑑𝑑𝑑′𝐴𝐴
�
𝑑𝑑𝑑𝑑
𝑐𝑐′𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 ∙ 𝑅𝑅𝑜𝑜 �
= −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐′𝐴𝐴
𝑟𝑟
𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 =
𝑅𝑅𝑖𝑖
𝐷𝐷𝐴𝐴𝐴𝐴 (𝑐𝑐′𝐴𝐴,∗ − 𝑐𝑐′𝐴𝐴𝐴𝐴 )
′,∗
𝑐𝑐𝐴𝐴
𝑅𝑅
𝑅𝑅𝑜𝑜 ln � 𝑜𝑜 �
𝑅𝑅𝑖𝑖
b. Determine NA at r = Ro
′
𝑐𝑐𝐴𝐴𝐴𝐴
= 𝐻𝐻 ∙ 𝑆𝑆 ∙ 𝑐𝑐𝐴𝐴∞ = �0.78
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
� �3.15
� �0.0050
� = 0.012
3
3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚
𝑚𝑚
𝑚𝑚3
DAB = 1.0 x 10-7 cm2/s (Problem 26.11)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1.0 𝑚𝑚3
𝑐𝑐𝑐𝑐2
−
0.012
�1.0 𝑥𝑥 10−7 𝑠𝑠 � �6.30
�
�
�
𝑚𝑚3
𝑚𝑚3
1.0 𝑥𝑥 106 𝑐𝑐𝑐𝑐3
𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 =
0.955 𝑐𝑐𝑐𝑐
0.955 𝑐𝑐𝑐𝑐 ∙ ln �
�
0.635 𝑐𝑐𝑐𝑐
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 1.613 𝑥𝑥 10−12
𝑐𝑐𝑐𝑐𝑠𝑠 ∙ 𝑠𝑠
26-17
26.13
A = H2O vapor, B = air
𝑇𝑇 = 303 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑃𝑃𝐴𝐴 = 0.042 𝑎𝑎𝑎𝑎𝑎𝑎, 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.266
𝑐𝑐𝑐𝑐2
𝑠𝑠
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 = 0.9952 𝑔𝑔/𝑐𝑐𝑐𝑐3
(0.042 𝑎𝑎𝑎𝑎𝑎𝑎)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑃𝑃𝐴𝐴
=
= 1.689 × 10−6
𝑐𝑐𝐴𝐴𝐴𝐴 =
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑅𝑅𝑅𝑅
𝑐𝑐𝑐𝑐3
�82.06
� ∙ (303 𝐾𝐾)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
𝑐𝑐𝐴𝐴∞ = 0
𝑅𝑅𝑜𝑜 = 0.5 𝑐𝑐𝑐𝑐
a. Determine WA
Develop model for WA
General Differential Equation for Mass Transfer (spherical system):
1 𝑑𝑑 2
�𝑟𝑟 𝑁𝑁𝐴𝐴,𝑟𝑟 � = 0
𝑟𝑟 2 𝑑𝑑𝑑𝑑
𝑁𝑁𝐴𝐴,𝑟𝑟 ∙ 𝑟𝑟 2 = 𝑁𝑁𝐴𝐴,𝑅𝑅 ∙ 𝑅𝑅2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Flux Equation:
𝑑𝑑𝑑𝑑𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
Combine
𝑅𝑅
𝑑𝑑𝑑𝑑𝐴𝐴
�
𝑑𝑑𝑑𝑑
∞
0
𝑁𝑁𝐴𝐴 ∙ 𝑅𝑅2 = 𝑟𝑟 2 �−𝐷𝐷𝐴𝐴𝐴𝐴
𝑐𝑐𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
𝑁𝑁𝐴𝐴 ∙ 𝑅𝑅2 � 2 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑑𝑑𝐴𝐴
𝑟𝑟
𝑁𝑁𝐴𝐴 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴
=
1 1
𝑅𝑅
𝑅𝑅2 � − �
𝑅𝑅 ∞
𝑊𝑊𝐴𝐴 = 2𝜋𝜋𝜋𝜋 ∙ 𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴
𝑐𝑐𝑐𝑐2
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 3600 𝑠𝑠 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
� �1.689 × 10−6
��
��
�
𝑠𝑠
𝑐𝑐𝑐𝑐3
ℎ𝑟𝑟
1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
= 5.08
ℎ𝑟𝑟
𝑊𝑊𝐴𝐴 = 2𝜋𝜋(0.50 𝑐𝑐𝑐𝑐) ∙ �0.266
26-18
b. Determine the time it takes for the water droplet to completely evaporate
Material balance on water (PSS mass transfer)
IN – OUT + GEN = ACC
0 − 𝑁𝑁𝐴𝐴,𝑟𝑟 𝑆𝑆 + 0 =
−
𝑑𝑑𝑚𝑚𝐴𝐴
𝑑𝑑𝑑𝑑
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑 2 3
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
∙ 2𝜋𝜋𝑅𝑅2 =
=
∙ 2𝜋𝜋𝑅𝑅2
� 𝜋𝜋𝑅𝑅 � =
𝑅𝑅
𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑
𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑 3
𝑀𝑀𝐴𝐴
𝑑𝑑𝑑𝑑
−𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑅𝑅 =
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 2 𝑑𝑑𝑑𝑑
∙ 𝑅𝑅
𝑀𝑀𝐴𝐴
𝑑𝑑𝑑𝑑
𝑡𝑡
0
0
𝑅𝑅
𝑀𝑀𝐴𝐴
� 𝑑𝑑𝑑𝑑 = � 𝑅𝑅𝑅𝑅𝑅𝑅
−𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑅𝑅2
𝑡𝑡 =
=
2𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴
𝑔𝑔
𝑐𝑐𝑐𝑐3
2
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔
𝑐𝑐𝑐𝑐
2 ∙ 0.266
∙ 1.689 × 10−6
∙ 18.02
3
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐
(0.5 𝑐𝑐𝑐𝑐)2 ∙ 0.995
t = 15,380 s = 4.3 hr
26-19
26.14
A = pheromone drug vapor, B = polymer diffusion layer
𝐿𝐿 = 0.15 𝑐𝑐𝑐𝑐, 𝑝𝑝𝐴𝐴 = 𝑐𝑐′𝐴𝐴 ∙ 𝑆𝑆, 𝑁𝑁𝐴𝐴 = 𝑘𝑘𝐺𝐺 (𝑝𝑝𝐴𝐴𝐴𝐴 − 𝑝𝑝𝐴𝐴∞ ), 𝑝𝑝𝐴𝐴∞ ≈ 0, 𝑝𝑝𝐴𝐴𝐴𝐴 = 𝑝𝑝𝐴𝐴∗
𝑘𝑘𝐺𝐺 = 1 × 10−5
10−6 𝑐𝑐𝑐𝑐2
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
,
𝐷𝐷
=
1.0
𝑥𝑥
, 𝑆𝑆 = 0.80 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎/𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝐴𝐴𝐴𝐴
𝑠𝑠
a. Develop model for NA
Assume: 1) Steady state, 2) 1-D flux along z, 3) Dilute system, 4) No homogenous reaction
Differential Equation for Mass Transfer
𝑑𝑑𝑁𝑁𝐴𝐴,𝑧𝑧
= 0 (𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑎𝑎𝑎𝑎𝑜𝑜𝑛𝑛𝑛𝑛 𝑧𝑧)
𝑑𝑑𝑑𝑑
Flux Equation
𝑑𝑑𝑑𝑑′𝐴𝐴
𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
𝑝𝑝
𝑐𝑐 ′ 𝐴𝐴𝐴𝐴 = 𝑆𝑆𝐴𝐴𝐴𝐴
𝐿𝐿
𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴
0
�
𝑐𝑐 ′ 𝐴𝐴𝐴𝐴
𝑝𝑝∗
= 𝐴𝐴
𝑆𝑆
𝑑𝑑𝑑𝑑𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴 ∗
(𝑝𝑝 − 𝑝𝑝𝐴𝐴𝐴𝐴 )
𝑁𝑁𝐴𝐴 =
𝐿𝐿𝐿𝐿 𝐴𝐴
𝑁𝑁𝐴𝐴 = 𝑘𝑘𝐺𝐺 (𝑝𝑝𝐴𝐴𝐴𝐴 − 𝑝𝑝𝐴𝐴∞ )
𝑁𝑁𝐴𝐴
𝑝𝑝𝐴𝐴𝐴𝐴 =
+ 𝑝𝑝𝐴𝐴∞
𝑘𝑘𝐺𝐺
𝐷𝐷𝐴𝐴𝐴𝐴 ∗ 𝑁𝑁𝐴𝐴
𝑁𝑁𝐴𝐴 =
− 𝑝𝑝𝐴𝐴∞ �
�𝑝𝑝 −
𝐿𝐿𝐿𝐿 𝐴𝐴 𝑘𝑘𝐺𝐺
(𝑝𝑝𝐴𝐴∗ − 𝑝𝑝𝐴𝐴∞ )
𝑁𝑁𝐴𝐴 =
𝐿𝐿𝐿𝐿
1
+
𝐷𝐷𝐴𝐴𝐴𝐴 𝑘𝑘𝐺𝐺
b. Determine NA
𝑁𝑁𝐴𝐴 =
(1.1 𝑎𝑎𝑎𝑎𝑎𝑎 − 0)
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇
=
9.82
𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠
0.15 𝑐𝑐𝑐𝑐 ∙ 0.80
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1
+
2
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐
1 × 10−5
1 × 10−6
𝑠𝑠
𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
26-20
26.15
bulk stomach fluid
(essentially water)
cA∞ ≈ 0
detail for one pore
cA∞ ≈ 0
0.4 mm
initially
NA
cross-section
view of pill
z
Lt =
2.0 mm
0.36 cm
Lo =
1.2 mm
at a
later time
solid
A
r
Given:
A = drug, B = water inside pore
T = 310 K
cA* = 2.0 ×10−4 gmole/cm3, ρA,solid = 1.10 g/cm3, MA = 120 g/gmole
DAB = 2.0 × 10−5 cm2/s
Pore dimensions: d = 0.04 cm, Lt = 0.20 cm, Lo = 0.12 cm
a. Transfer rate WA at L = Lt = 0.20 cm
Physical System: liquid inside pore (System I)
Source for A: solid A at base of pore (dissolves into liquid)
Sink for A: surrounding fluid at pore exit
Assumptions
1. Pseudo-steady state (PSS) process for A
2. No homogenous reaction of A
3. 1-D flux along z
26-21
4. Dilute UMD process with respect to A
5. Surrounding fluid is semi-infinite sink for A with cA∞ ≈ 0
General Differential Equation for Mass Transfer
∂c
∂c
−∇N A + RA =A RA = 0, A = 0,1-D flux along z
∂t
∂t
dN
∴ A, z =
0 ( flux is constant along z )
dz
General Flux Equation - for a dilute process with respect to A
dc
N A, z = − DAB A
dz
Since NA,z is constant along z, separate variables cA and z, pull NA,z outside of the integral, then
integrate with respect to limits shown below
z = 0, cA = cA*; z = L, cA = cA∞ = 0
L
0
L0
c*A
N A ∫ dz = − DAB ∫ dc A
NA =
DAB c*A
L − Lo
2

−5 cm  
−4 gmole 
2.0
10
×
2

  2.0 × 10

*
2
s 
cm3  π ( 0.04 cm )
DAB c A π d

WA= N A, z ⋅ S=
np =
(16 pores )
4
L − Lo 4
( 0.2 cm - 0.12 cm )
WA = 1.01 × 10−9
gmole
s
b. Time required for complete release of drug from pore
Material balance on solid drug A loaded into base of pore (System II)
IN – OUT + GEN = ACC
dm
0 − N AS + 0 = A
dt
0−
ρ
DAB c*A
dL
S + 0 = A,solid S
Lt − L
MA
dt
Let δ =
Lt − L,
dδ
dL
=
−
dt
dt
26-22
At complete release, L = Lt
DAB c*A M A
ρ A,solid
DAB c*A M A
ρ A,solid
t=
t
Lt
0
Lt − Lo
∫dt =
t=
∫ δ dδ
Lt 2 − ( Lt − Lo )
2
2
Lt 2 − ( Lt − Lo )
2
ρ A,solid
DAB c*A M A
2
1.0 h 
(( 0.2 cm ) − ( 0.2 cm − 0.12 cm ) ) 1.10 cmg   3600

s
2
2
3
t=

cm 2  
g 
−4 gmole  
2  2.0 ×10−5
120
  2.0 × 10
3 
s 
cm  
gmole 

= 10.7 h
c. Increase required time by a factor of 2.0
Increase Lt
 Lt 2 − ( Lt − Lo )2 
 Lt 2 − ( Lt − 0.12 cm )2 
tnew

 new

 new
= 2=
=
2
2
2
told
 Lt − ( Lt − Lo ) 
 0.2 cm ) − ( 0.2 cm − 0.12 cm )2 

 old (
 old
Lt ,new = 0.34 cm
26-23
26.16
bulk fluid
cA ∞ ≈ 0
detail for one pore
cA ∞ ≈ 0
0.1 cm
cross-section
view of pore
array
NA
Lt =
0.5 cm
z
at the
final
time
solid
A
4.0 cm
(not to scale)
Lo =0.20 cm
r
Assumptions
1.
2.
3.
4.
5.
Pseudo-steady state (PSS) process for A
No homogenous reaction of A
1-D flux along z
Dilute UMD process with respect to A
Surrounding fluid is semi-infinite sink for A with cA∞ ≈ 0
Total transfer rate, WA
General Differential Equation for Mass Transfer
∂c
−∇N A + RA =A
∂t
Simplification based on assumptions 1-3
∴
dN A, z
0 ( flux is constant along z )
=
dz
General Flux Equation - for a dilute process with respect to A
N A, z = − DAB
dc A
dz
L
0
L0
c*A
N A ∫ dz = − DAB ∫ dc A
26-24
NA =
DAB c*A
L − Lo
2

µ mole 
−5 cm  
1.6
10
×
2

 90

*
2
s 
cm3  π ( 0.19 cm )
DAB c A π d

WA= N A, z ⋅ S=
n p=
(16 pores )
4
L − Lo 4
( 0.25cm - 0.2 cm )
WA = 6.03 × 10−4
µ mole
s
b. Time required to completely dissolve solid in pore
Material balance on solid A loaded into base of pore (System II)
IN – OUT + GEN = ACC
dm
0 − N AS + 0 = A
dt
0−
ρ
DAB c*A
dL
S + 0 = A,solid S
Lt − L
MA
dt
Let δ =
Lt − L,
dδ
dL
=
−
dt
dt
At complete release, L = Lt
DAB c*A M A
ρ A,solid
DAB c*A M A
ρ A,solid
t=
t
Lt
0
Lt − Lo
∫dt = ∫ δ dδ
t=
Lt 2 − ( Lt − Lo )
2
2
Lt 2 − ( Lt − Lo )
2
ρ A,solid
DAB c*A M A
2
1.0 h 
(( 0.5 cm ) − ( 0.5 cm − 0.2 cm ) )  2.10 cmg  3600

s
2
2
3
t=

cm 2   µ mole  
g 
2 1.6 × 10−5
74
 90

3 
s 
cm   gmole 

= 438 h
26-25
26.17
Given:
A = O2, B = air, C = carbon (s), D = CO2
T = 1000 K, P = 2.0 atm
Electrode dimensions: L = 25 cm, d = 2.0 cm, δ = 0.5 cm boundary layer
Electrode materials: ρC,solid = 2.25 g/cm3, MC = 12 g/gmole
a. Determine WA
Model for WA
Physical System: gas space in boundary layer between carbon electrode surface and bulk gas
(System I)
Source for A: bulk gas
Sink for A: consumption by carbon electrode to CO2 gas
Assumptions
1.
2.
3.
4.
PSS process for A
No homogenous reaction of A in gas space
1-D flux along r
Instantaneous consumption of O2 gas (A) at electrode surface so that cAs = 0 at r = R
(instantaneous surface reaction)
5. NA,r = − ND,r by reaction C ( s ) + O 2 ( g ) → CO 2 ( g )
CO2
solid carbon
electrode
bulk gas
cA∞
O2 (A)
25 cm
NA,r
r = R+δ = 1.5 cm
r = R = 1.0 cm
cAs = 0
26-26
General Differential Equation for Mass Transfer (cylindrical system)
∂c A
=
RA 0,=
0 by assumptions 1 and 2
∂t
∂c A
1 d
rN A,r ) + RA =
(
∂t
r dr
1 d
∴
0
( r ⋅ N A, r ) =
r dr
d
( r ⋅ N A, r ) = 0
dr
−
By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r
Note WA = N A,r r = R ⋅ 2π RL = N A,r ⋅ 2π rL
Flux Equation
dc A c A
dc
c
+ ( N A,r + N B ,r + N C ,r + N D,r ) =− DAB A + A ( N A,r + 0 + 0 + − N A,r )
dr C
dr C
dc
N A,r = − DAB A
dr
dc 

N A,r ⋅ 2π rL =
2π rL  − DAB A 
=
Note: WA N=
A,r r = R ⋅ 2π RL
dr 

WA is not a function of r; separate variables cA, r, pull WA outside of the integral, then integrate
with respect to limits shown below
N A,r =− DAB
r = R, cA = cAs = 0 (rapid consumption of O2 at surface)
r = R + δ , cA = cA∞ (bulk gas composition beyond boundary layer)
R
0
dr
= −2π LDAB ∫ dc A
r
R +δ
c A∞
WA ∫
WA =
2π LDAB c A∞
 R +δ 
ln 

 R 
Calculate WA
26-27
c A∞ =
( 0.21)( 2.0 atm )
y A∞ P
gmole
=
=4.65 × 10−6
3
RT 
cm3
cm atm 
82.06
1100
K
(
)


gmole ⋅ K 

 P  T 
DAB (T , P) = DAB (Tref , Pref )  ref  

 P   Tref 
3/2
3/2
= 0.136
cm 2  1.0 atm   1100 K 
cm 2
=
0.55



s  2.0 atm   273 K 
s

cm 2  
−6 gmole 
2π ( 25 cm )  0.55
  4.65 × 10

s 
cm3 
gmole

= 1.80 × 10−3
WA =
s
 2.5 cm 
ln 

 2.0 cm 
b. Time it takes for the rod to completely disappear (t at R = 0)
Material balance on carbon (PSS mass transfer) to model “shrinkage” of carbon electrode
(System II)
IN – OUT + GEN = ACC
0 − N A, r ⋅
mC =
dm
1.0 mol C
⋅S +0 = C
1.0 mol O 2
dt
ρC π R 2 L
MC
ρC
DAB c A∞
dR
−
2π RL =
2π RL
MC
dt
 R +δ 
R ⋅ ln 

 R 
ρ A dR
DAB c A∞
R
−
=
 R + δ  M A dt
ln 

 R 
Separate variables R, t and integrate, then solve for time t at R = 0
t
ρC 0  R + δ 
− DAB c A∞ ∫dt =
ln 
 RdR
M C ∫R  R 
0
R
 R +δ 
1 
 R +δ 
 1
 R +δ 


=
R  R ⋅ ln 
 + δ  − δ ln 
∫ln  R  RdR
2 
 R 
 δ
 2
0
2
26-28

 2  R +δ 
 R +δ 
R  R ⋅ ln 
 + δ  − δ ln 

 R 
 δ  ρC

t= 
2 DAB c A∞
MC



2
 2.5 cm 
 2.5 cm   
g 
 2.0 cm  ( 2.0 cm ) ln 
 +0.5 cm  − ( 0.5 cm ) ln 
   2.25 3 
 2.0 cm 
 0.5 cm   
cm 


t=
= 5455 s
2


g 
cm  
−6 gmole 
2  0.55
  4.65 × 10
12.01 gmole 

s 
cm3 



26-29
26.18
Physical System: SiO2 solid layer (species B) containing “dissolved” O2 (species A)
SOURCE for O2: 100% O2 gas over SiO2 solid film
SINK for O2: unreacted Si at boundary surface
SYSTEM I: dissolved O2 in SiO2 film
SYSTEM II: unreacted Si
Heterogeneous Surface Reaction: Si ( s ) + O2 ( g ) → SiO2 ( s )
𝑇𝑇 = 1273 𝐾𝐾, 𝑃𝑃𝐴𝐴 = 1 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑐𝑐𝐴𝐴𝐴𝐴 = 9.6 × 10−8
Assumptions:
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑂𝑂2
𝑔𝑔
𝑔𝑔
, 𝑀𝑀𝐵𝐵 = 60.08
, 𝜌𝜌𝐵𝐵 = 2.27 3
3
𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐
1. The oxidation of Si to SiO2 occurs only at the Si/SiO2 interface. The unreacted Si at the
interface serves as the sink for molecular mass transfer of O2 through the film.
2. The O2 in the gas phase above the wafer represents an infinite source for O2 transfer. The O2
molecules “dissolve” into the nonporous SiO2 solid at the gas solid interface.
3. The rate of SiO2 formation is controlled by the rate of molecular diffusion of O2 (species A)
through the solid SiO2 layer (species B) to the unreacted Si layer. The reaction is very rapid,
so that the concentration of molecular O2 at the interface is equal to zero, i.e CA = 0.
Furthermore, there are no mass transfer resistances in the gas film above the wafer surface,
since O2 is a pure component in the gas phase.
4. The flux of O2 through the SiO2 layer is one-dimensional along coordinate z.
5. The rate of SiO2 film formation is slow enough so that at a given film thickness δ, there is no
accumulation of reactants or products within the SiO2 film. However, the thickness of the
film will still increase with time (the PSS Assumption).
6. The overall thickness of the wafer does not change as the result of the formation of the SiO2
layer.
7. The process is isothermal.
General Differential Equation for Mass Transfer
dN Az
= 0 (NA,z is constant along z, therefore can integrate flux equation directly)
dz
Flux Equation
dc
c
dc
c
dc
N Az =
− DAB A + A ( N Az + N Bz ) =
− DAB A + A N Az ≅ − DAB A
dz
c
dz
c
dz
δ
0
∫ N Az dz = − DAB ∫ dcA
0
N Az =
c As
DAB c As
δ
26-30
Mass balance on SiO2
IN – OUT + GENERATION = ACCUMULATION
1.0 mole SiO 2 (B) dmB
=
1.0 mole O 2 (A)
dt
δ
M B D AB c As t
d
δ
δ
=
∫
∫ dt or δ =
ρ
B
0
0
ρ Sδ 
d B

dmB
MB 

=
dt
dt
0 − 0 + N A, z ⋅ S ⋅
2 M B D AB C As
ρB
A plot of δ vs. t 0.5 data will be linear with least-squares slope =
note: δ ∝ t
t
2 M B D AB C As
ρB
and intercept = 0
SiO2 Film Thickness, δ (µm)
Measured
Predicted
Time
t (h)
t0.5 (h0.5)
(100)Si
(111)Si
(100)Si
(111)Si
1.0
2.0
4.0
7.0
16.0
1.000
1.414
2.000
2.646
4.000
0.049
0.078
0.124
0.180
0.298
0.070
0.105
0.154
0.212
0.339
0.069
0.098
0.138
0.183
0.277
0.081
0.115
0.163
0.215
0.326
(100)Si
(111)Si
slope =
0.0692
0.081
µm/h0.5
ρB =
2.270
2.270
g/cm3
MB =
60.0
60.0
g/mole
DAB cAs = 2.513E-16 3.486E-16 mol/cm-s
cAs = 9.550E-08 9.550E-08 mol/cm3
DAB = 2.631E-09 3.650E-09 cm2/s
Sample Calculation (with units)
2

ρB
2
0.5 2  1 cm   1 hr 
DAB = ( slope ) 
 = ( 0.0692 μm/h )  4

 
 10 μm   3600 s 
 2M B CAs 


3


( 2.27 g/cm )

 = 2.63 x 10-9 cm 2 /s


g 
-8 gmol 
  2*60
 9.550×10

3 
gmol  
cm  

26-31
0.4
Data (100)Si
SiO2 film thickness,δ (µm)
Best Fit (100)Si
Data (111)Si
0.3
Best Fit (111)Si
0.2
0.1
0.0
0
1
2
Time, t
3
0.5
4
5
0.5
(hr)
26-32
26.19
A = acetone, B = air
𝑑𝑑 = 0.3 𝑐𝑐𝑐𝑐, 𝑇𝑇 = 293.9 𝐾𝐾
𝑃𝑃𝐴𝐴∗ = 0.254 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑀𝑀𝐴𝐴 = 58
𝑔𝑔
𝑔𝑔
, 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 = 0.79 3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐
a. Determine DAB from Arnold Diffusion Cell data
Assume: 1) 1-D mass transfer along z, 2) No homogenous reaction, 3) PSS.
Material balance on acetone (PSS mass transfer)
IN – OUT + GEN = ACC
𝑑𝑑𝑚𝑚𝐴𝐴
0 − 𝑁𝑁𝐴𝐴,𝑧𝑧 𝑆𝑆 + 0 =
𝑑𝑑𝑑𝑑
𝑁𝑁𝐴𝐴 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐
1 − 𝑦𝑦𝐴𝐴2
ln �
�
1 − 𝑦𝑦𝐴𝐴1
𝑍𝑍
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐
1 − 𝑦𝑦𝐴𝐴2
ln �
�=
∙
𝑍𝑍
1 − 𝑦𝑦𝐴𝐴1
𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑
𝑡𝑡
𝑍𝑍
0
𝑍𝑍𝑜𝑜
𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴
� 𝑑𝑑𝑑𝑑 = � 𝑍𝑍𝑍𝑍𝑍𝑍
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙
Z2-Zo2 [cm2]
𝑍𝑍 2 − 𝑍𝑍0 2 =
2𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝑀𝑀𝐴𝐴
1 − 𝑦𝑦𝐴𝐴∞
ln �
� (𝑡𝑡 − 𝑡𝑡𝑜𝑜 )
𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙
1 − 𝑦𝑦𝐴𝐴𝐴𝐴
160
140
120
100
80
60
40
20
0
y = 0.6472x + 0.5537
0
50
100
150
200
250
time [hr]
26-33
Slope = 0.6472
𝑐𝑐 =
𝑃𝑃
=
𝑅𝑅𝑅𝑅
𝑦𝑦𝐴𝐴𝐴𝐴 =
𝐷𝐷𝐴𝐴𝐴𝐴 =
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
= 1.80 × 10−4
ℎ𝑟𝑟
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1.0 𝑎𝑎𝑎𝑎𝑎𝑎
= 4.15 × 10−5
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐3
82.06
∙ 293.9 𝐾𝐾
𝐾𝐾 ∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑃𝑃𝐴𝐴∗ 0.254 𝑎𝑎𝑎𝑎𝑎𝑎
=
= 0.254
𝑃𝑃
1.0 𝑎𝑎𝑎𝑎𝑎𝑎
𝜌𝜌𝐴𝐴 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
1−0
2 ∙ 𝑐𝑐 ∙ 𝑀𝑀𝐴𝐴 ln �
�
1 − 𝑦𝑦𝐴𝐴𝐴𝐴
𝑔𝑔
𝑐𝑐𝑐𝑐2
∙ 1.80 × 10−4
𝑐𝑐𝑐𝑐3
3
𝑠𝑠
𝑐𝑐𝑐𝑐
= 0.101
𝐷𝐷𝐴𝐴𝐴𝐴 =
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔
1−0
𝑠𝑠
2 ∙ 4.15 × 10−5
∙
58
∙
ln
�
�
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1 − 0.254
𝑐𝑐𝑐𝑐3
0.79
b. Determine DAB by correlation
Hirschfelder Correlation
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
1 2
0.001858 ∙ 𝑇𝑇 3/2 �𝑀𝑀 + 𝑀𝑀 �
𝜎𝜎𝐴𝐴 = 2.44 �
𝜀𝜀𝐴𝐴
𝜅𝜅
2
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴
ΩD
508 𝐾𝐾
47.4 𝑎𝑎𝑎𝑎𝑎𝑎
𝐴𝐴
𝐴𝐴
1
3
� = 5.380 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å,, 𝜎𝜎𝐴𝐴𝐴𝐴 =
5.380+3.617
𝜀𝜀
2
= 4.50 Å
= 0.77(508 𝐾𝐾) = 391.16 𝐾𝐾, 𝐵𝐵 = 97 𝐾𝐾,𝜀𝜀𝐴𝐴𝐴𝐴 = √391.16 ∙ 97 𝐾𝐾 = 194.79 𝐾𝐾
𝜅𝜅𝜅𝜅
293.9 𝐾𝐾
=
= 1.509,
𝜀𝜀𝐴𝐴𝐴𝐴 194.79 𝐾𝐾
𝐷𝐷𝐴𝐴𝐴𝐴 =
𝜅𝜅
∴ Ω𝐷𝐷 = 1.195
1
1 1/2
+
]
𝑐𝑐𝑐𝑐2
58.0 29.0
=
0.088
1.0 ∙ 4.50 2 ∙ 1.195
𝑠𝑠
0.001858 ∙ 293.93/2 [
26-34
26.20
A = O2 and B = CO2, 𝑦𝑦𝐴𝐴∞ = 1.0, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0, 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.1 𝑐𝑐𝑐𝑐, 𝐿𝐿𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.5 𝑐𝑐𝑐𝑐
𝑃𝑃 = 2.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑇𝑇 = 873 𝐾𝐾, 𝜌𝜌𝐶𝐶 = 2.25
𝑔𝑔
𝑔𝑔
, 𝑀𝑀𝐶𝐶 = 12.01
3
𝑐𝑐𝑐𝑐
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
Assume: 1) PSS system, 2) No homogenous reaction, 3) 1-D flux along z, EMCD
a. Determine WB at Z = 0.30 cm
Flux model for NA
d
( N A,z ) = 0 , constant flux along z
dz
𝑑𝑑𝑑𝑑𝐵𝐵
𝑁𝑁𝐴𝐴,𝑧𝑧 = −𝐷𝐷𝐵𝐵𝐵𝐵 𝑐𝑐
𝑑𝑑𝑑𝑑
𝑧𝑧 = 0, 𝑦𝑦𝐴𝐴𝐴𝐴 = 0, 𝑦𝑦𝐵𝐵𝐵𝐵 = 1.0
𝑍𝑍 = 0.3 𝑐𝑐𝑐𝑐, 𝑦𝑦𝐴𝐴𝐴𝐴 = 1.0, 𝑦𝑦𝐵𝐵𝐵𝐵 = 0
𝑍𝑍
𝑦𝑦𝐵𝐵∞ =0
0
𝑦𝑦𝐵𝐵𝐵𝐵 =1
𝑁𝑁𝐵𝐵 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐵𝐵𝐵𝐵 𝑐𝑐
𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐
𝑁𝑁𝐵𝐵 =
𝑍𝑍
�
𝑑𝑑𝑦𝑦𝐵𝐵
Hirschfelder Correlation
𝐷𝐷𝐴𝐴𝐴𝐴 =
1
1 2
0.001858 ∙ 𝑇𝑇 3/2 �𝑀𝑀 + 𝑀𝑀 �
2
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴
ΩD
𝐴𝐴
𝐴𝐴
𝜎𝜎𝐵𝐵 = 3.433 Å, 𝜎𝜎𝐴𝐴 = 3.996 Å,, 𝜎𝜎𝐵𝐵𝐵𝐵 =
𝜀𝜀𝐵𝐵
𝜅𝜅
𝜀𝜀
3.996+3.433
2
= 3.7145 Å
= 113 𝐾𝐾, 𝐴𝐴 = 190 𝐾𝐾,𝜀𝜀𝐵𝐵𝐵𝐵 = √113𝐾𝐾 ∙ 190 𝐾𝐾 = 146.53 𝐾𝐾
𝜅𝜅
𝜅𝜅𝜅𝜅
873 𝐾𝐾
=
= 5.958,
𝜀𝜀𝐵𝐵𝐵𝐵 146.53 𝐾𝐾
𝐷𝐷𝐴𝐴𝐴𝐴 = 𝐷𝐷𝐵𝐵𝐵𝐵 =
∴ Ω𝐷𝐷 = 0.8137
1
1 1/2
2
44.01 + 32.0] = 0.496 𝑐𝑐𝑐𝑐
2.0 ∙ 3.7145 2 ∙ 0.8137
𝑠𝑠
0.001858 ∙ 873 3/2 [
26-35
𝑐𝑐 =
𝑃𝑃
=
𝑅𝑅𝑅𝑅
2.0 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 2.79 × 10−5
3
𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐3
∙ 873 𝐾𝐾
82.06
𝐾𝐾 ∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑊𝑊𝐵𝐵 = 𝑁𝑁𝐵𝐵 𝑆𝑆 =
2
𝐷𝐷𝐵𝐵𝐵𝐵 𝑐𝑐 𝜋𝜋𝑑𝑑
=
𝑍𝑍
4
= 3.62 × 10−7
0.496
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐2
2
∙ 2.79 × 10−5
𝑠𝑠
𝑐𝑐𝑐𝑐3 𝜋𝜋(0.1 𝑐𝑐𝑐𝑐)
0.3 𝑐𝑐𝑐𝑐
4
b. Determine time t for carbon in pore to be completely oxidized
Material balance on solid carbon in pore
IN – OUT + GEN = ACC
0 − 𝑁𝑁𝐵𝐵,𝑧𝑧 𝑆𝑆 ∙
1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐶𝐶
𝑑𝑑𝑚𝑚𝐶𝐶
+0=
1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐶𝐶𝐶𝐶2
𝑑𝑑𝑑𝑑
𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐 𝜌𝜌𝐶𝐶 𝑑𝑑𝑑𝑑
=
𝑀𝑀𝐶𝐶 𝑑𝑑𝑑𝑑
𝑍𝑍
𝑡𝑡
𝑍𝑍
0
0
𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐
𝜌𝜌𝐶𝐶
� 𝑑𝑑𝑑𝑑 =
� 𝑑𝑑𝑑𝑑
𝑍𝑍
𝑀𝑀𝐶𝐶
𝑡𝑡 =
𝑍𝑍 2
𝜌𝜌𝐶𝐶
2𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐 𝑀𝑀𝐶𝐶
𝑔𝑔
3
𝑐𝑐𝑐𝑐
𝑡𝑡 =
= 609 𝑠𝑠𝑠𝑠𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 12.01 𝑔𝑔
𝑐𝑐𝑐𝑐2
2 ∙ 0.496
∙ 1.0 ∙ 2.79 × 10−5
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑠𝑠
𝑐𝑐𝑐𝑐3
(0.3 𝑐𝑐𝑐𝑐)2
2.25
26-36
26.21
drug patch (side view)
drug reservoir
impermeable
barrier
skin surface
diffusion barrier
(water-filled
micropores,
dpore = 100 Å)
L
infected tissue
(sink for drug)
SIDE VIEW
micropore array
(not to scale)
Given:
A = drug, B = water
T = 298 K
cAs = 1.0 × 10−6 gmole/cm3, DAB = 1.0 × 10−6 cm2/s
Pore dimensions: ds = 25 Å, dpore = 100 Å, S = 0.25 (4.0 cm2) = 1.0 cm2

μmol   1day 1 gmole 
−13 gmole
=
WA N=
 0.05

 = 5.79 × 10
AS
6
day   86,400 s 10 μmole 
s

a. Effective diffusion coefficient, DAe
Solute diffusion through solvent-filled pore
°
DAe = DAB
F1 (ϕ ) F2 (ϕ )
Reduced pore diameter
ds
25 A
=
ϕ =
= 0.25
d pore 100 A
F1 (ϕ ) =
(1 − ϕ ) 2 =
(1 − 0.25) 2 =
0.5625
F2 (ϕ ) =
1 − 2.104(ϕ ) + 2.09(ϕ )3 − 0.95(ϕ )5
F2 (ϕ ) =
1 − 2.104(0.25) + 2.09(0.25)3 − 0.95(0.25)5 =
0.5057
°
DAe = DAB
F1 (ϕ ) F2 (ϕ ) = 1.0 × 10−6
cm 2
cm 2
( 0.5625)( 0.5057 ) = 2.84 × 10−7
s
s
26-37
b. Thickness of the diffusion barrier, L
Physical System: liquid inside pore
Source for A: solid A at base of pore (dissolves into liquid)
Sink for A: surrounding fluid at pore exit
Assumptions
1.
2.
3.
4.
5.
6.
Steady-state (PSS) process for A
No homogenous reaction of A
1-D flux along z
Dilute UMD process with respect to A
Surrounding fluid is semi-infinite sink for A with cA∞ ≈ 0
Hindered diffusion of solute in solvent-filled pores
General Differential Equation for Mass Transfer
−∇N A + R=
A
∴
∂c A
∂t
R=
0,
A
∂c A
= 0,1-D flux along z
∂t
d
0 ( flux is constant along z )
( N A,z ) =
dz
General Flux Equation—for a dilute process with respect to A
dc
N A = − DAe A
dz
Since NA,z is constant along z, separate variables cA and z, pull NA,z outside of the integral, then
integrate with respect to limits shown below
z = 0, cA = cAs; z = L, cA = cA∞ = 0
L
0
0
c As
N A ∫dz = − DAe ∫ dc A NA =
DAe c As
L
WA N=
=
AS
L=
DAe c As
S
L
DAe c As
S
WA
2

−7 cm  
−6 gmole 
2.84
10
×

 1.0 × 10

s 
cm3 

L=
1.0 cm 2 = 0.568 cm

−13 gmole 
 5.79 × 10

s 

(
)
26-38
26.22
2.0 mol% TCE
in air
porous catalyst layer
degrades TCE
(k = 1.5 s-1)
well-mixed
gas phase
nonporous
barrier
CAo
z = 0 z = L = 0.5 cm
5.0 mol% TCE (A)
in air
a. Boundary Conditions
z = 0, CA = CAo
z = L,
dC A
=0
dz
b. yA at z = L
k
1.5s -1
L=
=0.5 cm
2.165
cm 2
DAe
0.080
s
at z = L
φ

k 
cosh  ( L − z )

DAe 
c Ao

c A c=
Ao
cosh(φ )
cosh(φ )
0.02
y Ao
=
= 0.0045
yA =
cosh(φ ) cosh(2.165)
c. Surface area S at WA = 0.20 gmole A/s
WA = N A S
26-39
=
NA
DAec Ao
DAe y AoC
=
φ tanh φ
φ tanh φ
δ
δ

cm 
gmole  1m 

 0.080
 ( 0.02 )  21.3


s 
m3  100 cm 


(2.165)tanh(2.165)
NA =
( 0.5 cm )
2
N A =1.44 × 10-3
=
S
gmole A
m2 ⋅ s
WA
0.20 gmole A/s
=
=139 m 2
-3
2
N A 1.44×10 gmoleA/m s
26-40
26.23
nonporous
support
well-mixed
liquid phase
enzyme
layer
cA∞= 1.0 mmol/m3
z=0
z = L = 0.2 cm
cAL = 0.25 mmol/m3
a. Boundary Conditions
z = 0, CA = CA∞
z = L,
dC A
=0
dz
b. Rate constant k
at z = L
c AL

k 
cosh  ( L − z )

DAe 
c A∞

c=
Ao
cosh(φ )
cosh(φ )
c 
 c AL 
φ = cosh −1  A∞ 
φ=L
k
DAe

cm 2 
2.0×10-5


3
 
s 

DAe 
-1  1.0 mmole/m
−1  c A∞ 
=
cosh
k = 2 cosh 



2
3 
L 
( 0.2 cm ) 
 0.25 mmole/m  
 c AL  
k 2.13 × 10−3 s −1
=
c. Total transfer rate of product B, WB
26-41
k
=
φ L= 0.2 cm
DAe
=
NA
2.13×10-3s -1
2.063
=
2
-5 cm
2.0×10
s
DAec Ao
DAe y AoC
=
φ tanh φ
φ tanh φ
δ
δ

mmole  1m 
−5 cm  
 2.0 × 10
1.0


s 
m3  100 cm 
(2.063)tanh(2.063)
NA = 
( 0.2 cm )
2
mmole A
m2 ⋅ s
mmole A
1 mol B
 1 mol B 
2
-6 mmole A 
4.0 × 10-6
=
WB =
NA S 
)
 2.0 × 10
 ( 2.0 m=
2
s
1 mol A
m ⋅s 
 1 mol A 
N A =2.0 × 10-6
d. At φ = 2.063, process is between diffusion control and reaction control
26-42
26.24
a. Develop differential model for cA(r) using shell balance
A = drug, B = gel material
System: gel bead
Source: the surrounding liquid
Sink: gel bead
Assume: 1) Steady State, 2) Uniform Distribution within bead, 3) 1-D Mass Transfer, 4) Dilute
with respect to solute A
Flux equation for first order reactions
dc
N A,r = − DAB A
dr
Material Shell Balance
IN – OUT + GEN = ACC = 0
4π r 2 N A,r |r −4π r 2 N A,r |r +∆r +4π r 2 ∆r =0
÷ 4𝜋𝜋 △ 𝑟𝑟, rearrange with lim ∆r → 0
1 d 2
− 2
0
( r N A , r ) + RA =
r dr
dc A 
1 d  2
0
 r DAB
 + RA =
2
r dr 
dr 
𝐷𝐷𝐴𝐴𝐴𝐴 �
2 𝑑𝑑𝑑𝑑𝐴𝐴 𝑑𝑑 2 𝑐𝑐𝐴𝐴
+
� − 𝑘𝑘1 𝑐𝑐𝐴𝐴 = 0
𝑟𝑟 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 2
Boundary Conditions
𝑟𝑟 = 0,
𝑟𝑟 = 𝑅𝑅,
𝑑𝑑𝑑𝑑𝐴𝐴
=0
𝑑𝑑𝑑𝑑
𝑐𝑐𝐴𝐴 = 𝑐𝑐𝐴𝐴∞
b. Determine WA
c A (r ) = c A∞
(
R sinh r k1 / DAB
r sinh R k / D
AB
1
( (
)
))
26-43
(
(

d 
R sinh r k1 / DAB
c A∞
dr 
r sinh R k1 / DAB

cosh ( r k / D )
)  =
c R sinh ( r k / D ) c R
k /D
−
+
sinh ( R k / D )
)  r sinh ( R k / D ) r
1
A∞
2
AB
A∞
1
1
1
AB
1
AB
AB
AB


R k1 / DAB
dc A
2
1 −

|
4
|
4
=
=
−
=
π
π
W=
SN
R
D
D
c
R
A
A, r r R
AB
r R
AB A∞
=
 tanh R k / D 
dr
1
AB




0.019 s -1 )
(


( 0.3 cm )
-6
2

( 2x10 cm /s ) 
-6
2
3

WA = 4π ( 2 x 10 cm /s )( 0.02 μmol/cm ) ( 0.3 cm ) 1

( 0.019 s-1 )  
 tanh  ( 0.3 cm )


( 2x10-6cm2 /s )  


μmol 3600 s
μmol
= 0.0153
WA = 4.26 x 10-6
s
hr
hr
(
)
Note
s )
( 0.019
=
φ R=
k /D
= 29.24
( 0.3 cm )
( 2x10 cm /s )
-1
1
AB
-6
2
26-44
26.25
Given:
A = TCE, B = biofilm
T = 20 C (293 K)
cAi = 0.25 mg TCE/L = 0.25 g TCE/m3
Biofilm: S = 800 m2, δ = 100 µm, DAB = 9.03 × 10−9 m2/s, k = 4.3 s−1
a. Volumetric inlet flow rate, vo
Develop process model:
Source for A: liquid inflow
Sink for A: consumption by biofilm
Assumptions
1.
2.
3.
4.
Constant source and sink for A—steady-state process
Well-mixed liquid phase
Constant liquid volume
No consumption of A in bulk liquid phase
Material balance in TCE in biofilm reactor, bulk liquid phase
IN – OUT + GEN =
ACC
v 0 c Ai − v 0 c Ao − S ⋅ N A =
0
v0 =
S ⋅ NA
c Ai − c A
Recall NA for 1-D flux along z, homogeneous first-order reaction with RA =−k ⋅ c A
D c
N A = AB Ao φ tanh φ
δ
Therefore,
v0
=
SDAB c Ao φ tanh φ
SN A
=
c Ai − c Ao
δ
( cAi − cAo )
Calculate φ and NA
26-45
 1.0 m 
k
4.3 s −1
=
φ δ= 100 μm  6 
= 6.909
2
DAB
 10 μm  9.03 × 10−10 m
s
2

g TCE 
−10 m  

 9.03 × 10
  0.05
s 
m3 
g TCE
NA 
( 6.909 ) tanh ( 6.909 ) = 3.12 × 10−6 2
−6
100 ×10 m
ms
Finally, back out vo

(800 m )  3.12 × 10 g mTCE

s   3600 s 
2
−6
2
v0 =
g TCE
g TCE  

− 0.05
 0.25

m3
m3 

h
m3
 = 44.9
h

b. TCE concentration at point biofilm is attached to surface, cA at z = δ
Concentration profile of A in biofilm

k1 
g TCE
c Ao cosh  (δ − z )

DAB  c A0 cosh ( 0 ) 0.25 m3 (1.0)
g TCE

= 4.99 × 10−4
c A ( z ) |z =δ
=
= =
cosh (φ )
cosh ( 6.909 )
m3

k1 
cosh  δ

DAB 

26-46
26.26
A = atrazine, B = water in water-saturated soil
a. Determine DAe
Wilkes-Chang correlation
1
T 7.4 x10 [ ΦBM B ] 2
DAB =
µB
VA0.6
−8
1
-8
2
293 ) 7.4 x 10 ( 2.6 )(18.02 )  2
(
-6 cm
=
6.859
x
10
DAB =
0.6
s
( 0.993 )
(170 )
2
=
DAe ε=
DAB
( 0.6 ) 6.859 x 10-6
2
cm 2
cm 2
= 2.47 x 10-6
s
s
b. Determine c A at z = δ
cA ( z ) =
c A* cosh (δ − z )
cosh (φ )
k1
DAB
hr 
( 5.0 x 10 hr )  3600

s
-4
k1
φ L=
=
10.0 cm
DAe
-1
cm 2
2.47 x 10
s
mmol
0.139
*
cA
L = 0.026 mmol
=
c A (δ ) =
cosh (φ )
cosh ( 2.371)
L
= 2.371
-6
26-47
26.27
A = CO, B = CO2
a. Determine total molar flow rate out n2
Material balance on reactor for CO
IN – OUT + GEN = ACC
y A,1n1 − y A,2 n2 + WA =
0
π d 2 DAe c A0
φ tan φ
WA =S ⋅ N A =
4
δ
k
=
φ δ= 1.0 cm
DAe
6.0 s -1
= 3.87
cm 2
0.4
s
yCO ,2 + yO2 ,2 + yCO2 ,2 =
1
∴ yCO2 ,2 =
1 − yCO ,2 − yO2 ,2 =
1.0 − 0.05 − 0.025 =
0.925 =
y A0
P
1.0 atm
gmole
=
c A0 y=
y A0 = 0.925
= 1.051 x 10-5
A 0C
3
RT
cm3

cm ⋅ atm 
82.06
1073
K
(
)


gmole ⋅ K 

 π (10.0 cm )2   0.40 cm 2 /s  
gmole 
1.051 x 10-5
3.87 ) tanh ( 3.87 )
WA = 


3 (

  1.0 cm  
4
cm



gmole
WA = 1.277 x 10-3
s
gmole CO 

(0.0)n1 − 0.05n2 + 1.277 x 10-3
0
=
s


gmole CO  60 s 
gmole CO
n2 = 0.0255

 = 1.53
s
min
 1.0 min 
b. Mole fraction y A on backside of catalyst layer at z = δ
=
yA ( z)
(
)
y A0 cosh (δ − z ) k / DAe
y A0
0.925
= =
= 0.0386
cosh (φ ) cosh ( 3.87 )
cosh δ k / DAe
(
)
26-48
26.28
A = glucose, B = gel
a. Develop model for cA(z)
Assumptions: 1) Steady state; 2) 1-D flux along z; 3) Sero order reaction with RA = -k; 4) Dilute
system
Flux Equation
dc
N A, z = − DAB A
dz
General Differential Equation for Mass Transfer
dN
− A, z + k =
0
dz
Plug Flux into Mass Transfer Equation and Integrate
dc
d
k
(− DAe A ) =
dz
dz
Integrate twice
k z2
=
cA
+ c1 z + c2
DAe 2
z L=
, c A c A0
=
dc A
z 0,=
0
=
dz
At z = 0, c1 = 0
At z = L, c2 = cA0
cA =
c A0 +
k
( z 2 − L2 )
2 DAe
b. Estimate DAe
mmole 

-0.05


mmole   mmole  
L-min 

(0.25 - 0)
 4.5
 =  50
+
L  
L 
2D Ae

𝑐𝑐𝑐𝑐2
−6
𝐷𝐷𝐴𝐴𝐴𝐴 = 2.29 × 10
𝑠𝑠
26-49
26.29
A = O2, B = tissue (approximates liquid water)
a. Model to predict cA(r)
From Chapter 25, Example 3, Differential Equation for Mass Transfer
DAB
d 2 c A 1 dc A
0
+
−m =
dr 2 r dr
d  dc A 
m
r
r
=
dr  dr  DAB
Boundary Conditions:
=
r R1 , c=
c=
c A=
*
A
As
pA
H
dc A
=
r R=
0 (net flux NA = 0 at r = R2)
2,
dr
Analytical Solution:
For this particular form of O.D.E., the general solution for cA( r) is obtained by “integrating
twice” with respect to r
First integration:
dc A
α
m
=
r+ 1
dr 2 DAB
r
Second integration:
m 2
cA (r ) =
r + α1 ln(r ) + α 2
4 DAB
Apply boundary conditions to solve for integration constants α1 and α2
at r = R2, α1 = −
mR12 mR2 2
mR 2 2
, at r = R1, α 2 =
c As −
+
ln( R1 )
2 DAB
4 DAB 2 DAB
Plug integration constants α1 and α2 back into general solution
c A (r ) =c As +
mR2 2  r 
m
(r 2 − R12 ) −
ln  
4 DAB
2 DAB  R1 
26-50
valid cA(r ) > 0 and R1 < r < R2
For a constant O2 consumption rate m, at some point cA( r) = 0. The “critical radius” where cA(
r) = 0 is found at R2 = Rc and r = Rc
0 c As +
c A ( Rc ) ==
mRc 2  Rc 
m
( Rc 2 − R12 ) −
ln  
4 DAB
2 DAB  R1 
Rc is found from the root of this nonlinear equation (m, R1, DAB known)
b. Estimate Rc and plot of cA(r)
0.35
Variable
Units
m = 6.94E-05 µmol O2/cm3-sec
0.25 µmol O2/cm3-sec
0.30
0.25
PA =
1.0 atm
H=
c As =
0.78 atm·m3/mole
D AB =
R1 =
Rc =
f(R c ) =
C A( r )
0.20
µmol O2 /
cm3
0.15
0.32 µmol O2/cm3
2
2.10E-05 cm /sec
0.25 cm
0.62 cm
0.0000
0.10
0.05
0.00
0.25
0.35
0.45
0.55
0.65
0.75
Radial position from center of tube, r (cm)
c. Calculate WA
dc A
R12 
m 
=
r
−


dr 2 DAB 
r 
26-51
dc


W
SN
2π RL  − DAB A |R2 
=
=
A |R2
A, r
dr


2
 m
R 
WA |R2 = 2π RL  −  R2 − 1  
R2  
 2
 

-11 mol 
2
  6.94 x 10 cm3s  


0.25
cm
(
)
mol
 0.75 cm 
= -1.63 x 10-9
= 2π ( 0.75 cm )(15 cm ) - 
(

)
2
s
( 0.75 cm )  




26-52
26.30
A = drug, B = tissue
a. Define system, source, and sink for mass transfer of species A, and five reasonable
assumptions
System: healthy tissue layer
Source: bulk liquid
Sink: tumor tissue
Assume:1) Steady State, 2) Dilute with respect to A, 3) 1-D Mass Transfer along z, 4) Drug A is
not very soluble in tissue, 5) Homogeneous first order reaction of drug in tissue.
b. Develop differential model for cA(z)
Differential Equation for Mass Transfer
dN
− A + RA =
0
dz
Combine Flux Equation
dc
N A, z = − DAe A
dz
dc 
d 
DAe A  + RA =
0

dz 
dz 
d 2cA
DAe
− k1c A =
0
dz 2
c. Boundary Conditions
z 0,=
c A Kc A∞
z = 0;=
z = L, N A, z = N A,0 = − DAe
dc A
dz
d. Analytical Model for cA(z)
General Solution
=
c A ( z ) α1 sinh( β z ) + α 2 cosh( β z )
β = k1 / DAe
Apply Boundary Conditions
26-53
At z = 0,
c A Kc
=
=
α1 sinh(0) + α 2 cosh(0)
A, ∞
α 2 = Kc A,∞
dc
At z = L, A β [α1 cosh( β L) + α 2 sinh( β L) ]
=
dz
N A, s =
− DAe β α1 cosh( β L) + Kc A,∞ sinh( β L) 
N

−  A, s + Kc A,∞ sinh( β L) 
D β

α1 =  Ae
cosh( β L)


N
A, s
−
+ Kc A,∞ tanh( β L) 
α1 =
 DAe β cosh( β L)

N A, s


−
+ Kc A,∞ tanh( β L)  sinh( β z ) + Kc A,∞ cosh( β z )
cA ( z ) =
 DAe β cosh( β L)

e. Determine N A,0
dc A
|z =0= β [α1 cosh(0) + α 2 sinh(0) ]= βα1
dz
N A, s


dc A
−
+ β Kc A,∞ tanh( β L) 
|z 0 =
dz
 DAe cosh( β L)

N A, s
=
+ β DAe KC A,∞ tanh( β L)
N A,0
cosh( β L)
=
β
=
k1 / DAe
( 4 x 10 s ) / (1.0 x 10 cm /s ) = 2.0 cm
-5
-1
-5
2
-1

-6 mg 
 2.0 x 10

cm 2s 

N A,0 =
+
cosh ( 2.0 cm -1 ) ( 0.5 cm )
(
)
2
cm3  
mg 
-1 
-5 cm 
2.0
cm
1.0
x
10
0.10
3.0 3  tanh ( 2.0 cm -1 ) ( 0.5 cm )
(
)

3 
s 
cm  
cm 

mg  86,400 s 
mg
N A,0 = 5.87 x 10-6
 = 0.507 2
2 
cm s  day 
cm day
(
)
26-54
26.31
Develop Model for Fick’s flux equation
General Flux Equation
dy
−CDAB A + y A ( N A, z + N B , z )
N A, z =
dz
Energy balance on adiabatic process
N A, z ( ∆H v , A ) + N B , z ( ∆H v , B ) =
0
( ∆H ) =
N
( ∆H )
( ∆H ) = 30 kJ/mol =0.909
( ∆H ) 33 kJ/mol
− N A, z
v, A
B,z
v,B
v, A
v,B
Combine

( ∆H v, A ) N 
dy
dy
−CDAB A + y A ( N A, z + N B , z ) =
−CDAB A + y A  N A, z −
N A, z =

dz
dz
( ∆H v,B ) A, z 

 ( ∆H v , A ) 
dy
−CDAB A + y A N A, z 1 −
N A, z =

 ( ∆H v , B ) 
dz


Simplify for N A, z
dy
N A, z − 0.091 y A N A, z =
−CDAB A
dz
δ
y As
dy A
1 − 0.091 y A )
y A∞ (
N A, z ∫ dz = −CDAB ∫
0
N A, zδ =
N A, z =
CDAB  1 − 0.091 y As 
ln 

0.091 1 − 0.091 y A∞ 
 1 − 0.091 y As 
CDAB
ln 

0.091δ 1 − 0.091 y A∞ 
26-55
26.32
L = 4.0 cm (along x), H = 6.0 cm (along y), W = 4.0 cm (along z)
PA
0.10 atm
=
c*A =
= 4.062 gmole/m3
3
RT 

-5 atm ⋅ m
 8.206 x 10
 (300 K)
gmole ⋅ K 

DAB =.020 cm2/sec
a. Flux out of y-z and x-z planes
Flux N A (0,y) on y-z plane, x = 0, 0 ≤ y ≤ H
∞
(−1)
 nπ x 
 nπ y 
−4 c*A ∑
c A ( x, y ) =
n=
sin 
sinh 
1,3,5

L 
L 
 nπ H 


n =1
nπ sinh 

 L 
∂c A ( x, y )
4 c* ∞
(−1) n
 nπ x 
 nπ y 
=
− A∑
n=
cos 
sinh 
1,3,5

∂x
L n =1
L 
L 
 nπ H 


sinh 

 L 
n
∂c A (0, y ) 4 DAB c*A ∞
(−1) n
 nπ y 
N A, x (0, y ) =
− DAB
=
sinh 
1,3,5
n=
∑
∂x
L
L 
 nπ H 

n =1
sinh 

 L 
Flux N A (x,0) on x-z plane, y = 0, 0 < x < L
4 c*A ∞
∂c A ( x, y )
(−1) n
 nπ x 
 nπ y 
sin 
cosh 
1,3,5
n=
=
−
∑

L n =1
L 
L 
∂y
 nπ H 


sinh 

 L 
4 c* ∞
∂c A ( x, y )
(−1) n
 nπ x 
 nπ y 
sin 
cosh 
1,3,5
n=
=
− A∑

L n =1
L 
L 
∂y
 nπ H 


sinh 

 L 
∂c ( x, 0) 4 DAB c*A ∞
(−1) n
 nπ x 
sin 
1,3,5
N A, y ( x, 0) =
n=
− DAB A
=
∑
L
L 
∂y
 nπ H 

n =1
sinh 

 L 
b. Average flux
N A, x x = 0 =
N A, x x = 0
1 y=H
N A, x (0, y ) dy
H ∫y = 0
1 y = H 4 DAB c*A ∞
(−1) n
 nπ y 
sinh 
=
∑
 dy n 1,3,5
∫
H y =0
L
L 
 nπ H 

n =1
sinh 

 L 
26-56
4 DAB c*A ∞
∑
H
n =1
N A, x x = 0

(−1) n
 nπ H  
=
cosh 
 − 1 n 1,3,5

L  
 nπ H  

nπ sinh 

 L 
N A, x x =0 = 6 x 10-4 gmole/m 2sec (slow convergence)
N A, y
y =0
N A, y
y =0
N A, y
y =0
N A, y
y =0
=
1 x=L
N A, y ( x, 0) dx
L ∫x =0
1 x = L 4 DAB c*A ∞
(−1) n
 nπ x 
sin 
=
∑
 dx n 1,3,5
∫
0
x
=
L
L
 nπ H   L 
n =1
sinh 

 L 
4 DAB c*A ∞
∑
L
n =1
(−1) n
=
( cos ( n π ) − 1) n 1,3,5
 nπ H 
n π sinh 

 L 
= 9.29 x 10-6 gmole/m 2 sec
3.5E-03
1.6E-05
3.0E-03
1.4E-05
- NA(x,0), gmol/m2-sec
- NA(0,y), gmole/m2-sec
Average flux values were calculated by implementation of convergent infinite series solution on
computer spreadsheet.
2.5E-03
2.0E-03
1.5E-03
1.0E-03
5.0E-04
1.2E-05
1.0E-05
8.0E-06
6.0E-06
4.0E-06
2.0E-06
0.0E+00
0.0E+00
0.0
1.0
2.0
3.0
4.0
5.0
0.0
6.0
2.0
3.0
4.0
x (cm)
y (cm)
Calculated local flux NA,x (0,y) through y-z
plane along y from y = 0 to y = H (source)
c. Total Release Rate, WA
WA =2 ⋅ W ⋅ H ⋅ N A, x x =0 +W ⋅ L ⋅ N A, y
1.0
Calculated local flux NA,y (x,0) through x-z
plane along x
y =0
(
)
(4.0 cm)(4.0 cm) ( 9.29 x 10 gmole/m sec ) (1m/100 cm)
=2(4.0 cm)(6.0 cm) 6 x 10-4 gmole/m 2 sec (1m/100 cm) 2 +
-6
2
2
= 2.9 x 10-5 gmole/sec
Note that flux is dominated by y-z plane, particularly near the source at y = H
26-57
26.33
Additional process information provided in Problem 25.13
Assumptions
1.
2.
3.
4.
5.
Steady state - constant source and sink for A
No homogeneous reaction of A within porous layer
1-D flux along position r - sealed top and bottom ends
Rapid reaction at catalyst surface (CA ≈ 0 at r = Ro)
Dilute with respect to solute A - H2 (B) is dominant species
Model for total transfer rate,WA
Flux Equation based on assumption 3
N A, r = − DAe
dC A
dr
General Differential Equation for Mass Transfer based on assumptions 1-3
d
( r N A, r ) = 0
dr
By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r
Boundary Conditions: r = Ro, CA =CAs ≈ 0, r = R1, CA = CA∞
=
WA 2=
π rL N A, r 2π R=
Ro WA is constant along r
o L N A, r r
C A∞
dr
= − DAe ∫ dC A
Ro r
C As
2π LDAe ( C As − C A∞ )
WA =
R 
ln  1 
 Ro 
WA ∫
R1
Aside: estimation of effective diffusion coefficient DAe
FSG Correlation
26-58
1/ 2
 1
1 
+
0.001T 

MA MB 

=
1/ 3 2
1/ 3
P ( Σv ) A + ( Σv ) B
1.75
DAB
(
)
1/2
0.001 ⋅ ( 473 )
1.75
 1 1
 + 
 28 2 
(1.5) ( ( 40.92 ) + ( 7.07 ) )
1/3
1/3 2
cm 2
DAB = 0.669
s
T
423
cm 2
4850(0.00010)
=
DKA 4850
=
d pore
= 1.885
MA
28
s






1
1
1
1
2
+
+
DAe = ε 2 
 = ( 0.40 ) 
2
2 
cm 
 DKA DAB 
1.885 cm
0.669

s
s 
cm 2
DAe = 0.079
s
−1
−1
Aside: =
C A y=
yA
AC
C A = ( 0.04 )
P
RT
(1.5 atm )

-5 m ⋅ atm 
 8.206×10
 ( 423 K )
gmole ⋅ K 

3
=1.73
gmole A
m3
Back out R1 for WA = 9.238 ×10-6 gmole A/s

cm 2 
gmole A  1m 
-2π(5.0 cm)  0.079
 ( 0-1.73)


s 
m3  100 cm 

−6 gmole A
=
−9.238 × 10
s
 R1 
ln 

 0.50 cm 
R1 = 0.8 cm
3
26-59
27.1
Concentration profile cA(z,t)
∞
𝑐𝑐𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴
4
1
𝑛𝑛𝑛𝑛𝑛𝑛 −�𝑛𝑛𝑛𝑛�2𝑋𝑋𝐷𝐷
= � sin �
� 𝑒𝑒 2
𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 𝜋𝜋
𝑛𝑛
𝐿𝐿
𝑛𝑛=1
∞
4
1
𝑛𝑛𝑛𝑛𝑛𝑛 −�𝑛𝑛𝑛𝑛�2 𝑋𝑋𝐷𝐷
� 𝑒𝑒 2
+ 𝑐𝑐𝐴𝐴𝐴𝐴
𝑐𝑐𝐴𝐴 = (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) � sin �
𝜋𝜋
𝑛𝑛
𝐿𝐿
𝑛𝑛=1
Average concentration:
∞
1 𝐿𝐿
4
1
𝑛𝑛𝑛𝑛𝑛𝑛 −�𝑛𝑛𝑛𝑛�2𝑋𝑋𝐷𝐷
𝑐𝑐�𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 = � �(𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) � � sin �
� 𝑒𝑒 2
�� 𝑑𝑑𝑑𝑑
𝐿𝐿 0
𝜋𝜋
𝑛𝑛
𝐿𝐿
∞
𝑛𝑛=1
𝑛𝑛𝑛𝑛 2
𝑛𝑛𝑛𝑛 2
4
1
𝑐𝑐�𝐴𝐴 = 𝑐𝑐𝐴𝐴𝐴𝐴 + (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) 2 � � 2 �𝑒𝑒 −� 2 � 𝑋𝑋𝐷𝐷 − cos(𝑛𝑛𝑛𝑛) 𝑒𝑒 −� 2 � 𝑋𝑋𝐷𝐷 ��
𝜋𝜋
𝑛𝑛
𝑛𝑛=1
Since cos(nπ) = -1
𝑛𝑛𝑛𝑛 2
∞
𝑒𝑒 −� 2 � 𝑋𝑋𝐷𝐷
8
𝑐𝑐�𝐴𝐴 = 𝑐𝑐𝐴𝐴𝐴𝐴 + (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) 2 � �
�
𝜋𝜋
𝑛𝑛2
𝑛𝑛=1
or
∞
𝑌𝑌� =
𝑐𝑐�𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴
8
𝑒𝑒
= 2��
𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴
𝜋𝜋
𝑛𝑛=1
𝑛𝑛𝑛𝑛 2
−� 2 � 𝑋𝑋𝐷𝐷
𝑛𝑛2
�
𝑛𝑛 = 1,3,5 …
See attached spreadsheet and plot
27-1
27.1 continued
XD
Y
1.00E-06
9.86E-01
1.00E-04
9.84E-01
1.00E-02
8.87E-01
1.00E-01
6.43E-01
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
8.11E-01
9.01E-02
3.24E-02
1.65E-02
1.00E-02
6.70E-03
4.79E-03
3.60E-03
2.80E-03
2.24E-03
1.84E-03
1.53E-03
1.29E-03
1.11E-03
9.62E-04
8.10E-01
8.99E-02
3.22E-02
1.63E-02
9.81E-03
6.50E-03
4.60E-03
3.41E-03
2.61E-03
2.05E-03
1.65E-03
1.34E-03
1.11E-03
9.29E-04
7.83E-04
7.91E-01
7.21E-02
1.75E-02
4.94E-03
1.36E-03
3.38E-04
7.41E-05
1.40E-05
2.24E-06
3.04E-07
3.46E-08
3.29E-09
2.60E-10
1.71E-11
9.38E-13
6.33E-01 4.95E-01 3.02E-01 2.36E-01 6.87E-02 5.83E-03 3.56E-06
9.78E-03 1.06E-03 1.25E-05 1.36E-06 2.04E-11 4.64E-21 5.41E-50
6.79E-05 1.42E-07 6.24E-13 1.31E-15 5.26E-29 8.55E-56 3.66E-136
9.29E-08 5.21E-13 1.64E-23 9.22E-29 5.14E-55 1.60E-107 4.81E-265
2.09E-11 4.37E-20 1.91E-37 3.99E-46 1.59E-89 2.54E-176 0.00E+00
7.24E-16 7.83E-29 9.15E-55 9.90E-68 1.46E-132 3.19E-262 0.00E+00
3.73E-21 2.89E-39 1.75E-75 1.36E-93 3.84E-184 0.00E+00 0.00E+00
2.79E-27 2.17E-51 1.30E-99 1.01E-123 2.83E-244 0.00E+00 0.00E+00
3.01E-34 3.24E-65 3.74E-127 4.03E-158 0.00E+00 0.00E+00 0.00E+00
4.65E-42 9.62E-81 4.12E-158 8.54E-197 0.00E+00 0.00E+00 0.00E+00
1.02E-50 5.64E-98 1.73E-192 9.58E-240 0.00E+00 0.00E+00 0.00E+00
3.15E-60 6.49E-117 2.75E-230 5.66E-287 0.00E+00 0.00E+00 0.00E+00
1.38E-70 1.46E-137 1.65E-271 0.00E+00 0.00E+00 0.00E+00 0.00E+00
8.47E-82 6.45E-160 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00
7.32E-94 5.55E-184 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00
Series Term n =
Series Term n =
2.00E-01
4.96E-01
4.00E-01
3.02E-01
5.00E-01 1.00E+00 2.00E+00 5.00E+00
2.36E-01 6.87E-02 5.83E-03 3.56E-06
1.00
0.80
Y
0.60
0.40
0.20
0.00
1.0E-06 1.0E-05 1.0E-04 1.0E-03 1.0E-02 1.0E-01 1.0E+00 1.0E+01
XD
27-2
27.2
A = aluminum (Al), B = solid silicon (Si)
T = 1250 K
L = 0.50 µm = 5.0 x 10-5 cm
wAS = 0.010
wA0 = 0
Assume back side is an impermeable barrier, so that x1 = L
𝐷𝐷𝐴𝐴𝐴𝐴 = Do e−Qo/RT
Do = 2.61 cm2/sec (Table 24.7)
Qo = 319,100 J/gmole (Table 24.11)
R = 8.314 J/gmole-K
𝐷𝐷𝐴𝐴𝐴𝐴 = (2.61 cm2 / sec ) exp[−(319,100 J/gmole )/(8.314 J/gmole ∙ K) (1250 K)]
DAB = 1.207 x 10-13 cm2/sec
Determine wA(z,t) for z = 0.25 µm at t = 10.0 hr (36,000 sec)
USS Diffusion in a slab
m = 0 (Al metal in direct contact silicon surface)
𝑛𝑛 =
𝑥𝑥
0.25 𝜇𝜇𝜇𝜇
=
= 0.50
𝑥𝑥1 0.50 𝜇𝜇𝜇𝜇
𝑋𝑋𝐷𝐷 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 (1.207 x 10−13 cm2 /sec)(36,000 sec)
=
= 1.738
(5.0 𝑥𝑥 10−5 𝑐𝑐𝑐𝑐)2
𝑥𝑥12
Figure F.1, Y ≈ 0.010
𝑌𝑌 =
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴 (𝑥𝑥, 𝑡𝑡)
=
= 0.01
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴0
𝑤𝑤𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑌𝑌 ∙ (𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴0 )
= 0.010 − 0.01 (0.010) = 0.0099
27-3
27.3
a. Determine temperature (T)
A = Boron, B = solid silicon (Si)
ρSi = 5.0 x 1022 atoms Silicon/cm3
CAs = 5.0 x 1020 atoms Boron/cm3 at z = 0
CA0 = 0.0 atoms Boron/cm3
CA(z,t) = 1.7 x 1019 atoms Boron/cm3 at z = 0.20 μm after t = 30 min
𝐷𝐷𝐴𝐴𝐴𝐴 = Do e−Qo/RT
Do = 0.019 cm2/sec
Qo = 2.74x105 J/gmole
R = 8.314 J/gmole-K
USS diffusion in semi-infinite medium
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴
𝑧𝑧
= erf �
� = erf(∅)
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
5.0 × 1020 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵/𝑐𝑐𝑚𝑚3 − 1.7 × 1019 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵/𝑐𝑐𝑚𝑚3
𝑧𝑧
= 0.966 = erf �
�
𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
5.0 × 1020 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
−
0
𝑐𝑐𝑚𝑚3
Appendix L, erf(φ) = 1.50
1.50 =
0.00002 𝑐𝑐𝑐𝑐
2�(0.0019 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠)(1800 𝑠𝑠𝑠𝑠𝑠𝑠)𝑒𝑒𝑒𝑒𝑒𝑒 � �−2.74 𝑥𝑥 105
T = 1204 K
b. Determine NA(0,t) at t = 10 min, 30 min
−𝑄𝑄
𝑐𝑐𝑚𝑚2
𝐷𝐷𝐴𝐴𝐴𝐴 = 𝐷𝐷𝑜𝑜 𝑒𝑒 𝑅𝑅𝑅𝑅 = 0.019
𝑒𝑒𝑒𝑒𝑒𝑒 �
𝑠𝑠𝑠𝑠𝑠𝑠
𝐷𝐷𝐴𝐴𝐴𝐴 = 8.455 × 10−13 𝑐𝑐𝑚𝑚2 /𝑠𝑠
�−2.74 𝑥𝑥 105
�8.314
𝐷𝐷𝐴𝐴𝐴𝐴
(C − CA0 )
𝜋𝜋𝜋𝜋 AS
𝐽𝐽
� /(8.314 𝐽𝐽/𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐾𝐾)𝑇𝑇�
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝐽𝐽
�
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
�
𝐽𝐽
� (1383 𝐾𝐾)
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
𝑁𝑁𝐴𝐴 (0, 𝑡𝑡) = �
𝑁𝑁𝐴𝐴 (0, 𝑡𝑡) = �
𝑁𝑁𝐴𝐴 (0, 𝑡𝑡) = �
8.455 × 10−13 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
(5.0 × 1020 − 0)𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎/𝑐𝑐𝑐𝑐3 = 1.059 × 1013
𝜋𝜋(600 𝑠𝑠𝑠𝑠𝑠𝑠)
𝑐𝑐𝑚𝑚2 𝑠𝑠𝑠𝑠𝑠𝑠
8.455 × 10−13 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
(5.0 × 1020 − 0)𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎/𝑐𝑐𝑐𝑐3 = 6.114 × 1012
𝜋𝜋(1800 𝑠𝑠𝑠𝑠𝑠𝑠)
𝑐𝑐𝑚𝑚2 𝑠𝑠𝑠𝑠𝑠𝑠
27-4
27.4
A = Phosphorous (P), B = solid silicon (Si)
T = 1000 °C (1273 K)
S = 100 cm2 square Si wafer
CAS = 1.0 x 1021 P atoms/cm3 Si at z = 0
CA0 = 0.0
Determine CA(z,t) (atoms P/ cm3) doped into Si at z = 0.468 μm and t = 40 min
USS diffusion in semi-infinite medium
𝑧𝑧
𝜙𝜙 =
2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑇𝑇
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑧𝑧, 𝑡𝑡)
= erf(𝜙𝜙)
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
Estimate DAB from data
4𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
(𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 )
𝜋𝜋
𝑚𝑚𝐴𝐴 (𝑡𝑡) − 𝑚𝑚𝐴𝐴0 = 𝑆𝑆�
6.0 𝑥𝑥 1018 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃
4𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
(𝐶𝐶𝐴𝐴𝐴𝐴 )
1
𝜋𝜋
100 sec 2
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃
6.0 ∗ 1016
1
2
(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) 𝜋𝜋
2
sec
=
𝐷𝐷𝐴𝐴𝐴𝐴 =
4(𝑆𝑆𝐶𝐶𝐴𝐴𝐴𝐴 )2
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃 2
4(100 𝑐𝑐𝑐𝑐2 ) �1.0 𝑥𝑥 1021
�
𝑐𝑐𝑚𝑚3
𝐷𝐷𝐴𝐴𝐴𝐴 = 2.827 𝑥𝑥 10−13 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠
1 𝑐𝑐𝑐𝑐
0.467 𝜇𝜇𝜇𝜇 � 4
�
10 𝜇𝜇𝜇𝜇
𝜙𝜙 =
= 0.90
2
𝑐𝑐𝑚𝑚
60
𝑠𝑠𝑠𝑠𝑠𝑠
2��2.827 𝑥𝑥 10−13
�40 𝑚𝑚𝑚𝑚𝑚𝑚
�
𝑠𝑠𝑠𝑠𝑠𝑠 �
1 𝑚𝑚𝑚𝑚𝑚𝑚
Appendix L, erf(φ) = 0.797
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
− 𝐶𝐶𝐴𝐴 (𝑧𝑧, 𝑡𝑡)
1.0 x 1021
𝑐𝑐𝑐𝑐3
erf(𝜙𝜙) = 0.797 =
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
1.0 x 1021
− 0
𝑐𝑐𝑐𝑐3
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝐶𝐶𝐴𝐴 (𝑧𝑧, 𝑡𝑡) = 2.03 𝑥𝑥 1020
𝑐𝑐𝑚𝑚3
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 =
= 𝑆𝑆�
27-5
27.5
Given:
A = Arsenic (As), B = solid silicon (Si)
T = 1050 C
Wafer dimensions: L = 1.0 mm, d = 10 cm
Solute A: cAs = cA* = 2.3 × 1021 atoms As/cm3 at z = 0
cAo = 2.3 × 1017 atoms As/cm3 initial As concentration in solid Si
DAB = 5.0 × 10−13 cm2/s
a. Time required and mA(t) to achieve CA(z,t) = 2.065 × 1020 atoms As/cm3 at z = 0.5 µm
USS diffusion in semi-infinite medium—concentration profile
atoms
atoms
− 2.065 10
× 20
3
cm
cm3 atoms
atoms
2.3 10
− 2.3 10
× 21
× 17
3
cm
cm3
2.3 10
× 21
c As − c A ( z , t )
= erf
=
(φ )
c As − c Ao
erf (φ ) = 0.9103
φ=
z
2 DAB t
From Appendix L, φ = 1.2
z2
t =
=
4φ 2 DAB
−5
4 (1.2 )
2
2
2

−13 cm 
5.0
10
×


s 

= 868 s
( 5.0 × 10 cm /s ) (868 s ) = 2.08 × 10 cm < < L
−13
DAB=
⋅t
Note
S=
( 5.0 ×10 cm )
2
−5
πd2
4
USS diffusion in semi-infinite medium—total mass loaded in medium
=
m
A ( t ) − m Ao
πd2
4
4 DAB t
π
( cAs − cAo )
27-6

cm 2 
4  5.0 10
× −13
 ( 868 s )
π (10 cm )
s 
atoms

2.3 10
× 21 − 2.3 10
× 17
mA ( t ) − mAo =
4
cm3
π
4.25 10
× 18 atoms As
mA ( t ) − mAo =
2
πd2
(
(
)
π (10cm )
)
2
mAo =
c Ao
L = 2.3 × 10 atoms As/cm
( 0.1cm ) = 1.8 ×1016 atoms As
4
4
× 18 atoms As = 4.25 10
× 18 atoms As
mA ( t ) = 1.8 × 1016 atoms As + 4.25 10
17
3
b. Plot of z and vs. t at φ = 1.2
1.20
1.00
0.80
z (µm)
z1/2 (µm)1/2
1.00
0.60
0.40
0.80
0.60
0.20
0.40
0.00
1000
0
2000
3000
Time, t (s)
20
4000
40
60
80
t1/2 (s)1/2
Note plot of z vs. t1/2 is linear
c. Transfer rate WA at t = 5.0 min, 10.0 min
π d 2 DAB
WA =
S ⋅ N A z =0 =
( c − c ) for t > 0
4
π t As Ao
At t = 5.0 min = 300 s
WA =
π (10 cm )
2
4
cm /s )
×
( 5.0 10
×
( 2.3 10
π ( 300 s )
−13
2
21
× 17
− 2.3 10
) atoms
cm
− 2.3 10
× 17
) atoms
cm
3
WA = 4.16 × 1015 atoms As / s
At t = 10.0 min = 600 s
WA =
π (10 cm )
4
2
cm /s )
×
( 5.0 10
×
( 2.3 10
π ( 600 s )
−13
2
21
3
WA = 2.94 ×1015 atoms As / s
Note as t increases, WA decreases. At t = 0, the flux is not defined because no concentration
gradient for As in the solid Si initially exists.
27-7
27.6
a. yA(0,t) at t = 30 s
Use Concentration-Time Charts
m = 0 (no convective mass transfer resistance)
=
n
r
0
=
=0
R 1.5 cm
DAet
=
XD =
R2
s)
( 0.0143 cm /s ) ( 30=
0.19
2
(1.5 cm )
2
From Figure F.9 (sphere, n = 0, m = 0), Y ≅ 0.28
=
Y
c A,1 − c A ( r, t ) c A∞ − c A ( 0, t ) y A∞ − y A ( 0, t )
= =
c A,1 − c Ao
c A∞ − c Ao
y A∞ − y Ao
=
y A ( 0, t ) y A∞ − Y ( y A∞ − y Ao )
y A ( 0, t )= 0.05 − (0.28)(0.05 − 0)= 0.036
b. Time t at yA(0,t) = 0.048
Y=
y A∞ − y A ( 0, t )
y A∞ − y Ao
=
( 0.050 − 0.048) =0.04
( 0.05 − 0.0 )
From Figure F.3 (sphere, n = 0, m = 0), XD ≅ 0.40
X R 2 ( 0.4 )(1.5 cm )
= 63 s
t= D =
DAe

cm 2 
 0.0143

s 

2
c. New time t if R = 3.0 cm
X R 2 ( 0.4 )( 3.0 cm )
t= D =
= 252 s (4 times longer)
DAe

cm 2 
 0.0143

s 

2
27-8
27.7
Comments: Without any insight to the patch shape or dimensions, we will assume the patch is
planar and it is capable of supplying a constant supply of drug. We will also assume that this is
the first patch applied, and that all tissue has no drug initially, with the tissue acting as semiinfinite medium for drug diffusion.
A = drug, B = tissue
CAS = CA* = 2.0 mole/m3
CA0 = 0.0 mole/m3
DAB = 1.0 x 10-6 cm2/sec
Determine time required (t) for cA(z,t) = 0.20 mole/m3 at z = 0.5 cm
USS diffusion in semi-infinite medium
𝑧𝑧
𝐶𝐶𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴
= erf �
� = 𝑒𝑒𝑒𝑒𝑒𝑒(∅)
𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴𝐴𝐴
2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
0.5 𝑐𝑐𝑐𝑐
0.20 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 − 2.0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
=
0.90
=
erf
�
�
0.0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 − 2.0 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
2�(1.0 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 ) 𝑡𝑡
From Appendix L, ∅ = 1.1631
1.1631 =
0.5 𝑐𝑐𝑐𝑐
2�(1.0 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 ) 𝑡𝑡
𝑡𝑡 = 46,201 𝑠𝑠𝑠𝑠𝑠𝑠 = 12.83 ℎ𝑟𝑟
27-9
27.8
dye solution (solute A)
z=0
sealed
edges
z = 0.05 cm
Solid (B)
CAo = 0
z = L = 1.0 cm
a. Time t required for CA(z,t) = 0.0062 gmole/m3 with z = 0.050 cm
Semi-infinite diffusion medium
Strategy: Back out DAB from observed flux NA(0,t) at t = 120 s, then use DAB to get CA(z,t)
N A (0, t)
=
DAB
( c − c ) for t > 0
π t As Ao
-9
2
N A (0, t ) ] π t ( 5.15×10 gmole/m ⋅ s ) π (120 s )
[
DAB =
=
2
2
( cAs − cAo )
( 0.01gmole/m3 - 0 )
2
2
-10 2
m /s 1.0×10-6 cm 2 /s
DAB 1.0×10
=
=
c As − c A ( z , t )
c As − c Ao
c As − c A ( z , t )
c As − c Ao
= erf (φ )
0.010-0.0062 ) gmole/m3
(
=
= 0.38
( 0.010-0.0 ) gmole/m3
Appendix L, at erf(φ) = 0.38, φ = 0.35
φ=
z
2 DAB t
z2
t =
=
4φ 2 DAB
( 0.050 cm )
4 ( 0.35 )
2
2
2

−6 cm 
×
1.0
10


s 

= 5102 s
b. The system is best represented by a semi-infinite medium
27-10
27.9
food (contains liquid water)
z = L = 0.2 cm
porous film (air + water vapor)
z=0
nonporous metal plate
Use Concentration-Time Charts
m = 0 (no convective mass transfer resistance)
n=
z
0
=0
=
L 0.20 cm
c A,1 − c A ( z, t ) PA* − PA ( 0, t )
=
Y
= =
c A,1 − c Ao
PA* − PAo
( 0.030-0.015) atm = 0.50
( 0.030-0.0 ) atm
Figure F.7 (slab, n = 0, m = 0), XD = 0.38
X L2 ( 0.38 )( 0.2 cm )
t= D =
= 117 s
DAe

cm 2 
 0.00013

s 

2
27-11
27.10
A = H2O vapor, B = air
R = 1.0 mm = 0.10 cm
𝐶𝐶𝐴𝐴𝐴𝐴 = 0.90 𝐶𝐶𝐴𝐴∗
CA0 = 0
DAB = 0.260 cm2/sec at T = 298 K and P = 1.0 atm (Table J.1)
Determine time required for center of air bubble to reach water vapor concentration of 90% of
saturation, 𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) = 0.90 𝐶𝐶𝐴𝐴∗
USS Diffusion in a sphere, use Concentration -Time charts
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) ∴ 𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴∞
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) 𝐶𝐶𝐴𝐴∗ −0.90𝐶𝐶𝐴𝐴∗
=
= 0.10
𝑌𝑌 =
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴∗ − 0
𝑛𝑛 =
𝑟𝑟
0 𝑐𝑐𝑐𝑐
=
=0
𝑅𝑅 0.10 𝑐𝑐𝑐𝑐
Figure F.3
𝑋𝑋𝐷𝐷 ≅ 0.40 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑅𝑅2
𝑋𝑋𝐷𝐷 𝑅𝑅2 (0.40)(0.1 𝑐𝑐𝑐𝑐)2
𝑡𝑡 =
=
= 0.015 𝑠𝑠𝑠𝑠𝑠𝑠
𝐷𝐷𝐴𝐴𝐴𝐴
0.260 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠
27-12
27.11
A = CO2, B = air
T = 25 °C, P = 1.0 atm
x1=L = 2.0 cm
CA∞ = 0
DAB = 0.161 cm2/sec
DAe = 0.010 cm2/sec
kc = 0.0025 cm/sec
Determine time required for yA(x,t) = 0.02 at 1.6 cm from the surface of the slab
USS Diffusion in a slab, use Concentration -Time charts
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝑦𝑦𝐴𝐴∞ − 𝑦𝑦𝐴𝐴 (𝑥𝑥, 𝑡𝑡)
=
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0
𝑦𝑦𝐴𝐴∞ − 𝑦𝑦𝐴𝐴0
𝑌𝑌 =
0 − 0.02
= 0.2
0 − 0.10
𝑌𝑌 =
𝑛𝑛 =
𝑚𝑚 =
2.0 𝑐𝑐𝑐𝑐 − 1.6 𝑐𝑐𝑐𝑐
𝑥𝑥
=
= 0.20
𝑥𝑥1
2.0 𝑐𝑐𝑐𝑐
𝐷𝐷𝐴𝐴𝐴𝐴
𝐷𝐷𝐴𝐴𝐴𝐴
0.010 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠
=
=
= 2.0
𝑘𝑘𝑐𝑐 𝑥𝑥1 𝑘𝑘𝑐𝑐 𝐿𝐿 �0.0025 𝑐𝑐𝑐𝑐 � (2.0 𝑐𝑐𝑐𝑐)
𝑠𝑠𝑠𝑠𝑠𝑠
Figure F.1
𝑋𝑋𝐷𝐷 = 4.0 =
𝑡𝑡 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑥𝑥12
𝑋𝑋𝐷𝐷 𝑥𝑥12
4.0(2.0 𝑐𝑐𝑐𝑐)2
=
= 1600 𝑠𝑠𝑒𝑒𝑐𝑐
𝐷𝐷𝐴𝐴𝐴𝐴
0.01 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠
27-13
27.12
A = solute, B = adsorbent material
D = 1.0 cm, R = 0.5 cm
L = 5.0 cm
K = 0.15 cm3 fluid, cm3 adsorbent
DAB = 4.0 x 10-7 cm2/sec
𝐶𝐶𝐴𝐴∞ ′ = 2.0 gmole A/m3 fluid
a. Determine time (t) required for CA (r,t) = 2.94 gmol/m3, r = 0.10 cm (0.40 cm from surface),
assuming no end effects
USS Diffusion in a cylinder, use Concentration -Time charts
′
𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴 (𝑅𝑅, 𝑡𝑡) = 𝐾𝐾 ∗ 𝐶𝐶𝐴𝐴∞
= 2.0
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐴𝐴 1.5 𝑚𝑚3 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐴𝐴
�
�
=
3.0
𝑚𝑚3 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑚𝑚3 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑚𝑚3
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟), ∴ 𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴∞
𝑌𝑌 =
𝑛𝑛 =
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) (3.0 − 2.94)𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑚𝑚3
=
= 0.020
(3.0 − 0)𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑚𝑚3
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴
𝑟𝑟 0.10 𝑐𝑐𝑐𝑐
=
= 0.20
𝑅𝑅 0.50 𝑐𝑐𝑐𝑐
Figure F.2
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑅𝑅2
𝑋𝑋𝐷𝐷 𝑅𝑅2
0.75(0.5 𝑐𝑐𝑐𝑐)2
𝑡𝑡 =
=
= 468,750 𝑠𝑠𝑠𝑠𝑠𝑠 = 130.2 ℎ𝑟𝑟
(4.0 𝑥𝑥10−7 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠)
𝐷𝐷𝐴𝐴𝐴𝐴
𝑋𝑋𝐷𝐷 = 0.75 =
b. Determine time (t) required for CA (r,z,t) = 2.94 gmol/m3, r = 0.10 cm (0.40 cm from surface),
z = 1.5 cm (1.0 cm from end)
USS Diffusion in a cylinder, use Concentration -Time charts
𝑌𝑌 = 𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑌𝑌𝑎𝑎
Since XD will be different for Ycylinder and Ya, an iterative solution is required were time (t) is
guessed, Ycylinder , Ya and Y are calculated, and CA (r,z,t) = 2.94 gmol/m3 is checked
Guess t = 100 hr (360,000 sec)
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 4.0 x 10−7 cm2 /sec(360,000 sec)
𝑋𝑋𝐷𝐷 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = 2 = 2 =
= 0.576
(0.5𝑐𝑐𝑐𝑐)2
𝑅𝑅
𝑥𝑥1
27-14
𝑋𝑋𝐷𝐷 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 4.0 x 10−7 cm2 /sec(360,000 sec)
= 2 =
= 0.023
2
𝑎𝑎
𝑥𝑥12
0.5
� 𝑐𝑐𝑐𝑐�
2
𝑛𝑛 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) =
𝑟𝑟 0.10 𝑐𝑐𝑐𝑐
=
= 0.20
𝑅𝑅 0.50 𝑐𝑐𝑐𝑐
𝑥𝑥
𝑥𝑥
2.5 𝑐𝑐𝑐𝑐 − 1.0 𝑐𝑐𝑐𝑐
𝑛𝑛(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) = � � =
=
= 0.60
𝑥𝑥1
2.5 𝑐𝑐𝑐𝑐
𝐿𝐿/2
𝑚𝑚 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 )
𝑚𝑚 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
Figures F.2 and F.1
𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.0541, 𝑌𝑌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.7073
𝑌𝑌 = 𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑒𝑒𝑒𝑒 𝑌𝑌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.03826
𝑌𝑌 =
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑧𝑧, 𝑡𝑡)
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴
𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑧𝑧, 𝑡𝑡) = 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝑌𝑌(𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴 ) = 3.00 − (0.0386)(3.00 − 0) = 2.88 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑚𝑚3
Guess t = 110 hr = 396,000 sec, CA (r,z,t) = 2.91 gmole/m3
Guess t = 120 hr = 432,000 sec, CA (r,z,t) = 2.94 gmole/m3 (done)
27-15
27.13
groundwater flow
water
flow
(v∞)
DAB = 1. 5x10-6 cm2/s
kc = 1.0 x 10-6 cm2/s
CAo= 2.0 mg/cm3
DAe = 5x10
DA-7 cm2/s
CAs
CA∞ ≈ 0
pesticide-filled polymer capsules
(1.0 cm diameter) buried in loose soil
a. cA(0,t) at time t = 173.6 h
Use Concentration-Time Charts
DAB
=
m =
kc R
=
n
(1.5×10 cm /s ) = 3.0
(1.0×10 cm/s ) ( 0.5 cm )
-6
-6
r
0
=
=0
R 0.5 cm
s
( 5.0×10 cm /s ) 173.6 h 3600

h 
-7
=
XD
2
DAet
=
R2
2
( 0.5 cm )
2
= 1.25
Figure F.9 (sphere, n = 0, m = 3.0, XD = 3.0), Y = 0.33
Y
=
c A,1 − c A ( r, t ) c A∞ − c A ( 0, t )
=
c A,1 − c Ao
c A∞ − c Ao
c A ( 0, t ) c A∞ − Y ( c A∞ − c Ao ) = 0 - 0.33 ( 0 - 2.0 ) mg/cm3
=
c A (0, t ) = 0.66 mg/cm3
b. Surface cA(R,t) and NA(R,t) at t = 173.6 h
At sphere surface, cA(R,t) = cAs
=
n
r R
=
=1
R R
Figure F.9 (sphere, n = 0, m = 3.0, XD = 3.0), Y = 0.30
27-16
=
Y
c A,1 − c A ( r, t ) c A∞ − c A ( R, t )
=
c A,1 − c Ao
c A∞ − c Ao
=
c A ( R, t ) c A∞ − 0.30 ( c A∞ − c Ao ) = 0 - 0.33 ( 0 - 2.0 ) mg/cm3
c A ( R, t ) = 0.60 mg/cm3
=
N A ( R, t ) kc [ c A ( R, t ) − c A∞ ]
N A ( R, t ) = 1.0×10-6
cm
mg
mg
( 0.60 - 0 ) 3 =6.0×10-7 2
s
cm
cm s
27-17
27.14
A = cadmium, B = biopolymer
D = 0.50 cm, R = 0.25 cm
DAB = 2.0 x 10–8 cm2/sec
K = 150 m3 aqueous phase/m3 biopolymer phase
CA∞ = 0.10 mol/m3
/
𝐶𝐶𝐴𝐴0 = 0 mol/m3
/
Determine time (t) required for 𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) = 12 mol/m3 at center of spherical absorbent bead (r
= 0)
USS Diffusion in a sphere, use Concentration -Time charts
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
/
𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐾𝐾 ∙ 𝐶𝐶𝐴𝐴∞ = (150 𝑚𝑚3 /𝑚𝑚3 )(0.10 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 ) = 15 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
/
/
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴 (𝑟𝑟, 𝑡𝑡)
𝑌𝑌 =
𝑛𝑛 =
/
/
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
=
𝑟𝑟
0 𝑐𝑐𝑐𝑐
=
=0
𝑅𝑅 0.25 𝑐𝑐𝑐𝑐
/
𝐾𝐾𝐾𝐾𝐴𝐴∞ − 𝐶𝐶𝐴𝐴𝐴𝐴 (𝑟𝑟, 𝑡𝑡)
/
𝐾𝐾𝐾𝐾𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0
(15 − 12)𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
=
= 0.20
15 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 − 0
Figure F.3
𝑋𝑋𝐷𝐷 ≅ 0.25 =
𝑡𝑡 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑅𝑅2
𝑋𝑋𝐷𝐷 𝑅𝑅2
0.25(0.25 𝑐𝑐𝑐𝑐)2
=
= 7.813 𝑥𝑥 105 𝑠𝑠𝑠𝑠𝑠𝑠
(2.0 𝑥𝑥10−8 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠)
𝐷𝐷𝐴𝐴𝐴𝐴
t = 217 hr
27-18
27.15
Given:
A = acetic acid, B = H2O (pickle)
T = 80 C(353 K)
Pickle Dimensions: D = 2.5 cm, R = 1.25 cm, L = 12 cm
cAo = 0
initial concentration of A inside pickle
cA∞ = 0.70 kgmole/m3 bulk fluid concentration of A
DAB = 1.21 × 10−5 cm2/s at T = 293 K
a. Time (t) required for cA(0,t) = 0.864 kgmole/m3
USS diffusion in a cylinder, use concentration–time charts
m = 0, no convection resistance assumed
∴cA,1 = cA∞ = 0900 kgmole/m3
=
n
r
0
=
= 0 relative position
R 1.25 cm
Relative concentration at r = 0 (n = 0) and m = 0
c A,1 − c A ( r, t ) c A∞ − c A ( r, t ) c A∞ − c A ( 0, t )
=
= =
Y =
c A∞ − c Ao
c A,1 − c Ao
c A∞ − c Ao
kgmole
m3
= 0.040
kgmole
( 0.900 − 0 ) 3
m
( 0.900 − 0.864 )
From Figure F.2 cylinder, at Y = 0.040, n = 0, m = 0, read off XD
=
X D 0.625
=
DAet DAB t
=
R2
R2
Scale diffusion coefficient from reference temperature (Tref) to process temperature (T) assuming
pickle approximates properties of liquid water
DAB (T )
(
(
)
)
2
× −6 Pa ⋅ s 
 T   µ (Tref )   353 K  993 10
−5 cm 
1.21 10
×
D=
T
(
)
 
 


AB
ref 

s 
× −6 Pa ⋅ s 
 Tref   µ (T )   293 K  352 10
× −5
DAB = 4.112 10
cm 2
s
27-19
X R 2 ( 0.625 )(1.25 cm )
=23,749 s = 6.6 h
t= D =
2
DAB

−5 cm 
 4.112 ×10

s 

2
b. Time (t) required for cA(0,t) = 0.864 kgmole/m3 for kc = 1.94 × 10−5 cm/s
USS diffusion in a cylinder, use concentration–time charts with convection resistance
Y = 0.040, n = 0 unchanged from part (a)
cm 2
4.112 10
× −5
DAB
s
=1.7
m =
=
kC R 
−5 cm 
×
1.94 10
 (1.25 cm )
s 

Use Figure F.5—cylinder, read off XD from Y = 0.040, n = 0, m = 1.7
3.0 (1.25 cm )
= 113,996 s = 31.7 h
2
−5 cm
4.112 10
×
s
Note that if m > 0 that convective resistance increases the time required.
X D R2
Read off XD = 3.0, and=
so t =
DAB
2
27-20
27.16
A = solvent, B = polymer
x1 = L = 6.0 mm = 0.60 cm
𝐶𝐶𝐴𝐴∞ = 0
wA0 = 0.010
DAB = 2.0 x 10-6 cm2/sec
Determine time (t) required for wA(x,t) = 0.00035 at 1.2 mm (0.12 cm) from the exposed surface
USS Diffusion in a slab, use Concentration -Time charts
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
𝑌𝑌 =
𝑛𝑛 =
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 0 − 0.00035
=
=
=
= 0.035
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝑜𝑜
0 − 0.010
𝑤𝑤𝐴𝐴𝑆𝑆 − 𝑤𝑤𝐴𝐴0
𝑥𝑥
0.60 𝑐𝑐𝑐𝑐 − 0.12 𝑐𝑐𝑐𝑐
=
= 0.80
𝑥𝑥1
0.60 𝑐𝑐𝑐𝑐
Figure F.1
𝑋𝑋𝐷𝐷 ≅ 1.0 =
𝑡𝑡 = 1.0
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑥𝑥12
(0.60 𝑐𝑐𝑐𝑐)2
𝑥𝑥12
= 1.0
= 180,000 sec (50 hr)
𝐷𝐷𝐴𝐴𝐴𝐴
2.0 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠
27-21
27.17
A = glucose, B = gel (liquid water)
L = 1.0 cm
x1 = L/2 = 0.50 cm (symmetry)
CA0 = 0
𝐶𝐶𝐴𝐴∞ = 50 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
Determine DAB for CA(x,t) = 48.5 mole/m3 at x = 0 (center, symmetry) and t = 42 hr (151,200 sec)
USS Diffusion in a slab, use Concentration -Time charts
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
𝑌𝑌 =
𝑛𝑛 =
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) (50.0 − 48.5)𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
=
=
= 0.030
(50.0 − 0.0)𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝑜𝑜
𝑥𝑥
0
=
=0
𝑥𝑥1 0.50 𝑐𝑐𝑐𝑐
Figure F.1
𝑋𝑋𝐷𝐷 ≅ 1.55 =
1.55 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑥𝑥12
𝐷𝐷𝐴𝐴𝐴𝐴 (151200 𝑠𝑠𝑠𝑠𝑠𝑠)
(0.5 𝑐𝑐𝑚𝑚)2
DAB = 2.56 x 10-6 cm2/sec
27-22
27.18
A = griseofulvin (drug), B = gel
D = 0.10 cm, R = 0.050 cm
CA0 = 0.200 mmol/L
𝐶𝐶𝐴𝐴∞ = 0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3
DAB = 1.5 x 10–7 cm2/sec
a. Determine time required (t) for CA(0,t) = 0.10CA0 at r = 0 (center of bead)
USS Diffusion in a sphere, use Concentration -Time charts
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
𝑌𝑌 =
𝑛𝑛 =
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) 0 − 0.10𝐶𝐶𝐴𝐴0
=
=
= 0.10
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
0 − 𝐶𝐶𝐴𝐴0
0 𝑐𝑐𝑐𝑐
𝑟𝑟
=
=0
𝑅𝑅 0.050 𝑐𝑐𝑐𝑐
Figure F.3 or F.9
𝑋𝑋𝐷𝐷 ≅ 0.30 =
𝑡𝑡 = 0.30
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑅𝑅2
(0.050 𝑐𝑐𝑐𝑐)2
𝑅𝑅2
= 0.30
= 5000 𝑠𝑠𝑠𝑠𝑠𝑠
𝐷𝐷𝐴𝐴𝐴𝐴
1.5 𝑥𝑥 10−7 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠
b. Determine time required (t) for CA(0,t) = 0.10CA0 at r = 0 (center of bead) using infinite series
solution, and plot out concentration profile CA(r,t)
At r = 0
∞
𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) − 𝐶𝐶𝐴𝐴0
2 2
2
𝑌𝑌 =
= 0.90 = 1 + 2 �(−1)𝑛𝑛 𝑒𝑒 −𝐷𝐷𝐴𝐴𝐴𝐴𝑛𝑛 𝜋𝜋 𝑡𝑡/𝑅𝑅
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴0
𝑛𝑛=1
𝑛𝑛 = 1,2,3 …
See spreadsheet and plot. The required time (t) was guessed iteratively until CA(0,t) = 0.020
mmol/L using the Solver function in Microsoft Excel, t = 5058 sec.
27-23
27.18 continued
2
D AB =
R=
C Ao =
1.50E-07 cm /s
0.050 cm
0.20 mmol/L
C As =
0 mmol/L
t=
t=
∆r =
1.405 hr
5058.635 sec
0.0083 cm (plot interval)
C A(0,t)/CAo =
Solver =
0.10
-7.44E-08
r (cm)
Y(r,t)
C A(r,t) (mmol/L)
0.00E+00
9.00E-01
2.00E-02
1
8.33E-03
9.05E-01
1.91E-02
2
1.67E-02
9.17E-01
1.65E-02
3
2.50E-02
9.36E-01
1.27E-02
4
3.33E-02
9.59E-01
8.27E-03
5
4.17E-02
9.81E-01
3.82E-03
6
5.00E-02
1.00E+00
0.00E+00
1
2
3
4
5
6
7
8
9
10
-5.00E-02
6.25E-06
-1.96E-12
1.53E-21
-2.99E-33
1.46E-47
-1.79E-64
5.47E-84
-4.18E-106
7.99E-131
-2.50E-02
2.71E-06
-6.52E-13
3.31E-22
-2.99E-34
2.99E-64
1.28E-65
-5.92E-85
4.64E-107
-6.92E-132
-4.33E-02
2.71E-06
-7.99E-29
-3.31E-22
5.18E-34
-5.97E-64
-2.21E-65
5.92E-85
-1.71E-122
-6.92E-132
-5.00E-02
3.83E-22
6.52E-13
-9.37E-38
-5.98E-34
8.96E-64
2.55E-65
-3.35E-100
-4.64E-107
4.89E-147
-4.33E-02
-2.71E-06
1.60E-28
3.31E-22
5.18E-34
-1.19E-63
-2.21E-65
-5.92E-85
3.41E-122
6.92E-132
-2.50E-02
-2.71E-06
-6.52E-13
-3.31E-22
-2.99E-34
1.49E-63
1.28E-65
5.92E-85
4.64E-107
6.92E-132
-6.13E-18
-7.66E-22
-2.40E-28
-1.87E-37
-3.66E-49
-1.79E-63
-2.19E-80
-6.70E-100
-5.12E-122
-9.79E-147
Series Term n =
Series Term n =
CA(r,t) (mmol/L)
0.03
0.02
0.01
0.00
0.00
0.01
0.02
0.03
0.04
Radial postion, r (cm)
0.05
0.06
27-24
27.19
A = drug (Dramamine), B = gel
L = 0.652 cm
x1 = L/2 = 0.326 cm (symmetry)
CA0 = 64 mg A/cm3
CA∞ = 0
DAB = 3.0 x 10-7 cm2/sec
a. Determine CA(x, t) at the center of the slab (x = 0, symmetry) at t = 96 hr, using Concentration
-Time charts
USS Diffusion in a slab
𝑋𝑋𝐷𝐷 =
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
=
𝑥𝑥12
3.0 𝑥𝑥 10−7
𝑐𝑐𝑚𝑚2
3600 𝑠𝑠𝑠𝑠𝑠𝑠
∗ 96 ℎ𝑟𝑟 �
�
𝑠𝑠𝑠𝑠𝑠𝑠
ℎ𝑟𝑟
= 0.976
(0.326 𝑐𝑐𝑐𝑐)2
𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
𝑛𝑛 =
𝑥𝑥
0
=
=0
𝑥𝑥1 0.326 𝑐𝑐𝑐𝑐
𝑌𝑌 =
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡)
=
=
≅ 0.12
𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴0
Figure F.1
𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = 𝑌𝑌 ∙ 𝐶𝐶𝐴𝐴0 = 0.12(64 𝑚𝑚𝑚𝑚 𝐴𝐴/𝑐𝑐𝑚𝑚3 ) = 7.7 𝑚𝑚𝑚𝑚 𝐴𝐴/𝑐𝑐𝑚𝑚3
b. Determine CA(x, t) at the center of the slab (x = L/2) at t = 96 hr, using the infinite series
solution approach (x = 0, surface; x = L/2 center)
C A − C AS 4 ∞ 1
 nπ x  −( nπ /2)2 X D
= ∑ sin 
,
e
C A0 − c AS π n =1 n
 L 
n = 1,3,5,
See Microsoft Excel spreadsheet, CA(x,t) = 7.34 mg A/cm3
27-25
27.19 continued
2
D AB =
3.00E-07 cm /sec
C AS =
0.0 mg/cm
3
64.00 mg/cm
345600 sec
t=
96.000 hr
x 1 = L/2 =
x=
C A (x,t) =
XD =
n
0.326 cm
0.326 cm (center of slab)
3
7.34 mg/cm
0.975573
Term
Summation
% Change
1
1.15E-01
1.147E-01
3
-1.66E-10
1.147E-01
0.000
5
1.87E-27
1.147E-01
0.000
7
-1.08E-52
1.147E-01
0.000
9
2.97E-86
1.147E-01
0.000
11
-3.71E-128
1.147E-01
0.000
13
2.08E-178
1.147E-01
0.000
15
-5.17E-237
1.147E-01
0.000
17
5.66E-304
1.147E-01
0.000
19
0.00E+00
1.147E-01
0.000
21
0.00E+00
1.147E-01
0.000
23
0.00E+00
1.147E-01
0.000
25
0.00E+00
1.147E-01
0.000
27
0.00E+00
1.147E-01
0.000
29
0.00E+00
1.147E-01
0.000
31
0.00E+00
1.147E-01
0.000
33
0.00E+00
1.147E-01
0.000
35
0.00E+00
1.147E-01
0.000
37
0.00E+00
1.147E-01
0.000
1.40
1.40
1.20
1.20
1.00
1.00
0.80
0.80
Summation
0.60
Term
0.60
0.40
0.40
0.20
0.20
0.00
0.00
-0.20
-0.20
-0.40
Term
t=
Summation of terms
C A0 =
3
-0.40
0 1 2 3 4 5 6 7 8 9 10111213141516171819202122232425
n
27-26
27.20
L = 0.125 cm, x1 = L/2 (symmetry)
W = 0.50 cm, S = W2
𝐶𝐶𝐴𝐴∞ = 𝐶𝐶𝐴𝐴𝐴𝐴 = 0
CA0 = 64 mg/cm3
DAB = 3.0 x 10-7 cm2/sec
USS Diffusion in a slab, total amount transferred mA(t) by infinite series solution, assuming
sealed edges
At surface (x = 0)
∞
2
4 DAB
=
N Az ( 0 ,t)
(C As − C Ao )∑ e − (n π / 2 ) X D
L
n =1
𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡
𝑋𝑋𝐷𝐷 =
𝑥𝑥12
m Ao = C Ao ⋅ S ⋅ L initial amount (moles A) loaded within slab at t = 0
mA∞= C AS ⋅ S ⋅ L
final (equilibrium) amount loaded within slab at t →∞
Flux is into both sides of the slab of thickness L
t
t
0
0
2S ∫ N
2S ∫
=
m=
A (t) − m Ao
Az ( 0 ,t)dt
 n 2π 2 DAB t 

L2

−
mA∞ − mA (t ) 8 ∞ 1
= 2 ∑ 2 exp 
mA∞ − mAo
π n =1 n

∞
− (n π / 2 )
4 DAB
(C AS − C A0 )∑ e 
L
n =1
2 DAB t 

x12 
dt
n = 1, 3, 5…∞
𝑚𝑚𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴𝐴𝐴 𝑊𝑊 2 𝐿𝐿 = (64 𝑚𝑚𝑚𝑚/𝑐𝑐𝑚𝑚3 )(0.5 𝑐𝑐𝑐𝑐)2 (0.125 𝑐𝑐𝑐𝑐) = 2.00 mg A
𝑚𝑚𝐴𝐴∞ = 0
See Microsoft Excel spreadsheet
At t = 1.0 hr (3600 sec)
𝑚𝑚𝐴𝐴0 − 𝑚𝑚𝐴𝐴 (𝑡𝑡) = 0.593 𝑚𝑚𝑚𝑚 𝐴𝐴
At t = 2.0 hr (7200 sec)
𝑚𝑚𝐴𝐴0 − 𝑚𝑚𝐴𝐴 (𝑡𝑡) = 0.839 𝑚𝑚𝑚𝑚 𝐴𝐴
27-27
27.20 continued
2
C AS =
3.00E-07 cm /sec
3
0.0 mg/cm
C Ao =
64.00 mg/cm
m Ao =
2.00 mg
m A,inf =
0.00 mg
m Ao - m A =
0.593 mg delivered
Term
Summation % Change
6.83E-01 6.835E-01
2.8
1.94E-02 7.029E-01
4.56E-04 7.033E-01
0.1
3.88E-06 7.033E-01
0.0
1.00E-08 7.033E-01
0.0
7.31E-12 7.033E-01
0.0
1.46E-15 7.033E-01
0.0
7.79E-20 7.033E-01
0.0
1.10E-24 7.033E-01
0.0
4.10E-30 7.033E-01
0.0
3.99E-36 7.033E-01
0.0
1.01E-42 7.033E-01
0.0
6.62E-50 7.033E-01
0.0
1.12E-57 7.033E-01
0.0
4.93E-66 7.033E-01
0.0
5.59E-75 7.033E-01
0.0
1.63E-84 7.033E-01
0.0
1.22E-94 7.033E-01
0.0
2.37E-105 7.033E-01
0.0
2
c As =
3.00E-07 cm /sec
3
0.0 mg/cm
c Ao =
64.00 mg/cm
mAo =
2.00 mg
mA,inf =
0.00 mg
7200 sec
2.00 hr
1.161 mg remaining
mAo - mA =
0.839 mg delivered
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
35
37
1.00
Summation
Term
Summation of terms
0.80
0.80
0.60
0.60
0.40
0.40
0.20
0.20
0.00
0.00
0
1
2
3
4
5
6
7
8
9
10
11
n
L=
0.125 cm
w=
0.50 cm
3
t=
t=
mA =
n
1.00
Term
Summation % Change
5.76E-01 5.763E-01
4.18E-03 5.805E-01
0.7
6.42E-06 5.805E-01
0.0
9.12E-10 5.805E-01
0.0
1.00E-14 5.805E-01
0.0
7.97E-21 5.805E-01
0.0
4.43E-28 5.805E-01
0.0
1.68E-36 5.805E-01
0.0
4.33E-46 5.805E-01
0.0
7.49E-57 5.805E-01
0.0
8.65E-69 5.805E-01
0.0
6.64E-82 5.805E-01
0.0
3.38E-96 5.805E-01
0.0
1.14E-111 5.805E-01
0.0
2.53E-128 5.805E-01
0.0
3.70E-146 5.805E-01
0.0
3.57E-165 5.805E-01
0.0
2.27E-185 5.805E-01
0.0
9.45E-207 5.805E-01
0.0
1.00
1.00
Summation
Term
0.80
0.80
0.60
0.60
0.40
0.40
0.20
0.20
0.00
Term
DAB =
0.50 cm
Term
3600 sec
1.00 hr
1.407 mg remaining
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
35
37
0.125 cm
3
t=
t=
mA =
n
L=
W=
Summation of terms
D AB =
0.00
0
1
2
3
4
5
6
7
8
9
10
11
n
27-28
27.21
Given:
A = drug, B = gel, C = surrounding fluid
Sphere dimensions: D = 0.5 cm, R = D/2 = 0.5 cm / 2 = 0.25 cm
Assume cA,1 = cA∞ = 0 bulk fluid, cAs > 0
DAe = 3.0 × 10−6 cm2/s effective diffusion coefficient of drug inside gel
DAC = 1.5 × 10−5 cm2/s molecular diffusion coefficient of drug in surrounding fluid
kc = 5.0 × 10−5 cm/s mass transfer coefficient for A around sphere
a. Biot number (Bi)
× −5cm 2 /s ( 0.25 cm )
5.0 10
kc R
=
Bi =
= 4.1
DAe
× −6 cm/s
3.0 10
(
)
(
)
b. CA(r,t) at center of sphere (r = 0) at t = 2.3 h
USS diffusion in a sphere
Get XD, n, m, then use to get Y from concentration–time charts
 3600 s 
× −6 cm 2 /s ( 2.3 h ) 
3.0 10

D t
 1h  = 0.40
X D = Ae2 =
2
R
( 0.25 cm )
(
)
n=
r
0
=
=0
R
0.25 cm
m=
DAe
3.0 10
× −6 cm 2 /s
=
= 0.24
kc R
5.0 10
× −5cm/s ( 0.25 cm )
(
)
Figure F.6 - sphere, for n = 0, m = 0.25, X= 0.40, read off Y = 0.15
c A,1 − c A ( r, t ) c A∞ − c A ( 0, t ) 0 − c A ( 0, t ) c A ( 0, t )
=
=
=
Y ≅=
0.15
c Ao
0 − c Ao
c A,1 − c Ao
c A∞ − c Ao
=
c Ao
mAo
mAo
0.005 mmol
mmole
=
=0.0764
=
4
4
3
V
cm3
π R3
π ( 0.25 cm )
3
3
mmole 
mmole

0.15  0.0764
= 0.0115 c A ( 0, t ) =
Y ⋅ c A,o =
3 
cm 
cm3

27-29
27.22
A = CO2, B = N2
T = 200 oC (473 K), P = 15.0 atm
K/ = 1776.2 cm3 gas/cm3 bulk adsorbent
ε = 0.60 cm3 gas/cm3 bulk adsorbent
DAe = 0.018 cm2/sec
S = 100 cm2
yA∞ = yAS = 0.10 bulk gas
CA0 = 0 inside porous adsorbent
Determine total CO2 loaded into adsorbent after t = 60 min (3600 sec)
USS diffusion in semi-infinite medium
4𝐷𝐷′𝐴𝐴𝐴𝐴 𝑡𝑡
(𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 )
𝜋𝜋
𝑚𝑚𝐴𝐴 (𝑡𝑡) − 𝑚𝑚𝐴𝐴0 = 𝑆𝑆�
′
=
𝐷𝐷𝐴𝐴𝐴𝐴
(0.60)2 (0.018 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠)
𝜀𝜀 2 𝐷𝐷𝐴𝐴𝐴𝐴
=
= 3.65 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠
(0.60 + 1776.2)𝑐𝑐𝑐𝑐3 /𝑐𝑐𝑐𝑐3
𝜀𝜀 + 𝐾𝐾′
𝐶𝐶𝐴𝐴𝐴𝐴 = 𝑦𝑦𝐴𝐴𝐴𝐴 ∙ 𝐶𝐶 = 𝑦𝑦𝐴𝐴𝐴𝐴
𝑚𝑚𝐴𝐴 (𝑡𝑡) = 𝑆𝑆�
𝑃𝑃
15 𝑎𝑎𝑎𝑎𝑎𝑎
= (0.10) �
� = 3.86 𝑥𝑥 10−5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑐𝑐𝑐𝑐3
(82.06 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎/𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾)(473 𝐾𝐾)
𝑅𝑅𝑅𝑅
4𝐷𝐷′𝐴𝐴𝐴𝐴 𝑡𝑡
4 ∙ (3.65 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠)(3600 sec)
(3.86 𝑥𝑥 10−5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑐𝑐𝑐𝑐3 )
𝐶𝐶𝐴𝐴𝐴𝐴 = 100 𝑐𝑐𝑐𝑐2 �
𝜋𝜋
𝜋𝜋
mA(t) = 5.0 x 10-4 gmole CO2
27-30
28.1
Given:
T = 300 K, P = 1.0 atm
µ
ν
Sc =
=
ρ DAB DAB
 P  T 
DAB ,T2 , P 2 = DAB ,T1 , P 1  1   2 
 P2   T1 
3/2
⋅
Ω D |T1
Ω D |T2
Neglect temperature dependency on Ω D
 P  T 
DAB ,T2 , P 2 = DAB ,T1 , P 1  1   2 
 P2   T1 
3/2
Appendix I: ν H2O,liquid (300K ) = 8.788 × 10−7
m2
m2
, ν air (300 K) = 1.5689 × 10−5
s
s
O2 gas (A) in Air (B)
Appendix J: DAB (273 K,1.0 atm) = 0.175
cm 2
s
3/2
cm 2  1 atm   300 K 
cm 2
=
0.202
DAB ,T2 , P 1 = 0.175



s  1 atm   273 K 
s
2
m
1.5689 × 10−5
vB
s = 0.786
Sc =
=
2
m
DAB
0.202 × 10−4
s
O2 (A) dissolved in liquid H2O (B)
cm3
µ L ,H O (300 K) =
8.76 ×10 Pa s =
0.876 cP , VA = 25.6
gmole
Determine DAB using Hayduk and Laudie correlation
−4
2
DAB = 13.26 × 10−5 µ L −1.14VA−0.589
(
)
DAB = 13.26×10 −5 ( 0.876 )
v
Sc = B =
DAB
8.788 × 10−3
2.28 × 10 −5
−1.14
( 25.6 )
−0.589
= 2.28 × 10 −5
cm 2
s
2
cm
s = 385
2
cm
s
CO2 gas (A) in air (B)
28-1
Appendix J: DAB (273 K,1 atm) = 0.136
DAB ,T2 , P 1 = 0.136
cm 2  1 atm   300 K 



s  1 atm   273 K 
cm 2
s
3/2
= 0.157
cm 2
s
CO2 (A) dissolved in liquid H2O (B)
Determine DAB using Hayduk and Laudie correlation
cm3
VCO2 = 34.0
gmole
DAB = 13.26 ⋅10−5 µ L −1.14VA−0.589
=
DAB
(13.26 × 10 )( 0.876
v
Sc = B =
DAB
−5
8.788 × 10−7
1.93 × 10 −9
−1.14
) ( 34.0 )
−0.589
=1.93 × 10 −5
cm 2
s
2
m
s =455
2
m
s
Schmidt numbers are higher in liquids relative to gases
28-2
28.2
St
=
kc  kc L   ν   DAB 
Sh
= 

 =

v ∞  DAB   Lv ∞   ν  Re⋅ Sc
Pe
=
v∞ L  v∞ L   ν 
= 
=
 Re⋅ Sc

DAB  ν   DAB 
28-3
28.3
Property
Char. Length = Diameter
Velocity
Fluid density
Fluid viscosity
Fluid Diffusivity
Mass Transfer Coefficient
Symbol
D
𝑣𝑣
𝜌𝜌
𝜇𝜇
DAB
kc
Units
L
L/t
M/L3
M/Lt
L2/t
L/t
2
L M  L 
a b c
=
v ρ DAB ( L) a    3   
π 1 D=
t L   t 
=
M : 0 c=
c 0
t : 0 =−b − 1
b =−1
L : 0 =a + b − 3c + 1
a =−1
D
π 1 = AB
Dv
b
c
e
f
L M   M 
=
v ρ µ ( L)    3  
π 2 D=

 t   L   L ⋅t 
t : 0 =−e − 1
e =−1
−1
M :0 =
f +1
f =
L : 0 =d + e − 3 f − 1
d =−1
µ
1
=
π2 =
D ρ v Re
d
e
f
d
h
l
L M  L
g h l
=
v ρ kc ( L ) g    3   
π 1 D=
t L  t
=
M : 0 l=
l 0
t : 0 =−h − 1
h =−1
L : 0 = g + h − 3l + 1
g=0
k
π3 = c
v
k
π2
µ
=
= Sc → c = f [Re, Sc]
v∞
π 1 ρ DAB
28-4
28.4
gas distributor
Drying Chamber
cA,∞ ≈ 0
60 m3/min air H = 1.0 m
27°C, 1.0 atm (W = 1.5 m)
L = 1.5 m
painted
steel plate
heated surface (27°C)
Given:
A = solvent, B = air
T = 300 K, P = 1.0 atm
g
,
cm3
Plate dimensions: L = 150 cm, H = 100 cm, W = 150 cm
m3
1.0
3
3
vo
m
m
s= 0.667 m
v∞ =
=
vo 60
= 1.0 =
W ⋅ L 1.0 m ⋅1.5 m
s
min
s
Paint coating:  = 0.10 cm , wA = 0.30 , ρ A,liq = 1.5
Physical parameters:
g
m2
cm 2
M A = 78
, p*A = 0.138 atm , DAB = 0.097
, ν air (300K)
= 1.5689 ×10−5
gmole
s
s
Physical System: boundary layer between surface and bulk gas
Source for A: solvent coating surface
Sink for A: bulk flowing gas
a. ShL over entire plate
Re and Sc:
m

0.667  (1.5 m )

v∞ L
s 

=
=
Re =
63, 771 laminar flow
2
ν

−5 m 
1.5689 × 10

s 

28-5
m2
1.5689 × 10
vair
s = 1.617
Sc =
=
2
m
DAB
0.097 × 10−4
s
−5
Laminar flow over a flat plate:
1/3
=
ShL 0.664
=
Re1/2
L Sc
=
) (1.617 ) 197
( 0.664 )( 63, 771
1/2
1/3
b. Emissions rate WA
Transfer rate across boundary layer
=
WA N=
kc (c As − c A,∞ )S
AS
c A,∞ ≈ 0 in bulk air flow
S =L ⋅ W =(150 cm)(150 cm) = (150 cm) 2
cm 2
197 ⋅ 0.097
Sh L DAB
cm
s
kc =
=
=
0.127
L
150 cm
s
c=
As
( 0.138 atm )
p*A
mol
=5.61 × 10−6
=
3
RT 
cm3
cm ⋅ atm 
 82.06
 ( 300 K )
mol ⋅ K 

cm  
g   60 s 
g
2

−6 mol 
150 cm )  78
=
WA  0.127
  5.61 × 10

 =75.0
3 (
s 
cm 
min

 mol   min 
c. Time required for complete drying of paint
Material balance on solvent in paint layer
IN – OUT + GEN = ACC
dm
0 − WA + 0 = A
dt
0
t
m Ao
0
∫ dmA = ∫ WAdt
t=
mAo
WA
28-6
g 

mA,0 = V ⋅ ρ ⋅ x A =  ⋅ L2 ⋅ ρ ⋅ wA = ( 0.1 cm ) (150 cm) 2 1.5 3  (0.30) = 1012.5 g
 cm 
mA,0 1012.5g
=
t =
= 13.5 min
WA 75 g/min
d. Boundary layer thickness at x = L = 1.5 m
( 5)(1.5 m ) = 2.97 cm
5⋅ x
=
Re x
63771
δ
2.97 cm
δc =
=2.53 cm
=
1/3
1/3
Sc
(1.617 )
δ
=
These are both much smaller than the height of the drying chamber.
e. Volumetric flowrate of air needed for WA = 150 g A/min, all other variables unchanged
WA,new kc ,new
= =
WA,old
kc ,old
1/2
 v ∞ ,new 
=


 v ∞ ,old 
2
1/2
 vo ,new 


 vo ,old 
 WA,new  
m3  150 g/min 
m3
vo ,new v=
=
=
60
240




o ,old 
 
min
 WA,old   min   75 g/min 
2
28-7
28.5
W = 0.50 m, L = 2.5 m, 𝑣𝑣∞ = 1.5
0.080
𝑐𝑐𝑐𝑐2
𝑠𝑠
, 𝑀𝑀𝐴𝐴 = 86
υ air = 1.505 × 10 −5
𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
m2
s
𝑚𝑚
𝑠𝑠
, 𝑇𝑇 = 293 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑝𝑝𝐴𝐴∗ = 0.159 𝑎𝑎𝑎𝑎𝑎𝑎, 𝐷𝐷𝐴𝐴𝐴𝐴 =
, 𝑝𝑝𝐴𝐴,∞ ≈ 0, ∴ 𝑐𝑐𝐴𝐴,∞ ≈ 0
a. Sc and Sh
2

−5 m 
×
1.505
10


s 
υair 
Sc =
=
=
1.88
2
DAB 
−4 m 
 0.080 ×10

s 

m
1.5 ⋅ 0.5m
v∞ L
s
=
=
=
Re
4.98 x 104 laminar
L
2
m
ν
1.505 ×10−5
s
1/3
ShL =
=
0.664 Re1/2
0.664 ( 4.98 ×104 )
L Sc
1/2
(1.88)
1/3
=
183
b. Solvent evaporation rate, WA

cm 2 
(183)  0.08

s 
Sh ⋅ DAB
cm

=
kc =
= 0.293
W
s
( 50 cm )
( 0.159 atm )
p*A
gmole
=
= 6.61 x 10-6
3
RT 
cm3
cm ⋅ atm 
82.06
293
K
)
(


gmole ⋅ K 

c A, ∞ = 0
c=
As
WA = N A S = kc ( c As − c A,∞ ) ⋅ 2WL =
2
cm  
gmole


 100 cm 
-6 gmole
-0  ( 2 ⋅ 0.5 m ⋅ 2.5 m ) 
 0.293
  6.61 x 10
 = 0.048
3
s  
cm
s


 1m 
c. Determine 𝑋𝑋𝐴𝐴𝐴𝐴
𝑔𝑔 𝐴𝐴
𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑋𝑋𝐴𝐴,𝑂𝑂 = 0.10
, 𝑚𝑚̇𝑠𝑠 = 50.0
𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑠𝑠
Mass balance on solvent in moving sheet
IN – OUT – GEN = ACC
28-8
•
•
X A,0 m s − X A, f m s − WA M A + 0 =
0
gmole  
gA 

 0.048
  86
s   gmole 
WA M A
gA
gA

X A, f =
X A,0 − •
=
−
=
0.10
0.017
g poly
g poly
 g poly 
ms
 50

s 

28-9
28.6
a. Mass transfer coefficients for AsH3 and Ga(CH3)3
𝐴𝐴 = 𝐴𝐴𝐴𝐴𝐻𝐻3 , 𝐵𝐵 = 𝐺𝐺𝐺𝐺(𝐶𝐶𝐶𝐶3 )3 , 𝐶𝐶 = 𝐻𝐻2, , 𝑇𝑇 = 800 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎
𝑚𝑚2
𝑐𝑐𝑐𝑐
𝐿𝐿 = 15 𝑐𝑐𝑐𝑐, 𝑆𝑆 = 225 𝑐𝑐𝑐𝑐2 , 𝑣𝑣∞ = 100
, 𝜈𝜈𝐻𝐻2 = 5.6863 × 10−4
𝑠𝑠
𝑠𝑠
𝑐𝑐𝑐𝑐2
𝑦𝑦𝐴𝐴∞ = 0.001, 𝑦𝑦𝐵𝐵∞ = 0.001, 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝑐𝑐𝐵𝐵𝐵𝐵 = 0, 𝐷𝐷𝐵𝐵𝐵𝐵 = 1.55
𝑠𝑠
𝑃𝑃
1 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐 =
=
= 1.52 × 10−5
𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎
𝑅𝑅𝑅𝑅
𝑐𝑐𝑐𝑐3
82.06
∙ 800 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
cm 

100

 (15 cm )
v∞ L 
s 
Re L =
=
= 263.8 Re , 2.0 x 105 laminar flow
2
νC


cm
 5.6863

s 

Ga(CH3)3

cm 2 
5.6863


s 
vC

=
=
Sc =
3.67
DBC

cm 2 
1.55

s 

DBC
1.55 cm 2 /s
cm
1/2
1/3
1/3
0.664 Re1/2
=
( 0.664 )( 263.8) ( 3.67 ) = 1.72
L Sc
L
15 cm
s
cm
gmole
gmole
N B= kc ( cB∞ − cBs =
) kc yB∞ c= 1.72  ( 0.001) 1.52 x 10-5 3  = 2.61 x 10-8 2
cm 
cm s
s 


AsH3
1 1/2
1.5 1
𝑇𝑇 � +
�
𝑀𝑀𝐴𝐴 𝑀𝑀𝐶𝐶
𝐷𝐷𝐴𝐴𝐴𝐴 = 0.001858
2
𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴
Ω𝐷𝐷
1
1 1/2
0.001858 ∙ 8001.5 �
+
�
𝑐𝑐𝑐𝑐2
77.95 2.02
=
=
4.54
1 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ (3.514) 2 ∙ 0.535
𝑠𝑠
2

cm 
 5.6863

s 
vC

=
=
Sc =
1.252
DAC

cm 2 
 4.54

s 

D
4.54 cm 2 /s
cm
1/2
1/3
1/3
=
kc = AC 0.664 Re1/2
Sc
( 0.664 )( 263.8) (1.25) = 3.52
L
15 cm
s
L
cm
gmole
gmole
N A= kc ( c A∞ − c As =
) kc y A∞ c=  3.52  ( 0.001) 1.52 x 10-5 3  = 5.35 x 10-8 2
s 
cm 
cm s


kc =
28-10
b. Ratio of AsH3 and Ga(CH3)3
gmole
5.35 x 10-8
NA
cm 2s 2.05
=
=
N B 2.61 x 10-8 gmole
cm 2s
y c⋅k
NA
1.0
Let =
= A∞ c , A
NB
y B∞ c ⋅ k c , B
kc , A
yB∞ y=
=
A∞
kc , B
( 0.001)
3.52 cm/s
= 0.00205
1.72 cm/s
28-11
28.7
A = polymer, B = MEK (liquid solvent)
L = 20 cm, W = 10 cm, 𝑙𝑙𝑜𝑜 = 0.02 𝑐𝑐𝑐𝑐, 𝑣𝑣𝑜𝑜 = 30
𝐷𝐷𝐴𝐴𝐴𝐴 = 3 × 10−6
𝑐𝑐𝑐𝑐2
2
, 𝜈𝜈𝐵𝐵 = 6.0 × 10−3
a. Sc and Sh
2

−3 cm 
⋅
6
10


s 
υB 
=
=
Sc =
2000
2
DAb 

cm
−6
 3 ⋅10

s 

𝑐𝑐𝑐𝑐2
𝑠𝑠
𝑐𝑐𝑐𝑐3
𝑠𝑠
, 𝑐𝑐𝐴𝐴∞ ≈ 0
, 𝑐𝑐𝐴𝐴∗ = 0.04
𝑔𝑔
𝑐𝑐𝑐𝑐3
1/2


cm
⋅ 20cm 
1.5

1/3
s
=
=
=
Sh 0.664
Re1/2 Sc1/3 ( 0.664 ) 
( 2000 ) 591.6
2 
 6 ⋅10−3 cm 


s 

b. Flux NA
=
N A kc ( c As − c A,∞ )
2

-6 cm 
( 591.6 )  3.0 x 10

s 
Sh ⋅ DAB

kc =
=
= 8.87 x 10-5 cm/s
20
cm
L
(
)
gA
gA


N A = ( 8.87 x 10-5 cm/s )  0.04 3 - 0  = 3.55 x 10-6
cm
cm 2s


c. Local kc,x vs. position x
kc,x [cm/s] *104
2.1
1.8
1.5
1.2
0.9
0.6
0.3
0
5
10
15
20
x [cm]
28-12
d. Polymer coating thickness vs. position and time
Material balance on solute
IN – OUT + GEN = ACC
mA
t
dm
0 − WA + 0 = A , ∫ dmA = ∫ WA dt
dt
m A ,0
0
mA (t ) = mA,0 − WA ⋅ t
mA (t )= (t ) ⋅ S ⋅ ρ A, solid
kc ( c As − c A,∞ ) S ⋅ t
kc ( c As − c A,∞ ) ⋅ t
(t ) =
(to ) −
(to ) −
=
S ⋅ ρ A, solid
ρ A, solid
e. Boundary layer thickness at x = 10 cm
=
δ
5⋅ x
=
Re x
( 5)(10 cm ) = 1.0 cm
2500
=
/ Sc1/3 1.0 cm / ( 2000 ) = 0.079 cm
δ c δ=
1/3
28-13
28.8
Shx =
kc x
= 0.0292 Re 4x / 5 Sc1 / 3 , where Re L ≥ 3.0 ⋅106
D AB
L
kc =
0.0292 Re 4x / 5 Sc1/ 3 DAB
dx
∫0
x
L
∫ dx
=
0.0365 Re 4L/ 5 Sc1/ 3 DAB
L
0
=
ShL
kc L
1/3
= 0.0365 Re 4/5
L Sc
DAB
28-14
28.9
a. Determine α, β, η, ξ
v= α + β y1/7
BC for velocity profile:
v( y = 0) = 0, α = 0
v( y = δ ) = v∞ , β =
 y
v x = v∞  
δ 
v
δ 1/ 7
1/ 7
Given that:
c A − c A,∞ = η + ξy1/ 7
BC for concentration profile:
y = 0; c A − c A,∞ = c A, s − c A,∞
y = δ c ; c A − c A,∞ = 0
η = c A, s − c A,∞
c
−c
ξ = − A, s 1 / 7 A,∞
δc
c A − c A,∞
 y
= 1 −  
c A, s − c A,∞
δc 
1/ 7
b. Local mass transfer coefficient, kc
δ
d c
(c A − c A,∞ )v x dy = k c (c A, s − c A,∞ )
dx ∫0
δ
k
d c (c A − c A , ∞ ) v x
dy = c
∫
dx 0 (c A, s − c A,∞ ) v ∞
v∞
28-15
d  c   y 
 1−  
=
υ ∞ dx  ∫0   δ c 
 
kc
δ
1/ 7
δc
1/ 7
2/7

 
 y   d    y   y
dy 
 ∫  1 / 7  −
 dy =
1/ 7 1/ 7 



δ
dx
δ
  
 0  
  δ c δ  

δc
d  7 y8/ 7 7 y9/ 7 
d  7 δ c8 / 7 
=
−
=




dx  8 δ 1/ 7 9 δ c1/ 7δ 1/ 7  0 dx  92 δ 1/ 7 
If δ c =δ →
kc
7 d
=
[δ ]
υ∞ 72 dx
For turbulent flow:
υ 
δ=
= 0.371 
1/ 5
 v∞ 
 xv ∞ 


 υ 
0.371x
υ 
dδ
= 0.371 
dx
 v∞ 
1/ 5
1/ 5
x4/5
k
4 −1 / 5 0.291
7  0.291  0.0289
= 1 / 5 → c =  1 / 5  =
x
5
v ∞ f 72  Re x  Re 1x/ 5
Re x
k c = 0.0289v ∞ Re −x 1 / 5
28-16
28.10
𝑐𝑐𝐴𝐴 − 𝑐𝑐𝐴𝐴,𝑠𝑠 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐𝑦𝑦 2 + 𝑑𝑑𝑑𝑑 3
BC for concentration profile:
=
y 0,
=
c A c A, s
=
a 0
y =δ , c A =c A,∞
y=
δc ,
c A,∞ − c A, s =bδ + cδ 2 + d δ 3
d
0
( c A − c A, s ) =
dy
d2
y=
0
( cA − cA,s ) =0
dy 2
d =−
c A, ∞ − c A, s
2δ
c A − c A, s
3
c
b=
b + 2cδ c + 3d δ c 3
0=
0=
2c + 6dy , y =
0, c =
0
3  c A, ∞ − c A, s 


δc
2 

3 y  1 y 
=   −  
c A, ∞ − c A, s 2  δ c  2  δ c 
3
28-17
28.11
A = methylene chloride, B = air
𝑇𝑇 = 300 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑣𝑣∞ = 7.5
𝑚𝑚
𝑠𝑠
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
, 𝜐𝜐𝑎𝑎𝑎𝑎𝑎𝑎 = 0.15
𝐷𝐷𝐴𝐴𝐴𝐴 = 0.085
𝑠𝑠
𝑠𝑠
= 750
𝑐𝑐𝑐𝑐
𝑠𝑠
, L = 100 m = 100 x 102 cm
A = methylene chloride, B = water (liquid)
𝐷𝐷𝐴𝐴𝐴𝐴 = 1.07 × 10−5
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
, 𝜐𝜐𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 0.010
𝑠𝑠
𝑠𝑠
a. Position for laminar flow transition, Lt
=
Ret 2.0
=
x 105
v∞ Lt
υ
𝑐𝑐𝑐𝑐2
5
𝑅𝑅𝑅𝑅𝑡𝑡 𝜈𝜈 2.0 × 10 ∙ 0.15 𝑠𝑠
𝐿𝐿𝑡𝑡 =
=
= 40 𝑐𝑐𝑐𝑐
𝑐𝑐𝑐𝑐
𝑣𝑣∞
750
𝑠𝑠
Since the laminar portion is so small compared to the length of the pond, it is assumed that
turbulent flow mass transfer will dominate.
b. Average gas phase film mass transfer coefficient, kc
𝑐𝑐𝑐𝑐
2
𝑣𝑣∞ 𝐿𝐿 750 𝑠𝑠 ∙ 100 × 10 𝑐𝑐𝑐𝑐
𝑅𝑅𝑅𝑅 =
=
= 5.0 𝑥𝑥106
𝑐𝑐𝑐𝑐2
𝜈𝜈𝑎𝑎𝑎𝑎𝑎𝑎
0.15
𝑠𝑠
𝑐𝑐𝑐𝑐2
0.15 𝑠𝑠
𝜈𝜈𝑎𝑎𝑎𝑎𝑎𝑎
𝑆𝑆𝑆𝑆 =
=
= 1.765
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴
0.085
𝑠𝑠
6
Re > 3.0 x 10 , fully turbulent
1
𝐷𝐷𝐴𝐴𝐴𝐴
∙ 0.0365𝑅𝑅𝑅𝑅𝐿𝐿0.80 𝑆𝑆𝑆𝑆 3 =
𝐿𝐿
𝑐𝑐𝑐𝑐2
0.085 𝑠𝑠
1
𝑐𝑐𝑐𝑐
0.80 (1.765)3
∙
0.0365(50,000,000)
= 0.541
2
100 × 10 𝑐𝑐𝑐𝑐
𝑠𝑠
𝑘𝑘𝑐𝑐 =
28-18
𝑘𝑘𝑐𝑐
𝑘𝑘𝐺𝐺 =
=
𝑅𝑅𝑅𝑅
82.06
𝑐𝑐𝑐𝑐
0.541 𝑠𝑠
𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐾𝐾
∙ 300 𝐾𝐾
= 2.2 × 10−5
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
c. Schmidt number for gas and liquid phase
𝑐𝑐𝑐𝑐2
0.15 𝑠𝑠
= 1.76
𝑆𝑆𝑆𝑆𝑔𝑔𝑔𝑔𝑔𝑔 =
𝑐𝑐𝑐𝑐2
0.085
𝑠𝑠
𝑐𝑐𝑐𝑐2
0.010 𝑠𝑠
𝑆𝑆𝑆𝑆𝑙𝑙𝑙𝑙𝑙𝑙 =
= 935
𝑐𝑐𝑐𝑐2
1.07 × 10−5
𝑠𝑠
28-19
28.12
A = octane, B = air
𝑊𝑊 = 10.0 𝑚𝑚 , 𝐿𝐿 = 1.0 𝑚𝑚, 𝑇𝑇 = 288 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑝𝑝𝐴𝐴 = 0.01025 𝑎𝑎𝑎𝑎𝑎𝑎, 𝜀𝜀 = 0.40
𝑚𝑚2
𝑐𝑐𝑐𝑐
𝑐𝑐𝑐𝑐2
, 𝑣𝑣∞ = 2.0
, 𝐷𝐷𝐴𝐴𝐴𝐴 (298 𝐾𝐾, 1 𝑎𝑎𝑎𝑎𝑎𝑎) = 0.0602
𝜐𝜐𝑎𝑎𝑎𝑎𝑎𝑎 (288 𝐾𝐾) = 1.46 × 10−5
𝑠𝑠
𝑠𝑠
𝑠𝑠
𝑐𝑐𝑐𝑐2 288 𝐾𝐾 3/2
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴 (𝑇𝑇, 𝑃𝑃) = 0.0602
�
� = 0.0572
𝑠𝑠 298 𝐾𝐾
𝑠𝑠
a. Determine yAs
v∞W ( 2.0 cm/s )(1000 cm )
=
Re L =
=1.37 x 104
2
υ
0.146 cm /s
(
)
υ
0.146 cm 2 /s
=
= 2.55
Sc =
DAB 0.0572 cm 2 /s
Re < 2.0 x 105, therefore laminar flow
1/2
D
0.0572 cm 2 /s
cm
1/3
1/3
=
kc = AB ( 0.664 ) Re1/2
Sc
( 0.664 ) (1.37 x 104 ) ( 2.55) = 0.00607
L
s
L
1000 cm
(1039 Pa )
pA
mol
= 4.34 x 10-7
c*A =
=
3
3
cm3
RT 
m Pa 
 100 cm 
8.314
288
K
)

(

mol ⋅ K 
 1m 

101,325 Pa
mol
P
= 4.23 x 10-5
=
c =
3
3
cm3
RT 
m Pa 
 100 cm 
 8.314
 ( 288 K ) 

mol ⋅ K 
 1m 

Convection through gas phase boundary layer
=
N A kc ( c As − c A∞ )
Diffusion flux through porous rock layer
DAe *
=
NA
( cA − cAs )
L
Equate flux
ε 2 DAB *
kc ( c As − c=
)
( cA − cAs )
A∞
L
c A∞ ≈ 0
(0.40) 2 ( 0.0572 cm 2 /s ) 
-7 mol 
c
 4.34 x 10

100 cm
cm3 

L
c As =
=
2
2
ε 2 DAB
cm (0.40) ( 0.0572 cm /s )
kc +
0.00607
+
L
s
100 cm
mol
c As = 6.45 x 10-9
cm3
ε 2 DAB
*
A
28-20
mol
3
c As
cm=
y=
=
1.52 x 10-4
As
mol
c
4.23 x 10-5
cm3
6.45 x 10-9
cm 
mol


-9 mol
− 0  = 3.97 x 10-11
N A = kc ( c As − c A∞ )=  0.00607
 6.45 x 10
3
s 
cm
cm 2s


b. If 𝑣𝑣∞ = 50
𝑐𝑐𝑐𝑐
𝑠𝑠
, Determine yAs
 cm 
 50
 (1000 cm )
s 

Re L =
=342,466

cm 2 
 0.146

s 

Re > Ret = 2.0 x 105 transition region
D
DAB 1/3
4/5

0.664 AB Sc1/3 Re1/2
kc =
Sc  Re 4/5
t + 0.0365
L − Ret 
L
L


cm 2 
cm 2 
0.0572
 0.0572



4/5
s 
s 
1/3
1/3
4/5

5 1/2
kc = ( 0.664 ) 
2.55
2.0
x
10
+
0.0365
( ) (
(
)
( 2.55) ( 342,466 ) - ( 2.0 x 105 ) 
)
(1000 cm )
(1000 cm )
= 0.050
cm
s
(0.40) 2 ( 0.0572 cm 2 /s ) 
-7 mol 
c
 4.34 x 10

100 cm
cm3 

L
=
c As =
2
2
ε 2 DAB
cm (0.40) ( 0.0572 cm /s )
kc +
0.050
+
L
s
100 cm
mol
c As = 7.93 x 10-10
cm3
mol
7.93 x 10-10
c As
cm3 = 1.88 x 10-5
=
y=
As
mol
c
4.23 x 10-5
cm3
ε 2 DAB
*
A
cm 
mol


-10 mol
− 0  = 3.97 x 10-11
N A = kc ( c As − c A∞ )=  0.050
 7.93 x 10
3
2
s
cm
cm
s



c. Biot number
At v∞ = 2.0 cm/s,=
Bi
kc L
kL
= 2 c=
DAe ε DAB
( 0.00607 cm/s ) (100 cm)
= 66.3
(0.40) 2 (0.0572 cm 2 /s)
( 0.050 cm/s ) (100 cm)
kc L
kL
= 2 c=
= 546
DAe ε DAB (0.40) 2 (0.0572 cm 2 /s)
Even at low gas flowrate, the Bi number is high and so the process is limited by the diffusion of
gasoline vapor through the porous rock layer
At v∞ = 50 cm/s,=
Bi
28-21
28.13
A = lysozyme, B = water (liquid)
𝑐𝑐𝑐𝑐
, 𝐿𝐿 = 10 𝑐𝑐𝑐𝑐, 𝑙𝑙 = 0.10 𝑐𝑐𝑐𝑐, 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 30 × 10−7 𝑐𝑐𝑐𝑐
𝑇𝑇 = 298 𝐾𝐾, 𝑣𝑣∞ = 5.0
𝑠𝑠
𝜈𝜈𝐵𝐵 = 9.12 𝑥𝑥 10−3 𝑐𝑐𝑐𝑐2 /𝑠𝑠
𝑐𝑐𝑐𝑐2
𝑔𝑔
−6
, 𝑀𝑀𝐴𝐴 = 14,100
, 𝑑𝑑 = 4.12 × 10−7 𝑐𝑐𝑐𝑐
𝐷𝐷𝐴𝐴𝐴𝐴 = 1.04 × 10
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐴𝐴
𝑐𝑐𝑐𝑐2
−7
𝐷𝐷𝐴𝐴𝐴𝐴 = 5.54 × 10
𝑠𝑠
a. Flux model
Flux through membrane
N A kc ( c Ao − c As )
=
Flux through boundary layer over membrane
DAe
=
NA
( cAs − cA∞ )

Eliminate cAs
NA
= c Ao − c As
kc
N A
= c As − c A∞
DAe
1
 
NA  +
c Ao − c A∞
=
k
D
Ae 
 c
(c − c )
N A = Ao A∞

1
+
kc DAe
b. Estimate NA
cm 

5.0

 (10.0 cm )
v∞ L 
s 
Re L =
=
= 5482 Re < 2.0 x 105, laminar flow
2
νB

-3 cm 
 9.12 x 10

s 

cm 2
νB
s = 8769
=
Sc =
cm 2
DAB
1.04 x 10-6
s
Estimate kc
-6
2
D
1/2
1/3
1/3  1.04 x 10 cm /s 
-4
kc = AB 0.664 Re1/2
Sc
=

 0.664 ( 5482 ) ( 8769 ) =1.05 x 10 cm/s
L
L
10 cm


9.12 x 10-3
28-22
N=
N=
A
A
( cAo − cA∞ )
1

+
kc DAe
mmol   1 m   gmol 

 (1.0 − 0.40 )1.0



gmol
m3   100 cm   1000 mmol 
= 3.16 x 10-11 2
NA= 
1
0.10 cm
ms
+
2
cm
cm
1.05 x 10-4
5.54 x 10-7
s
s
Limiting case: all mass transfer resistance in boundary layer
N A ≈=
kc ( c Ao − c A∞ ) 1.05 x 10-4
cm 
mmol   1 m   gmol 
 (1.0 − 0.4 )1.0



m3   100 cm   1000 mmol 
s 
gmol
m 2s
Limiting case: all mass transfer resistance in membrane
cm 2
5.54 x 10-7
DAe
s  (1.0 − 0.4 )1.0 mmol   1 m   gmol 
NA ≈=
( cAo − cA∞ )





0.10 cm
m3   100 cm   1000 mmol 

gmol
= 3.32 x 10-10 2
ms
= 6.326 x 10-10
c. Biot number
cm 

1.05 x 10-4

 (0.10 cm)
kc  
s 
Bi =
=
= 19.03
2
DAe

-7 cm 
 5.54 x 10

s 

The analysis suggests most of the resistance to mass transfer is through membrane, not through
the boundary layer
d. Concentration boundary layer thickness
5(10 cm) ( 8769 )
5x
= 3.3 x 10-2 cm
δc =
δ ⋅ Sc = 1/2 Sc −1/3 =
1/2
 v∞ x 


cm 
10.0
cm


  5.0

(
)

s 
 νB 


2
 


cm
-3
  9.12 x 10
 
s  
 
∴δ c <  =
0.10 cm
-1/3
−1/3
28-23
28.14
𝐴𝐴 = 𝐶𝐶𝐶𝐶, 𝐵𝐵 = 𝑂𝑂2 , 𝐶𝐶 = 𝐶𝐶𝐶𝐶2 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑔𝑔𝑔𝑔𝑔𝑔)
𝑐𝑐𝑐𝑐2
𝑠𝑠
600 𝐾𝐾 1.5 Ω𝐷𝐷,300 𝐾𝐾
𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 𝐷𝐷𝐴𝐴𝐴𝐴 (300 𝐾𝐾) �
� ∙
300 𝐾𝐾
Ω𝐷𝐷,600𝐾𝐾
Appendix K
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
0.978
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴 (300 𝐾𝐾) = 0.213
, 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 0.213
∙ 2.828 ∙
= 0.709
𝑠𝑠
𝑠𝑠
𝑠𝑠
0.8308
1.0615
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
, 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 0.155
∙ 2.828 ∙
= 0.531
𝐷𝐷𝐴𝐴𝐴𝐴 (300 𝐾𝐾) = 0.155
0.8764
𝑠𝑠
𝑠𝑠
𝑠𝑠
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐2
1.0665
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐵𝐵𝐵𝐵 (300 𝐾𝐾) = 0.166
, 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 0.166
∙ 2.828 ∙
= 0.570
𝑠𝑠
𝑠𝑠
0.879
𝑠𝑠
𝑇𝑇 = 600 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝜈𝜈𝐶𝐶𝐶𝐶2 = 0.30004
a. Sc for CO and O2 mass transfer
CO
=
Sc
ν
=
DAC
cm 2
s
0.566
=
cm 2
0.531
s
0.30004
O2
cm 2
ν
s
=
=
0.527
Sc =
cm 2
DBC
0.570
s
b. Average kc for CO mass transfer, local kc at x=L = 50 cm
cm 

4000

 ( 50 cm )
v∞ L 
s 
=
=
Re
= 6.67 ⋅105 transition region
L
2
υ

cm 
 0.30004

s 

Average kc
D
DAC 1/3
4/5

0.664 AC Sc1/3 Re1/2
kc =
Sc  Re 4/5
tr + 0.0365
L − Retr 
L
L
cm 2
cm 2
0.531
0.531
s ( 0.566 )1/3 2.0 x 105 1/2 + 0.0365
s ( 0.566 )1/3
kc = 0.664
(
)
50 cm
50 cm
4/5
4/5
cm
⋅ ( 6.67 x 105 ) - ( 2.0 x 105 )  = 11.7


s
0.30004
28-24
Local kc
DAC
0.0292 Re0.80
Sc1/3
x
x

cm 2 
0.534


s  ( 0.0292 ) 6.67 x 105 0.8 ( 0.562 )1/3 = 11.7 cm
kc , x = 
(
)
s
 50 cm 




kc , x =
c. Scale kc for CO mass transfer to kc for O2 mass transfer
2/3
2/3
 DBC 
cm   0.570 cm 2 /s 
cm

=
11.7
kc , B k=

 = 12.3
c, A 


2
s   0.531 cm /s 
s

 DAC 
d. Flux NA
For a first-order reaction on a convective surface
kk
N A = c A∞ c s
kc + k s
c=
A, ∞
( 0.001)(1.0 atm )
yA P
mol
=
= 2.03 x 10-8
3
RT 
cm3
cm atm 
82.06
600
K
)

(
mol ⋅ K 

 
cm   cm  
11.7

 1.5


mol
s 
s  


-8 mol 

 = 2.7 x 10-8
N A =  2.03 x 10
3 
cm   cm  
cm   
cm 2s

11.7
+
1.5

 

s  
s  
 
28-25
28.15
𝑇𝑇 = 600 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, ℎ = 50
Appendix I, for CO2
𝑊𝑊
𝑚𝑚2 ∙ 𝐾𝐾
𝑘𝑘𝑘𝑘
𝐽𝐽
𝑚𝑚2
𝜌𝜌𝐶𝐶𝐶𝐶2 = 0.8941 3 , 𝑐𝑐𝑝𝑝,𝐶𝐶𝐶𝐶2 = 1076.1
, 𝜈𝜈
= 3.004
, 𝑃𝑃𝑃𝑃𝐶𝐶𝐶𝐶2 = 0.668
𝑚𝑚
𝑘𝑘𝑘𝑘 ∙ 𝐾𝐾 𝐶𝐶𝐶𝐶2
𝑠𝑠
a. Estimate kc for CO in CO2 by Reynolds and Chilton-Colburn analogies
Reynolds Analogy
J 

50 2


v∞ C f
h
m
m sK

=
kc =
=
= 0.0520
ρcp 
2
s
kg  
J 
 0.8941 3 1076.1

m 
kg K 

Chilton Colburn Analogy
kc =
h  Pr 
ρ ⋅ CP  Sc 
2/3
cm 2
ν
s
=
=
0.566
Sc =
cm 2
DAC
0.531
s
J 

2/3
 50 2

m
 0.668 
 m sK
kc =

 = 0.0581
s
kg  
J   0.566 

 0.8941 3 1076.1

m 
kg K 

0.30004
b. Scale kc for CO mass transfer to kc for O2 mass transfer
2/3
2/3
 DBC 
m   0.570 cm 2 /s 
m

=
kc , B k=
0.0581


 =0.0691
c, A


2
s   0.531 cm /s 
s

 DAC 
28-26
28.16
A = n-octane, B = air, T = 27 C = 300 K, P = 1.0 atm
νB = 0.155 cm2/s = 1.55 x 10-5 m2/s, Appendix I
FSG correlation for DAB
1/2
 1
1 
0.001T 
+

MA MB 

=
1/3
1/3 2
P ( Σv ) A + ( Σv ) B
1.75
DAB
(
DAB = 0.066
)
1/2
0.001 ⋅ ( 300 )
1.75
1 
 1
 +

 29 114 
( 1.0 ) ( ( 20.1) + (167.64 ) )
1/3
1/3 2
cm 2
s
Antoine Equation for PA*
B
1351.99
1351.99
= 6.91868 −
= 6.91868 −
= 1.194
C +T
209.155 + T
209.155 + 27
 1.0 atm 
PA* = 15.6 mm Hg 
 = 0.021 atm
 760mm Hg 
log10 PA*= A −
( 0.021 atm )
PA*
gmole
= 0.853
=
3
m3
RT 
-5 m ⋅ atm 
8.206
10
300
K
×
)

(
gmole ⋅ K 

a. Point of transition, hydrodynamic thickness, (δ) and boundary layer thickness (δc)
=
C A*
Ret = 2.0 x 105 for end of laminar flow
Ret =
v ∞ Lt
νB
2

−5 m 
5
×
1.55
10

 ( 2 ×10 )
s 
ν Re
=
L t = B t =
0.39 m
m
v∞

 8.0 
s 

δ and δc at x = L
m

8.0  (10 m )

v∞ L
s 
=
= 
Re
= 5.16 ×106
x
2
νB


m
−5
1.55 × 10

s 

28-27
cm 2
0.155
vB
s = 2.35
=
Sc =
cm 2
DAB
0.066
s
( 5)(10 m ) = 0.022 m = 2.20 cm
5⋅ x
δ =
=
Re x
5.16 ×106
δc
=
δ
=
Sc1/3
2.2 cm
( 2.348)
1/3
= 1.66 cm
b. Re and flow regime
Flow over a flat surface
m

8.0  (10 m )

v∞ L
s 
=
= 
Re
= 5.16 ×106 fully-developed turbulent flow (ReL > 3 x 106)
L
2
νB


−5 m
1.55 × 10

s 

c. n-octane vapor emissions rate WA
=
WA N=
kc (C A* − C A,∞ )S
AS
S =L ⋅ W =(10 m)(8 m) = 80 m 2
kc for fully-developed turbulent flow over a flat surface
0.8
kc L
1/3
1/3
ShL =
=
=
0.0365 Re0.8
1.138 ×104
( 0.0365) ( 5.16 ×106 ) ( 2.35) =
L Sc
DAB
cm 2 
4 
1.138×10
0.066
(
)

s 
Sh L DAB
m

kc
=
=
= 7.51×10−3
s
L
 100 cm 
10 m 

 1m 
m 
gmole


- 0  ( 80 m 2 )
WA =  7.51×10-3   0.853
3
s 
m


WA = 0.51 gmole/s
d. This is a combined heat and mass transfer process subject to both a material and energy
balance. The liquid must be vaporized. If the heat transfer rate is not sufficient to sustain the
surface temperature at the bulk gas stream temperature, then the liquid temperature will be
lowered to a provide a vapor pressure where the simultaneous material and energy balance is
satisfied.
28-28
28.17
A = H2O vapor, B = air
𝑘𝑘𝑘𝑘
, 𝑃𝑃 = 2.34 × 103 𝑃𝑃𝑃𝑃, 𝑐𝑐𝐴𝐴∞ ≈ 0
𝑔𝑔 𝐴𝐴
𝑐𝑐𝑐𝑐2
𝑔𝑔
𝐽𝐽
𝑃𝑃𝑃𝑃𝐵𝐵 = 0.708, 𝜈𝜈𝐵𝐵 = 0.1505
, 𝜌𝜌𝐵𝐵 = 0.001207 3 , 𝑐𝑐𝑝𝑝,𝐵𝐵 = 1.006
𝑠𝑠
𝑐𝑐𝑐𝑐
𝑔𝑔 ∙ 𝐾𝐾
Appendix J:
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴 = 0.260
𝑠𝑠
𝑐𝑐𝑐𝑐2
0.1505 𝑠𝑠
𝜈𝜈
𝑆𝑆𝑆𝑆 =
=
= 0.579
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴
0.260
𝑠𝑠
𝑃𝑃𝐴𝐴
2.34 × 103 𝑃𝑃𝑃𝑃
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
−7
=
=
9.60
×
10
𝑐𝑐𝐴𝐴𝐴𝐴 =
𝑐𝑐𝑐𝑐3 𝑃𝑃𝑃𝑃
𝑅𝑅𝑅𝑅
𝑐𝑐𝑐𝑐3
8.314 × 106
∙ 293 𝐾𝐾
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾
𝑞𝑞
= ℎ(𝑇𝑇∞ − 𝑇𝑇𝑠𝑠 ) = ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 𝑁𝑁𝐻𝐻2𝑂𝑂
𝐴𝐴
∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 𝑁𝑁𝐻𝐻2𝑂𝑂
+ 𝑇𝑇𝑠𝑠
𝑇𝑇∞ =
ℎ
𝑘𝑘𝑐𝑐
𝑇𝑇∞ = ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 ∙ (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴∞ ) + 𝑇𝑇𝑠𝑠 ,
ℎ
2/3
𝑘𝑘𝑐𝑐
1 𝑃𝑃𝑃𝑃
=
� �
ℎ
𝜌𝜌𝑐𝑐𝑝𝑝 𝑆𝑆𝑆𝑆
1 𝑃𝑃𝑃𝑃 2/3
𝑇𝑇∞ = ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 ∙
� � (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴∞ ) + 𝑇𝑇𝑠𝑠
𝜌𝜌𝑐𝑐𝑝𝑝 𝑆𝑆𝑆𝑆
𝑇𝑇𝑆𝑆 = 293 𝐾𝐾, ∆𝐻𝐻𝑣𝑣,𝐴𝐴 = 2.45
2
𝐽𝐽
𝑔𝑔
1
0.708 3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 2450 ∙ 18.02
∙
− 0�
�
� �9.60 × 10−7
𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 0.001207 𝑔𝑔 ∙ 1.006 𝐽𝐽
0.579
𝑐𝑐𝑐𝑐3
𝑔𝑔 ∙ 𝐾𝐾
𝑐𝑐𝑐𝑐3
+ 293 𝐾𝐾
𝑇𝑇∞ = 333 𝐾𝐾
28-29
28.18
Set L = Lt = 40 cm
Ret =
v ∞ Lt
νB

cm 2 
5
0.00
20

 ( 2×10 )
s 
ν Re
v∞ = B t = 
= 10 cm/s
Lt
( 40 cm )
cm 2
vB
s = 133.3
Sc =
=
2
cm
DAB
1.5 ×10−5
s
0.0020
At x = 0, the local kc(x) is not defined. As x increases, kc(x) decreases with kc ( x) ∝ x −1/2
At x = L, the local kc(x) for laminar flow is
1/3
Shx =
0.332 Re1/2
=
( 0.332 ) ( 2 ×105 )
x Sc
1/2

kc=
(x)
Sh x DAB
=
x
( 758.5) 1.5 ×10−5

40 cm
(133.3)
1/3
=
758.5
cm 2 

s 
cm
= 2.84 ×10−4
s
28-30
28.19
Laminar flow over a flat plate
𝑆𝑆ℎ = 0.664𝑅𝑅𝑅𝑅1/2 𝑆𝑆𝑆𝑆1/3
𝑐𝑐𝑐𝑐2
0.15869 𝑠𝑠
𝜈𝜈
𝑆𝑆𝑆𝑆 =
=
= 1.743
𝑐𝑐𝑐𝑐2
𝐷𝐷𝐴𝐴𝐴𝐴
0.090
𝑠𝑠
𝑅𝑅𝑅𝑅 =
𝑣𝑣∞ 𝐿𝐿
=�
𝜈𝜈
𝑘𝑘𝑐𝑐 𝐿𝐿
1 =�
1
0.664 ∙ 𝑆𝑆𝑆𝑆 3
𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0.664 ∙ 𝑆𝑆𝑆𝑆 3
𝑣𝑣∞ = 𝜈𝜈�
= 0.046
𝑆𝑆ℎ
𝑘𝑘𝑐𝑐
1
𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0.664 ∙ 𝑆𝑆𝑆𝑆 3 ∙ 𝐿𝐿
𝑐𝑐𝑐𝑐
𝑠𝑠
𝑐𝑐𝑐𝑐2
= 0.15869
∙�
𝑠𝑠
0.090
𝑐𝑐𝑐𝑐
0.090 𝑠𝑠
1
𝑐𝑐𝑐𝑐2
∙ 0.664 ∙ 1.7433 ∙ 15 𝑐𝑐𝑐𝑐
𝑠𝑠
28-31
28.20
𝐿𝐿 = 150 𝑐𝑐𝑐𝑐, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑇𝑇 = 298 𝐾𝐾, 𝑆𝑆 = 0.50 𝑐𝑐𝑐𝑐2 , 𝛿𝛿 = 0.2 𝑐𝑐𝑐𝑐, 𝑦𝑦𝐴𝐴∞ = 1.0, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0
𝑚𝑚3 𝑎𝑎𝑎𝑎𝑎𝑎
𝑐𝑐𝑐𝑐2
𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎
−5
𝐻𝐻 = 29.5
= 29500
, 𝐷𝐷𝐴𝐴𝐴𝐴 = 2.0 × 10
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑠𝑠
a. Average molar flux of A into falling liquid film
At 25 oC, for liquid water, ρ = kg/m3, µ =
kg/m∙s
=
N A kc ( c As − c A0 )
4 DAB vmax
πL
kg  
m
2

996.7 3   9.81 2  ( 0.002 m )

2
ρ gδ 
m
m 
s 
vmax =
=
= 21.5
-6
2µ
s
( 2 ) ( 909.25 x 10 kg/m ⋅ s )
kc =
2

cm 
-5 cm  
4
2.0
x
10
( )
  2150

s 
s 
cm

kc =
= 0.0191
s
( π )(150 cm )
c=
As
p A y A P (1.0)(1.0 atm)
kmol
mol
=
=
= 0.03389 3 = 3.39 x 10-5
3
H
H
m
cm3

m atm 
29.5


kmol 

cm  
mol
mol


N A =  0.0191   3.39 ⋅10−5 3 − 0  = 6.48 ×10−7
s 
cm
cm 2 s


b. Average concentration in liquid film
N A,ave = vmax c A,ave
mol 

6.48 ⋅10−7


N A,ave 
mol
cm 2 s 
c A,ave
=
=
= 3.0 ⋅10−10
cm 
vmax
cm3

2150


s 

28-32
29.1
a.
Equilibrium distribution curves
Cl2 − Water, T = 293 K,
𝑘𝑘𝑘𝑘
, 𝑀𝑀
= 0.01802 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚3 𝐻𝐻2 𝑂𝑂
yA - xA
0.25
0.25
0.20
0.20
0.15
0.15
yA
pA [atm]
pA - cAL
𝜌𝜌𝐻𝐻2 𝑂𝑂(𝑙𝑙) = 998.2
0.10
0.10
0.05
0.05
0.00
0.00
0
0.01
0.02
0.03
0
0.0003
xA
cAL [gmol/L]
ClO2 − Water, T = 293 K
pA - cAL
yA - xA
0.200
0.200
0.150
0.150
yA
pA [atm]
0.100
0.050
0.100
0.050
0.000
0.000
0
0.1
0.2
0
0.001 0.002 0.003 0.004
xA
cAL [gmol/L]
NH3 − Water, T = 303 K, 𝜌𝜌𝐻𝐻2 𝑂𝑂(𝑙𝑙) = 995.2
pA - cAL
𝑘𝑘𝑘𝑘
𝑚𝑚3
, 𝑀𝑀𝐻𝐻2 𝑂𝑂 = 0.01802 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
yA - xA
1.0
1.0
0.8
0.8
0.6
0.6
yA
pA [atm]
0.0006
0.4
0.4
0.2
0.2
0.0
0.0
0
10
20
cAL [gmol/L]
0
0.2
0.4
xA
29-1
SO2 − Water, T = 303 K
yA - xA
1.0
1.0
0.8
0.8
0.6
0.6
yA
pA [atm]
pA - cAL
0.4
0.4
0.2
0.2
0.0
0.0
0
0.5
1
1.5
0
0.01
0.02
xA
cAL [gmol/L]
Ammonia, 𝑁𝑁𝑁𝑁3 , is the most soluble in water. It has a lower range of gas mole fractions
than the other species.
Chlorine, 𝐶𝐶𝐶𝐶2 , is the easiest to operate liquid stripping. As seen in the 𝑦𝑦𝐴𝐴 𝑣𝑣𝑣𝑣. 𝑥𝑥𝐴𝐴 graph,
the gas mole fraction is 20% when the liquid mole fraction is less than a tenth of a
percent, 𝑥𝑥𝐴𝐴 ~0.0006, so there is a huge driving force to the gas phase.
b. Henry’s law constants
Chlorine, 𝐶𝐶𝐶𝐶2
Chlorine dioxide, 𝐶𝐶𝐶𝐶𝐶𝐶2
Ammonia, 𝑁𝑁𝑁𝑁3
Sulfur dioxide, 𝑆𝑆𝑆𝑆2
H, based on 𝑝𝑝𝐴𝐴 = 𝐻𝐻 ∙ 𝐶𝐶 ∗𝐴𝐴𝐴𝐴
𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
�
�
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
m, based on 𝑦𝑦𝐴𝐴 = 𝑚𝑚 ∙ 𝑥𝑥𝐴𝐴∗
6.45
358
0.764
42.3
0.0378
2.09
0.779
43.0
29-2
29.2
a. Effect of temperature on the solubility of H2S gas in water
At 40 oC, to maintain a given mole fraction of H2S dissolved in the liquid, the
equilibrium mole fraction of H2S must be higher relative to 20 oC. Therefore, as
temperature increases, H2S is less soluble in water.
0.25
40 C
20 C
0.20
0.15
yA
0.10
0.05
0.00
0.0000
0.0001
0.0002
0.0003
mole fraction A in solution, xA
0.0004
b. yA-xA equilibrium line at 40 oC for H2S in water vs. H2S in 15.9 wt% MEA solution
Convert liquid phase units to mole fraction using the following conversion factor
15.9 𝑤𝑤𝑤𝑤% 𝑀𝑀𝑀𝑀𝑀𝑀 =
=
15.9 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀
100 𝑔𝑔 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀𝑀𝑀
15.9 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 ∙ 61.08 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀
1 𝑚𝑚𝑜𝑜𝑜𝑜 𝑀𝑀𝑀𝑀𝑀𝑀
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻 𝑂𝑂
�15.9 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 ∙ 61.08 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 + 84.1 𝑔𝑔 𝐻𝐻2 𝑂𝑂 ∙ 18.02 𝑔𝑔 𝐻𝐻2 𝑂𝑂 �
2
= 0.053
𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀𝑀𝑀
𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
29-3
PA
yA
mol H2S/
(mm Hg)
0.00
0.96
3.00
9.10
43.10
59.70
106.00
0.0000
0.0013
0.0039
0.0120
0.0567
0.0786
0.1395
mol MEA
0.000
0.125
0.208
0.362
0.643
0.729
0.814
mol H2S/
mol H2S +
MEA
0.0000
0.1111
0.1722
0.2658
0.3914
0.4216
0.4487
xA
0.0000
0.0063
0.0094
0.0183
0.0325
0.0369
0.0412
0.25
water
0.20
15.9 wt%
MEA
0.15
yA
0.10
0.05
0.00
0.000
0.010
0.020
0.030
0.040
mole fraction A in solution, xA
When compared to water, the solubility of H2S in 15.9 wt% MEA solution drastically
increases at a given mole fraction of H2S in the gas phase.
29-4
29.3
Given:
A = O2 (solute), B = H2O (solvent)
T = 293 K, P = 2.0 atm
Gas: y A = 0.21, yN2 = 0.78, yAr = 0.010
Solvent: ρB,liq = 1000 kg/m3, MB = 18 kg/kgmole
Solute: H = 40,000 atm with p A,i = Hx A,i
a. Determine xA*
x*A
=
pA yA P
=
=
H
H
( 0.21)( 2.0 atm ) = 1.05 × 10−5
( 40,100 atm )
b. Determine cAL*
kg 

1000 3 

ρ H O,liq
ρ B ,liq
kgmole
m 
c* AL =
x*AcL  x*A 2
x*A
=
=
1.05 × 10−5 
=5.83 × 10−4
M H2O
MB
m3

kg 
18 kgmol 


c. Determine cAL* if P = 4.0 atm
x=
A*
pA yA P
=
=
H
H
( 0.21)( 4.0atm ) = 2.10 × 10−5
( 40,100 atm )
kg 

1000 3 

ρ H O,liq
kgmole
m 
=
2.10 × 10−5 
= 1.17 × 10−3
c* AL =
x*AcL =
x*A 2
m3
M H2O

kg 
18.0

kgmole 

(
)
29-5
29.4
Given:
A = ClO2 (solute), B = H2O (solvent)
T = 293 K, P 1.5 atm
Operating point: yA = 0.040, xA = 0.00040
Solute: MA = 67.5 kg/gmole
Solvent: ρB,liq = 992.3 kg/m3, MB = 18 kg/kgmole
Film Mass Transfer Coefficients:
kgmole
kgmole
k x =1.0
(liquid film), kG = 0.010 (gas
film)
2
m ⋅s
m 2 ⋅ s ⋅ atm
a.
pA vs. CAL equilibrium line and operating point
Operating point:
c AL = x AcL  x A
ρ H O ,liq
2
M H2O
= xA
ρ B ,liq
MB
( 992.3 kg/m ) = 0.022 kgmole
= ( 0.00040 )
3
(18 kg/kgmole )
m3
=
p A y=
(0.06)(1.5 atm) = 0.090 atm
AP
0.12
cAL , pA
0.10
pA (atm)
0.08
0.06
0.04
cAL,i , pA,i
0.02
pA*
cAL*
0.00
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
cAL (kgmole/m 3)
Gas Absorption process since the operating point is above the equilibrium line
29-6
b. Determine m
Get H from slope of PA vs. CAL data, H = 0.772 atm∙m3/kgmole
(
)(
)
0.772 atm ⋅ m3 /kgmole 55.1 kgmole/m3
HcL
=
= 28.4
P
(1.5 atm )
c. Determine kL

kg 
18 kgmole 
kx
MB 
m
gmole 


= 1.81 × 10−5
=
k L =≅
kx
1.0

2
ρ B ,liq 
s
m s 
CL
kg   1000 gmole 
 992.3 3  

m   kgmole 

∗
d. Determine 𝑐𝑐𝐴𝐴𝐴𝐴
m=
*
c=
AL
pA
=
H
( 0.06 atm )
( 0.772 atm ⋅ m /kgmole )
3
= 0.078
kgmole
m3
e. Compositions at the gas-liquid interface, pA,i and cAL,i
Recall −
k L ( p A − p A ,i )
=
kG ( c AL − c AL ,i )
For linear equilibrium line, p A,i = Hc AL ,i


atm ⋅ m3
m

−
c AL ,i 
0.090
atm
0.772
− 1.81 × 10−5 

kgmole
( pA − HcAL,i ) ⇒ 
k
s 


=
− L
=
kgmole
kgmole
kG



( cAL − cAL,i ) 1.0 × 10−5 2
− c AL ,i 

 0.022
3
m ⋅ s ⋅ atm 
m



kgmole
m3

atm ⋅ m3  
kgmole 
0.772
=
p A,i Hc
=

  0.0503
AL ,i
 = 0.039 atm
kgmole  
m3 

c AL ,i = 0.0503
f. Determine Ky by two approaches
Path 1: Convert kG to ky, get Ky from kx, ky, m
kgmole 

−5
−5 kgmole
ky =
kG P =
1.0 × 10
 (1.5 atm ) =1.5 × 10
2
m ⋅ s ⋅ atm 
m 2s

29-7
-1


 1 m


1
28.4
kgmole
+
= 1.05 × 10−5
K y = +  =

kgmole
kgmole
m 2s
 k y k x 
1.5 × 10−5 2

1.0 × 10−3
2
m s ⋅ atm
ms 

−1
Path 2: KG from kL kG, H, then convert KG to Ky
-1

atm ⋅ m3 
−1
0.772

 1 H
kgmole
1
kgmole 
K G = +  =
+
= 7.10 × 10−6 2
m ⋅ s ⋅ atm
1.0 × 10−5 kgmole 1.81 × 10−5 m 
 kG k L 
2


ms
s 

kgmole 

−5 kgmole
K y K G P =  7.10 × 10−6 2
=
 (1.5 atm) = 1.05 × 10
m ⋅ s ⋅ atm 
m 2s

g.
Flux NA
Base on overall gas phase mole fraction driving force as an example

atm ⋅ m3  
kgmole 
0.772
=
p A * Hc
=

  0.022
AL
 = 0.017 atm
kgmole  
m3 

p A * 0.017 atm
=
yA* =
= 0.0113
P
1.5 atm
kgmole 

N A = K y ( y A − y* A ) = 1.05 × 10−5
 ( 0.040 − 0.0113)
m 2s 

kgmole
N A = 3.0 × 10−5
m 2s
29-8
29.5
P = 2.20 atm, T = 293 K, 𝜌𝜌𝐻𝐻2 𝑂𝑂 = 992.3
𝑥𝑥𝐴𝐴 = 0.0040, 𝑦𝑦𝐴𝐴 = 0.10
𝑘𝑘𝑥𝑥 = 0.125
𝑘𝑘𝑘𝑘
𝑚𝑚3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
,
𝑘𝑘
=
0.010
𝑦𝑦
𝑚𝑚2 ∙ 𝑠𝑠
𝑚𝑚2 ∙ 𝑠𝑠
Equilibrium: 𝑦𝑦𝐴𝐴,𝑖𝑖 = 𝑚𝑚𝑥𝑥𝐴𝐴,𝑖𝑖 , 𝑚𝑚 = 50.0
a. xA vs. yA equilibrium line and operating point
0.5
0.4
yA
0.3
0.2
(Operating Point)
0.1
0
0
0.002
0.004
0.006
0.008
0.01
xA
The operation is liquid stripping, because at the specified liquid mole fraction, the
gas mole fraction is below the equilibrium line.
b. Estimate H for linear equilibrium line
𝐻𝐻 =
𝑚𝑚𝑚𝑚
(50)(2.2 𝑎𝑎𝑎𝑎𝑎𝑎)
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3
=
=
1.995
55.1 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘/𝑚𝑚3
𝐶𝐶𝐿𝐿
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘
29-9
c. Determine 𝐾𝐾𝐿𝐿
𝑘𝑘𝑘𝑘
0.018
𝑘𝑘𝑥𝑥 𝑀𝑀𝐵𝐵
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚
𝑘𝑘𝐿𝐿 =
≅
𝑘𝑘𝑥𝑥 =
∙ 0.125 2 = 2.27 ∙ 10−6
𝑘𝑘𝑘𝑘
𝐶𝐶𝐿𝐿
𝜌𝜌𝐵𝐵
𝑚𝑚 ∙ 𝑠𝑠
𝑠𝑠
992.3 3
𝑚𝑚
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑘𝑘𝑦𝑦 0.010 𝑚𝑚2 ∙ 𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑘𝑘𝐺𝐺 =
=
= 4.55 ∙ 10−3 2
𝑃𝑃
2.2 𝑎𝑎𝑎𝑎𝑎𝑎
𝑚𝑚 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝐾𝐾𝐿𝐿 = �
1
𝑘𝑘𝐿𝐿
𝐾𝐾𝑥𝑥 = �
1
+
𝑘𝑘𝑥𝑥
1
𝐻𝐻𝑘𝑘𝐺𝐺
+
1
�
−1
𝑚𝑚𝑚𝑚𝑦𝑦
�
=�
−1
1
𝑚𝑚
2.27∙10−6
𝑠𝑠
=�
d. Determine 𝑁𝑁𝐴𝐴
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
0.125 2
𝑚𝑚 ∙𝑠𝑠
𝑁𝑁𝐴𝐴 = 𝐾𝐾𝑥𝑥 (𝑥𝑥𝐴𝐴 − 𝑥𝑥 ∗𝐴𝐴 ) = 0.1
e. Determine xA,i and yA,i
1
+
1
�
𝑎𝑎𝑎𝑎𝑎𝑎∙𝑚𝑚3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
2.0×10−3
∙ 4.55∙10−3 2
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚 ∙𝑠𝑠∙𝑎𝑎𝑎𝑎𝑎𝑎
+
1
−1
�
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
50.0 ∙0.010 2
𝑚𝑚 ∙𝑠𝑠
= 0.1
−1
= 1.82 × 10−6
𝑚𝑚
𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚2 ∙𝑠𝑠
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
(0.004
−
0.002)
=
0.0002
𝑚𝑚2 ∙ 𝑠𝑠
𝑚𝑚2 ∙ 𝑠𝑠
𝑘𝑘
Equate the equilibrium line 𝑦𝑦 𝐴𝐴 = 𝑚𝑚𝑥𝑥 𝐴𝐴 to a line with slope − 𝑥𝑥 through the operating
point to find the intersection (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ).
𝑘𝑘
− 𝑥𝑥 = −12.5, 𝑦𝑦 𝐴𝐴 = −12.5 𝑥𝑥𝐴𝐴 + 𝑏𝑏,
𝑘𝑘𝑦𝑦
𝑘𝑘
𝑘𝑘𝑦𝑦
0.1 = −12.5 ∗ 0.004 + 𝑏𝑏, 𝑏𝑏 = 0.15
𝑚𝑚𝑥𝑥𝐴𝐴,𝑖𝑖 = − 𝑥𝑥 𝑥𝑥𝐴𝐴,𝑖𝑖 + 0.15, 50𝑥𝑥𝐴𝐴,𝑖𝑖 = −12.5𝑥𝑥𝐴𝐴,𝑖𝑖 + 0.15, (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) = (0.0024 , 0.12)
𝑘𝑘𝑦𝑦
29-10
29.6
a. Determine 𝐾𝐾𝑦𝑦 and % resistance in the gas phase
𝑘𝑘𝑥𝑥 = 0.01
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
,
𝑚𝑚2 ∙ 𝑠𝑠
𝑘𝑘𝑦𝑦 = 0.02
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚2 ∙ 𝑠𝑠
𝑘𝑘
From the operating point, use the slope − 𝑥𝑥 = −0.5 towards the equilibrium line to find
𝑘𝑘𝑦𝑦
(𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) on the intersection. (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) = (0.027 , 0.016)
The operating point is below the equilibrium line (liquid stripping).
𝑦𝑦𝐴𝐴∗ − 𝑦𝑦𝐴𝐴,𝑖𝑖
0.0425 − 0.016
𝑚𝑚 =
=
= 2.04
𝑥𝑥𝐴𝐴 − 𝑥𝑥𝐴𝐴𝐴𝐴
0.04 − 0.027
′′
1 𝑚𝑚′′
𝐾𝐾𝑦𝑦 = � +
�
𝑘𝑘𝑥𝑥
𝑘𝑘𝑦𝑦
−1
1
2.04 −1
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
=�
+
� = 0.0039 2
0.02 0.01
𝑚𝑚 ∙ 𝑠𝑠
Finally, the gas phase percent resistance is
1
1
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
� �
0.02 2
𝑘𝑘𝑦𝑦
𝑔𝑔𝑔𝑔𝑔𝑔 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑚𝑚 ∙ 𝑠𝑠 ∗ 100% = 19.5%
=
=
1
1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
�𝐾𝐾 �
. 0039 𝑚𝑚2 ∙ 𝑠𝑠
𝑦𝑦
b. Determine ky for 10% gas film controlling resistance
1/𝑘𝑘𝑦𝑦
= 0.10 =
1/𝐾𝐾𝑦𝑦
1
�
𝑘𝑘𝑦𝑦
,
1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
�. 0039 2 �
𝑚𝑚 ∙ 𝑠𝑠
�
∴ 𝑘𝑘𝑦𝑦 = 0.039
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚2 ∙ 𝑠𝑠
c. Determine xAi and yAi
−
𝑘𝑘𝑥𝑥
𝑘𝑘𝑦𝑦,𝑛𝑛𝑛𝑛𝑛𝑛
= −0.256, (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) = (0.0253 , 0.0138)
The interface point moves down along the equilibrium line indicating that the liquid
resistance begins to control more of the system.
29-11
29.7
𝑙𝑙𝑏𝑏
𝑤𝑤𝐴𝐴 = 0.003, 𝜌𝜌𝐿𝐿 = 61.8 𝑚𝑚3 , 𝑇𝑇 = 303 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎
𝑓𝑓𝑓𝑓
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑘𝑘𝐿𝐿 = 2.5
, 𝑘𝑘𝐺𝐺 = 0.125 2
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ (
)
𝑓𝑓𝑓𝑓 3
a. pA vs. CAL equilibrium line and operating point
𝐶𝐶𝐴𝐴𝐴𝐴 ≅ 𝑥𝑥𝐴𝐴 𝐶𝐶𝐿𝐿 = (8.46 𝑥𝑥 10−4 ) �3.43
𝑦𝑦𝐴𝐴 =
𝑝𝑝𝐴𝐴
𝑃𝑃
, 𝑃𝑃𝐴𝐴 = 0.1 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 3
� = 2.9 𝑥𝑥 10−3
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 3
pA - cAL
0.14
0.12
pA [atm]
0.10
0.08
0.06
0.04
Equilibrium
0.02
Operating Point
0.00
0
0.005
0.01
0.015
cAL [lbmole/ft3]
∗
𝑐𝑐𝐴𝐴𝐴𝐴
= 0.00944
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓3
,
𝑝𝑝𝐴𝐴∗ = 0.0261 𝑎𝑎𝑎𝑎𝑎𝑎
b. Determine pA,i and cAL,i
𝑘𝑘
− 𝐿𝐿 is the slope of the line going from the operating point to �𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 , 𝑝𝑝𝐴𝐴,𝑖𝑖 �.
𝑘𝑘𝐺𝐺
Use this slope backwards from the operating point to find the y-intercept (b):
𝑝𝑝 𝐴𝐴 = −
𝑘𝑘𝐿𝐿
𝑐𝑐 + 𝑏𝑏,
𝑘𝑘𝐺𝐺 𝐴𝐴𝐴𝐴
𝑏𝑏 = 0.1 𝑎𝑎𝑎𝑎𝑎𝑎 +
2.5
0.00289 = 0.158 𝑎𝑎𝑎𝑎𝑎𝑎
0.125
Polynomial fit of equilibrium line, equate to operating point line to find 𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 :
3
4
2
−20𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 + 0.158 = 12422𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖
− 2033.5𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖
+ 113.97𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖
+ 10.121𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 − 0.0041
29-12
𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 = 0.00529
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
,
𝑓𝑓𝑓𝑓 3
𝑝𝑝𝐴𝐴,𝑖𝑖 = 0.0523 𝑎𝑎𝑎𝑎𝑎𝑎
c. Determine 𝐾𝐾𝐺𝐺 , 𝐾𝐾𝐿𝐿 , 𝐾𝐾𝑦𝑦 , 𝐾𝐾𝑥𝑥, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑁𝑁𝐴𝐴
Assume linear equilibrium, 𝐻𝐻 =
𝑝𝑝𝐴𝐴𝐴𝐴
𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴
= 9.887
𝑎𝑎𝑎𝑎𝑎𝑎∙𝑓𝑓𝑓𝑓 3
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑓𝑓𝑓𝑓 3
⎞
9.887
1
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
+
⎟
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
⎟
2.5
2
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ (
)
𝑓𝑓𝑓𝑓 3
⎝
⎠
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
= 0.0836 2
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
1
𝐻𝐻 −1 ⎛
𝐾𝐾𝐺𝐺 = � + � = ⎜
⎜
𝑘𝑘𝐺𝐺 𝑘𝑘𝐿𝐿
0.125
1
1 −1
𝐾𝐾𝐿𝐿 = � +
�
𝑘𝑘𝐿𝐿 𝐻𝐻𝑘𝑘𝐺𝐺
⎛
=⎜
⎜
−1
⎞
1
1
+
⎟
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑓𝑓𝑓𝑓 3
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ⎟
2.5
9.887 �
� ∙ 0.125 2
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ (
)
3
𝑓𝑓𝑓𝑓
⎝
⎠
= 0.827 𝑓𝑓𝑓𝑓/ℎ𝑟𝑟
𝐾𝐾𝑦𝑦 = 𝐾𝐾𝐺𝐺 𝑃𝑃 = 0.0836
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓2 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
∙ 1 𝑎𝑎𝑎𝑎𝑎𝑎 = 0.0836
−1
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑡𝑡2 ∙ ℎ𝑟𝑟
𝑙𝑙𝑙𝑙
0.827 𝑓𝑓𝑓𝑓/ℎ𝑟𝑟 ∙ 61.8 � 𝑚𝑚3 �
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓
𝐾𝐾𝑥𝑥 = 𝐾𝐾𝐿𝐿 𝐶𝐶𝐿𝐿 =
= 0.798 2
𝑙𝑙𝑙𝑙𝑚𝑚
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
64.07 �
�
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑁𝑁𝐴𝐴 = 𝐾𝐾𝐺𝐺 (𝑝𝑝𝐴𝐴 − 𝑝𝑝𝐴𝐴∗ ) = 0.0836
= 0.00618
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
∙ (0.1 − 0.0261) 𝑎𝑎𝑎𝑎𝑎𝑎
29-13
d. Gas phase resistance
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
1
𝐾𝐾𝐺𝐺 0.0836 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
𝑘𝑘𝐺𝐺
=
=
∙ 100%
1
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑘𝑘𝐺𝐺
0.125 2
𝐾𝐾𝐺𝐺
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎
= 66.9% 𝑔𝑔𝑔𝑔𝑔𝑔 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
29-14
29.8
𝑦𝑦𝐴𝐴 = 0.030, 𝑥𝑥𝐴𝐴 = 0.010
𝑘𝑘𝑦𝑦 = 1.0
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
a. xA vs. yA equilibrium line and operating point
0.05
Equilibrium
0.04
Operating Point
yA
0.03
0.02
0.01
yA *
xA*
0
0
From graph 𝑥𝑥 ∗𝐴𝐴 ≈
0.01
0.036,
b. Determine xAi, yA,i
0.02
∗
0.03
0.04
xA
𝑦𝑦 𝐴𝐴 ≈ 0.003
1
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎
𝐾𝐾𝑥𝑥 �𝑥𝑥𝐴𝐴,𝑖𝑖 − 𝑥𝑥𝐴𝐴 �
𝑘𝑘𝑥𝑥
= 0.80 =
=
=
1
(𝑥𝑥𝐴𝐴∗ − 𝑥𝑥𝐴𝐴 )
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑘𝑘𝑥𝑥
𝐾𝐾𝑥𝑥
0.80 =
(𝑥𝑥𝐴𝐴𝐴𝐴 − 0.010)
�𝑥𝑥𝐴𝐴,𝑖𝑖 − 𝑥𝑥𝐴𝐴 �
=
(𝑥𝑥𝐴𝐴∗ − 𝑥𝑥𝐴𝐴 )
(0.036 − .0.010)
xAi = 0.031, from graph, yAi = 0.021
c. Determine Ky and NA
−
�𝑦𝑦𝐴𝐴 − 𝑦𝑦𝐴𝐴,𝑖𝑖 � (0.030 − 0.021)
𝑘𝑘𝑥𝑥
=
=
= −0.43
𝑘𝑘𝑦𝑦 �𝑥𝑥𝐴𝐴 − 𝑥𝑥𝐴𝐴,𝑖𝑖 � (0.010 − 0.031)
29-15
𝑘𝑘𝑥𝑥 = 0.43𝑘𝑘𝑦𝑦 = 0.43
𝑚𝑚′ =
(𝑦𝑦𝐴𝐴 − 𝑦𝑦𝐴𝐴∗ )
�𝑥𝑥𝐴𝐴,𝑖𝑖 − 𝑥𝑥𝐴𝐴, �
1 𝑚𝑚′
𝐾𝐾𝑦𝑦 = � +
�
𝑘𝑘𝑦𝑦 𝑘𝑘𝑥𝑥
=
−1
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
0.021 − 0.0030
= 0.86
0.031 − 0.010
1 0.86 −1 𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
=� +
= 0.33 2
�
2
1 0.43
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
𝑁𝑁𝐴𝐴 = 𝐾𝐾𝑦𝑦 �𝑦𝑦𝐴𝐴 − 𝑦𝑦 ∗ 𝐴𝐴 � = 0.33
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
∙ (0.03 − 0.003) = 9.0 ∙ 10−3 2
2
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
29-16
29.9
𝑇𝑇 = 293 𝐾𝐾,
𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎,
a. Diagram of tower
𝐶𝐶𝐶𝐶2 = 𝐴𝐴,
𝑘𝑘𝑦𝑦 = 5
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
,
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
𝑘𝑘𝑥𝑥 = 5
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
b. Plot of equilibrium line and terminal stream operating points
Gas Absorption Tower
0.25
Equilibrium
0.20
2 - Top
yA
0.15
1 - Bottom
0.10
0.05
0.00
0.0000
0.0002
0.0004
0.0006
xA
c. Determine overall mass transfer coefficients 𝐾𝐾𝑦𝑦 for top and bottom of the column
3
Polynomial fit of equilibrium line, 𝑦𝑦𝐴𝐴 = −6 ∙ 108 𝑥𝑥𝐴𝐴 + 106 𝑥𝑥𝐴𝐴 2 − 34.998𝑥𝑥𝐴𝐴 − 0.0013
29-17
𝑘𝑘
Using a slope of − 𝑥𝑥 = −0.25 , move from the operating points at 1 and 2 to find
𝑘𝑘𝑦𝑦
�𝑥𝑥𝐴𝐴1,𝑖𝑖 , 𝑦𝑦𝐴𝐴1,𝑖𝑖 � 𝑎𝑎𝑎𝑎𝑎𝑎 �𝑥𝑥𝐴𝐴2,𝑖𝑖 , 𝑦𝑦𝐴𝐴2,𝑖𝑖 � located on the equilibrium line.
Find (𝑥𝑥𝐴𝐴2𝑖𝑖 , 𝑦𝑦𝐴𝐴2𝑖𝑖 ) for Top of Tower:
3
−0.25𝑥𝑥𝐴𝐴2,𝑖𝑖 + 0.05 = −6 ∙ 108 𝑥𝑥𝐴𝐴2,𝑖𝑖 + 106 𝑥𝑥𝐴𝐴2,𝑖𝑖 2 − 34.998𝑥𝑥𝐴𝐴2,𝑖𝑖 − 0.0013,
∴ 𝑥𝑥𝐴𝐴2,𝑖𝑖 = 0.000269, 𝑦𝑦𝐴𝐴2,𝑖𝑖 = 0.0499
Find (𝑥𝑥𝐴𝐴1𝑖𝑖 , 𝑦𝑦𝐴𝐴1𝑖𝑖 ) for Bottom of Tower:
3
−0.25𝑥𝑥𝐴𝐴1,𝑖𝑖 + 0.20 = −6 ∙ 108 𝑥𝑥𝐴𝐴1,𝑖𝑖 + 106 𝑥𝑥𝐴𝐴1,𝑖𝑖 2 − 34.998𝑥𝑥𝐴𝐴1,𝑖𝑖 − 0.0013,
∴ 𝑥𝑥𝐴𝐴1,𝑖𝑖 = 0.000584, 𝑦𝑦𝐴𝐴1,𝑖𝑖 = 0.1998
Find m/ for Top and Bottom using 𝑚𝑚′ =
∗
(𝑥𝑥𝐴𝐴2 = 0) = −0.0013
𝑦𝑦𝐴𝐴2
∗ (𝑥𝑥
𝑦𝑦𝐴𝐴1 𝐴𝐴1 = 0.0005) = 0.156
𝑦𝑦𝐴𝐴,𝑖𝑖 −𝑦𝑦𝐴𝐴 ∗
𝑥𝑥𝐴𝐴,𝑖𝑖 −𝑥𝑥𝐴𝐴
𝑦𝑦𝐴𝐴2,𝑖𝑖 − 𝑦𝑦𝐴𝐴2 ∗ 0.0499 + 0.0013
𝑚𝑚 𝑡𝑡𝑡𝑡𝑡𝑡 =
=
= 190.3
𝑥𝑥𝐴𝐴2,𝑖𝑖 − 𝑥𝑥𝐴𝐴2
0.000269 − 0
′
𝑚𝑚′ 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 =
𝑦𝑦𝐴𝐴1,𝑖𝑖 − 𝑦𝑦𝐴𝐴1 ∗
0.1998 − 0.156
=
= 521.4
𝑥𝑥𝐴𝐴1,𝑖𝑖 − 𝑥𝑥𝐴𝐴1
0.000584 − 0.0005
Find 𝐾𝐾𝑦𝑦 for Top and Bottom
𝐾𝐾𝑦𝑦,𝑡𝑡𝑡𝑡𝑡𝑡 = �
1 𝑚𝑚′ 𝑡𝑡𝑡𝑡𝑡𝑡
+
�
𝑘𝑘𝑦𝑦
𝑘𝑘𝑥𝑥
𝐾𝐾𝑦𝑦,𝑏𝑏𝑏𝑏𝑏𝑏 = �
′
−1
= �
1 𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
+
�
𝑘𝑘𝑦𝑦
𝑘𝑘𝑥𝑥
−1
1
190.3
+
�
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
5 2
20 2
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
= �
−1
= 0.103
1
521.4
+
�
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
5 2
20 2
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟
𝐾𝐾𝑦𝑦 varies due to nonlinearity of the equilibrium line
−1
𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
= 0.0381
𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚
𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟
29-18
29.10
a. xA* and yA*
From equilibrium distribution plot at P = 2.0 atm
At xA = 0.020, yA* = 0.008; at yA = 0.060, xA* = 0.040
0.07
xA, yA
0.06
0.05
0.04
yA
0.03
xAi, yAi
0.02
yA*
0.01
xA*
0.00
0.000
0.010
0.020
0.030
mole fraction H2S in solution, xA
0.040
b. xA,i and yA,i
(
)
N A= K y y A − y*A = k y ( y A − y A,i )
Ky
ky
=
(y − y )
A
A,i
A
*
A
(y − y )
gmole
m 2s = ( 0.06 − y A,i )
gmole ( 0.06 − 0.008 )
0.040 2
ms
0.030
yAi = 0.021, xAi = 0.030 (from equilibrium distribution plot)
29-19
c. KG
K=
G
K y 0.030 gmole/m 2s
gmole
=
= 0.015 2
P
2.0 atm
m s ⋅ atm
d. NA
(
)
N=
K y y A − y*A= 0.030
A
gmole
gmole
( 0.060 − 0.008=) 1.56×10-3 2
2
ms
ms
29-20
29.11
a. cAL* and pA*
Plot not provided
H = 4.0 m3atm/kgmole = 0.0040 m3atm/kgmole
*
c=
AL
pA
=
H
( 0.0 atm )
( 0.0040 m atm/gmole )
3
= 0.0
gmole
m3

m3atm  
gmole 
p*A =H ⋅ c AL = 0.0040
  3.0
 = 0.012 atm
gmole  
m3 

b. cAL* < cAL, therefore stripping mass transfer process with operating point (pA, cAL)
below the equilibrium line
c. KG
1
1 H
=
+
K G kG k L
cL 
ρ B,liq
MB
(1000 kg/m ) = 55.6 kgmole
=
3
(18 kg/kgmole )
m3
gmole 

2.0


kx
m
m 2s 

=
kL =
= 3.6 ×10-5
CL 
s
kgmole   1000 gmole 
 55.6


3
m   kgmole 

-1

m3atm 
−1
0.0040

 1 H
1
kgmole
gmole 
K G = +  =
+
= 2.25 × 10−3 2
m ⋅ s ⋅ atm
 3.0 × 10−3 kgmole 1.81 × 10−5 m 
 kG k L 
2


m s ⋅ atm
s 

d. NA
29-21
gmole 

N A = K G p A − p* A =  2.25 × 10−3 2
 ( 0.012 atm - 0.0 atm )
m s ⋅ atm 

gmole
N A = 2.7 × 10−5
m 2s
(
)
e. pAi
m
3.6 ×10-5
kL
m3atm
s
= =
0.012
kG 3.0 × 10−3 kgmole
gmole
m 2s ⋅ atm
− p A ,i )
( pA =
k
=
− L
and p A,i Hc AL ,i
kG ( c AL − c AL ,i )
kL
c AL
kG
=
k
1+ L
kG H
pA +
=
∴ p A ,i

m3atm   3.0 gmole 


gmole  
m3


= 0.0090 atm
3

m atm 
 0.012

gmole 

1+

m3atm 
0.0040


gmole 

( 0.0 atm ) +  0.012
29-22
29.12
A = NH3
p A = 0.20 atm, x A = 0.04 atm, P = 2.0 atm, T = 303 K, CL = 55.6
kG = 1.0
kgmole
m3
kgmole
m
, k L = 0.045
2
s
m ⋅ s ⋅ atm
a. pA vs. xA equilibrium line and operating point
0.5
0.4
pA
0.3
(Operating
Point)
0.2
0.1
0
0.00
0.05
0.10
0.15
0.20
0.25
xA
b. Determine k x
(55.6
=
k x C=
L kL
kgmole
m
kgmole
)(0.045=
) 2.50 2
3
m
s
m ⋅s
c. Determine x A,i , p A,i
−
kx
=
−2.50 atm
kG
Find intercept through the equilibrium plot to find interface compositions.
x A,i = 0.0761 , p A,i = 0.110 atm
29-23
Determine ( x*A , p*A )
Use plot to find x*A = 0.1216 , p*A = 0.053 atm
d. Determine K G
H = mP
H'
=
p A,i − p*A 0.110 atm − 0.053 atm
=
= 1.58 atm
0.0761 − 0.0400
x A ,i − x A
1
1 H'
=
+
K G kG k x
−1



1
1.58 atm 
kgmole
+
= 0.613 2
KG = 

kgmole
kgmole
m ⋅ s ⋅ atm
1.0 2

2.50
2
m ⋅ s ⋅ atm
m ⋅s 

e. Determine N A
N=
K G ( p A − p*A=
) 0.613
A
kgmole
kgmole
= 0.0901
(0.20 atm − 0.053 atm)
2
m ⋅ s ⋅ atm
m2 ⋅ s
29-24
29.13
waste water
A = H2S
vo = 20 m3/h
cAL,o = 2.5 gmole/m3
Open Tank
yA ≈ 0.005, P = 1.0 atm
D = 5.0 m
cAL
well-mixed
effluent
waste water
cAL = ?
Given;
A = H2S (solute), B = H2O (solvent)
T = 293 K, P = 1.0 atm
Solute: H = 9.34
m3 ⋅ atm
kgmole
Solvent: ρB,liq = 1000 kg/m3, MB = 18 kg/kgmole
m3
gmole
Stirred Tank: c AL ,o = 2.50
, vo = 20
, D = 5.0 m (diameter), depth = 1.0 m
hr
m3
Operating point: yA = 0.0050; xA and cAL unknown
Film Mass Transfer Coefficients:
k L = 2.0 × 10−4
m
kgmole
(liquid film), kG = 5.0 × 10−4 2
s
m ⋅ s ⋅ atm
a. Estimate m
kg 

1000 3 

ρ
kgmole
m 
cL = B ,liq = 
= 55.56
MB 
m3
kg 
18

kgmole 

kgmole  
m3 ⋅ atm 

55.56
9.34



m3  
kgmole 
C H 
m= L =
= 518.9
P
(1.0 atm )
Operating point in mole fraction coordinates (yA, xA)
29-25
0.035
0.030
0.025
yA 0.020
yA*
0.015
0.010
xA, yA
0.005
xA*
0.000
0.0E+00
2.0E-05
4.0E-05
Mole fraction of H2S in water, xA
6.0E-05
Need cAL from part (c) to estimate xA
 gmole 
1.7

c AL
m3 

=
= 3.06 × 10−5
xA =
cL 
kgmole   1000 gmole 
 55.5

m3   kgmole 

b. Determine K y and KL
kgmole 

−4 kgmole
ky =
kG P =  5.0 × 10−4 2
 (1.0 atm) = 5.0 × 10
m ⋅ s ⋅ atm 
m2 ⋅ s

kgmole  
kgmole

−4 m 
=
k x C=
L kL
 55.56
  2.0 × 10
 = 0.011
3
m
s 
m 2 ⟨s


−1
 1 m
K y = + 
 k y k x 
−1




1
518.9
kgmole
=
+
= 2.03 × 10−5

kgmole
kgmole
m2 ⋅ s
 5.0 × 10−4

0.011
m2 ⋅ s
m2 ⋅ s 

-1


−1


1
1 
1
1

 =1.92 × 10−4 m
+
KL =
 +
 =
3

s

−4 m
m ⋅ atm  
−4 kgmole 
 k L HkG 
 2.0 × 10 s  9.34

  5.0 × 10
2
kgmole  
m ⋅ s ⋅ atm  


29-26
c. Determine cAL
Physical System: liquid in tank with liquid inflow and outflow
Source for A: liquid containing H2S dissolved in liquid
Sink for A: gas space over liquid
Assumptions
1. Constant source and sink for A—steady-state process
2. Well-mixed liquid phase, inflow or outflow of liquid
3. Constant liquid volume with constant inflow and outflow of liquid
4. No homogeneous reaction of A in liquid
5. Dilute UMD process with respect to dissolved A
Mass balance on dissolved H2S, liquid phase
IN – OUT + GENERATION = ACCUMULATION (moles A/time)
S = surface area between the two phases
c AL ,o v o − c AL v o − N A S =
0
c AL ,o vo − c AL vo − SK L (c AL −c AL *) =
0
c AL =
S=
*
c=
AL
c AL ,o v o + SK L c AL *
v o + SK L
π D 2 π (5.0 m) 2
=
=19.6 m 2
4
4
pA yA P
=
=
H
H
(0.005)(1.0 atm)
gmole
= 0.53
3
−3
m3
( 9.34 × 10 atm ⋅ m /gmole )
kgmole   m3   h 
gmole 

−4 m  
2 
 0.00250
  20
 +(19.6 m ) 1.92 × 10
  0.53


3
m 
h   3600 s 
s 
m3 


c AL =

m3   h 
−4 m 
2 
20
 + (19.6 m ) 1.92 × 10



h   3600 s 
s 


gmole
c AL = 1.7
m3
29-27
29.14
gmole
gmole
m3
c AL ,o = 0.5
, vo = 2.0
, c AL = 0.35
, L = 10.0 m, p A ≈ 0 , c*AL ≈ 0
3
3
m
m
s
m3 ⋅ atm
m3 ⋅ atm
=
= 500
T=
293 K, H 0.50
, P = 1.0 atm
gmole
kgmole
kc 2.67 ×10−4
=
m
m
5.5 ×10−3
, k=
L
s
s
a. Determine K L
m
kc
gmole
s
kG =
=
= 0.0111 2
3
m ⋅ atm
RT
m ⋅ s ⋅ atm
(8.206 x 10−5
)(293 K)
gmole ⋅ K
2.67 x 10−4
1
1
1
=
+
K L k L H kG
−1




1
1
m

 =
KL =
0.00276
+
3
m ⋅ atm
gmole 
s
 5.5 x 10−3 m
(0.50
)(0.0111 2
)

s
gmole
m ⋅ s ⋅ atm 

b. Develop material balance model and determine surface area S
Assume: 1) steady state, 2) no reaction, 3) constant T and P, 4) liquid stripping process.
A = Species A, B = Air, C = Water
IN – OUT – FLUX OUT = 0
c AL ,o vo − c AL vo − N A S =
0
c AL ,o vo − c AL vo − SK L (c AL − c*AL ) =
0
Determine S , the surface area between the two phases
29-28
m3
gmole
gmole
2.0 (0.5
)
− 0.35
3
vo (c AL ,o − c AL )
s
m
m3
311 m 2
S =
=
=
*
m
gmole
gmole
K L (c AL − c AL ) 0.00276 (0.35
)
−0
s
m3
m3
c. Determine v∞
Laminar flow over a flat plate:
ν=
B
µB
=
ρB
kg
2
m ⋅=
s 1.503 x 10−5 m
kg
s
1.206 3
m
1.813 x 10−5
νB 
13
kc L  DAB 

v∞ =


L  0.664 DAB  ν B 




2
2
13
2

 
m2 
cm 2
−4 m
−4 m
1.503 x 10
(2.67 x 10
)(10 m)
)(1 x 10
) 

 (0.08
2 
m
s
s
s
cm

v∞ =

  0.249
2
2
2
cm
cm
s
10 m

−4 m

 
1.503 x 10−5
 
 0.664(0.08 s )(1 x 10 cm 2 ) 
s
 

d. Determine p A,i
−5
p A − p A ,i
kL
m3 ⋅ atm
− =
=
−0.495
kG c AL − c AL ,i
gmole
0 atm − p A,i
m3 ⋅ atm
=
gmole 0.35 gmole − c
AL ,i
m3
m3 ⋅ atm
0.50
p A,i Hc
c AL ,i
=
=
AL ,i
gmole
−0.495
atm, c AL ,i 0.174
=
=
p A,i 0.087
Two equations, two unknowns:
gmole
m3
29-29
29.15
A = ozone, B = water, P = 1.5 atm, S = 4.0 m2, T = 293 K, ρ = 992.3
kg
m3
gmole
mg
gmole
m3
1.0
c
=
c AL 3.0
=
=
238
,
,
=
v
0.050
AL , o
o
m3
m3
L
hr
m3 ⋅ atm
m
−3 m
k=
5.0 ×10
3.0 ×10
, k=
, H = 68.2
, p A=
H ⋅ c AL ,i
L
c
,i
kgmole
s
s
−6
Assume: 1) dilute solution, 2) constant P and T, 3) linear equilibrium.
a. Determine m
kg
m3 ⋅ atm
(68.2
)
CL H
ρH
m3
kgmole
m
=
=
=
= 2.51×106
P
M B P (0.018 kg ) ⋅1.5 atm
kgmole
992.3
m is large, so ozone is sparingly soluble in water.
b. Determine K G
1
1 H RT H
=
+ =
+
K G kG k L k c k L
−1

m3 ⋅ atm
m3 ⋅ atm 
−8
(8.206
x
10
)(293
K)
68.2

kgmole
K ⋅ kgmole
kgmole 
KG 
4.40 x 10−8 2
=
+=
m
m
m ⋅ s ⋅ atm

5.0 x 10−3
3.0 x 10−6


s
s


c. Determine K L
1
1
1
1 RT
= +
= +
K L k L HkG k L Hkc
−1


m3 ⋅ atm
−8
(8.206
x
10
)(293 K) 

1
m
K ⋅ kgmole

 =
3.00 x 10−6
KL =
+
3
m ⋅ atm
m 
s
 3.0 x 10−6 m
(68.2
)(5.0 x 10−3 ) 

s
kgmole
s 

29-30
d. Develop material balance model
Assume: 1) steady state, 2) no reaction, 3) gas absorption process.
IN – OUT + FLUX IN = 0
c AL ,o vo − c AL vo + N A S =
0
vo (c AL ,o − c AL ) + SK G ( p A − p*A=
) vo (c AL ,o − c AL ) + SK G ( p A − Hc AL=
) 0
e. Determine WA
Determine p A and y A
=
pA
vo (c AL ,o − c AL )
SK G
+ Hc AL
m3
gmole
gmole
− 1.0
(0.050
)(3.0
)
3
m3 ⋅ atm
kgmole
hr
m
m3
=
+ (68.2
pA
)(0.001=
) 0.226 atm
3
gmole
s
kgmole
m
−5
2
(4.0 m )(4.40 x 10
)(3600 )
m 2 ⋅ s ⋅ atm
hr
p A 0.226 atm
=
= 0.151
yA =
P
1.5 atm
=
WA SN
=
SK G ( p A − Hc AL )
A
WA = (4.0 m 2 )(4.40 x 10−5
gmole 
m3 ⋅ atm
gmole 
)
0.226
atm-(0.0682
)(1.0
)

2
m ⋅ s ⋅ atm 
gmole
m3 
gmole
s
f. Resistance in liquid phase
= 2.78 x 10−5
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾𝐿𝐿 3.0 𝑥𝑥 10−6 𝑚𝑚/𝑠𝑠
=
=
= 1.0
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝐿𝐿 3.0 𝑥𝑥 10−6 𝑚𝑚/𝑠𝑠
Therefore the process is liquid phase controlling
29-31
29.16
H = 170 cm3 aqueous phase/cm3 organic phase
/
H ⋅ cA
At equilibrium, c=
A
m
m
/
3.5 ×10−6 , k=
2.5 ×10−5
k=
L
L
s
s
a. Determine K L/
=
Aqueous Film:
N A k L ( c AL − c AL ,i )
/
/
=
Organic Film: N=
k L / ( c AL
) kL / ( HcAL,i − cAL/ )
A
,i − c AL
(
)
/
/
/
Organic Overall:=
N A K L / ( c AL
K L / ( Hc AL − c AL
)
) − c=
AL
*
Combine:
1
1 H
=
+
/
K L k L/ k L
−1




1
170
m
K L/ =
+
1.41 x 10−7

 =
m
m
s
 3.5 x 10−6
(2.5 x 10−5 ) 
s
s 

b. Determine K L
1
1
1
=
+
K L k L H ⋅ k L/
−1




1
1
m
+
=
KL =
2.40 x 10−5


m
m
s
 2.5 x 10−5
170 ⋅ 3.5 x 10−6 
s
s 

c. Determine percent resistance to mass transfer encountered in the aqueous liquid film
1k
K
2.40 x 10−5 m/s
⋅100% =
Percent Resistance = L = L =
96.0%
1 K L kL
2.5 x 10−5 m/s
29-32
29.17
gmole
, D = 20 m, P = 1 atm, T = 293 K
m3
gmole
gmole
gmole
, k x = 200 2
, k y = 0.1 2 , p A=
y A = 0.04 , c AL = 10.0
H ⋅ x A ,i ,
,i
3
m
m ⋅s
m ⋅s
c AL ,o = 50
H = 550 atm
a. Determine K L
A = TCE, B = water
C C P
ρ
ρB P
1
1
1
= +
=L + L = B +
K L k L HkG k x Hk y M B k x M B Hk y
−1


g
g
(998, 200 3 )
(998, 200 3 )(1 atm)


m
m
m
KL =
7.78 x 10−4
+

 =
g
gmole
g
gmole
s
 (18
)(200 2 ) (18
)(550 atm)(0.1 2 ) 
 gmole

m ⋅s
gmole
m ⋅s 
b. Determine N A

p c 
yA PρB 

N=
K L (c AL − c*AL=
) K L  c AL − A L=

A
 K L  c AL −
H 
MBH 



g 
(0.04)(1 atm)(998, 200 3 ) 

gmole
m  = 4.64 x 10−3 gmole
−
N A = 7.78 x 10−4 m/s 10.0
3
g
m
m2 ⋅ s


(18
)(550
atm)


gmole


c. Develop material balance model and determine vo
Assume: 1) steady state, 2) well-mixed, 3) no reaction, 4) liquid stripping.
IN – OUT – FLUX OUT = 0
c AL ,o vo − c AL vo − N A ⋅ S =
0
(c AL ,o − c AL )vo − N A
vo
=
π D2 N A
=
4(c AL ,o − c AL )
π D2
4
=
0
gmole
)
m2 ⋅ s
0.0364 m3 /s
=
gmole
gmole
4(50
)
− 10
m3
m3
π (20 m) 2 (4.64 x 10−3
29-33
29.18
D = 4 m, p A = 0.020 atm, P = 1.0 atm, T = 293 K
m3 ⋅ atm
L
m3
0.020
H
=
, c AL ,o = 0 , p A=
,
=
vo 200
= 0.20
H
⋅
c
,i
AL ,i
kgmole
hr
hr
kgmole
m
kG = 1.25 2
, k L = 0.05
m ⋅ hr ⋅ atm
hr
a. Develop material balance and determine cAL
c AL ,o vo − c AL vo + N A S =
0
c AL ,o vo − c AL vo + K L (c*AL − c AL )π
c AL ,o vo − c AL vo +
D2
=
0
4
π D 2 K L  pA

− c AL  =
0

4 H

1
1
1
=
+
K L k L HkG
−1


 1

1
m

 =
+
KL =
0.0167
3
m
m ⋅ atm
kgmole 
hr
 0.05
(0.020
)(1.25 2
)

hr
kgmole
m ⋅ hr ⋅ atm 

Aside:
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾𝐿𝐿 0.0167 𝑚𝑚/ℎ𝑟𝑟
=
=
= 0.334
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝐿𝐿
0.050 𝑚𝑚/ℎ𝑟𝑟
Therefore 33% resistance in liquid, 67% resistance in gas
Determine c AL
m
)(0.020 atm)
hr
0+
m3 ⋅ atm
π D 2 K L pA
4(0.020
)
vo c AL ,o +
kgmole
kgmole
H
4
c AL =
=
= 0.512
2
m
π D KL
m3
(4 m) 2 (0.0167 )
π
3
vo +
m
hr
4
(0.20
)+
hr
4
π (4 m) 2 (0.0167
29-34
b. Determine p A,i and c AL ,i
p A − p A ,i
k
=
− L
=
kG c AL − c AL ,i
p A − p A ,i
P
c AL − A,i
H
3
0.020 atm − p A,i
k
m ⋅ atm
− L =
−0.04
=
p A ,i
kgmole
kG
kgmole
0.512
−
3
m3 ⋅ atm
m
0.020
kgmole
p A,i = 0.0135 atm
c AL ,i
=
p AL ,i
=
H
0.0135 atm
kgmole
= 0.675
3
m ⋅ atm
m3
0.020
kgmole
c. Determine WA
WA = SN A =
WA = 0.102
π D K L  PA
2
4

 − c AL  =
H

π (4 m) 2 (0.0167
4

m 
)

hr  (0.020 atm) − 0.512 kgmole 
m3 ⋅ atm
m3 

 (0.020 kgmole )



kgmole
hr
d. Would the flux NA increase by the following:
Increasing volume level of tank with fixed surface area: does not affect flux.
Increasing the agitation intensity of the bulk liquid: will increase the flux since kL will
increase.
Increasing the agitation intensity of the bulk gas: will increase the flux since kG will
increase and the overall mass transfer coefficient is determined by contributing
resistances in both the gas and liquid films.
Increasing the system temperature: will likely increase the flux since the Henry’s
constant will increase with increasing temperature, which will lower the solubility of the
solute in the gas and lower the concentration driving force for mass transfer.
29-35
29.19
bulk gas
y A∞ = y A ≈ 0
gas film
kc = 5.0 x 10-3 m/s
liquid film
kL = 4.0 x 10-5 m/s
bulk liquid
CAL = 0.15 gmole A/m3
bulk air flow
v∞ = ? m/s
Waste water IN
CAL,o = 0.25 gmole A/m3
Vo = ? m3/h
1.0 m
wwopen pond
well-mixed
Waste water OUT
CAL = 0.15 gmole A/m3
Vo = ? m3/h
x=0
10 m
a. H, cAL*, pA* , cAL,i, pA,i
Plot not provided
H
3

m3 Pa   1.0 atm 
-4 m atm
= 3.6 ×10
 36.6

gmole   101,250 Pa 
gmole

Alternatively, calculate analytically using value for H
At pA = 0.0 atm, cAL = 0.15 gmole/m3
( 0.0 atm )
pA
gmole
=
= 0.0
3
H 
m3
-4 m atm 
 3.6 ×10

gmole 

3

gmole 
*
-4 m atm  
-5
p A =H ⋅ c AL = 3.6 ×10
  0.15
 = 5.4 ×10 atm
3
gmole  
m 

*
c=
AL
(
)
5.0×10-3 m/s
kc
gmole
=
kG =
= 0.208 2
3
RT 
m s ⋅ atm
-5 m atm 
 8.206×10
 ( 293K )
gmole K 

29-36
m
3
kL
s= 1.92 ×10-4 m atm
=
gmole
kG 0.208 kgmole
2
m s ⋅ atm
− p A ,i )
( pA =
k
and p A,i Hc AL ,i
=
− L
kG ( c AL − c AL ,i )
4.0 ×10-5
kL
c AL
kG
=
k
1+ L
kG H
pA +
=
∴ p A ,i

m3atm   0.15 gmole 


gmole  
m3


= 1.9 ×10-5 atm
3

-4 m atm 
1.92 ×10

gmole 
1+ 
3

-4 m atm 
 3.6 ×10

gmole 

( 0.0 atm ) + 1.92 ×10-4
b. KL


−1


1
1 
1
1


=
+
K
 +
 =
L
3
m

k
Hk


-5
m
atm
gmole


-4
G 
 L
 4.0 ×10 s  3.6 ×10

  0.208 2
gmole  
m s ⋅ atm  


m
K L = 2.61×10−5
s
-1
c. v∞, Sc, Re for air flow over pond
cm 2
vB
s = 2.32
=
Sc =
2
cm
DAB
0.065
s
m

5.0×10-3  (10 m )

kc L
s 

=
=
Sh =
7692
2
DAB 
cm 2   1m 
 0.065


s   100 cm 

Assume that flow is turbulent, check ReL
0.151
ShL
=
kc L
1/3
= 0.0365 Re0.8
L Sc
DAB
5/4


ShL
Re L =
=
1/3 
 0.0365Sc 
 ( 7692 ) 


 0.0365 ( 2.32 ) 
5/4
=3.16×106 (> 3.0×106 , ∴ turbulent)
29-37
Re L =
v∞ L
νB

cm 2   1m 
3.16×10 )  0.151
(


s   100 cm 
Re L ν B
m

=
v∞ =
= 4.8
s
L
(10 m )
2
6
d. vo based on mass balance
S = surface area between the two phases
0
c AL ,o v o − c AL v o − N A S =
0
c AL ,o v o − c AL v o −  K L ( c AL − c*AL )  S =
vo =
vo
K L ( c AL − c*AL ) S
c AL ,o − c AL
gmole 

-5 m  
 2.61×10
  ( 0.15 - 0 )
 (10 m x10 m )
m3
m3
s 
m3 
=
=3.9×10-3
14.0
gmole
s
h
( 0.25 - 0.15) 3
m
29-38
30.1
Given:
A = solvent, B = air
T = 298 K, P = 1.0 atm
Sphere: D = 1.0 cm, 0.12 g A/cm2 liquid solvent on sphere at t = 0
Physical parameters:
MA = 78 g/gmole, pA* = 1.17 × 104 Pa (298 K), DAB = 0.0962 cm2/s, νB = 0.156 cm2/s
Physical System: boundary layer between surface of sphere and bulk gas
Source for A: solvent coating surface of sphere
Sink for A: bulk flowing gas
a. Time to dry for still air
Mass balance on solvent evaporating from paint-coated sphere
Initial mass of solvent on sphere
π D 2 0.12 g
=
mAo
=
M A cm 2
π(1.0 cm) 2  0.12 g 
=
4.83 x 10-3 gmole

2 
( 78 g/gmole )  cm 
IN – OUT + GEN = ACCUMULATION (moles A/time)
dm
0 − N AS + 0 = A
dt
t
mA
0
m Ao
−WA ∫ dt =
∫ dmA
WA ⋅=
t mAo − mA
=
WA kc ( c*A − c A∞ ) π D 2
Determine kc, WA, then t
For still air, Sh = 2.0
DAB
0.0962 cm 2 /s
=
kc Sh
=
2.0
=0.192 cm/s
D
1.0 cm
30-1
(1.17 x 10 Pa )
= 4.72 gmole/m = 4.72x10 gmole/cm
(8.314 m ⋅ Pa/gmole ⋅ K ) (298 K)
4
pA
=
c
=
RT
*
A
3
-6
3
3
WA =π D 2 kc (c*A − c A∞ ) =
π (1.0 cm ) ( 0.192 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.85 x 10-6 gmole/s
2
Final mass of solvent on sphere − mA = 0 for dried paint coating on sphere
t
=
mAo
4.83 x 10-3 gmole
=1697 s
=
WA 2.85 x 10-6 gmole/s
b. Time to dry for flowing air with v∞ = 1.0 m/s
Determine kc, WA, then t
For air at 298 K and 1.0 atm, νair = 1.56 × 10−5 m2/s (Appendix I)
Re =
Sc =
v∞ D
ν air
ν air
DAB
=
=
(1.0 m/s )( 0.01m ) = 641
1.56 x 10-5 m 2 /s
0.156 cm 2 /s
=1.62
0.0962 cm 2 /s
Froessling equation for gas flow around a single sphere
Sh =
kc D
= 2 + 0.552 Re1/2 Sc1/3 =2.0+0.552(641)1/2 (1.62)1/3 =18.4
DAB
DAB
0.0962 cm/s
18.4
= 1.77 cm/s
=
=
kc Sh
1.0 cm
D
WA =π D 2 kc (c*A − c A∞ ) =
π (1.0 cm ) (1.77 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.63 x 10-5 gmole/s
2
=
t
mAo
4.83 x 10-3 gmole
= 184 s
=
WA 2.63 x 10-5 gmole/s
30-2
30.2
A = naphthalene, B = air
a. Initial evaporation rate (as flux NA)
, ν air 1.3876 ×10−5 m 2 /s
T = 280 K=
From Appendix
J, T 298
K , DAB P 0.619 m 2 -Pa/s
=
=
3
DAB
0.619 m 2 -Pa/s  280  2
5.57×10-6 m 2 /s
=


1.013×105 Pa  298 
Re
ν air 1.3876 x 10-5 m 2 /s
Dv ∞ ( 0.0175 m )(1.4 m/s )
Sc
=
=
1766,
=
= 2.49
=
ν air
DAB
1.3876 × 10-5 m 2 /s
5.57×10-6 m 2 /s
Sh =
kc D
= 2.0 + 0.552 Re1/2 Sc1/3 = 2.0+0.552(1766)1/2 (2.49)1/3 =33.4
DAB
33.4 DAB 33.4(5.57×10-6 m 2 /s)
=
= 0.011m/s
kC =
D
( 0.0175 m )
PA
2.8 Pa
=
= 1.203 x 10-3 mol/m3
3
RT 
Pa-m 
 8.314
 (280 K)
mol-K 

N A = kc (C As − C A∞ ) = (0.011m/s)(1.203 x 10-3 mol/m3 ) = 1.32 x 10-5 mol/m 2 -s
C=
As
=1.32 x 10-8 kgmol/m 2 -s
b. Time to reach half of the original diameter
-2
-3
-3
m, R1 8.75×10
m, D2 8.75×10
m, R2 4.38×10-3 m
=
D1 1.75×10
=
=
=
4
V = π R3
3
dV
dR
= 4π R 2
dt
dt
let "A" be Naphthalene, M A = 128 lb/lbmol
ρ A dV
= W=
kc C As 4π R 2
A
M A dt
ρA
dR
4π R 2
= kc C As 4π R 2
MA
dt
ρ A 71.136 lb/ft 3
= 0.556 lbmol/ft 3 = 8.90 kgmol/m3
=
M A 128 lb/lbmol
R2 1
t
ρA
=
dR ∫=
dt t
∫
0
M A C As R1 kc
30-3
ρA
8.90 × 103 mol/m3
=
=
7.4 × 106
-3
3
M A C As 1.203 x 10 mol/m
R2 1
dR =
t
(7.4 × 106 ) ∫
R1 k
c
Evaluate by graphical integration, plot 1/k c vs. R (R = D/2)
=
Re
Sh =
( D )(1.4 m/s )
Dv∞
=
=
( D)(1.009 × 105 ) m −1
-5 2
ν air 1.3876 x 10 m /s
kc D
= 2.0 + 0.552 Re1/2 Sc1/3
DAB
k c as a function of D = 2R
(5.57×10-6 m 2 /s)
 2.0 + 0.552 (( D)(1.009 × 105 ) m −1 )1/2 (2.49)1/3 
D
6
(7.4 × 10 )(0.3505) =
2.59 × 106 sec =
720 hr
t=
kc
=
30-4
30.3
z = L = 25 m
water
CAL,o = 0 gmole A/m3
v∞ = 0.50 m/s
293 K
D = 2.0 cm
CAL
solid contaminant layer
lining inside bottom half of tube
CAL*
tube
cross section
a. Solute A undergoes mass transfer. The SOURCE for A is the solid A in the contaminant layer,
and dissolved solute A carried in with the inlet water (zero). The SINK for A is the flowing
fluid. A = solute A, B = water
b. Mass flowrate of inlet water
at 20 C, ρB = 998.2 kg/m3
π D2 
m 
kg   π(0.02 m) 2 
kg
wB v=
ρ
0.50
998.2
=
 = 0.157
∞ B ,liq


3 
4
s 
m 
4
s


c. kL
2
µ B 9.93×10-4 kg/m ⋅ s
-7 m
=
9.95×10
=
ν=
B
998 kg/m3
s
ρB
m2
ν B 9.95×10 s
199
=
Sc =
=
2
DAB
-9 m
5.0×10
s
-7
=
Re
v ∞ D (0.5m/s)(0.02m)
=
= 10, 050 laminar regime
(1.47 cm 2 /s)
νB
0.83
=
=
Sh 0.023Re
Sc1/3 0.023 (10,=
050 ) (199 )
281.7
0.83
Sh =
1/3
kL D
DAB
2

-9 m 
281.7
5.0×10
(
)

s 
Sh ⋅ DAB
m

=
kL =
= 7.04 ×10-5
D
s
( 0.02 m )
30-5
d. Differential model for cAL(z)
Differential material balance on dissolved solute A in liquid phase
IN – OUT + GEN = ACCUMULATION (moles A/time)
v∞
π D2
4
c AL z − v∞
π D2
4
(
=
N A k L c*AL − c AL
Note
c AL z +∆z − N A
π D∆z
2
+0=
0
)
Rearrange, lim Δz→0
c AL z +∆z − c AL z  2k
−
− L c*AL − c AL =
0
∆z
v∞ D
(
)
Differential model
−
(
)
dc AL 2k L *
c AL − c AL =
0
−
dz
v∞ D
z = 0, cAL = cAL,o (entrance); z = L, cAL = cAL
Integrated model
c AL
L
dc A
2k
= − L ∫ dz
∫
*
v∞ D 0
c AL ,o c AL − c AL
(
)
 c* − c AL ,o  2 L k L
ln  AL*
 =
−
c
c
 AL AL  D v ∞
e. cAL
 2L kL 
c AL =c*AL − ( c*AL − c AL ,o ) exp  −

 D v∞ 
 2(25 m) ( 7.04×10-5 m/s ) 
mmole  mmole


c AL = 10
- 10
- 0  exp m3 
m3
(0.02 m)(0.5 m/s) 



mmole
c AL = 2.97
m3
30-6
30.4
stent
vo = 10 cm3 /sec
1.0 cm
blood vessel
stent detail
support ring
drug (Taxol)
loaded on surface
of stent
1.0 cm
D = 0.2 cm
Given:
A = taxol, B = H2O (liquid bodily fluid)
taxol: cA* = 2.5 × 10–4 mg/cm3, DAB = 1.0 × 10–6 cm2/s
Body fluid: µB = 0.040 g/cm∙s, ρB =1.05 g/cm3, MB = 18 g/gmole
Stent: Dcyl = 0.2 cm, L = 1.0 cm, mAo = 5.0 mg
Physical System: boundary layer between surface of cylindrical stent and body fluid
Source for A: drug coating external surface of cylindrical stent
Sink for A: body fluid
a. Convective mass transfer coefficient kL
External flow of liquid around a cylinder
k x Sc 0.56
= 0.281( Re ') −0.4
GM
4vo
4 ⋅10.0 cm3 / s
=
= 12.73 cm/s
v∞ =
π Dtube 2
π (1.0 cm) 2
=
GM
v ∞ ρ B (12.73 cm/s)(1.05g/cm3 )
=
=0.743 gmol/cm 2 ⋅ s
MB
18g/gmol
30-7
v ∞ Dcyl v ∞ Dcyl ρ L (12.73cm/s)(0.2 cm)(1.05 g/cm3 )
=66.83
Re ' =
=
=
µL
vL
( 0.040 g/cm ⋅ s )
µL
0.040 g/(cm-s)
=
= 38, 095
ρ L DAB (1.05 g/cm3 )(1.0×10-6 cm 2 /s)
=
Sc
kx
0.281(
=
Re ') −0.4 ( Sc) −0.56 GM 0.281(66.83) −0.4 (38, 095) −0.56 (0.743gmole/cm 2s)
k x =1.06 x 10-4 gmole/cm 2s
 18 g/gmol 
kx
M
=
=
1.06 ×10−4 gmol/cm 2s ) 
1.81 x 10-3 cm/s
kL =
kx L =
(
3 
ρL
CL
 1.05 g/cm 
b. Time required for complete Taxol release
Mass balance on Taxol coating stent
IN – OUT + GEN = ACCUMULATION (moles A/time)
dm
0 − N AS + 0 = A
dt
t
mA
0
m Ao
−WA ∫ dt =
∫ dmA
WA ⋅=
t mAo − mA
Determine WA, then t
WA =
k L (c*AL − c AL )π Dcyl L =
k L (c*AL )π Dcyl L
WA = (1.81×10-3 cm/s )( 2.5×10-4 mg/cm3 ) ( π )( 0.2 cm )(1.0 cm )
WA = 2.84 x 10-7 mg/s
t
=
mAo − mA mAo
5.0 mg
= =
WA
WA 2.84 x 10-7 mg/s
= 1.76 x 107 sec = 4890 h
30-8
30.5
Let A = O2, B = H2O (liquid)
a. Material balance
In − Out + Generation =
Accumulation
[ S ⋅ N A + v∞ LWC Alo ] − [v∞ LWC Al ] + 0 =
0
*
[( N tπ DL)(k L (C AL
0, combineterms to solve for C AL
− C AL )) + v∞ LWC Alo ] − [v∞ LWC Al ] =
*
(v∞ LW )C Alo + ( N tπ DLk L )C AL
C AL =
(v∞ LW ) + ( N tπ DLk L )
b. Mass transfer coefficient, kL
=
ShD
kL D
= 0.281( Re ') 0.6 ( Sc ) 0.44
DAB
v∞
=
 w
nM
=
LW ρ
0.44
 ν 
 DAB 




 D 
 DAB 
( 50 mol/s )( 0.018 kg/mol )
= 1.80×10-3 m/s
(1 m )( 0.50 m ) ( 998.2 kg/m3 )
v D
k L = 0.281 ∞ 
 ν 
0.6
 (1.80×10-3 m/s)(0.02 m) 
k L = 0.281

(0.995×10-6 m 2 /s)


0.6
 (0.995×10-6 m 2 /s) 


-9
2
 (2×10 m /s) 
0.44
 (2×10-9 m 2 /s) 
−6
=

 3.72 × 10 m/s
(0.02
m)


c. Outlet concentration, CAL
For CAL,o = 0
C AL
*
( N tπ DLk L )C AL
(10)π(0.02 m)(1 m)(3.72×10-6 m/s)(5 mol O 2 /m3 )
=
(v∞ LW ) + ( N tπ DLk L ) (1.80×10-3 m/s)(0.02 m)(1 m)+(10)π(0.02 m)(1 m)(3.72×10-6 m/s)
C AL = 0.0129 mol/m3
Increase CAL* by increasing pressure on the oxygen side to increase CAL.
30-9
30.6
Let A = H2O vapor, B = air
Attempt to use Concentration-Time Charts with m, n, XD, and Y
For m, Estimate kc first
Using Froessling Equation for liquid flow around a sphere with 2 ≤ Re ≤ 800 and 0.6 ≤ Sc ≤ 2.7
Re
=
v∞ D (50 cm/ s)(0.20 cm)
=
= 63.74
(0.15689 cm 2 / s )
ν
Sc
=
=
DAB
=
Sh
kc D
= 2.0 + 0.552 Re1/2 Sc1/3
DA− air
ν
0.15689cm 2 / s
= 0.603
0.260cm 2 / s
=
kc  2.0 + 0.552 Re1/2 Sc1/3 
DA− air
D
kc =  2.0+0.552(63.74)1/2 (0.603)1/3 
=
m
0.260 cm 2 /sec
= 7.049 cm/sec
0.20 cm
DAB
1.0 x 10-6 cm 2 /sec
=
=1.42 x 10-6 ≈ 0
kc R (7.049 cm/sec)(0.10 cm)
n = 0 (center of the sphere)
sec 
(1.0 x 10 cm /sec )  2.78 hr 3600

1 hr 
-6
DAB t
=
XD =
R2
2
(0.10 cm) 2
= 1.00
Off the range of the Charts for a sphere. Therefore, use Infinite Series Solution for center of
sphere
C As= C A* =
S ·PA=
S ⋅ yA P
=2.0 gmole O 2 /(cm3 silica)(atm H 2 O vapor)(0.01)(1.0 atm) = 0.02 gmole/cm3
∞
2 2
c −c
n
Y = A Ao = 1 + 2∑ ( −1) e − n π X D , r = 0, n = 1, 2, 3,
c As − c Ao
n =1
c −0
Y= A
 1.0 , CA = 0.02 gmole/cm3
0.02 − 0
30-10
30.7
Let A = Dicyclomine, B = intestinal fluid (H2O liquid)
a. Concentration of A in center of sphere
Use Concentration-Time charts for a sphere, Fig. F.3
DAet (4.0×10-6 cm 2 /sec)(15,625 sec)
=1.00
=
(0.25 cm) 2
R2
0 cm
r
n =
=
= 0
R 0.25 cm
1 1
DAe
m=
=
= = 0.20
kc R Bi 5
XD
=
0.002= Y=
C A (0, t ) − C A∞
C Ao − C A∞
2.0 mg
mAo
mAo
=
=
= 30.56 mg/cm3
4 3 4
V
π(0.25 cm)3
πR
3
3
C A (0, t )= Y (C Ao − C A∞ ) + C A∞ = 0.002(30.56 - 0.20) mg/cm 2 +0.20 mg/cm3 = 0.26 mg/cm3
C
=
Ao
b. Estimate of Bi
 ρ v D   µB 
=
Pe (=
Re)( Sc)  B ∞   =

 µ B   ρ B DAB 
 (1.0 g/cm3 )(0.50 cm/sec)(0.50 cm)  

(0.007 g/cm-sec)

=

3
-5
2
(0.007 g/cm-sec)

  (1.0 g/cm )(1.0×10 cm /sec) 
=
= 25, 000
Pe (35.7)(700)
kR kD
= c , c = Sh
= (4.0 + 1.21Pe 2/ 3 )1/2
Bi
DAe DAB
 R   DAB 
2/3 1/2
=
Bi 

 (4.0 + 1.21Pe )
D
D

 Ae  

  (1.0×10-5 cm 2 /sec) 
(0.25 cm)
2/3 1/2
Bi 
40.3
=
 (4.0 + 1.21(25, 000) )

-6
2
(0.50 cm)
 (4.0×10 cm /sec)  

30-11
30.8
Let A = solute, B = solvent
a. Estimate the film convective mass transfer coefficient, kL
v ∞ D (10 cm/sec)(1.0 cm)
Re =
= 1005
=
ν
(9.95 x 10-3 cm 2 /sec)
ν
9.95 x 10-3 cm 2 /sec
=
= 829.17
Sc =
DAB 1.2 x 10-5 cm 2 /sec
=
Pe (Re)(
=
Sc) 8.33 x105 > 10, 000
Using Levich Equation for liquid flow around a sphere with Pe > 10,000
1/3
1/3
v D υ 
 v∞ D 
k D
1/3
Sh = L =1.01Pe1/3
=1.01 ∞
 1.01

AB =1.01(Re ⋅ Sc)
DAB
 υ DAB 
 DAB 
1.01( Pe ) DAB 1.01(8.33 x 105 )1/3 (1.2 x 10-5 cm 2 /sec)
= 1.14 x 10-3 cm/sec
=
kL
=
1.0 cm
D
b. Dissolution rate of the pellet at initial diameter of D = 1.0 cm (D = 2R)
1/3
*
=
W A 4π R 2 ( C AL
− C AL ,∞ )
= 4π ( 0.50 cm ) (1.14 x 10-3 cm/sec )( 7.0 x 10-4 - 1.0 x 10-4 ) gmol/cm3 = 2.15 x 10-6 /gmol/sec
2
c. Time t at D = 0.50 cm
In – Out +Generation = Accumulation (moles A/time)
ρ
 ρA
d  A V
4π R 2 dR
M
dmA dmA
A
=
 MA
0 − N=
,
=
A ⋅S +0
dt
dt
dt
dt
ρ
dR
A
4π R 2
−k L (C AL* − C AL∞ )4π R 2 =
MA
dt
M
dR
*
=
−k L (C AL
− C AL ,∞ ) A
ρA
dt
1/3
kL
−
 v ∞ 2 R  DAB
(v ∞ )1/3 ( DAB ) 2/3
1.01
=
1.01


(2 R) 2/3
 DAB  2 R
M
1
*
dR =(C AL
− C AL ,∞ ) A dt
kL
ρA
30-12
R
t
1.01(v ∞ )1/3 ( DAB ) 2/3 *
M
=
− ∫ R dR
(C AL − C AL ,∞ ) A ∫ dt
2/3
2
ρA 0
Ro
2/3
Let D = 2R, integration and simplification yields
ρA
5
t
Do5/3 − D 5/3
1/3
3
v

*
− C AL ,∞ )
2.02 DAB M A  ∞ Sc  ( C AL
ν

(
t
5
3
5
t=
3
ρA
1/3
v

2.02 DAB M A  ∞ Sc 
ν

(C − C )
*
AL
(D
5/3
o
)
− D 5/3
)
AL , ∞
( 2.0 g/cm )
3
1/3
10 cm/sec


2.02 (1.2 x 10 cm /sec ) (110 g/gmol ) 
⋅ 829 
-3
2
 9.95 x 10 cm /sec

1
(1.0 cm)5/3 -(0.5 cm)5/3
⋅
-4
-4
3
( 7.0 x 10 - 1.0 x 10 ) gmol/cm
5
2
(
)
t = 7583 sec = 2.1 hr
30-13
30.9
sunlight
wastewater
CAL,o = 10 g A/m3
Re = 5000
313 K
z = L = 25 m
CAL
D = 1.5 cm
NA
transparent TiO2 catalyst layer
lining inside of glass tube
CAL,s ≈ 0
catalyst surface
a. Solute A (benzene) undergoes mass transfer. The SOURCE for benzene is the benzene
dissolved in the inlet liquid. The SINK for benzene is the consumption by photocatalytic
reaction at the catalyst surface lining the inner wall of the tube. A = benzene, B = water
b. The photocatalytic decomposition of benzene is a heterogeneous reaction at a boundary
surface.
c. kL at Re = 5000
DAB 13.26 ×10−5 µ B−1.14VA−0.589
=
VA = 6 ⋅ VC + 6 ⋅ VH − 15 = 6 ⋅14.8 + 6 ⋅ 3.7 − 15 = 96cm3 /gmole
DAB = 13.26×10-5 ( 0.658 )
Sc
=
µ
B
=
ρ B DAB
-1.14
( 96 )
-0.589
=1.45×10-5
cm 2
s
( 0.00658g/cm ⋅ s ) = 457
( 0.993 g/cm )(1.45×10 cm /s )
3
-5
2
Re > 2000 therefore turbulent flow inside tube
ρ v D
Re = B ∞
µB
v∞ =
µ B Re ( 0.00658 g/cm ⋅ s )( 5000 )
cm
=
= 22.1
3
ρB D
s
( 0.993 g/cm ) (1.5 cm )
0.83
Sh 0.023Re
Sc1/3 0.023 ( 5000
=
=
=
) ( 457 ) 208.2
0.83
Sh =
kL D
DAB

kL
=
1/3
Sh ⋅ DAB
=
D
( 208.2 ) 1.45×10-5

(1.5 cm )
cm 2 

s 
= 2.013 ×10-3
cm
s
30-14
d. Differential model for cAL(z) rapid surface reaction
Differential material balance on dissolved solute A in liquid phase
IN – OUT + GEN = ACCUMULATION (moles A/time)
v∞
π D2
4
c AL z − v∞
π D2
4
(
c AL z +∆z − N Aπ D∆z + 0 =0
)
Note N A = k L c AL − c AL , s ≈ k L c AL assuming rapid reaction of A at catalyst surface
Rearrange, lim Δz→0
c AL z +∆z − c AL z  4k
−
− L c AL =
0
∆z
v∞ D
Differential model
dc
4k
0
− AL − L c AL =
v∞ D
dz
( )
( )
z = 0, cAL = cAL,o (entrance); z = L, cAL = cAL
Integration (needed for part e)
c AL
L
dc A
4k
= − L ∫ dz
∫
c
v∞ D 0
c AL ,o AL
Integrated model
 c  4L kL
ln  AL ,o  =
 c AL  D v ∞
e. cAL for rapid surface reaction
 4L kL 
c AL c AL ,o exp  −
=

 D v∞ 
 4(2500 cm) ( 2.013×10-3 cm/s ) 
gA

c AL = 10 3 exp m
(1.5 cm)(22.1 cm/s)


gA
c AL = 5.45 3
m
30-15
f. cAL for surface reaction with ks = 0.010 cm/s (cAL,s > 0)
At surface
(
)
N A = k L c AL − c AL , s = ks c AL , s
c AL , s =
k L c AL
ks + kL
∴N
=
A
ks k L c AL
= k ' c AL
ks + kL
ks kL
=
k' =
ks + kL
( 0.10 cm/s ) ( 2.013×10-3 cm/s )
= 6.68×10-4 cm/s
-3
( 0.10 cm/s ) + ( 2.013×10 cm/s )
Modified model
 4L k ' 
c AL c AL ,o exp  −
=

 D v∞ 
 4(2500 cm) ( 6.68×10-4 cm/s ) 
gA

c AL = 10 3 exp m
(1.5 cm)(22.1 cm/s)


gA
c AL = 8.17 3
m
30-16
30.10
Let A = O2, B = H2O (liquid)
a. Material balance on O2 (component A) in the liquid phase
Steady state material balance for species A on well mixed liquid phase of constant volume:
C AL ,o vo + N A Ai − C AL vo + RAV =
0
*
=
− C AL )
N A K L (C AL
a=
Ai
V
v
*
− C AL ) + RA =
(C AL ,o − C AL ) o + k L a (C AL
0
V
b. kL for db = 2 mm
µL
825×10-6 kg/m-sec
=
= 412.5
Sc =
ρ L DAB (1000 kg/m3 )(2.0×10-9 m 2 /sec)
Gr =
d b 3 ρ L g ( ρ L − ρG )
µL2
3
1m 

3
2
3
 2 mm ⋅
 (1000 kg/m )(9.81 m/s )( (1000-1.2 ) kg/m )
1000 mm 
= 
= 1.153 × 105
-6
2
(825 × 10 kg/m-sec)
=
=
112.3
Sh 0.31(
Sc)1/3 (Gr )1/3 0.31(412.5)1/3 (1.153 × 105 )1/3 =
Sh ⋅ DAB (112.3)(2.0×10-9 m 2 /sec)
=
=
kL
=1.123×10-4 m/sec
db
 1m 
(2 mm) 

 1000 mm 
c. Outlet concentrations CAL
PA
0.21 atm
*
=262.5 mmol/m3
C=
=
AL
3
H
m -atm
8×10-4
mmol
30-17
vo
*
+ k L aC AL
+ RA
V
C AL =
v
kL + o
V
3
mmol O 2
3  0.05 m /sec 
-4
2
3
3
(10.0 mmol/m ) 
 + (1.123×10 m/sec)(10 m /m )(262.5 mmol/m ) - 0.2 3
3
m -sec
 1000 m 
=
3
 0.05 m /sec 
(1.123×10-4 m/sec)(10 m 2 /m3 ) + 

3
 1000 m 
C AL ,o
= 81.23 mmol/m3
30-18
30.11
flue gas
10 mole% CO2 (A)
90 mole% N2
Open Pond
(no inflow or outflow of liquid)
well-mixed
pond water + algae + dissolved CO2
RA
CAL
Given:
A = CO2, B = H2O (liquid)
Cells: k/1 = 0.06435 m3/g cells∙h, X = 50 g cells/m3
Liquid: νB = 0.995 x 10-6 m2/s, ρB =1000 kg/m3
Gas: db = 7.0 mm, a = 5.0 m2/m3 , ρG = 1.2 kg/m3
a. DAB for dissolved CO2 (solute) in water (solvent)
Wilke-Chang Correlation (assume infinite dilute solution)
DAB µ B 7.4 x10 [φB M B ]
=
T
VA0.6
−8
1/2
7.4x10-8 ( 2.6 )(18 )  (293)
7.4 x10−8 [φB M B ] T
=
=1.8 x 10-5cm 2 /s
VA0.6
µB
(34.0)0.6 (993x10-3 )
1/2
1/2
DAB
b. Determine kL for CO2 transfer through liquid film surrounding gas bubble
=
Gr
db3 ρ L g ∆ρ (7x10-3 m)3 (998.2 kg/m3 )(9.81m/s 2 )(998.2 kg/m3 - 1.7967 kg/m3 )
= 3.39 x 106
=
2
-6
2
(993 x 10 kg/m ⋅ s)
µL
( 0.995 x 10 m /s ) = 553
=
Sc =
D
(1.8 x 10 m /s )
ν
-6
-9
2
2
AB
30-19
=
Sh
k L db
= 0.42Gr1/3 Sc1/2 =0.42(3.39 x 106 )1/3 (553)1/2 = 1483
DAB
D 
k L = 0.42Gr1/3 Sc1/2  AB 
 db 
 1.8 x 10-9 m/s 
-4
k L = (0.42)(3.39x106 )(553)1/2 
 = 3.82 x 10 m/s
-3
 7.0 x 10 m 
c. Material balance model for predicting dissolved CO2 concentration cAL
Physical System: liquid in pond—no liquid inflow or outflow
Source for A: gas containing CO2 bubbled into liquid
Sink for A: consumption of CO2 by cells suspended in pond
Assumptions
1. Constant source and sink for CO2—steady-state process
2. First-order homogeneous reaction of CO2 in pond by cells suspended in liquid
3. Well-mixed liquid phase, but no inflow or outflow of liquid
4. Constant liquid volume
5. Dilute UMD process with respect to dissolved CO2
6. Liquid-phase controlling mass transfer process with kL ≈ KL
Mass balance on dissolved CO2, liquid phase
IN – OUT + GEN = ACCUMULATION (moles A/time)
N A ⋅ Ai − 0 + RA ⋅ V =
0
K L a (c*AL − c AL ) ⋅ V − k1c AL ⋅ V =
0
K L a(
pA
− c AL ) − k1c AL =
0
H
Solve for outlet concentration cAL
=
c AL
K L a ⋅ PA
k L a ⋅ PA
=
H ( K L a + k1 ) H (k L a + k1 )
d. Determine cAL

m3  
g cells 
-1
-4 -1
=
k1 k=
' X  0.06435
 50
 = 3.22 hr = 8.938 x 010 s
3
g cells hr  
m 

30-20
c AL =
kL a ⋅ pA
H (k L a + k1 )
( 3.82 x 10 m/s )( 5.0 m /m ) (0.10 atm)
= 2.64 gmole/m
c =
( 0.025 atm·m /gmole ) (3.82 x 10 m/sec×5.0 m /m + 8.938 x 10 s )
-4
AL
3
2
-4
3
2
3
-4
3
-1
e. Determine the total CO2 removal rate in kg CO2/h if the total pond volume is 1000 m3
WA =
N A Ai =
RA ⋅ V =
−k1C AL ⋅ V =
−k1' X ⋅ C AL ⋅ V
gmole 

3  0.044 g CO 2 
WA = − ( 3.22 hr -1 )  2.64

 (1000 m ) 
3
m 

 1 gmole 
WA = −274 g CO 2 /hr
30-21
30.12
Let A = TCE, and B = H2O (liq)
a. Determine kL
DAB = 8.9x10-10 m2/sec
ρL = 998.2 kg/m3, µL = 9.93x10-4 kg/m-sec, νL = 0.995x10-6 m2/sec
ρG = 1.19 kg/m3
Gr
=
Sc =
-3
3
3
2
3
db3 ρ L g ∆ρ (5.0 x 10 m) (998.2 kg/m )(9.81m/sec )( ( 998.2-1.19 ) kg/m )
= 1.24 x 106
=
2
-6
2
(993 x 10 kg/m ⋅ sec)
µL
ν
=
DAB
db = 5 mm
Sh
=
0.995 x 10-6 m 2 /sec
=1118
8.9 x 10-10 m 2 /sec
k L db
= 0.42Gr1/3 Sc1/2
DAB
D 
k L = 0.42Gr1/3 Sc1/2  AB 
 db 
 8.9 x 10-10 m 2 /sec 
-4
k L = (0.42)(1.24x106 )1/3 (1118)1/2 
 =2.69 x 10 m/sec
-3
5
x
10
m


3
b. Time required for CAL = 0.005 g/m , no inflow or outflow
Material balance model (A = TCE)
Assumptions: 1) PA ≈ 0, 2) no reaction occurs, 3) unsteady state, 4) dilute system, 5) kL≈KL
In – Out +Generation = Accumulation (liquid phase – mole A/time)
d ( C ⋅V )
0 − N A ⋅ Ai + 0 = AL
dt
A
dC
*
) ⋅ i = AL
−k L (C AL − C AL
V
dt
P
*
A
C=
≈0
AL
H
C AL
A t
1
− ∫
dC AL =
k L i ∫ dt
C AL
V 0
C AL ,0
 C AL ,o 
Ai
ln 
 = kL t
V
 C AL 
30-22
a=
Ai Va 6
6
= ⋅
= 0.015 m 2 /m 2
= 18 m 2 /m3
V
V db
0.005 m
 C AL ,0  1
 50g TCE/m3 
1
1hr
=
×
×
=0.528 hr
t ln 
 ⋅ = ln 
3 
-4
2
3
 0.005g TCE/m  (2.69 x 10 m/sec)(18m /m ) 3600 sec
 C AL  k L a
c. Determine CAL at steady state
Material balance model
In – Out +Generation = Accumulation (liquid phase – mole A/time)
C AL ,0V0 − C ALV0 − k L a ⋅ V ⋅ C AL + 0 =
0
C AL ,o ⋅ V0
(50gTCE/m3 )(1.0m3 /sec)
C AL
=
=
V0 + k L a ⋅ V (1.0 m3 /sec)+(2.69 x 10-4 m/sec)(18m 2 /m3 )(2 m ⋅ 10 m ⋅ 10 m)
= 25.4 g TCE/m3
30-23
30.13
Let A = CO2, B = H2O (liquid)
a. Determine kL
=
Gr
-3
3
3
2
3
db3 ρ L g ∆ρ (2.0 x 10 m) (998.2 kg/m )(9.81m/sec )( ( 998.2-1.7967 ) kg/m )
= 79161
=
(993 x 10-6 kg/m ⋅ sec) 2
µ L2
ν
0.995 x10−6 m 2 / sec
=
= 562
DAB 1.77 x10−9 m 2 / sec
db = 2.0 mm
=
Sc
=
Sh
k L db
= 0.31Gr1/3 Sc1/3
DAB
D 
k L = 0.31Gr1/3 Sc1/3  AB 
 db 
kL
 1.77 x10−9 m 2 / sec 
−5
=
(0.31)(79161)
(562) 
 9.721x10 m / sec
−3
2 x10 m


1/3
1/3
b. Determine if the inlet flow rate of CO2 gas is sufficient
PA = 2.0 atm
PA
2.0 atm
=0.06757 kmol/m3
=
3
H 29.6 atm ⋅ m /kmol
6φg
A
*
− C AL ,o )=
WA= N A i V= k L
V ( C AL
V
db
*
C=
AL
(9.721 x 10-5 m/sec)
6(0.05)
2.0 m3 ) (0.06757 kmol /m3 ) = 1.97 x 10-3 kmol CO 2 /sec
(
-3
2.0 x 10 m
Convert WA in kmol/sec to volumetric flowrate of STD m3 CO2/ min
VCO 2 = WA M A / ρ A = (1.97 x 10-3 kmol CO 2 /sec)(44kg/kmol)(1m3 /1.7967 kg)(60 s/min)
= 2.89 m3 CO 2 /min
The inlet flow rate of CO2 from the problem statement (4.0m3/min) is larger than 2.89 m3/min,
thus the inlet flow rate of CO2 gas is sufficient to ensure that the CO2 dissolution is mass transfer
limited.
c. Determine CAL,out
Assumptions : 1 ) no chemical reaction, 2) steady state, 3) dilute system
30-24
In – Out +Generation = Accumulation (liquid phase – mole A/time)
N A ⋅ Ai − C ALV0 + 0 =
0
−C ALV0 + k L a ⋅ V ⋅ (C*AL − C AL ) + 0 =
0
6(0.05)
(0.06757 kmolCO 2 /m3 )
-3
k L a ⋅ V ⋅ (C )
2.0
x10
m
C AL ,out
=
=
(0.45m3 /min)(1min/60 sec)
6(0.05)
V0
+(9.721 x 10-5 m/sec)
+ kL a
3
2.0 m
2.0 x 10-3 m
V
C AL ,out = 0.0537 kmol/m3
*
AL
(9.721 x 10-5 m/sec)
30-25
30.14
Let A = TCE, B = water
Determine kL
DAB = 8.9x10-10 m2/sec
ρL = 998.2 kg/m3, µL = 9.93x10-4 kg/m∙sec
ρG = 1.19 kg/m3
=
Gr
db3 ρ L g ∆ρ (0.01m)3 (998.2kg / m3 )(9.81m / s 2 )(998.2 − 1.19kg / m3 )
=
= 9.89 x106
−6
2
2
(993 x10 kg / m − sec)
µL
µL
993 x10−6 kg / m 2 sec
=
= 1118
ρ L DAB (998.2 kg/ m3 )(8.9 x10−10 m 2 / sec)
db = 10 mm
Sc
=
=
Sh
k L db
= 0.42Gr1/3 Sc1/2
DAB
D 
k L = 0.42Gr1/3 Sc1/2  AB 
 db 
 8.9 x10−10 m 2 / sec 
−4
(0.42)(9.89
x106 )1/3 (1118)1/2 
=
 2.683 x10 m / sec
0.01m


Material balance model (A = TCE)
kL
Assumptions: 1) steady state, 2) dilute system, 3) concentration along z-direction only, 4) No
reaction, 5) PA≈0
In – Out +Generation = Accumulation (liquid phase – mole A/time)
A
v∞ C AL − N A ⋅ i ⋅ ∆z − v∞ C AL
+ 0 =0
z
z +∆z
V
÷ Δz, rearrangement, ∆z → 0
A
dC AL
v∞
+ NA ⋅ i =
0
dz
V
A
PA
dC AL
*
*
) i =
0 , also C=
≈0
+ k L (C AL − C AL
AL
H
dz
V
A
dC AL
+ k L C AL i =
0
dz
V
C AL ,out dC
k L Ai L
AL
−∫
=
dz
C AL ,o
C AL
v∞ V ∫0
 C AL ,o  k L Ai
L
ln 
 =
C
v
V
∞
 AL ,out 
30-26
Determine L
Ai Va 6 0.02 m3 TCE
6
= ⋅
=
a=
×
=12 m 2 /m3
3
V
V db
0.01 m
m water
 C AL ,o  v∞
 50 mg/L 
0.05 m/sec
L ln=
ln
= 107 m
=



-4
2
3
C
k
a
0.05
mg/L
(2.683x10
m/sec)(12
m
/m
)


,
AL
out
L


30-27
30.15
Let A = POCl3, B = He
a. Develop material balance model
Assumptions: 1) Steady state; 2) Concentration profile along z-direction only; 3) No reaction; 4)
Dilute solution and constant fluid velocity.
Let A = POCl3, B = He (g)
Differential material balance
π D2
π D2
v∞ c A + N A π D ∆z −
v∞ cA
+ 0 =0
4
4
z
z +∆z
÷ ∆z, rearrangement, ∆z → 0
4k
dc
− A + c c*A − c A =
0
dz
v∞
Integration:
c A ,out − dc
4k L
∫cA,o c*A − cAA = − v∞ Dc ∫0 dz
(
)
 c*A − c A,o 
4 kc L
ln  *
 = −
 c −c
v∞ D
 A A,out 
b. Determine mass transfer coefficient, kc
4Q g 4 (110 cm3 /s)
v∞ =
= 35 cm/s
=
π (2.0 cm) 2
π D2
v∞ D ( 35 cm/s )( 2.0 cm )
=
= 50 (laminar)
Re =
νB
1.4 cm 2 /s
(
)
νB
1.4 cm 2 /s
=
= 3.78
DAB 0.37 cm 2 /s
Sieder-Tate Correlation for laminar flow in tube
1/3
1/3
D

 2.0 cm

=
Sh 1.86
=
1.86 
( 50 )( 3.78)  = 3.44
 Re Sc 
 L

 60 cm

Sh
3.44
=
kc =
DAB
0.37 cm 2 /s = 0.636 cm/s
D
2.0 cm
=
Sc
c. Determine yA,out
*
=
y A,out c=
PA / P =0.15 atm/1.0 atm = 0.15
A / c, y A
30-28


 4k L  
 -4(0.636 cm/s)(60 cm)  
y A,out = y*A 1 − exp  − c   = 0.15 1 − exp 
  = 0.133
v
(35
cm/s)(2.0
cm)
D


∞





d. kc and yA,out at D = 4.0 cm
v∞ = 8.75 cm/s, Re = 25; k c ∝ D −1 , D1/3 , Re1/3
−1
+1/3
+1/3
 Dnew   Dnew   Re new 
cm  4.0 cm   4.0 cm   25 
kc ,new k=
0.636
 
 

c , old 

 
  
s  2.0 cm   2.0 cm   50 
 Dold   Dold   Reold 
cm
= 0.318
s


 4k L  
 -4(0.318 cm/s)(60 cm)  
y A,out = y*A 1 − exp  − c   = 0.15 1 − exp 
  = 0.133
 (8.75 cm/s)(4.0 cm)  
 v∞ D  


-1
+1/3
+1/3
No net change from part c.
e. yA,out →yA* as z→∞
30-29
30.16
Liquid TEOS
flow In
TEOS vapor
+He gas Out
liquid film
L = 2.0 m
D = 5 cm
cA
z+Δz
NA
cAs
z
cA
v∞
Liquid TEOS
Out ( = 0)
100% He flow in
333 K, 1.0 atm
2000 cm3/s
Given:
A = TEOS, B = He
T = 333 K, P = 1.0 atm
TEOS: PA = 2133 Pa, DAB = 1.315 cm2/s
He gas: νB = 1.47 cm2/s, Vo = 2000 cm3/s
Wetted wall column: D = 5.0 cm, L = 2.0 m
Physical System: gas phase inside tube
Source for A: volatile liquid TEOS flowing down inner surface of tube which vaporizes into the
gas phase
Sink for A: He carrier gas flowing up tube
a. Mass transfer coefficient for TEOS vapor in He gas, kc
Estimate Reynolds number to establish flow regime inside tube
Vo
2000 cm3 /s
=
= 101.86 cm/s
π 2 π
2
D
(5.0 cm)
4
4
v ∞ D (5.0 cm)(101.86 cm/s)
=
= 346.4, laminar regime
Re =
(1.47 cm 2 /s)
νB
=
v∞
30-30
For laminar flow inside tube
1/3
kc D
D

= 1.86  ReSc 
DAB
L

νB
=
Sc
1.47 cm 2 /s
=
= 1.118
DAB 1.315 cm 2 /s
kc
 1.315 cm 2 /s 
 DAB 
D

 5.0 cm

(346.4)(1.118) 
1.042 cm/s

1.86  =
=
1.86  ReSc 
L

 200 cm

 D 
 5.0 cm 
1/3
1/3
Convert kc to kG
kG =
kc
RT
( 0.01042 m/s )
(8.314 J/mol ⋅ K)(333 K)
= 3.76 x 10-6 mol/m 2s ⋅ Pa
b. Mole fraction TEOS in outlet gas
Develop process model for prediction of cA(z), then input kc and other parameters posed by
model
Assumptions:
1.
2.
3.
4.
5.
6.
Constant source and sink for TEOS vapor—steady-state process
No radial concentration gradient of TEOS vapor in gas phase
Constant temperature and pressure, therefore cAs constant
Dilute process with respect to TEOS vapor in He
TEOS transfer from liquid to gas limited by boundary layer mass transfer through gas
Falling liquid film of pure TEOS is thin (<< D) and uniformly distributed
Differential material balance on TEOS in gas phase
IN – OUT + GEN = ACCUMULATION (moles A/time)
v∞
π D2
4
cA z − v∞
π D2
4
c A z +∆z − N Aπ D∆z + 0 =0
N A kc ( c As − c A )
Note=
Rearrange, lim Δz→0
cA
− c A z  4kc
−  z +∆z
−
c As − c A =
0
∆z
v∞ D
(
)
30-31
−
(
)
dc A 4kc
−
c As − c A =
0
dz v∞ D
Separate variables cA, z and integrate with limits shown below
z = 0, cA = cAo (entrance); z = L, cA = cA,out
c A ,out
∫ (
c Ao
L
4k
dc A
= − c ∫ dz
v∞ D 0
c As − c A
)
 c − c  4 L kc
ln  As Ao  =
c −c

 As A,out  D v ∞
Determine cA,out, then yA,out
PA
2133 Pa
=
= 0.770 mol/m3
RT (8.314 J/mol-K)(333 K)
c Ao = 0
c=
As

 4 L kc  
c A,out = c As 1 − exp  −
 
 D v∞  

4 L kc 4(200 cm) (1.042 cm/s)
= 1.6368
D v∞
( 5.0 cm ) (101.86 cm/s)
=
c A,out
c=
P
RT
y=
A, out
( 0.770 mol/m ) ⋅ (1 − exp ( −1.6368) ) = 0.620 mol/m
3
(101,250 Pa )
=
(8.314 J/mol-K)(333 K)
3
36.6 mol/m3
c A,out 0.620 mol/m3
=
= 0.017
c
36.6 mol/m3
c. Transfer rate of TEOS (g/s), WA
Overall gas phase material balance for TEOS—the rate of input of TEOS liquid flowing down
the inner wall of the tube must balance its mass-transfer limited evaporation to the He carrier gas
W
=
Vo (c A,out −c Ao
=
) Vo c A,out
A
WA = ( 2000 cm3 /s )( 0.620 mol/m3 )(1.0m3 /106 cm3 ) = 1.24 x 10-3 mol/sec
WA M A = (1.24 x 10-3 mol/s)(208.33 g/mol) = 0.258 g TEOS/s
30-32
30.17
Let A = O3, B = H2O
a. Determine kL
Vo
V
100cm3 / sec
v=
=
=
= 50 cm/sec
∞
A hw (1.0cm)(2.0cm)
4hw
4(1.0cm)(2.0cm)
d eq
= 1.33 cm
=
=
2h + 2 w 2(1.0cm) + 2(2.0cm)
v ∞ d eq (50 cm / sec)(1.33 cm)
=
Re =
= 6650
ν
0.01 cm 2 / sec
νL
0.01 cm 2 / sec
=
= 575
DAB 1.74 x10−5 cm 2 / sec
Gilliland-Sherwood correlation for turbulent flow of liquid flowing inside of a tube
kL D
=
Sh =
0.023Re0.83 Sc1/3
DAB
Sc
=
D 
k L = 0.023Re0.83 Sc1/3  AB 
 D 
 1.74 x10−5 cm 2 / sec 
−3
0.83
=
k L 0.023(6650)
(575)1/3 
 3.73 x10 cm / sec
1.33cm


b. Determine L
Assumptions: 1) Steady state, 2) Dilute system, 3) Concentration along z-direction only, 4) No
homogeneous chemical reaction
Material balance
In – Out +Generation = Accumulation (liquid phase – mole A/time)
 C
 C
V
+ N A (2 w ⋅ ∆z ) − V
0 AL
0
AL
z
z +∆z
+ 0 =0
÷ Δz, rearrangement, ∆z → 0
dC
2w
- AL + k L (C*AL − C AL )
0
=

V
dz
0
∫
C AL ,out
C AL ,o
 C*AL − C AL ,o  2 wk L
dC AL
2 wk L L
=
dz → ln  *
 =  L
 ∫0
C −C
C*AL − C AL
V
V0
0
AL ,out 
 AL
PA
15 atm
⋅ CL =
(0.056 mol/cm3 ) = 223.40 gmole/m3
H
3760 atm
100 cm3 /sec


223.4 gmole/m3 -0
L = ln
= 60.3cm
3
3 
-3
223.4
gmole/m
-2.0
gmole/m
2
2.0cm
3.73x10
cm/sec
(
)


*
C AL
=
(
(
)
)
30-33
c. Determine new L when PA = 30atm
DAB will not change with respect to P in this case because this is diffusion in liquid phase.
*
will change because it is dependent on P
However, C AL
*
=
C AL
PA
30 atm
⋅ CL =
(0.056 mol/cm3 ) = 446.81 gmole/m3
H
3760 atm
(
)
100 cm3 /sec


446.81 gmole/m3 -0
L = ln 
=30 m
3
3 
-3
 446.81 gmole/m -2.0 gmole/m  2 ( 2.0 cm ) 3.73x10 cm/sec
(
)
30-34
30.18
Let A = CO2, and B = H2O (liquid)
DAB = 1.77x10-9 m2/sec
w
36 g / sec

V
=
=
= 3.61x10−5 m3 / sec
0
ρ L 998.2 x103 g / m3
a. Determine CAL*
PA
2.54 atm
*
=0.10 kgmol/m3
C=
=
AL
3
H
atm ⋅ m
25.4
kgmol
b. Determine kL
4 w
4(36 g / sec)
Re =
=
= 769
π Dµ L π (6.0cm)(9.93x10−3 g/ cm − sec)
Sc
=
µL
993 x10−6 kg / m 2 sec
=
= 562
ρ L DAB (998.2 kg/ m3 )(1.77 x10−9 m 2 / sec)
1/6
 ρ 2 g ⋅ z3 
kL z
=
Sh =
0.433( Sc)1/2  L 2  (Re L )0.4
DAB
 µL 
1/6
 ρ 2 g ⋅ z3 
D 
k L = 0.433( Sc)  L 2  (Re L )0.4  AB 
 z 
 µL 
1/2
1/6
-9
2
 (998.2 kg/m3 ) 2 (9.81 m/sec 2 )(2m)3 
0.4  1.77 x 10 m /sec 
(769)
k L = 0.433(562) 



2.0 m
(993 x 10-6 kg/m-sec) 2




-5
=2.69 x 10 m/sec
1/2
c. Determine CAL,out
Assumption: 1) no reaction, 2) Steady state, 3) dilute system
Material balance on differential volume element (A = CO2)
In – Out +Generation = Accumulation (liquid phase – mole A/time)
V0C AL + N A ⋅ π D ⋅ ∆z − V0C AL
z
÷ Δz, rearrangement, ∆z → 0
 dC AL + N ⋅ π D =
V
0
A
0
dz
z +∆z
+ 0 =0
30-35
dC AL π D
*
k L (C AL
+
− C AL ) =
0

dz
V0
C AL ,out
L
dC AL
kL
−∫
=
π
D
∫0 dz
C AL ,o C * − C
V0
AL
AL
*
 C AL
− C AL ,o  k L
ln  *
 =  π DL
C −C
AL
AL
out
,

 V0


 −kL

 -π(0.06 m)(2.69x10-5 m/sec)(2 m)  
*
3
π
−
1
exp
=
0.10
kgmol/m
1-exp
C=
C
DL
) 
 
 (

AL , out
AL 
3.61x10-5 m3 /sec



 V0
 

3
C AL ,out = 0.0245 kgmol/m
30-36
30.19
Let A = O2, B = blood
a. Determine kL
4 w
4V
4(300 cm3 /min)(1min/60 sec)
Re =
=
= 79.6
=
π D µ L π Dν L
π(2.0 cm)(0.040 cm 2 /sec)
νL
0.040 cm 2 /sec
= 2000
=
Sc =
DAB 2.0 x 10-5 cm 2 /sec
1/6
 g ⋅ z3 
kL z
Sh =
=
0.433( Sc)1/2  2  (Re L )0.4
DAB
 νL 
1/6
 g ⋅ z3 
D 
k L = 0.433( Sc)  2  (Re L )0.4  AB 
 z 
 νL 
1/2
1/6
-9
 (9.81 m/s 2 )×(0.5m)3 
m 2 /sec 
0.4  2.0 x 10
-5
(79.6)
k L = 0.433(2000) 

 = 2.91 x 10 m/sec
-6
2
2 
0.5 m


 (4 x 10 m /sec) 
1/2
b. Determine CAL,out
Assumption: 1) No reaction, 2) Steady state, 3) Dilute system, 4) kL≈KL
Material balance on differential volume element (A = CO2)
In – Out +Generation = Accumulation (liquid phase – mole A/time)
V0C AL + N A ⋅ π D ⋅ ∆z − V0C AL
z
z +∆z
+ 0 =0
÷ Δz, rearrangement, ∆z → 0
 dC AL + N ⋅ π D =
V
0
A
0
dz
dC AL π D
*
+
− C AL ) =
0
k L (C AL

dz
V0
C AL ,out
L
dC AL
kL
−∫
=
π D ∫ dz

C AL ,o C * − C
0
V0
AL
AL
 C * − C AL ,0  k L
ln  *AL
 =  π DL
 C −C
AL
AL
out
,

 V0
 −k

*
*
− (C AL
− C AL ,o ) exp  L π DL 
C AL ,out =C AL

 V0

30-37
Also,
PA
k ⋅ PA
*
C AL
C AL ,max =
=
+
H 1 + k ⋅ PA
1atm
(28 atm -1 )(1 atm)
+
9.3 mol O 2 /m3 ) =10.23 mol O 2 /m3
(
3
-1
0.80 atm ⋅ m /mol 1 + (28 atm )(1 atm)
C AL ,out = 10.23 mol O 2 /m3 - (10.23 mol O 2 /m3 -1.0 mol O 2 /m3 )
 -π(0.02 m)(2.91 x 10-5 m/sec)(0.5m) 
exp 

-4
3
 (3 x 10 m /min)(1min/60 sec) 
C AL ,out = 2.54 mol O 2 /m3
30-38
30.20
Model 1
kc
1.17
( Sc) 2/3
=
=
jD
0.415
v∞
Re
3
At 311 K,
,υ 1.673 × 10−5 m 2 /s
=
ρ 1.136 kg/m=
3/2
DAB
2.634 m 2 -Pa/ssec  311 K 
2.772×10-5 m 2 /s
=


-5
1.013 x 10 Pa  298 K 
v=
∞
G 0.816 kg/m 2 -s
= 0.72 m/s
=
ρ
1.136 kg/m3
=
Re
Dv∞ (5.71×10-3 m)(0.72 m/s)
=
= 246
υ
(1.673×10-5 m 2 /s)
Re0.415 = 9.82
=
Sc
1.673 x 10-5 m 2 /sec
υ
=
= 0.60
DAB 2.772 x 10-5 m 2 /sec
Sc 2/3 = 0.711
1.17v∞
1.17(0.72 m/sec)
=
=
= 0.12 m/sec
kc
0.415
2/3
(9.82)(0.711)
Re Sc
kc
0.12 m/sec
=
=
kG
=4.64×10-8 kgmol/m 2 -sec-Pa = 4.70×10-3 kgmol/m 2 -sec-atm
RT (8.314 J/mol-K)(311 K)
Model 2
k
2.06
=
ε jD =
ε c ( Sc) 2/3
0.575
v∞
Re
0.575
=
=
23.7
Re0.575 (246)
2.06v∞
2.06(0.72 m/sec)
=
=0.117 m/sec
kc =
0.575
2/3
(0.75)(23.7)(0.711)
ε Re Sc
k
0.117 m/sec
= 4.52×10-8 kgmol/m 2 -sec-Pa = 4.58 x 10-3 kgmol/m 2 -sec-atm
kG = c =
RT (8.314 J/mol-K)(311 K)
30-39
30.21
Let A = TCE, B = air
a. Determine kc, gas-film mass transfer coefficient for TCE vapor in air
d p G (3 x 10-3 m)(0.10 kg/m 2 -sec)
=
= 16.55
Re =
µ
(1.813 x 10-5 kg/m-sec)
Re0.31 = 2.39
Sc =
ν
DAB
=
1.505 x 10-5 m 2 /sec
= 1.863
8.08 x 10-6 m 2 /sec
Sc 2/3 = 1.514
G 0.1 kg/m 2 -sec
=
= 0.083 m/sec
ρ
1.206 kg/m3
k
0.25
= ε c ( Sc) 2/3
0.31
v∞
Re
v=
∞
kc
0.25  0.083 m/sec   1 
=


 0.011m/sec
2.39  1.514
  0.5 
b. Length required, L
Material balance on a differential volume element (moles A/time)
 A
v∞=
C AWD z kc   ( C A* − C A ) WD∆z + v∞C AWD z +∆z
V 
dC A
 A
kc   ( C A* − C A ) =
v∞
dz
V 
A π D2 π
where= =
V
D3
D
π  L
0.999 C *A
dC A
kc   ∫ dz = v∞ ∫
*
0
d  0
(CA − CA )
 p
L
v∞ d p 
 (0.083 m/sec)(3 x 10-3m)  1 
C A*
=
ln  *
ln 

 = 0.050 m
π kc  C A − 0.999C A 
π(0.011m/sec)
 1-0.999 
30-40
30.22
Film theory, k L α DAB1.0
1.0
 DCO2 − H 2O 
 1.46 x 10-5 cm 2 /sec 
-1
(k L a )CO2 (=
(300 hr -1 ) 
=
k L a )O2 

 =158 hr
-5
2
 DO − H O 
2.77
x
10
cm
/sec


 2 2 
1.0
Penetration theory, k L α DAB1/2
=
(k L a )CO2 (300
=
hr -1 ) ( 0.527 )
217.8 hr -1
1/2
Boundary theory, k L α DAB 2/3
=
(k L a )CO2 (300
=
hr -1 ) ( 0.527 )
195.8 hr -1
2/3
30-41
30.23
Let A = O2, B = H2O (liquid)
a. Volumetric mass transfer coefficient for O2, ( k L a )O
2
=
Rei
2
di N
=
υ
Po= 6=
2
(0.3 m) (4 rev/s)
= 4.83 x 105
-6
2
(0.746 x 10 m /s)
Pg c
ρ L N 3 di5
6(993.7 kg/m3 )(4 rev/sec)3 (0.3 m)5
= 927.2 N ⋅ m/sec
1kg ⋅ m/sec 2 ⋅ N
Aerated power input, Pg
P
 Pg 
d 
log10   = −192  i 
 dT 
 P
 0.3 m 
= −192 

 1m 
4.38
4.38
 di2 N 


 υL 
0.115
d 
1.96 i 
 dT 
 di N 2 


 g 
 (0.3 m) 2 (4.0 rev/sec) 


-6
2
 (0.746 x 10 m /sec) 
0.115
 Qg 
 3 
 di N 
 0.3 m 
1.96

 1m 
 (0.3 m)(4 rev/sec) 2 


9.8 m/sec 2




0.02 m3 /sec


3
 (0.3 m) (4 rev/sec) 
=-0.539
Pg =10(-0.539) (P)=(0.289)(927.2 N ⋅ m/sec)=268 N ⋅ m/sec
Volumetric mass transfer coefficient, kLa
0.4
P
−2  g 
( k L a=
)O2 (2.6 × 10 )   (u gs )0.5
V 




P
Q


g
g

= (2.6 × 10−2 )   
2
 V   π dt 


 4 
b. O2 transfer rate, WA
0.4
0.5


0.4 
3
0.02 m /sec 
 268 N ⋅ m/sec 
=(2.6 x 10-2 ) 

 
3
2
2.0 m

  π(1.0 m) 




4
0.5
= 0.0294 sec-1
 PO

WA = k L a (C A* − C A∞ )V = k L a  2 CL − 0  V
 H

 0.21 atm 
3 
3
=(0.0294 sec-1 ) 
 (55.6 mol/L)(1000 L/m )  (2.0 m )
4
5.05
x
10
atm



= 0.01359 mol/sec = 0.82 mol/min
30-42
30.24
Let A = CO2, B = H2O (liquid)
a. Determine kLa
=
Sc
=
S
L
993×10-6 kg/m-s
µ
=
= 562
ρ DAB (998.2 kg/m3 )(1.77×10-9 m 2 /s)
π D2
π (0.25) 2
=
= 0.0491m 2 = 0.534 ft 2
4
4
1
 5.0 kgmol   18 kg   2.204 lb   60 min  

=22,288 lb/ft 2 ⋅ hr





2 
 min   kgmol   kg   hr   0.534 ft 
µ = 993 x 10-6 kg/m-sec = 2.40 lb/hr ⋅ ft
-9
=
=
m 2 /s 6.86×10-5 ft 2 /hr
DAB 1.77×10
1− n
1− 0.22
 22,288 lb/hr ⋅ ft 2 
L
1/2
-5
2
k L a D=

AB 
 ( Sc) (α ) (6.86×10 ft /hr) 
µ
 2.40 b/hr ⋅ ft 
= 202
=
hr -1 0.056 sec-1
(562)1/2 (100)
b. Determine length required, L

m3
1
 5.0 kgmol  
  1min 
v∞ = 


 =0.0306 m/s
2 
 min   55.5 kgmol   0.0491 m  60 s 
dC A
k L a ( C A* − C A ) =
v∞
dz
v C AL dC A
L= ∞ ∫
k L a 0 ( C A* − C A )
L
 ( 0.0306 m/sec)  1 
v∞ 
C A*
ln  *
ln 
=
 = 1.64 m
* 
k L a  C A − 0.95C A 
(0.056 sec-1 )
 1-0.95 
30-43
30.25
Let A = O2, B = blood
a. Determine kLa
Liquid mass flow rate per cross-sectional area of the empty tower (L), lbm/ft2-hr
νL
0.040cm 2 / sec
=
= 2000
Sc =
DAB 2.0 x10−5 cm 2 / sec
w ρ LV0 (1.025 g/cm3 )(300 cm 2 /min)(1.0 lb m /454g)(60 min/1 hr)
= =
=
L
= 12017 lb m /ft ⋅ hr
A π D2
 π(2.0 cm) 2 
2

 (1ft/30.48 cm)
4
4


1lb m 30.48 cm 3600 sec
(1.025 g/cm3 )(0.040 cm 2 /sec)
= 9.9 lb m /ft ⋅ hr
=
µ L ρ=
Lν L
454 g
1ft
1hr
DAB = (2.0 x 10-5 cm 2 /sec)(1ft/30.48 cm) 2 (3600 sec/1 hr) = 7.75 x 10-5 ft 2 /hr
1− n
 L 
kL a
=α

DAB
 µL 
Sc 0.5
1− n
 L 
kL a α 
=

 µL 
Sc 0.5 ⋅ DAB
1-0.46
 12017 lb m /ft 2 hr 
k L a = 550 

 9.91 lb m /ft-hr 
20000.5 ( 7.75 x 10-5 ft 2 /hr ) = 88.2 hr -1
b. Determine CAL,out
Assumption: 1) No reaction, 2) Steady state, 3) Dilute system, 4) kL ≈ KL
Material balance for species A on a differential volume element
In – Out +Generation = Accumulation (liquid phase – mole A/time)
π D2
V0 C AL + N A ⋅ a ⋅
⋅ ∆z − V0 C AL
+ 0 =0
z
z +∆z
4
÷ Δz, rearrangement, ∆z → 0
2
 dC AL + N ⋅ a ⋅ π D =
V
0
0
A
dz
4
dC AL π D 2
*
k L a (C AL
+
− C AL ) =
0

dz
4V0
−∫
C AL ,out
C AL ,o
L
dC AL
π D2
k L a ∫ dz
=
*
0
C AL − C AL
4V0
30-44
 C * − C AL ,0  π D 2
ln  *AL =
k a⋅L

 C −C
 L
4
V
AL
AL
out
,
0


 −π D 2

*
*
C AL ,out =C AL
kL a ⋅ L 
− (C AL
− C AL ,o ) exp 

 4V0

PA
k ⋅ PA
*
C AL
=
+
C AL ,max =
H 1 + k ⋅ PA
Also,
1atm
(28 atm -1 )(1 atm)
+
( 9.3 molO2 /m3 ) =10.23 molO2 /m3
0.8 atm-m3 /mol 1+(28 atm -1 )(1 atm)
 -π(0.02 m) 2 (88.2 hr -1 )(0.5 m) 
C AL ,out =10.23 mol O 2 /m3 - (10.23 mol O 2 /m3 -1.0 mol O 2 /m3 ) exp 

-4
3
 4(3 x 10 m /min)(60 min/1hr) 
C AL ,out = 5.96 mol O 2 /m3
c. Performance comparison
The packed bed tower performs better than the wetted wall column in terms of overall mass
transfer because of the increase in surface area for interphase mass transfer when using packed
bed.
30-45
30.26
Let A = Cu+2, B = H2O
a. Determine kc
2
d 2ω ( 8 cm ) ( 2.0 rev/sec ) 2π
Re =
= 80425
=
v
0.01 cm 2 /sec
ν
0.01 cm 2 /sec
= 833.33
Sc = L =
DAB
1.20 x 10-5 cm 2 /sec
kc d
0.62 Re1/2 Sc1/3
Sh =
=
DAB
DAB
d
D
1.2 x 10-5 cm 2 /sec
0.62
Re1/2 Sc1/3 AB 0.62(80425)1/2 (833.33)1/3
=2.48 x 10-3 cm/sec
=
8 cm
d
kc = 0.62 Re1/2 Sc1/3
kc
b. Determine NA
=
N A kc (C A,∞ − C As ) (1)
NA =
ks C As
− RA =
(2)
Combine (1) and (2)
kc (C A,∞ − C As ) =
ks C As
kc
C A, ∞
kc + k s
Combine (2), (3)
kk
N A = c s C A, ∞
kc + k s
C As =
(3)
( 2.48 x 10 cm/sec ) ( 3.2 cm/sec ) (0.005 gmole/L)(1L/1000 cm ) = 1.24 x 10 mol
N =
cm sec
( 2.48 x 10 cm/sec ) + ( 3.2 cm/sec )
-3
3
A
-8
2
-3
Since ks >> kc, boundary layer diffusion controls
c. Determine time required, t
Mass Balance on Cu2+:
VC A,0 + mA,0 = VC A,t + mA,t , also mA,0 = 0 and mA,t =
ρA π d 2
MA 4
lA
30-46
VC A,0 −
C A, f =
ρA π d 2
MA
V
4
lA
 8.96 g/cm3   π(8 cm) 2 
( 500 cm )( 5 x 10 mole/cm ) -  64 g/gmole   4  ( 2 x 10-4 cm )



=
500 cm3
C A, f = 2.185 x 10-3 mol/L
3
-6
3
Assumptions: 1) Unsteady state, 2) Dilute system, 3) Well-mixed liquid phase
Unsteady-state material balance for species A on well-mixed liquid phase
In – Out + Generation = Accumulation (liquid phase – mole A/time)
dC A
πd2
V
0 − NA ⋅
=
dt
4
CA, f
t
2
π d kc k s
1
−
dt =
dC A
4V kc + ks ∫0
C A,∞ C∫A ,0
ln
C A, f
C A,0
=
−π d 2 kc k s
t
4V kc + k s
 C A, f  4V kc + k s
t = − ln 
 C  π d 2 k k
c s
 A,0 
 2.185 x 10-3 mol/L  4(500 cm3 ) ( 2.48 x 10 cm/sec ) + ( 3.2 cm/sec )
t = -ln 

2
-3
 0.005 mol/L  π ( 8.0 cm ) ( 2.48 x 10 cm/sec ) ( 3.2 cm/sec )
-3
= 3323 sec = 55 min
30-47
30.27
Let A = O2, B = H2O
Liquid film mass transfer coefficient, kL
Re
=
v∞ D (5.0 cm/sec)(1.0 cm)
= 548
=
νL
9.12 x 10-3 cm 2 /sec
νL
9.12x10−3cm 2 / sec
=
= 434
2.1x10−5 cm 2 / sec
DAB
Sieder-Tate correlation for laminar flow of liquid flowing inside of a tube (function of tubing
length L)
Sc
=
1/3
1/3
kL D
D

D
 D 
=
→ k L 1.86  Re⋅ Sc   AB 
Sh =
1.86  Re⋅ Sc  =
DAB
L

L
  D 
-5
2
 1.0cm
  2.1 x 10 cm /sec 
-3 -1/3
(548)(434)  
k L = 1.86 
 =2.42 x 10 L cm/sec
L
1.0
cm

 

1/3
Determine the overall mass transfer coefficient KL
1
1 1
=
+
K L k L km

atm ⋅ m3 
-6
2
0.78

 ( 5.0 x 10 cm /sec )
gmole
H ⋅ DAe

= 
= 6.72 x 10-4 cm/sec
km =
3
S m ⋅ lm (0.029 atm ⋅ m silicone/gmole)(0.20 cm)
−1
 1


1 
1
1
+
KL =
 +  =


−3 −1/3
−4
 2.42 x10 L (cm / sec) 6.72 x10 cm / sec 
 k L km 
1
(cm / sec)
KL =
1/3
413.223L + 1488.1
−1
Mass transfer model
Assumptions: 1) Steady state, 2) Dilute system, 3) Concentration along z-direction only, 4) No
reaction
Material balance on differential volume element
In – Out +Generation = Accumulation (liquid phase – mole A/time)
π D2
v∞ C AL + N A ⋅ π D ⋅ ∆z −
π D2
4
4
z
÷ Δz, rearrangement, ∆z → 0
+ 0 =0
v∞ C AL
z +∆z
30-48
dC AL
4
+ NA ⋅
=
0
dz
v∞ D
dC
4
*
K L (C AL
− AL +
− C AL ) =
0
dz
v∞ D
−
−∫
C AL , out
C AL , o
L
dC AL
4 L
4
1
(cm / sec)dz
K
dz
=
=
L
*
1/3
∫
∫
v∞ D 0 413.223 z + 1488.1
C AL − C AL v∞ D 0
*

 C AL
 3 L + 3.6012  
− C AL ,0   v∞ D 
2/3
3
f
L
L
L
−
=
=
−
+
ln  *
(
)
0.00363
0.0261
0.0942
ln
 


 
C −C
4 

AL , out  
 3.6012  
 AL
Determine the length of tubing (L)
*
C AL
=
PA
=
H
2.0 atm
= 2.56 gmol/m3
3
atm-m
0.78
gmol
*
 C AL
− C AL ,0   v∞ D 
 (2.56 - 0) gmole/m3   5.0 cm × 1.0 cm 
2
ln  *
ln
−
=

 = 0.446 cm

3 
C −C
  4 
4
(2.560
.768)
gmole/m



AL , out 
 AL

 3 L +3.6012  
0.446cm 2 = 0.00363L2/3 -0.02613 L +0.0942ln 
 

 3.6012  
Using Solver, L = 23.9 m
30-49
30.28
NA
yA(z)
v∞
yAs = 1.0
100% O2
yAo = 0
shell side
100% H2
H2 permeable tube wall
tube side
yAL
v∞
Di = 1.5 cm Do = 2.5 cm
Re = 10,000
z = 0 cm
z = 10 cm
a. v∞ on shell side (annular space) at Re =10,000
D=
Do − Di = 2.5 cm - 1.5 cm = 1.0 cm
e
Re =
v∞
v ∞ De
νB
( 0.159 cm /s ) (10,000 ) = 1590 cm 15.9 m
=
=
ν B Re
De
2
(1.0 cm )
s
s
b. Sc for shell side
A = H2 (g), B = O2 (g)
DAB = 0.697 cm2/s at 273 K, 1.0 atm (Appendix J)
1.5
1.5
 T 

cm 2   300 K 
cm 2
=
=
(
,
)
0.697
0.803
DAB (T , P) D=
T
P




AB
ref

 Tref 
s   273 K 
s



cm 2
0.159
νB
s = 0.198
=
Sc =
cm 2
DAB
0.803
s
c. kc for shell side
Re > 2000 turbulent flow through tube
30-50
0.83
=
=
=
Sh 0.023Re
Sc1/3 0.023 (10, 000
) ( 0.198) 28
0.83
Sh =
1/3
kc De
DAB

cm 2 
28
0.803
( )

s 
Sh ⋅ DAB
cm

= 22.5
=
kc =
s
De
(1.0 cm )
d. Model for yA(z) shell side
Differential material balance on dissolved solute A in liquid phase
IN – OUT + GEN = ACCUMULATION (moles A/time)
π ( Do2 − Di2 )
π ( Do2 − Di2 )
cA z − v∞
c A z +∆z + N Aπ Di ∆z + 0 =0
4
4
=
N A k y y As − y A assume that yAs ≈1.0 at outer surface of inner tube (constant)
Note
v∞
(
)
cA = yAC if constant T and P, then C is constant
ky = kcC
v∞
π ( Do2 − Di2 ) C
4
y A z − v∞
π ( Do2 − Di2 ) C
4
(
)
y A z +∆z + k y y As − y A π Di ∆z + 0 =0
Rearrange, lim Δz→0
 y A z +∆z − y A z 
4kc CDi
−
−
0
y As − y A =
∆z
v∞ Do2 − Di2 C
(
)
(
)
Differential model
−
4kc Di
dy A
−
dz v∞ Do2 − Di2
β=
(
4kc Di
(
v∞ Do2 − Di2
)
0
(y − y )=
As
A
)
z = 0, yA = yAo (entrance); z = L, yA = yAL
Integration
y AL
L
dy A
= − β ∫ dz
∫
y Ao ( y As − y A )
0
30-51
Integrated model
 y − y Ao 
ln  As
 = −β L
 y As − y AL 
e. yAL


4kc Di

y AL =y As − ( y As − y Ao ) exp  −
L
2
2
 v ∞ ( Do − Di ) 


cm 

4  22.5
 (1.5 cm )(10 cm ) 

s 


y AL =1.0 − (1.0 − 0) exp  −
2
2
 ( 2.5 cm ) - (1.5 cm ) (1590 cm/s ) 


y AL = 0.191
(
)
30-52
31.1
a. Determine KLa for O3 transfer
3
m3  3.28 ft   1.0 hr 
ft 3
Qg = 17.8

 
 =10.5
hr  1.0 m   60 min 
min
At depth = 10.5 ft, from Eckenfelder plot
KL
A
ft 3
V ≅ 400
V
hr
K L ( A / V )V
K La =
n
=
V
( 400 ft /hr ) (8 spargers ) =1.132 hr
ft 
(80 m )  3.28

1.0 m 
3
-1
3
3
1/ 2
1/ 2
K L a O  DO − H O 
 1.7 x 10-5 cm 2 /sec 
3
3
2
=
0.90
=

=

-5
2
K La O
2.1
x
10
cm
/sec
D




−
O
H
O
 2 2 
2
K L a O = (1.132 hr -1 ) ( 0.9 ) =1.02 hr -1
3
b. Estimate time required for CA = 0.25 CA* (A = O3)
Ptop + Pbottom
1
P=
= Ptop + ρ L gh
ave
2
2


1  1000 kg  9.81 m 
1.0 atm
Pave = 1.0atm + 
3.2 m ) 
=1.155 atm

3
2 (
5
2 
2 m
 sec 
 1.0123 x 10 kg/m ⋅ sec 
=
PA y=
AP
( 0.04 )(1.155atm ) = 0.0462 atm
gmol
mg
PA
0.0462 atm
=
= 0.683 3 = 327
3
m
L
H 0.0667atm ⋅ m /gmol
gmol
mg
C A,t = 0.150 3 =7.2
m
L
 C A* − C Ao   1 
 ( 0.683 - 0.0 ) gmol/m3  
1

t ln=
ln
= 0.246 hr = 886 sec
 *



3 
-1 
 ( 0.683 - 0.150 ) gmol/m   1.02 hr 
 C A − C A,t   K L a 
*
C=
A
31-1
31.2
a. Required KLa values for H2S transfer and O2 transfer at t = 2.5 hr and CA = 0.050 gmole/m3
(A = H2S)
 C * − C A0 
ln  A*

CA − CA 

K La =
t
PA
*
=
=
C A* x=
C 0 (=
PA 0)
AC
H
 0.0 - 0.30 gmol/m3 
ln 
0.0 - 0.05 gmol/m3 

= 0.717 hr -1
KLa H S =
2
2.5 hr
KLa H S
0.717 hr -1
2
KLa O
=
=
= 1.075 hr -1
1/ 2
1/2
2
-5
2
 DH 2 S − H 2 O 
 1.4 x 10 cm /sec 


 2.1 x 10-5cm 2 /sec 


 DO2 − H 2 O 
b. Aeration rate to each sparger (based on O2 trasnfer)
 A  K L ( A / V )V
KL   =
n
V
V 
K L ( A / V ) V (15 spargers)
1.075 hr -1 =
(425 m3 ) (3.28 ft/1.0 m)3
K L ( A / V ) V = 1075 ft 3 /hr
Using the Eckenfelder plot, at 3.2 m (10.5 ft) depth, Qg = 20 SCFM
31-2
31.3
Given:
Counter-Current Flow
Co-Current Flow
XA2 = 0
xA2 = 0
L2
Ls = 1.4 Ls,min
YA2
yA2
G2
Gs
R = 0.80
XA1
xA1
L1
Ls = 1.4 Ls,min
YA1 = 0.05
yA1
G1 = 10 lbmole/ft2-hr
Gs
YA2
yA2
G2
Gs
R = 0.80
YA1 = 0.05
yA1
G1 = 10 lbmole/ft2-hr
Gs
XA2
xA2
L2
Ls = 1.4 Ls,min
XA1 = 0
xA1 = 0
L1
Ls = 1.4 Ls,min
a. See YA-XA plots after parts (b) and (c) are completed
b. Determine Ls,min, Ls, XA1 for counter-current flow
YA 2= YA1 (1 − R )= 0.05 (1 − 0.8 )= 0.010
y A1
=
YA1
0.05
= = 0.0476
1 + YA1 1.05
Gs = G1 (1 − y A1 ) = 10 (1 − 0.0476 ) = 9.52
lbmol
ft 2 h
X A1,min 0.23 from YA − X A plot
=
YA1Gs + X A 2 Ls ,min =
YA 2Gs + X A1,min Ls ,min
 YA1 − YA 2 
 0.050 − 0.010 
Ls ,min =
Gs 
=

 ( 9.52 )
X

 0.23 − 0.0 
 A1,min − X A 2 
lbmol
Ls ,min = 1.66 2
ft h
Ls
=
=
(1.4
)(1.66 ) 2.32
lbmol
ft 2 h
31-3
YA1Gs + X A 2 Ls =YA 2 Gs + X A1 Ls
9.52 ) + 0 ( 0.01)( 9.52 ) + X A1 ( 2.32 )
( 0.05)(=
X A1 = 0.164
0.06
YA1
0.05
0.04
0.03
YA
0.02
0.01
YA2
XA1
0.00
0.00
0.05
0.10
0.15
0.20
XA1,min
0.25
0.30
XA (mole A / mole solvent)
c. Determine Ls,min, Ls, XA1 for cocurrent flow
=
At YA1 0.01,
=
X A1,min 0.03
YA 2Gs + X A 2 Ls ,min =
YA1Gs + X A1,min Ls ,min
9.52 ) + 0 ( 0.01)( 9.52 ) + ( 0.03) ( Ls ,min )
( 0.05)(=
Ls ,min = 12.7
=
Ls
lbmol
ft 2 h
=
(1.4
)(12.7 ) 17.78
lbmol
ft 2 h
YA 2Gs + X A 2 Ls =YA1Gs + X A1 Ls
9.52 ) + 0 ( 0.01)( 9.52 ) + X A1 (17.78 )
( 0.05)(=
X A1 = 0.0214
31-4
31.3 cont.
0.06
0.05
XA2,YA2
0.04
0.03
YA
0.02
0.01
YA1
XA1,min
0.00
0.00 XA1 0.05
0.10
0.15
0.20
0.25
0.30
XA (mole A / mole solvent)
31-5
31.4
a. Determine AGs,min
x A1 AL1
(1 − R ) xA2 AL2 =
lbmol
hr
0.2
x A1 AL1 =
(1 − 0.8)( 0.01)(100 ) =
ALs = AL2 (1 − x A 2 ) =
(100 )(1 − 0.01) = 99
lbmol
hr
 x

x A1 AL1 = ALs  A1 
 1 − x A1 
 x

0.20 = ( 99 )  A1 
 1 − x A1 
x A1 2.02 ⋅ 10−4
=
AL1
=
AL s
99
lbmol
99.02
=
=
−4
1 − x A1 1 − 2.02 ⋅ 10
hr
y A1 AG1,min + x A 2 AL2 = x A1 AL1 + y A 2,min AG2,min
*
y=
A 2,min
H
 515 atm 
xA2 
=
 ( 0.01) = 0.412
PT
 12.5 atm 
0 + ( 0.01)(100 ) =
( 0.20 ) + ( 0.412 ) ( AG2,min )
AG2,min = 1.94
lbmol
hr
lbmol
1.14
AGs ,min =
(1 − 0.412 )(1.94 ) =
(1 − y A2,min ) AG2,min =
hr
b. Determine yA2,min
From part (a) above,
y*A 2,min = 0.412
31-6
31.5
a. Determine yA2,min, see YA-XA plot
xA2
0.107
=
= 0.12
X A2 =
1 − x A 2 1 − 0.107
kgmol
ALS =
1.786
(1 − xA2 ) AL2 =
(1 − 0.107 )( 2.0 ) =
sec
X A1 =
0.020
(1 − R ) X A2 =
(1 − 0.833)( 0.12 ) =
YA 2,min = 0.195 (see YA -X A plot)
=
YA1 AGs ,min + X A 2 AL
YA 2,min AGS ,min + X A1 ALs
s
0 + ( 0.12 )(1.786=
)
AGs ,min = 0.916
=
y A 2,min
( 0.195) ( AGS ,min ) + ( 0.02 )(1.786 )
kgmol
sec
YA 2,min
0.195
= = 0.163
1 + YA 2,min 1 + 0.195
b. Determine AG1 and yA2
YA1 = 0
=
=
=
AG
AG
1.5 ( AGs ,min
)
s
1
=
) 1.374
(1.5)( 0.916
kgmol
sec
YA1 AGs + X A 2 ALs =YA 2 AGs + X A1 ALs
YA 2 (1.374 ) + ( 0.02 )(1.786 )
0 + ( 0.12 )(1.786 ) =
YA 2 = 0.13
=
y A2
YA 2
0.130
=
= 0.115
1 + YA 2 1.0 + 0.130
AG2 =
AGs
1.374
=
= 1.55 kgmol/sec
1 − y A 2 1.0-0.115
31-7
31.5 continued
0.25
0.20
YA2,min
0.15
YA2
YA
0.10
0.05
0.00
0.00
XA2
XA1, YA1
0.05
0.10
0.15
XA (mole dissolved NH3 / mole water)
31-8
31.6
a.
Determine L2 and mole fraction compositions of terminal streams
(1 − y A1 ) G1 ( 5)(1 − 0.12 )
kgmol
=
= 4.44 2
1 − y A2
m hr
(1 − 0.01)
=
G2
y A1G1 + x A 2 L2 = y A 2 G2 + x A1 L1
( 0.12=
)( 5) + 0 ( 0.01)( 4.44 ) + ( 0.03) L1
L1 = 18.519
kgmol
m 2 hr
Ls = L1 (1 − x A1 ) = (18.519 )(1 − 0.03) = 17.96
kgmol
m 2 hr
Ls = L2
=
y A 2 0.01,=
x A 2 0,=
y A1 0.12,=
x A1 0.03
b. See YA-XA plot, all mole fraction values scaled to mole ratios
c. Determine packing height z
z = H OG N OG
G1 + G2
kgmol
kgmol
= 4.72 2 , K y' a = K G' a ⋅ PT = 2.0 3
2
m hr
m hr
kgmol 

4.72 2 

G
m hr 

=
=
H
= 2.36 m
OG
'
kgmol 
Kya 
 2.0 3 
m hr 

y −y
N OG = A1 * A 2
( yA − yA )
G=
lm
y
*
A2
= 0.0 at x A2 = 0.0
y*A1 = 0.09 at x A1 = 0.090
( y − y ) − ( y − y ) ( 0.12 − 0.090 ) − ( 0.01 − 0 )
= = 0.018
(y − y ) =
(y − y )
 ( 0.12 − 0.090 ) 
ln 


ln 
 ( 0.010 − 0 ) 
 ( y − y ) 
A
*
A lm
A1
*
A1
A2
A1
*
A1
A2
*
A2
*
A2
0.12 − 0.01
= 6.042
0.018
z = ( 6.042 )( 2.36 m ) = 14.26 m
=
N OG
31-9
31.6 continued
0.15
XA1, YA1
0.10
YA
0.05
XA2, YA2
0.00
0.00
0.01
0.02
0.03
0.04
0.05
XA (mole A / mole solvent)
31-10
31.7
Given:
XA2 = 0
xA2 = 0
L2
Ls = 2.0 Ls,min
YA2
yA2 = 0.005
G2
Gs
R
XA1
xA1
L1
Ls = 2.0 Ls,min
YA1
yA1 = 0.06
G1 = 10 lbmole/ft2-hr
Gs
a. Molar flowrate and mole fraction composition of all terminal streams
x A 2 0,=
y A 2 0.005,
y A1 0.06
=
=
x A1,min 0.00085 from y A − x A plot
=
lbmol
ft 2 h
Gs
9.4
lbmol
G2 =
=
= 9.45 2
ft h
(1 − y A2 ) (1 − 0.005)
Gs = G1 (1 − y A1 ) = 10 (1 − 0.06 ) = 9.4
y A1G1 + x A 2 L2 = y A 2G2 + x A1,min L1,min
L1,min =
L1,min
y A1G1 + x A 2 L2 − y A 2G2
x A1,min
0.06 )(10 ) + 0 − ( 0.005 )( 9.45 )
(=
lbmol
650.3
0.00085
ft 2 h
Ls ,min = L1,min (1 − x A1,min ) = ( 650.3)(1 − 0.00085 ) = 649.74
lbmol
ft 2 h
31-11
Equilibrium Line
xA
kg SO2 /
PA
100 kg H2O
(mm Hg, SO2)
0.00
0
0.20
0.30
yA
T=
PT =
o
30 C
1 atm
0.00000
0.000
Mw, SO2 =
64 g/gmole
29
0.00056
0.038
Mw, H2O =
18 g/gmole
46
0.00084
0.061
0.50
83
0.00140
0.109
0.70
119
0.00196
0.157
0.16
0.14
0.12
0.10
yA0.08
xA1, yA1
0.06
0.04
0.02
xA1,min
xA2,yA2
0.00
0.00E+00
5.00E-04
1.00E-03
1.50E-03
2.00E-03
xA (mole fraction SO2 in water)
lbmol
ft 2 h
y A1G1 + x A 2 L2 = y A 2G2 + x A1 L1
=
Ls 2=
Ls ,min 1299.48
=
( 0.06
)(10 ) + 0 ( 0.005)( 9.45) + ( xA1 )( L1 )
x A1 L1 = 0.5528
G1 + L2 = G2 + L1
L1 =
10 + 1299.48 − 9.45 =
1300
lbmol
ft 2 h
x A1 L1 0.5528
=
= 4.25 ⋅10−4
L1
1300
Material Balance Summary:
G2 (lbmol/ft2 h)
9.45
G1 (lbmol/ft2 h)
10
2
L2 (lbmol/ft h)
1299.48
L1 (lbmol/ft2 h)
1300
x1
0.000425
x2
0
y1
0.06
y2
0.005
=
x A1
31-12
b. Packing height z
z = H OG N OG
( y A1 − y A2 )
=
( y A1 − y*A1 ) − ( y A2 − y*A2 )
N OG
 ( y A1 − y*A1 ) 

ln 
*
 ( y A 2 − y A 2 ) 
H OG =
=
G
( 0.06 − 0.005)
= 3.94
( 0.06 − 0.03) − ( 0.005 − 0 )
 ( 0.06 − 0.03) 
ln 

 ( 0.005 − 0 ) 
G
K y' a
G1 + G2
lbmol
= 9.73 2
2
ft h
m ' 1 ( 70.5 )
1
1
= +
= +
K y a k y a k x a 15
250
K y a = 2.87
lbmol
ft 3 h
lbmol
lbmol
(1 − 0.06 )= 2.70 3
3
ft h
ft h
lbmol
lbmol
Top: K y' a = K y a (1 − y A 2 ) = 2.87 3 (1 − 0.005 ) = 2.85 3
ft hr
ft h
lbmol
Average: K y' a = 2.78 3
ft h
lbmol
9.73 2
ft h =3.51 ft
H OG =
lbmol
2.78 3
ft h
z = ( 3.51 ft )( 3.94 ) =13.8 ft
Bottom: K y' a= K y a (1 − y A1 )= 2.87
31-13
31.8
a. Determine xA1

std m3   1.0 kmol 
kmol
AG1 =  880
= 39.3

3 
hr   22.4 std m 
hr

AGs = AG1 (1 − y A1 )=
AG2 =
( 39.3)(1-0.01) =38.9
( 38.9 )
AGs
kmol
=
=38.9
1 − y A 2 (1-0.0005 )
hr
kmol
hr
AG1 + AL2 = AG2 + AL1
39.3 + 50 = 38.9 + AL1
kmol
hr
y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1 AL1
AL1 = 50.4
=
( 39.3
)( 0.01) + 0 ( 38.9 )( 0.0005) + ( 50.4 )( xA1 )
x A1 = 0.0074
Aside: (1 − R) y A1G1 =
y A 2 G2
( 0.0005)( 38.9 )
(1 − R )( 0.01)( 39.3) =
R = 0.95 (95%)
b. See yA-xA plot
c. Determine Ls,min
=
ALs ,min AL1,min (1 − x A1,min )
y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1,min AL1,min
x A1,min = 0.028
( 39.3)( 0.01) + 0 = ( 38.9 )( 0.0005) + ( AL1,min ) ( 0.028)
AL1,min = 13.3
kmol
hr
ALs ,min = (13.3)(1-0.028 ) =12.9
A=
kmol
hr
π D2
4
Ls ,min =
ALs ,min
A
kmol 

12.9

kmol
hr 

=
= 4.1 2
2
m hr
 π ( 2.0 m ) 




4


31-14
d. Determine KGa a z = 10 m
z = H OG N OG
N OG
( y A1 − y A2 )
=
( y A1 − y*A1 ) − ( y A2 − y*A2 )
 ( y A1 − y*A1 ) 

ln 
 ( y A 2 − y*A 2 ) 
z
10 m
=
H=
= 3.22 m
OG
N OG 3.06
H OG =
( 0.01 − 0.005)
= 3.06
( 0.01 − 0.0005) − ( 0.005 − 0 )
 ( 0.01 − 0.0005 ) 
ln 

 ( 0.005 − 0 ) 
G
K y' a
G1 + G2 39.3+38.9
kgmol
=
=39.1 2
2
2
m hr
kgmol
39.1 2
G
'
m hr 12.15 kgmol
=
K=
ya =
3.22 m
H OG
m3 hr
G=
K y' a
'
K=
=
Ga
P
kgmol
kgmol
m3 hr
= 3.04 3
1.0 atm 
m ⋅ hr ⋅ atm
( 405 kPa ) 

 101.25 kPa 
12.15
31-15
31.8 continued
Equilibrium Line
T=
P=
MEA Conc. =
MW MEA =
PA
mm Hg
40 oC
4.0 atm
15.3
61
wt% MEA =
g/gmole
mol
H2S/
mol
MEA
mol H2S/
mol
H2S+MEA
5.061
mole%
MEA
xA
yA*
XA
YA*
0.000
0.000
0.000
0.0000
0.0000
0.0000
0.0000
0.960
3.000
0.125
0.208
0.111
0.172
0.0063
0.0094
0.0003
0.0010
0.0064
0.0095
0.0003
0.0010
9.100
0.362
0.266
0.0183
0.0030
0.0187
0.0030
43.100
0.643
0.391
0.0325
0.0142
0.0336
0.0144
59.700
0.729
0.422
0.0369
0.0196
0.0383
0.0200
106.000
0.814
0.449
0.0412
0.0349
0.0430
0.0361
0.04
0.03
0.02
yA
xA1, yA1
0.01
xA2, yA2
0.00
0.00
XA1,min
0.01
0.02
0.03
0.04
0.05
xA
31-16
31.9
a. Characterize molar flowate and mole fraction composition of all terminal streams
kgmol
kgmol
AGs = AG2 (1 − y A1 )= 2.0
(1 − 0.10)= 1.8
sec
sec
AGS 1.8 kgmol/sec
kgmol
=
=1.836
AG2 =
1 − y A2
1-0.02
sec
AG1 + AL2 = AG2 + AL1
kgmol
AL1 = AG1 + AL2 − AG2 = 2.0 - 1.837 + 3.0 = 3.163
sec
AG1 y A1 + AL2 x A 2 = AG2 y A 2 + AL1 x A1
(2.0)(0.10) + (3.0)(0.01)
= (1.837)(0.020) + (3.163) x A1
x A1 = 0.061
Material Balance Summary:
AG1 (kgmol/sec) 2.0
AG2 (kgmol/sec) 1.84
AL1 (kgmol/sec) 3.163
AL2 (kgmol/sec) 3.0
yA1
0.1
yA2
0.02
xA1
0.061
xA2
0.01
b. Tower diameter (D) at 50% of flooding condition
Cf = 98 for 1.0 inch ceramic Intalox saddles (Table 31.2)
=
AL '1 AL1 ( x A1 M A + (1 − x=
A1 ) M B )
( 0.061(17)+(1-0.061)(18) )( 3.163) =56.75 kg/sec
1/2
kg 

kg


1/2
2.8 3
 56.75



AL1'  ρG 
sec 

m
0.056
x − axis
=
=


=

AG1  ρ L − ρG 
kgmol  
kg  1000 kg − 2.8 kg 


 2.0
  26.9
m3
m3 
kgmol  
sec  

0.25
y − axis =
Y=
1/2

  ρ ( ρ − ρ ) g 
( 2.8)(1000 − 2.8)(1.0 ) 
kg
0.25 =

3.77 2
G
Y  G L 0.1 G c  
=
0.1


C f µL J
m sec

 
( 98)( 0.001) (1) 
 
kg
kg
53.8
53.8
AG1'
sec =14.27 m 2 at flooding,
sec = 28.54 m 2 at 50% of flooding
=
=
A =
A
'
kg
kg
Gf
3.77 2
1.885 2
m sec
m sec
1/2
'
f
31-17
1/2
 4A 
D=

 π 
1/2
 4 ⋅ 14.27 m 2 
=
 = 4.25 m at flooding
π


1/2
 4 ⋅ 28.54 m 2 
D=
 = 6.03 m at 50% of flooding
π


c. Log-mean mass transfer driving force (yA – yA*)lm
y A1 − y*A1 ) − ( y A 2 − y*A 2 )
(
*
( y A − y A )lm = 
y A1 − y*A1 ) 
(

ln 
 ( y A 2 − y*A 2 ) 
x A1 0.061,
y*A1 0.035
=
=
x A 2 0.01,
y*A 2 0.005
=
=
( y − y ) = ( 0.10 − 0.035) = 0.065
( y − y ) = ( 0.02 − 0.005) = 0.015
A1
*
A1
A2
*
A2
−y )
(y =
A
*
A lm
( 0.065 − 0.015)
= 0.034
 0.065 
ln 

 0.015 
0.12
XA1, YA1
0.10
0.08
YA0.06
0.04
XA2, YA2
0.02
0.00
0.00
0.02
0.04
0.06
0.08
XA (mole dissolved NH3 / mole water)
31-18
31.10
a. Determine L2
kgmol 
kgmol

AGs = AG1 (1 − y A1 )= 1.25
 (1 − 0.08 )= 1.15
hr 
hr

AGs
kgmol
1.1512
AG2 =
=
hr
(1 − y A2 )
y A1 AG1 + x A 2 AL2 = x A1 AL1 + y A 2 AG2
1.25 ) + 0 ( 0.02 )( AL1 ) + ( 0.001)(1.1512 )
( 0.08)(=
AL1 = 4.942
kgmol
hr
ALs = AL1 (1 − x A1 ) =
( 4.942 )(1 − 0.02 ) = 4.845
kgmol
hr
4.843
ALs
kgmol
=
= 4.845
(1 − x A 2 ) (1 − 0)
hr
b. See yA-xA plot
=
AL2
x A1,min = 0.35 from y A -x A plot
x A1 AL1
=
x A1,min
=
AL1,min
( 0.02 )( 4.942 )
0.35
= 0.282
kgmol
hr
ALs ,min = AL1,min (1 − x A1,min )= 0.282(1 − 0.35)= 0.184
kgmol
hr
c. Log-mean mass transfer driving force (yA – yA*)lm
(y − y )−(y − y )
(y − y ) = 
( y − y ) 
ln 
 ( y − y ) 
A
*
A lm
A1
*
A1
A2
A1
*
A1
A2
*
A2
*
A2
x A1 0.02,
y*A1 0.0
=
=
x A 2 0.00,
y*A 2 0.0
=
=
( y − y ) = ( 0.08 − 0.0 ) = 0.080
( y − y =) ( 0.0010 − 0.0=) 0.0010
A1
*
A1
A2
*
A2
−y )
( y=
A
*
A lm
( 0.080 − 0.0010 )
= 0.018
 0.080 
ln 
 0.0010 
31-19
Aside:
y A1
dy A
*
yA 2 y A − y A
N OG = ∫
m ≅ 0, y*A =mx A =0
N OG
=
 y A1 
dy A
 0.08 
ln
4.38
=
=
∫y y A − 0  y A2  ln =
0.001 
A2
y A1
d. Determine ∆P/z for tower diameter D = 0.50 m
Cf = 52 for 1.5 inch ceramic Intalox saddles (Table 31.2)
(
) (1.25) ( 0.08)( 36.5) + (1 − 0.08)( 2 )
=
AG1' AG1 y A1 M HCl + (1 − =
y A1 ) M H 2
AG1' = 5.95
(
kg
hr
) ( 4.94 ) ( 0.02 )( 36.5) + (1 − 0.02 )(18)
=
AL1' AL1 x A1 M HCl + (1 − x=
A1 ) M H 2O
AL1' = 90.75
kg
hr

(1.0 atm )  4.76
kg 

kgmol 
kg
 PT 

= 0.195 3
M w,ave =

3
m

m atm 
 RT 
 0.08206
 ( 298 K )
kgmol-K 

Flooding Correlation
ρG = 
1./2
0.195
AL'  ρG 
 90.75  

x − axis :
 = 

 = 0.21
' 
AG  ρ L − ρG 
 5.95   1033 − 0.195 
1/2
kg 

5.95 

AG
kg
hr   1hr 

8.418 ⋅10−3 2
y − axis : G ' = =
=
2 
A
m sec
 π ( 0.5m )   3600sec 


4


'
1
( G ') C f µ L0.1 J (8.418 ⋅10−3 ) ( 52 ) (1.2 ⋅10−3 )
= 9.33 ⋅10−6
=
Y =
ρG ( ρ L − ρG ) g c
( 0.195)(1033 − 0.195)(1)
2
2
0.1
∴∆P / Z << 50 Pa / m tower will not flood
31-20
31.10 contined
0.15
0.10
xA1, yA1
yA1
yA
0.05
0.00
0.00
xA1,min
xA2, yA2
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
xA (mole fraction HCl in water)
31-21
31.11
Given:
XA2
xA2 = 0.0020
L2 = 100 kgmole/hr
Ls
YA2
yA2 = 0.015
G2 = 12 kgmole/hr
Gs
1.5 inch
plastic
Pall rings
XA1
xA1 = 0.00020
L1
Ls
YA1
yA1
G1
Gs
a. AGs,min
Get yA2,min from equilibrium curve
H
 41.25 atm 
−3
y A 2,min =m ⋅ x A 2 = ⋅ x A 2 =
 ( 2.0 × 10 ) = 0.050
P
1.65
atm


Terminal stream material balance for AG2,min
x A 2 AL2 + y A1 AG1,min =
x A1 AL1 + y A 2,min AG 2,min
3
100 (1 − 2 ⋅10−=
AL
=
AL2 (1 − x A=
) 99.8
s
2)
=
AL1
ALs
=
(1 − xA1 )
kgmol
h
99.8
kgmol
99.82
=
−4
h
(1 − 2 ⋅10 )
x A 2 AL2 + y A1 AG 1,min =
x A1 AL1 + y A 2,min AG 2,min
( 2 ⋅10 ) (100 ) + 0 = ( 2 ⋅10 ) ( 99.82 ) + ( 0.05) ( AG )
−3
−4
2,min
AG2,min = 3.60
kgmol
hr
kgmol

AGs ,min
= AG2,min (1 − y A 2,min=
) 3.42
)  3.60 kgmol
 (1 − 0.05=
h 
h

31-22
b. Gf’ and tower diameter (D) at flooding for AG2 = 12 kgmol/h and yA2 = 0.015
Use Flooding Correlation for packed towers in counter-current gas–liquid flow
Cf = 39 for 1.5 inch plastic Pall rings (Table 31.2)
AL'2
kgmol  
kg 
kg

100
 1800
=
 18
h   kgmol 
h

kg 
kg

'
AG2=
AG2 ⋅ M w,ave= AG2 ( y A 2 M W , A + (1 − y A 2 ) M w,air =
) 12 kgmol
= 366.8
  30.56
hr  
kgmol 
hr





kg
kg
P
kg 
1.65atm
 2.1
=
=
, ρ L 1000 3
=
=
30.56
ρG M
w ,ave
3
3


m
m
RT
kgmol 
m atm 
  0.0821
 ( 293K ) 
kgmol K 


Flooding Correlation
1/2
kg  
kg


2.1 3
 1800 h  

AL'2  ρG 
m
x − axis
0.225
=
=
=




' 
kg
kg 
AG2  ρ L − ρG 
 366.8   (1000 − 2.1) 3 
h 
m 

y − axis, Y ≅ 0.13
1/2
Gas flooding velocity Gf’
 Y ⋅ ρG ( ρ L − ρG ) g c 
G =

C f ⋅ µ L0.1 ⋅ J


1/2
'
f
J = g c =1 for SI units
1/2
G
'
f
 0.13 ⋅ 2.1(1000 − 2.1)1 
kg
3.7 2
=

0.1
ms
 39 ⋅ ( 0.001) ⋅1 
Tower diameter D at the flooding condition
kg   h 

366.8  


AG
h   3600 s 
=0.027 m 2
A= ' =
kg 
Gf

 3.7 2 
m sec 

'
2
1/2
 4 ( 0.027 m 2 ) 
 4A 
 =0.2 m
D=
 = 

π
 π 


1/2
31-23
31.12
a. Determine xA1
AG1 + AL2 = AG2 + AL1
AL1 = AG1 + AL2 − AG2 = 32 + 18 − 30.7 = 19.3
lbmol
hr
y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1 AL1
( 0.05)( 32=
) + ( 0 ) ( 0.01)( 30.7 ) + ( xA1 )(19.3)
x A1 = 0.067
b. Determine packing height, z
z = H OG N OG
=
A
π D2
π(2.0 ft) 2
=3.1415 ft 2
4
=
4
( AG1 + AG2 ) / 2 A=
G
H
=
=
OG
K ya
 y + y A2 
K G a ⋅ P 1 − A1
2


( 32+30.7 ) lbmol/hr/2 ( 3.1415 ft 2 )

lbmol 
0.05 + 0.01  


 2.15 ft 3hr ⋅ atm  (1.4 matm ) 1 2




H OG = 3.41 ft
=
N OG
y A1 − y A 2
y A1 − y A 2
=
=
*
( y A − y A ) ( y A1 − y*A1 ) − ( y A2 − y*A2 )
lm
 ( y A1 − y*A1 ) 

ln 
*
 ( y A 2 − y A 2 ) 
z = (1.76 )( 3.41 ft ) = 6.01 ft
( 0.05 − 0.01)
= 1.76
( 0.05 − 0.0067 ) − ( 0.01 − 0 )
 0.05 − 0.0067 
ln 
 0.01 − 0 
c. Pressure drop (∆P/z) for 0.25-inch ceramic Raschig rings
Flooding Correlation
∆P / z = 300 N/m 5/8-inch ceramic Raschig rings
For 0.25" Raschig Rings, C f = 1600
G'
=
AG '1 1006 lb m /hr
=320.2 lb m /ft 2 hr
=
2
A
π (2.0 ft) /4
( G ) C µ J ( 320.2 ) ( 580 )( 2 ) (1.503) = 0.038
Y =
=
ρ (ρ − ρ ) g
( 0.11)( 55 − 0.11) ( 4.18 ⋅ 10 )
' 2
f
2
0.1
L
0.1
8
G
L
G
c
C f ,new
=
scaled y-axis,
Ynew Y=
old
C f ,old

( 0.038=
) 
 0.104
1600
 580 
∆P / z =
1200 N / m
31-24
31.13
a. Determine ALs,min
y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1,min AL1,min
H 0.46 atm
=0.023
=
2.0 atm
P
y A1 0.03
= =
= 0.130
x A1,min
m 0.23
AG1 (1 − y A1 ) 2.0(1-0.03)
lbmol
= 1.95
=
AG2 =
1-0.005
hr
(1 − y A2 )
=
m
=
( 0.03
)( 2 ) + 0 ( 0.005)(1.95) + ( 0.130 ) AL1,min
AL1,min = 0.387
lbmol
hr
ALs ,min = AL1,min (1 − x A1,min ) =
( 0.387 )(1-0.130 ) =0.336
lbmol
hr
b. Estimate required K’ya for z = 6.0 ft
z = H OG N OG
H OG =
Z
N OG
AL2 + AG1 = AL1 + AG2
1.0 + 2.0 = AL1 + 1.95
lbmol
hr
x A 2 AL2 + y A1 AG1 = x A1 AL1 + y A 2 AG2
AL1 = 1.05
0 + 0.03(2.0) = x A1 (1.05) + 0.005(1.95)
x A1 = 0.048
( AG1 + AG2 )
lbmol
= 1.975
2
hr
( AL1 + AL2 )
lbmol
=
AL = 1.025
2
hr
*
0.23(0.048) = 0.011
=
=
y A1 mx
A1
AG
=
0.23(0)
=
y*A 2 mx
=
= 0.0
A2
31-25
( y A1 − y A2 )
=
( y A1 − y*A1 ) − ( y A2 − y*A2 )
N OG
 ( y A1 − y*A1 ) 

ln 
 ( y A 2 − y*A 2 ) 
H OG =
( 0.03 − 0.005)
= 2.39
( 0.03 − 0.011) − ( 0.005 − 0 )
 ( 0.030 − 0.011) 
ln 

 ( 0.005 − 0 ) 
G
6.0 ft
=2.51ft = '
2.39
Kya
( AG1 + AG2 )
lbmol
= 10.07 2
2A
ft hr
lbmol
10.07 2
G
'
ft hr = 4.0 lbmol
=
Kya =
H OG
2.51 ft
ft 3 hr
=
G
c. New size of packing at 45% of flooding
 G1' 
 '  = 0.26
 G f old
G 'f ,old =
G1'
0.26
 G1' 
 '  = 0.45
 G f new
G1'
0.45
 G1' 


G 'f ,new  0.45 
=
=
G 'f ,old  G1' 


 0.26 
G 'f ,new =
1/2
 380 


 C f ,new 
C f ,new = 1138
Therefore 3/8 inch packing size is suitable
31-26
31.14
a. See yA-xA plot
b. Determine the overall mass transfer driving force (xA* – xA) and NOL
x A,2 − x A,1
N OL =
( xA − x*A )
lm
(x − x ) −(x − x )
(x − x ) = 
( x − x ) 
ln 
 ( x − x ) 
x − x = (1.6 ⋅ 10 − 8 ⋅ 10 )= 8.0 ⋅ 10
x − x = ( 2 ⋅ 10 − 0 ) = 2.0 ⋅ 10
(8.0 ⋅10 ) − ( 2.0 ⋅10 =) 6.0 ⋅10 = 4.33 ⋅10
( x − x )=
1.39
 ( 8.0 ⋅ 10 ) 

ln 
 ( 2.0 ⋅ 10 ) 
1.6 ⋅ 10 − 2 ⋅ 10 )
(=
=
3.235
N
A
A2
*
A2
A1
*
A1
A
*
A lm
*
A2
A2
*
A lm
*
A1
A1
A2
*
A2
A1
*
A1
−4
−5
−5
−5
−5
−5
−5
−5
−5
−5
−5
−4
−5
.433 ⋅ 10−5
OL
c. Determine packing height, z
z = H OL N OL
H OL =
L
K x' a
K x' a ≅ K x a
L=
( L1 + L2 )
2
L2 (1 − x A 2 )
Ls
lbmol
L1 =
=
= 99.99 2
ft hr
(1 − xA1 ) (1 − xA1 )
lbmol
ft 2 hr
1
1
1
=
+
K x a k x a mk y a
L = 100
31-27
lb 

62.4 3 

=
k x a k=
(17.4 hr -1 )  lbft  =60.32 lbmol
L aC L
ft 3 hr
 18

 lbmol 
lbmol
lbmol
k y a= kG a ⋅ P= 7.4 3
1.2 atm = 8.88 3
ft hr atm
ft hr
H 150 atm
=
= 125
m =
P 1.2 atm
1
1
1
=
+
K x a 60.32 (125 )( 8.88 )
lbmol
ft 3 hr
lbmol
100 2
ft hr =1.75ft
H OL =
lbmol
57.21 3
ft hr
z = (1.75ft )( 3.235 ) =5.65 ft
K x a = 57.21
2.00E-02
1.50E-02
xA2, yA2
1.00E-02
yA
5.00E-03
0.00E+00
0.00E+00
xA1, yA1
5.00E-05
1.00E-04
1.50E-04
2.00E-04
xA (mole fraction benzene in water)
31-28
31.15
a. See yA-xA plot
b. Determine xA2 for z = 4.0 ft
( 5.0 atm )
PA
*
=
=1.142 x 10-4
x=
A
4
H ( 4.38 x 10 atm )
lb   lbmol   min 
lbmol

AL1 ≈ AL2 ≈ AL ≈  200 m  
  60
 =666.7
min   18 lb m  
hr 
hr

lbmol
'
K=
k=
k=
194 3
(no gas phase mixture)
xa
xa
xa
ft hr
lbmol
666.7
AL
hr = 3.346 ft 3
=
AH=
OL
'
lbmol
kx a
194 3
ft hr
3
3.346 ft
=1.094 ft
H OL =
3.1415 ft 2
xA 2
 x*A − x A1 
dx A
4.0 ft
Z
ln  * =
=3.657
N
=
 =
OL
∫x x*A −=
xA
 x A − x A 2  H OL 1.094 ft
A1
 x*A − x A1   1.142 ⋅ 10−4 − 0 
e
=
=
 

−4
*
 x A − x A 2  1.142 ⋅ 10 − x A 2 
x A 2 1.113 ⋅ 10−4
=
3.657
1.20
yA1 = yA2 = 1.0
1.00
0.80
0.60
yA
0.40
0.20
xA1 = 1.113 x 10-4
0.00
0.0E+00
2.0E-05
4.0E-05
6.0E-05
8.0E-05
1.0E-04
1.2E-04
xA (mole fraction O2 in water)
31-29
31.16
a. Tower diameter (D) at ∆P/z = 200 Pa/m
Cf = 98 for 1.0 inch ceramic Intalox saddles (Table 31.2)
kg 
kg
 kgmol  
AG1' = 1
 =44
  44
min
 min   kgmol 
kgmol  
kg 
kg

AL'1 =  4.0
 = 72
 18
min   kgmol 
min





P 
kg  
2.0 atm
 = 3.66 kg
ρG M
=
=
 44

A
3

RT  kgmol  
m3
m atm 
293
K
  0.08206

(
)


kgmol K 


Flooding Correlation
1/2
AL'  ρG 
3.66
 72  

:
0.10
x − axis=

=


' 
AG  ρ L − ρG 
 44   998.2 − 3.66 
y − axis= 0.04 at ∆P/z = 200 Pa/m
(G ) C µ J
' 2
=
=
Y 0.04
1/2
f
0.1
L
ρG ( ρ L − ρG ) g c
 ρG ( ρ L − ρG ) g c 
G' = Y


C f µ L0.1 J


1/2
1/2
 0.04 3.66 998.2-3.66 
[ ][ ][
]  =1.722 kg

=
0.1
 [98] 993×10-6  [ 0.01] 
m 2 sec




kg   1min 

44

AG '  min   60 sec 
=
A =
= 0.426 m 2
kg 
G'

1.722 2

m sec 

1/2
 4 ⋅ 0.426 m 2 
 4A 
=
D =

 = 0.74 m

π
 π 


1/2
31-30
b. Estimate kLa by Sherwood and Holloway correlation
1− n
0.5
 µL 


 ρ L DAB 
cm 2
m2
ft 2
kg
lb
DAB = 1.92 x 10-5
= 2.4
= 1.92 x 10-9
=7.43 x 10-5
, μ L = 993 x 10-6
sec
sec
hr
m×sec
ft ×hr
α = 170, n = 0.28
 L 
kL a
=α

DAB
 µL 
lb   min 
kg  

 72
  2.2 kg   60 hr 
lb
min  


=2073 2
×
L= 
2
2
ft hr
ft 
( 0.426 m )

 3.28 
m

1-0.28
 2073 
k L a = 170 

 2.4 


993 x 10-6 kg/m ⋅ sec )
(


 ( 998.2 kg/m3 )(1.92 x 10-9 m 2 /sec ) 


1/2
( 7.43 x 10 ) = 37.4 hr
-5
-1
c. Estimate volume of packing required for separation
V =
A ⋅ H OL ⋅ N OL
kgmol   min 

4.0

  60

AL2
min  
hr 

A ⋅ H OL =
=
=0.1156 m3
kgmol
k L a ⋅ CL , r
( 37.4 hr -1 )  55.5 m3 
 x*A − x A 2 
 1.419 ⋅ 10−3

dx A
=
=
=
N OL ∫=
ln
ln
1.22



−3
−3 
*
*
1.419 ⋅ 10 − 10 
 x A − x A1 
xA 2 x A − x A
x A1
V = (1.22 ) ( 0.1156 m3 ) = 0.141 m3
31-31
30.17
a. G2
G2
=
G1 (1 − y A1 ) 10.0(1-0.05)
kgmol
= 9.6
=
m 2 hr
(1-0.01)
(1 − y A2 )
b. Ls,min at 40 oC and 100 oC
Develop equilibrium distribution plot in yA vs. xA coordinates
Sample calculation for xA (A = H2S)
For 15.3 wt% MEA, wMEA = 0.153
wMEA / M w, MEA
(1 − wMEA ) / M w,H O
2
xH 2 S
=
kg MEA
kg MEA+kg H 2 O
0.153/61
mol MEA
= 0.053
(1-0.153)/18
mol H 2 O
mol H 2S
=
mol H 2S+mol MEA+mol H 2 O
mol H 2S
mol MEA
mol H 2S
mol H 2 O
+1+
mol MEA
mol MEA
For 0.365 mol H2S/mol MEA
xA
mol H 2S
0.365
mol MEA
=
0.018
mol H 2S
mol H 2 O
0.365
+1+
mol MEA
0053 mol MEA
(see plot next page at 40 oC and 100 oC)
At 40 oC, xA1,min = 0.0315
=
Ls ,min L1,min (1 − x A1,min )
y A1G1 + x A 2 L2 = y A 2G2 + x A1,min L1,min
( 0.05)(10.0 ) + 0 = ( 0.01)( 9.6 ) + ( 0.0315) ( L1,min )
L1,min = 12.8
kgmol
m 2 hr
Ls ,min = (12.8 )(1 - 0.0315 ) =12.4
kgmol
m 2 hr
At 100 oC, xA1,min = 0.0090
31-32
=
Ls ,min L1,min (1 − x A1,min )
y A1G1 + x A 2 L2 = y A 2G2 + x A1,min L1,min
( 0.05)(10.0 ) + 0 = ( 0.01)( 9.6 ) + ( 0.009 ) ( L1,min )
L1,min = 44.9
kgmol
m 2 hr
Ls ,min = ( 44.9 )(1 - 0.009 ) =44.5
kgmol
m 2 hr
As temperature is lowered, minimum solvent rate goes down due to higher solubility of solute in
solvent.
T=
40 C
PA
yA
mol H2S/
xA
CAL
mm Hg
0.00
0.96
0.0000
0.0013
mol MEA
0.000
0.125
0.0000
0.0063
(kgmol/m3)
0.00
0.33
3.00
9.10
43.10
59.70
106.00
0.0039
0.0120
0.0567
0.0786
0.1395
0.208
0.362
0.643
0.729
0.814
0.0104
0.0180
0.0315
0.0356
0.0396
0.54
0.94
1.65
1.86
2.07
T=
100 C
1.00
3.00
5.00
10.00
0.0013
0.0039
0.0066
0.0132
0.029
0.050
0.065
0.091
0.0015
0.0025
0.0033
0.0046
0.08
0.13
0.17
0.24
30.00
50.00
70.00
100.00
0.0395
0.0658
0.0921
0.1316
0.160
0.203
0.238
0.279
0.0080
0.0102
0.0119
0.0139
0.42
0.53
0.62
0.73
0.060
40 C
yA1
100 C
Mole fraction H2S in gas, yA
0.050
0.040
0.030
0.020
xA1,min
xA1,min
0.010
0.000
0.000
0.010
0.020
0.030
0.040
Mole fraction H2S in solvent (water+15 wt% MEA), xA
30.18
31-33
a. Tower diameter D
A = H2S, B = N2
Cf = 52 for 1.5 inch ceramic Intalox saddles (Table 31.2)
kgmol  
kg 

=
AG '1 AG1=
M w,ave AG1 ( y A1 M A + (1 − y A=

1)MB )
 360
  ( 0.10(64)+(1-0.10)(28) )
hr  
kgmol 

kg
hr

(1.5 atm )
kg 
kg
 P 
ρG =  T  M w,ave =
 31.6
 = 1.9 3
3
kgmol 
m

m atm 
 RT 

 0.08206
 ( 305 K )
kgmol-K 

Flooding Correlation
1/2
kg 

kg

1/2
18, 000  
1.9

'

AL1  ρG 
hr  

m3
=
=
x − axis
0.066
=


kg
kg 
kg  
AG1  ρ L − ρG 


−
1090
1.9
11,376  
m3
m3 
hr 

0.25
y − axis =
Y=
= 11,376
1/2
1/2

  ρ ( ρ − ρ ) g 
(1.9 )(1090 − 1.9 )(1.0 )  4.4 kg
G 'f =
Y  G L 0.1 G c  
0.25 =

0.1

C f µL J
m 2 sec
52
0.0012
1


(
)(
)
(
)
 


kg   1hr 

11,376  
'


AG1
hr   3600sec 

A =
=1.19 m 2 at 60% flooding
=
0.6G 'f

kg 
0.6  4.4 2

m sec 

1/2
 4 ⋅1.19 m 2 
 4A 
D=
=

 = 1.23 m

π
 π 


1/2
b. Will tower “flood” if AL1/ = 54,000 kg/hr (3x old AL1/)
Get new xaxis; gas flow unchanged but now need yaxis at 0.6Gf/
xaxis ,new =
3 ⋅ xaxis ,old =
(0.066(3)) =
0.20
2
 G' 
2
yaxis yaxis,f=
0.090
=
=

 0.25(0.6)
G
'
 f 
Locus of intersection is below flooding line with ∆P/z between 50 and 100 Pa/m, therefore tower
will not flood.
31-34
Part 2: Instructor Only Problems
Chapter 1
Instructor Only Problems
1.23
A clean glass capillary tube contains water at 40oC. Please calculate how high (in millimeters)
the water will rise in the capillary tube if the diameter of the tube is 0.25 cm.
Solution
H2O at 60oC = 40 + 273 = 313 K
Equation 1-17: σ = 0.123(1 − 0.00139 T)
= 0.123 [1-0.00139 (313)]
= 0.0695 N/m
For a clean capillary: θ = 0o
Height, using Equation 1-20: h =
2 cos θ
ρgr
σ = 0.0695 N/m
ρ = 992.2 kg/m3 (from data in Appendix I)
ℎ=
2𝜎𝑐𝑜𝑠𝜃
2(0.0695 𝑁/𝑚)cos (0)
=
= 0.01144 𝑚 = 𝟏𝟏. 𝟒𝟒 𝒎𝒎
0.25 𝑐𝑚
𝑚
𝜌𝑔𝑟
992.2 𝑘𝑔/𝑚3 (9.8𝑚/𝑠 2 ) ( 2
𝑥 100 𝑐𝑚)
1.33
A beaker of water with a density of 987 kg/m3 has a capillary tube inserted into it. The water is
rising in the capillary tube to a height of 1.88 cm. The capillary tube is very clean and has a
diameter of 1.5 mm. What is the temperature of the water?
Solution
First, calculate the surface tension of water using the equation for the height:
2𝜎𝑐𝑜𝑠𝜃
ℎ=
𝜌𝑔𝑟
Rearrange, solving for 𝜎,
𝑚
1.5 𝑚𝑚
𝑚
(1.88 𝑐𝑚) (
) (987 𝑘𝑔/𝑚3 )(9.81 𝑚/𝑠 2 ) ( 2 ) (1000 𝑚𝑚)
ℎ𝜌𝑔𝑟
100
𝑐𝑚
𝜎=
=
2𝑐𝑜𝑠𝜃
2 cos 0
𝑘𝑔
= 0.0683 2 = 0.0683 𝑁/𝑚
𝑠
Now use the surface tension to calculate the temperature:
𝜎 = 0.123(1 − 0.00139𝑇)
0.0683 𝑁/𝑚 = 0.123(1 − 0.00139𝑇)
𝑇 = 320.2 𝐾 = 47.02 ℃
1.34
The water in a lake has an average temperature of 60oF. If the barometric pressure of the
atmosphere is 760 mm Hg (which is equal to 2.36 feet). Determine the gage pressure and the
absolute pressure at a water depth of 46 feet.
Solution
From Appendix I, 𝜌 = 62.3 𝑙𝑏𝑚 /𝑓𝑡 3 at 60oF.
(62.3
𝑃𝑔 = 𝜌𝑤 𝑔ℎ =
(847
𝑃𝑎𝑡𝑚 = 𝜌𝐻𝑔 𝑔ℎ =
𝑓𝑡
𝑙𝑏𝑚
) (32.2 2 ) (46 𝑓𝑡)
3
𝑙𝑏𝑓
𝑓𝑡
𝑠
= 2868 2
𝑙𝑏 𝑓𝑡
𝑓𝑡
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
𝑓𝑡
3.28 𝑓𝑡
𝑙𝑏𝑚
760 𝑚𝑚
) (32.2 2 ) (
)(
)
𝑙𝑏𝑓
𝑚
1000 𝑚𝑚/𝑚
𝑓𝑡 3
𝑠
= 2113
𝑙𝑏 𝑓𝑡
𝑓𝑡 2
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔 = 644.24
𝑙𝑏𝑓
𝑙𝑏𝑓
𝑙𝑏𝑓
+ 2868 2 = 3512 2
2
𝑓𝑡
𝑓𝑡
𝑓𝑡
1.35
Air initially fills a very long vertical capillary tube of inside diameter D. The tube is suddenly
immersed in a large body of water, still in the vertical position. Water wets the tube surface and as
soon as the ends of the tube are submerged, water enters the tube. When equilibrium is reached,
what is the length, if any, of the air column that remains in the tube? For this problem we know
that the tube diameter is 0.1 cm and the surface tension of water is 0.072 N/m and the density of
water is 1000 kg/m3. We can assume that the figure is not drawn to scale and that the capillary
tube is drawn much bigger than reality, such that D<<H1, and that the maximum bubble pressure
occurs when the radius of curvature equals the tube radius so that R1=R2=D/2. Assume the system
is at 273 K.
After immersion, the system looks like this:
Air/water interface
Air Pocket
H1
H2
L
D
Solution
Begin with the Young-Laplace Equation:
1 1
𝑃𝐻2 − 𝑃𝐻2 = 𝜎 ( + )
𝑅1 𝑅 2
Since R1 = R2 = D/2
𝑃𝐻2 − 𝑃𝐻2 =
2𝜎
4𝜎
=
𝐷/2
𝐷
Realizing that ∆𝑃 = 𝜌𝑔ℎ
𝜌𝑔𝐻2 − 𝜌𝑔𝐻1 =
4𝜎
𝐷
Realizing that
𝐻2 − 𝐻1 = 𝐿
and
0.072
𝑁
𝑘𝑔
𝑔
= 0.072 2 = 72 2
𝑚
𝑠
𝑠
So,
𝐻2 − 𝐻1 = 𝐿 =
4𝜎
4(72 𝑔/𝑠 2 )
=
= 2.94 𝑐𝑚
𝜌𝑔𝐷 (1 𝑔/𝑐𝑚3 )(980 𝑐𝑚/𝑠 2 )(0.1 𝑐𝑚)
1.36
Hydrometers are inexpensive and easy to use to measure specific gravity and then density. You
want to measure the density of an unknown liquid but all you have is a test tube that is 1 cm in
diameter. You fill the test tube with water and drop it into a known sample of water with a
temperature of 20℃, and measure the distance from the bottom of the test tube to the surface of
the water to be 8 cm. Then you take this same test tube (after cleaning on the outside) and drop
into an unknown liquid which is also at 20℃ and measure the distance submerged to be 7.4 cm.
Hydrometers floating in liquids are in static equilibrium. What is the density of the unknown
liquid?
8 cm
submerged
7.4 cm
submerged
Water
Unknown
Solution
A hydrometer floating in water is in static equilibrium and the buoyant force FB exerted by the
liquid must always be equal to the weight W of the hydrometer, so that F B = W.
F = PA = ρghA = ρgW
FB = ρgVsubmerged = ρghATT
Where h is the height of the submerged portion of the test tube and A TT is the cross-sectional
area of the test tube which is constant.
For the pure water: 𝑊 = 𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇
For unknown: 𝑊 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇
Setting the equations equal since the weight of the test tube does not change,
𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇
Solve for the density of the unknown liquid,
ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 =
ℎ𝑤
ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛
𝜌𝑤 =
8 𝑐𝑚
(998.2 𝑘𝑔/𝑚3 ) = 1079 𝑘𝑔/𝑚3
7.4 𝑐𝑚
Chapter 2
Instructor Only
2.30
A large industrial waste collection tank contains butyl alcohol, benzene and water at 80℉ that have
separated into three distinct phases as shown in the figure. The diameter of the circular tank is 10
feet and it has total depth is 95 feet. The gauge at the top of the tank reads 2116 lbf /ft2. Please
calculate (a) the pressure at the butyl alcohol/benzene interface, (b) the pressure at the
benzene/water interface and (c) the pressure at the bottom of the tank.
Solution
(a) the pressure at the butyl alcohol/benzene interface
∆𝑃 = 𝜌𝑔ℎ
𝑃𝑡𝑜𝑝 − 𝑃𝐴𝑖𝑟−𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = −𝜌𝑎𝑖𝑟 𝑔(17𝑓𝑒𝑒𝑡)
𝑃𝐴𝑖𝑟−𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 − 𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = −𝜌𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝑔(19𝑓𝑒𝑒𝑡)
𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑃𝑡𝑜𝑝 + 𝜌𝑎𝑖𝑟 𝑔(17𝑓𝑒𝑒𝑡) + 𝜌𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝑔(19𝑓𝑒𝑒𝑡)
𝑙𝑏𝑚
2
𝑙𝑏𝑓 (0.0735 𝑓𝑡 3 ) (32.2 𝑓𝑡/𝑠 )(17 𝑓𝑡)
= 2116 2 +
𝑙𝑏 𝑓𝑡
𝑓𝑡
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
𝑙𝑏𝑚
(50.0 3 ) (32.2 𝑓𝑡/𝑠 2 )(19 𝑓𝑡)
𝑙𝑏𝑓
𝑓𝑡
+
= 3068.0 2 = 21.3 𝑝𝑠𝑖
𝑙𝑏 𝑓𝑡
𝑓𝑡
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
(b) the pressure at the benzene/water interface
𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 − 𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = −𝜌𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝑔(34𝑓𝑒𝑒𝑡)
𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 + 𝜌𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝑔(34𝑓𝑒𝑒𝑡)
𝑙𝑏𝑚
2
𝑙𝑏𝑓 (54.6 𝑓𝑡 3 ) (32.2 𝑓𝑡/𝑠 )(34 𝑓𝑡)
𝑙𝑏𝑓
= 3068.0 2 +
= 4925.9 2 = 34.2 𝑝𝑠𝑖
𝑙𝑏 𝑓𝑡
𝑓𝑡
𝑓𝑡
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
(c) the pressure at the bottom of the tank
𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 − 𝑃𝐵𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑇𝑎𝑛𝑘 = −𝜌𝑊𝑎𝑡𝑒𝑟 𝑔(25𝑓𝑒𝑒𝑡)
𝑃𝐵𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑇𝑎𝑛𝑘 = 𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 + 𝜌𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝑔(25𝑓𝑒𝑒𝑡)
𝑙𝑏𝑚
2
𝑙𝑏𝑓 (62.2 𝑓𝑡 3 ) (32.2 𝑓𝑡/𝑠 )(25 𝑓𝑡)
𝑙𝑏𝑓
= 4925.9 2 +
= 6482.16 2 = 45 𝑝𝑠𝑖
𝑙𝑏 𝑓𝑡
𝑓𝑡
𝑓𝑡
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
2.31
The maximum blood pressure in the upper arm of a healthy person is about 120 mm Hg (this is a
gauge pressure). If a vertical tube open to the atmosphere is connected to the vein in the arm of a
person, determine how high the blood will rise in the tube. Take the density of the blood to be
constant and equal to 1050 kg/m3. (The fact that blood can rise in a tube explains why IV tubes
must be placed high to force fluid into the vein of a patient.) Assume the system is at 80oF.
Solution
∆𝑃 = 𝜌𝑔ℎ
For Blood:
𝑃 = 𝜌𝐵 𝑔ℎ𝐵
For Mercury: 𝑃 = 𝜌𝑀 𝑔ℎ𝑀
So,
∆𝑃 = 𝜌𝐵 𝑔ℎ𝐵 = 𝜌𝑀 𝑔ℎ𝑀
Solve for the height of the blood,
𝜌𝐵 𝑔ℎ𝐵 = 𝜌𝑀 𝑔ℎ𝑀
Eliminate the gravity terms,
𝜌𝐵 ℎ𝐵 = 𝜌𝑀 ℎ𝑀
ℎ𝐵 =
𝜌𝑀
ℎ
𝜌𝐵 𝑀
We take 120 mm Hg as the height here, and 120 mm = 0.12 m,
and the density of mercury is 845 𝑙𝑏𝑚 /ft 3 .
(13535.6 𝑘𝑔/𝑚3 )
(0.12 𝑚) = 1.55 𝑚
ℎ𝐵 =
(1050 𝑘𝑔/𝑚3 )
2.32
If the nitrogen tank in the figure below is at a pressure of 4500 lbf/ft2 and the entire system is at
100℉, please calculate the pressure at the bottom of the tank of glycerin.
2 ft
Nitrogen
12 ft
4 ft
7 ft
2.5 ft
Benzene
4 ft
7.5 ft
10 ft
5 ft
8 ft
5 ft
11 ft
10 ft
7 ft
7.5 ft
Water
5.8 ft
Mercury
Glycerin
Solution
PN − P1 = −ρNitrogen g(5 ft) so P1 = PN + ρNitrogen g(5 ft)
P2 − P1 = −ρH2O g(8 − 5 ft) so P2 = P1 − ρH2O g(8 − 5 ft)
P2 − P3 = −ρFreon g(7 − 4 ft) so P3 = P2 + ρFreon g(7 − 4 ft)
P3 − P4 = −ρMercury g(10 − 7 ft) so P4 = P3 + ρMercury g(10 − 7 ft)
P5 − P4 = −ρBenzene g(10 − 7.5 ft) so P5 = P4 − ρBenzene g(10 − 7.5 ft)
P5 − PA = −ρA g(11 ft) so PA = P6 + ρA g(11 ft)
PA = ρA g(11 ft) − ρBenzene g(10 − 7.5 ft) + ρMercury g(10 − 7 ft) + +ρFreon g(7 − 4 ft)
− ρH2O g(8 − 5 ft) + ρNitrogen g(5 ft) +PN
(78.2 lb /ft3 )
(53.6 lb /ft3 )
PA = (32.174 lb mft/lb s2) (32.2 ft/s2 )(11 ft) − (32.174 lb mft/lb s2) (32.2 ft/s2 )(10 − 7.5 ft) +
m
(843 lbm
/ft3 )
(32.174 lbm ft/lbf s2 )
lb
(62.1 m
3)
ft
lb ft
(32.174 m 2 )
lbf s
m
f
(32.2 ft/s 2 )(10 − 7 ft) +
ft
(32.2 s2) (8 − 5 ft) +
lb
(78.7 m
3)
ft
lb ft
(32.174 m 2 )
lbf s
lbm
(0.0685 3 )
ft
ft
lbm ft
s2
(32.174
)
lbf s2
f
ft
(32.2 s2) (7 − 4 ft) −
lb
(32.2 ) (5 ft) + 4500 ft2f = 7808.0 lbf /ft 2
2.33
The water in a lake has an average temperature of 60oF. If the barometric pressure of the
atmosphere is 760 mm Hg (which is equal to 2.36 feet). Determine the gage pressure and the
absolute pressure at a water depth of 46 feet.
Solution
From Appendix I, 𝜌 = 62.3 𝑙𝑏𝑚 /𝑓𝑡 3 at 60oF.
(62.3
𝑃𝑔 = 𝜌𝑤 𝑔ℎ =
(847
𝑃𝑎𝑡𝑚 = 𝜌𝐻𝑔 𝑔ℎ =
𝑓𝑡
𝑙𝑏𝑚
) (32.2 2 ) (46 𝑓𝑡)
3
𝑙𝑏𝑓
𝑓𝑡
𝑠
= 2868 2
𝑙𝑏 𝑓𝑡
𝑓𝑡
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
𝑓𝑡
3.28 𝑓𝑡
𝑙𝑏𝑚
760 𝑚𝑚
) (32.2 2 ) (
)(
)
𝑙𝑏𝑓
𝑚
1000 𝑚𝑚/𝑚
𝑓𝑡 3
𝑠
= 2113
𝑙𝑏 𝑓𝑡
𝑓𝑡 2
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔 = 644.24
𝑙𝑏𝑓
𝑙𝑏𝑓
𝑙𝑏𝑓
+ 2868 2 = 3512 2
2
𝑓𝑡
𝑓𝑡
𝑓𝑡
2.34
Air initially fills a very long vertical capillary tube of inside diameter D. The tube is suddenly
immersed in a large body of water, still in the vertical position. Water wets the tube surface and as
soon as the ends of the tube are submerged, water enters the tube. When equilibrium is reached,
what is the length, if any, of the air column that remains in the tube? For this problem we know
that the tube diameter is 0.1 cm and the surface tension of water is 0.072 N/m and the density of
water is 1000 kg/m3. We can assume that the figure is not drawn to scale and that the capillary
tube is drawn much bigger than reality, such that D<<H1, and that the maximum bubble pressure
occurs when the radius of curvature equals the tube radius so that R1=R2=D/2. Assume the system
is at 273 K.
After immersion, the system looks like this:
Air/water interface
Air Pocket
H1
H2
L
D
Solution
Begin with the Young-Laplace Equation:
1 1
𝑃𝐻2 − 𝑃𝐻2 = 𝜎 ( + )
𝑅1 𝑅 2
Since R1 = R2 = D/2
𝑃𝐻2 − 𝑃𝐻2 =
2𝜎
4𝜎
=
𝐷/2
𝐷
Realizing that ∆𝑃 = 𝜌𝑔ℎ
𝜌𝑔𝐻2 − 𝜌𝑔𝐻1 =
4𝜎
𝐷
Realizing that
𝐻2 − 𝐻1 = 𝐿
and
0.072
𝑁
𝑘𝑔
𝑔
= 0.072 2 = 72 2
𝑚
𝑠
𝑠
So,
𝐻2 − 𝐻1 = 𝐿 =
4𝜎
4(72 𝑔/𝑠 2 )
=
= 2.94 𝑐𝑚
𝜌𝑔𝐷 (1 𝑔/𝑐𝑚3 )(980 𝑐𝑚/𝑠 2 )(0.1 𝑐𝑚)
2.35
Hydrometers are inexpensive and easy to use to measure specific gravity and then density. You
want to measure the density of an unknown liquid but all you have is a test tube that is 1 cm in
diameter. You fill the test tube with water and drop it into a known sample of water with a
temperature of 20℃, and measure the distance from the bottom of the test tube to the surface of
the water to be 8 cm. Then you take this same test tube (after cleaning on the outside) and drop
into an unknown liquid which is also at 20℃ and measure the distance submerged to be 7.4 cm.
Hydrometers floating in liquids are in static equilibrium. What is the density of the unknown
liquid?
8 cm
submerged
7.4 cm
submerged
Water
Unknown
Solution
A hydrometer floating in water is in static equilibrium and the buoyant force F B exerted by the
liquid must always be equal to the weight W of the hydrometer, so that F B = W.
F = PA = ρghA = ρgW
FB = ρgVsubmerged = ρghATT
Where h is the height of the submerged portion of the test tube and A TT is the cross-sectional
area of the test tube which is constant.
For the pure water: 𝑊 = 𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇
For unknown: 𝑊 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇
Setting the equations equal since the weight of the test tube does not change,
𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇
Solve for the density of the unknown liquid,
ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 =
ℎ𝑤
ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛
𝜌𝑤 =
8 𝑐𝑚
(998.2 𝑘𝑔/𝑚3 ) = 1079 𝑘𝑔/𝑚3
7.4 𝑐𝑚
Chapter 4
Instructor Only Problems
4.27
A well-mixed tank initially contains pure water. The tank has inlet and outlet ports both with a
diameter of 1 centimeter, and the inlet port is 20 centimeters above the outlet port. Into the tank
through the inlet port flows a 10% solution of sodium sulfate dissolved in water with a velocity of
0.1 meters/second that has a density of 1000 kg/m3. (a) After 60 seconds of flow, the solution at
the outlet port is measured to contain 14 grams of sodium sulfate per liter of solution at a constant
flow rate of 0.01 liters per second. Please calculate the number of grams of sodium sulfate in the
tank at this time. (b) What is the mass flow rate at the exit port once the system has reached steady
state?
Solution
(a) Grams of sodium sulfate in the tank after 60 seconds
𝑑𝑆
𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 +
=0
𝑑𝑡
𝜋(0.01𝑚)2
𝑑𝑆
3
(14 𝑔/𝐿)(0.01 𝐿/𝑠) − (1000 𝑘𝑔/𝑚 )(1000 𝑔/𝑘𝑔)(0.1 𝑚/𝑠) (
) (0.1) +
=0
4
𝑑𝑡
𝑑𝑆
(0.14 𝑔/𝑠) − (0.7854 𝑔/𝑠 ) +
=0
𝑑𝑡
𝑑𝑆
= 0.6454 𝑔/𝑠
𝑑𝑡
𝑆
60
∫ 𝑑𝑆 = 0.6454 𝑔/𝑠 ∫ 𝑑𝑡
0
0
𝑆 = 38.7 𝑔𝑟𝑎𝑚𝑠
(b) What is the mass flow rate at the exit port once the system has reached steady state?
At Steady state, 𝑚̇𝑜𝑢𝑡 = 𝑚̇𝑖𝑛 = 0.7854 grams/sec
4.28
A cylindrical water tank is 4 feet high, with a 3-foot diameter, open to the atmosphere at the top is
initially filled with water with a temperature of 60oF. The outlet at the bottom has a diameter of
0.5 in. is opened and the tank empties. The average velocity of the water exiting the tank is given
by the equation, v = √2gh, where h is the height of the water in the tank measured from the center
of the outlet port and g is the gravitational acceleration. Determine: (a) how long will it take for
the water level in the tank to drop to 2 feet from the bottom, and (b) how long will it take to drain
the entire tank.
Solution
𝑑𝑆
𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 +
=0
𝑑𝑡
Since there is no fluid entering the tank: 𝑚̇𝑖𝑛 = 0
𝑑𝑆
= −𝑚̇𝑜𝑢𝑡
𝑑𝑡
𝑑ℎ 𝜋𝑑𝑡2
𝜋𝑑𝑜2
𝜌 (
) = −𝜌√2gh (
)
𝑑𝑡 4
4
𝑑ℎ 𝑑𝑜2
=
√2gh
𝑑𝑡 𝑑𝑡2
𝑑𝑡2 𝑑ℎ 1
𝑑𝑡 = − 2
𝑑𝑜 √ℎ √2𝑔
𝑡
∫ 𝑑𝑡 = −
0
ℎ2
𝑑𝑡2 1
𝑑ℎ
∫
2
𝑑𝑜 √2𝑔 ℎ0 √ℎ
1/2
1/2
𝑑𝑡2 1 ℎ2 − ℎ0
𝑡=− 2
𝑑𝑜 √2𝑔 (1/2)
(a) How long will it take for the water level in the tank to drop to 2 feet from the bottom?
𝑡=−
(3 𝑓𝑡)2
1
21/2 − 41/2
= 745 𝑠𝑒𝑐 = 12.4 𝑚𝑖𝑛
(0.042𝑓𝑡)2 √2(32.2𝑓𝑡/𝑠 2 ) (1/2)
(b) How long will it take to drain the entire tank?
(3 𝑓𝑡)2
1
01/2 − 41/2
𝑡=−
= 2543 𝑠𝑒𝑐 = 42.4 𝑚𝑖𝑛
(0.042𝑓𝑡)2 √2(32.2𝑓𝑡/𝑠 2 ) (1/2)
4.29
A tank with a total volume of 75 liters is initially empty. Into the tank flows pure water with a
temperature of 273K, density of 999.3 kg/m3 and surface tension of 0.0763 N/m. This pure water
is being delivered from an inlet port with a diameter of 0.01m at a velocity of 2.0 m/s. The exit
port is initially closed but it too has a diameter of 0.01m.
(a) How long does it take to fill the tank with water (assuming the exit port remains closed)?
Solution
∬ ρ(v ∙ n) dA +
∂
∭ ρdV = 0
∂t
Since exit port is closed
−ṁin +
∂ V
∫ ρdV = 0
∂t 0
The tank has a total volume of 75 liters,
(75 liters)
0.028317 m3
= 0.075m3
28.32 liters
3
kg
π(0.01 m)2
kg ∂ 0.075 m
− (999.3 3 ) (2 m/s) (
) + (999.3 3 ) ∫
dV = 0
m
4
m ∂t 0
1.57x10−4 m3 t
∫ dt
s
0
t = 447.5 sec
0.075 m3 =
(b) Once the tank is completely filled with pure water, into the inlet port (still with a diameter of
0.01m) flows at 2.0 m/s a 60% solution of sodium chloride with a specific gravity of 1.07. What
is the total mass (water plus sodium chloride) in the tank after 600 seconds if a sample at the exit
port at that time contains 50 g/liter of sodium chloride and the flow rate is 0.5 liters/s?
Solution
The tank is now filled with pure water which has a mass of:
(75 liters)
0.028317 m3 999.3 𝑘𝑔
(
) = 74.93 kgs
28.32 liters
𝑚3
𝑀
𝑑𝑀
=0
𝑀0 𝑑𝑡
𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 + ∫
𝑀
𝑘𝑔
𝑚 π(0.01 m)2
𝑔
𝑑𝑀
)
(1000
)
(2
)
(
)
+
∫
=0
3
𝑚
𝑠
4
𝑘𝑔
𝑀0 𝑑𝑡
𝑀
𝑔
𝑑𝑀
25 − 168 𝑔/𝑠 + ∫
𝑠
74.93 𝑘𝑔 𝑑𝑡
(0.5 𝐿/𝑠)(50 𝑔/𝐿) − (1.07) (999.3
𝑚 = 160.7 𝑘𝑔
4.30
A large tank of unknown total volume is initially filled with 6000 grams of a 10% by mass sodium
sulfate solution. Into this tank a 50% sodium sulfate solution is added at a rate of 40 g/min. At the
single outlet to the tank flows a 20 g/Liter solution at a rate of 0.01667 liters/sec. Please calculate
(a) the total mass in the tank after 2 hours and (b) the amount of sodium sulfate in the tank after 2
hours.
Solution
(a) Total mass in the tank after 2 hours
Total Mass = M
𝜕
∭ 𝜌𝑑𝑉 = 0
𝜕𝑡
𝑑𝑀
𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 +
=0
𝑑𝑡
𝑀
𝑔
𝑑𝑀
(20 𝑔/𝐿)(0.01667 𝐿/𝑠)(60 𝑠/𝑚) − 40
+∫
=0
𝑚𝑖𝑛
6000 𝑑𝑡
𝑀
𝑑𝑀
20 𝑔/𝑚𝑖𝑛 − 40 𝑔/𝑚𝑖𝑛 + ∫
=0
6000 𝑑𝑡
𝑔
𝑑
−20
+ (𝑀 − 6000) = 0
𝑚𝑖𝑛
𝑑𝑡
𝑀 = 6000 + 20𝑡
At the 2 hour point, which is 120 minutes,
𝑀 = 6000 + 20𝑡 = 6000 𝑔𝑟𝑎𝑚𝑠 + 20 𝑔𝑟𝑎𝑚𝑠/𝑚𝑖𝑛(120𝑚𝑖𝑛) = 8400.5 𝑔𝑟𝑎𝑚𝑠
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +
(b) The amount of sodium sulfate in the tank after 2 hours
𝑑𝑆
𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 +
=0
𝑑𝑡
𝑆
𝑆
=
𝑀 6000 + 20𝑡
𝑆
𝑑𝑆
(20 𝑔/𝐿)(0.01667 𝐿/𝑠)(60 𝑠/𝑚) (
) − (0.50)(40 𝑔/𝑚𝑖𝑛) +
=0
6000 + 20𝑡
𝑑𝑡
𝑆
𝑑𝑆
(20 𝑔/𝑚𝑖𝑛) (
) − (0.50)(40 𝑔/𝑚𝑖𝑛) +
=0
6000 + 20𝑡
𝑑𝑡
𝑆
𝑑𝑆
− 20 𝑔/𝑚𝑖𝑛 +
=0
300 + 𝑡
𝑑𝑡
Rearrange,
𝑑𝑆
𝑆
+
= 20
𝑑𝑡 300 + 𝑡
Integrating,
𝑑𝑆
(300 + 𝑡) = (300 + 𝑡)20
𝑑𝑡
20𝑡 2
𝑆(300 + 𝑡) = 6000𝑡 +
+𝐶
2
20𝑡 2
6000𝑡 + 2
𝐶
𝑆=
+
(300 + 𝑡)
(300 + 𝑡)
6000𝑡 + 10𝑡 2
𝐶
𝑆=
+
(300 + 𝑡)
(300 + 𝑡)
The initial condition given in the problem was that at time = 0, there is 6000 grams of a 10%
solution of sodium sulfate solution.
6000(0) + 10(0)2
𝐶
(6000 𝑔𝑟𝑎𝑚𝑠)(0.1) =
+
(300 + 0)
(300 + 0)
𝐶 = 180,000
6000𝑡 + 10𝑡 2
180,000
𝑆=
+
(300 + 𝑡)
(300 + 𝑡)
At the 2 hour point (2 hours = 120 minutes):
6000(120) + 10(120)2
180,000
𝑆=
+
= 2485.7 𝑔𝑟𝑎𝑚𝑠
(300 + 120)
(300 + 120)
4.31
A spherical metal tank has a constant volume of 450 in3 and contains air at an absolute pressure
of 85 psi and a temperature of 20oC. You are told that the air is leaking through a 0.1 inches in
diameter whole in the side. The air exits at 900 ft/s with a density of 0.075 lbm/ft3. Calculate the
change in density (loss) with respect to time.
Solution
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +
𝜕
∭ 𝜌𝑑𝑉 = 0
𝜕𝑡
𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 +
𝜕𝑀
=0
𝜕𝑡
Since nothing is going into the tank, 𝑚̇𝑖𝑛 = 0
𝑚̇𝑜𝑢𝑡 +
𝜕𝑀
=0
𝜕𝑡
𝑚̇𝑜𝑢𝑡 +
𝜕𝜌𝑉
=0
𝜕𝑡
𝑓𝑡 0.1
𝑙𝑏𝑚
𝜕𝜌
𝑚̇𝑜𝑢𝑡
𝜌𝑣𝐴 0.075 𝑓𝑡 3 (900 𝑠 ) ( 12 𝑓𝑡) 0.5625
=−
=−
=
=
= 2.16 𝑙𝑏𝑚 𝑓𝑡 2 /𝑠
𝑓𝑡 3
𝜕𝑡
𝑉
𝑉
0.260
3
450 𝑖𝑛 ( 3 3 )
12 𝑖𝑛
4.32
You are working on a project for lab when you realize that you forgot to measure the volume of
your mixing tank. Luckily, you took excellent notes in lab and have a lot of information about the
system. You were running an experiment to determine the concentration of salt in a mixing tank
at unsteady state. You measured the flow rate in and out of the mixing tank to be 1 L/min. The
25 𝑔 𝑠𝑎𝑙𝑡
tank was supplied with salt water at a concentration of 𝐶 = 100 𝑚𝑙 𝑠𝑎𝑡𝑒𝑟. You also observed that the
solution in the tank was uniform and well mixed. Use a computational model (equation 1) and
your data in the table below from the experiment to determine the volume of the mixing tank.
Time
0 2
4
6
8
10
12
14
16
18
20
22
24
Concentration 0 0.059 0.106 0121 0.187 0.170 0.205 0.202 0.237 0.219 0.242 0.238 0.243
The equation describing the concentration is:
𝑡
𝐶(𝑡) = 𝐶𝑠𝑜𝑢𝑟𝑐𝑒 (1 − 𝑒 −𝜏 )
(equation 1)
In the above equation, t is time and 𝜏 is the volume of the tank.
(Hint: A residual is the difference between a measured value, as in the data in the table, and the
predicted value of a regression model, which is what you will calculate with the equation. It is
important to understand residuals because they show how accurate a mathematical function, such
as a line, is in representing a set of data. You can find the volume of the tank by minimizing the
sum of the squares of all the residuals of all data points. One way to do this is to put in guesses of
the tank volume to find when the sum of the squares of the residuals is the smallest. There are
other more elegant ways to do this, but a guess a tank value and calculate the sum of the squares
of the residuals is a fine way. This is easy to do in MSExcel)
Solution
For this problem the data in the table was put into MSExcel. The sum of the residuals squared
was calculated and then minimized using a solver by varying 𝜏 to determine the volume of the
tank.
𝜏 = 7.45 𝐿, based on the graph and data.
Time
Concentration
0
0.059
0.106
0.121
0.187
0.17
0.205
0.202
0.237
0.219
0.242
0.238
0.243
0
2
4
6
8
10
12
14
16
18
20
22
24
Co
Tau
Fitted Line
(using given
equation)
0.000
0.059
0.104
0.138
0.165
0.185
0.200
0.212
0.221
0.228
0.233
0.237
0.240
0.25
7.45
Residual
(difference
between
actual
concentration
and that of
the fitted
line).
0.000
0.000
0.002
-0.017
0.022
-0.015
0.005
-0.010
0.016
-0.009
0.009
0.001
0.003
Total:
Residual
Squared
0.00000
0.00000
0.00000
0.00030
0.00050
0.00022
0.00002
0.00010
0.00026
0.00008
0.00008
0.00000
0.00001
0.0015718
Problem 3
Concentration [g/ml]
0.3
0.25
0.2
0.15
Data
0.1
Fitted Line
0.05
0
0
5
10
15
Time [min]
20
25
30
Chapter 5
Instructor Only Problems
5.5
(a) Determine the magnitude of the x and y
components of the force exerted on the fixed blade
shown by a 3 ft3/s jet of water flowing at 25 ft/s.
Assume the water is at 100oF.
(b) If the blade is moving to the right at 15 ft/s, find
the magnitude and velocity of the water jet leaving
the blade. Assume steady state.
Solution
(a)
Q = vA = 3 ft3/s, v1 = v2 = 25 ft/s
∑ 𝐹 = ∬ 𝑣𝑥 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 +
𝜕
∭ 𝑣𝑥 𝜌𝑑𝑉
𝜕𝑡
𝐹𝑥 = 𝜌𝑣2 𝐴2 𝑣2 𝑐𝑜𝑠𝜃 − 𝜌𝑣1 𝐴1 (−𝑣1 )
Where the negative sign on the v1 is because the blade is moving in the negative x-direction.
𝐹𝑥 = 𝜌𝑄𝑣2 𝑐𝑜𝑠𝜃 − 𝜌𝑄(−𝑣1 )
(62.1 𝑙𝑏𝑚 /𝑓𝑡 3 )
(3 𝑓𝑡 3 /𝑠)(25 𝑓𝑡/𝑠)𝑐𝑜𝑠30
=
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
(62.1 𝑙𝑏𝑚 /𝑓𝑡 3 )
(3 𝑓𝑡 3 /𝑠)(−25 𝑓𝑡/𝑠) = 270.12 𝑙𝑏𝑓
−
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
y-direction, remembering that the water velocity is flowing out in the minus y-direction,
(62.1 𝑙𝑏𝑚 /𝑓𝑡 3 )
(3 𝑓𝑡 3 /𝑠)(−25 𝑓𝑡/𝑠)𝑠𝑖𝑛30 = −72.4𝑙𝑏𝑓
𝐹𝑦 = 𝜌𝑣2 𝐴2 𝑣2 𝑠𝑖𝑛𝜃 =
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
(b) If the blade moves to the right at 15 ft/s then,
𝑣1 = 25 + 15 = 40 𝑓𝑡/𝑠
𝑣2 = −40 𝑓𝑡/𝑠
So the magnitude of the velocity leaving the blade is:
𝑣 = 𝑣2 (𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑏𝑙𝑎𝑑𝑒) + 𝑣𝑏𝑙𝑎𝑑𝑒
𝑣 = [𝑣𝑥 𝑐𝑜𝑠𝜃𝑒𝑥 + 𝑣𝑦 𝑠𝑖𝑛𝜃𝑒𝑦 ] + 𝑣𝑥 𝑒𝑥 = [40𝑐𝑜𝑠(30)𝑒𝑥 + 40 sin(30) 𝑒𝑦 ] + 15𝑒𝑥
= [34.64𝑒𝑥 + 15𝑒𝑦𝑥 ] − 20𝑒𝑦 = 49.64𝑒𝑥 − 20𝑒𝑦
𝑣 ∙ 𝑒𝑠 = (49.64𝑒𝑥 − 20𝑒𝑦 ) ∙ (𝑐𝑜𝑠𝜃𝑒𝑥 + 𝑠𝑖𝑛𝜃𝑒𝑦 ) = 49.64 cos(−30) − 20 sin(−30)
= 52.98 𝑓𝑡/𝑠
5.34
You have been given a project where you must attach a particular reducing pipe fitting into a
Freon-12 delivery system. Freon-12 is a potentially hazardous material so you must be certain
that the reducing pipefitting is attached properly and will withstand the force due to the
reduction in pipe diameter. The direction of the flow, the inlet and outlet positions are labeled in
the figure below. The inlet has a diameter of 1 foot, and a pressure of 1000 lb f /ft2. The outlet
port has a diameter of 0.2 feet and a pressure of 200 lbf /ft2. The system must maintain a constant
flow rate of 4.5 ft3/s and constant temperature of 80℉. The weight of the Freon in the fitting is
6 lbf. In your analysis you may assume that the system is at steady state and that the fluid is
incompressible. Please calculate the forces in all directions necessary to hold the fitting
stationary.
Outlet from
Reducing Fitting
y
x
60 degrees
Inlet to
Reducing
Fitting
Solution
𝐹𝑥 + 𝑃2 𝐴2 cos 𝜃 = 𝜌𝑣22 𝐴2 cos 𝜃
𝐹𝑥 = 𝜌𝑣22 𝐴2 cos 𝜃 − 𝑃2 𝐴2 cos 𝜃
𝐹𝑥 =
81.3 𝑙𝑏𝑚 /𝑓𝑡 3
𝜋(0.2𝑓𝑡)2
(143.3 𝑓𝑡/𝑠)2 (
) cos(60)
𝑙𝑏𝑚 𝑓𝑡
4
32.174
𝑙𝑏𝑓 𝑠 2
𝜋(0.2𝑓𝑡)2
2
− (200 𝑙𝑏𝑓 /𝑓𝑡 ) (
) cos(60) = 811.93 𝑙𝑏𝑓
4
𝐹𝑦 + 𝑃1 𝐴1 − 𝑃2 𝐴2 sin 𝜃 − 𝑊 = 𝜌𝑣22 𝐴2 sin 𝜃 − 𝜌𝑣12 𝐴1
𝐹𝑦 = 𝜌𝑣22 𝐴2 sin 𝜃 − 𝜌𝑣12 𝐴1 − 𝑃1 𝐴1 + 𝑃2 𝐴2 sin 𝜃 + 𝑊
81.3 𝑙𝑏𝑚 /𝑓𝑡 3
𝜋(0.2𝑓𝑡)2
2
(143.3 𝑓𝑡/𝑠) (
𝐹𝑦 =
) sin 60
𝑙𝑏 𝑓𝑡
4
32.174 𝑚 2
𝑙𝑏𝑓 𝑠
81.3 𝑙𝑏𝑚 /𝑓𝑡 3
𝜋(1𝑓𝑡)2
𝜋(1𝑓𝑡)2
(5.73𝑓𝑡/𝑠)2 (
−
) − 1000 (
)
𝑙𝑏𝑚 𝑓𝑡
4
4
32.174
𝑙𝑏𝑓 𝑠 2
𝜋(0.2𝑓𝑡)2
+ 200 (
) sin 𝜃 + 6 𝑙𝑏𝑓 = 572.6 𝑙𝑏𝑓 𝑑𝑜𝑤𝑛
4
5.35
The horizontal Y-fitting splits 80℉ water into two
equal parts. The first part goes through Exit 1 and the
second part goes through Exit 2. The volumetric flow
rate at the entrance is 8 ft3/sec, and the gage
pressures at the three positions are 25 lbf/in2 at the
entrance, 1713 lbf/ft2 at exit 1, and 3433 lbf/ft2 at exit
2. The diameters at the entrance, exit 1, and exit 2 are
6.5, 4 and 3.5 inches respectively. Please determine
the forces in all directions required to keep the fitting
in place.
Solution
ventry =
v1 =
Q
8 ft 3 /sec
=
= 34.72 ft/sec
A1 π 6.5 in 2
(
)
4 12in/ft
Q
4 ft 3 /sec
=
= 45.83 ft/sec
A2 π 4in 2
(
)
4 12in/ft
Q
4 ft 3 /sec
v2 =
=
= 59.87 ft/sec
A2 π 3.5 in 2
(
)
4 12in/ft
First, calculate the force in the x-direction:
Fx + Pentry Aentry
2
− P1 A1 cos 50 − P2 A2 cos 45 = ρv12 A1 cos 50 + ρv22 A2 cos 45 − ρ ventry
Aentry
Fx = −Pentry Aentry
2
+ P1 A1 cos 40 + P2 A2 cos 45 + ρv12 A1 cos 40 + ρv22 A2 cos 45 − ρ ventry
Aentry
2
2
= − (25 + 14.7
lbf 144 in
π 6.5in
)(
) )
)( (
2
2
in
ft
4 12in
ft
2
lbf π 4in
) cos 40 + (3433
+ (1713 + 2116.2 2 ) (
ft 4 12in
ft
2
lbf π 3.5in
) (
) cos 45
ft 2 4 12in
ft
2
lbm
ft 2 62.2 ft 3
π 4in
)
) ] cos 40
+ (45.83
[ (
sec 32.174 lbm ft 4 12in
ft
lbf s2
2
lb
m
2
62.2
ft
ft 3 [π ( 3.5in ) ] cos 45
)
+ (59.87
sec 32.174 lbm ft 4 12in
ft
lbf s2
2
lbm
ft 2 62.2 ft 3
π 6.5in
)
) ]
− (34.72
[ (
sec 32.174 lbm ft 4 12in
ft
lbf s2
Fx = −1317.37 + 255.98 + 262.17 + 271.45 + 327.38 − 537.03 = −737.42 lbf (to left)
+ 2116.2
Next, calculate the force in the y-direction:
Fy − P1 A1 sin 40 + P2 A2 sin 45 = v1 ρv1 A1 sin 40 + v2 ρ(−v2 )A2 sin 45
Fy = v1 ρv1 A1 sin 40 + v2 ρ(−v2 )A2 sin 45 + P1 A1 sin 40 − P2 A2 sin 45
Fy = (45.83
ft 2 62.2 lbm /ft 3
π 4in 2
) (
)[ (
) ] sin 40
lbm ft 4 12in/ft
sec
32.174
lbf s2
ft 2 62.2 lbm /ft 3
π 3.5in 2
) (
)[ (
) ] sin 45
− (59.87
lb ft
sec
4 12in/ft
32.174 m 2
lbf s
lbf π 4in 2
) sin 40 − (3433
+ (1713 + 2116.2 2 ) (
ft 4 12in/ft
lbf π 3.5in 2
) sin 45 = 227.77 − 327.38 + 214.8 − 262.17
+ 2116.2 2 ) (
ft 4 12in/ft
= −146.98 lbf
5.36
Water at 100℃ flows through a 10 cm diameter
pipe that has a 180 degree vertical turn as
shown in the figure below. The volumetric flow
rate is constant at 0.2 m3/s. The absolute
pressure at position 1 is 64,000 Pa and the
absolute pressure at position 2 is 33,000 Pa.
The weight of the fluid and pipe together is
10 kg. Please calculate the total force that
vertical turn must withstand to remain in place
assuming the system is at steady state.
Solution
𝑄
0.2 𝑚3 /𝑠
𝑣1 = 𝑣2 = = 𝜋
= 25.46 𝑚/𝑠
𝐴
(0.10 𝑚)2
4
𝜕
∑ 𝐹 = ∬ 𝑣𝑥 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 + ∭ 𝑣𝑥 𝜌𝑑𝑉
𝜕𝑡
Since we are working at steady state,
∑ 𝐹 = ∬ 𝑣𝑥 𝜌(𝑣 ∙ 𝑛)𝑑𝐴
𝐹𝑥 + 𝑃1 𝐴1 − 𝑃2 𝐴2 cos 180 = 𝜌2 𝑣22 𝐴2 cos 180 − 𝜌1 𝑣12 𝐴1
𝐹𝑥 = −𝑃1 𝐴1 + 𝑃2 𝐴2 cos 180 + 𝜌2 𝑣22 𝐴2 cos 180 − 𝜌1 𝑣12 𝐴1
𝜋
𝜋
𝐹𝑥 = −(64,000 𝑃𝑎) ( (0.1 𝑚)2 ) + (33,000 𝑃𝑎) ( (0.1 𝑚)2 ) cos 180
4
4
𝑘𝑔
𝑚 2 𝜋
+ (958.4 3 ) (25.46 ) ( (0.1 𝑚)2 ) cos (180)
𝑚
𝑠
4
𝑘𝑔
𝑚 2 𝜋
− (958.4 3 ) (25.46 ) ( (0.1 𝑚)2 ) = −10,520.3 𝑁
𝑚
𝑠
4
𝑘𝑔∙𝑚
𝐹𝑦 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑃𝑖𝑝𝑒 + 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐹𝑙𝑢𝑖𝑑 = (10 𝑘𝑔)(9.8 𝑚/𝑠 2 ) = 98 𝑠2 = 98 𝑁
5.37
Benzene is flowing at steady state
through the nozzle shown below at
80oF and at a constant flow rate of 3
ft3/sec. Note that the nozzle exit is at
an angle of 50 degrees as shown. At
the inlet the diameter is 5 inches and
the liquid is at a pressure of
500 lbf /ft2. At the outlet, the diameter
is 2 inches and the pressure is 300
lbf/ft2. For this problem the weight of
the benzene in the fitting is 30 lbf. Calculate the forces in all directions necessary to hold the pipe
bend stationary.
Solution
𝐹𝑥 + 𝑃1 𝐴1 − 𝑃2 𝐴2 cos 𝜃 = 𝜌2 𝑣22 𝐴2 cos 𝜃 − 𝜌1 𝑣12 𝐴1
𝐹𝑥 = 𝑃2 𝐴2 cos 𝜃 − 𝑃1 𝐴1 + 𝜌2 𝑣22 𝐴2 cos 𝜃 − 𝜌1 𝑣12 𝐴1
𝜋 2 2
𝜋 5 2
2
2
𝐹𝑥 = (300 𝑙𝑏𝑓 /𝑓𝑡 ) ( ( ) ) cos 𝜃 − (500 𝑙𝑏𝑓 /𝑓𝑡 ) ( ( ) )
4 12
4 12
2
54.6 𝑙𝑏𝑚 /𝑓𝑡
3 𝑓𝑡 /𝑠
𝜋 2 2
(
)
( ) ) cos 𝜃
(
𝑙𝑏𝑚 𝑓𝑡 𝜋 2 2
4 12
32.174
𝑙𝑏𝑓 𝑠 2 4 (12)
3
+
3
2
54.6 𝑙𝑏𝑚 /𝑓𝑡 3 3 𝑓𝑡 3 /𝑠
𝜋 5 2
(
) ( ( ) ) = 501.6 𝑙𝑏𝑓 𝑡𝑜 𝑙𝑒𝑓𝑡
−
𝑙𝑏𝑚 𝑓𝑡 𝜋 5 2
4 12
32.174
𝑙𝑏𝑓 𝑠 2 4 (12)
𝐹𝑦 + 𝑃2 𝐴2 sin 𝜃 − 𝑊 = −𝜌2 𝑣22 𝐴2 𝑠𝑖𝑛 𝜃
𝐹𝑦 = −𝑃2 𝐴2 sin 𝜃−𝜌2 𝑣22 𝐴2 𝑠𝑖𝑛 𝜃 + 𝑊
2
2
𝜋 2
54.6 𝑙𝑏𝑚 /𝑓𝑡
3 𝑓𝑡 /𝑠
𝜋 2 2
(
𝐹𝑦 = −(300 𝑙𝑏𝑓 /𝑓𝑡 ) ( ( ) ) sin 𝜃 −
2 ) ( 4 (12) ) sin 𝜃 + 30 𝑙𝑏𝑓
𝑙𝑏 𝑓𝑡
4 12
32.174 𝑚 2 𝜋 ( 2 )
𝑙𝑏𝑓 𝑠
4 12
= −215.24 𝑙𝑏𝑓 𝑢𝑝
2
3
3
5.39
A rectangular plate 5 feet long and 3 feet deep, hangs in the air from a hinge at the top as shown
in the figure. The plate is struck in the center by a horizontal jet of water at 60oF that is 4 inches in
diameter and moving at 20 ft/s, causing the plate to swing outward at an angle 𝜃. If the plate weighs
12 lbs., calculate the angle 𝜃 that the plate will hang from the vertical.
hinge
𝜃
Fjet
Solution
From Appendix I: Water at 60oF, ρ = 62.3 lbm /ft 3
∑ F = ∬ vx ρ(v ∙ n)dA +
∂
∭ vx ρdV
∂t
Steady state reduces to,
2
Fx = ∬ vx ρ(v ∙ n)dA = ρin vin
Ain =
2
(62.3 lbm /ft 3 )
π 4
2
(20 ft/s) ( ft) cosθ
32.174 lbm ft/lbf s2
4 12
= 67.6(cos θ) lbf
5ft
5ft
∑ M = 0 = ∑ F X r = −67.6(cos θ) ( ) + (12 lbs)(32.2 ft/s) ( ) sin θ
2
2
0 = −67.6(cos θ) (
5ft
5ft
) + (12 lbs)(32.2 ft/s) ( ) sin θ
2
2
5ft
5ft
67.6(cos θ) ( ) = (12 lbs)(32.2 ft/s) ( ) sin θ
2
2
5ft
67.6 ( 2 )
sin θ
=
= 0.175
cos θ (12 lbs)(32.2 ft/s) (5ft)
2
tan θ = 0.175
θ = 9.92o
5.40
A circular water jet 2 inches in diameter strikes a concrete (SG = 2.6) slab as shown which rests
freely on a level floor. The slab is 40 inches tall, 10 inches deep and 5 inches wide into the paper.
Please calculate the jet velocity which will just begin to tip the slab over if the water is at 62.4
lbm/ft3 and the system is at 273K.
10 inches
Water jet
Concrete
Slab
2 inches
40 inches
30 inches
B
Need to find the water force and then the moments about the lower left corner of the slab at B.
The slab will tip when the center of mass is ½ the width at the tipping point B.
∂
∭ vx ρdV
∂t
∂
∑ M = ∬(rxv)z ρ(v ∙ n)dA + ∭(rxv)z ρdV
∂t
∑ F = ∬ vx ρ(v ∙ n)dA +
∑ Fx = Fjet = ∑ ṁout vout − ṁin vin = 0 − ṁin vin = −ρAv(−v) = ρv 2 A
Wslab = ρVg =
2.6(62.4 lbm /ft 3 ) 10
40
5
( ft) ( ft) ( ft) (32.2 ft/s2 ) = 187.9 lbf
2
32.174 lbm ft/lbf s 12
12
12
∑ MB = ρv 2 A (
30
5
ft) − Wslab ( ft)
12
12
2
62.4 lbm /ft 3
π 2
30 + 2/2
10 1
( ft) v 2 (
ft) = (187.9 lbf ) ( ft)
2
32.174 lbm ft/lbf s 4 12
12
12 2
Notes
2
2
30+( )
1. The center of the water jet is at a height of 30 inches + 2 inch/2 or (
10
1
2. The center of the slab is ½ it’s width or (12 ft) 2
0.1093 v 2 = 78.3 lbf ft
v = 26.7 ft/s
12
ft)
Chapter 6
Instructor Only Problems
6.41
You have been asked to do an analysis of a steady state pump that will transfer Aniline, an organic
material from one area to another in a chemical plant. Please determine the temperature change
that the fluid undergoes during this process. The flow rate is constant in the system at 1.0 ft3/sec.
The inlet to the pump has a diameter of 8 inches and the pressure to the pump is 250 lb f /ft2. The
outlet from the pump has a diameter of 5 inches and a pressure of 600 lbf /ft2. The fluid that will
be pumped has a density of 63.0 lbm/ft3, a heat capacity of 0.490 BTU/lbmoF, viscosity of 1.8x103
lbm/ft sec, and kinematic viscosity of 2.86x10-5 ft2/s. The outlet from the pump is located 75 feet
higher than the inlet to the pump. The pump provides work to the fluid at a rate of 8.3x106 lbmft2/s3
and the heat transferred to the CV from the pump is 1.40x106 BTU/hr. In this analysis you may
assume steady state and ignore viscous work.
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
δQ
BTU
lbm ft
hr
lbm ft 2
6
) (778.17 ft lbf /BTU) (32.174
)(
) = 9736549.5
= (1.4x10
∂t
hr
lbf s2 3600 s
s2
2
δWs
lbm ft
= 8.3x106
∂t
s2
δWμ
= 0 since there air is a very low viscosity fluid
∂t
∂
∭ eρdV = 0 since the system is a steady state
∂t
δQ δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
∂t
ρ
δQ δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
∂t
ρ
δQ δWs
p
p
−
= (e + ) ṁ − (e + ) ṁ
∂t
∂t
ρ 2
ρ 1
2
2
δQ δWs
v2 − v1
P2 P1
−
= ρQ [U2 − U1 +
+ ( − ) + g(z2 − z1 )]
∂t
∂t
2
ρ2 ρ1
Q
1 ft 3 /s
v1 =
=
2 = 2.865 ft/s
A1 π 8
4 (12 ft)
v2 =
Q
1 ft 3 /s
=
2 = 7.33 ft/s
A2 π 5
4 (12 ft)
δQ δWs
v22 − v12
P2 P1
−
= ρQ [U2 − U1 +
+ ( − ) + g(z2 − z1 )]
∂t
∂t
2
ρ2 ρ1
9736549.5
lbm ft 2
lb ft 2
6 m
+
8.3x10
s2
s2
= (63 lbm /ft 3 )(1 ft 3 /s) [U2 − U1 +
+(
(7.33ft/s)2 − (2.865ft/s)2
2
600 − 250 lbm /ft 2
) (32.174 lbm ft/lbf s 2 ) + 32.2 ft/s2 (75ft)]
63 lbm /ft 3
ft 2
U2 − U1 = 283666.0 2
s
∆U = Cv dT
Solve for dT
283666.0
dT =
ft 2
s2
BTU
ft lb
0.49
(778.17 BTUf ) (32.174 lbm ft/lbf s2 )
lbm ℉
= 23.12℉
6.42
Air flows steadily through a turbine that produces 3.5x105 ft-lbf/s of work. Using the data below
at the inlet and outlet, where the inlet is 10 feet below the outlet, please calculate the heat
transferred in units of BTU/hr. You may assume steady flow and ignore viscous work.
Inlet
diameter = 0.962 ft
pressure = 150 lbf/in2
Turbine
temperature = 300oF
density = 0.0534 lbm/ft3
heat capacity = 0.243 BTU/lbmoF
viscosity = 1.6x10-5 lbm/ft-s
kinematic vicsocity = 3.06x10-4 ft2/s
velocity = 100 ft/s
Outlet
diameter = 0.5 ft
pressure = 400 lbf/in2
temperature = 35oF
density = 0.0810 lbm/ft3
heat capacity = 0.240 BTU/lbmoF
viscosity = 1.5x10-5 lbm/ft-s
kinematic vicsocity = 1.42x10-4 ft2/s
velocity = 244 ft/s
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
δWμ
= 0 since there air is a very low viscosity fluid
∂t
∂
∭ eρdV = 0 since the system is a steady state
∂t
δQ δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
∂t
ρ
δQ δWs
p
p
−
= (e + ) ṁ − (e + ) ṁ
∂t
∂t
ρ 2
ρ 1
2
2
δQ δWs
v2 − v1
P2 P1
−
= ṁ [U2 − U1 +
+ ( − ) + g(z2 − z1 )]
∂t
∂t
2
ρ2 ρ1
U2 − U1 = Cv2 T2 − Cv1 T1
= [(0.240 BTU/lbm ℉)(35℉)
− (0.243 BTU/lbm ℉)(300℉)](778.17 ft lbf /BTU)(32.174 lbm ft/lbf s 2 )
= −1.61x106 ft 2 /s2
v22 − v12 (244 ft/s)2 − (100 ft/s)2
=
= 24768 ft 2 /s 2
2
2
lb
lb
2
400 2f
150 2f
P2 P1
in −
in ) (32.174 lbm ft) (144 in ) = 9.865x106 ft 2 /s 2
( − )=(
lb
lbm
ρ2 ρ1
lbf s2
ft 2
0.0810 m
0.0534
ft 3
ft 3
ft
g(z2 − z1 ) = 32.2 2
= 322 ft 2 /s2
s (10 ft)
Plug back into original equation,
δQ δWs
v22 − v12
P2 P1
−
= ṁ [U2 − U1 +
+ ( − ) + g(z2 − z1 )]
∂t
∂t
2
ρ2 ρ1
δQ
− 3.5x105 ft ∙ lbf /s
∂t
π(0.962 ft)2
(0.0534 lbm /ft 3 )(100 ft/s) (
)
4
[−1.61x106 ft 2 /s2
=
lbf s2
(
)
32.174 lbm ft
+ 24768 ft 2 /s 2 + 9.865x106 ft 2 /s2 + 322 ft 2 /s2 ]
δQ
lbf
− 3.5x105 ft ∙
= 9.9887x105 ft ∙ lbf /s
∂t
s
δQ
ft ∙ lbf
= 1.348x106
∂t
s
Convert to BTU/hr
δQ
ft ∙ lbf 3600 s
1
)(
)
= (1.348x106
= 6.24x106 BTU/hr
∂t
s
hr
778.17 ft lbf /BTU
6.43
You have been asked to do an analysis of a pump that will transfer a fluid. The inlet to the pump
has a diameter of 0.35 m and the pressure to the pump is 2500 kg/m-sec2. The outlet from the pump
has a diameter of 0.15 m and a pressure of 6000 kg/m-sec2. The fluid that will be pumped has a
viscosity of 1.09 x 10-3 kg / meter-sec and a density of 600 kg/m3. The outlet from the pump is
located 5 m higher than the inlet to the pump. In your analysis assume that there is no temperature
change between the inlet and outlet and that the system is running at steady state. Please determine
the amount of power that the pump must add to the fluid to maintain a constant flow rate of 10
m3/sec.
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
δWμ
= 0 since there is no viscous work
∂t
∂
∭ eρdV = 0 since the system is a steady state
∂t
δQ
= 0 from problem statement
∂t
δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
ρ
δWs
p
p
v22 − v12
P2 − P1
) + g(y2 − y1 )]
−
= ρQ (e + ) − ρQ (e + ) = ρQ [
+(
∂t
ρ 2
ρ 1
2
ρ
Q
10 m3 /s
v1 =
=
= 104 m/s
A1 π(0.35m)2
4
Q
10 m3 /s
v2 =
=
= 565.85 m/s
A2 π(0.15m)2
4
δWs
v22 − v12
P2 − P1
) + g(y2 − y1 )]
−
= ρQ [
+(
∂t
2
ρ
(565.85 m/s)2 − (104 m/s)2
3 )(10 3
(600
=
kg/m
m /s) [
2
6000 − 2500 kg/ms 2
+
+ (9.8 m/s2 )(5 m)] = 9.28x108 J/s
600 kg/m3
6.44
You have just been hired into a new manufacturing facility where you are the process engineer.
You are touring the facility with your new boss and she mentions to you that there is some concern
in the plant about the change in pressure (in units of lbf /ft2) in a new system you will be working
on, and asks you calculate this pressure change as soon as possible before the process is started.
The system will be pumping a fluid at steady state from Tank#1, which is at 30℉ with a viscosity
of 7.2 lbm/ft-s, heat capacity of 0.54 BTU/lbm℉ and kinematic viscosity of 9.03 ft2/s, up to Tank#2
that is at 100℉ with a viscosity of 7.10 lbm/ft-s, heat capacity of 0.598 BTU/lbm℉ and kinematic
viscosity of 0.128 ft2/s. The bottom tank is connected to the pump with a pipe that is 10 inches in
diameter. The outlet to the pump is connected to a pipe that is 4 inches in diameter. The flow rate
through the system is constant at 0.3 ft3/sec. At steady state the pump provides work to the fluid
of 4.3x107 lbmft2/s3 and the heat transfer in the system is 4.6 x106 BTU/hr. You may assume that
viscous work is negligible, and that Tank #1 is 40 feet above Tank #2, and that the system is at a
constant density of density of 79.7 lbm/ft3 (the assumption of constant density is a simplification
to allow the problem to be solved more easily).
Solution
From the problem statement, this requires the energy equation to solve.
𝛿𝑊𝜇
𝛿𝑄 𝛿𝑊𝑠
𝑃
𝜕
−
= ∬ (𝑒 + ) 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 + ∭ 𝑒𝜌𝑑𝑉 +
𝜕𝑡
𝜕𝑡
𝜌
𝜕𝑡
𝜕𝑡
𝛿𝑄
𝐵𝑇𝑈
𝑙𝑏𝑚 𝑓𝑡
ℎ𝑟
𝑙𝑏𝑚 𝑓𝑡 2
6
7
) (778.17 𝑓𝑡 𝑙𝑏𝑓 /𝐵𝑇𝑈) (32.174
) = 3.2𝑥10
= (4.6𝑥10
)(
𝜕𝑡
ℎ𝑟
𝑙𝑏𝑓 𝑠 2 3600 𝑠
𝑠3
𝜕
∭ 𝑒𝜌𝑑𝑉 = 0 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑎 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒
𝜕𝑡
𝛿𝑄 𝛿𝑊𝑠
𝑃
−
= ∬ (𝑒 + ) 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴
𝜕𝑡
𝜕𝑡
𝜌
2
𝛿𝑄 𝛿𝑊𝑠
𝑣22
𝑃2
𝑣21
𝑃1
)]
−
= 𝜌𝑄 [𝑈2 + + ( ) + 𝑔(𝑦2 − 𝜌𝑄 [𝑈1 +
+ ( ) + 𝑔(𝑦1 )]
𝜕𝑡
𝜕𝑡
2
𝜌
2
𝜌
𝒗𝟐𝟐
𝑷𝟐
𝝆𝟐 𝑸𝟐 [𝑼𝟐 +
+ ( ) + 𝒈(𝒚𝟐 )]
𝟐
𝜌
𝑙𝑏𝑓
𝑙𝑏𝑚
𝑓𝑡 3
𝐵𝑇𝑈
𝑙𝑏𝑚 𝑓𝑡
) (100℉) (778.17 𝑓𝑡
= 79.7 3 (0.3
) (0.598
) (32.174
)
𝑓𝑡
𝑠𝑒𝑐
𝑙𝑏𝑚 ℉
𝐵𝑇𝑈
𝑙𝑏𝑓 𝑠 2
[
2
𝑓𝑡 3
(0.3
)
𝑠𝑒𝑐
(
)
𝜋 4 2 2
𝑃2
𝑓𝑡
4 (12) 𝑓𝑡
+
+ + 32.2 2 (40 𝑓𝑡)
2
𝜌
𝑠
]
𝑙𝑏𝑚
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 4
𝑓𝑡
= 23.91
[1497203 2 + 5.91 2 + 𝑷𝟐 (0.4114)
+ 1288 2 ]
2
𝑠
𝑠
𝑠
𝑙𝑏𝑓 𝑠
𝑠
2
𝑙𝑏𝑚
𝑓𝑡
𝑃2
= 23.91
[1.5𝑥106 2 + ]
𝑠
𝑠
𝜌2
𝒗𝟐𝟏
𝑷𝟏
𝝆𝟏 𝑸𝟏 [𝑼𝟏 +
+ ( ) + 𝒈(𝒚𝟏 )]
𝟐
𝜌
𝑙𝑏𝑓
𝑙𝑏𝑚
𝑓𝑡 3
𝐵𝑇𝑈
𝑙𝑏𝑚 𝑓𝑡
) (30℉) (778.17 𝑓𝑡
= 79.7 3 (0.3
) (0.54
) (32.174
)
𝑓𝑡
𝑠𝑒𝑐
𝑙𝑏𝑚 ℉
𝐵𝑇𝑈
𝑙𝑏𝑓 𝑠 2
[
2
𝑓𝑡 3
(0.3
)
𝑠𝑒𝑐
(
)
𝜋 10 2 2
𝑷𝟏
𝑓𝑡
4 (12) 𝑓𝑡
+
+
+ 32.2 2 (0)
2
𝜌
𝑠
]
𝑙𝑏𝑚
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 4
= 23.91
[405596.8 2 + 0.1513 2 + 𝑷𝟏 (0.4036)
+0]
𝑠
𝑠
𝑠
𝑙𝑏𝑓 𝑠 2
𝑙𝑏𝑚
𝑓𝑡 2 𝑷𝟏
= 23.91
[4.05𝑥105 2 + ]
𝑠
𝑠
𝝆𝟏
Combine everything,
2
𝛿𝑄 𝛿𝑊𝑠
𝑣22
𝑃2
𝑣21
𝑃1
−
= 𝜌2 𝑄 [𝑈2 +
+ ( ) + 𝑔(𝑧2 )] − 𝜌1 𝑄 [𝑈1 +
+ ( ) + 𝑔(𝑧1 )]
𝜕𝑡
𝜕𝑡
2
𝜌2
2
𝜌1
7
3.2𝑥10
7.5𝑥107
𝑙𝑏𝑚 𝑓𝑡 2
𝑙𝑏 𝑓𝑡 2
7 𝑚
+ 4.3𝑥10
𝑠3
𝑠3
𝑙𝑏𝑚
𝑓𝑡 2
𝑃2
𝑙𝑏𝑚
𝑓𝑡 2
𝑃1
= 23.91
[1.5𝑥106 2 + ( )] − 23.91
[4.05𝑥105 2 + ( )]
𝑠
𝑠
𝜌
𝑠
𝑠
𝜌
𝑙𝑏𝑚 𝑓𝑡 2
𝑙𝑏𝑚
𝑓𝑡 2
𝑃2
𝑙𝑏𝑚
𝑓𝑡 2
𝑃1
6
5
(
)]
(
)]
=
23.91
[1.5𝑥10
+
−
23.91
[4.05𝑥10
+
𝑠3
𝑠
𝑠2
𝜌
𝑠
𝑠2
𝜌
3.14𝑥106
𝑓𝑡 2
𝑓𝑡 2
𝑃2
𝑓𝑡 2
𝑃1
6
5
(
)
(
)
=
1.5𝑥10
+
−
4.05𝑥10
+
𝑠2
𝑠2
𝜌
𝑠2
𝜌
𝑓𝑡 2 𝑃2 − 𝑃1
2.04𝑥10
=
𝑠2
𝜌
6
𝑃2 − 𝑃1 = 2.04𝑥106
𝑙𝑏𝑓
𝑓𝑡 2
𝑙𝑏𝑚
lbf s2
(79.7
)
(
) = 5.05𝑥106 2
2
3
𝑠
𝑓𝑡
32.174 lbm ft
𝑓𝑡
This is a high pressure, so it is a good thing that your boss asked you to calculate it so that proper
safety measures could be taken, the system probably should not be operating at this pressure and
changes will need to be made to reduce it!
6.45
As part of your responsibilities as a new hire in a chemical plant, you have been asked to buy a
pump. So that you know which pump to purchase, you need to calculate the work being done on
the system and they you will contact the supplier to discuss the size needed. The important data is
included in the table below:
Position
Density
(lbm/ft3)
Inlet
Outlet 1
Outlet 2
55.2
53.6
51.8
Heat
Capacity
(Btu/lbmoF)
0.395
0.420
0.450
Area of
pipe
(ft2)
0.34
1.00
0.25
Volumetric
Flow Rate
(ft3/s)
85.7
40.0
50.0
Temperature
(oF)
Pressure
(lbf/ft2)
60
100
150
Distance pipe is
from floor (ft)
20.0
30.0
21.5
Please calculate the work done by the system assuming steady flow of a Newtonian fluid, that
viscous work is negligible and that the heat transferred to the CV from the pump is 1.40x106 Btu/hr.
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
δQ
BTU
lbm ft
hr
lbm ft 2
) (778.17 ft lbf /BTU) (32.174
)
(
)
= (1.4x106
=
9736549.5
∂t
hr
lbf s2 3600 s
s3
δQ δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
∂t
ρ
δQ δWs
p
p
p
−
= (e + ) ṁ2 + (e + ) ṁ3 − (e + ) ṁ 1
∂t
∂t
ρ 2
ρ 3
ρ 1
δQ δWs
v22
P2
v32
P3
−
= (U2 + + gz2 + ) ṁ2 + (U3 + + gz3 + ) ṁ3
∂t
∂t
2
ρ2 2
2
ρ3 3
2
v1
P1
− (U1 + + gz1 + ) ṁ 1
2
ρ1 1
v1 =
Q1 85.7 ft 3 /s
=
= 252 ft/s
A1
0.34 ft 2
v2 =
Q2 40 ft 3 /s
=
= 40 ft/s
A2
1.0 ft 2
Q3 50 ft 3 /s
v3 =
=
= 200 ft/s
A3 0.25 ft 2
0.0
4.0
1.5
Position
Density
(lbm/ft3)
Inlet
Outlet 1
Outlet 2
55.2
53.6
51.8
Heat
Capacity
(Btu/lbmoF)
0.395
0.420
0.450
Area of
pipe
(ft2)
0.34
1.00
0.25
Volumetric
Flow Rate
(ft3/s)
132.5
40.0
50.0
Temperature
(oF)
60
100
150
Pressure
(lbf/ft2)
Distance pipe is
from floor (ft)
20.0
30.0
21.5
𝑣12
𝑃1
𝑣12
𝑃1
(𝑈1 + + 𝑔𝑧1 + ) 𝑚̇1 = (𝐶𝑣1 𝑇1 + + 𝑔𝑧1 + ) 𝜌1 𝑣1 𝐴1
2
𝜌1 1
2
𝜌1 1
(252 𝑓𝑡/𝑠)2
𝑓𝑡 𝑙𝑏𝑓
𝐵𝑇𝑈
𝑙𝑏𝑚 𝑓𝑡
) (60℉) (778.17
= [(0.395
) (32.174
)
+
𝑙𝑏𝑚 ℉
𝐵𝑇𝑈
𝑙𝑏𝑓 𝑠 2
2
2
2
20 𝑙𝑏𝑓 /𝑓𝑡 (32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 )
+ (32.2 𝑓𝑡/𝑠 2 )(0 𝑓𝑡) +
] (55.2 𝑙𝑏𝑚
55.2 𝑙𝑏𝑚 /𝑓𝑡 3
1
/𝑓𝑡 3 )(252 𝑓𝑡/𝑠)(0.34 𝑓𝑡 2 )
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 2
𝑙𝑏𝑚
)
= [593373.15 2 + 31752 2 + 0 + 11.66 2 ] (4729.6
𝑠
𝑠
𝑠
𝑠
𝑙𝑏𝑚 𝑓𝑡 2
= 2.95655𝑥109
𝑠3
𝑣22
𝑃2
𝑣22
𝑃2
(𝑈2 +
+ 𝑔𝑧2 + ) 𝑚̇2 = (𝐶𝑣2 𝑇2 +
+ 𝑔𝑧2 + ) 𝜌2 𝑣2 𝐴2
2
𝜌2 2
2
𝜌2 2
(40 𝑓𝑡/𝑠)2
𝑓𝑡 𝑙𝑏𝑓
𝐵𝑇𝑈
𝑙𝑏𝑚 𝑓𝑡
) (100℉) (778.17
= [(0.420
) (32.174
)
+
𝑙𝑏𝑚 ℉
𝐵𝑇𝑈
𝑙𝑏𝑓 𝑠 2
2
2
2
30 𝑙𝑏𝑓 /𝑓𝑡 (32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 )
+ (32.2 𝑓𝑡/𝑠 2 )(4 𝑓𝑡) +
] (53.6 𝑙𝑏𝑚
53.6 𝑙𝑏𝑚 /𝑓𝑡 3
1
/𝑓𝑡 3 )(40 𝑓𝑡/𝑠)(1.0 𝑓𝑡 2 )
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 2
𝑙𝑏𝑚
)
= [1051547.3 2 + 800 2 + 128.8 2 + 18.0 2 ] (2144
𝑠
𝑠
𝑠
𝑠
𝑠
𝑙𝑏𝑚 𝑓𝑡 2
= 2.2565𝑥109
𝑠3
0.0
4.0
1.5
𝑣32
𝑃3
𝑣32
𝑃3
(𝑈3 +
+ 𝑔𝑧3 + ) 𝑚̇3 = (𝐶𝑣3 𝑇3 +
+ 𝑔𝑧3 + ) 𝜌3 𝑣3 𝐴3
2
𝜌3 3
2
𝜌3 3
(200 𝑓𝑡/𝑠)2
𝑓𝑡 𝑙𝑏𝑓
𝐵𝑇𝑈
𝑙𝑏𝑚 𝑓𝑡
) (150℉) (778.17
= [(0.450
) (32.174
)
+
𝑙𝑏𝑚 ℉
𝐵𝑇𝑈
𝑙𝑏𝑓 𝑠 2
2
2
2
21.5 𝑙𝑏𝑓 /𝑓𝑡 (32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 )
+ (32.2 𝑓𝑡/𝑠 2 )(1.5 𝑓𝑡) +
] (51.8 𝑙𝑏𝑚
51.8 𝑙𝑏𝑚 /𝑓𝑡 3
1
/𝑓𝑡 3 )(200 𝑓𝑡/𝑠)(0.25 𝑓𝑡 2 )
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 2
𝑙𝑏𝑚
)
= [1689986.8 2 + 20,000 2 + 48.3 2 + 13.35 2 ] (2590
𝑠
𝑠
𝑠
𝑠
𝑠
𝑙𝑏𝑚 𝑓𝑡 2
= 4.43𝑥109
𝑠3
9736549.5
𝑙𝑏𝑚 𝑓𝑡 2 𝛿𝑊𝑠
𝑙𝑏 𝑓𝑡 2
𝑙𝑏 𝑓𝑡 2
𝑙𝑏 𝑓𝑡 2
9 𝑚
9 𝑚
9 𝑚
−
=
4.43𝑥10
+
2.2565𝑥10
−
2.95655𝑥10
𝑠3
𝜕𝑡
𝑠3
𝑠3
𝑠3
𝛿𝑊𝑠
𝑙𝑏 𝑓𝑡 2
9 𝑚
−
= 3.73𝑥10
𝜕𝑡
𝑠3
𝛿𝑊𝑠
𝑙𝑏𝑚 𝑓𝑡 2
= −3.73𝑥109
𝜕𝑡
𝑠3
Could have also expressed as,
𝑙𝑏𝑓 𝑓𝑡
𝛿𝑊𝑠
𝐵𝑡𝑢
= −1.16𝑥108
= −1.5𝑥105
= −5.38𝑥108 𝐵𝑡𝑢/ℎ𝑟
𝜕𝑡
𝑠
𝑠
6.46
A Newtonian fluid is being pumped through a pumping station that has fluid in inlet and outlet
pipes. The inlet fluid is at a temperature of 80oF, density of 50 lbm/ft3, viscosity of 1.80 x 10-3 lbm/ft
sec, heat capacity of 0.58 Btu/lbmoF and kinematic viscosity of 3.60 x 10-5 ft2/sec. The pressure at
the inlet is 300 psi. The outlet fluid is at a temperature of 100oF, density of 49.6 lbm/ft3, viscosity
of 1.30 x 10-3 lbm/ft sec, heat capacity of 0.61Btu/lbmoF and kinematic viscosity of 2.62 x 10-5
ft2/sec. The pressure at the outlet is 750 psi. The outlet is 167 ft above the inlet, and the pump
provides work to the fluid at 1 x 108 lbm ft2/s3. The inlet has a diameter of 0.5 feet, but the outlet
diameter is unknown. The inlet velocity is 86 ft/s and the outlet velocity is 184 ft/s. For this analysis
you may assume that the fluid undergoes no viscous work, and that the system is at steady state.
Find the number of Btu’s added to the system after 4 hours.
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
∂
∭ eρdV = 0 since the system is at steady state
∂t
δWs
−
= 1x108 lbm ft 2 /s3
∂t
δQ δWs
p
p
−
= (e + ) ṁ − (e + ) ṁ
∂t
∂t
ρ 2
ρ 1
δQ δWs
v22
P2
v12
P1
−
= ρ2 v2 A2 [U2 + + ( ) + g(z2 )] − ρ1 v1 A1 [U1 + + ( ) + g(z1 )]
∂t
∂t
2
ρ2
2
ρ1
We are not given the exit diameter, we could calculate it, but we don’t need to since we are told
that the system is at steady state so we could just use,
ρ2 v2 A2 = ρ1 v1 A1
Or we could calculate it,
πD22
π(0.5ft)2
) = (50 lbm /ft 3 )(86 ft/s) (
)
4
4
πD22
= 0.0925 ft 2
4
(49.6 lbm /ft 3 )(184 ft/s) (
D2 = 0.343 ft
Or you could simply say that ṁ1 = ṁ2 = ṁ, and not bother to calculate D2.
δQ δWs
v22 − v12
P2 P1
−
= ṁ [U2 − U1 +
+ ( − ) + g(z2 − z1 )]
∂t
∂t
2
ρ2 ρ1
Plugging in the values,
δQ
+ 1x108 lbm ft 2 /s3
∂t
= (50 lbm /ft 3 )(86 ft
π(0.5ft)2
𝐵𝑡𝑢
) (100℉)
/s) (
) [{(0.61
4
𝑙𝑏𝑚 ℉
𝑓𝑡 𝑙𝑏𝑓
𝐵𝑡𝑢
lbm ft
) (80℉)} (778.17
)
− (0.58
) (32.174
𝑙𝑏𝑚 ℉
𝐵𝑡𝑢
lbf s 2
(184 ft/s)2 − (86 ft/s)2
+
2
750 lbf /in2 300 lbf /in2 144 in2
lbm ft
(32.174
)
+(
−
)
(
)
49.6 lbm /ft 3 50 lbm /ft 3
ft 2
lbf s 2
+ 32.2 ft/s 2 (167 ft)]
δQ
lbm ft 2
lbm
ft 2
ft 2
ft 2
ft 2
+ 1x108
=
844.3
[365537.88
+
13230
+
42257.95
+
5377.4
]
∂t
s3
s
s2
s2
s2
s2
δQ
lb ft 2
lbm
ft 2
8 m
+ 1x10
= 844.3
[426403.23 2 ]
∂t
s3
s
s
δQ
lbm ft 2
lb ft 2
8 m
+ 1x108
=
3.6
x
10
∂t
s3
s3
2
2
δQ
lbm ft
lbf s
BTU
Btu
) = 10380
= 2.6x108
(
)(
3
∂t
s
32.174 lbm ft 778.17 ft lbf
s
10380
Btu 60 sec 60 min
x
x
x 4 hr = 1.49x108 Btu
s
min
hr
6.47
Venturi meters are commonly used to measure the flow rate of gases and liquids where the gas or
liquid is considered frictionless and incompressible under the conditions of the flow. Air at a
constant temperature of 75℉, is flowing steadily from left to right (as illustrated in the figure by
the arrows). At position labeled 1 the diameter is 3 inches and the absolute pressure is 14 psi. At
the position labeled 2, the diameter is 2 inches and the absolute pressure is 12 psi. Assuming that
the flow is Newtonian, isothermal and inviscid with no change in internal energy, please calculate
the volumetric flow rate of the air. (Note: at these pressures, air can be considered an
incompressible fluid.)
1
2
Solution
P1 v12
P2 v22
+ + gy1 =
+ + gy2
ρ
2
ρ
2
Q = v1 A1 = v2 A2
A1 =
A2 =
v1 =
Q
A1
v2 =
Q
A2
1 ft 2
π (3in x 12 in)
4
1 ft 2
π (2 in x 12 in)
4
= 0.05 ft 2
= 0.022 ft 2
𝑙𝑏
144 𝑖𝑛2
𝑙𝑏
𝑃1 = 14 2 𝑥
= 2016 2
2
𝑖𝑛
𝑓𝑡
𝑓𝑡
𝑃2 = 12
𝑙𝑏
144 𝑖𝑛2
𝑙𝑏
𝑥
= 1728 2
2
2
𝑖𝑛
𝑓𝑡
𝑓𝑡
P1 v12
P2 v22
+ + gy1 =
+ + gy2
ρ
2
ρ
2
y1 = y2
P1 v12
P2 v22
+
=
+
ρ
2
ρ
2
v22 − v12
P1 − P2 = ρ (
)
2
Plug in values for v1 and v2 in terms of Q which is what we are solving for,
Q 2
Q 2
(A ) − (A )
1
)
P1 − P2 = ρ ( 2
2
Solve for Q:
P1 − P2
Q 2
Q 2
)=( ) −( )
2(
ρ
A2
A1
𝑙𝑏
𝑙𝑏
− 1728 2
2
1
1
𝑓𝑡
𝑓𝑡
2
= 𝑄2 [ 2 − 2 ]
2
𝑙𝑏𝑓 𝑠
𝐴2 𝐴1
𝑙𝑏
)
0.075 𝑚3 (
32.174
𝑙𝑏
𝑓𝑡
𝑓𝑡
(
)
𝑚
2016
𝑙𝑏
𝑙𝑏
− 1728 2
𝑓𝑡 2
𝑓𝑡
)
2(
𝑙𝑏𝑓 𝑠 2
𝑙𝑏𝑚
𝑓𝑡 2
)
0.075 3 (
247096 2
𝑓𝑡 32.174 𝑙𝑏𝑚 𝑓𝑡
𝑠 = 12.2 𝑓𝑡 3⁄
𝑄=
=√
𝑠
1
1
1
1666
[
−
]
2 2
(0.05 𝑓𝑡 2 )2
𝑓𝑡 4
√ (0.022 𝑓𝑡 )
2016
6.48
Water at 293 K is flowing steadily from a hose with a diameter of 1.6 cm. You cover the hose
outlet with your thumb causing the area of the opening to be 25% of what it was originally,
resulting a thin jet of high-speed water to emerge at a velocity of 3 m/s. If the hose is held upward
(perfectly vertical), and the pressure of the water in the hose just before it exits is 40 kPa gauge,
what is the height above the nozzle opening that the water jet will achieve? You may assume
inviscid, isothermal flow with no change in internal energy and no shaft work occurs in the control
volume.
Solution
Assumptions: incompressible, Newtonian fluid, inviscid, isothermal flow with no change in
internal energy and no shaft work
P1 v12
P2 v22
+ + gy1 =
+ + gy2
ρ
2
ρ
2
v2 = 0 at the top of the jet, y1 = 0 a the top of hose and bottome of jet
𝑃𝑔𝑎𝑢𝑔𝑒 = 𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 − 𝑃𝐴𝑇𝑀 = 40 𝑘𝑃𝑎 (𝑔𝑎𝑢𝑔𝑒)
𝑦2 =
(3 𝑚/𝑠)2
𝑃1 − 𝑃2 𝑣12
40𝑥103 𝑘𝑔/𝑚𝑠 2
+
=
+
= 4.5 𝑚
𝜌𝑔
2𝑔 (998.2 𝑘𝑔/𝑚3 )(9.8 𝑚/𝑠 2 ) 2(9.8 𝑚/𝑠 2 )
6.49
A system is pumping a fluid at steady state from tank #1 to tank #2. The data for the two tanks is
included in the table below. Tank #1 is connected to the pump with a pipe that is 10 inches in
diameter and is at a pressure of 300 lbf/ft2 as it enters the pump. The outlet to the pump is connected
to a pipe that is 4 inches in diameter and at a pressure of 850 lbf/ft2. The flow rate through the
system is constant at 0.3 ft3/min. Please calculate the work that the pump provides to the fluid if
the heat transfer in the system is 4.65x103 BTU/hr, and tank #2 is 100 ft above tank #1. Viscous
work is assumed negligible. (You may assume that the pump and tank#1 are at the same height.)
Tank
Density
(lbm/ft3)
Viscosity
(lbm/ft s)
Heat
Capacity
(BTU/lbmoF)
0.563
Temperature
(oF)
1.4
Kinematic
Viscosity
(ft2/s)
1.77x10-2
#1
79.1
#2
78.7
0.6
0.762x10-2
0.580
80
60
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
δQ
BTU
lb
ft
hr
lbm ft 2
m
= (4.65x103
) (778.17 ft lbf /BTU) (32.174
)(
) = 3.23x104
2
∂t
hr
lbf s
3600 s
s3
∂
∭ eρdV = 0 since the system is a steady state
∂t
δQ δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
∂t
ρ
δQ δWs
v22
P2
v22
P1
−
= ρ2 Q [U2 + + ( ) + g(z2 )] − ρ1 Q [U1 + + ( ) + g(z1 )]
∂t
∂t
2
ρ2
2
ρ1
𝛒𝟐 𝐯𝟐 𝐀𝟐 [𝐔𝟐 +
𝐯𝟐𝟐
𝐏𝟐
+ ( ) + 𝐠(𝐳𝟐 )]
𝟐
𝛒𝟐
= 78.7
lbm
ft 3
min
BTU
lbf
lbm ft
(0.3
)(
) (0.598
) (80℉) (778.17 ft
) (32.174
)
ft 3
min 60 sec
lbm ℉
BTU
lbf s2
[
2
ft 3
min
)
(0.3
)(
min 60 sec
(
)
2
π 4
850lbf
( ) ft 2
lbm ft
ft
4 12
ft 2
+
+
(32.174
) + 32.2 2 (y2 )
2
lb
2
lbf s
s
78.7 m
ft 3
]
lbm
ft 2
ft 2
ft 2
ft
lbm ft 2
6
−3
(100
= 0.3935
[1.2x10 2 + 1.64x10
+ 347.5 2 + 32.2 2
ft)] = 471319.5
2
s
s
s
s
s
s s2
𝛒𝟏 𝐯𝟏 𝐀𝟏 [𝐔𝟏 +
𝐯𝟏𝟐
𝐏𝟏
+ ( ) + 𝐠(𝐳𝟏 )]
𝟐
𝛒𝟏
= 79.1
lbm
ft 3
min
BTU
lbf
lbm ft
(0.3
)(
) (0.563
) (60℉) (778.17 ft
) (32.174
)
ft 3
min 60 sec
lbm ℉
BTU
lbf s2
[
2
ft 3
min
)
(0.3
)(
min 60 sec
(
)
2
π 10
300lbf
( ) ft 2
lbm ft
ft
4 12
ft 2
(0 ft)
+
+
(32.174
2 ) + 32.2 s 2
lbm
2
lb
s
f
79.1 3
ft
]
lbm
ft 2
ft 2
ft 2
lbm ft 2
−5
= 0.3955
[845744.5 2 + 4.2x10
+
122
+
0]
=
334491.95
s
s
s2
s2
s s2
Combine everything,
3.23x104
lbm ft 2 δWs
lbm ft 2
lbm ft 2
lbm ft 2
+
= (471319.5
− 334491.95
) = 136827.55
3
2
2
s
∂t
s s
s s
s s2
δWs
lb ft 2
= 1.045x105 m3
∂t
s
Chapter 7
Instructor Only Problems
7.21
A thin coating is to be applied to both sides of a piece of thin plastic that is being mechanically
transported. The plastic is 4.5 μm thick and 0.0254 meters wide and is very long. We want to coat
a specific length of the plastic that is 1 meter. This thin plastic will break if the force applied
exceeds 20 lbf. The thin coating is made of transporting it mechanically through a narrow gap
that determines the thickness of the coating. The plastic is centered in the gap with a clearance
of 1. 0 μm on the top and bottom (as a result the coating is 1. 0 μm thick on both the top and
bottom). The incompressible Newtonian fluid coating fluid a temperature of 80oF, density of 52.5
lbm/ft3, heat capacity of 0.453 BTU/lbmoF, viscosity of 6.95x10-3 lbm/ft-s, and kinematic viscosity
of 1.33x10-4 ft2/s and completely fills the space between the plastic and gap for a length of 0.75
inches along the plastic. Calculate the velocity with which the tape can be transported through
the gap using the maximum force (so that the tape does not break).
Solution
Area = length x width = (1 meter)(0.0254 meter) = 0.0254 m2 = 0.2734 ft2
Since we have a Newtonian fluid we can use Newton’s law:
F
dv
=τ=μ
A
dy
Rearrange, solving for dv and since the tape has two sides we must multiply the area by 2, and
integrating,
v
∫ dv =
0
y
F
∫ dy
2μA 0
1.0 μm = 3.28x10-6 ft on each side
v=
(20 lbf )(3.28x10−6 ft)(32.174 lbm ft/lbf s2 )
F
ft
(y) =
= 0.55 = 33.08 ft/min
−3
2
2μA
2(6.95x10 lbm /ft s)(0.2754 ft )
s
7.22
A viscometer is used to measure viscosity when the fluid in common laboratory settings. A
rheometer is more sophisticated instrement in that it measures the way a liquid flows in response
to an applied force. A rheometer is used to study fluids which cannot be defined by a single
viscosity value and thus measures the rheology of the fluid. In a concentric cylinder (couette
geometry) rheometer, the rotating of the cup produces a strain on the sample, which in turn
applies a torque on the center cylinder. A transducer measures the magnitude of the torque, and
(based on the geometry of the system) determines the resultant shear stress from the applied
strain. Figure 1 shows the geometry of the experiment. The fluid lies between R 0 and R i where
R i = 16.00 mm and R 0 = 17.00 mm. The gap between the two cylinders can be modeled as two
parallel plates separated by a fluid. The torque is the force times the moment arm, which is the
radius 𝑅𝑖 of the inner cylinder. The tangential velocity is equal to the angular velocity times the
radius. The inner cylinder is rotated at 6000 rpm and the torque is measured to be 0.03 Nm. The
length of the inner cylinder is 33.4 mm. Determine the viscosity of the fluid. We can assume that
the inner cylinder is completely submerged in the fluid, that the fluid is Newtonian and that the
viscous end effects of the two ends of the inner cylinder are negligible.
Solution
We are given that “The torque is the force times the moment arm, which is the radius R i of the
inner cylinder. The tangential velocity is equal to the angular velocity times the radius.”
In equation format these statements become:
Torque = T = FR i
Velocity = v = ωR
We know that the wetted surface area of the inner cylinder is: A = 2πRL.
As a result, we can express the torque as:
μ(2πRL)(ωR)
T = FR =
ℓ
We can rewrite this equation noting that, if the cylinder length is L and the number of revolutions
per unit time is ṙ , which is expressed here as revolutions per minute (rpm). One note, the angular
distance traveled during one rotation is (2π rad) and the rpm is ω = 2πṙ . Thus,
T = FR = τAR =
μv(2πRL)R μωR(2πRL)R μ(2πṙ )R(2πRL)R μ4π2 R3
=
=
=
ṙ L
ℓ
ℓ
ℓ
ℓ
Rearranging and solving for viscosity,
Tℓ
μ= 2 3
4π R ṙ L
m
(0.03 Nm)(1 mm) (
1000
mm)
=
3
1m
rotations
min
m
(33.4 mm) (
) (6000
4π2 (16 mm)3 (
)
(
)
3
3
minute
60 sec
1000 mm)
1000 mm
Ns
= 0.0555 2 = 55.54 centipoise
m
7.23
A Non-Newtonian fluid is being sheared between two parallel plates. The top plate is moving at a
velocity of 0.09 ft/sec and the bottom plate is stationary. The two plates are separated by a distance
of 0.008 ft. The fluid is at 100℉, with a density of 84.3 lbm/ft3, heat capacity of 0.0331 BTU/lbm℉,
and surface tension of 2.5x10-3 lbf /inch. If the force on the fluid by the moving plate is 2.3 𝑙𝑏𝑓 and
the area of contact is 0.05 ft2, and the yield stress of 38.45 lbf/ft2, please calculate the viscosity of
this fluid in this situation.
Solution
Begin with the appropriate equation for a Non-Newtonian fluid, incorporating the yield stress.
τ=μ
dv
+ τ0
dy
F
dv
= μ + τ0
A
dy
2.3 𝑙𝑏𝑓
𝑙𝑏𝑓 𝑠 2
𝑙𝑏𝑓
0.09 ft/s
=
μ
(
) + 38.45 2
2
0.05 ft
0.008 ft 32.174 𝑙𝑏𝑚 𝑓𝑡
ft
46
𝑙𝑏𝑓
𝑙𝑏𝑓 𝑠
𝑙𝑏𝑓
= μ(0.35)
+ 38.45 2
2
ft
𝑙𝑏𝑚 𝑓𝑡
ft
μ = 21.57
𝑙𝑏𝑚
ft − s
7.24
A Non-Newtonian fluid is being sheared between two parallel plates. The top plate is moving at
a velocity of 0.09 ft/sec and the bottom plate is stationary. The two plates are separated by a
distance of 0.008 ft. The fluid is at 100℉, with a density of 84.3 lbm/ft3, heat capacity of 0.0331
BTU/lbm℉, surface tension of 2.5x10-3 lbf /inch, and kinematic viscosity of 0.256 ft2/sec. If the
force on the fluid by the moving plate is 6 𝑙𝑏𝑓 and the area of contact is 0.05 ft2, please calculate
the yield stress in this situation.
Solution
Begin with the appropriate equation for a Non-Newtonian fluid,
dv
τ = μ + τ0
dy
F
dv
= μ + τ0
A
dy
6 𝑙𝑏𝑓
𝑙𝑏𝑓 𝑠 2
0.09 ft/s
3 )(3.56 2
(84.3
=
lbm /ft
ft /s)
(
) + τ0
0.05 ft 2
0.008 ft 32.174 𝑙𝑏𝑚 𝑓𝑡
120
𝑙𝑏𝑓
𝑙𝑏𝑓
= 104.93 2 + τ0
2
ft
ft
τ0 = 15.07
𝑙𝑏𝑓
ft 2
7.25
A Bingham plastic is being sheared between two parallel plates. This material has a yield stress of
90 lbf/ft2, a density of 71 lbf/ft3. If the velocity of the upper plate is 0.041 ft/s and weight 0.5 lbf.
The plates are 0.5 inches apart and have an area of 1.2 ft2 what is the viscosity of this fluid? (Hint:
remember that the units of viscosity are lbm/ft s).
Solution
𝜏𝑦𝑥 = 𝜇
𝑑𝑣𝑥
+ 𝜏0
𝑑𝑦
𝐹
𝑑𝑣𝑥
=𝜇
+ 𝜏0
𝐴
𝑑𝑦
7 𝑙𝑏𝑓
𝑙𝑏𝑓 𝑠 2
𝑙𝑏𝑓
0.041 𝑓𝑡/𝑠
=
𝜇
+ 3.3 2
2
0.5
0.75 𝑓𝑡
32.174 𝑙𝑏𝑚 𝑓𝑡
𝑓𝑡
12 𝑓𝑡
0.5
7 𝑙𝑏𝑓
𝑙𝑏𝑓
12 𝑓𝑡 32.174 𝑙𝑏𝑚 𝑓𝑡 = 33.7 𝑙𝑏𝑚
𝜇=(
−
3.3
)
0.75 𝑓𝑡 2
𝑓𝑡 2 0.24 𝑓𝑡/𝑠
𝑙𝑏𝑓 𝑠 2
𝑓𝑡 𝑠
Chapter 8
Instructor Only Problems
8.20
You have been asked to calculate the density of an incompressible Newtonian fluid in steady flow
that is flowing continuously at 250℉ down a 2500 foot pipe with a constant diameter of 4 inches
and a volumetric flow rate of 2.5 ft3/s. The only fluid properties known is that the kinematic
viscosity is 7.14x10-6 ft2/sec and a surface tension is 0.0435 N/m. Assuming that the flow is laminar
and fully developed with a pressure drop of 256 lbf/ft2, please calculate the density of this fluid if
the no-slip boundary condition applies. (Hint: the units of density are lbm/ft3.)
Solution
dP 32μvavg
=
dx
D2
∆P 32μvavg
=
L
D2
3
Q
2.5 ft /s
vavg = =
2 = 28.65 ft/s
A π 4
4 (12 ft)
Rearrange the Hagen-Poiseuille Equation and solve for viscosity:
2
4
(256 lbf /ft 2 ) ( ft)
∆PD2
lbf s
12
μ=
=
= 1.24x10−5 2
32Lvavg 32(2500 ft)(28.65 ft/s)
ft
Realizing the relationship between density, viscosity and kinematic viscosity:
lb s
1.24x10−5 f2
μ
ft (32.174 lbm ft) = 55.93 lb /ft 3
ρ= =
m
ν 7.14x10−6 ft 2 /s
lbf s2
−
8.21
Determine the pressure drop across the ends of a pipe that is 10 meters long, 0.1 meters in
diameter through which is flowing a Newtonian oil with a viscosity of 6 Pa-s and a flow rate of
8.5 cubic meters per hour.
Solution
−
𝑑𝑃 8𝜇𝑣𝑎𝑣𝑔
=
𝑑𝑥
𝑅2
∆𝑃 8𝜇𝑣𝑎𝑣𝑔
=
𝐿
𝑅2
𝜋𝐷2
𝑄 = 𝑣𝐴 = 𝑣 (
) = 𝑣𝜋𝑅 2
4
∆𝑃 =
8𝐿𝜇𝑄 8(10 𝑚)(6 𝑃𝑎 ∙ 𝑠)(8.5 𝑚3 ⁄ℎ𝑟)(1 ℎ𝑟 ⁄3600𝑠𝑒𝑐 )
=
= 5.77𝑥104 𝑃𝑎
𝜋𝑅 4
𝜋(0.05𝑚)4
8.22
Please calculate the density of an incompressible, Newtonian fluid in steady flow that is flowing
continuously at 250℉ down a 2500 foot pipe with a constant diameter of 4 inches and a volumetric
flow rate of 2.5 ft3/s. This fluid has a heat capacity of 0.527 BTU/lbm℉, a kinematic viscosity of
7.14x10-6 ft2/sec and a surface tension of 0.0435 N/m. Assuming that the flow is laminar and fully
developed with a pressure drop of 256 lbf/ft2, please calculate the density of this fluid if the noslip boundary condition applies.
Solution
Assumptions: incompressible, Newtonian, Steady, Continuous, Fully Developed, Laminar and
the No-Slip Boundary Condition suggests the use of the Hagen-Poiseuille Equation.
−
dP 32μvavg
=
dx
D2
∆P 32μvavg
=
L
D2
Q
2.5 ft 3 /s
vavg = =
2 = 28.65 ft/s
A π 4
4 (12 ft)
Rearrange the Hagen-Poiseuille Equation and solve for viscosity:
2
2) 4
2
(256
lb
/ft
(
ft)
f
∆PD
lbf s
12
μ=
=
= 1.24x10−5 2
32Lvavg 32(2500 ft)(28.65 ft/s)
ft
Realizing the relationship between density, viscosity and kinematic viscosity:
lb s
1.24x10−5 f2
μ
ft (32.174 lbm ft) = 55.93 lb /ft 3
ρ= =
m
−6
ν 7.14x10 ft 2 /s
lbf s2
8.23.
Water at 293 K, heat capacity of 4182 J/kg-K, density of 998.2 kg/m3, kinematic viscosity of 9.95
x10-7 m2/s and viscosity of 9.93 x10-4 Pa-s in fully developed, laminar flow is flowing continuously
and steadily through a tube with a diameter of 10 cm and velocity of 45 m/s where the pressure
drop is measured to be 15,000 Pa, please estimate the length of the pipe over which the pressure
drop was determined.
Solution
−
dP 32μvavg
=
dx
D2
∆P 32μvavg
=
L
D2
2
m
)
∆PD
100cm = 104.9 m
L=
=
−4
32μvavg
32(9.93 x10 Pa ∙ s)(45 m/s)
2
(15,000 Pa) (10 cm x
8.24
Benzene flows steadily and continuously at 100oF through a 4500 ft horizontal pipe with a
constant cross-sectional area of 12.56 in2. The pressure drop across the pipe under these
conditions is 300 lbf/ft2. Assuming fully developed, parabolic, laminar, incompressible flow,
calculate the maximum velocity and volumetric flow rate of this fluid.
Solution
𝜋𝐷2
𝐴=
4
So,
𝐷=√
4𝐴
4(12.56 𝑖𝑛2 )
√
=
= 4 𝑖𝑛
𝜋
𝜋
−
dP 32μvavg
=
dx
D2
∆P 32μvavg
=
L
D2
2
vavg =
2
4
(300 𝑙𝑏𝑓 /𝑓𝑡 2 ) (12 𝑓𝑡)
∆PD
lbm ft
=
32.174
= 22.56 𝑓𝑡/𝑠
32𝜇𝐿 32 (3.3𝑥10−4 𝑙𝑏𝑚 ) (4500 𝑓𝑡)
lbf s 2
𝑓𝑡 𝑠
12.56 𝑖𝑛2
(22.56
Q = vavg 𝐴 =
𝑓𝑡/𝑠) (
) = 1.97 𝑓𝑡 3 /𝑠
2
2
144 𝑖𝑛 /𝑓𝑡
v𝑚𝑎𝑥 = 2vavg = 45 𝑓𝑡/𝑠
Chapter 9
Instructor Only Problems
9.25
A Newtonian fluid is flowing through a cylindrical conduit. Beginning with the appropriate
form of the Navier-Stokes Equations (see Appendix E), derive the following equation,
0 = −r
∂P ∂
+ (r τrz )
∂z ∂r
State the reason why you eliminated any terms in your derivation.
9.26
A film in free stream flow is flowing down
an inclined plan as shown in the figure.
Beginning with the appropriate form of the
Navier-Stokes equations, please derive the
following equation:
∂
τ + ρgsin(θ) = 0
∂y yx
As always, please state all assumptions you
make in your derivation.
9.27
An incompressible Newtonian fluid is confined between two concentric circular cylinders of
infinite length, a solid inner cylinder of radius R, and a hollow stationary outer cylinder of radius
Ro. See the figure below where the z-axis is coming out of the page. The inner cylinder rotates
with an angular velocity ωi . The flow is steady, laminar and two-dimensional in the rθ-plane. The
flow is also rotationally symmetric, meaning that nothing is a function of coordinate θ (so vθ and
P are functions of the radius, r only) and that gravity can be ignored. We can also assume that the
flow is circular, meaning that the velocity component vr = 0 everywhere. Please derive an exact
expression for the velocity component vθ as a function of the radius r and the other parameters in
the problem.
i
R0
Liquid: ,
Ri
Rotating inner cylinder
Stationary outer cylinder
9.28
Beginning with the appropriate form of the Navier-Stokes Equation, please derive an expression
for the velocity of a fluid flowing through a vertical pipe with diameter equal to D. For this
derivation, please assume steady, incompressible, fully developed flow and that the fluid is not
open to the atmosphere. Please be sure to indicate the reason why you eliminate any terms in the
original equation as you proceed with your derivation. (Hint: assume that at the center of the pipe,
r = 0)
9.29
Fluid
Direction
y
𝜃
x
h
VB
A Newtonian fluid is flowing down an incline
between two vertical, infinitely long parallel plates
separated by a distance h as shown in the figure. The
plate on the right is moving up with a velocity of VT
and the plate on the left is moving up with a velocity
of VB . The velocities of the plates are different such
that VT ≠ VB . The fluid is not open to the atmosphere
and is flowing in one direction only as shown in the
figure at an angle 𝜃 with the y-axis. Assuming steady,
incompressible, fully developed, laminar flow, please
derive an equation for the velocity of the fluid. You
must use the coordinate system shown in the figure in
your solution to the problem (which means that you
cannot draw or make another coordinate system).
VT
9.30
Beginning with the appropriate Navier-Stokes equation, please derive the appropriate equation to
describe the flux of 𝐳-momentum in the y-direction assuming assume steady, fully developed flow
between two infinitely long vertical parallel plates separated by a distance L, where the fluid is not
open to the atmosphere and parabolic flow can be assumed. Please indicate the reason why you
eliminate any terms in the equations as you proceed with your derivation. [Hint: we want an
equation to describe 𝐳-momentum in the y-direction and NOT 𝐲-momentum in the 𝐳-direction.]
9.31
Two different Newtonian fluids, so that they have different densities and viscosities (Fluid 1 and
Fluid 2 in the figure), are flowing steadily down between infinitely long vertical plates as shown
in the figure below. The plate at the left is moving down at a velocity of VA, the center plate is
moving up at a velocity of VB and the plate at the right is stationary. In the figure, the distances
between the plates are not equal so that L1≠L2.
(1) Using the appropriate form of the Navier-Stokes Equations, please derive equations for the
velocity of Fluid 1 and the velocity of Fluid 2 assuming that both fluids are in fully developed,
continuous, incompressible, adiabatic, isothermal flow. You may also assume that the no-slip
boundary condition applies.
(2) Please draw the velocity profiles for both Fluid 1 and Fluid 2 in the figure provided below. Be
careful in your drawing so it is clear.
9.32.
Consider steady, incompressible, laminar, fully developed flow of a Newtonian fluid in an
infinitely long, stationary pipe of diameter D inclined at an angle of 𝛽 to the horizontal plane as
shown in the figure for problem 9.24. You may assume that the No-Slip boundary condition
applies.
(a) Show that the continuity equation holds in this system,
(b) Using the appropriate form of the Navier-Stokes Equation, derive an expression to describe the
velocity of this fluid. Be sure to indicate the reason why you eliminate any terms in the original
equation as your proceed with your derivation.
9.33
V0
Fluid 2
h2
Interface
Fluid 1
h1
Two immiscible viscous fluids are sandwiched between two infinitely long and wide parallel flat
plates as shown in the figure above. Because the fluids are immiscible they do not mix at any
point. At time t = 0, the fluids and the plates on either side are stationary. The upper plate is set
in motion at a velocity V0, and the bottom plate remains stationary at all times. The flow is
incompressible, parallel, fully developed and laminar. The flow is caused solely by the viscous
effects created by the moving upper plate. You may ignore gravity, and assume that the fluids
are not open to the atmosphere and that the interface between the fluids is horizontal at all
times.
(a) In the figure above please draw the velocity profiles for both Fluid 1 and Fluid 2 in this system.
(b) Please derive equations for the velocity profiles of both fluids when the flow is at steady state.
9.25
A Newtonian fluid is flowing through a cylindrical conduit. Beginning with the appropriate
form of the Navier-Stokes Equations (see Appendix E), derive the following equation,
0 = −r
∂P ∂
+ (r τrz )
∂z ∂r
State the reason why you eliminated any terms in your derivation.
Solution
Begin with the z-direction equation in cylindrical coordinates,
∂vz
∂vz vθ ∂vz
∂vz
∂P
1 ∂ ∂vz
1 ∂2 vz ∂2 vz
)=−
(r
)+ 2 2 + 2)
ρ(
+ vr
+
+ vz
+ ρg z + μ (
∂t
∂r
r ∂θ
∂z
∂z
r ∂r ∂r
r ∂θ
∂z
To obtain the desired equation we must make the following assumptions:
∂vz
Steady State:
=0
∂t
∂vz vθ ∂vz
No velocity in the r or θ directions: vr
=
=0
∂r
r ∂θ
∂vz ∂2 vz
Fully developed flow: vz
=
=0
∂z
∂z 2
Gravity is negligable: ρg z = 0
1 ∂2 vz
There is no θ direction contribution to the flow: 2 2 = 0
r ∂θ
0=−
∂P
1 ∂ ∂vz
(r
))
+ μ(
∂z
r ∂r ∂r
∂v
If we assume that the fluid is Newtonian, so τrz = μ ∂rz,
Then after multiplying through by r,
0 = −r
∂P ∂
+ (rτ )
∂z ∂r rz
9.26
A film in free stream flow is flowing down
an inclined plan as shown in the figure.
Beginning with the appropriate form of the
Navier-Stokes equations, please derive the
following equation:
∂
τ + ρgsin(θ) = 0
∂y yx
As always, please state all assumptions you
make in your derivation.
Solution
ρ(
∂vx
∂vx
∂vx
∂vx
∂P
∂2 vx ∂2 vx ∂2 vx
)=−
+ vx
+ vy
+ vz
+ ρg x + μ ( 2 +
+ 2)
∂t
∂x
∂y
∂z
∂x
∂x
∂y 2
∂z
Evaluating the equation term by term:
∂vx
= 0 since the flow is steady
∂t
∂vx ∂2 vx
vx
=
= 0 since the flow is fully developed
∂x
∂x 2
∂vx
vy
= 0 since there is no y direction velocity
∂y
∂P
= 0 since this is free stream flow
∂x
∂2 vx
= 0 since we have two dimensional flow
∂z 2
Taking the angle of the flow into consideration, ρg x = ρg x sin(θ) we now have,
∂2 vx
0 = ρg x sin(θ) + μ 2
∂y
Next, we must assume that we have a Newtonian Fluid,
∂τyx
0 = ρg x sin(θ) +
∂y
Rearranging,
∂τyx
+ ρg x sin(θ) = 0
∂y
9.27
An incompressible Newtonian fluid is confined between two concentric circular cylinders of
infinite length, a solid inner cylinder of radius R, and a hollow stationary outer cylinder of radius
Ro. See the figure below where the z-axis is coming out of the page. The inner cylinder rotates
with an angular velocity ωi . The flow is steady, laminar and two-dimensional in the rθ-plane. The
flow is also rotationally symmetric, meaning that nothing is a function of coordinate θ (so vθ and
P are functions of the radius, r only) and that gravity can be ignored. We can also assume that the
flow is circular, meaning that the velocity component vr = 0 everywhere. Please derive an exact
expression for the velocity component vθ as a function of the radius r and the other parameters in
the problem.
i
R0
Liquid: ,
Ri
Rotating inner cylinder
Solution
Stationary outer cylinder
We begin with the Navier-Stokes Equation for cylindrical coordinates in the θ-direction.
∂vθ
∂vθ vθ ∂vθ vr vθ
∂vθ
)
ρ(
+ vr
+
+
+ vz
∂t
∂r
r ∂θ
r
∂z
1 ∂P
∂ 1∂
1 ∂2 vθ 2 ∂vr ∂2 vθ
(rvθ )) + 2
=−
+ ρg θ + μ ( (
+
+
)
r ∂θ
∂r r ∂r
r ∂θ2 r 2 ∂θ
∂z 2
Evaluation of the terms,
∂vθ
= 0 since we have steady state
∂t
∂vθ vr vθ
∂vθ
vr
=
= vz
= 0 since there is no z or r direction velocity
∂r
r
∂z
vθ ∂vθ
= 0 since we have fully developed flow
r ∂θ
1 ∂P 1 ∂2 vθ
=
= 0 since we have rotational symmey
r ∂θ r 2 ∂θ2
2 ∂vr
= 0 since we have no r direction velocity change in the θ direction
r 2 ∂θ
ρg θ = 0 since we are told we can neglect gravity
Thus,
∂ 1∂
(rv )))
0 = μ( (
∂r r ∂r θ
We can convert from partial differentials to ordinary differentials since vθ = f(r) only, and
integrate,
μd
(rv ) = C1
r dr θ
Rearrange and integrate again,
vθ = C1
r
C2
+
2μ r
Determine the boundary conditions to find values for C1 and C2,
BC#1: at the inner wall the No-Slip boundary condition applies: r = R i at vθ = ωi R i
BC#2: at the outer wall the No-Slip boundary condition applies: r = R o at vθ = 0
Apply BC#2:
0 = C1
R 0 C2
+
2μ R 0
C1 R20
C2 = −
2μ
Apply BC#1:
R i C1 R20
Ri
R20
ωi R i = C1
−
= C1 ( −
)
2μ 2μR i
2μ 2μR i
So,
2μωi R2i
C1 = − 2
R 0 − R2i
And,
C1 R20
2μωi R2i R20
ωi R2i R20
=
=
2μ
2μ(R20 − R2i ) (R20 − R2i )
Plug back into the original equation,
2μωi R2i
r
ωi R2i R20
rωi R2i
ωi R2i R20
( )+
vθ = − 2
=− 2
+
(R 0 − R2i ) 2μ
(R 0 − R2i ) r(R20 − R2i )
r(R20 − R2i )
C2 = −
ωi R2i
R20
vθ = 2
( − r)
(R 0 − R2i ) r
9.28
Beginning with the appropriate form of the Navier-Stokes Equation, please derive an expression
for the velocity of a fluid flowing through a vertical pipe with diameter equal to D. For this
derivation, please assume steady, incompressible, fully developed flow and that the fluid is not
open to the atmosphere. Please be sure to indicate the reason why you eliminate any terms in the
original equation as you proceed with your derivation. (Hint: assume that at the center of the pipe,
r = 0)
Solution
∂vz
∂vz vθ ∂vz
∂vz
∂P
1 ∂ ∂vz
1 ∂2 vz ∂2 vz
)=−
(r
)+ 2 2 + 2)
ρ(
+ vr
+
+ vz
+ ρg z + μ (
∂t
∂r
r ∂θ
∂z
∂z
r ∂r ∂r
r ∂θ
∂z
Evaluating term by term:
Steady State:
∂vz
=0
∂t
∂vz vθ ∂vz
=
=0
∂r
r ∂θ
∂vz ∂2 vz
Fully developed flow: vz
=
=0
∂z
∂z 2
No velocity in the r or θ directions: vr
1 ∂2 vz
There is no θ direction contribution to the flow: 2 2 = 0
r ∂θ
Since the pipe is vertical, the gravity acts in the negative z-direction: ρg z = −ρg z
∂P
1 ∂ ∂vz
(r
))
0=−
− ρg z + μ (
∂z
r ∂r ∂r
Rearrange,
∂ ∂vz
r ∂P
(r
) = ( + ρg z )
∂r ∂r
μ ∂z
Integrate,
∂vz r 2 ∂P
( + ρg z ) + C1
r
=
∂r
2μ ∂z
Rearrange (divide through by r and by μ),
μ
∂vz r ∂P
C1
= ( + ρg z ) + μ
∂r
2 ∂z
r
𝜕𝑣
Boundary Condition #1: r = 0 in center of tube would result in 𝜇 𝜕𝑟𝑧 = 𝜏𝑟𝑧 = 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒, which is
impossible, so 𝐶1 = 0 (we have seen this boundary condition before). Could have also said that
𝜕𝑣𝑧
cannot be infinite.
𝜕𝑟
Integrate again,
vz =
r 2 ∂P
( + ρg z ) + C2
4μ ∂z
BC#2: At vz = 0, r = D/2 = R (No Slip Boundary Condition)
Apply BC#2:
(−R)2 ∂P
( + ρg z ) + +C2
0=
4μ
∂z
Thus,
C2 = −
R2 ∂P
( + ρg z )
4μ ∂z
Resulting in,
r 2 ∂P
R2 ∂P
1 ∂P
( + ρg z ) − ( + ρg z ) =
( + ρg z ) (r 2 − R2 )
vz =
4μ ∂z
4μ ∂z
4μ ∂z
9.29.
Fluid
Direction
y
𝜃
x
h
VB
VT
A Newtonian fluid is flowing down an incline between
two vertical, infinitely long parallel plates separated
by a distance h as shown in the figure. The plate on
the right is moving up with a velocity of VT and the
plate on the left is moving up with a velocity of VB .
The velocities of the plates are different such that
VT ≠ VB . The fluid is not open to the atmosphere and
is flowing in one direction only as shown in the figure
at an angle 𝜃 with the y-axis. Assuming steady,
incompressible, fully developed, laminar flow, please
derive an equation for the velocity of the fluid. You
must use the coordinate system shown in the figure
in your solution to the problem (which means that
you cannot draw or make another coordinate
system).
Solution 1
Here the key points in the problem statement were that the flow is in one direction only and that
it was vertical flow.
Need x-direction Navier-Stokes Equation
𝜕𝑣𝑦
𝜕𝑣𝑦
𝜕𝑣𝑦
𝜕𝑣𝑦
𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦
𝜕𝑃
𝜌(
+ 𝑣𝑥
+ 𝑣𝑦
+ 𝑣𝑧
)=−
+ 𝜌𝑔𝑦 + 𝜇 ( 2 +
+
)
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑦
𝜕𝑥
𝜕𝑦 2
𝜕𝑧 2
Using the conditions in the problem statement this equation becomes,
𝜕 2 𝑣𝑦
𝜕𝑃
0=−
− 𝜌𝑔𝑦 cos 𝜃 + 𝜇 ( 2 )
𝜕𝑦
𝜕𝑥
Integrate twice solving for velocity,
𝜕 2 𝑣𝑦 1 𝜕𝑃
= ( + 𝜌𝑔𝑦 cos 𝜃)
𝜕𝑥 2
𝜇 𝜕𝑦
𝜕𝑣𝑦 𝑥 𝜕𝑃
= ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1
𝜕𝑥
𝜇 𝜕𝑦
𝑣𝑦 =
Boundary Conditions:
𝑥 2 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 𝑥 + 𝐶2
2𝜇 𝜕𝑦
BC#1: 𝑣𝑦 = −VB 𝑎𝑡 𝑥 = 0
BC#2: 𝑣𝑦 = −VT 𝑎𝑡 𝑥 = ℎ
Apply Boundary Condition #1
(0)2 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 (0) + 𝐶2
−VB =
2𝜇 𝜕𝑦
𝐶2 = −VB
Apply Boundary Condition #2
ℎ2 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 ℎ + VB
−VT =
2𝜇 𝜕𝑦
VB − VT
ℎ 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃)
−
h
2𝜇 𝜕𝑦
𝐶1 =
And Finally
𝑥 2 𝜕𝑃
VB − VT
ℎ 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + 𝑥 [
( + 𝜌𝑔𝑦 cos 𝜃)] − VB
𝑣𝑦 =
−
2𝜇 𝜕𝑦
h
2𝜇 𝜕𝑦
Alternative Solution (and perhaps a better solution):
𝑣𝑥 =
𝑥 2 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 𝑥 + 𝐶2
2𝜇 𝜕𝑦
Boundary Conditions:
BC#1: 𝑣𝑥 = −VB cos 𝜃 𝑎𝑡 𝑥 = 0
BC#2: 𝑣𝑥 = −VT cos 𝜃 𝑎𝑡 𝑥 = ℎ
Apply B.C. #1:
−VB sin 𝜃 = 𝐶2
Apply B.C. #2:
−V𝑇 sin 𝜃 =
𝐶1 =
ℎ2 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 ℎ − VB cos 𝜃
2𝜇 𝜕𝑦
VB cos 𝜃 − VT cos 𝜃 ℎ 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃)
−
h
2𝜇 𝜕𝑦
𝑥 2 𝜕𝑃
VB cos 𝜃 − VT cos 𝜃 ℎ 𝜕𝑃
( + 𝜌𝑔𝑦 cos 𝜃) + [
( + 𝜌𝑔𝑥 cos 𝜃)] 𝑥 − VB cos 𝜃
𝑣𝑦 =
−
2𝜇 𝜕𝑦
h
2𝜇 𝜕𝑦
9.30
Beginning with the appropriate Navier-Stokes equation, please derive the appropriate equation to
describe the flux of 𝐳-momentum in the y-direction assuming assume steady, fully developed flow
between two infinitely long vertical parallel plates separated by a distance L, where the fluid is not
open to the atmosphere and parabolic flow can be assumed. Please indicate the reason why you
eliminate any terms in the equations as you proceed with your derivation. [Hint: we want an
equation to describe 𝐳-momentum in the y-direction and NOT 𝐲-momentum in the 𝐳-direction.]
Solution
dv
The flux of z-momentum in the y-direction is defined as τyz = μ dyz , so we need to begin with the
Navier-Stokes equation in the z-direction of rectangular coordinates. As a result, the coordinate
system is chosen for us in the problem statement. Since the problem states that the flow is vertical,
we keep the gravity term, and since it is not open to the atmosphere, we keep the pressure term.
∂vz
∂vz
∂vz
∂vz
∂P
∂2 vz ∂2 vz ∂2 vz
)=−
ρ(
+ vx
+ vy
+ vz
+ ρg z + μ ( 2 + 2 + 2 )
∂t
∂x
∂y
∂z
∂z
∂x
∂y
∂z
Applying all the simplifying assumptions of the problem statement this equation simplifies to:
∂P
∂2 vz
0=−
− ρg z + μ ( 2 )
∂z
∂y
2
∂ vz
∂P
μ( 2 ) =
+ ρg z
∂y
∂z
dτyz ∂P
=
+ ρg z
dy
∂z
∂P
τyz = ( + ρg z ) y + C1
∂z
Next, we need to determine the boundary condition. We know that the shear is negligible in the
center, away from the boundary, and are given that a distance, L, separates the plates. So at the
center,
L
Boundary condition: τyz = 0 at y =
2
Substitute and solve for C1 ,
∂P
L
0 = ( + ρg z ) + C1
∂z
2
C1 = − (
Plug this into the original equation,
∂P
L
+ ρg z )
∂z
2
∂P
∂P
L
+ ρg z ) y − ( + ρg z )
∂z
∂z
2
∂P
L
τyz = ( + ρg z ) (y − )
∂z
2
τyz = (
9.31
Two different Newtonian fluids, so that they have different densities and viscosities (Fluid 1 and
Fluid 2 in the figure), are flowing steadily down between infinitely long vertical plates as shown
in the figure below. The plate at the left is moving down at a velocity of VA, the center plate is
moving up at a velocity of VB and the plate at the right is stationary. In the figure, the distances
between the plates are not equal so that L1≠L2.
(1) Using the appropriate form of the Navier-Stokes Equations, please derive equations for the
velocity of Fluid 1 and the velocity of Fluid 2 assuming that both fluids are in fully developed,
continuous, incompressible, adiabatic, isothermal flow. You may also assume that the no-slip
boundary condition applies.
(2) Please draw the velocity profiles for both Fluid 1 and Fluid 2 in the figure provided below. Be
careful in your drawing so it is clear.
Solution
We first assign an appropriate coordinate system:
VA
VB
Fluid 1
Fluid 2
y
x
L1
L2
Need y-direction Navier-Stokes Equation
∂vy
∂vy
∂vy
∂vy
∂2 vy ∂2 vy ∂2 vy
∂P
ρ(
+ vx
+ vy
+ vz
)=−
+ ρg y + μ ( 2 +
+ 2)
∂t
∂x
∂y
∂z
∂y
∂x
∂y 2
∂z
Using the conditions in the problem statement this equation becomes,
∂2 vy
∂P
0=−
− ρg y + μ ( 2 )
∂y
∂x
Integrate twice solving for velocity,
∂2 vy 1 ∂P
= ( + ρg y )
∂x 2
μ ∂y
∂vy x ∂P
= ( + ρg y ) + C1
∂x
μ ∂y
2
x ∂P
( + ρg y ) + C1 x + C2
vy =
2μ ∂y
This equation can be used for both sides using appropriate boundary conditions.
There are two ways to do this problem depending on how you choose the boundary conditions.
METHOD #1
For Fluid #1:
v1 =
x 2 ∂P
( + ρg y ) + C1 x + C2
2μ ∂y
Boundary Conditions:
BC#1: v1 = −V𝐴 at x = −L1
BC#2: v1 = V𝐵 at x = 0
Apply BC#1:
−V𝐴 =
(−L1 )2 ∂P
( + ρg y ) + C1 (−L1 ) + C2
2μ
∂y
Apply BC#2:
V𝐵 = 0 + 0 + C2 → C2 = V𝐵
Plug the result of BC2 into BC1 equation and solve for C1
(−L1 )2 ∂P
( + ρg y ) + C1 (−L1 ) + V𝐵
−V𝐴 =
2μ
∂y
C1 =
V𝐴 + V𝐵 𝐿1 2 ∂P
( + ρg y )
+
𝐿1
2μ ∂y
x 2 ∂P
V𝐴 + V𝐵 𝐿1 2 ∂P
( + ρg y ) + [
( + ρg y )] x + V𝐵
v1 =
+
2μ ∂y
𝐿1
2μ ∂y
For Fluid #2:
x 2 ∂P
( + ρg y ) + C1 x + C2
v2 =
2μ ∂y
Boundary Conditions:
BC#1: v2 = V𝐵 at x = 0
BC#2: v2 = 0 at x = L2
Apply BC#1:
V𝐵 = 0 + 0 + C2
→
C2 = V𝐵
Apply BC#2 (and incorporating the result of BC#1):
0=
L22 ∂P
( + ρg y ) + C1 L2 + V𝐵
2μ ∂y
C1 = −
L2 ∂P
V𝐵
( + ρg y ) −
2μ ∂y
L2
Substituting this back to obtain the final equation for v2 :
v2 =
x 2 ∂P
L2 ∂P
V𝐵
( + ρg y ) − [ ( + ρg y ) + ] x + V𝐵
2μ ∂y
2μ ∂y
L2
METHOD #2
For Fluid #1:
v1 =
x 2 ∂P
( + ρg y ) + C1 x + C2
2μ ∂y
Boundary Conditions:
BC#1: v1 = −V𝐴 at x = 0
BC#2: v1 = V𝐵 at x = L1
Apply BC#1:
−V𝐴 = 0 + 0 + C2
→
𝐶2 = −𝑉𝐴
Apply BC#2:
V𝐵 =
L21 ∂P
𝑉𝐴 + V𝐵 L1 ∂P
( + ρg y ) + C1 L1 − 𝑉𝐴 → 𝐶1 =
− ( + ρg y )
2μ ∂y
𝐿1
2μ ∂y
So, the equation for the velocity of Fluid 1 is,
x 2 ∂P
𝑉𝐴 + V𝐵 L1 ∂P
( + ρg y ) + x [
v1 =
− ( + ρg y )] − 𝑉𝐴
2μ ∂y
𝐿1
2μ ∂y
For Fluid #2:
x 2 ∂P
( + ρg y ) + C1 x + C2
v2 =
2μ ∂y
Boundary Conditions:
BC#1: v2 = V𝐵 at x = L1
BC#2: v2 = 0 at x = L1 + L2
Apply BC#1:
V2 =
𝐿21 ∂P
( + ρg y ) + C1 L1 + C2
2μ ∂y
Apply BC#2:
0=
Set C2 = C2
(L1 + L2 )2 ∂P
( + ρg y ) + 𝐶1 (L1 + L2 ) + C2
2μ
∂y
(L1 + L2 )2 ∂P
𝐿21 ∂P
( + ρg y ) − C1 L1 = −
( + ρg y ) − 𝐶1 (L1 + L2 )
2μ ∂y
2μ
∂y
2
(L1 + L2 )2 ∂P
𝐿1 ∂P
( + ρg y )
𝐶1 (L1 + L2 ) − C1 L1 = −VB + ( + ρg y ) −
2μ ∂y
2μ
∂y
VB −
C1 L2 = −VB +
(L1 + L2 )2 ∂P
𝐿21 ∂P
( + ρg y ) −
( + ρg y )
2μ ∂y
2μ
∂y
(L1 + L2 )2 ∂P
−VB
𝐿21 ∂P
( + ρg y ) −
( + ρg y )
C1 =
+
L2
2μL2 ∂y
2μL2
∂y
So,
(L1 + L2 )2 ∂P
x 2 ∂P
−VB
𝐿21 ∂P
( + ρg y ) + [
( + ρg y ) −
( + ρg y )] x + C2
v2 =
+
2μ ∂y
L2
2μL2 ∂y
2μL2
∂y
Next, solve for C2 , could pick either result form application of BC#1 or BC#2, we’ll use the result
of BC#2 here since velocity is zero in that boundary condition,
(L1 + L2 )2 ∂P
(L1 + L2 )2 ∂P
−VB
𝐿21 ∂P
( + ρg y ) + [
( + ρg y ) −
( + ρg y )] (L1 + L2 )
0=
+
2μ
∂y
L2
2μL2 ∂y
2μL2
∂y
+ C2
(L1 + L2 )2 ∂P
( + ρg y )
C2 = −
2μ
∂y
(L1 + L2 )2 ∂P
−VB
𝐿21 ∂P
( + ρg y ) +
( + ρg y )] (L1 + L2 )
−[
+
L2
2μL2 ∂y
2μL2
∂y
And the final equation for the velocity is,
v2 =
(L1 + L2 )2 ∂P
x 2 ∂P
−VB
𝐿21 ∂P
( + ρg y ) + [
( + ρg y ) −
( + ρg y )] x+
+
2μ ∂y
L2
2μL2 ∂y
2μL2
∂y
2
(L1 + L2 ) ∂P
( + ρg y )
=−
2μ
∂y
2
(L1 + L2 )2 ∂P
−VB
𝐿1 ∂P
( + ρg y ) +
( + ρg y )] (L1 + L2 )
−[
+
L2
2μL2 ∂y
2μL2
∂y
This could be further simplified if desired.
9.32
Consider steady, incompressible, laminar, fully developed flow of a Newtonian fluid in an
infinitely long, stationary pipe of diameter D inclined at an angle of 𝛽 to the horizontal plane as
shown in the figure for problem 9.24. You may assume that the No-Slip boundary condition
applies.
(a) Show that the continuity equation holds in this system,
(b) Using the appropriate form of the Navier-Stokes Equation, derive an expression to describe the
velocity of this fluid. Be sure to indicate the reason why you eliminate any terms in the original
equation as your proceed with your derivation.
Solution
(a) We begin with the general form of continuity given in the equation sheets (or something
similar),
∇ ∙ ρv +
∂ρ
=0
∂t
which expands, in cylindrical coordinates to,
∂ρ 1 ∂
1 ∂
∂
(ρrvr ) +
(ρvθ ) + (ρvz ) = 0
+
∂t r ∂r
r ∂θ
∂z
∂ρ
= 0 since we have steady flow
∂t
1∂
(ρrvr ) = 0 since we do not have any r − direction flow
r ∂r
1 ∂
(ρvθ ) = 0 since we don′ thave any θ − direction flow
r ∂θ
∂
(ρvz ) = 0 since we have fully developed flow
∂z
Thus,
0=0
And we have proved that continuity holds.
(b) For flow down a pipe we need the z-direction of the Navier-Stokes equations in cylindrical
coordinates:
∂vz
∂vz vθ ∂vz
∂vz
∂P
1 ∂ ∂vz
1 ∂2 vz ∂2 vz
)=−
(r
)+ 2 2 + 2)
ρ(
+ vr
+
+ vz
+ ρg z + μ (
∂t
∂r
r ∂θ
∂z
∂z
r ∂r ∂r
r ∂θ
∂z
Applying the assumptions given in the problem statement including that it is on an angle, this
equation simplifies to,
0=−
∂P
1 ∂ ∂vz
(r
))
+ ρg z sinβ + μ (
∂z
r ∂r ∂r
Rearrange and integrate,
1 ∂ ∂vz
1 ∂P
(r
) = ( + ρg z sinβ)
r ∂r ∂r
μ ∂z
∂ ∂vz
r ∂P
(r
) = ( + ρg z sinβ)
∂r ∂r
μ ∂z
r
∂vz r 2 ∂P
( + ρg z sinβ) + C1
=
∂r
2μ ∂z
∂vz
r ∂P
C1
( + ρg z sinβ) +
=
∂r
2μ ∂z
r
The first boundary condition is that, because of fully developed flow, there is no change in the
z-direction velocity with respect to r in the center of the pipe where r = 0, so,
B.C. #1:
∂vz
∂r
= 0 at r = 0
Application of this boundary condition gives,
C1
0= 0+
0
→ C1 = 0
So the equation is now,
∂vz
r ∂P
( + ρg z sinβ)
=
∂r
2μ ∂z
Integrate,
vz =
r 2 ∂P
( + ρg z sinβ) + C2
2μ ∂z
For the second boundary condition, the No-Slip boundary conditions applies,
B.C. #2: vz = 0 at r = R
Apply this boundary condition,
R2 ∂P
( + ρg z sinβ) + C2
0=
2μ ∂z
Giving,
R2 ∂P
C2 = − ( + ρg z sinβ)
2μ ∂z
So the final equation for the velocity is
r 2 ∂P
R2 ∂P
( + ρg z sinβ) − ( + ρg z sinβ)
vz =
2μ ∂z
2μ ∂z
This can be simplified if desired.
9.33
V0
Fluid 2
h2
Interface
Fluid 1
h1
Two immiscible viscous fluids are sandwiched between two infinitely long and wide parallel flat
plates as shown in the figure above. Because the fluids are immiscible they do not mix at any
point. At time t = 0, the fluids and the plates on either side are stationary. The upper plate is set
in motion at a velocity V0, and the bottom plate remains stationary at all times. The flow is
incompressible, parallel, fully developed and laminar. The flow is caused solely by the viscous
effects created by the moving upper plate. You may ignore gravity, and assume that the fluids
are not open to the atmosphere and that the interface between the fluids is horizontal at all
times.
(c) In the figure above please draw the velocity profiles for both Fluid 1 and Fluid 2 in this system.
(d) Please derive equations for the velocity profiles of both fluids when the flow is at steady state.
Chapter 10
Instructor Only Problems
10.25
The stream function for steady, incompressible flow is given by, Ψ = y 2 − xy − x 2 . Determine
the velocity components for this flow.
Solution
We define the stream function by,
∂Ψ
= vx
∂y
and
−
Thus,
∂Ψ
= vy
∂x
∂Ψ
= 2y − x
∂y
∂Ψ
vy = −
= y + 2x
∂x
vx =
10.26
The stream function for steady, incompressible flow is given by
Ψ(x, y) = 2x 2 − 2y 2 − xy.
Determine the velocity potential for this flow.
Solution
Since a condition for the velocity potential to exist is irrotational flow, we must first determine
whether the flow in this problem is irrotational.
We first find the components of the velocity using the definition of the stream function.
We define the stream function by,
∂Ψ
= vx
∂y
and
∂Ψ
−
= vy
∂x
Thus,
∂Ψ
vx =
= −4y − x
∂y
∂Ψ
vy = −
= −4x + y
∂x
Next, we want to determine whether the flow is rotational or irrotational. To do this we must
satisfy Equation 10.1
1 ∂vy ∂vx
1
wz = (
−
) = (−4 − (−4)) = 0
2 ∂x
∂y
2
So the flow is irrotational as desired.
Next we determine the equation for the velocity potential. The velocity potential is defined by
Equation 10-15.
v = ∇∅
or
∂ϕ
= vx = −4y − x
∂x
x2
ϕ = −4xy − + f(y)
2
Differentiating with respect to y and equating to vy
∂ϕ
d
= −4x + y = −4x + f(y)
∂y
dy
Thus,
d
f(y) = y
dy
y2
f(y) =
2
So that the final equation for the velocity potential is
x2 y2
ϕ = −4xy − +
2
2
10.27
Determine the velocity components for the flow in Problem 10.23.
Solution
We defined the stream function as
∂Ψ
= −vy
∂x
And
∂Ψ
= vx
∂y
Thus,
∂Ψ
= −12y
∂y
∂Ψ
vy = −
= −12x
∂x
vx =
The equations for the velocity components are vx = −12y and vy = −12x.
10.28
The steady, incompressible flow field for two-dimensional flow is given by the following velocity
components, 𝑣𝑥 = 16𝑦 − 𝑥 and 𝑣𝑦 = 16𝑥 + 𝑦. Determine the equation for the stream function
and the velocity potential.
Solution
First, let’s check to make sure continuity is satisfied.
𝜕𝑣𝑥 𝜕𝑣𝑦
𝜕
𝜕
(16𝑦 − 𝑥) +
(16𝑥 + 𝑦) = −1 + 1 = 0
+
=
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦
So, continuity is satisfied which is a necessary condition for us to proceed.
(1)
We defined the stream function as
∂Ψ
= vx
∂y
(2)
and
−
∂Ψ
= vy
∂x
(3)
Thus,
∂Ψ
= 16𝑦 − 𝑥
(4)
∂y
∂Ψ
vy = −
= 16x + y (5)
∂x
We can begin by integrating equation (4) or equation (5). Either will result in the same answer.
(Problem 2 of the Homework will let you verify this.) We will choose to begin by integrating
equation (4) partially with respect to y,
Ψ = 8y 2 − xy + f1 (x)
(6)
Where f1 (x) is an arbitrary function of x.
Next, we take the other part of the definition of the stream function, equation (3), and
differentiate equation (6) with respect to x,
∂Ψ
vy = −
= y − f2 (x) (7)
∂x
df
Here, f2 (x) is dx since f is a function of the variable x.
The result is that we now have two equations for vy , equations (5) and (7). We can now equate
these and solve for f2 (x).
vy = y − f2 (x) = 16x + y
Solving for f2 (x),
f2 (x) = −16x
So
x2
f1 (x) = −16 = −8x 2 + C
2
The integration constant C is added to the above equation since f is a function of x only. The
final equation for the stream function is,
Ψ = 8y 2 − xy − 8x 2 + C
(6)
The constant C is generally dropped from the equation because the value of a constant in this
equation is of no significance. The final equation for the stream function is,
vx =
Ψ = 8y 2 − xy − 8x 2
One interesting point is that the difference in the value of one stream line in the flow to another
is the volume flow rate per unit width between the two streamlines.
Next we want to find the equation for the velocity potential. Since a condition for the velocity
potential to exist is irrotational flow, we must first determine whether the flow in this example is
irrotational.
To do this we must satisfy Equation 10.1
1 ∂vy ∂vx
1
(
−
) = (16 − 16) = 0
2 ∂x
∂y
2
So the flow is irrotational as required.
We now want to determine the equation for the velocity potential. The velocity potential is
defined by Equation 10-15.
v = ∇∅
or
∂ϕ
= vx = 16y − x
∂x
x2
ϕ = 16xy − + f(y)
2
Differentiating with respect to y and equating to vx
∂ϕ
d
= 16x +
f(y) = 16x + y
∂y
dy
Thus,
d
f(y) = y
dy
And,
y2
f(y) =
2
So that the final equation for the velocity potential is
x2 y2
ϕ = 16xy − +
2
2
wz =
10.29
Repeat Problem 2 but instead begin by integrating the following equation,
∂Ψ
vx =
.
∂y
and show that it does not matter which equation you begin with; the results are identical.
Solution
We defined the stream function as
∂Ψ
= vx
∂y
(2)
and
−
∂Ψ
= vy
∂x
(3)
Thus,
∂Ψ
= 16𝑦 − 𝑥
(4)
∂y
∂Ψ
vy = −
= 16x + y (5)
∂x
We will choose to begin by integrating equation (5) partially with respect to x,
Ψ = −8x 2 − xy + f1 (x)
(6)
Where f1 (y) is an arbitrary function of y.
Next, we take the other part of the definition of the stream function, equation (2), and
differentiate equation (6) with respect to y,
∂Ψ
vx =
= −x + f2 (y) (7)
∂y
df
Here, f2 (y) is dy since f is a function of the variable y.
vx =
The result is that we now have two equations for vx , equations (4) and (7). We can now equate
these and solve for f2 (y).
vy = −x + f2 (y) = 16𝑦 − 𝑥
Solving for f2 (y),
f2 (y) = 16y
So
y2
f1 (x) = 16 = 8y 2 + C
2
The integration constant C is added to the above equation since f is a function of y only. The
final equation for the stream function is,
Ψ = −8x 2 − xy + 8y 2 + C
(6)
The constant C is generally dropped from the equation because the value of a constant in this
equation is of no significance. The final equation for the stream function is,
Ψ = 8y 2 − xy − 8x 2
10.30
In 2-dimensional potential flow the stream function Ψ(x,y) and the velocity potential Φ(x.y) are
geometrically related. Sketch this relationship on an x-y plane.
What is the physical significance of Ψ(x,y)=C1 and Φ(x.y)=C2 where C1 and C2 are arbitrary
constants?
Solution
vy =
 ( x, y ) −  ( x, y )
=
y
x
vx =
 ( x, y ) 
=
x
y
dy
1
=
dx  =cons tan t − dy / dx  =cons tan t
These equations suggest that the stream lines intersect the velocity potential lines at 90o at all
times and that they are perpendicular to each other at the intersection.








 = C2
y
90o
angle
 = C1
x
Chapter 11
Instructor Only Problems
11.4
The maximum pitching moment that is developed by the water on a flying boat as it lands is
noted as Cmax. The following are the variables involved in this action:
= angle made by flight path of plane with horizontal
=angle defining attitude of plane
M=mass of plane
L=length of hull
=density of water
g=acceleration due to gravity
R=radius of gyration of plane about axis of pitching
(a) According to the Buckingham  theorem, how many independent dimensionless groups
should there be which characterize this problem?
Cmax = ML2t-2
= angle
=angle
M = Mass
L = Length
 = density
g = gravity
R = radius
dimensionless
dimensionless
M
L
ML-3
Lt-2
L
Dimension  0 and the rank =3
i = n – r = 8 – 3 = 5 independent dimensionless groups
(b) What is the dimensional matrix of this problem? What is its rank?
M
L
t
Cmax
1
2
-2

0
0
0

0
0
0
M
1
0
0
L
0
1
0

1
-3
0
g
0
1
-2
R
0
1
0
(c) Evaluate the appropriate dimensionless parameters for this problem.
 1= 
2 = 
c
3=MaLbgcCmax
2
 L  ML
MoLoto = 1 = (M )a (L )b  2 
t  t2
M: 0 = a+1
L: 0 = b+c+2
T: 0 = -2c-2
Thus: a= -1, b= -1, c= -1
1=-1D-2-2 →
 C max 

1= 
 M Lg 
c
4=MaLbgc
L M
MoLoto = 1 = (M )a (L )b  2  2
t  L
M: 0 = a+1
L: 0 = b+c-3
T: 0 = -2c
Thus: a= -1, b= 3, c= 0
1=-1D-2-2 →
 L3  

1= 

 M 
c
5=MaLbgcR
L
MoLoto = 1 = (M )a (L )b  2  L
t 
M: 0 = a
L: 0 = b+c+1
T: 0 = -2c
Thus: a= 0, b= 3, c= 0
1=-1D-2-2 →
R
1=  
L 
11.12
Identify the variables associated with Problems 8.13 and find the dimensionless
parameters.
Solution
Variable
ΔP
Q
h
Ω
L
μ
R
Dimensions
M/Lt2
L3/t
L
1/t
L
M/Lt
L
i = 7 − 3 = 4 , Choose core as h, Ω, μ
L
π1 = ha Ωb μc L ∴ =
h
R
π2 = hd Ωe μf R ∴ =
h
ΔP
π3 = hg Ωh μi ΔP ∴ =
Ωμ
Q
π4 = hj Ωk μl Q ∴ = 3
h Ω
11.23
A pump in a manufacturing plant is transferring viscous fluids to a series of delivery tanks.
This critical transfer requires careful monitoring of the solution mass flow rate, the power
(work) that the pump adds to the fluid, the internal energy of the system, the viscosity and
density of the solution. Please determine the dimensionless groups formed from the
variables involved using the Buckingham method. Carefully choose your core group
based on the description of the system.
Solution
Variable
Mass flow rate
Work of pump
Internal energy
viscosity
Density
M
L
t
ṁ
δWs / ∂t
1
0
-1
1
2
-3
Symbol
ṁ
δWs / ∂t
U
μ
ρ
Dimensions
M/t
ML2/t3
L2/s2
M/Lt
M/L3
U
0
2
-2
μ
ρ
1
-1
-1
1
-3
0
The rank is found to be 3 (rank of a matrix is the number of rows (columns) in the
largest nonzero determinant that can be formed from it). Using Equation 11-4, the
number of dimensionless parameters is
𝑖 =5−3=2
Thus there are 2 dimensionless parameters to be found.
Choose the core group to be δWs / ∂t and viscosity.
Find π1 :
b
π1 = ρa ṁ U c
δWs
∂t
c
M a M b L2 ML2
M L T = 1 = ( 3) ( ) ( 2 ) 3
L
L
t
t
0 0 0
Evaluate for M, L and t
M:
0=a+b+1
L:
0 = −3a − b + 2c + 2
t:
0 = −2c − 3
1
1
Thus, a = − 2 , b = − 2 and c = −3/2 and
−1/2
π1 = π1 = ρ−1/2 ṁ
U −3/2
δWs
∂t
=
1/2 3/2
ρ1/2 ṁ
U
Find π2 :
δWs
∂t
e
π2 = ρd ṁ U f μ
f
M d M e L2 ML
0 0 0
M L T = 1 = ( 3) ( ) ( 2 )
L
L
t
t
Evaluate for M, L and t
M:
0=d+e+1
L:
0 = −3d − e + 2f + 1
t:
0 = −2f − 1
1
2
1
Thus, d = − 3 , e = − 3 , f = − 2 and
1
−
2
1
π1 = ρ−3 ṁ 3 U −2 μ =
μ
2/3 1/2
ρ1/3 ṁ
U
11.24
The performance of a fluid flow system is affected by variables of pressure, density, angular
velocity, impeller diameter, volumetric flow rate and viscosity. Determine the dimensionless
groups formed from the variables involved using the Buckingham Method. Choose the groups
so that , Q and P each appear in one group only, so that the core group is , D and .
Solution
Core Group = ρ, D, ω
M
L
t
P
1
-1
-2
ρ
1
-3
0
ω
0
0
1
D
0
1
0
𝜋1 = 𝜌𝑎 𝐷𝑏 ω𝑐 𝑃
𝑀 𝑎
1 𝑐 𝑀
𝑏
𝑀 𝐿 𝑡 = 1 = ( 3 ) (𝐿) ( )
𝐿
𝑡 𝐿𝑡 2
0 0 0
𝑀: 0 = 𝑎 + 1
𝐿: 0 = −3𝑎 + 𝑏 − 1
𝑡: 0 = −𝑐 − 2
𝑐 = −2,
𝑎 = −1,
𝜋1 = 𝜌−1 𝐷−2 ω−2 𝑃 =
𝑏 = −2
𝑃
𝜌𝐷2 𝜔 2
𝜋2 = 𝜌𝑎 𝐷𝑏 ω𝑐 𝑄
𝑀 𝑎
1 𝑐 𝐿3
𝑀0 𝐿0 𝑡 0 = 1 = ( 3 ) (𝐿)𝑏 ( )
𝐿
𝑡 𝑡
𝑀: 0 = 𝑎
𝐿: 0 = −3𝑎 + 𝑏 + 3
𝑡: 0 = −𝑐 − 1
𝑐 = −1,
𝑎 = 0,
𝜋2 = 𝜌0 𝐷 −3 ω−1 𝑄 =
𝑏 = −3
𝑄
𝐷3 𝜔
Q
0
3
-1

1
-1
-1
𝜋3 = 𝜌𝑎 𝐷𝑏 ω𝑐 𝜇
𝑀 𝑎
1 𝑐𝑀
𝑏
𝑀 𝐿 𝑡 = 1 = ( 3 ) (𝐿) ( )
𝐿
𝑡 𝐿𝑡
0 0 0
𝑀: 0 = 𝑎 + 1
𝐿: 0 = −3𝑎 + 𝑏 − 1
𝑡: 0 = −𝑐 − 1
𝑐 = −1,
𝑎 = −1,
𝜋3 = 𝜌0 𝐷−3 ω−1 𝑄 =
𝑏 = −2
μ
𝜌𝐷2 𝜔
Units check:
𝑀
3
μ
𝐿𝑡 = 𝑀𝐿 𝑡 = 1
=
𝜌𝐷2 𝜔 𝑀 𝐿2 1 𝐿𝑡𝑀𝐿2
𝐿3 𝑡
11.25
The performance of a fluid flow system is affected by the following variables:
force, F (units are MLt-2),
fluid density,  (units are ML-3),
velocity, v (units are L/t),
coating diameter, D (units are L)
fluid viscosity,  (units are ML-1t-1).
Determine the dimensionless groups formed from the variables involved using the Buckingham
Method.
Solution
Core Group = ρ, D, 
M
L
t
F
1
1
-2
ρ
1
-3
0
v
0
1
1
D
0
1
0
5 variables, rank is 3 so i=2.
π1 = ρa Db c F
M a
M c ML
b
(L)
( ) 2
M L t = 1 = ( 3)
L
Lt t
0 0 0
𝑀: 0 = 𝑎 + 𝑐 + 1
𝐿: 0 = −3𝑎 + 𝑏 − 𝑐 + 1
𝑡: 0 = −𝑐 − 2
𝑐 = −2, 𝑎 = 1, 𝑏 = 0
π1 = ρ1 D0 −2 P =
𝑃ρ
2
Units Check:
ML M
𝑃ρ
2 3
𝐿𝑡𝐿𝑡
= t L = 2 2 =1
2
MM

t 𝐿
Lt Lt
π2 = ρa Db c v
M a
M cL
M 0 L0 t 0 = 1 = ( 3 ) (L)b ( )
L
Lt t

1
-1
-1
M: 0 = a + c
L: 0 = −3a + b − c + 1
t: 0 = −c − 1
c = −1,
a = 1,
π1 = ρ1 D1 −1 P =
b=1
vρD

Unit Check
LM
vρD
t 3 𝐿 𝐿𝐿𝐿𝑡
= L =
=1
M

𝑡𝐿𝐿𝐿
Lt
Chapter 12
Instructor Only Problems
12.34
Air, water and glycerin are flowing through separate tubes with diameters of 0.5 inches and a
velocity of 40 ft/sec and are all at 80℉. Determine in each case whether the flow is laminar or
turbulent.
Solution
We must determine the Reynolds Number for each situation. Appendix I contains all the values
for density and viscosity that we need.
Air
lbm
ft
ρvD (0.0735 ft 3 ) (40 sec) (0.5 in) ft
Re =
=
= 9879
lbm
μ
12
in
−5
(1.24x10
s)
ft
Turbulent Flow (Re > 2300)
Water
lbm
ft
ρvD (62.2 ft 3 ) (40 sec) (0.5 in) ft
Re =
=
= 1.8 x 105
lb
μ
12
in
(0.578x10−3 m s)
ft
Turbulent Flow (Re > 2300)
Glycerin
lbm
ft
ρvD (78.7 ft 3 ) (40 sec) (0.5 in) ft
Re =
=
= 218
lbm
μ
12
in
(0.6
s)
ft
Laminar Flow (Re < 2300)
12.35
The determination of the boundary layer thickness is important in many applications. Aniline at
100℉ is flowing down a flat plate at 2.0 ft/s. When the fluid is 0.1 ft beyond the leading edge,
what is the thickness of the boundary layer?
Solution
Re =
ρvx (63.0 lbm /ft 3 )(2 ft/s)(0.1 ft)
=
= 7000
(180x10−5 lbm /ft s)
μ
Since Re is in the laminar region, we choose the laminar equation for the boundary layer
thickness.
𝛿
5
5
=
=
0.1 𝑓𝑡 √𝑅𝑒𝑥 √7000
δ = 0.006 ft = 0.072 inches
12.36
Water at 20oC is flowing along a 5-m-long flat plate with a velocity of 50 m/s. What will be the
boundary layer thickness at a position 5 m from the leading edge? Is the boundary-layer flow at
this location laminar or turbulent? If each of the plate surfaces measures 500 m2, determine the
drag force if both sides are exposed to the flow.
Solution
(a) Is this laminar or turbulent flow?
ρvx (998.2 kg/m3 )(50 m/s)(5 m)
Re =
=
= 2.5 x 108
kg
μ
(9.93 x10−4 Pa − s) (
)
Pa m s2
The flow is turbulent because Re > 3 x 106
(b) Calculate the boundary layer thickness.
δ 0.376
=
x Re1/5
0.376(5 m)
δ=
= 0.039 m
(2.5x108 )1/5
(c) Calculate the drag force on the surface.
0.0576
Cfx =
= 0.0012
Re1/5
FD = ACfx ρ
v∞2 (500 m2 )(0.0012)(998.2 kg/m3 )(50 m/s)2
=
x 2 sides = 1.5x106 kg m/s 2
2
2
12.37
In the case of carbon monoxide at 100oF flowing along a plane surface at a velocity of 4 ft/s, will
the boundary-layer flow at a position 0.5 ft from the leading edge be laminar or turbulent? What
will be the boundary-layer thickness at this location? For a plane surface 0.5 ft long with an
effective area of 200 ft2 on each side, determine the total drag force exerted if both sides are
exposed to the flow of CO.
Solution
(a) Is this laminar or turbulent flow?
Re =
ρvx (0.0684 lbm /ft 3 )(4 ft/s)(0.5 ft)
=
= 44487.8
lbm
μ
−5
(1.23x10
)
ft s
The flow is laminar because Re < 2 x 105
(b) Calculate the boundary layer thickness.
δ
5
= 1/2
x Re
δ=
5(0.5 ft)
= 0.012 ft
(44487.8)1/2
(c) Calculate the drag force on the surface.
CfL =
1.328
= 0.0063
Re1/2
FD = ACfL ρ
v∞2 (200 ft 2 )(0.0063)(0.0684 lbm /ft 3 )(4ft/s)2
=
x 2 sides = 1.38 lbm ft/s 2
2
2
12.38
Graph the results of Problem 12.8 and 12.18 and in one sentence describe what the data
describes.
Solution
The sphere without the dimples has significantly more drag than that of the golf ball with dimples,
so the dimples on the golf ball significantly reduce drag.
0.5
0.45
0.4
Drag (lbf)
0.35
0.3
0.25
Drag for Sphere
0.2
Drag for Golf Ball
0.15
0.1
0.05
0
0
100
200
300
Velocity (ft/s)
400
12.39
Using the data in Appendix E, calculate the Reynolds number for water flowing through a 0.25inch pipe at 0.5 and 1 ft/s in the temperature range from 32 to 600℉. Graph the Re vs.
temperature and state approximately where the flow goes from laminar or turbulent.
Solution
We must determine the Reynolds Number for each situation. Appendix I contains all the values
for density and viscosity that we need.
Re =
ρvD
μ
Re at
Temperature Density
Viscosity
0.5 ft/s Re at 1 ft/s
32
62.4
1.20E-03
541.7
1083.3
60
62.3
7.60E-04
853.9
1707.8
80
62.2
5.78E-04 1121.0
2241.9
100
62.1
4.58E-04 1412.4
2824.8
150
61.3
2.90E-04 2201.9
4403.7
200
60.1
2.06E-04 3039.0
6078.1
250
58.9
1.60E-04 3834.6
7669.3
300
57.3
1.30E-04 4591.3
9182.7
400
53.6
9.30E-05 6003.6
12007.2
500
49
7.00E-05 7291.7
14583.3
600
42.4
5.79E-05 7628.1
15256.2
For 0.5 ft/s, the flow reaches the transition point just above 150oF, and just about 80oF for the 1
ft/s velocity.
Temperature effect on Re
18000.0
Reyolds Number
16000.0
14000.0
12000.0
10000.0
8000.0
Re at 0.5 ft/s
6000.0
Re at 1 ft/s
4000.0
2000.0
0.0
0
100
200
300
400
500
TemperatureAxis Title
600
700
12.40
Water, air and benzene all at 60oF are flowing over a flat plate at 2.5 ft/s. For each fluid,
determine the boundary layer thickness at a point 0.25 ft beyond the leading edge.
Solution
Water
Re =
ρvx (62.3 lbm /ft 3 )(1.5 ft/s)(0.25 ft)
=
= 30740
(0.76x10−3 lbm /ft s)
μ
Since Re is in the laminar region, we choose the laminar equation for the boundary layer
thickness.
𝛿
5
5
=
=
0.25 𝑓𝑡 √𝑅𝑒𝑥 √30740
δ = 0.007 ft = 0.085 inches
Air
ρvx (0.0764 lbm /ft 3 )(1.5 ft/s)(0.25 ft)
e=
=
= 2368
(1.21x10−5 lbm /ft s)
μ
Since Re is in the laminar region, we choose the laminar equation for the boundary layer
thickness.
𝛿
5
5
=
=
0.25 𝑓𝑡 √𝑅𝑒𝑥 √2368
δ = 0.026 ft = 0.31 inches
Benzene
Re =
ρvx (55.2 lbm /ft 3 )(1.5 ft/s)(0.25 ft)
=
= 46517
(44.5x10−5 lbm /ft s)
μ
Since Re is in the laminar region, we choose the laminar equation for the boundary layer
thickness.
𝛿
5
5
=
=
0.25 𝑓𝑡 √𝑅𝑒𝑥 √46517
δ = 0.0058 ft = 0.07 inches
Chapter 13
Instructor Only Problems
13.19
A 2.20-in diameter pipe carries water at 15oC. The head loss due to friction is 0.500 m per 300 m
of pipe. Determine the volumetric flow rate of the water leaving the pipe.
Solution
H2 O@15℃:
ΔP
= 0.5m,
ρ
L = 300m,
D = 2.2m,
hL = 2ff
Re =
ff v 2 = 0.01799
ν = 1.195x10−6
L 2
v
D
(2.2)(v)
Dv
=
= 1.841x106 v,
−6
ν
1.195x10
300 2
hL = 9.81(0.5) = 2ff
v ,
2.2
Trial & Error − Assume turbulent flow smooth pipe, ff = 0.003
v = 2.448
v = 2.86
m
, Re = 4.508x106 , ff = 0.0022
s
m
m
, Re = 5.26x106 , ff = 0.0021, v = 2.93 − CLOSE ENOUGH
s
s
π
m3
2
v = 2.93 ( ) (2.2) = 11.13
4
s
m2
,
s
13.29
You have been hired to consult on a pilot plant project that requires the installation of a rough
horizontal pipe made from commercial steel into a flow system. The pipe has a diameter of 4 inches
and a length of 30 feet. A Newtonian fluid (heat capacity = 0.149 BTU / lbmºF,
density=0.161 lbm/ft3, viscosity=8.88x10-4 lbm/ft-sec and kinematic viscosity = 5.52x10-3 ft2/sec)
will flow through the pipe at 300 ft3/min. Assume that the flow is without edge effects and that the
no-slip boundary condition applies. Determine the pressure drop of the fluid as it travels through
the pipe.
Solution
ft 3
min
ft 3
Q = 300
x
=5
min 60 sec
s
ft 3
5 s
Q
v= =π
= 58.5 ft/s
A
(0.33 ft)3
4
ft
Dv (0.33 ft) (58.5 s )
Re =
=
= 3497.3
ft 2
ν
−3
5.52x10 s
This Reynolds number indicates that the flow is turbulent.
Figure 13.2 for commercial steel give a value for e of 0.00015. Calculating the relative
roughness,
e 0.00015
=
= 4.54x10−4
D
0.33
Now we use Equation 13-14 for turbulent flow in a rough pipe,
1
D
0.33
) + 2.28 = 15.65
= 4 log ( ) + 2.28 = 4 log (
e
0.00015
√ff
ff = 0.00408
Leq v 2
ΔP
hL =
= 2ff
ρg
D g
Rearranging,
ΔP = 2ρff
Leq 2
(0.166 lbm /ft 3 )(0.00408) 30 ft
(
) (58.5 ft/s)2 = 13.0 lbf /ft 2
v =2
D
32.174 lbm ft/lbf s2
0.33 ft
13.30
Your boss comes to you with an important project where you are to determine the pressure drop
in a rough horizontal high pressure pipe made of cast iron. The pipe has a diameter of 0.25 feet
and a length of 10 feet. Benzene at 150℉ flows through the pipe at 250 ft3/min.
Solution
This problem will illustrate the use of figures 13.1 and 13.2.
We begin by making some basic calculations for flow rate, velocity, and most importantly the
Reynolds Number.
ft 3
min
ft 3
Q = 250
x
= 4.17
min 60 sec
sec
ft 3
4.17
Q
sec = 85 ft/s
v= =π
A
(0.25 ft)2
4
ρvD (51.8 lbm /ft 3 )(85 ft/s)(0.25 ft)
Re =
=
= 4.5x106
lbm
μ
−5
24.5 x 10
ft s
Figure 13.2 contains roughness parameters for various materials including cast iron which the
figure gives as e=0.00085. We will use Figure 13.1 for that.
We next calculate the relative roughness,
𝑒 0.00085
=
= 0.0034
D
0.25 ft
Next we turn to Figure 13.1. Along the right hand side y-axis are values for the relative roughness.
We find 0.0034 on the axis. This is obviously an approximation and with some practice is
straightforward. The value for 0.0034 is between 0.003 and 0.004 so we pick a point approximately
half way between and we follow an imaginary line in between the line for 0.003 and 0.004 until
we cross the line for the Reynolds Number we calculated above to be 4.5x106 . We find that point
and then look to the y-axis on the left to find a value for the friction factor to be 0.0068. Since the
flow is Turbulent, we use Equation 13-3 and 13-16 in combination to calculate the pressure drop.
Leq v 2
∆P
hL =
= 2ff
ρg
D g
Rearrange and solve for the pressure change,
Leq 2
(51.8 lbm /ft 3 )
10 ft
(0.0068)
(85 ft/s)2 = 6328 lbf /ft 2
∆P = 2ρff
v =2
2
D
32.174 lbm ft/lbf s
0.25 ft
Problem 13.31
Freon-12 is flowing at 10 ft/sec through a pipe made of galvanized iron at 100℉. If the pipe is
100 feet long at 4 inches in diameter, what is the headloss in this system?
Solution
From Figure 13.2, e=0.0005 for galvanized iron.
ρvD (78.7 lbm /ft 3 )(10 ft/s)(4/12 ft)
Re =
=
= 3.0x105
lbm
μ
−5
88.4 x 10
ft s
𝑒 0.0005
=
= 0.0015
D 4/12 ft
From Figure 13.1, 𝑓𝑓 = 0.0056.
Leq v 2
100 ft (10 ft/s)2
)
hL = 2ff
= 2(0.0056) (
= 10.4 ft
D g
4/12 ft 32.2 ft
s2
13.32
Water at a rate of 0.1 ft3/min (1.6x10-3 ft3/sec) and 60oF flows through a smooth horizontal tube
200 ft long. The pressure drop is 300 lbf/ft2. Determine the pipe diameter.
Solution
Δ𝑃
𝐿 𝑣2
= ℎ𝐿 = 2𝑓𝑓
𝜌𝑔
𝐷 𝑔
Δ𝑃
𝐿
= 2𝑓𝑓 𝑣 2
𝜌
𝐷
300 𝑙𝑏𝑓 /𝑓𝑡 3 (32.174
62.3 𝑙𝑏𝑚 /𝑓𝑡 3
𝑙𝑏𝑚 𝑓𝑡
)
𝑙𝑏𝑓 𝑠 2
2
200𝑓𝑡 (1.6𝑥10−3 𝑓𝑡 3 /𝑠)
= 2𝑓𝑓
[
]
𝜋 2
𝐷
𝐷
4
𝑓𝑓
= 99329
𝐷5
(1.6𝑥10−3 𝑓𝑡 3 /𝑠) 167
𝐷𝑣
𝐷
𝑅𝑒 =
=
=
𝜋 2
𝜈
1.22𝑥10−5 𝑓𝑡 2 /𝑠
𝐷
𝐷
4
Use trial and error to solve for D:
167
1
𝑓𝑓
Guess
𝑅𝑒 =
= 4 log10 (𝑅𝑒√𝑓𝑓 ) − 0.4
𝐷5 =
𝐷
99329
√𝑓𝑓
0.025
6680
0.00862
𝑓𝑓 = 0.001
0.0387
4315
0.00976
𝑓𝑓 = 0.00862
0.0397
4209
0.00984
𝑓𝑓 = 0.00976
0.0397
4202
0.00984
𝑓𝑓 = 0.00984
Note: A solver was used to do the calculation in the far-right column (MatLab, Python or a
calculator with a solver will all work for this).
D=0.04 feet.
13.33
You want to understand how the piping material will affect the pressure drop for benzene flowing
at 80oF with a flow rate of 1𝑥10−3 𝑓𝑡 3 /𝑠. All the pipes are 100 ft long with a diameter of 0.25
inches and considered rough. Calculate the pressure drop for Commercial Steel, Galvanized Iron,
and Cast Iron. Make a table with the material, relative roughness and pressure drop to compare
the results.
Solution
Q = 1𝑥10−3 𝑓𝑡 3 /𝑠
Q 1𝑥10−3 𝑓𝑡 3 /𝑠
v= =
2 = 2.93 ft/s
A
π 0.25
4 ( 12 ft)
0.25
Dv ( 12 ft) (2.93 ft/s)
Re =
=
= 8783
ft 2
ν
0.695x10−5 s
This Reynolds number indicates that the flow is turbulent.
Commercial Steel
Figure 13.2 for commercial steel give a value for e of 0.00015. Calculating the relative
roughness,
e 0.00015
=
= 7.2x10−3
D
0.0208
Now we use Equation 13-14 for turbulent flow in a rough pipe,
1
D
0.0208
) + 2.28 = 10.84
= 4 log ( ) + 2.28 = 4 log (
e
0.00015
√ff
ff = 0.0085
Leq v 2
ΔP
hL =
= 2ff
ρg
D g
Rearranging,
Leq 2
(54.6 lbm /ft 3 )(0.0085) 100 ft
(
) (2.93 ft/s )2 = 1190.7 lbf /ft 2
v =2
D
32.174 lbm ft/lbf s2
0.0208 ft
Galvanized Iron
Figure 13.2 for commercial steel give a value for e of 0.0005. Calculating the relative roughness,
e 0.0005
=
= 2.4x10−2
D 0.0208
Now we use Equation 13-14 for turbulent flow in a rough pipe,
1
D
0.0208
) + 2.28 = 8.75
= 4 log ( ) + 2.28 = 4 log (
e
0.0005
√ff
ff = 0.01306
Leq v 2
ΔP
hL =
= 2ff
ρg
D g
ΔP = 2ρff
Rearranging,
ΔP = 2ρff
Leq 2
(54.6 lbm /ft 3 )(0.01306) 100 ft
(
) (2.93 ft/s )2 = 1829.5 lbf /ft 2
v =2
D
32.174 lbm ft/lbf s2
0.0208 ft
Cast Iron
Figure 13.2 for commercial steel give a value for e of 0.0005. Calculating the relative roughness,
e 0.00085
=
= 4.0x10−2
D
0.0208
Now we use Equation 13-14 for turbulent flow in a rough pipe,
1
D
0.0208
) + 2.28 = 7.83
= 4 log ( ) + 2.28 = 4 log (
e
0.00085
√ff
ff = 0.0163
Leq v 2
ΔP
hL =
= 2ff
ρg
D g
Rearranging,
Leq 2
(54.6 lbm /ft 3 )(0.0163) 100 ft
(
) (2.93 ft/s )2 = 2283.4 lbf /ft 2
ΔP = 2ρff
v =2
D
32.174 lbm ft/lbf s2
0.0208 ft
Material
Relative Roughness
Pressure drop (lbf /ft 2 )
−3
Commercial Steel
1190.7
7.2x10
−2
Galvanized Iron
1829.5
2.4x10
−2
Cast Iron
2283.4
4.0x10
13.34
A fluid moves through a 50 meter horizontal tube with a 3 cm diameter at a temperature of 20oC,
density of 870 kg/m3 and viscosity of 0.10 kg/m-s. What is the pressure drop if the velocity is 3
m/s?
Solution
𝐿 v2
𝐷𝑔
∆𝑃
ℎ𝐿 =
𝜌𝑔
ℎ𝐿 = 2𝑓𝑓
Equating these gives,
∆𝑃
𝐿 v2
= 2𝑓𝑓
𝜌𝑔
𝐷𝑔
Solving for ∆𝑃,
∆𝑃 = 2𝜌𝑓𝑓
𝐿 2
50 𝑚
(3 𝑚/𝑠)2 = 2.6𝑥107 𝑓𝑓
v = 2(870 𝑘𝑔/𝑚3 )𝑓𝑓
𝐷
3/100 𝑚
Next, need to find 𝑓𝑓 , and to do this we must determine whether the flow is laminar or turbulent.
3
3
𝜌v𝐷 (870 𝑘𝑔/𝑚 )(3 𝑚/𝑠) (100 𝑚)
𝑅𝑒 =
=
= 783
𝑘𝑔
𝜇
(0.1
)
𝑚∙𝑠
16
𝑓𝑓 =
= 0.02043
𝑅𝑒
Thus,
∆𝑃 = 2.6𝑥107 𝑓𝑓 = 2.6𝑥107 (0.02043) = 5.5𝑥105
𝑘𝑔
= 5.5𝑥105 𝑃𝑎
𝑚𝑠 2
Chapter 14
Instructor Only Problems
14.25
A centrifugal pump with an impeller diameter of 0.18 m is to be used to pump water (density =
1000 kg/m3) with the pump inlet located 3.8 m above the surface of the supply reservoir. At a
flow rate of 0.760 m3/s, the head loss between the reservoir surface and the pump inlet is 1.8 m
of water. The performance curves are shown below. Would you expect cavitation to occur?
Solution
Pump: D = 0.18 m, inlet at y = 3.8m above supply reservoir, v̇ = 0.760
m3
s
Between reservoir survace & 𝑝𝑢𝑚𝑝 𝑖𝑛𝑙𝑒𝑡: hL = 1.8 m H2 O
Patm − Pv
Energy Equation: NPSH =
− y2 − hL at 20℃ PV = 2.34 kPa
ρg
Patm − Pv (101.3 − 2.34)(103 )
=
= 10.09m
ρg
1000(9.81)
NPSH = 10.09 − 3.8 − 1.8 = 4.49 mH2 O
m3
From Performance curve @ v̇ = 0.760
, NPSH ≅ 3.9m, Cavitation should NOT occur
s
14.26
Pumps used in an aqueduct operate at 400 rpm and deliver a flow rate of 220 m3/s against a head
of 420 m. What types of pumps are they?
Solution
m3
= 3.487x106 gpm, h = 420 m = 1378 ft
s
1
(400)(3.487x106 )2
NS =
= 3302
3
(1378)4
According to Fig 14.11 This is probably a high capacity Centrifugal Pump
v̇ = 220
14.27
A pump is required to deliver 60,000 gpm against a head of 300 m when operating at 2000 rpm.
What type of pump should be specified?
Solution
Pump to deliver 60,000 gpm with h = 300m @ 2000 rpm
1
NS =
(2000)(6x104 )2
3
≅ 2790
300 4
(0.3048)
Accoding to Fig 14.11 This is probably a high capacity Centrifugal Pump
14.28
An axial-flow pump has a specified specific speed of 6.0. The pump must deliver 2400 gpm
against a head of 18 m. Determine the required operating rpm of the pump.
Solution
1
Axial Flow Pump − NS = 6.0 = NS =
CQ2
1
v2w
3 = 3 3
4
h4 g 4
CH
This ratio is (obviously)dimensionless − by converting to units on abscissa of Figure 14.11
− the ratio of NS is given by this equation to the value on Figure 14.11 is 2733
− So A value of 6 for the equation is equivalent to 6(2733)
= 1.64x104 on abscissa of Figure 14.11
1
∴ 1.64x104 =
n(2400)2
3
(18)4
∴ n = 2925 rpm
14.29
A pump operating at 520 rpm has the capability of producing 3.3m3/s of water flow against a head
of 16 m. What type of pump is this?
Solutions
Pump @ 520 rpm,
m3
v̇ = 3.3
,
s
h = 16m
3
1
) (7.48)(60) = 52302 gpm
v̇ = (3.3) (
0.3048
h=
16
= 42.65 ft
0.3048
1
Ns =
(520)(5.23x105 )2
3
(42.65)4
= 22532 ∴ from Fig. 14.11 − Axial Flow
14.30
A pump operating at 2400 rpm delivers 3.2 m3/s of water against a head of 21 m. Is this pump an
axial-flow, mixed-flow, or radial-flow machine?
Solution
Pump @ 2400 rpm, v̇ = 3.2
m3
, h = 21m
s
3
1
) (7.48)(60) = 50720 gpm
v̇ = (3.2) (
0.3048
h=
21
= 68.9 ft
0.3048
1
Ns =
(2400)(5.072x104 )2
= 22601
3
(68.9)4
∴ from Fig. 14.11 − Axial Flow
Chapter 15
Instructor Only Problems
15.28
Water at 40°F is to flow through a 11=2-in: schedule 40 steel pipe. The outside surface of the
pipe is to be insulated with a 1-in.-thick layer of 85% magnesia and a 1-in.-thick layer of packed
glass wool, k = 0.022 Btu/h ft °F. The surrounding air is at 100°F.
a. Which material should be placed next to the pipe surface to produce the maximum insulating
effect?
b. What will be the heat flux on the basis of the outside pipe surface area? The convective heattransfer coefficients for the inner and outer surfaces are 100 and 5Btu/hft°F, respectively.
Solution
Per unit length:
1
R ins =
= 0.0238
2πri hi
R1 =
R2 =
R3 =
R4 =
r
ln (r2 )
1
2πk 2
r
ln (r3 )
2
2πk 2
r
ln (r4 )
3
2πk 3
= 0.00105
=
0.115
,
k2
=
0.0659
,
k3
1
= 0.1296
2πr4 ho
ΣR = 0.1545 +
0.115 0.0659
+
k2
k3
Case 1: k 2 for Magnesia: ΣR = 6.134
Case 2: k 2 for Glass Wool: ΣR = 7.096 (BEST)
q=
60
BTU
= 8.47
7.096
hrft
q
8.47
BTU
=
= 5.55
A 2π (2.95)
hrft 2
12
15.29
A 1-in.-nominal-diameter steel pipe with its outside surface at 400°F is located in air at 90°F with
the convective heat-transfer coefficient between the surface of the pipe and the air equal to 1.5
Btu/h-ft-°F. It is proposed to add insulation having a thermal conductivity of 0.06 Btu/h ft °F to
the pipe to reduce the heat loss to one-half that for the bare pipe. What thickness of insulation is
necessary if the surface temperature of the steel pipe and ho remain constant?
Solution
For Base Pipe: q = πD1 hΔT = π (
For Insulated Pipe: 80 =
Ts − T∞
D
ln (D2 )
1
1
+
2πk
πD2 h
𝜋ℎ 𝐷2
1
𝜋ℎ
ln +
=
∆𝑇
2𝜋𝑘 𝐷1 𝐷2 80
D2
1
)+
12.5 ln (
= 18.25
0.1905
D2
By Trial & Error: D2 = 0.382ft,
2t = D2 − D1 = 3.265in,
t = 1.63in
1.315
BTU
) (1.5)(310) = 160
per ft
12
hr
15.30
1/4
If, for the conditions of Problem 15.29, ho in Btu/h ft °F varies according to ℎ𝑜 = 0.575/𝐷𝑜 ,
where Do is the outside diameter of the insulation in feet, determine the thickness of insulation
that will reduce the heat flux to one-half that of the value for the bare pipe.
Solution
For Base Pipe, Per Foot: q = hAΔT =
ΔT
D
ln ( Do )
i
2πk
53.3 =
+
1
πDo ho
310
D
ln ( o )
1.315 +
2π(0.06)
1
D 4
(12o )
D
0.575π 12o
By trial and error: Do = 9.22in
Insulation thickness =
1.315
BTU
) (310) = 106.7
12
hr
1π(
1.315 4
( 12 )
With Insulation: q = 53.3
q=
0.575
9.22 − 1.315
= 3.95in
2
15.31
Liquid nitrogen at 77 K is stored in a cylindrical container having an inside diameter of 25 cm.
The cylinder is made of stainless steel and has a wall thickness of 1.2 cm. Insulation is to be
added to the outside surface of the cylinder to reduce the nitrogen boil-off rate to 25% of its
value without insulation. The insulation to be used has a thermal conductivity of 0.13 W/m K.
Energy loss through the top and bottom ends of the cylinder may be presumed negligible.
Neglecting radiation effects, determine the thickness of insulation when the inner surface of the
cylinder is at 77 K, the convective heat-transfer coefficient at the insulation surface has a value
of 12 W/m2 K, and the surrounding air is at 25°C.
Solution
qo =
ΔT
ΣR
13.7
1 ln (12.5)
1
0.6136 K
without insulation: ΣR =
[
+
]=
(12)(0.137)
2πL
17.3
2πL W
13.7
ro
ln (13.7
)
1 ln (12.5)
1
With insulation: ΣR =
[
+
+
]
(12)(ro )
2πL
17.3
0.13
q wo
If q w =
4
2.454
ΣR w = 4ΣR wo =
2πL
13.7
ro
ln (
)
ln (13.7
)
1
12.5 +
+
= 2.454
(12)(ro )
17.3
0.13
By trial and error, ro = 0.177m,
insulation thickness = ro − ri = 0.177 − 0.137 = 0.04m = 4 cm
15.32
Air at 100oC is moving over a stainless-steel plate. Thermocouples are positioned at 15 and 25 mm
below the surface and measure temperatures at 60 and 50oC respectively. Calculate the convective
heat transfer coefficient (assume negligible radiation).
Solution
Stainless Steel: k=17.3 𝑊/𝑚𝐾 (Appendix H).
𝑞𝑖𝑛 = 𝑞𝑜𝑢𝑡
ℎ𝐴∆𝑇 = −𝑘𝐴𝛻⃑𝑇
ℎ(𝑇∞ − 𝑇𝑠𝑠 ) = −𝑘
At steady state, 𝑞 is constant in the plate.
𝑑𝑇
𝑑𝑥
𝑑𝑇
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑𝑥
𝑑𝑇
𝑞
=−
𝑑𝑥
𝑘
𝑑𝑇 𝑇1 − 𝑇2 𝑇𝑆𝑆 − 𝑇1
=
=
𝑑𝑥 𝑥1 − 𝑥2
0 − 𝑥1
𝑇𝑆𝑆 = 75℃
𝑞 = −𝑘
Back to equation to solve for h:
ℎ(𝑇∞ − 𝑇𝑠𝑠 ) = −𝑘
ℎ=−
𝑊
(17.3 𝑚 ∙ 𝐾)
𝑑𝑇
𝑑𝑥
60 − 50
= 692 𝑊/𝑚 ∙ 𝐾
(100 − 75) 0.015 − 0.025𝑚
15.33
A steel rod at a temperature of 450oF with an outside diameter of 2 inches is suspended in air.
The air temperature is 100oF and the heat transfer coefficient between the rod and the air is 2.0
BTU/h ft2 oF. The rod is coated with insulation with a thermal conductivity of 0.05 BTU/h ft oF
in an attempt to minimize heat loss from the rod. Determine the thickness of the insulation
necessary to reduce the heat loss by 40%.
Solution
q ins
= 0.4 (= 40%)
q bare
r0 = ri + ∆r (where the unknown is ∆r and represents the insulation thickness)
Tp − T∞
(Tp − T∞ )L
∆T
q ins =
=
=
r
r
0
IR
ln( 0⁄ri )
1 ln( ⁄ri )
1
+
+
hA
2πkL
2πk ins
h(2πr0 )
q bare = hA∆T = h(2πri L)(Tp − T∞ )
q ins
= 0.4 =
q bare
1
r
ln( 0⁄ri )
1
(
) (h(2πri L))
+
2πk ins
h(2πr0 )
2
0.167 ft
D = 2 in =
ft = 0.167 ft,
R=
= 0.083 ft
12
2
1
0.4 =
r
ln( 0⁄0.083 ft)
1
BTU
(
+
BTU
BTU ) (2 h ft 2 ℉ (2π(0.083 ft)))
(2
) (2πr0 ) 2π (0.05
)
h ft ℉
h ft 2 ℉
r
ln( 0⁄0.083 ft)
1
BTU
(
) (2
+
(2π(0.083 ft))) (0.4) = 1
BTU
BTU
h ft 2 ℉
(2πr
)
(2
)
2π
(0.05
)
0
h ft ℉
h ft 2 ℉
ro = 0.15 ft
r0 = ri + ∆r
∆r = 0.15 ft − 0.083 ft = 0.067ft = 0.8 in
15.34
Derive an expression for the thermal resistance, as described in Equation 15-16, for onedimensional heat conduction through a spherical shell assuming steady state.
Solution
⃑T
q = −kA∇
q r = −k(4πr 2 )
dT
dr
r2
T2
dr
=
−4πk
∫
dT
2
r1 r
T1
qr ∫
q r [(−
1
1
) − (− )] = −4πk(T2 − T1 )
r2
r1
qr =
4πk(T2 − T1 )
1
1
(r ) − (r )
1
2
Using the form of Equation 15-16,
R=
1
1
(r ) − (r )
1
2
4πk
Chapter 16 Problem Assignments
Show/Hide problems: 16.1, 16.5 (solutions provided)
Instructor Only Problems: (16.15, 16.16, 16.17)
(3 new problems)
Chapter 17
Instructor Only Problems
17.43
Circular fins are employed around the cylinder of a lawn mower engine to dissipate heat. The fins are
made of aluminum are 0.3-m thick, and extend 2 cm from base to tip. The outside diameter of the
engine cylinder is 0.3 m. Design operating conditions are T∞ = 30°C and h = 12 W/m2 K. The
maximum allowable cylinder temperature is 300°C. Estimate the amount of heat transfer from a
single fin. How many fins are required to cool a 3-kW engine, operating at 30% thermal efficiency, if
50% of the total heat given off is transferred by the fins?
Solution
rL =
t=
0.3
= 0.15 m r0 = 0.15 + 0.02 = 0.17 m
2
0.003
= 0.0015 m
2
A = 2π(r02 − ri2 ) + Aend = 2π(0.172 − 0.152 ) + 2π(0.17)(0.003) = 0.0434 m2
1
1
2
h 2
12
(r0 − ri ) ( ) = 0.02 [
] = 0.263
(46.4)(0.0015)
kt
r0
= 1.13
ri
From fig 17.11 ηf ≅ 0.96
q = Af ηf h(T0 − T∞ ) = 0.0434(0.96)(12)(270) = 135 W per fin
For a 30% 3 kW engine
(3 kW)
Qin =
= 10 kW
0.3
Qout = Qin − W = 7 kW
Amount Tx from fins = 0.5(7) = 3.5 kW
Number of fins required:
n=
3500 W
= 25.92 26 fins required
W
135
fin
17.44
Heat from a flat wall is to be enhanced by adding straight fins, of constant thickness, made of
stainless steel. The following specifications apply:
h = 60 W⁄m2 K
Tb (base) = 120℃
T∞ (air) = 20℃
Fin base thickness, t = 6mm
Fin length, L = 20 m m
Determine the fin efficiency and heat loss per unit width for the finned surface.
Solution
a) L = 20 mm t = 6 mm
For both cases: Tb = 120℃ T∞ = 20℃
h = 60
W
,
m2 K
k s.s. = 15.3
W
mK
For case a) straight fin
q = ηf hAf θb
Using text – Fig 17.11
3
2
1
2
1
3
2
t
h
60
2[
(0.02
(L + ) [
=
+
0.003)
= 0.588
]
]
t
(15.3)(0.006)(0.023)
2
kt (L + 2)
ηf ≅ 0.80
Per meter of width {neglecting ends}:
q f = 0.8(60)(100)(2)(0.020) = 192
For case b) triangular
W
m
1
3
(L)2 [
1
3
2
h 2
60
] = 0.022 [
] = 0.723
(15.3)(0.003)(0.02)
kt(L)
ηf ≅ 0.81
q = ηf Af hθb = (0.81)(2)(0.02)(60)(100) = 194.4
W
m
17.49
Find the rate of heat transfer from a 3-in.-OD pipe placed eccentrically inside a 6-in.-ID cylinder with
the axis of the smaller pipe displaced 1 in. from the axis of the large cylinder. The space between the
cylindrical surfaces is filled with rock wool (k = 0.023 Btu/h ft °F). The surface temperatures at the
inside and outside surfaces are 400°F and 100°F, respectively.
Solution
Using table 17.1
S=
2π
1 + ρ2 − ϵ 2
)
cosh−1 (
2ρ
ρ = 0.5 ϵ =
S=
1
6
2π
1 1
1 + 4 − 36
−1
)
cosh (
1
=
2π
11
cosh−1 ( )
= 9.6
9
q
Btu
= kS∆T = (0.023)(9.6)(300) = 66.3
L
hr ft
17.50
A cylindrical tunnel with a diameter of 2 m is dug in permafrost (k = -0.341 W/m2 K) with its axis
parallel to the permafrost surface at the depth of 2.5 m. Determine the rate of heat loss from the
cylinder walls, at 280 K, to the permafrost surface at 220 K.
Solution
Table 17.1
S=
2π
2π
=
= 1.311
−1
ρ
cosh−1 ( r ) cosh (2.5)
q
W
= kS∆T = (0.341)(1.311)(60) = 26.83
L
m
17.51
Determine the heat flow per foot for the configuration shown, using the numerical procedure for a
grid size of 1.5 ft. The material has a thermal conductivity of 0.15 Btu/h ft °F. The inside and outside
temperatures are at the uniform values of 200°F and 0°F, respectively.
Solution
200 + 0 + 2T2 − 4T1 = 0
200 + 0 + T1 − T3 − 4T2 = 0
0 + 0 + 2T2 − 4T3 = 0
T1 = 91.7 ℉ T2 = 83.3 ℉ T4 = 41.7℉
q
200 − 91.7
Btu
= 8k [
+ 200 − 83.3] = 205
L
2
hr ft
17.52
Repeat the previous problem, using a grid size of 1 ft.
Solution
By iteration:
Node
1
2
3
4
5
6
7
T,℉
123
112
74
58
51.5
36
18
q
Btu
= 8k[(200 − 123) + (200 − 112)] = 198
L
hr ft
17.53
A 5-in. standard-steel angle is attached to a wall with a surface temperature of 600°F. The angle
supports a 4.375-in. by 4.375-in. section of building bring whose mean thermal conductivity may be
taken as 0.38 Btu/h ft °F. The convective heat- transfer coefficient between all surfaces and the
surrounding air is 8 Btu/h ft2 °F. The air temperature is 80°F. Using numerical methods, determine
(a) the total heat loss to the surrounding air
(b) the location and value of the minimum temperature in the brick
Solution
Numerical solution using a 12x12 mesh yields the following:
q
Btu
= 1400
L
hr ft 2
Tmin = 91.8 ℉ @ i, j = 12,12
17.54
Saturated steam at 400°F is transported through the 1-ft pipe shown in the figure, which may be
assumed to be at the steam temperature. The pipe is centered in the 2-ft-square duct, whose surface
is at 100°F. If the space between the pipe and duct is filled with powdered 85% magnesia insulation,
how much steam will condense in a 50-ft length of pipe?
Solution
S=
2π
= 8.17
r0
ln ( ri ) − 0.271
q = kSL∆T = (0.037)(8.17)(300)(50) = 2180
q
2180
2
lb
Steam condensed = h = 826 = 65 ( hrm )
fg
Btu
hr
17.55
A 32.4-cm-OD pipe, 145-cm long, is buried with its centerline 1.2 m below the surface of the ground.
The ground surface is at 280 K and the mean thermal conductivity of the soil is 0.66 W/m ? K. If the
pipe surface is at 370 K, what is the heat loss per day from the pipe?
Solution
S=
2π
ρ =
cosh−1 ( r )
q = kS∆T = (0.66
2π
2π
= 3.17
1.2
cosh−1 (0.324)
W
) (3.17)(90 K)(145 m) = 273 W
mK
Chapter 18
Instructor Only Problems
18.32
A thick wall of oak, initially at a uniform temperature of 25°C, is suddenly exposed to combustion
exhaust at 800°C. Determine the time of exposure required for the surface to reach its ignition
temperature of 400°C, when the surface coefficient between the wall and combustion gas is 20
W/m2 K.
Solution
T − T0
x
hy h2 αt
x
h √αt
) − [exp ( + 2 )] [erfc (
= erfc (
+
)]
T∞ − T0
k
k
k
2√αt
2√αt
T − T0
400 − 25
=
− 0.484
T∞ − T0 800 − 25
@ surface (x=0)
x
2√αt
= 0 erf(0) = 0 erfc(0) = 1
h √αt
20
=
(1.07x10−7 )2 = 0.0311t 2 = z
k
0.21
2
0.484 = 1 − ez (1 − erf(z))
1
Trial and error z ≅ 0.73 = 0.0311t ⁄2
t = 551 s = 9.18 min
18.33
Air at 65°F is blown against a pane of glass 1/8 in. thick. If the glass is initially at 30°F, and has frost
on the outside, estimate the length of time required for the frost to begin to melt.
Solution
For glass:
α=
k
ρĈp
=
0.45
= 0.0132 ft 2 ⁄hr
(170)(0.2)
Using semi-infinite wall expression
T − T0 32 − 30
x
)
=
= 0.0572 = erfc (
Ts − T0 65 − 30
2√αt
x
2√αt
= 1.38
t = 3.9 s
18.34
18.34 How long will a 1-ft-thick concrete wall subject to a surface temperature of 1500°F on one side
maintain the other side below 130°F? The wall is initially at 70°F.
Solution
Charts apply but are difficult to read – check validity of infinite wall solution
L
2√αt
=
1 ft
>2
2(0.0231t)2
Works for t > 27 ℎ𝑟𝑠
x
2√αt
=
1
3.29
= 1
2
2(0.0231t)
t ⁄2
Ts − T
3.29
= erf ( 1 )
Ts − T0
t ⁄2
t ≅ 5.2 hrs
18.35
A stainless-steel bar is initially at a temperature of 25°C. Its upper surface is suddenly exposed to an
air stream at 200°C, with a corresponding convective coefficient of 22 W/m2 K. If the bar is
considered semi-infinite, how long will it take for the temperature at a distance of 50 mm from the
surface to reach 100°C?
Solution
Applicable expression is
T∞ − T
hx
= erfc(A) + exp ( + B2 ) [1 − erf(A + B)]
T∞ − T0
k
A=
x
2√αt
=
0.05
1
2(0.444x10−5 t) ⁄2
= 11.92t
−1⁄
2
hx 22(0.05)
=
= 0.0636
k
17.3
B=
h √αt
22
1
1
=
(0.444x10−5 t) ⁄2 = 0.00267t ⁄2
k
17.3
B2 = 7.11x10−6 t
Trial and error
t ≅ 49000 s ≅ 13.6 hrs
18.39
IF the heat flux into a solid is given as F(t), show that the penetration depth 𝛿 for a semi-infinite
solid is of the form
𝑡
1/2
∫ 𝐹(𝑡)𝑑𝑡
𝛿 = (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)√𝛼 [ 0
]
𝐹(𝑡)
Solution
Eqn. (18-33)
qx
d δ
dδ
= ∫ ρĈp T dx − ρĈp T0
A dt 0
dt
T − T0
x
= Φ( )
Ts − T0
δ
∂T
1 ∂Φ
|
= (Ts − T0 )
|
∂x x=0
δ ∂x x=0
∂Φ
∂x
(0) = K (a constant)
qx
∂T
= −K (0) = F(t)
A
∂x
 Ts − T0 =
F(t)
ρĈp
=
F(t)δ
kK
d δ
dδ
∫ T dx − T0
dt 0
dt
δ
d δ
dδ d
dδ d
∫ T dx = T0 + ( Ts − T0 ) ∫ Φ dx = T0
+ [(T − T0 )Bδ]
dt 0
dt dt
dt dt s
0
(B is a constant)
d
d Bδ2 F(t)
[(T
)Bδ]
=
= [
]
s − T0
dt
kK
ρĈp dt
F(t)
F(t)kK
ρĈp B
=
d 2
[δ F(t)]
dt
t
K
δ F(t) = α ∫ F(t) dt
B 0
2
t
∫ F(t) dt 1⁄
δ = (constant)√α( 0
) 2
F(t)
18.40
If the temperature profile through the ground is linear, increasing from 35°F at the surface by 0.5°F
per foot of depth, how long will it take for a pipe buried 10 ft below the surface to reach 32°F if the
outside air temperature is suddenly dropped to 0°F. The thermal diffusivity of soil may be taken as
0.02 ft2/h, its thermal conductivity is 0.8 Btu/h ft °F, and the convective heat- transfer coefficient
between the soil and the surrounding air is 1.5 Btu/h ft2 °F.
Solution
Numerical solution required
Initial temperature profile –
T = 35 + 0.5x T in ℉, x in ft
Algorithms
For all nodes except surface:
Tit+1 =
Ti+1 + Ti−1
2
For surface node:
T0t+1 = T1t −
~
α∆t 1
= ;
∆x 2 2
h∆x t
T
k 0
2h∆t
ρĈp ∆x
=
h∆x
;
k
T∞ = 0
Result – Using a spreadsheet or program
t ≅ 1800 hours
18.41
A brick wall (α=0.016ft2/h) with a thickness of 1.51 ft is initially at a uniform temperature of 80°F.
How long, after the wall surfaces are raised to 300°F and 600°F, respectively, will it take for the
temperature at the center of the wall to reach 300°F?
Solution
Numerical solution required
Algorithms:
Node 1:
T1t+1 =
T0 + T2n
2
t+1
Tn−1
=
t
Tn + Tn−2
2
Tit+1 =
Ti+1 + Ti−1
2
α∆t
1
For ∆x2 = 2
If ∆x = 0.25 ft; ∆t = 1.95 hr
No of increments ≅7.4
t = 7.4(1.95) ≅ 144 hrs
18.42
A masonry brick wall 0.45 m thick has a temperature distribution at time, t=0 which may be
approximated by the expression T(K)=520+330sinπ(x/L) where L is the wall width and x is the
distance from either surface. How long after both surfaces of this wall are exposed to air at 280 K
will the center temperature of the wall be 360 K? The convective coefficient at both surfaces of the
wall may be taken as 14 W/m2 K. What will the surface temperature be at this time?
Solution
T = 520 + 330 sin (
α
k
ρĈp
=
πx
)
L
0.66
= 4.72x10−5 m2 ⁄s
(1670)(8.38)
α∆t 1
=
∆x 2 2
Same as problem (18.34)
α∆t
1
For ∆x2 = 2 If ∆x = 0.225 m; ∆t = 536 s
No of increments ≅2.4
Time ≅ 2.4(536) =1286 s =21.4 min
At this time Tsurface ≅ 360 K
Chapter 19
Instructor Only Problems
19.20
Determine the total heat transfer from the vertical wall described in Problem 19.19 to the
surrounding air per meter of width if the wall is 2.5 m high.
Solution
L
L
k
βg 4 3
5
1
−1
q = ∫ hx ∆T dx = ∫ Nux ∆T dx = k∆T ⁄4 (0.508)Pr ⁄2 (Pr + 0.954) ⁄4 ( 2 ) ( L ⁄4 )
x
ν
3
0
0
= 995 W
19.21
Simplified relations for natural convection in air are of the form
ℎ = 𝛼(∆𝑇/𝐿)𝛽
where α, β are constants; L is a significant length, in ft; ΔT is Ts - T∞, in °F; and h is the
convective heat-transfer coefficient, Btu/h ft2 °F. Determine the values for α and β for the plane
vertical wall, using the equation from Problem 19.14.
Solution
hx x
1
−1
1
Nux =
= 0.508Pr ⁄2 (Pr + 0.954) ⁄4 Gr ⁄4
k
For air: Pr = 0.72, k = 0.015 Btu⁄hr ft ℉
1
1
∆T 4
∆T 4
h = k(0.38) (2x106 ) = 0.445 ( )
x
x
L
q = hL A∆T = ∫ hx ∆T dx
0
1
1 L
∆T 4
∆T β
hL = ∫ hx dx = 0.286 ( ) = α ( )
L 0
L
L
α = 0.286 β = 1⁄4
19.23
Repeat Problem 19.22 for velocity and temperature profiles of the from,
𝑣 = 𝑎 + 𝑏𝑦 + 𝑐𝑦 2
T − Ts = α + βy + γy 2
Solution
For v = a + by + cy 2
B.C.
v(0) = 0
v(δ) = v∞
dv
(δ) = 0
dy
v
y
y
= 2 − ( )2
v∞
δ
δ
For T − Ts = α + βy + γy 2
B.C.
(T − Ts )|0 = 0
(T − Ts )|δt = 0
∂
(T − Ts )| = 0
∂y
δ
t
T − Ts
y
y
= 2 − ( )2
T∞ − Ts
δt
δt
Into momentum equation to get:
νx
(1)
δ2 = 30
v∞
Into energy equation to get:
1
α
δξd (δξ2 − δξ3 ) = 12
dx
5
v∞
δ
Where ξ = δt
Solution gives ξ ≅ Pr
∴ δt = Pr
−1⁄
3δ
−1⁄
3
(2)
q
dT
Since A = −k dy (0) = h(Ts − T∞ )
−1
1
h 2 2Pr ⁄3
v∞ ⁄2
−1
= =
= 0.365Pr ⁄3 ( )
k δt
δ
νx
hx
Or: Nux = k = 0.365Pr
.
−1⁄
1
3 (Rex ) ⁄2
19.26
For the case of a turbulent boundary layer on a flat plate, the velocity profile has been shown to
follow closely the form
𝑣
𝑦 1/7
=( )
𝑣∞
𝛿
Assuming a temperature profile of the same form – that is,
1
T − Ts
y 7
=( )
T∞ − Ts
δt
And assuming that 𝛿 = 𝛿1 , use the integral relation for the boundary layer to solve for hx and
Nux. The temperature gradient at the surface may be considered similar to the velodity gradient
at y=0 given by equation (13-26).
Solution
1
v
y 7
=( )
v∞
δ
1
T − Ts
y 7
=( )
T∞ − Ts
δt
Energy equation:
∂T
d δt vx
T − Ts
(1 −
) dy
α (0) = (T∞ − Ts )v∞ ∫
∂y
dx 0 v∞
T∞ − Ts
1
0.0225(T∞ − Ts )v∞ ν 4
(
)
LHS =
ν⁄
δv∞
α
RHS = (T∞ − Ts )v∞
7 dδ
72 dx
{Assumes δ = δt for integration}
Equating & some algebra:
δ
−1⁄
4
= 0.371Rex 5 Pr − ⁄5
x
1
q
∂T
k(0.0225)(∆T)v∞ ν 4
(
) = h∆T
= −k (0) = −
A
∂y
ν
δv∞
Nux =
19⁄
hx
1
= 0.0288Rex 20 Pr ⁄5
k
19.32
Work Problem 19.29 for the case in which the flowing fluid is sodium entering the tube at 200oF.
Solution
q
For contant ⁄A
q
⁄
500
T = T0 + A = 200 +
≅ 200℉
(58.1)(12.25 ∗ 3600)(0.332)
ρvCp
Chapter 20
Instructor Only Problems
20.38
A tube bank employs tubes that are 1.30 cm in outside diameter at ST = SL = 1.625 cm. There are
eight rows of tubes, which are held at a surface temperature of 90°C. Air, at atmospheric pressure
and a bulk temperature 27°C, flows normal to the tubes with a free-stream velocity of 1.25 m/s. The
tube bank is eight rows deep, and the tubes are 1.8 m long. Estimate the heat-transfer coefficient.
Solution
Using figure 20.12
Same caveats as for problem 20.51
Dequi v =
Re =
4
π
[(0.032)(0.032) − (0.013)2 ] = 0.0873 m
π(0.013)
4
(0.0873)(1.25)
= 6.95x103
−5
1.569x10
Out of laminar range, must use Fig. 20.13
Re =
(0.013)(1.25)
= 1.04x103
1.569x10−5
For in-line configuration
j ≅ 0.017
h = 0.017(1006.3)(1.177)(1.25)(0.708)
−
2 2.143 −0.14
3(
)
= 30.95
A = 64π(0.013)(1.8) = 4.70 m2
q = hA∆T = 30.95(4.70)(63) = 9.164 kW
1.813
W
m2 K
20.39
Rework Problem 20.38 for a staggered arrangement. All other conditions remain the same.
Solution
Same conditions as Prob. 20.53 except tubes are in staggered configuration
All calculations the same as in prob. 20.54 except j = 0.035
Giving h = 63.7 W/m2 K
& q = (63.7)(4.70)(63) = 18.87 kW
20.40
Air at 60°F and atmospheric pressure flows inside a 1-in., 16-BWG copper tube whose surface is
maintained at 240°F by condensing steam. Find the temperature of the air after passing through 20
ft of tubing if its entering velocity is 40 fps.
Solution
Assume TL = 235℉ Tavg = 148℉
0.87
Dv ( 12 ) (40)
Re =
=
= 1.39x104
ν
0.209x10−3
{turbulent}
−0.2
St = 0.023Re−0.8
= 4.33x10−3
D Pr
L
TL − Ts
= e−4(D)St
T0 − Ts
TL = 240 − 180(0.0083) = 239 ℉
Close enough
20.41
A valve on a hot-water line is opened just enough to allow a flow of 0.06 fps. The water is maintained
at 180°F, and the inside wall of the 1/2-in. schedule-40 water line is at 80°F. What is the total heat
loss through 5 ft of water line under these conditions? What is the exit water temperature?
Solution
G = ρν = 3.64
GD
Re = μ =
Lbm
s ft 2
0.622
)
12
0.29x10−3
3.64(
= 650 {Laminar}
Use Sieder-Tate eqn. assume Tb avg = 150℉
1
0.14
D 3 μ
Nu = 1.86 (RePr ) ( b )
L
μs
1
k
0.383(1.86)
0.622 3 0.29 0.14
Btu
)] (
)
h = ( ) Nu =
[(650)(2.71) (
= 32.9
0.622
D
12(5)
0.578
hrft 2 ℉
12
St =
Nu
= 0.00252
RePr
L
TL − Ts
= e−4(D)St = e−0.972 = 0.378
T0 − Ts
T = 80 + 0.378(100) = 117.8 ℉
Tb avg =
117.8 + 180
= 149 ℉
2
{using steam tables}
q = ṁCp ∆T = 3.64(0.0021)(149.7 − 85.7) = 1710 Btu/hr
20.42
When the valve on the water line in Problem 20.41 is opened wide, the water velocity is 35 fps. What
is the heat loss per 5 ft of water line in this case if the water and pipe tempera- tures are the same as
specified in Problem 20.41?
Solution
G = 60.6(35) = 2120
Lbm
s ft 2
Re = 307000 {turbulent}
Tf ≅ 130℉ use Colburn eq:
2
St = 0.023(307000)−0.2 (3.44)−3 = 0.000807
St(4)(5)
T = 80 + 100e
− 0.622
12
= 153.3℉ {First guess}
80 + 154
+ 180]
2
Tf =
≅ 149℉
2
[
At this temp: Re=443000 Pr=4.51 St=0.000625
T ≅ 155℉
20.43
Steam at 400 psi, 800°F flows through a 1-in. schedule- 40 steel pipe at a rate of 10,000 lbm/h.
Estimate the value of h that applies at the inside pipe surface.
Solution
GA = 10000
G=
Lbm
hr
10000
Lbm
= 37400
0.276
hrft 2
Re =
7.001
37400 ( 12 )
1.63x10−5 (3600)
= 3.72x105 {Turbulent}
Use Dittus-Boelter Eqn.:
k
0.0321
Btu
(0.023)(3.72x105 )0.8 (0.912)0.3 = 35.2
h = ( ) (0.023)Re0.8 Pr 0.3 =
7.001
D
hrft 2 ℉
12
20.49
Air at 25 psia is to be heated from 60°F to 100°F in a smooth, 3/4-in.-ID tube whose surface is held
at a constant temperature of 120°F. What is the length of the tube required for an air velocity of 25
fps? At 15 fps?
Solution
L
T − Ts
= e−4(D)St
T0 − Ts
St =
D
60 − 120
0.017171
)=
ln (
4L
100 − 120
L
a) v = 25 ft/s
Re =
0.75
( 12 ) (25)
0.181x10−3
= 8640
C
2
Use Colburn analogy: St = 2f Pr −3
St =
L=
0.0078
(1.257) = 0.0049
2
0.01717
= 3.5 ft
0.0049
ft
b) v = 15 s
Re = 5190
St = 0.00566 L = 3.03 ft
20.50
Air is transported through a rectangular duct measuring 2 ft by 4 ft. The air enters at 120°F and flows
with a mass velocity of 6 lbm/s ? ft2. If the duct walls are at a temperature of 80°F, how much heat
is lost by the air per foot of duct length? What is the corresponding temperature decrease of the air
per foot?
Solution
Dequiv =
Re =
4(2)(4) 16
=
ft
(2)(6)
6
16
( 6 ) (6)
1.28x10−5
= 1.25x106
2
2
St = 0.023Re−0.2 Pr −3 = 0.023(1.25x106 )−0.2 (0.703)−3 = 0.00176
For the short distance involved:
h = St(ρvCp ) = (0.00176)(6)(924) = 2.53x10−3
q = hA∆T = (2.53x10−3 )(12)(40) = 1.215
∆T =
q
1.215
=
= 0.105 ℉ per ft
ṁCp (0.24)(6)(8)
Btu
s ft 2 ℉
Btu
Btu
per ft = 4370
per ft
s
hr
20.51
Cooling water flows through thin-walled tubes in a condenser with a velocity of 1.5 m/s. The tubes
are 25.4 mm in diameter. The tube-wall temperature is maintained constant at 370 K by condensing
steam on the outer surface. The tubes are 5 m long and the water enters at 290 K.
Estimate the exiting water temperature and the heat- transfer rate per tube.
Solution
Assume
Tin = 290 K
Tout = 350 K
Tsurf = 370 K Tb avg = 320 K
ν = 0.596x10
−6
m2
Pr = 3.87
s
L
T − Ts
= e−4(D)St
T0 − Ts
Re =
(0.0254)(1.5)
= 63900
0.596x10−6
Use Dittus-Boelter equation:
St = 0.023 = 0.023(63900)−0.2 (3.87)−0.6 = 0.00112
L
e−4(D)St = 0.414Re−0.2 Pr −0.6
Tout = 370 − (0.414)(80) ≅ 337
Second try
Tout = 337 Tb avg = 313.5
ν=0.663x10−6 Pr = 4.33
Re = 57460 St = 0.00107
L
e−4(D)St = 0.432
Tout = 370 − (0.432)(80) = 335 K
π
q = ṁCp ∆T = (992) ( ) (0.0254)2 (1.5)(4175)(45) = 141.6 kW
4
20.52
Air, at 322 K, enters a rectangular duct with a mass velocity of 29.4 kg/s m2. The duct measures 0.61
m by 1.22 m and its walls are at 300 K. Determine the rate of heat loss by the air per meter of duct
length and the corresponding decrease in air temperature per meter.
Solution
Rectangular duct: 0.61m x 1.22m
Dequi v =
4(0.61)(1.22)
= 0.813 m
2(0.61 + 1.22)
q = hA∆T
Use Dittus-Boelter Equation
Re =
DG (0.813)(29.4)
=
= 1.227x106
−5
μ
1.948x10
Pr = 0.703
h=
k
0.0279
W
(0.023)Re0.8 Pr 0.3 =
(0.023)(1.227x106 )0.8 (0.703)0.3 = 52.8 2
D
0.813
m K
q = hA∆T = 52.8(2)(0.61 + 1.22)(22) = 4250
q = ṁCp ∆T = 6ACp ∆T
∆T =
4250
= 0.193 K per m
29.4(0.61)(1.22)(1007)
W
m
Chapter 21
Instructor Only Problems
21.16
If eight tubes of the size designated in Problem 21.13 are arranged in a vertical bank and the flow is
assumed laminar, determine a. the average heat-transfer coefficient for the bank b. the heat-transfer
coefficient for the first, third, and eighth tubes
Solution
1
1 4
h
havg = h̅ ( ) =
8
1.681
W
hhoriz = 2250 m2K {from prob. 21.16}
2250
W
For bank: havg = 1.681 = 1341 m2K
For n tubes
q = havg,n nAtube ∆T = havg,(n−1) (n − 1)Atube ∆T
̅̅̅n − (n − 1)h
̅̅̅̅̅̅
nth tube: hn = nh
n−1
W
top tube: hi = 2250 m2 K
W
̅̅̅3 = 1710 W
3rd tube: ̅̅̅
h2 = 1890 m2K h
m2 K
h3 = 3(1710) − 2(1890) = 1350
W
m2 K
W
̅̅̅7 = 1383 W
8th tube: ̅̅̅
h8 = 1341 m2K h
m2 K
h8 = 8(1341) − 7(1383) = 1047
W
m2 K
21.17
Given the conditions of Problem 21.16, what height of vertical wall will cause the film at the bottom
of the tube to be turbulent?
Solution
4AΓc
= Rec = 2000
Pμf
4AΓc
A ṁ 1
4
hPL∆T
= 4( )( )( ) = ( )(
)
Pμf
P A μ
Pμ
hfg
1
μhfg
(0.0206x10−3 )(970)(2000)L4 (3600)
L=
=
1
4h∆T
4(100) [2250(1.3)(0.02)4 ]
3
L4 = 3.27 L = 4.85 ft
21.18
A vertical flat surface 2 ft high is maintained at 60°F. If saturated ammonia at 85°F is adjacent to the
surface, what heat-transfer coefficient will apply to the condensation process? What total heat
transfer will occur?
Solution
1
1
4
3
ρL (∆ρ)g (hfg + 8 Cp ∆T)
37.2(32.2)(37.2)(0.294)3 (505) 4
L
h = 0.943 [
] = 0.943 [
]
Lμ∆T
2(14x10−5 )(25)
= 694
Btu
hr ft 2 ℉
q = hA∆T = 694(2)(25) = 34700
Btu
per foot of witdh
hr
Chapter 22
Instructor Only
22.10
Consider the exchanger in Problem 22.9. After 4 years of operation, the outlet of the oil reaches 28°C
instead of 30°C with all other conditions remaining the same. Determine the fouling resistance on
the oil side of the exchanger.
Solution
Same exchanger & entrance conditions as problem 22.8 {Problem statement should say T0 out =
28℃}
q = ṁ0 Cp ∆T0 = (12)(2200)(8) = ṁw Cp ∆Tw = 3.26(4180)∆Tw
0
w
∆Tw ≅ 16℃ T0 out ≅ 59℃
∆TLM =
47 − 39
= 42.9℃
47
ln (39)
To find F:
Y=
8
= 0.145
55
Z=
16
=2
8
Fig 22.9 a F≅1
U=
(12)(2200)(8)
q
W
=
= 802 2
UF∆TLM
6.14(42.9)
m K
UA|0 =
Fouling Resistance
1
∑ R0
UA|1 =
1
∑ R1
∑ R0 =
1
= 1.508x10−4
(1080)(6.14)
∑ R1 =
1
= 2.031x10−4
(802)(6.14)
= ∑ R1 − ∑ R 0 = 0.5230x10−4
K
W
22.12
A shell-and-tube exchanger having one shell pass and eight tube passes is to heat kerosene from
80°F to 130°F. The kerosene enters at a rate of 2500 lbm/h. Water entering at 200°F and at a rate of
900 lbm/h is to flow on the shell side. The overall heat-transfer coefficient is 260 Btu/h ft2 °F.
Determine the required heat-transfer area.
Solution
ṁw = 900
U = 260
∆Tw =
Lbm
Lbm
ṁk = 2500
hr
hr
Btu
hr ft 2 ℉
2500(130 − 80)(0.51)
= 70.8
900(1)
Tw,out = 129℉ ∆TLM =
AF =
70 − 49
= 58.9℉
70
ln (49)
q
2500(50)(0.51)
=
= 4.16
U∆TLM
58.9(260)
Figure 22.9a F≅
Y=
130 − 80
= 0.416
200 − 80
Z=
200 − 129
= 1.4
130 − 80
A=
4.16
= 5.01 ft 2
0.83
22.13
A condenser unit is of a shell-and-tube configuration with steam condensing at 85°C in the shell. The
coefficient on the condensate side is 10,600 W/m2 ? K. Water at 20°C enters the tubes, which make
two passes through the single-shell unit. The water leaves the unit at a temperature of 38°C. An
overall heat- transfer coefficient of 4600 W/m2 ? K may be assumed to apply. The heat-transfer rate
is 0.2 106 kW. What must be the required length of tubes for this case?
Solution
2x108
q = ṁw Cp ∆Tw = q = Cw ∆Tw Cw = ρAvCp =
= 1.11x107
w
18
UA
4600A
=
= 4.14x10−4 A
Cmin 1.11x107
A = nπDL = nπ (
1.37
) L = 0.359nL
12
UA
= 1.485x10−4 nL
Cmin
Neglect tube resistance:
U=
1
1
1
1
=
−
hi 4600 10600
1
1
+( )
hi
h0
hi = 8130
Nui =
1.37
8130 ( 12 )
0.615
= 1509
Using Dittus Boelter equation:
1.37 0.8
ρv ( 12 )
Nu = 1509 = 0.023 [
] (5.65)0.4
−6
825x10
q
2x108
ρv =
=
= 3192
nπD2
ACp ∆T
(
) (4179)(18)
4
n = 81 tubes
q = εCmin (65)
ε = 0.277
Fig. 22.2c
UA
Cmin
≅ 0.38
L=
0.38
= 31.7 m
(1.48x10−4 )(81)
22.25
Determine the required heat-transfer surface area fro a heat exchanger constructed from 10-cm OD
tubes. A 95% ethanol solution (cp = 3.810 kJ/kg K), flowing at 6.93 kg/s is cooled from 340 to 312 K
by 6.30 kg/s of water that is available at 283 K. The overall heat-transfer coefficient based on
outside tube area is 568 W/m2 K. Three different exchanger configurations are of interest:
a. Counterflow, single pass
b. Parallel flow, single pass
c. Crossflow with one tube pass and one shell pass, shell-side fluid mixed
Solution
ṁEth = 6.93
kg
s
q = ṁCp ∆T|Eth = (6.93)(3810)(28) = ṁCp ∆T|w = 6.30(4182)∆T
∆Tw = 28.1
(a) Counter flow:
∆TLM ≅ 29℉
A=
q
6.93(3810)(28)
=
= 44.9 m2
U∆TLM
568(29)
(b) Parallel flow:
∆TLM =
A=
57 − 0.9
= 13.52
57
ln (0.9)
q
= 96.3 m2
568(13.52)
(c) Cross flow:
Cmixed = ṁCp = 26350
w
Cunmixed = ṁCp = 26403
E
Y=
312 − 340
= 0.491
283 − 340
Z=
28.1
≅1
28
F ≅ 0.85
A=
44.9
= 52.8 m2
0.85
22.26
Water flowing at a rate of 10 kg/s through 50 tubes in a double-pass shell-and-tube heat exchanger
heats air that flows on the shell side. The tubes are made of brass with outside diameters of 2.6 cm
and are 6.7 m long. Surface coefficients on the inside and outside tube surfaces are 470 and 210
W/m2 ? K, respec- tively. Air enters the unit at 15°C with a flow rate of 16 kg/s. The entering water
temperature is 350K. Determine the following:
(a) Heat-exchanger effectiveness
(b) Heat-exchanger rate to the air
(c) Exiting temperatures of the water and air systems
If, after a long period of operation, a scale has been built up inside the tubes resulting in an added
fouling resistance of 0.0021 m2 K/W, determine the new results for parts (a), (b), and (c), above.
Solution
hw = 470
W
m2 K
ṁw = 10
hA = 210
W
m2 K
ṁA = 16
q = (10
kg
s
kg
s
kg
J
kg
J
) (4181
) (350 − Tw out ) = (16 ) (1007
) (TA out − 288)
s
kgK
s
kgK
Cw = 4180
CA = 16112 = Cmin
Cmin
= 0.385
Cmax
A = πDL(50) = π(0.026)(6.7)(50) = 27.36 m2
{Assumes total length of each tube is 6.7 m}
U=
1
1
1
( ) + R cond + ( )
hi
h0
=
1
1
1
(470) + (210)
UA
145(27.36)
=
= 0.246
Cmin
16112
ε ≅ 0.20
= 145
q = εCmin (Tw i − TA i ) = 0.2(16112)(62) = 199800 W
Tw out = 347.6
TA out = 294.2
22.28
For the heat exchanger described in Problem 22.27, it is observed, after a long period of operation,
that the cold stream leaves at 184°F instead of at the design value of 220°F. This is for the same
flow rates and entering temperatures of both streams. Evaluate the fouling factor that exists at the
new conditions.
Solution
If counter flow:
AUold =
1
1
1
)]
[(
) + RT + (
Ai hi
A0 h0
= 36.7(300)
For new operating conditions:
q = 5000(1)(184 − 75) = 2400(1)∆TH = 545000
∆TH = 227
∆TLM =
216 − 98
= 149.3
216
ln ( 98 )
AUnew =
AUnew =
[
TH out = 173
q
545000
1
=
= 3650 =
1
1
∆TLM
149.3
)] + R F
[(
) + RT + (
Ai hi
A0 h0
1
1
[36.7 (300)] + R F
1
1
(300)] + R F =
36.7
3650
R F = 1.83x10−4
K
W
Or
−3
AR F = 36.7R = 6.27x10
m2 K
W
Chapter 23
Instructor Only Problems
23.21
A circular duct 2 ft long with a diameter of 3 in. has a thermocouple in its center with a surface area
of 0.3in.2. The duct walls are at 200°F, and the thermocouple indicates 310°F. Assuming the
convective heat-transfer coefficient between the thermocouple and gas in the duct to be 30 Btu/h ft2
°F, estimate the actual temperature of the gas. The emissivity of the duct walls may be taken as 0.8
and that of the thermocouple as 0.6.
Solution
Assuming thermocouple at geometric center of duct
Solid angle of duct opening
1.5 2
( 12 )
πR2
duct area
Ω≅ 2=
=
= 0.0156
12
πR 0 hemisphere surface
{Thermocouple sees duct primarily}
For thermocouple: q rad = q conv
Aℱcw (Ebc − Ebw ) = hA(TG − Tc )
Ac ℱcw =
ℱcw =
1
ρ
ρ
1
(A cϵ ) + (A F ) + (A wϵ )
c c
c cw
w w
1
ρ
Ac ρ
1
(ϵc ) + (F ) + (A ϵw )
c
ℱcw ≅ 1
cw
w w
Ac
≅0
Aw
∴ ℱcw =
1
≅ ϵc = 0.6
ϵ
1 − (ϵc ) + 1
c
30(TG − Tc ) = ϵc (0.1714) [(
Tc = 316℉
Tc 4
Tw 4
) −(
) ]
100
100
23.26
A room measuring 12 ft by 20 ft by 8 ft high has its floor and ceiling temperatures maintained at
85°F and 65°F, respectively. Assuming the walls to be reradiating and all surfaces to have an
emissivity of 0.8, determine the net-energy exchange between the floor and ceiling.
Solutions
{Walls assumed to be at a uniform temperature}
R1 =
0.2
= 0.00104
12(20)(0.8)
R 2 = 0.00104
R3 =
1
1
=
= 0.00903
AF FF−c 12(20)(0.45)
R4 =
1
1
=
= 0.0076
AF FF−w 12(20)(0.55)
R 5 = R 4 = 0.0076
1
R equiv =
(
1
1
)+(
)
R3
R4 + R5
= 0.0058
∑ R = R1 + R 2 + R equiv = 0.00785
q=
σ(TF4 − Tc4 ) 0.1714(5.454 − 5.254 )
Btu
=
= 2680
∑R
0.00785
hr
23.27
A dewar flask, used to contain liquid nitrogen, is made of two concentric spheres separated by an
evacuated space. The inner sphere has an outside diameter of 1 m and the outer sphere has an
inside diameter of 1.3 m. These surfaces are both diffuse-gray with ε = 0.2. Nitrogen, at 1
atmosphere, has a saturation temperature of 78 K and a latent heat of vaporization of 200 kJ/kg.
Under conditions when the inner sphere is full of liquid nitrogen and the outer sphere is at a
temperature of 300 K, estimate the boil-off rate of nitrogen.
Solution
{Equivalent circuit}
R1 =
ρ1
ρ
1
1
R2 =
=
R3 = 2
A1 ϵ1
A1 F12 A2 F21
A 2 ϵ2
T1 = 300 K T2 = 78 K
A1 = πD12 = π(1.3)2 = 1.69π m2
A2 = πD22 = π(1)2 = π m2
R1 =
0.8
2.37 −1
=
m
(1.69π)(0.2)
π
R2 =
1
1
= m−1
π(1) π
(23.7 continued)
R3 =
0.8
4
= m−1
π(0.2) π
∑R =
7.37
= 2.35 m−1
π
Eb1 − Eb2 σ(T14 − T24 ) 5.676(34 − 0.784 )
=
=
= 194.8 W
∑R
∑R
2.35
q=
q
Boil off rate: ṁ = h
fg
ṁ =
1948
kg
kg
= 9.74x10−4 = 3.51
5
2x10
s
hr
23.31
Two parallel rectangles have emissivities of 0.6 and 0.9, respectively. These rectangles are 1.2 m
wide and 2.4 m high and are 0.6 m apart. The plate having ε = 0.6 is maintained at 1000 K and the
other is at 420 K. The surroundings may be considered to absorb all energy that escapes the twoplate system. Determine
a. The total energy lost from the hot plate
b. The radiant-energy interchange between the two plates
Solution
ϵ1 = 0.6 T1 = 1000 K A1 = 2.88 m2
ϵ2 = 0.9 T2 = 400 K
A2 = 2.88 m2
R1 =
ρ1
= 0.231
A1 ϵ1
R3 =
ρ2
1
= 0.039 R 4 = R 5 =
= 0.694
A 2 ϵ2
A1 F13
R2 =
1
= 0.694
A1 F12
Writing equations for loops as shown:
Eb1 − 0 = (I1 + I2 )R1 + I1 R 4
Eb2 − 0 = (I2 + I3 )R 3 + I2 R 5
Eb1 − Eb2 = (I1 + I3 )R1 + I3 R 2 + (I3 −I2 )R 3
Substituting values & solving simultaneous equations
I1 = 59550 I2 = 4695 I3 = 42970
q1 net = I1 + I3 = 102.5 kW
q12 = I3 = 42.97 kW
{These results presume no HT TX from other sides of plates}
23.34
Evaluate the net heat transfer between the disks described in Problem 23.33 if they are bases of a
cylinder with the side wall considered a nonconducting, reradiating surface. How much energy will
be lost through the hole?
Solution
R1 = 470 R 2 = 11.94 R 3 = 30.6
R 4 = 735 R 5 = 19.6
Eb1 = 1460 Eb2 = 345 Eb3 = 0
4 is adiabatic
For black surfaces
q12 =
Eb1 − Eb2
R equiv,12
R equiv,12 =
q12 =
1
1
1
(R ) + R + R
1
2
4
1460 − 345
Btu
= 3.87
288
hr
q lost through hole = q13 =
R equiv =
Eb1 − 0
R equiv
1
1
1
(R ) + (R + R )
5
q lost =
1
2
1460
Btu
= 77.5
18.83
hr
For grey surfaces:
ϵ1 = 0.6 ϵ2 = 0.3
Additional resistance RA, RB
= 288
= 18.83
RA =
RB =
q12 =
ρ1
0.4
=
= 7.64
A1 ϵ1 π 4 2
4 (12) 0.6
0.3
π 4
2.5 2
[(
)
−
(
4 12
12 ) ] 0.7
2
= 8.06
1460 − 345
Btu
= 3.67
288 + 7.64 + 8.06
hr
q lost =
1460
Btu
= 55.2
18.83 + 7.64
hr
23.35
Evaluate the heat transfer leaving disk 1 for the geometry shown in Problem 23.33. In this case the
two disks comprise the bases of a cylinder with side wall at constant temperature of 350°F. Evaluate
for the case where
a. The side wall is black
b. The side wall is gray with ε=0.2
Determine the rate of heat loss through the hole in each case.
Solution
R1 = 470 R 2 = 11.94 R 3 = 30.6
R 4 = 735 R 5 = 19.6
Eb1 = 1460 Eb2 = 345
Eb3 = 0 Eb4 = 738
Writing loop equations:
Eb1 − Eb4 = R 2 (I1 + I4 + I5 )
Eb4 − 0 = R 3 (I3 − I5 )
Eb1 − Eb2 = R1 (I2 − I4 )
0 = R1 (I4 − I2 ) + R 2 (I4 + I1 − I5 ) + I4 R 4
0 = R 3 (I5 −I3 ) + R 2 (I5 − I1 − I4 ) + I5 R 5
Solving:
I1 = 134.8
I2 = 2.84
I4 = 0.47
I3 = 98.7
I5 = 74.6
q12 = I2 = 2.84
Btu
hr
q lost = I3 + I5 = 173.3
Btu
hr
23.39
A duct with square cross section measuring 20 cm by 20 cm has water vapor at 1 atmosphere and
600 K flowing through it. One wall of the duct is held at 420 K and has an emissivity of 0.8. The other
three walls may be considered refractory surfaces. Determine the rate of radiant-energy transfer to
the cold wall from the water vapor.
Solution
q gas−wall direct = A1 F1G αG σ(TG4 − T14 )
q gas to reradiating walls
= A2 F2G αG σ(TG4 − T24 )
q gas reradiating walls to (1)
= A1 F1G τG σ(TG4 − T14 )
q G2 = q 21 = q R =
σ(TG4 − T14 )
1
1
(A F τ ) + (A F α )
1 1G G
2 2G G
q total to 1 = q G1 + q R
L = 3.4(0.2)(0.2)(1)4(0.2)(1) = 0.17 m
p = 1 atm
αG = 0.22
PL = 0.558 atm ft
τG = 0.78
R1 =
1
= 22.7
0.2(1)(0.22)
R2 =
0.2
= 1.25
0.2(1)(0.8)
R3 =
1
= 7.58
3(0.2)(1)(1)(0.22)
R4 =
1
= 6.41
0.2(1)(1)(0.78)
R equiv =
1
1
1
(R ) + (R + R )
1
3
4
= 8.66
∑ R = 8.66 + 1.25 = 9.91
q=
5.676(64 − 4. 24 )
W
= 564
9.91
m
23.40
A gas of mixture at 1000 K and a pressure of 5 atm is introduced into an evacuated spherical cavity
with a diameter of 3 m. The cavity walls are black and initially at a temperature of 600 K. What initial
rate of heat transfer will occur between the gas and spherical walls if the gas contains 15% CO2 with
the remainder of the gas being nonradiating?
Solution
q net = σA(ϵG TG4 − αG Tw4 )
A = 4πr 2 = π(3 m)2 = 28.27
TG = 1000 K Tw = 600 K
L=
2
D=2m
3
pL = 0.15(5)(6.56) = 4.92 atm ft
αG = 0.18 ϵG = 0.22
q net = 5.676(9π)[0.22(104 ) − 0.18(64 )] = 316 kW
24.1
A = H2S, B = N2, C = SO2
"
34.08
0.03,
g
g‰Š‹Œ
,"
0.92, Q
28.01 g‰Š‹Œ, "Q
0.05
T = 350 K, P = 1.0 atm.
64.07
g
a. Determine c, cA, and ρA
Z
1.0 atm
mM · atm
ab
u8.206 > 10AB
v 350 K
gmole · K
gmole
v
mM
0.03 u34.8
"
u1.04
y
€
•
373.2K, N+
0.841 NQ
¥jn
cmM */M
0.841 98.5
gmole
*/M
0.77 bQ
0.77 373.2K
y
3.884 Å
•b
€
2
y
• • */K
u · v b
€ €
2
287.4K
3.681 Å, ~
3.681 Å
}
3.78 Å
1
"
Zy K Ω{
0.001858 b M⁄K w
1 *⁄K
x
"
.
gmole
mM
g
m3
3.884 Å
91.5 K .
1
1 *⁄K
u
·
v · 350 K
287.4K 91.5K
From Table K.1, interpolate to get ΩD = 1.047.
g‰Š‹Œ
34.8
35.4
98.5 ijklm.
From Appendix K, Table K.2, y
y
gmole
mM
gmole
g
v
u34.08
v
m3
gmole
b. Estimate DAB for H2S in N2
b+
1.04
g
2.16
0.001858 · 350 K M/K •
34.08
1
g
gmole
28.01
1.0 atm 93.78 Å; 1.047
K
*/K
1
g ‚
gmole
0.207 cmK ⁄s
c. Estimate DA-m for H2S in mixture gas
Find DAC for H2S in SO2
From Appendix K, Table K.2, yQ
y
yQ
•b
€Q
u
2
yQ
3.884 Å
• • */K
· v b
€ €Q
2
4.290 Å
1
~
4.09 Å
1 *⁄K
x
"Q
1
"
Zy Q K Ω{
0.001858 b M⁄K w
1
0.001858 · 350 K M/K u
34.08 g/gmole
0.03,
A
0.03
A
¢
1
252 K
}
1
1 *⁄K
u
·
v · 350 K
287.4K 252K
From Table K.1, ΩD = 1.273.
Q
4.290 Å, ¡
0.92,
¢Q
Q
,
1.30
*/K
1
v
64.07 g/gmole
0.12 cmK ⁄s
1.0 atm 94.09 Å; 1.273
0.92
0.21 cmK ⁄s
Q
0.05
¢
K
1
0.05
,
0.12 cmK ⁄s
,
¢Q
§
A
1
Q
,
ŽK
0.20
s
•¦
1
A‰
Q
Q
This diffusion coefficient is slightly different than that of H2S-N2 because H2S diffuses with N2
and SO2 in the mixture gas. But, since N2 gas dominates, the diffusion coefficient is very similar.
24.2
A = O2, B = N2
P = 1.0 atm, T = 293 K
D¨©
0.181
M¨
32.00
D¨©
dªk«m
0.181
¥jS
Ÿ
at T = 273 K, P = 1.0 atm
cmK 293 K
u
v
s 273 K
g
,M
gmole ©
10 > 10A• cm
*.B
0.201
28.01
cmK
s
g
gmole
a. Effective diffusion coefficient for straight 10 nm pores in parallel array
D¹¨
T
4850 · dªk«m · º
M¨
D¨m
0.085 cmK /s
1
D¨m
1
u
D¨©
1
v
D¹¨
293 K
4850 · 10 > 10A• cm · â
g
32.00
gmole
1
u
0.201 cmK /s
1
v
0.147 cmK /s
0.147 cmK /s
b. Effective diffusion coefficient for random pores of 10 nm diameter with void fraction of
ε 0.40
See part (a)
D¢¨m
εK D¨m
0.40K · 0.085
cmK
s
0.0136
cmK
s
c. Effective diffusion coefficient for random pores of 1.0 µm diameter with void fraction of
ε 0.40
dªk«m
1 > 10A¤ cm
D¹¨
T
4850 · dªk«m · º
M¨
D¨m
0.177 cmK /s
1
D¨m
D¢¨m
1
u
D¨©
εK D¨m
1
v
D¹¨
4850 · 1 > 10A¤ cm · â
1
u
0.201 cmK /s
0.40K · 0.177
cmK
s
293 K
g
32.00
gmole
1
v
1.47 cmK /s
0.0283
cmK
s
1.468
cmK
s
24.3
T = 288 K
Physical property information
Interpolate and convert Pa·s to cP from Appendix I to find viscosity of liquid water at 288 K:
µìKí 288 K
1.193 cP (water)
For viscosity of liquid ethanol and n-butanol, From Perry’s Chemical Engineering Handbook,
interpolate and convert to cP:
µîì¤í 288 K
0.641 cP (ethanol)
µî¤ì*•í 288 K
3.41 cP (n-butanol)
From Table 24.4 and Table 24.5
VìKí 18.9 cmM /gmole (water)
Vîì¤í
14.8 4 · 3.7 7.4 37.0 cmM /gmole (ethanol)
Vî¤ì*•í
4 · 14.8 10 · 3.7 7.4 103.6 cmM /gmole (n-butanol)
ΦìKí = 2.6 (water)
Φîì¤í = 1.9 (ethanol)
Φî¤ì*•í = 1.0 (n-butanol)
MìKí 18 g/gmole (water)
Mîì¤í 32.04 g/gmole (ethanol)
Mî¤ì*•í 74.12 g/gmole (n-butanol)
Hayduk and Laudie Equation (used for nonelectrolyte solutes in H2O water solvent)
D¨©
13.26 > 10AB µA*.*¤
· V Ä•.BÌ¿
©
D¨©
7.4 > 10AÌ Φ© M© */K T
·
µ©
V¨•.•
Wilke Chang Equation
a. Molecular diffusion coefficient for liquid methanol solute (A) in solvent H2O (B)
Appendix J.2 Value: DAB = 1.28 > 10AB cmK /s
D¨©
13.26 > 10AB µA*.*¤
V Ä•.BÌ¿
©
D¨©
13.26 > 10AB 1.193 A*.*¤ · 37.0 A•.BÌ¿
1.29 > 10AB
cmK
b. Molecular diffusion coefficient for liquid solute H2O (A) in solvent ethanol (B)
D¨©
D¨©
7.4 > 10AÌ Φ© M© */K T
·
µ©
V¨•.•
7.4 > 10AÌ 1.9 · 18 */K 288
·
18.9•.•
0.641
4.45 > 10
AB
cmK
s
c. Molecular diffusion coefficient for liquid n-butanol solute (A) in solvent H2O (B)
Appendix J.2 Value: D¨©
D¨©
D¨©
7.7 > 10A• cmK /s
13.26 > 10AB ” µ©A*.*¤ · V Ä•.BÌ¿
13.26 > 10AB 1.193 A*.*¤ · 103.6 A•.BÌ¿
7.05 > 10A•
cmK
s
d. Molecular diffusion coefficient for liquid H2O solute (A) in solvent n-butanol (B)
D¨©
D¨©
7.4 > 10AÌ Φ© M© */K T
·
µ©
V¨•.•
7.4 > 10AÌ 1.0 · 74.12 */K 288
·
18.9•.•
3.41
9.23 > 10A•
cmK
s
25.1
Control-volume expression for the conservation of mass
{net rate of efflux of A } + {rate of accumulation} − {rate of production} = 0
Net rate of A efflux
N A , x ∆y ∆z | x + ∆x − N A , x ∆y ∆z | x
(x-direction)
N A, y ∆x∆z | y +∆y − N A, y ∆x∆z | y
(y-direction)
N A, z ∆x∆y |z +∆z − N A, z ∆x∆y |z
(z-direction)
Rate of accumulation in control volume
∂c A
∆x∆y∆z
∂t
Rate of production in control volume
RA ∆x∆y ∆z
N A, x ∆y∆z |x +∆x − N A, x ∆y∆z |x + N A, y ∆x∆z | y +∆y − N A, y ∆x∆z | y + N A, z ∆x∆y |z +∆z − N A, z ∆x∆y |z
+
∂cA
∆x∆y∆z − RA ∆x∆y∆z = 0
∂t
Divide by volume, ∆x∆y∆z, to obtain
N A, x ∆y∆z |x +∆x − N A, x ∆y∆z |x
∆x ∆y ∆z
+
+
N A, y ∆x∆z | y +∆y − N A, y ∆x∆z | y
∆x∆y∆z
+
N A, z ∆x∆y |z +∆z − N A, z ∆x∆y |z
∂cA
− RA = 0
∂t
This equation reduces to
N A, x |x +∆x − N A, x |x
∆x
+
+
N A, y | y +∆y − N A, y | y
∆y
+
N A, z |z +∆z − N A, z |z
∆z
∂c A
− RA = 0
∂t
Evaluated in the limit as ∆x, ∆y, and ∆z approach zero
∂N A, x
∂x
+
∂N A , y
∂y
+
∂N A, z
∂z
+
∂c A
− RA = 0
∂t
or
∇ ⋅ NA +
∂c A
− RA = 0
∂t
∆x ∆y ∆z
25.2
Spherical coordinates
Assumptions: 1) 1D mass transfer (in r-direction only), 2) Steady-state, 3) No homogeneous
reaction in gas phase, RA = 0, 4) Temperature and pressure are constant, 5) Constant
concentration at r = R
General differential equation (r-direction only) for spherical coordinates
∂c A 1 ∂ 2
+ 2 ( r N A , r ) = RA
∂t
r ∂r
∂c
Since A =0 and RA = 0
∂t
1 ∂ 2
( r N A, r ) = 0
r 2 ∂r
r 2 N A,r is constant
Fick’s flux equation for A
N A , r = − cD AB
(
N A , r = − cD AB
dy A
+ y A ( N A ,r + 0 )
dr
dy A
+ y A N A ,r + N B ,r
dr
Since N B , r = 0
or
N A, r =
−cDAB dy A
1 − y A dr
Boundary conditions
PAo
At r = R, y A =
RT
At r = ∞, y A = y A,∞
)
25.3
System: Air Bubble
P|}||~•
0.2 cm, T = 298 K
M@B € 298 ‚!
0.03 6 1, M|}||~• 298 ‚!
16 1
A = H2O, B = Air
Assume: 1) 1-D (radial) flux, 2) Unsteady State, 3) Constant bubble diameter, 4) Slow bubble velocity
(diffusion dominates), 5) Dilute water in air, 6) No homogeneous reaction,
0.
a. Fick’s Flux Equation
P
,
P5
,d
P7O3 2 4/O3 7/.:
Differential Model, Spherical Geometry
8
g o l
dB
^5 %
d
^ dB
B
d
,
_
%
·
d
0
,
d
_
,
b. Boundary and Initial Conditions
BC:
5
5
,
0,
, !
0, !
5
IC: t = 0, cA(r,0) = 0
M@B €
„
0
k1
26.1
A = MeOH vapor, B = O2 gas
T = 293 K, P = 1.0 atm, G€ [w
5M
a. Determine effective diffusion coefficient, DAe
1 3
1
0.001858 · )/3 45 6 5 7
2"
3
" Ω9
).LXLp). ))
3.585 , " 3.433 ,, "
qb
r
!
507 , r
293
239.36
qs
"
2"
113 ,
1.224,
√113
"
> Ω?
3
· 507
1.308
1
1 B/3
0.001858 · 293 )/3 @
6
A
32.04 32.00
1.0 · 3.5093 · 1.308
Knudsen Diffusion Coefficient
3.509
239.36
0.145
3
293
7.33
32.04
C
Effective Diffusion Coefficient
1
1
1
1
1
6
6
3
3 >2 w
2 w 2 " 2t
0.145
7.33
C
C
2t
4850 · 5 1 10
C
3
v
0.142
C
3
b. Boundary conditions for system I and II:
•
For system I:
„ h
0,
,
h
nB ,
B
For system II:
At z = L1, yA = yA1, at z = L2, yA = yA2
c. Model for WA
Assumptions: 1) Steady state, 2) Dilute UMD process, 3) No reaction, 4) 1-D transfer along z, 5)
Constant T and P, 6) No sink for O2
D
T2 "
G
Gh
d
( N A,z ) = 0 , constant flux along z
dz
System I:
i…
D ] Gh
D
Z
j
T2 "
2"
T
nB
DU
System II:
2 "
D ] Gh
T2 w
D
i†
i…
Z
2w
BT
DU
kb…
] G
kb_
B
kb†
T
nB
B
] G
\G3
4
kb…
n3 T nB
3
2 w
BT
n3 T nB
3
„€
Eliminate yA1 and solve for for WA
T 3
4nB
n Tn
6 „3 2 B
3
\G 2 "
€
w
Z
1.0
)·
82.06
( ./0 ·
d. Estimate WA
„€
/O
Z
,
0.5
0.005
3
‡
„T
,„
ˆ6
0.129
1.0
T
3
4nB
n T nB
6 3
3
„€ 2 w
\G 2 "
· 293
3
, „B
4.16 1 10 L
\G3
4
11.9673, ‡
0.129
\ 0.1
\ 10
)
3
4
3626.55, ˆ
4.16 1 10 L
4 · 0.3
3 · 0.145
WA = 9.55 x 10-9 gmole/sec = 35 µmol/hr
( ./0
3
T34.29
( ./0
)
C
7.854 1 10 u
36
0.129 T 0.020
0.45 T 0.30
0.005
3 · 0.142
C
3
26.2
A = naphthalene, B = air
347
,
% ,* |'Ÿ
*
,'
1.145
,
1. 0
(
,
)
666
8.314 1 10o
1.0
,,
666
0.25
)·
,5
( ./0 ·
· 347
128
(
,2 "
( ./0
2.31 1 10
0.0819
( ./0
u
)
C
General Differential Equation for Mass Transfer (spherical system):
1 G 3
”Y D ,[ – 0
Y 3 GY
D ,[ · Y 3
D ,^ · , 3
Combine
Y 3 RT2 "
Flux Equation:
G
D
T2 "
GY
D · ,3
^
GY
D · ,3 ] 3
Y
e
G
S
GY
abc
T2 " ] G
2"
1 1
, 3 • , T ∞•
D
.OC O
j
2"
,
Material balance on naphthalene (PSS mass transfer)
IN – OUT + GEN = ACC
0 T D ,[ U 6 0
T
2"
,
T2 "
*
*
· 4\, 3
G
G
% ,* |'Ÿ G’
5
G
% ,* |'Ÿ
G,
·,
5
G
% ,* |'Ÿ G 4 )
R \, S
5 G 3
% ,* |'Ÿ
G,
· 4\, 3
5
G
3
T2 "
*
5
% ,* |'Ÿ
¢
]G
j
% ,* |'Ÿ ”,'3 T , 3 –
22 " * 5
t = 221,636 s = 61.6 hr
^£
] ,G,
^`
1.145
(
3
)
1.0
3
T 0.25
2 · 0.0819 C · 2.31 1 10 u
( ./0
)
3
· 128
(
( ./0
26.3
A = TCE, B = biofilm
a. Determine volumetric flow rate vo
Material balance in TCE in biofilm reactor
IN – OUT + GEN = ACC = 0
v0cAi − v0 cA0 − SN A = 0
NA =
v0 =
DAB cA0
δ
φ tanh φ
SN A
SDAB cA0 φ tanh φ
=
c Ai − c A0
δ
( cAi − cA0 )
φ =δ
 1m 
k1
= 100 µm  6

DAB
 10 µm 
4.3 s -1
9.03 x 10
-10
m2
s
= 6.909
2

g TCE 
-10 m  
 9.03 x 10
  0.05

s 
m3 
g TCE

NA =
( 6.909 ) tanh ( 6.909 ) = 3.12 x 10-6 2
-6
100 x 10 m
ms

(800 m2 )  3.12 x 10-6 g mTCE

2
m3
s   3600 s 
v0 =
=
44.9
g TCE
g TCE  hr 
hr
0.25
0.05
3
3
m
m
b. Determine TCE concentration at the postion biofilm is attached to surface (z = δ)

k1 
g TCE
c A0 cosh  (δ − z )

DAB  c A0 cosh ( 0 ) 0.25 m3 (1.0)
g TCE

c A ( z ) |z =δ =
=
=
= 4.99 x 10-4
cosh (φ )
cosh ( 6.909 )
m3

k1 
cosh  δ

DAB 

27.1
A = H atoms, B = solid iron (Fe)
56
‹6 W X
124 N 10 f P /c
CA0 = 0
‹6 W |
Determine ‹6 W
2.2 N 10 €
Pbk Œ
j•
Ž ••
1.76N 10 € ‘ ’“ aS
0.1 P
USS diffusion in semi-infinite medium
W X W ,S
^
erf _
erf \
WX W)
2]5 6 S
Or
‹6 W , S
^
1
e” \
‹6 W |
2]5 6 S
Pbk Œ
1.76 N 10 € j •
2.2 N10 €
erf _
Pbk Œ
j•
0.200
From Appendix L, _
S
S
1
$ % $
%
_
45 6
0.8
1
0.17925
e” _
0.1 P
1
1 •e
$
% q
r$
%
P
0.17925
3600 cec
f
4 $124 N 10
%
sec
174.3 •e
27.2
A = solvent, B = polymer
D = 0.3 cm, R = 0.15 cm
W• 0
wA0 = 0.2 wt% solvent in polymer
DAe = 4.0 x 10-7 cm2/sec
a. Determine time (t) for wA(r,t) = 0.002 wt% for center of bead (r = 0)
USS Diffusion in a sphere, use Concentration -Time charts
P 0 b b › S•b e c•cSa
/
W ∞ W e, S
W∞ W)
e
ž
0 P
1.25 P
0.55
ž
5“
QR Ÿ 0.55
Figure F.3
S
W
0
5 “S
ž
0.55
W
W e, S
W
0.15 P
4.0 N 10 €
P
sec
Y X Y e, S
YX Y)
0
0
0.002
0.200
30,937.5 sec
8.6 •e
0.010
b. Determine time (t) for wA(r,t) = 0.18 wt% for center of bead 0.01 cm from the surface
/
YX
Y e, S
Y) YX
e
ž
0.18
0.200
1.25 P 0.01 P
1.25 P
0
0
0.90
0.99 Ÿ 1.0
Figure F.9, off the chart. The charts will not work, but penetration depth is small, therefore the
process can be approximated as USS diffusion in semi-infinite medium
Y
,S
Y)
YX
YX
erf _
From Appendix L, erf _
_
S
2]5 6 S
\
1
^
2]5 6 _
\
0.90
0.90; _
0.01 P
1.17
2]4.0 N 10 € P /c
1
^
1.17
46 c
28.1
Local mass transfer coefficient at position x
W ,&X YZX[
W ,^_[`_&'Z^
\
K
\
0.332;GH] =K/> @* 8
K
0.0292;GH] =E/J @* 8
Average mass transfer coefficient
kc =
4/5

0.664 DAB Ret1/2 Sc1/3 + 0.0365DAB Sc1/3 Re4/5
L − Ret 
L
 0.664 DAB Re1/ 2 Sc1/3 


kc ,lam
L


=
1/2
1/3
1/3
4/5
kc
 0.664 DAB Ret Sc + 0.0365 DAB Sc  Re 4/5

 L − Ret  

L


kc ,lam
kc
=
0.664 Re1/2
t
1/2
4/5


0.664 Ret + 0.0365  Re 4/5
L − Ret 
ReL = 3.0 x 106, Ret = 2.0 x 105
kc ,lam
kc
=
0.664(2.0 x 105 )1/2
=0.0606 (6.06%)
0.664(2.0 x 105 )1/2 +0.0365 (3.0 x 106 )4/5 -(2.0 x 105 ) 4/5 
28.2
?
200 * , j
1.97 C 10DE
0.065
*
6
>
200 * , k
,
0.01 * ,
900
, cXY[ ;400 =
W.
,"
8
400 ,
300
2.5909 C 10
1.0
.
, ∆7m,
. 34H
DJ
a. Position for transition from laminar flow, Lt
50 · ?^
6
GH 2 C 10J
10.4 *
> ) ?^
2.5909 C 10DJ
6
50 6 · 2.0
3.86 C 10V
GH
>
2.5909 C 10DJ 6
turbulent flow
>
6
,
200
L = 2.0 m >> Lt = 0.104 m, therefore laminar portion can be neglected
b. Average mass transfer coefficient, kc
W
E
?
0.25909
A
@*
K
0.0365GHe J @* 8
*
0.065
>
*
*
6
6
>
>
3.986
0.065 6
E
*
0.0365;3.86 C 10V =J ;3.986=K/8 3.50
200 *
6
c. Time required for local evaporation at x = 1.2 cm
W
GH]
W ,]
50 6 · 1.2
2.5909 C 10DJ
\
>
E
6
K
0.0292GH] J @* 8
2.32 C 10V
* >
0.065 6
E
0.0292;2.32 C 10V =J ;3.986=K/8
120 *
3.1
*
6
n
.
5,000
*
6
*
p
G
W ,] ;*
1.86 C 10Dg
-
p@
900
W.
1.97 C 10DE
* 8·
82.06
· 400
. 34H ·
a*
. 34H
* >·6
@ℓ"
p@
=
W ,] · *
ℓ"
p
6.0 C 10Do
3.1
*
. 34H
q6.0 C 10Do
r
6
* 8
W. 34H 1000 . 34H
·
300 W. 1 W. 34H
. 34H
1.86 C 10DE >
·6
8 · 100 C 10
DV
* 8
. 34H
·
1613 6
27
d. Heat transfer coefficient (h) and surface temperature Ts
Chilton-Colburn analogy
@* >/8
k
*v W q r
1
*v,XY[ ;400 =
k
900
W.
1.0142 C 10D8
· 0.5 C 10
8
D8
n
W. ·
n
W. ·
>
3.986 8
0.035 q
r
6 0.689
0.0506
n
>·6·
Energy Balance
Q
= N A ∆H v , A M A = h(Ts − T∞ )
A
gmole   200 J  300 g 

1.86 x 10-4 2  

N A ∆H v , A M A
m ⋅ s   g  gmole 

Ts = T∞ +
= 400 K +
J
h


 0.0506 2

m ⋅s⋅K 

Ts = 621 K
tu
29.1
𝑔 1003 𝑐𝑚3
𝑔
5
𝜌𝐵 = 0.8 3 ∙
=
8
×
10
𝑐𝑚
𝑚3
𝑚3
𝑀𝐵 = 180
𝑔
,
𝑚𝑜𝑙𝑒
𝑇 = 293 𝐾,
𝑝𝐴 = 0.015 𝑎𝑡𝑚,
𝑦𝐴 =
𝑘𝑔𝑚𝑜𝑙𝑒
𝑃 = 1.5 𝑎𝑡𝑚
0.015 𝑎𝑡𝑚
= 0.01,
1.5 𝑎𝑡𝑚
𝑘𝑔𝑚𝑜𝑙𝑒
𝑘𝑥 = 0.01 𝑚2 ∙𝑠 , 𝑘𝑦 = 0.02 𝑚2 ∙𝑠
𝑝𝐴,𝑖 = 𝐻𝑥𝐴,𝑖 ,
𝐻 = 0.15 𝑎𝑡𝑚
a. Determine 𝐾𝐿
𝑘
𝑀
𝑘𝐿 = 𝐶𝑥 ≅ 𝜌 𝐵 𝑘𝑥 =
𝐿
𝑥𝐴 = 0.05
𝐵
𝑔
𝑔𝑚𝑜𝑙
𝑔
8×105 3
𝑚
180
𝑔𝑚𝑜𝑙
𝑚
∙ 10 𝑚2 ∙𝑠 = 2.25 ∙ 10−3 𝑠
𝑘𝑔𝑚𝑜𝑙
𝑘𝑦 0.02 � 𝑚2 ∙ 𝑠 �
𝑘𝑔𝑚𝑜𝑙
𝑘𝐺 =
=
= 1.33 ∙ 10−2 2
1.5(𝑎𝑡𝑚)
𝑃
𝑚 ∙ 𝑠 ∙ 𝑎𝑡𝑚
−1
1
1
𝐾𝐿 = � +
�
𝑘𝐿 𝐻𝑘𝐺
= 1.06 ∙ 10−3
𝑚
𝑠
1
1
=�
+
𝑚
2.25 ∙ 10−3 𝑠 0.15 𝑎𝑡𝑚 ∙ 1.33 ∙ 10−2
𝑃𝐴
𝑁𝐴 = 𝐾𝐿 (𝑐 ∗𝐴𝐿 − 𝑐𝐴𝐿 ) = 𝐾𝐿 � ∙ 𝑐𝐿 − 𝑥𝐴 𝑐𝐿 �
𝐻
𝑚
𝑔
𝑔
𝑔𝑚𝑜𝑙
𝑔𝑚𝑜𝑙
𝑘𝑔𝑚𝑜𝑙
𝑚2 ∙ 𝑠 ∙ 𝑎𝑡𝑚
−1
�
5
8×105 3
0.015 𝑎𝑡𝑚 8×10 𝑚3
𝑔𝑚𝑜𝑙
𝑚
−
0.05
∙
𝑔
𝑔 � = 0.236
𝑚2 ∙𝑠
180
180
= 1.06 ∙ 10−3 𝑠 ∙ � 0.15 𝑎𝑡𝑚 ∙
b. Determine 𝑥𝐴𝑖 , and 𝑝𝐴𝑖 .
𝑘
Equate the equilibrium line 𝑝𝐴,𝑖 = 𝐻𝑥𝐴,𝑖 to a line with slope − 𝑘𝑥 through the operating
point to find the intersection (𝑥𝐴𝑖 , 𝑦𝐴𝑖 ).
𝐻
Equilibrium line: 𝑦 𝐴𝑖 = 𝑃 ∙ 𝑥 𝐴𝑖
𝑦
𝑘
Operating point line: 𝑦 𝐴 = − 𝑘𝑥 𝑥𝐴 + 𝑏, 0.01 = −0.5 ∗ 0.05 + 𝑏, ∴ 𝑏 = 0.035
𝐻
𝑃
𝑦
∙ 𝑥 𝐴𝑖 = −0.5𝑥𝐴,𝑖 + 0.035,
(𝑥𝐴𝑖 , 𝑦𝐴𝑖 ) = (0.0583 , 0.00585)
and 𝑃𝐴𝑖 = 𝑦𝐴𝑖 ∙ 𝑃 = 0.00585 ∙ 1.5 𝑎𝑡𝑚 = 0.00878 𝑎𝑡𝑚
29.2
gmole
m3
c A=
1.0 ×10
, vo = 0.20
, p A ≈ 0 , P = 1 atm, D = 4 m, depth = 1.0 m
,o
m3
min
−3
k L = 5 ×10−4
m3 ⋅ atm
m
kgmole
H
=
10.0
k
=
0.01
H ⋅ c AL ,i
, G
,
, p A=
,i
s
m 2 ⋅ s ⋅ atm
kgmole
a. Determine % resistance to mass transfer in the liquid film
1
1
1
=
+
K L k L HkG
−1




1
1

 =
4.98 x 10−4 m/s
+
KL =
3
m
m ⋅ atm
kgmole 
 5 ×10−4
(10.0
)(0.01 2
)

s
kgmole
m ⋅ s ⋅ atm 

% resistance (liquid) =
1 k L K L 4.98 x 10−4 m/s
= =
⋅100% =
99.5%
1 K L kL
5 x 10−4 m/s
b. Develop material balance model and determine c AL and c AL,i
Assume: 1) steady state, 2) no reaction, 3) liquid stripping.
IN – OUT – FLUX OUT = 0
c AL ,o vo − c AL vo − N A S =
0
vo (c AL ,o − c AL ) − SK L (c AL − c*AL ) =
0
vo (c AL ,o − c AL ) − π
p
D2
0
K L (c AL − A ) =
4
H
π D2 K L pA
vo c AL ,o +
4H
=
c AL =
π D2 K L
vo +
4
m
)(0 atm)
s
m3 ⋅ atm
4(10.0
)
kgmole
m
s
π
(4 m) 2 (4.98 x 10−4
)(60
)
3
m
s
min
(0.20
)+
min
4
m3
gmole
(0.20
)(0.001
)+
min
m3
π (4 m) 2 (4.98 x 10−4
c AL = 3.48 x 10−4
gmole
m3
Determine c AL ,i
m
5 x 10−4
p A − p A ,i
p − Hc AL ,i
kL
s =
−
=
−
=A
kgmole
kG
c AL − c AL ,i
c AL − c AL ,i
0.01 2
m ⋅ s ⋅ atm
m3 ⋅ atm
−
0
(10.0
) ⋅ c AL ,i
atm ⋅ m3
kgmole
−0.05
=
kgmole 3.48 x 10−7 kgmole − c
AL ,i
m3
c AL ,i 1.73
x 10−9 kgmole/m3 1.73 x 10−6 gmole/m3
=
=
c. Would the flux N A increase by the following:
Increasing volume level of tank with fixed surface area: does not affect flux.
Increasing the agitation intensity of the bulk liquid: will increase the flux since k L will
increase.
Increasing the agitation intensity of the bulk gas: will not increase the flux since the mass
transfer is liquid film controlling.
Increasing the system temperature: will likely increase the flux since the Henry’s
constant will increase with increasing temperature, which will lower the solubility of the
solute in the gas and lower the concentration driving force for mass transfer.
30.1
Diffusion through stagnant air ( N B = 0 )
cDAB dy A
N Ar = −
1 − y A dr
∞
A∞
−dy A
dr
( N Ar ⋅ R ) ∫ 2 = cDAB ∫
1 − yA
R r
y As
y
2
1 − y A∞ 
1 1
( N Ar ⋅ R 2 )  −  = cDAB ln 
 , at the surface
R ∞
 1 − y As 
cD
N Ar R = AB ( y As − y A∞ )
yB ,lm
N Ar
D DAB
=
( C As − C A∞ )
2 yB ,lm
N Ar =
2 DAB
( C As − C A∞ )
yB ,lm D
Recognize that N Ar = kc ( C As − C A∞ ) such that kc =
kc D
2
=
DAB yB ,lm
Sh ≈ 2 , when yB ,lm ≈ 1.0
2 DAB
yB ,lm D
30.2
Let A = O3, B = H2O (liquid)
a. Determine kLa
P
4.0 atm
C A* = A =
=14.29 gmole/m3
3
H
atm ⋅ m
0.07
gmole
DAB = 1.74 x10 −9 m 2 / s at T = 20oC and P = 1.0 atm
ρL = 998 kg/m3, ρA = 2.14 kg/m3, µ L = 993x10-6 kg/m-sec, νL = 0.995x10-6 m2/sec
Material balance model on dissolved O2
Assumptions: 1) unsteady state, 2) dilute system, 3) kL≈KL, 4) No reaction occurs.
In – Out +Generation = Accumulation (moles A/time)
dC A
N A ⋅ Ai − 0 + 0 = V
dt
dC A
k L a (C A* − C A ) =
dt
CA
t
dC
∫0 CA* − ACA = kL a ∫0 dt
 C* 
ln  * A  = kL a ⋅ t
 CA − CA 
 C* 
 14.29 gmole/m 3 
ln  * A  ln 

CA − CA 
14.29-4.00 gmole/m 3 


kL a =
=
= 0.033 min -1 = 5.473 x 10-4 sec -1
t
10.0 min
b. Bubble diameter, db
3
3
2
3
db3 ρ L g ∆ρ d b (998 kg/m )(9.81 m/sec )( ( 998-1.2 ) kg/m )
Gr =
=
=9.8x1012 d 3b
2
-6
2
µL
(993 x 10 kg/m ⋅ sec)
Sc =
ν
0.995 x 10-6 m 2 /sec
=
= 572
DAB
1.74 x 10-9 m 2 /sec
Assume db < 2.5 mm
k d
Sh = L b = 0.31Gr1/3 Sc1/3
DAB
D 
kL = 0.31Gr1/3 Sc1/3  AB 
 db 
 1.71x10−9 m2 / sec 
−5
k L = (0.31)(9.8 x1012 db3 )1/3 (572)1/3 
 = 9.6 x10 m / sec
db


−4 −1
k a
5.473 x10 s
a= L =
= 5.7 m 2 /m3
−5
kL
9.6 x10 m / sec
6φg
a=
db
db =
6φg
=
3
3
6(0.005mgas
/ m liquid
)
−1
= 5.3 x10−3 m = 5.3 mm
a
5.7 m
5.3 mm > 2.5 mm, therefore assume db ≥ 2.5 mm
D 
k d
Sh = L b = 0.42Gr1/3 Sc1/2 → kL = 0.42Gr1/3 Sc1/2  AB 
DAB
 db 
 1.74 x10−9 m 2 / sec 
−4
k L = (0.42)(9.8 x1012 db3 )1/3 (572)1/2 
 = 3.75 x10 m / sec
db


−4 −1
k a
5.473 x10 s
= 1.46m −1
a= L =
−4
kL
3.75 x10 m / sec
6φg
a=
db
db =
6φg
a
=
6(0.005 m 3gas /m 3liquid )
1.46 m -1
= 0.0205 m = 20.5 mm
20.5 mm > 2.5 mm, therefore correct correlation used
c. Determine how bubble diameter, liquid viscosity, and aeration change in order to increase the
kLa:
6φ
A
k L a = k L i = k L g , k L a ↑⇒ d b ↓
V
db
1/3
 1 
k L ∝  2  µ L1/3 = µ L −1/3 for db < 2.5mm
 µL 
1/3
 1 
k L ∝  2  µ L1/2 = µ L −1/6 for db ≥ 2.5mm
 µL 
k L a ↑⇒ µ L ↓
WA = k L aV ∆C AL , k L a ↑⇒ WA ↑
30.3
Let A = CO2, B = N2
P
 1atm 
2
DAB = DAB , ref ref = ( 0.166cm 2 /sec ) 
 = 0.0415 cm /sec
P
 4atm 
wɺ (0.54 kgmol/hr)(1000 gmol/1kgmol)(1hr/3600 sec)
Vɺ0 = =
= 923.175 cm 3 /sec
4atm
C
 82.06atm-cm 3 

 (300 K)
 gmol-K

4 ( 923.175 cm /sec )
4Vɺ0
=
v∞ =
= 1.88 cm/sec
π D2
π(25 cm 2 )
3
a. Determine kc
Re =
Sc =
v∞ D ρ L
µL
µL
ρ L DAB
=
=
(1.88 cm/sec)(25 cm)(4.68x10-3 g/cm3 )
= 1236
1.78x10-4 g/cm-sec
1.78x10-4 g/cm-sec
(4.68x10-3 g/cm3 ) ( 0.0415 cm 2 /sec )
= 0.916
Sieder-Tate correlation for laminar flow of liquid flowing inside of a tube
1/3
kc D
D

= 1.86  Re⋅ Sc 
Sh =
DAB
L

1/3
D
 D 
kc = 1.86  Re⋅ Sc   AB 
L
  D 
1/3
2
 25 cm
  0.0415 cm /sec 
kc = 1.86 
(1236)(0.916)  

25 cm
 3000 cm
 

= 0.0941(3000 cm)-1/3 = 6.52 x 10-3 cm/sec
b. Determine yA,out
Assumptions: 1) Steady state, 2) Dilute system, 3) Concentration along z-direction only, 4) No
reaction, 5) PA ≈ 0, 6) Kc ≈ kc
Material balance on differential volume element (A = CO2)
In – Out +Generation = Accumulation (liquid phase – mole A/time)
π D2
4
v∞C A − N A ⋅ π D ⋅ ∆z −
z
π D2
4
+0=0
v∞C A
z +∆z
÷ ∆z, rearrangement, ∆z → 0
dC
4
− A − NA ⋅
=0
dz
vɺ∞ D
dC
4
− A−
kc C A = 0
dz
v∞ D
∫
C A ,out
C A ,o
dC A
4k L
= − c ∫ dz
CA
v∞ D 0
C

4k
ln  A, out  = − c L
 C 
v∞ D
 A, o 
 4k 
y A, out = y A, o ⋅ exp  − c L 
 v∞ D 
 4 ( 6.52 x 10-3 cm/sec )

y A, out = ( 0.05 ) ⋅ exp  ( 3000 cm )  = 9.5 x 10-3
 (1.88 cm/sec)(25 cm)

c. Determine DAe
T
300 K
DKA = 4850d pore
=4850(0.8x10-4 cm)
=1.013 cm 2 /sec
MA
44 g/gmole
-1
1
1

−1 −1 
2
DAe = (D −AB1 + DKA
) =
+
 = 0.0399 cm /sec
2
2
 0.0415 cm /sec 1.013 cm /sec 
'
2
2
DAe = ε DAe = (0.6) (0.0399 cm 2 /sec) = 0.0144 cm 2 /sec
d. Determine Kc
D
0.0144 cm 2 /sec
km = Ae =
= 0.012 cm/sec
lm
1.2 cm
−1
1


1 
1
1
Kc =  +
+
 =

-1/3
 0.0941 L (cm/sec) 0.012 cm/sec 
 kc k m 
1
Kc =
= 4.23x10-3 cm/sec
1/3
10.63(3000cm) +83.33
-1
e. Determine new yA,out
 4 ( 4.23 x 10-3 cm/sec )

y A, out = ( 0.05 ) ⋅ exp ( 3000 cm ) = 0.017
 (1.88 cm/sec)(25 cm)


0.141(3000 cm) 2/3 



-4
3
y A,out = 0.05 ⋅ exp 


3000
cm
+7.839


3
  
 (1.88 cm/sec)(25 cm)  2.212 3000 cm +17.343ln 
7.839

 


y A,out = 0.013
31.1
a. Determine CA at t = 5.0 min (A = O2)
3
m3  3.28 ft   60 sec 
ft 3
=16.5
sec  1.0 m   1min 
min
At depth = 4.55 m (15 ft), from Eckenfelder plot
Qg = 0.0078
A
ft 3
V ≅ 1200
V
hr
K L ( A / V )V

ft 3  ( 6 spargers )
K La =
n =  1200
=0.72 hr -1

3
V
hr
10,000
ft
(
)


KL
Ptop + Pbottom
1
ρ L gh
2


1  1000 kg  9.81 m 
1.0 atm
Pave = 1.0atm + 
=1.22 atm

 ( 4.55 m ) 
5
2 
2  m3  sec 2 
1.0123
x
10
kg/m
sec
⋅


Pave =
2
= Ptop +
PA y A P 0.21(1.22 atm )
= 7.83 x 10-6
=
=
H
H
3.27 x 10 4atm
ρ
1000 g/L
gmol
C= B =
= 55.56
M B 18 g/gmole
L
x*A =
gmol 
gmol

C *A = x*AC = ( 7.83 ⋅ 10−6 )  55.56
= 4.35 ⋅ 10−4

L 
L

t = 5.0 min = 0.083 hr
C A = C A* − ( C A* − C A0 ) e(
− K L a ⋅t )
gmol 
gmol
gmol 
-1
-  4.35 x 10-4
- 0.05 x 10-3
 exp ( -0.72 hr ) ( 0.083hr ) 
L
L
L 

gmol
mmol
C A = 7.23 x 10-5
= 0.0723
L
L
C A = 4.35 x 10-4
b. Determine time (t) required for CA = 0.20 mmole/L
 ( C A* − C A ) 

ln  *
 ( C A − C A0 ) 
t=
( −K La )
 ( 0.435 mmol/L - 0.20 mmol/L ) 
ln 

( 0.435 mmol/L - 0.05 mmol/L ) 

t=
= 0.685 hr
( -0.72 hr -1 )
31.2
a. Determine AG1 and YA2
From YA -X A plot, YA2,min = 0.152
x A2 =
X A2
0.20
=
= 0.167
1 + X A2 1.0 + 0.20
x A1 =
X A1
0.05
=
= 0.0476
1 + X A1 1.0 + 0.05
ALs = AL2 (1 − x A2 ) = 100 (1.0 − 0.167 ) = 83.3
lbmol
hr
YA1 AGs ,min + X A 2 ALs = YA2 AGs ,min + X A1 ALs
0 + ( 0.20 )( 83.3) = ( 0.152 ) ( AGs ,min ) + ( 0.05 )( 83.3)
AGs ,min = 82.2
lbmol
hr
lbmol
hr
YA1 AGs + X A2 ALs = YA2 AGs + X A1 ALs
AGs = (1.4 )( 82.2 ) = 115.1
0 + ( 0.20 )( 83.3) = (YA 2 )(115.1) + ( 0.05 )( 83.3)
YA 2 = 0.109
b. See YA-XA plot
0.16
YA2,min
0.14
0.12
YA2
0.10
0.08
YA
0.06
0.04
0.02
0.00
0.00
XA2
XA1, YA1
0.05
0.10
0.15
XA (mole A / mole solvent)
0.20
0.25
31.3
a. Determine AL1

std m3   kgmol 
kgmol
AG1 =  224
= 10.0

3 
hr   22.4 std m 
hr

AGs = AG1 (1 − y A1 ) = 10(1-0.09)=9.1
(1 − R ) yA1 AG1 = y A2 AG2 =
kgmol
hr
y A1
AGs
1 − y A1
 y A1 
 ( 9.10 )
 1 − y A1 
(1 − 0.898)( 0.09 )(10 ) = 
y A2 = 0.01
AG2 =
AGs
9.1
kgmol
=
=9.19
1 − y A 2 1-0.01
hr
y A1 AG1 + x A2 AL2 = y A 2 AG2 + xA1 AL1
( 0.09 )(10 ) + 0 = ( 0.01)( 9.19 ) + ( 0.02 ) AL1
AL1 = 40.4
kgmol
hr
b. Determine ALs,min
y A2 AG2 + 0 = y A 2 AG2 + xA1,min AL1,min
x A1,min = 0.092 at y A1 = 0.090
( 0.09 )(10 ) = ( 0.01)( 9.19 ) + ( 0.092 ) AL1,min
AL1,min = 8.78
kgmol
hr
ALs ,min = AL1,min (1 − x A1,min ) = 8.78 (1 − 0.092 ) = 7.97
c. Determine packing height, z
kgmol
10
AG1
hr = 0.50 m 2
A=
=
kgmol
G1
20 2
m hr
H OG =
G=
G
K 'y a
AG1 + AG2 (10+9.19 ) kgmol/hr
kgmol
=
=19.19 2
2
2A
2 ⋅ 0.5 m
m hr
kgmol
hr
1
1
H
= ' + '
'
K G a kG a k L a
mP (1.2 )(1.0 atm )
m3atm
=
=0.0216
kgmol 
CL 
kgmol
 55.6

3
m 

1
1 0.0216
=
+
'
K G a 20
50
H=
kgmol
m ⋅ hr ⋅ atm
kgmol 
kgmol

K y' a = K G' a ⋅ P =  19.58 3
 (1.0 atm ) = 19.58 3
m ⋅ hr ⋅ atm 
m ⋅ hr ⋅ atm

kgmol
19.19 2
m hr
H OG =
= 0.98 m
kgmol
19.58 3
m ⋅ hr ⋅ atm
y −y
N OG = A1 *A2
( yA − yA )
K G' a = 19.58
3
lm
*
A1
y = 0.023 at x A1 = 0.020
y*A 2 = 0.0 at x A2 = 0.0
( yA − y*A ) =
lm
( y − y ) − ( y − y ) = ( 0.09 − 0.023) − ( 0.01 − 0.0 ) = 0.030
(y − y )
 ( 0.090 − 0.023) 
ln 
ln 


 ( y − y ) 
 ( 0.010 − 0.0 ) 
A1
*
A1
A2
A1
*
A1
A2
*
A2
0.09 − 0.01
= 2.67
0.030
z = H OG N OG
N OG =
z = ( 0.98 m )( 2.67 ) = 2.6 m
*
A2
d. Gas flooding velocity, Gf


kg 
kg  
 kmol  
AG1' = AG1 (1 − y A1 ) M W ,air + y A1 M W , A  = 10
 + ( 0.09 )  58

 (1-0.09 )  29
hr  

 kgmol 
 kgmol  
AG1' = 316
kg
hr


kg 
kg  
 kmol  
AL1' = AL1 (1 − x A1 ) M W , H 2O + xA1 M W , A  =  40
 + ( 0.02 )  58

 (1-0.02 ) 18
hr  

 kgmol 
 kgmol  
AL1' = 759.5
kg
hr




P 
kg  
1.0 atm
 = 1.34 kg
=  31.6
ρG = M W ,ave

3

RT 
kgmol  
m3
m atm 
  0.08206
 ( 288 K ) 
kgmol ⋅ K 


Flooding Correlation
1/2
AL1'  ρG 


AG1'  ρ L − ρG 
x − axis:
1/2
759.5 
1.34



316  1000 − 1.34 
=
= 0.09
( G ) C µ J = ( G ) (100)(1.0) ⋅ 1
y − axis: Y = 0.22 =
'
f
2
f
0.1
L
ρ G ( ρ L − ρG ) g c
G 'f = 1.7
'
f
2
0.1
1.34 (1000 − 1.34 ) ⋅ 1
kg
m 2 sec
1/2
1/2
 C f ,old 
kg   100 
kg

G =G
=
1.7




 = 1.2 2
2
m sec   200 
m sec

 C f ,new 
Intalox Saddles have lower G'f , and so the packing is not better
'
f
'
f , old
e. The operating mole fraction inlet gas is above the saturation mole fraction, so some acetone
will condense to lower the outlet mole fraction.
31.3 continued
0.10
0.08
xA1, yA1
yA1
0.06
yA
0.04
0.02
xA2,
yA2
0.00
0.00
xA1,min
0.02
0.04
0.06
0.08
xA (mole fraction acetone in water)
0.10
Part 3: Show/Hide Problems
Chapter 1
Show/Hide Problems
1.2
Which of the quantities listed below are flow properties and which are fluid properties?
Solution
Flow Properties: velocity, pressure gradient, stress
Fluid Properties: pressure, temperature, density, speed of sound, specific heat.
1.18
A vertical cylindrical tank having a base diameter of 10 meters and a height of 5 meters is filled
to the top with water at 20℃. How much water will overflow if the water is heated to 80℃?
Solution
Diameter = 10 m
Height = 5 m
V=
π 2
π
d h = (10m)2 (5m) = 392.7 m3
4
4
At 20℃, ρWater = 998.2 kg/m3
So the mass of this is: m = ρW V = (998.2 kg/m3 )(392.7 m3 ) = 391992.2 kg
At 80℃, ρWater = 971.8 kg/m3
So the mass of this is: m = ρW V = (971.8 kg/m3 )(392.7 m3 ) = 381625.86 kg
So the mass of water at 80℃ that can be in the tank is lower than at 20℃.
The difference is: ∆m = 10366.34 kg
1.21
The bulk modulus of elasticity for water is 2.205 GPa. Determine the change in pressure required
to reduce a given volume of 0.75%.
Solution
𝛽𝐻2𝑂 = 2.205 𝐺𝑃𝑎
∆𝑉
= −0.75% = −0.0075
𝑉
dP
∆P
𝛽𝐻2𝑂 = V ( ) ≅ −V
dV
∆V
Rearrange,
∆𝑃 = 𝛽𝐻2𝑂
∆𝑉
= −2.205 𝐺𝑃𝑎(−0.0075) = 0.0165𝐺𝑃𝑎 = 16.5𝑀𝑃𝑎
𝑉
1.30
A colleague is trying to measure the diameter of a capillary tube, something that is very difficult
to physically accomplish. Since you are a Fluid Dynamics student, you know that the diameter
can be easily calculated after doing a simple experiment. You take a clean glass capillary tube
and place it in a container of pure water and observe that the water rises in the tube to a height of
17.5 millimeters. You take a sample of the water and measure the mass of 100 mls to be 97.18
grams and you measure the temperature of the water to be 80℃. Please calculate the diameter of
your colleague’s capillary tube.
Solution
First, calculate the surface tension of water using the temperature of the water:
𝑇 = 80℃ = 353𝐾
𝜎 = 0.123(1 − 0.00139𝑇)
𝜎 = 0.123(1 − 0.00139(353)) = 0.06265 𝑁/𝑚
Next, using the equation for the height of a fluid in a capillary,
2𝜎𝑐𝑜𝑠𝜃
ℎ=
𝜌𝑔𝑟
Rearranging and solving for the radius:
2𝜎𝑐𝑜𝑠𝜃
𝑟=
𝜌𝑔ℎ
2(0.06265 𝑁/𝑚) cos(0)
=
97.18 𝑔
𝑘𝑔
1000 𝑚𝑙
28.32 𝑙
𝑚
)(
((
)(
)(
)) (9.81 𝑚/𝑠 2 )(17.5𝑚𝑚) (1000𝑚𝑚 )
100 𝑚𝑙𝑠 1000𝑔
𝑙𝑖𝑡𝑒𝑟
0.028317 𝑚3
= 7.509𝑥10−4 𝑚𝑒𝑡𝑒𝑟𝑠 = 0.7509 𝑚𝑚
Diameter is 2r so D = 1.50 mm
Chapter 2
Show/Hide Problems
2.6
The practical depth limit for a suited diver is about 185 meters. What is the gage pressure of sea
water at that depth? Assume that the specific gravity of sea water is constant at 1.025.
Solution
𝑑𝑃
= −𝜌𝑔
𝑑𝑦
𝑃
0
∫
𝑑𝑃 = −𝜌𝑔 ∫ 𝑑𝑦
𝑃𝑎𝑡𝑚
ℎ
𝑃 − 𝑃𝑎𝑡𝑚 = − 𝜌𝑔(0 − ℎ) = 𝜌𝑔ℎ
𝑃 − 𝑃𝑎𝑡𝑚 = 1.025(1000 𝑘𝑔/𝑚3 )(9.8 𝑚/𝑠 2 )(185 𝑚𝑒𝑡𝑒𝑟𝑠) = 1.86𝑥106
𝑘𝑔
= 1.86𝑥106 𝑃𝑎
𝑚𝑠 2
2.9
Determine the depth change to cause a pressure increase of 1 atm for (a) water, (b) sea water
(SG=1.0250) and (c) mercury (SG=13.6). (Assume the temperature is 30℃.)
Solution
𝑃 − 𝑃𝑎𝑡𝑚 = 𝜌𝑔ℎ
𝜌𝑤𝑎𝑡𝑒𝑟 = 995.2 𝑘𝑔/𝑚3
𝑁
1 𝑎𝑡𝑚 = 101325 2 = 101325 𝑘𝑔/𝑠 2 𝑚
𝑚
ℎ=
𝑃 − 𝑃𝑎𝑡𝑚
𝜌𝑔
(a) water
ℎ=
𝑃 − 𝑃𝑎𝑡𝑚
101325 𝑘𝑔/𝑠 2 𝑚
=
= 10.4 𝑚
(995.2 𝑘𝑔/𝑚3 )(9.8 𝑚/𝑠 2 )
𝜌𝑔
(b) sea water (SG=1.0250)
𝑃 − 𝑃𝑎𝑡𝑚
101325 𝑘𝑔/𝑠 2 𝑚
ℎ=
=
= 10.13 𝑚
(995.2 𝑘𝑔/𝑚3 )(1.0250)(9.8 𝑚/𝑠 2 )
𝜌𝑔
(c) mercury (SG=13.6)
ℎ=
𝑃 − 𝑃𝑎𝑡𝑚
101325 𝑘𝑔/𝑠 2 𝑚
=
= 0.764 𝑚
(995.2 𝑘𝑔/𝑚3 )(13.6)(9.8 𝑚/𝑠 2 )
𝜌𝑔
2.12
What is the pressure pA in the figure? The oil in the
middle tank has a specific gravity of 0.8. Assume that
the entire system is at 80oF.
D
Solution
𝑃𝐸 = 𝑃𝐵 − 𝜌𝑜𝑖𝑙 𝑔(10 𝑓𝑒𝑒𝑡)
(1)
𝑃𝐶 = 𝑃𝐵 + 𝜌𝐻2𝑂 𝑔(15 − 10 𝑓𝑒𝑒𝑡) (2)
𝑃𝐷 = 𝑃𝐶 − 𝜌𝐻𝑔 𝑔(1 𝑓𝑜𝑜𝑡)
(3)
Insert (2) into (3) then into (1):
𝑃𝐷 = 𝑃𝐵 + 𝜌𝐻2𝑂 𝑔(15 − 10 𝑓𝑒𝑒𝑡) − 𝜌𝐻𝑔 𝑔(1 𝑓𝑜𝑜𝑡)
𝑃𝐸 = 𝑃𝐷 − 𝜌𝐻2𝑂 𝑔(15 − 10 𝑓𝑒𝑒𝑡) + 𝜌𝐻𝑔 𝑔(1 𝑓𝑜𝑜𝑡) − 𝜌𝑜𝑖𝑙 𝑔(10 𝑓𝑒𝑒𝑡)
From Appendix I, at 80oF: ρH2O=62.2 lbm/ft3, ρHg=845 lbm/ft3
Thus,
(62.2 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(15 − 10 𝑓𝑡)
𝑃𝐴 = 𝑃𝐸 −𝑃𝐷 = −
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
(845 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(1 𝑓𝑡) 0.8(62.2 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(10 𝑓𝑡)
+
−
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
𝑙𝑏𝑓
= 36.45 2 (𝑔𝑎𝑢𝑔𝑒)
𝑓𝑡
PA is a gauge pressure because PD is open to the atmosphere.
2.13
Assume that the entire system is at 80oF.
Referring to the figure at left, please find
the pressure difference between tanks A
and B if d1=2 feet, d2=6 inches, d3=2.4
inches and d4=4 inches.
3
Solution
𝑃𝐴 − 𝑃3 = −𝜌𝐻2 𝑂 𝑔𝑑1
(1)
𝑃𝐵 − 𝑃3 = −𝜌𝐻𝑔 𝑔𝑑3 − 𝜌𝐻𝑔 𝑔𝑑4 𝑠𝑖𝑛𝜃
(2)
Subtract (1) -(2):
𝑃𝐴 − 𝑃𝐵 = −𝜌𝐻2 𝑂 𝑔𝑑1 + 𝜌𝐻𝑔 𝑔𝑑3 + 𝜌𝐻𝑔 𝑔𝑑4 sin (45)
From Appendix I, at 80oF: ρH2O=62.2 lbm/ft3, ρHg=845 lbm/ft3
𝑃𝐴 −𝑃𝐵 =
−(62.2 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(2 𝑓𝑡) (845 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(0.2 𝑓𝑡)
+
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
(845 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(0.333 𝑓𝑡)sin (45)
𝑙𝑏𝑓
𝑙𝑏𝑓
+
= 244 2 = 1.70 2
2
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠
𝑓𝑡
𝑖𝑛
= 1.7 𝑝𝑠𝑖
2.15
Assume the system is at 80oF.
Points A and 3 are not necessarily at the same height.
4
3
1
X
2
A differential manometer is used to measure the pressure change caused by a flow constriction in
a piping system as shown. Determine the pressure difference between points A and B in psi.
Which section has the higher pressure?
Solution
𝑃1 − 𝑃𝐴 = 𝜌𝐻2 𝑂 𝑔(𝑑𝑥 + 𝑑1 ) → 𝑃1 = 𝑃𝐴 + 𝜌𝐻2 𝑂 𝑔(𝑑1 + 𝑑1 ) = 𝑃𝐴 + 𝜌𝐻2 𝑂 𝑔(10 + 𝑥 𝑖𝑛)
𝑃2 − 𝑃𝐵 = 𝜌𝐻2 𝑂 𝑔𝑑1 + 𝜌𝐻2 𝑂 𝑔𝑑𝑥 + 𝜌𝐻𝑔 𝑔𝑑2 → 𝑃2 = 𝑃𝐵 + 𝜌𝐻2 𝑂 𝑔𝑑1 + 𝜌𝐻2 𝑂 𝑔𝑑𝑥 + 𝜌𝐻𝑔 𝑔𝑑2
= 𝑃𝐵 + 𝜌𝐻2 𝑂 𝑔(4 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛)
Equate the two equations by realizing that 𝑃1 = 𝑃2
𝑃𝐴 + 𝜌𝐻2 𝑂 𝑔(10 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛) = 𝑃𝐵 + 𝜌𝐻2 𝑂 𝑔(4 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛)
Rearrange and cancel 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛),
𝑃𝐴 − 𝑃𝐵 = −𝜌𝐻2 𝑂 𝑔(10 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(4 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛) = −𝜌𝐻2 𝑂 𝑔(6 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛)
From Appendix I:
𝜌𝐻2 𝑂 𝑎𝑡 80℉ = 62.2 𝑙𝑏𝑚 /𝑓𝑡 3
𝜌𝐻𝑔 𝑎𝑡 80℉ = 845 𝑙𝑏𝑚 /𝑓𝑡 3
In addition, we know that 10 inches = 0.83 ft and 6 inches = 0.5 ft.
−(62.2 𝑙𝑏𝑚 /𝑓𝑡 3 )(32.2 𝑓𝑡/𝑠 2 )(0.5 𝑓𝑡) + (845 𝑙𝑏𝑚 /𝑓𝑡 3 )(32.2 𝑓𝑡/𝑠 2 )(0.83 𝑓𝑡)
𝑃𝐴 − 𝑃𝐵 =
32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2
2
= 670.82 𝑙𝑏𝑓 /𝑓𝑡
We were asked to state the problem in terms of psi (pounds per square inch):
𝑃𝐴 − 𝑃𝐵 = 670.82
𝑙𝑏𝑓
𝑙𝑏𝑓
𝑓𝑡 2
𝑥
= 4.66 2 = 4.66 𝑝𝑠𝑖
2
2
𝑓𝑡 144 𝑖𝑛
𝑖𝑛
Since
𝑃𝐴 − 𝑃𝐵 = 4.66 𝑝𝑠𝑖
The pressure at A must be greater than the pressure at B.
Chapter 4
Show/Hide Problems
4.3
Water is flowing through a large circular
conduit with a velocity profile given by the
𝑟2
equation 𝑣 = 9 (1 − 16) fps. What is the
average velocity in the 1.5 ft. pipe?
Solution
The control volume is the dashed red line in the figure above.
For the above control volume,
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +
𝜕
∭ 𝜌𝑑𝑉 = 0
𝜕𝑡
Assume steady flow,
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 = 0
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴2 − ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴1 = 0
𝜌2 𝑣2,𝑎𝑣𝑔 𝐴2 − 𝜌1 𝑣1,𝑎𝑣𝑔 𝐴1 = 0
𝑟2
Assuming incompressible flow and substituting 𝑣 = 9 (1 − 16) and integrating from 0 → 𝑅,
𝑣2,𝑎𝑣𝑔 𝐴2 − 𝑣1,𝑎𝑣𝑔 𝐴1 = 0
𝑅
𝑟2
𝑣2,𝑎𝑣𝑔 𝐴2 − ∫ 9 (1 − ) 2𝜋𝑟𝑑𝑟 = 0
16
0
𝐴2 =
𝑣2 (
𝜋𝐷2 𝜋(1.5)2
=
4
4
𝜋(1.5)2
9
) − [9𝜋𝑅 2 − 𝜋𝑅 4 ] = 0
4
32
Solve for 𝑣2 ,
𝑣2 =
9
[9𝜋42 − 32 𝜋44 ]
𝜋(1.5)2
(
)
4
= 128 𝑓𝑡/𝑠
4.10
Beginning with the integral mass balance equation,
𝜕
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 + ∭ 𝜌𝑑𝑉 = 0
𝜕𝑡
and using the symbol M for the mass in the control volume, show that equation (1) above can be
written as:
𝜕𝑀
+ ∬ 𝑑𝑚̇ = 0
𝜕𝑡
Solution
We can solve this problem in pieces.
First, let's examine: ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴. If we let 𝑚̇ = 𝜌𝑣𝐴 = 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒.
We can now write:
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 = ∬ 𝑑(𝜌𝑣𝐴) = ∬ 𝑑𝑚̇
𝜕
Next, let's examine the accumulation term: 𝜕𝑡 ∭ 𝜌𝑑𝑉.
If we rearrange this term,
𝜕
𝜕
𝜕
∭ 𝜌𝑑𝑉 = ∭ 𝑑(𝜌𝑉) = 𝑀
𝜕𝑡
𝜕𝑡
𝜕𝑡
As a result,
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +
𝜕
∭ 𝜌𝑑𝑉 = 0
𝜕𝑡
Becomes,
𝜕𝑀
+ ∬ 𝑑𝑚̇ = 0
𝜕𝑡
4.18
Water flows steadily through a piping junction shown in the figure below. It enters section 1 at
0.0013 m3/s. The average velocity at section 2 is 2.1 m/s. A portion of the flow is diverted through
the showerhead that contains 100 holes of 1-mm diameter. Assuming uniform shower flow,
estimate the exit velocity from the showerhead jets.
j
Solution
The Control Volume for this problem is shown as a dotted red line in the figure above.
Since we are told that we have steady flow, we can do the following mass balance:
Mass Flow In = Mass Flow Out
Mass flow in = 𝜌1 𝑣1 𝐴1
Mass flow out = 𝜌2 𝑣2 𝐴2 + 𝜌𝑗 𝑣𝑗 𝐴𝑗 (where j stands for the flow from the showerhead jets).
So,
𝜌1 𝑣1 𝐴1 = 𝜌2 𝑣2 𝐴2 + 𝜌𝑗 𝑣𝑗 𝐴𝑗
𝜌1 𝑄 = 𝜌2 𝑣2 𝐴2 + 𝜌𝑗 𝑣𝑗 𝐴𝑗
Since in this system the fluid is water, we can assume incompressible flow and that density is
constant.
Thus,
𝑄 = 𝑣2 𝐴2 + 𝑣𝑗 𝐴𝑗
𝜋
1.3𝑥10−3 𝑚3 /𝑠 = (2.1 𝑚/𝑠)(𝜋(10−2 𝑚)2 ) + (𝑣𝑗 )(100) ( (10−3 )2 )
4
Rearrange and solve for 𝑣𝑗 ,
1.3𝑥10−3 𝑚3 /𝑠 − (2.1 𝑚/𝑠)(𝜋(10−2 𝑚)2 )
𝑣𝑗 =
𝜋
(100) ( (10−3 )2 )
4
𝑣𝑗 = 8.2 𝑚/𝑠
4.19
The jet pump injects water at V1=40 m/s through a
7.6 cm pipe and entrains a secondary flow of water
V2=3 m/s in the annular region around the small
pipe. The two flows become fully mixed
downstream, where V3 is approximately constant.
For steady incompressible flow, compute V3.
Solution
The control volume for the problem is in the red dashed line in the figure.
Mass flow in = Mass flow out
𝜌1 𝑣1 𝐴1 + 𝜌2 𝑣2 𝐴2 = 𝜌3 𝑣3 𝐴3
Since all the fluid is water,
𝜌1 = 𝜌2 = 𝜌3
𝑣1 𝐴1 + 𝑣2 𝐴2 = 𝑣3 𝐴3
2
𝜋
1𝑚
(40 𝑚/𝑠) [(7.6 𝑐𝑚) (
)]
4
100 𝑐𝑚
2
2
𝜋
1𝑚
𝜋
1𝑚
)] − (3 𝑚/𝑠) [(7.6 𝑐𝑚) (
)] }
+ {(3 𝑚/𝑠) [(25 𝑐𝑚) (
4
100 𝑐𝑚
4
100 𝑐𝑚
2
𝜋
1𝑚
)]
= (𝑣3 ) [(25 𝑐𝑚) (
4
100 𝑐𝑚
0.1814
𝑚
𝑚
+ {0.1473 − 0.0136 } = 0.0490(𝑣3 )
𝑠
𝑠
𝑣3 = 6.48 𝑚/𝑠
4.21
The hypodermic needle shown in the figure below contains an incompressible liquid serum with
a density of 1 g/cm3. If the serum is to be injected steadily at 6 cm3/s, please calculate how fast
the plunger must be advanced: (a) if leakage in the plunger clearance is neglected and (b) if
leakage is 10% of the needle flow.
Solution
(a) if leakage in the plunger clearance is neglected
𝜕
∭ 𝜌𝑑𝑉 = 0
𝜕𝑡
Rearrange and insert mass and 𝑚̇ into the equation,
∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +
𝜕𝑀
+ ∬ 𝑑𝑚̇ = 0
𝜕𝑡
−𝜚𝑣𝐴 + 𝜚𝑄 = 0
𝜚𝑣𝐴 = 𝜚𝑄
Since the serum is incompressible,
𝑄
6 cm3 /s
𝑣= =
= 1.91 𝑐𝑚/𝑠
𝐴 𝜋(1 𝑐𝑚)2
(b) if leakage is 10% of the needle flow.
𝜕𝑀
+ ∬ 𝑑𝑚̇ = 0
𝜕𝑡
−𝜚𝑣𝐴 + 𝜚𝑄 + 𝜚𝑄𝑙𝑒𝑎𝑘 = 0
Since the serum is incompressible, and 10% = 0.10,
𝑄 + 𝑄𝑙𝑒𝑎𝑘 = 𝑣𝐴
cm
cm3
(6
) + (0.10) (6
) = v(π(1 cm2 ))
s
s
3
𝑣 = 2.10 𝑐𝑚/𝑠
(So leakage from the plunger end necessitates a higher velocity.)
Chapter 5
Show/Hide Problems
5.1
A two-dimensional object is placed in a 4-ft-wide water tunnel as shown. The upstream velocity,
v1, is uniform across the cross section. For the downstream velocity profile as shown, find the
value of v2. (Assume steady state.)
Solution
! 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 +
𝜕
- 𝜌𝑑𝑉 = 0
𝜕𝑡
If have steady state (a.k.a. steady flow),
! 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 = 0
4
2
4
1 6
1 9
1 𝜌𝑣2 𝑑𝑥 = 1 𝜌𝑣6 𝑑𝑥 + 1 𝜌𝑣6 𝑑𝑥 + 1 𝜌𝑣6 𝑑𝑥 + 1 𝜌𝑣6 𝑑𝑥
2 2
2 6
5
5
9
1
1
4𝑣2 𝜌 = 1𝜌𝑣6 + (2 − 1)𝜌𝑣6 + (3 − 2)𝜌𝑣6 + (4 − 3)𝜌𝑣6
2
2
Divide through by 𝜌 and simplify,
4𝑣2 = 3𝑣6
Solve for 𝑣6
4
4
𝑓𝑡
𝑓𝑡
𝑣6 = 𝑣2 = =20 @ = 26.67
3
3
𝑠
𝑠
5.7
Fluid is H20
P1= 60 Psig
P2= 14.7 Psia (units?)
D1= 0.25 ft.
D2 = ft.
= 400 gal/m= 0.892 ft.3/s
Steady, Incompressible Flow
dA =0
=
=
=
= 18.17
Fx-P1A1-P2A2=
= 18.17 ft./s
= 72.7 ft./s
( - )
Fx= ( - )-P1A1+P2A2
=[62.4(0.892)(72.7-18.17)]/32.2-(60+14.7)(144) (0.25)2+(14.7)(144)
=94.3-528.0+25.4
=-408 lbf
2
5.16
Steady incompressible flow:
=
dA
A1= (1.5)2=1.767 ft.2
A2= (1)2=0.785 ft.2
A3= (2)2=3.142 ft.2
For C.V. between (1)&(2) of the figure:
=
dA
Fx+F2+P1A1-P2A2=
Fx
( 2-
1)
+(530-14.7)(144)(0.785)-(990-14.7)(144)(1.767)-F2 = -130,000 lbf –F2
For C.V. between (2)&(3) of the figure:
Fx+P2A2-P3A3= ( 3- 2)
Fx=
(6700-3400)+(26-14.7)(144)(3.14)-(530-14.7)(144)(0.785) = 25777 lbf
Stress at position 2 in the figure:
= =
= 1823 psi Compression
At 1: F1= -130,000+25777 = -104220 lbf
=
= 4915 psi Tension
5.23
x momentum:
=
+
dV
Fx= - wAj ( cos )+ wAj
= wAj ( - cos )
=1000( )(0.1)2(20)[4.5-20cos45 ]
= -1515 N
Force on car by jet: Fx=1515 N
y momentum:
=
Fy= - sin
+
dV
( Aj)+ 0 = -(20sin45)(1000)(-20)* (0.1)2 = +2220 N
Force exerted by H20:
Fy= -2220 N
Total Force = = 1515 - 2220
N
5.29
=
x )z
=
dA
( )
Mz=2()
ry= Rsin
= sin - R
Mz=2 [-Rsin ( sin - r )]
=
+
Chapter 6
Show/Hide Problems
6.2
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3/s. The pump
inlet is 0.25 m in diameter. At the inlet the pressure is 81326 N/m2. The pump outlet is 0.152 m
in diameter and is 1.8 meters above the inlet. The outlet pressure is 175 kPa. If the inlet and exit
temperatures are equal, how much power does the pump add to the fluid?
(Note, inlet pressure changed for this problem relative to book problem.)
Solution
Begin with the overall energy balance equation,
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
∂
∭ eρdV = 0 because we have steady state
∂t
δQ
= 0 since we have no heat transfer out of the system
∂t
δWμ
= 0 since there is no viscous work being done
∂t
δWs
p
p
−
= ∬ eρ(v ∙ n) dA = (e + ) ṁ − (e + ) ṁ
∂t
ρ 2
ρ 1
2
2
v2 − v1 P2 − P1
= ṁ [U2 − U1 +
+
+ g(z2 − z1 )]
2
ρ
Since the inlet and exit temperatures are equal, U2 = U1.
Next, we calculate some quantities,
ṁ = ρQ = (1025 kg/m3 )(0.21 m3 /s) = 215.25 kg/s
Q
0.21 m3 /s
v1 =
=
= 4.28 ft/s
A1 π(0.25 m)2
4
Q
0.21 m3 /s
v2 =
=
= 11.57 ft/s
A2 π(0.15 m)2
4
δW
Substituting into the above equation for − ∂t , and remembering that Pa=N/m2 and N=kg m/s2,
−
δWs
kg (11.57 ft/s)2 − (4.28 ft/s)2 175000 − 81326 N/m2
= 215.25
[
+
∂t
s
2
1025 kg/m3
+ (9.81 m/s2 )(1.8m)] = 35,908 Nm/s
δW
The negative sign on − ∂t s indicates work done to the fluid.
6.3
Air at 70oF flows into a 10 ft3 resevoir at a velocity of 110 fps. If the reservoir pressure is 14 psig
and the reservoir temperature is 70oF, find the rate of temperature increase in the reservoir.
Assume the incoming air is at reservoir pressure and flows through an 8 in diameter pipe.
Solution
δQ
dt
δQ
δW
δWμ
δW
δWμ
P
d
− dts − dt = ∬ (e + ρ ) ρ(v
⃑ ∙n
⃑ )dA + dt ∭ eρdV
= dts = dt = 0
dt
v2
d
- ṁ[h1+ 21 + gz1]+ dt[mu]sys=0
gz1=negligible
v2
dT
ρvcv dt = ρA v( 2 )
dT
Av v2
= vc
dt
v
=
2
π 8 2 2
ft.
( ) ft. (110 )3
4 12
s
778 ft.LBf 32.2 LBm ft.
BTU
3
)(0.17
)
2(10ft.
)(
)(
LBm F
BTU
s2 LBf
= 21.8 F/s
6.4
Water flows through a 2 inch diameter horizontal pipe at a flow rate of 35 gal/min. Assume steady
flow, that the heat transfer to the pipe is negligible, and that frictional forces cause a pressure drop
of 10 psi. What is the temperature change of the water?
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
δWs
= 0 since there is no shaft work
∂t
∂
∭ eρdV = 0 since we assume steady state
∂t
δWμ
= 0 since there is no viscous work
∂t
δQ
= 0 since no heat transfer
∂t
P
∬ (e + ) ρ(v ∙ n) dA = 0
ρ
P1 v12
P2 v22
(U1 + + + gy1 ) − (U2 + + + gy2 ) = 0
ρ
2
ρ
2
v2
v2
Since we have a horizontal pipe, gy1 = gy2 , and we are given, 21 = 22
(U1 +
P1
P2
) − (U2 + ) = 0
ρ
ρ
Recalling that dU = Cv dT,
(10 lb/in2 )(144 in2 /ft 2 )
U2 − U1 P1 − P2
∆T =
=
=
= 0.030℉
(62.4 lbm /ft 3 )(1 BTU/lbm ℉)(778.18 ft ∙ lb/BTU)
Cv
ρCv
6.7
A fan draws air from the atmosphere at an inlet
velocity of 5.0 m/s through a 0.30 meter diameter
round duct that has a smoothly rounded entrance. A
differential manometer connected to an opening in
the wall of the duct shows a vacuum pressure of 2.50
cm of water. The density of air is 1.22 kg/m3.
Assume steady flow and that there is no change in
internal energy and no shaft or viscous work. (a)
Calculate the volume rate of air flow in the duct in
cubic feet per second. (b) Calculate the horsepower output of the fan.
Solution
Begin with Bernoulli's Equation:
P1 v12
P2 v22
+ + gy1 = hL + + + gy2
ρ
2
ρ
2
Since the body is horizontal, y1 = y2, and assume no head loss
P1 v12 P2 v22
+
= +
ρ
2
ρ
2
P1 − P2 v22 − v12
=
ρ
2
Given,
P1 − P2 = 2.5 cm H2 O = 0.025 m H2 O
atm
101325 Pa
x
= 245.22 Pa = 245.22 kg m/s 2
10.33 m H2 O
atm
m 1/2
1/2
2(245.22 kg m/s 2 ) + (5.0 s )2
2(P1 − P2 ) + v12
v2 = [
] =[
] = 20.55 m/s
ρ
1.22 kg/m3
0.025 m H2 O x
Flow Rate is:
Q = Av =
πD2
π(0.3m)2
m3
(v2 ) =
(20.55 m/s) = 1.452
= 51.3 ft 3 /s
4
4
s
(b) Calculate the horsepower output of the fan.
The energy equation simplifies to:
δWs
P
= ∬ (e + ) dṁ
∂t
ρ
m
With vout = 20.05 s , vin = 0 and Pout = Patm
−
δWs
v22 − v12 P2 − P1
ṁ 2 ρQv22
−
= ṁ [U2 − U1 +
+
+ g(y2 − y1 )] = v2 =
∂t
2
ρ
2
2
2
(20.55
m/s)
kgm2
3
3
= (1.22 kg/m )(1.452 m /s) (
) = 374.0 3 = 374.0 J/s
2
s
J
sec
hp
−
hr
347.5 x 3600
x 3.725 x10−7
= 0.501 hp
s
hr
J
6.9
A liquid flows from point A to point B in a horizontal
pipe line shown in the figure. The liquid flows at a rate
of 3 ft3/s with a friction loss of 0.45 ft of flowing fluid.
If the pressure head at point B is 24 inches, what will
be the pressure head at point A?
Solution
U −U
We are given that friction loss = 0.45 ft = B g A
We want to find the pressure at A.
P1 v12
P2 v22
(U1 + + + gy1 ) = (U2 + + + gy2 )
ρ
2
ρ
2
UB − UA vB2 − vA2 PB − PA
+
+
=0
g
2g
ρg
Q
3 ft 3 /s
vA =
=
= 3.83 ft/s
π(1)2
AA
4
Q
3 ft 3 /s
v2 =
=
= 15.28 ft/s
A2 π(0.5 m)2
4
(15.28 ft/s)2 − (3.83 ft/s)2
PB − PA UB − UA vB2 − vA2
=
+
= 0.45 ft +
= 3.85 ft of fluid
ρg
g
2g
2(32.2 ft/s2 )
PA = 3.85 ft + PB = 3.85 + (24 in x
∆P
Remember, ∆P = ρgh so that h = ρg
1 ft
) = 5.85 feet of fluid
12 in
6.11
A Venturi meter with an inlet diameter of 0.6 m is designed to handle 6 m3/s of standard air. What
is the required throat diameter if this flow is to give a reading of 0.10 m of alcohol in a differential
manometer connected to the inlet and the throat? The specific gravity of alcohol may be taken as
0.8.
Solution
V̇= 6 m3/s
∆P= 0.10 m Alcohol (S.G.= 0.8) = 0.08 m H20= 785 Pa
π
A1= 4(0.6)2= 0.283 m2
V̇
6
v1=A = 0.283= 21.2 m/s
1
Energy Eqn. reduces to:
P2 −P1
ρ
v2 −v2
+ 2 2 1 + g∆y=0
P1 −P2 v22 −v21
ρ
-
2
g∆y=0
v2
V̇2 1
1
2
1
A
= 2 [A2 - A2]= 21[(A1)2 − 1]
P1 −P2
2
785
= 1.226= 640 m2/s2
ρ
(21.2)2
640=
2
A
[(A1)2 − 1]
2
A2= 0.144 m2
D2=0.428m
6.39
A client has asked you to find the pressure change in a pumping station. The outlet from the pump
is 20 feet above the inlet. A Newtonian fluid is being pumped at steady state. At the inlet to the
pump where the diameter of 6 inches, the temperature of the fluid is 80℉, the viscosity is 1.80x103
lbm/ft sec, the density is 50 lbm/ft3, the heat capacity is 0.580 BTU/lbm F, and the kinematic
viscosity is 3.60 x 10-5 ft2/sec. At the outlet to the pump where the diameter of 4 inches, the
temperature of the fluid is 100℉, the viscosity is 1.30x10-3 lbm/ft sec, the density is 49.6 lbm/ft3,
the heat capacity is 0.610 BTU/lbm F, and the kinematic viscosity is 2.62 x 10-5 ft2/sec. The flow
rate through the system is constant at 20 ft3/sec. The pump provides work to the fluid at 3.85 x 108
lbm ft2/s3 and the heat transferred is 2.32x106 BTU/hr. You may neglect viscous work in your
analysis. Under these circumstances, please calculate the pressure change between the inlet and
the outlet of the pumping station.
Solution
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
First, since the shaft work and heat must be added, let’s make them have the same units:
δWs
lbm ft 2
8
−
= 3.85x10
∂t
s3
δQ
BTU
ft
−
lb
lb
ft
hr
lbm ft 2
m
6
(778.17
) (32.174
)(
) = 16134853.46 3
= 2.32x10
∂t
hr
BTU
lbf s2 3600 sec
s
δWμ
δQ δWs
P
∂
−
= ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV +
∂t
∂t
ρ
∂t
∂t
Assuming no viscous work and steady state,
δQ δWs
P
−
= ∬ (e + ) ρ(v ∙ n) dA
∂t
∂t
ρ
Writing in terms of energy efflux,
δQ δWs
P
P
−
= (e + ) ρvA − (e + ) ρvA
∂t
∂t
ρ 2
ρ 1
Expanding the energy term, and using ρvA = ρQ,
δQ δWs
v22 − v12 P2 − P1
−
= ρQ (U2 − U1 +
+
+ g(y2 − y1 ))
(equation 1)
∂t
∂t
2
ρ
Next, let’s calculate each of the parts individually and then combine them.
U2 − U1 = Cv2 T2 − Cv1 T1
BTU
) (100℉)
= [(0.61
lbm ℉
BTU
ft − lb
lbm ft
ft 2
) (80℉)] (778.17
) (32.174
)
− (0.58
=
365537.89
lbm ℉
BTU
lbf s2
s2
2
3
2
3
20ft /s
20ft /s
Q 2
Q 2 ( (4/12)2 ) − ( (0.5)2 )
π
π
(229.18 ft/s)2 − (101.86 ft/s)2
v22 − v12 (A2 ) − (A1 )
4
4
=
=
=
2
2
2
2
ft 2
= 21074.0 2
s
ft 2
2
g(y2 − y1 ) = (32.2 ft/s )(20 ft) = 644 2
s
Plug these values back into equation 1,
lbm ft 2
lbm ft 2
16134853.46 3 + 3.85x108
s
s3
lbm
ft 2
ft 2
P2 − P1
ft 2
3
(20ft
)
/s)
[365537.89
+
21074.0
+
+
644
]
ft 3
s2
s 2 (50 lbm )
s2
ft 3
Note that the same value of ρ was used in both places in the above equation, it could have been
left out since it cancels.
= (50
lbm ft 2
lbm
ft 2
P2 − P1
=
1000
[387255.89
+
]
3
2
lbm
s
s
s
(50 3 )
ft
2
P2 − P1
ft
= 13878.96 2
lb
s
(50 m
)
ft 3
ft 2
lb
13878.96 2 (50 m
)
s
ft 3 = 21568.6 lbf
P2 − P1 =
lb ft
ft 2
32.174 m 2
lbf s
40113485305
Chapter 7
Show/Hide Problems
7.6
Calculate the viscosity of oxygen at 350 K and compare with the value given in Appendix I.
Solution
Begin with the equation (7-10) for viscosity of a pure gas,
𝜇 = 2.6693𝑥10−6
√𝑀𝑇
𝜎 2 Ω𝜇
Appendix K, Table K.2 has for O2 at 350 𝐾:
𝜖𝐴⁄
𝜅 = 113 𝐾 𝑎𝑛𝑑 𝜎 = 3.433 Å
𝜅𝑇
𝑇
350
=𝜖 =
= 3.097
⁄𝜅 113
𝜖
Go to Table K.1 on and find the value for Ω𝜇 .
Ω𝜇 = 1.030
The molecular weight of O2 is 16+16 = 32.
Plug in the data,
√(32)(350)
√𝑀𝑇
𝜇 = 2.6693𝑥10−6 2 = 2.6693𝑥10−6
= 2.327𝑥10−5 Pa − s
(3.433)2 (1.030)
𝜎 Ω𝜇
For comparison, Appendix I on page 736 gives viscosity results for Oxygen at 350K as 2.3176 x
10-5 Pa-s, which is in reasonable agreement.
7.10
Repeat the preceding problem for air.
Solution
For Air:
@ 140oF
@ 32oF
μ= 1.34x10-5 LBm/s*ft.
μ= 1.15x10-5 LBm/s*ft.
For V̇~1/μ
V̇140
1.15
=
= 0.852
V̇32
1.34
Percent change =
1.15−1.34
1.34
= 0.852-1= -0.148 = -14.8 %
7.20
A Newtonian oil with a density of 60 lbm/ft3, viscosity of 0.206 x 10-3 lbm/ft-sec and kinematic
viscosity of 0.342 x 10-5 ft2/sec undergoes steady shear between a horizontal fixed lower plate
and a moving horizontal upper plate. The upper plate is moving with a velocity of 3 feet per
second. The distance between the plates is 0.03 inches, and the area of the upper plate in
contact with the fluid is 0.1 ft2. Assume incompressible, isothermal, inviscid, frictionless flow.
MOVING AT 3 ft/sec
0.03 inches
(a) What is shear stress exerted on the fluid under these conditions?
(b) What is the force of the upper plate on the fluid?
Solution
(a) What is shear stress exerted on the fluid under these conditions?
𝜏𝑦𝑥 = 𝜇
𝑑𝑣𝑥
𝑑𝑦
0.03 𝑖𝑛𝑐ℎ𝑒𝑠
3 𝑓𝑡/𝑠
𝜏𝑦𝑥 ∫
𝑑𝑦 = 𝜇 ∫
0
0
𝑑𝑣𝑥
𝜏𝑦𝑥 (0.03 𝑖𝑛𝑐ℎ𝑒𝑠) = 𝜇(3 𝑓𝑡/𝑠)
Solving for shear stress,
𝜏𝑦𝑥 =
(0.206 𝑥10−3 𝑙𝑏𝑚 /𝑓𝑡 ∙ 𝑠)(3 𝑓𝑡/𝑠)
𝑙𝑏𝑓
= 7.68𝑥10−3 2
𝑓𝑡
𝑙𝑏 𝑓𝑡
𝑓𝑡
(0.03 𝑖𝑛) (
) (32.174 𝑚 2 )
12 𝑖𝑛
𝑙𝑏𝑓 𝑠
(b) What is the force of the upper plate on the fluid?
𝜏𝑦𝑥 =
𝐹 = 𝜏𝑦𝑥 𝐴 = (7.68𝑥10−3
𝐹
𝐴
𝑙𝑏𝑓
) (0.1 𝑓𝑡 2 ) = 7.68𝑥10−4 𝑙𝑏𝑓
𝑓𝑡 2
FIXED
Chapter 8
Show/Hide Problems
8.1
Express equation 8-9 in terms of the flow rate and the pipe diameter. If the pipe diameter is doubled
at constant pressure drop, what percentage change will occur in the flow rate?
Solution
Equation 8-9:
−
Substitute: 𝑄 = 𝑣𝐴 𝑎𝑛𝑑 𝐴 =
𝜋𝐷 2
dP 32μvavg
=
dx
D2
4
−
dP 128𝜇𝑄
=
dx
𝜋𝐷4
Which is an equation in terms of flow rate and diameter.
Rearrange and solve for Q:
dP 𝜋𝐷4
𝑄=−
dx 128𝜇
We are given that 𝐷2 = 2𝐷1 , and it follows that 𝑄2 = 16𝑄1
The percentage change will be:
𝑄2 − 𝑄1
16𝑄1 − 𝑄1
𝑥100 =
𝑥100 = 15𝑥100 = 1500%
𝑄1
𝑄1
So the flow rate increases 1500% by doubling the pipe diameter.
8.9
Determine the velocity profile for fluid flowing between two parallel plates separate by a
distance 2h. The pressure drop is constant.
Solution
dp
d
(Hint: Begin with the following equation: dx = dy τyx )
This analysis is similar to what is done in the text in section 8.1 except in rectangular
coordinates.
dp
d
=
τ
dx dy yx
Substitute Newton's Law of viscosity,
dp
d2 vx
=μ 2
dx
dy
Rearrange,
d2 vx 1 dp
=
dy 2
μ dx
Integrate,
dvx 1 dp
=
y + C1
dy
μ dx
integrate again,
vx =
1 dp 2
y + C1 y + C2
2μ dx
Boundary conditions for this problem,
1. y = +h at vx = 0 (No Slip Boundary Condition)
2. y = −h at vx = 0 (No Slip Boundary Condition)
Substitute in the first boundary condition and determine that C1 = 0.
1 dp
Next, substitute in the second boundary condition to determine C2 = − 2μ dx ℎ2
Thus,
vx =
1 dp 2
1 dp 2
y −
h
2μ dx
2μ dx
8.18
A Newtonian fluid in continuous, incompressible, laminar flow is moving steadily through a very
long 700 meter, horizontal pipe. The inside radius is 0.25 meter for the entire length and the
pressure drop across the pipe is 1000 Pa. The average velocity of the fluid is 0.5 m/s. What is
the viscosity of this fluid?
Solution
𝑑𝑃 8𝜇𝑣𝑎𝑣𝑔
=
𝑑𝑥
𝑅2
∆𝑃 8𝜇𝑣𝑎𝑣𝑔
=
𝐿
𝑅2
(1000 𝑃𝑎)(0.25 𝑚)2
∆𝑃𝑅 2
𝜇=
=
= 0.022 𝑃𝑎 ∙ 𝑠
8𝑣𝑎𝑣𝑔 𝐿 (8)(0.5 𝑚/𝑠)(700 𝑚)
−
8.19
Benzene, which is an incompressible Newtonian Fluid, flows steadily and continuously at 150oF
through a 3000 foot pipe with a constant diameter of 4 inches with a volumetric flow rate of 3.5
ft3/s. Assuming fully developed, laminar flow and that the no-slip boundary condition applies,
calculate the change in pressure across this pipe system.
Solution
dP 32μvavg
=
dx
D2
∆P 32μvavg
=
L
D2
−
Calculate vavg ,
vavg =
Q 3.5 ft 3 /s
=
= 40.11 ft/s
A
4 2
π (12)
4
Calculate viscosity,
32μvavg L 32(2.45x10−4 lbm /ft s)(40.11 ft/s)(3000 ft)
1
∆P =
=
2
2
lb ft
D
4
32.174 m 2
(12) ft 2
lbf s
2
= 263.89 lbf /ft
Chapter 9
Show/Hide Problems
9.2
∂
∂
∂
In Cartesian coordinates, show that vx ∂x + vy ∂y + vz ∂z may be written as (v ∙ ∇). What is the
physical meaning of the term (v ∙ ∇)?
Solution
Start with the equations for ∇ and v,
v = vx ex + vy ey + vz ez
∇=
v ∙ ∇= vx
∂
∂
∂
ex + e y + ez
∂x
∂y
∂z
∂
∂
∂
(ex ∙ ex ) + vy (ey ∙ ey ) + vz (ez ∙ ez )
∂x
∂y
∂z
Thus,
v ∙ ∇= vx
∂
∂
∂
+ vy + vz
∂x
∂y
∂z
Physical Meaning: v ∙ ∇ is the rate of change due to motion.
9.4
𝐷𝑣
Find an equation for 𝐷𝑡 in polar coordinates by taking the derivative of the velocity. This can be
done by using
𝑣 = 𝑣𝑟 (𝑟, 𝜃, 𝑡)𝑒𝑟 + 𝑣𝜃 (𝑟, 𝜃, 𝑡)𝑒𝜃
And
𝑣 = 𝑣𝑟 𝑒𝑟 + 𝑣𝜃 𝑒𝜃
Remember that unit vectors have derivatives. It might be helpful to consult Appendix A of the
text.
Solution
𝜕𝑣
𝜕𝑣
𝜕𝑣
𝑑𝑟 +
𝑑𝜃 +
𝑑𝑡
𝜕𝑟
𝜕𝜃
𝜕𝑡
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)
𝑑𝑣 𝜕𝑣 𝑑𝑟 𝜕𝑣 𝑑𝜃 𝜕𝑣 𝑑𝑡
=
+
+
𝑑𝑡 𝜕𝑟 𝑑𝑡 𝜕𝜃 𝑑𝑡 𝜕𝑡 𝑑𝑡
Partial differentiating 𝑣 by 𝜕𝑟:
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)
𝑑𝑣 =
Divide by dt
𝜕𝑒𝜃
𝜕𝑒𝑟
𝜕𝑣 𝜕𝑣𝑟
𝜕𝑣𝜃
=
𝑒𝑟 +
𝑒𝜃 + 𝑣𝑟
+ 𝑣𝜃
𝜕𝑟
𝜕𝑟
𝜕𝑟
𝜕𝑟
𝜕𝑟
Partial differentiating 𝑣 by 𝜕𝜃:
𝜕𝑒𝜃
𝜕𝑒𝑟
𝜕𝑣 𝜕𝑣𝑟
𝜕𝑣𝜃
=
𝑒𝑟 +
𝑒𝜃 + 𝑣𝑟
+ 𝑣𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4)
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 5)
Realizing that 𝑒𝑟 and 𝑒𝜃 are related to 𝑒𝑥 and 𝑒𝑦 by:
𝑒𝑟 = 𝑒𝑥 cos 𝜃 + 𝑒𝑦 sin 𝜃
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 6)
𝑒𝜃 = −𝑒𝑥 sin 𝜃 + 𝑒𝑦 cos 𝜃
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 7)
Differentiate (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 6) with respect to 𝑟 𝑎𝑛𝑑 𝜃 (see Appendix A of text):
𝜕𝑒𝑟
𝜕𝑟
=
𝜕𝑒𝑟 𝜕𝜃
𝜕𝜃 𝜕𝑟
𝜕𝑒𝑟
𝜕𝜃
= 𝑒𝜃
= 𝑒𝜃
𝜕𝜃
=0
𝜕𝑟
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 8)
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 9)
Differentiate (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 7) with respect to 𝑟 𝑎𝑛𝑑 𝜃 (see Appendix A of text):
𝜕𝑒𝜃
𝜕𝑟
=
𝜕𝑒𝜃 𝜕𝜃
𝜕𝑒𝜃
𝜕𝜃
𝜕𝜃 𝜕𝑟
= −𝑒𝑟
=0
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 10)
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 11)
Now, substitute the values from equations 8, 9, 10, and 11 into 4 and 5:
𝜕𝑒𝜃
𝜕𝑒𝑟
𝜕𝑣 𝜕𝑣𝑟
𝜕𝑣𝜃
=
𝑒𝑟 +
𝑒𝜃 + 𝑣𝑟
+ 𝑣𝜃
𝜕𝑟
𝜕𝑟
𝜕𝑟
𝜕𝑟
𝜕𝑟
So,
𝜕𝑣 𝜕𝑣𝑟
𝜕𝑣𝜃
=
𝑒𝑟 +
𝑒
𝜕𝑟
𝜕𝑟
𝜕𝑟 𝜃
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 12)
𝜕𝑒𝜃
𝜕𝑒𝑟
𝜕𝑣 𝜕𝑣𝑟
𝜕𝑣𝜃
=
𝑒𝑟 +
𝑒𝜃 + 𝑣𝑟
+ 𝑣𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
Becomes,
𝜕𝑣 𝜕𝑣𝑟
𝜕𝑣𝜃
=
𝑒𝑟 +
𝑒 + 𝑣𝑟 𝑒𝜃 − 𝑣𝜃 𝑒𝑟
𝜕𝜃
𝜕𝜃
𝜕𝜃 𝜃
Which simplifies to
𝜕𝑣
𝜕𝑣𝑟
𝜕𝑣𝜃
=(
− 𝑣𝜃 ) 𝑒𝑟 + (
+ 𝑣𝑟 ) 𝑒𝜃
𝜕𝜃
𝜕𝜃
𝜕𝜃
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 13)
From the hint given in the problem statement:
𝑑𝑟
= 𝑣𝑟
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 14)
𝑑𝑡
𝑑𝜃 𝑣𝜃
=
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 15)
𝑑𝑡
𝑟
From equation 3,
𝑑𝑣 𝜕𝑣 𝑑𝑟 𝜕𝑣 𝑑𝜃 𝜕𝑣
=
+
+
𝑑𝑡 𝜕𝑟 𝑑𝑡 𝜕𝜃 𝑑𝑡 𝜕𝑡
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)
Substitute in equations 14, 15, 12 and 13 and rearrange:
𝐷𝑣 𝑑𝑣
𝜕𝑣𝑟 𝑣𝜃 𝜕𝑣𝑟 𝑣𝜃2
𝜕𝑣𝜃 𝑣𝜃 𝜕𝑣𝜃 𝑣𝜃 𝑣𝑟
) 𝑒𝜃
=
+ (𝑣𝑟
+
− ) 𝑒𝑟 + (𝑣𝑟
+
+
𝐷𝑡
𝑑𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝑟
𝜕𝑟
𝑟 𝜕𝜃
𝑟
9.5
For flow at very low speeds and with large viscosity (the so-called creeping flows) such as occur
in lubrication, it is possible to delete the inertia terms, Dv/Dt from the Navier-Stokes equation.
For flows at high velocity and small viscosity, it is not proper to delete the viscous term 𝜈∇2 𝐯.
Explain this.
Solution
Navier-Stokes Eqn.- Incompressible form:
Dυ⃑
Dt
∇P
= ⃑g - ρ +ν∇2 υ⃑
dυ⃑
(a) For υ⃑ small- all terms involving υ⃑ (~ dt & ν∇2 ⃑υ) are small relative to other terms.
(b) For ν small & υ⃑ large the product νυ
⃑ cannot be considered small relative to other terms
9.6
Using the Navier-Stokes equations and the continuity equation, obtain an expression for the
velocity profile between two flat, parallel plates.
Solution
First we sketch a picture of the problem.
L
flow direction
We begin with the Navier-Stokes equation in rectangular coordinates for the x-direction.
∂vx
∂vx
∂vx
∂vx
∂P
∂2 vx ∂2 vx ∂2 vx
)=−
ρ(
+ vx
+ vy
+ vz
+ ρg x + μ ( 2 +
+ 2)
∂t
∂x
∂y
∂z
∂x
∂x
∂y 2
∂z
Evaluating the equation term by term:
∂vx
= 0 since the flow is steady
∂t
∂vx ∂2 vx
vx
=
= 0 since the flow is fully developed
∂x
∂x 2
∂vx
vy
= 0 since there is no y direction velocity
∂y
∂vx
vz
= 0 since there is no z direction velocity
∂z
ρg x = 0 since there is no gravity in the x direction
∂2 vx
= 0 since we have two dimensional flow
∂z 2
Thus,
∂P
∂2 vx
0=−
+μ( 2 )
∂x
∂y
Rearrange,
∂2 vx 1 ∂P
=
∂y 2
μ ∂x
Integrate,
∂vx 1 ∂P
=
y + C1
∂y
μ ∂x
Integrate again,
1 ∂P 2
vx =
y + C1 y + C2
2μ ∂x
Boundary conditions:
BC#1: when v=0, y=0 (No-Slip boundary condition)
BC#2: when v=0, y=L (No-Slip boundary condition)
From BC#1: C2=0.
L ∂P
From BC#2: C1 = − 2μ ∂x
Thus,
vx =
1 ∂P 2
1 ∂P
1 ∂P 2
(y − Ly)
y −
Ly =
2μ ∂x
2μ ∂x
2μ ∂x
9.21
Beginning with the appropriate form of the Navier-Stokes equations, develop an equation in the
appropriate coordinate system to describe the velocity of a fluid that is flowing in the annular
space as shown in the figure. The fluid is Newtonian, and is flowing in steady, incompressible,
fully developed, laminar flow through an infinitely long vertical round pipe annulus of inner radius
RI and outer radius RO. The inner cylinder (shown in the figure as a dotted line) is solid and the
fluid flows between the inner and outer walls as shown in the figure. The center cylinder moves
downward in the same direction as the fluid with a velocity V0. The outside wall of the annulus is
stationary. In developing your equation, please specifically state the reason for eliminating any
terms in the original equation.
Fluid flow around center section
RO
RI
r
V0
Solution
Need to begin with z-direction Navier-Stokes equation in cylindrical coordinates:
𝜌(
𝜕𝑣𝑧
𝜕𝑣𝑧 𝑣𝜃 𝜕𝑣𝑧
𝜕𝑣𝑧
+ 𝑣𝑟
+
+ 𝑣𝑧
)
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝜕𝑃
1𝜕
𝜕𝑣𝑧
1 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧
=−
+ 𝜌𝑔𝑧 + 𝜇 (
+
(𝑟
)+ 2
)
𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝜕𝜃 2
𝜕𝑧 2
Applying the assumptions given in the problem statement:
0=−
𝜕𝑃
1𝜕
𝜕𝑣𝑧
− 𝜌𝑔𝑧 + 𝜇 (
(𝑟
))
𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
1𝜕
𝜕𝑣𝑧
𝜕𝑃
𝜇(
+ 𝜌𝑔𝑧
(𝑟
)) =
𝑟 𝜕𝑟
𝜕𝑟
𝜕𝑧
𝜕
𝜕𝑣𝑧
𝑟 𝜕𝑃
(𝑟
) = ( + 𝜌𝑔𝑧 )
𝜕𝑟
𝜕𝑟
𝜇 𝜕𝑧
2
𝜕𝑣𝑧 𝑟 𝜕𝑃
𝑟
= ( + 𝜌𝑔𝑧 ) + 𝐶1
𝜕𝑟
𝜇 𝜕𝑧
𝜕𝑣𝑧
𝑟 𝜕𝑃
𝐶1
=
( + 𝜌𝑔𝑧 ) +
𝜕𝑟
2𝜇 𝜕𝑧
𝑟
2
𝑟 𝜕𝑃
𝑣𝑧 =
( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑟 + 𝐶2
4𝜇 𝜕𝑧
Boundary Conditions:
1. 𝑣𝑧 = 0 𝑎𝑡 𝑟 = 𝑅0
2. 𝑣𝑧 = −𝑣0 𝑎𝑡 𝑟 = 𝑅𝐼
Apply Boundary Condition #1:
0=
𝑅0 2 𝜕𝑃
( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅0 + 𝐶2
4𝜇 𝜕𝑧
Apply Boundary Condition #2:
𝑅𝐼 2 𝜕𝑃
−𝑣0 =
(
+ 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅𝐼 + 𝐶2
4𝜇 𝜕𝑧
Solve for 𝐶1 ,
𝑅0 2 𝜕𝑃
𝑅𝐼 2 𝜕𝑃
( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅0 =
( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅𝐼 + 𝑣0
4𝜇 𝜕𝑧
4𝜇 𝜕𝑧
𝐶1 =
𝜕𝑃
+ 𝜌𝑔𝑧 ) + 𝑣0
𝜕𝑧
4𝜇(ln 𝑅0 − ln 𝑅𝐼 )
𝑅𝐼 2 − 𝑅0 2 (
Solve for 𝐶2 by plugging back into equation for Boundary Condition #1 (if you plug
into the Boundary Condition #2 equation your result will be slightly different and
no less correct),
𝜕𝑃
𝑅𝐼 2 − 𝑅0 2 (
+ 𝜌𝑔𝑧 ) + 𝑣0
𝑅0 2 𝜕𝑃
𝜕𝑧
0=
ln 𝑅0 + 𝐶2
( + 𝜌𝑔𝑧 ) +
4𝜇 𝜕𝑧
4𝜇(ln 𝑅0 − ln 𝑅𝐼 )
𝜕𝑃
𝑅𝐼 2 − 𝑅0 2 (
+ 𝜌𝑔𝑧 ) + 𝑣0
𝑅0 2 𝜕𝑃
𝜕𝑧
𝐶2 = −
ln 𝑅0
( + 𝜌𝑔𝑧 ) −
4𝜇 𝜕𝑧
4𝜇(ln 𝑅0 − ln 𝑅𝐼 )
Thus,
𝜕𝑃
𝑅𝐼 2 − 𝑅0 2 (
+ 𝜌𝑔𝑧 ) + 𝑣0
𝑟 2 𝜕𝑃
𝑅0 2 𝜕𝑃
𝜕𝑧
𝑣𝑧 =
ln 𝑟 −
( + 𝜌𝑔𝑧 ) +
( + 𝜌𝑔𝑧 )
4𝜇 𝜕𝑧
4𝜇(ln 𝑅0 − ln 𝑅𝐼 )
4𝜇 𝜕𝑧
𝜕𝑃
𝑅𝐼 2 − 𝑅0 2 (
+ 𝜌𝑔𝑧 ) + 𝑣0
𝜕𝑧
−
ln 𝑅0
4𝜇(ln 𝑅0 − ln 𝑅𝐼 )
Chapter 10
Show/Hide Problems
10.3
Find the stream function for a flow with a uniform free-stream velocity v∞ . The free-stream
velocity intersects the x-axis at an angle α.
Solution
dψ = −vy dx + vx dy = (−v∞ sinα)dx + (v∞ cosα)dy
ψ = (−v∞ sinα)x + (v∞ cosα)y + ψ0
10.5
5
The velocity potential for a given two-dimensional flow field is ϕ = 3 x 3 − 5xy 2 . Show that the
continuity equation is satisfied and determine the corresponding stream function.
Solution
5 3
x − 5xy 2
3
since v = ∇ψ, continuity can be expressed as ∇ ∙ v
⃑ = 0 or ∇2 ϕ = 0
ϕ=
using ∇2 ϕ = 0
∂2 ϕ ∂2 ϕ
+
=0
∂x 2 ∂y 2
10x − 10x = 0
vx =
∂ϕ ∂ψ
=
= 5x 2
∂x
∂y
or ψ = 5x 2 y
∂ϕ
∂ψ
~ ∂y = vy = − ∂x ~ Check
10.17
Sketch the streamlines and potential lines of the flow due to a line source at (a,0) plus an equivalent
sink at (-a,0)
Solution
10.23
The stream function is given by
ψ = 6x 2 − 6y 2
Determine whether this flow is rotational or irrotational.
Solution
To be irrotational, the flow must satisfy the equation, ∇ x v = 0 and as a result, Equation 10.1
must be equal to zero.
1 𝜕𝑣𝑦 𝜕𝑣𝑥
𝑤𝑧 = (
−
)=0
2 𝜕𝑥
𝜕𝑦
For the equation of the stream function we solve for 𝑣𝑥 and 𝑣𝑦 using the definition of the
stream function,
𝜕𝜓
𝑣𝑥 =
= −12𝑦
𝜕𝑦
𝑣𝑦 = −
And then,
𝜕𝜓
= −12𝑥
𝜕𝑥
𝜕𝑣𝑥
= −12
𝜕𝑦
𝜕𝑣𝑦
= −12
𝜕𝑥
1
𝑤𝑧 = (−12 − (−12)) = 0
2
So the condition of irrotational flow is satisfied.
Chapter 11
Show/Hide Problems
11.1
The power output of a hydraulic turbine depends on the diameter D of the turbine, the density
 of water, the height H of water surface above the turbine, the gravitational acceleration g, the
angular velocity  of the turbine wheel, the discharge Q of water through the turbine, and the
efficiency  of the turbine. By dimensional analysis, generate a set of appropriate
dimensionless groups.
Solution
D = diameter
 = density
H = height
g = gravity
 = angular velocity
Q = flow rate
 = efficiency
P = power
M
L
t
D
0
1
0

1
-3
0
L
ML-3
L
Lt-2
t-1
L3t-1
unitless
ML2t-3
H
0
1
0
g
0
1
-2

0
0
-1

0
0
0
Q
0
3
-1
P
1
2
-3
i = n-r = 8 - 3 = 5
core group = , D, 
a
1=aDbcH
H
 1=  
D
1=0D-10H →
a
2=aDbcQ
c
M 
1
MoLoto = 1 =  3  (L )b   L
L 
t 
M: 0 = a+0
L: 0 = -3a+b+1
T: 0 = -c+0
Thus: a= 0, b= -1, c= 0
c
3
M 
1 L
MoLoto = 1 =  3  (L )b  
L 
t  t
M: 0 = a
L: 0 = b+3
T: 0 = -c+1
Thus: a= 0, b= -3, c= -1
 Q 
 1=  3 
D 
2=0D-10H →
3=aDbcg
M 
MoLoto = 1 =  3 
a
c
(L )   L2
L 
t  t
M: 0 = a
L: 0 = -3a+b+1
T: 0 = c-2
Thus: a= 0, b= -1, c= -2
 g 
 1= 
2 
 D 
3=0D-10g →
a
4=aDbcP
b1
c
2
M 
 1  ML
MoLoto = 1 =  3  (L )b   3
L 
t  t
M: 0 = a+1
L: 0 = -3a+b+2
T: 0 = -c-3
Thus: a= -1, b= -5, c= -3
4=0D-10P →
5=  (by inspection)
 P 

1= 
5 3
 D  
11.3
The pressure rise across a pump P (this term is proportional to the head developed by the
pump) may be considered to be affected by the fluid density , the angular velocity , the
impeller diameter D, the volumetric rate of flow Q, and the fluid viscosity . Find the pertinent
dimensionless groups, choosing them so that P, Q and  each appear in one group only.
Solution
Variable
ΔP
ρ
ω
D
Q
μ
Dimensions
M/Lt2
M/L3
1/t
L
L3/t
M/Lt
i = 6 − 3 = 3 , Choose core as ρ, D, ω
ΔP
π1 = ρa Db ωc ΔP ∴ = 2 2
ρD ω
Q
π2 = ρd De ωf Q ∴ = 3
D ω
μ
g h i
π3 = ρ D ω μ ∴ = 2
ρD ω
11.7
The functional frequency n of a stretched string is a function of the string length L, it’s diameter
D, the mass density ρ, and the applied tensile force T. Suggest a set of dimensionless parameters
relating these variables.
Solution
Variable
n
d
L
T
ρ
Dimensions
1/t
L
L
ML/t2
M/L3
i = 5 − 3 = 2 , Choose core as d, n, ρ
L
π1 = da nb ρc L ∴ =
d
T
π2 = dd ne ρf T ∴ = 2 4
n Lρ
11.8
The power P required to run a compressor varies with compressor diameter D, angular velocity
ω, volume flow rate Q, fluid density ρ, and fluid viscosity μ. Develop a relation between these
variable by dimensional analysis, where fluid viscosity and angular velocity appear in only one
dimensionless parameter.
Solution
Variable
Q
d
P
ω
ρ
μ
Dimensions
L3/t
L
ML2/t3
1/t
M/L3
M/Lt
i = 6 − 3 = 3 , Choose core as d, Q, ρ
d3 ω
π1 = da Qb ρc ω ∴ =
Q
dμ
π2 = dd Qe ρf μ ∴ =
ρQ
d4 P
g h i
π3 = d Q ρ P ∴ = 3
ρQ
π3 = f(π1 , π2 )
Chapter 12
Show/Hide Problems
12.1
If Reynolds’s experiment was performed with a 38-mm ID pipe, what flow velocity would occur
at transition?
Solution
At transiton Red =
Dv
= 2300
ν
H2 O@20℃: ν = 0.995x10−6
v=
m2
s
2300(0.995x10−6 )
m
cm
= 0.060 = 6
0.038m
s
s
12.2
Modern subsonic aircraft have been refined to such an extent that 75% of the parasite drag
(portion of total aircraft drag not directly associated with producing lift) can be attribute to
friction along the external surfaces. For a typical subsonic jet, the parasite drag coefficient based
on wing area is 0.011. Determine the friction drag on such an aircraft.
(a) at 500 mph at 35,000 ft
(b) at 200 mph at sea level
Solution
v2
fD = CD Aρ
2
For 35000 ft: ρ = 0.0237
lbm
lbm
, for S. L. , ρ = 0.0766 3
3
ft
ft
a) @35000 ft, 500mph = 733ft/s
0.011(2400)(0.0237)(733)2
fDp =
= 5220 lb𝑓
2(32.2)
b)@ Sea Level, 200mph = 293ft/s
0.011(2400)(0.0766)(293)2
fD =
= 2700 lb𝑓
2(32.2)
12.3
Consider the flow of air at 60oF and 30 m/s along a flat plate. At what distance from the leading
edge will transition flow occur?
Solution
Transition over a flat plate begins at Re=2x105
𝑅𝑒 =
𝐷𝜌v
𝜇
From Appendix I, Air at 60oF:
kinematic viscosity =  = 0.159x10-3 ft2/s
Density =  = 0.0764 lbm/ft3
viscosity =  = 1.21x10-5 lbm/ft-s
Transition occurs,
𝑅𝑒𝜇 (2𝑥105 )(1.21𝑥10−5 𝑙𝑏𝑚 /𝑓𝑡 ∙ 𝑠)
𝑚2
𝑥=
=
(
) = 0.1 𝑚
(0.0765 𝑙𝑏𝑚 /𝑓𝑡 3 )(30 𝑚/𝑠)
𝜌v
10.76 𝑓𝑡 2
12.9
For what wind velocities will a 12.7-mm-diameter cable be in the unsteady wake region of Figure
12.2?
Solution
In the unsteady wake region, 1 < 𝑅𝑒 < 103 , Re =
@ 20℃, ν = 1.505x10−5
@Re = 1 =
m2
s
0.0127v
m
∴
v
=
0.00119
1.505x10−5
s
@Re = 103 =
0.0127v
m
∴ v = 1.185
−5
1.505x10
s
These are the lower & 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑𝑠 𝑓𝑜𝑟 𝑣
Dv
ν
12.10
Estimate the drag force on a 3-ft radio antenna with an average diameter of 0.2 in at a speed of 40
mph.
Solution
ft 2
lbm
For Air @ 80℉, ν = 0.169x10−3 , ρ = 0.0735 3
s
ft
0.2
( 12 ) (88)
Re =
= 8680
0.169x10−3
From 12.2, CD ≅ 1.2
FD = CD Aρ
v2
0.2
0.0735 882
)(
= 1.2 ( ) (3) (
) = 0.530 lbf
2
12
32.2
2
12.11
A 2007 Toyota Prius has a drag coefficient of 0.26 at road speeds, using a reference area of 2.33
m2. Determine the horsepower required to overcome drag at a velocity of 30 m/s. Compare this
figure with the case of head and tail winds of 6 m/s.
Solution
100mph = 44.7
m
s
v 2 0.21(2.33)(1.2048)(44.7)2
FL = CL Aρ =
= 589 N
2
2
Chapter 13
Show/Hide Problems
13.1
An oil with a kinematic viscosity of 0.08 x 10-3 ft2/s, viscosity of 0.00456 lbm/ft-s and a density of
57 lbm/ft3 flows through a horizontal tube 0.24 inches in diameter at the rate of 10 gallons per hour.
Determine the pressure drop in 50 feet of tube.
Solution
Q = 10
gal
ft 3
min
hr
x
x
x
= 3.714x10−4 ft 3 /s
hr 7.480 gal 60 sec 60 min
v=
Q
3.714x10−4 ft 3 /s
=
= 1.18 ft/s
A
ft 2
π (0.24in x 12 in)
4
Next, we need to determine the change in pressure in a tube with a length of 50 ft. We can do this
by equating the head loss equations,
L v2
hL = 2ff
Dg
∆P
hL =
ρg
Equating these gives,
∆P
L v2
= 2ff
ρg
Dg
Solving for ∆P,
(57 lbm /ft 3 )
L
50 ft
(1.18 ft/s)2 = 12334ff
∆P = 2ρff v2 = 2
ff
2
(32.174 lbm ft/lbf s ) 0.24/12 ft
D
Next, need to find ff , and to do this we must determine whether the flow is laminar or turbulent.
Re =
ρvD (57 lbm /ft 3 )(1.18 ft/s)(0.02 ft)
=
= 295
(0.00456 lbm /ft ∙ s)
μ
16
ff =
= 0.054
Re
Thus,
∆P = 12334ff = 12334(0.054) = 666.0 lbf /ft 2
13.2
A lubricating line carrying oil has an inside diameter of 0.1 inches and is 30 inches long. If the
pressure drop is 15 psi, determine the flow rate of the oil if the oil has a kinematic viscosity of 0.08
x10-3 ft2/s and a density of 57 lbm/ft3. (Hint, you many assume laminar flow to begin the problem
but you must confirm this once you have the velocity calculated.)
Solution
hL =
∆P 32μvavg L
=
ρg
ρgD2
Solve for vavg ,
vavg =
∆PρgD2 ∆P D2 ∆P D2
=
=
32μLρg
32μL 32νρL
Where ν is the kinematic viscosity.
2
vavg =
∆P D2
=
32νρL
lb
144 in2
ft
) (0.1 in (
(15 2f ) (
12 in))
ft
in
32 (57
lbm
ft 2
ft
) (0.08x10−3 s ) (30 in (12 in))
3
ft
(32.174
lbm ft
) = 13.23 ft/s
lbf s2
Before we proceed further, we need to confirm that the flow is laminar.
ft
vD (13.23 ft/s) [0.1 in (12 in)]
Re =
=
= 1378 (so we confirm laminar flow)
ft 2
ν
0.08x10−3 s
2
ft
π (0.1 in (12 in))
πD2
Q = Av = (
) (13.23 ft/s) =
4
(13.23 ft/s) = 7.2x10−4
4
(
)
ft 3
s
13.3
The pressure drop in a section of a pipe is determined from tests with water. A pressure drop of
13 psi is obtained at a flow rate of 28.3 lbm/s. If the flow is fully turbulent, what will be the
pressure drop when liquid oxygen (density = 70 lbm/ft3) flows through the pipe at the rate of 35
lbm/s.
Solution
L 2
v
D
e
For a specified pipe: ΔP~fF v 2 , and if fully turbulent, fF ~ only ∴ ΔP = v 2
D
lbm
For H2 O: ΔP = 13 PSI for ṁ = 28.3
s
ΔP = 2fF
For Lox: ρ = 70
lbm
lbm
and
m
̇
=
35
ft 3
s
m 2
(ρA)
ΔPlox
35 2 62.4 2
Lox
(
) (
) = 1.21
=
=
m 2
ΔPH2O
70
28.3
(ρA)
H O
2
For Lox: ΔP = 13(1.21) = 158 PSI
13.8
Water at a rate of 118ft3/min flows through a smooth horizontal tube 250 ft long. The pressure
drop is 4.55 psi. Determine the tube diameter.
Solution
P2 − P1 v22 − v12
0=
+
+ gΔy + hL
ρ
2
ΔP
4.55(144)(32.2)
ft 2
−
=−
= −338.1 ,
ρ
62.4
s
Δv 2 = 0,
gΔy = 0,
L
hL = 2ff v 2
D
118
2.50
v=
=
,
π
D2
(60) ( ) (D2 )
4
250 2.5 2 3125
( ) = 5 ff
hL = 2ff
D D2
D
3125
Governing Eqn is − 338.1 + 5 ff = 0 ∴ ff = 0.1082D5 , other constraint is ff (Re)
D
Dv
118
2.052x105
Re =
=
=
πD
ν
D
60 ( 4 ) (1.22x10−5 )
Trial & Error − Assume ff = 0.004
0.004
] = 0.517ft,
0.1082
2.052x10−5
Re =
= 3.97x10−5 ,
0.517
∴ ff = 0.00325
using this value,
D = 0.496 ft,
Re = 4.137x10−5 ,
ff = 0.0031
→ D = 0.491 ft (5.9 in)
D=[
13.12
A galvanized rectangular duct 8 in. square is 25 ft long and carries 600 ft 3/min of standard air.
Determine the pressure drop in inches of water.
Solution
Rectangular duct: 8”x8”x25 ft
v̇ = 600ft 3 /m Standard Air
4(8)(8)
Deq =
= 8in
4
8
600/60
v=
= 22.5 ft/s
(8)8/144
Energy equation reduces to:
∆P
L
= 2ff v 2
ρ
D
8
( ) (25)
Re = 32
= 9.59x104
1.56x10−5
e 0.005
=
= 0.00075
D 8/12
From Figure 13.1: ff = 0.0054
∆P
25
) (22.5)2 = 205 ft 2 /s2
= 2(0.0054) (
ρ
8/12
∆P =
205(0.0766)
0.0766
= 0.4876 psf = 6.366 ft air = 76.4 in air = 76.4
= 0.0938 in H2 O
32.2
62.4
Chapter 14
Show/Hide Problems
14.1
A centrifugal pump delivers 02 m3/s of water when operating at 850 rpm. Relevant impeller
dimensions are as follows: Outside diameter = 0.45 m, blade length = 50 cm, and blade exit
angle = 24o. Determine (a) the torque and power required to drive the pump and (b) the
maximum pressure increase across the pump.
Solution
Centrifugal Pump:
m3
v̇ = 0.2 ,
s
ω = 850rpm (89.0
r2 = 0.225m,
kg
ρ = 1000 3 ,
m
L = 0.05m,
β2 = 24°
rad
),
s
Torque − Eqn. 14.9:
mz = ρv̇ r2 [r2 ω −
v̇
cotβ2 ]
2πr2 L
2π
= 89.0 rad/s
60
v̇
0.2cot24
mz = ρv̇ r2 [r2 ω −
cotβ2 ] = (1000)(0.2)(0.225)x [(0.225)(89) −
]
2πr2 L
2π(0.225)(0.05)
= 615Nm
w = 850
ẇ = mz ω = 615 ∗ 89 = 54.75kW
Nm
54.75x103 s
ΔP
ẇ
ẇ
|
= − = − ∴ ΔPmax = −
= −274kPa
m2
ρ max
ṁ
ρv̇
0.2 s
14.5
A centrifugal pump is being used to pump water at a flow rate of 0.018 m3/s and the required
power is measured to be 4.5 kW. If the pump efficiency is 63%, determine the head generated by
the pump.
Solution
kg
Centrifugal Pump: ρ = 1000 3 ,
m
η=
ṁΔP
ρẇ
∴ ΔP =
ẇ = 4.5kW,
m3
v̇ = 0.018
,
s
η = 63%
ηρẇ
ηẇ
0.63(4500)
=
=
= 157.5 kPa
ṁ
v̇
0.018
ΔP
157500
=
= 16.05 m H2 O
ρg 1000(9.81)
14.17
The pump having the characteristics shown in Problem 14.14 is to be operated at 800 rpm. What
discharge rate is to be expected if the head developed is 410 m?
Solution
Same Pump family as Prob 14.14
New Pump: η = 800 rpm = 83.8
CH =
rad
, h = 410m
s
gh
9.81(410)
=
= 4.161
2
2
(83.8)2 (0.371)2
n D
At this value of CH ,
CQ = 0.16 =
CQ ≅ 0.16
v
nD3
∴ v̇ = 0.16n1 D13 = 0.16(83.8)(0.371)2 = 0.685
m3
s
14.18
If the pump having the characteristics shown in Problem 14.14 is tripled in size but halved in
rotational speed, what will be the discharge rate and head when operating at maximum
efficiency?
Solution
Same Pump family as Prob 14.14
New Pump: D2 = 3D1 , n2 = 0.5n1
@ηmax , CQ ≅ 0.12, CH ≅ 5.3
h2
ω2 2 D2 2
= ( ) ( ) = 0.52 (3)2 = 2.25
h1
ω1
D1
v2̇
n2 D2 3 1
= ( ) ( ) = (3)3 = 13.5
v1̇
n1 D1
2
2 (0.371)2
5.3(37.70)
h1 =
= 105.7 m
9.81
∴ h2 = 105.7(2.25) = 238m
m3
2
v1̇ = 0.12(37.70)(0.371) = 0.231
s
m3
∴ v2 = 0.231(13.5) = 3.12
s
Chapter 15
Show/Hide Problems
15.5
A sheet of insulating material, with thermal conductivity of 0.22W/m K, is 2 cm thick and has a
surface area of 2.97m2. If 4 kW of heat are conducted through this sheet and the outer (cooler)
surface temperature is measured at 55°C (328K), what will be the temperature on the inner (hot)
surface?
Solution
kA
ΔT
L
4000W(0.02m)
ΔT =
= 122.4K
0.22W
( m ∙ K ) (297m2 )
q=
Ti = 55 + 122.4 = 177.4 C
15.7
Plate glass, k = 1.35 W/m K, initially at 850 K, is cooled by blowing air past both surfaces with an
effective surface coefficient of 5 W/m2 ? K. It is necessary, in order that the glass not crack, to limit
the maximum temperature gradient in the glass to 15 K/cm during the cooling process. At the start
of the cooling process, what is the lowest temperature of the cooling air that can be used?
Solution
dt
1.35W 15K 100cm
W ΔT
)(
)(
) = 2025 2 =
|
=(
dx max
m∙K
cm
m
m
R
1
ΔT = 2025 ( ) = 405K
5
Tmin = 850 − 405 = 445K
q max = −k
15.8
Solve Problem 15.7 if all specified conditions remain the same but radiant-energy exchange from
glass to the surround- ings at the air temperature is also considered.
Solution
W ΔT
4
=
+ σ(Tsurf
− TA4 )
2
m
R
4
850 − T
T
4
) ]
=
+ 5.676 [8.5 − (
1
100
5
by Trial and Error, T = 836K
q max (From previous problem) = 2025
15.9
The heat loss from a boiler is to be held at a maximum of 900 Btu/h ft2 of wall area. What thickness
of asbestos (k = 0.10 Btu/h ft °F) is required if the inner and outer surfaces of the insulation are to be
1600 and 500°F, respectively?
Solution
q kΔT
=
A
L
L=
(0.10
or
kΔT
L= q
A
Btu
) (1000F)
hrftF
= 0.122ft = 1.47in
Btu
900
hrft 2
15.14
If, in Problem 15.13, the plate is made of asbestos, k = 0.10 Btu/h ft °F, what will be the temperature
of the top of the asbestos if the hot plate is rated at 800 W?
Solution
BTU
hr
If all heat leaves top surface
800W = 2730
q = hAΔT + σAϵ[T 4 − Ts4 ]
2700
T 4
) − 5.44 ]
= 5(T − 40) + 0.1714(1) [(
A
100
By trial & error: T = 1086R = 626F
Chapter 16
Show’Hide Problems
16.1
The Fourier Field Equatio in cylindrical coordinate is
𝜕𝑇
𝜕 2 𝑇 1 𝜕𝑇 1 𝜕 2 𝑇 𝜕 2 𝑇
= 𝛼( 2 +
+
+
)
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2 𝜕𝑧 2
(a) What form does this equation reduce to for the case of steady-state, radial heat transfer?
(b) Given the boundary conditions
𝑇 = 𝑇𝑖 𝑎𝑡 𝑟 = 𝑟𝑖
𝑇 = 𝑇𝑜 𝑎𝑡 𝑟 = 𝑟𝑜
(c) Generate an expression for the heat flow rate, 𝑞𝑟 , using the result from part (b).
Solution
d2 T 1 dT
1 d dT
(r ) = 0
In Cylindrical Coordinates: 2 +
= 0 or
dr
r dr
r dr dr
r
dT
= c1
dr
T = c1 lnr + c2
BC: Ti = c1 lnri + c2 ,
To = c1 lnro + c2
Ti − To
c2 = TL − c1 lnri
r
ln ( ro )
i
r
ln (r )
i
T = Ti − (Ti − To )
ro
ln ( r )
c1 = −
i
q = −kA
dT
dT
2πkL
= −k(2πrL)
= −2πkLc1 =
r (Ti − To )
dr
dr
ln ( ro )
i
16.5
Solve equation (16-19) for the temperature distribution in a plane wall if the internal heat
generation per unit volume varies according to 𝑞̇ = 𝑞̇ 𝑜 𝑒 −𝛽/𝐿 . The boundary conditions that
apply are
𝑇 = 𝑇𝑜 𝑎𝑡 𝑥 = 0
𝑇 = 𝑇𝐿 𝑎𝑡 𝑥 = 𝐿
Solution
@ Steady State ∇2 T +
q̇
=0
k
d2 T q̇ −βx
+ e L =0
dx 2 k
dT q ȯ L −βx
=
e L + c1
dx
k β
q ȯ L2 −βx
T=−
e L + c1 x + c2
k β2
q ȯ L2 −βx
BC: T(O) = To ,
To = −
e L + c2 ,
k β2
q ȯ L2 −β
T(L) = TL ,
TL = −
e + c1 L + c2
k β2
βx
x q ȯ L2
x
−
L − (1 − e−β )]
(T
)
T = To + L − T0 +
[1
−
e
L k β2
L
Chapter 17
Instructor Only Problems
17.1
Solution
dT
qx= -kAdx = kA/L (T1-T2)
For T1-T2= 75 K
q=
(30 W/m∗K)(1 m2 )(75 K)
0.30 m
∆T
= 7500 w/m2
75 K
dT/dx = L = 0.30 m = -250 K/m
For T1=300 K
q
q= -2000 W/m
dT/dx= − kA =
qL
∆T = kA =
2000 W/m2
W
)
m∗K
(30
−2000 (0.3)
30
= 66.7 K/m
= -20 K
T2= 320 K
For T2=350 K
dT/dx= -300 K/m
q= -(30)(-300)= 9000 W/m2
∆T= -300 K/m (0.3m)= 90 K
T1= 440 K
For T1=250 K
dT/dx= 200 K/m
q = -(30)(200 K/m)= -6000 W/m2
∆T= -200(0.3m)= -60 K
T2= 310 K
17.2
Solution
̅
kA
k
r −r
2πk
ln
ln o
q = r −r ∆T = r −r 2π o roi ∆T =
o
o
i
% Error=
=
ALM −AAM
ALM
i
ri
x100
2π(ro −ri )
− π (ro +ri )
r
ln o
ri
2π(ro − ri )
r
ln o
ri
r
(ro +ri ) ln o
ri
=|1 −
2(ro − ri )
r
ri
r
ri
r
ri
x 100
| x 100
( o +1) ln o
=|1 −
r
r
ri
2( o −1)
| x 100
For ro =1.5
% error= 1.3 %
=3
=5
% error= 10.0%
% error= 20.7%
i
∆T (a)
17.3
Solution
4πkr r
̅=4πro ri
(a) A
q = r −or i ∆T
o
Am =
i
4π(r2o +r2i )
2
% Error =
= 2π(ro2 + ri2 )
4πro ri −2π (r2o +r2i )
4πro ri
r
r
= 1-(1/2)( ro − r i )
i
o
(b)
ro
ri
= 1.5
% error=8.3 %
=3
=5
% error= 66.6%
% error= 160 %
17.4
Solution
dT
dT
q”= -k dx = -ko(1+bT) dx
From 0 to 12:
1.2
T
q” ∫0 dx= -ko∫925(1 + bT)dT
k
b
q”= 1.2o [T+2T2]T925 = 23(T-300)
Solving: TRH wall= 307.1 K
q”= 163.3 W/m2
From 0 to L:
L
650
q”∫0 = −k o ∫925 (1 + bT)dT
k
b
L= q"o [T + 2 T 2 ]925
650
& Solving: L=0.646 m
17.8
Solution
Brick Size= 9”x 4.5”x 3”
Brick #1
Brick # 2
BTU
k=0.44 Hr Ft F
Tmax=1500 F
k=0.94 Hr Ft F
Tmax=2200 F
BTU
Most economical arrangement is to use as much of #1 as possible (low k). Use #2 next to high
temp such that its cooler surface has T≤ 1500 F.
q”=
2000−T
Lz =
L/k
LActual = 28.5 in
k(2000−Tm )
200
= 2.35 Ft.
= 28.2 in.
(9x2+4.5+2x3)
q′′
TInterface = TH − k2 = 1495 F
L2
Lmin =
0.44(1495−300)
200
= 2.63 Ft
= 31.6 in.
LAct = 33 in
Most economical:
L1 = 33 in
L2 = 28.5 "
17.9
Solution
Ri=1/115= 8.696x10-5 K/W
R1= 0.25/1.13= 0.221 K/W
R2= 0.20/1.45= 0.138 K/W
R3= 0.10/0.66= 0.152 K/W
Ro= 1/23 = 0.0435 K/W
∑ R = 0.563
1070
q= ∆T/ ∑ R= 0.563 = 1900 W/m2 = 176.5 W/ft2
q= 23(To-300)
To= 383 K
17.10
Solution
To= 325 K
q=23(325-300)
= 575 W/m2
Ri=1/115= 8.696x10-5 K/W
R1= 0.25/1.13= 0.221 K/W
R2= L/1.45
R3= 0.10/0.166= 0.152 K/W
∑ R= 0.416 + L/1.45= ∆T/q
1045
0.416+ L/1.45= 573
L= 2.03 m
17.12
Solution
0.125
12
R1 = 0.15 = 0.0694
R2 =
R3 =
0.125/12
=0.00104
10
0.25/12
π
4
(22)(2)( )(
0.75 2
)
12
RConduction, Equiv =
= 0.154
1
1
1
+
R1 +R2 R3
=
1
1
1
+
0.07044 0.154
= 0.0483
∑ R per side = 0.0483+ 1/3= 0.3817
New Ht. Flux =
Increase =
930
0.3817
2437−2305
2305
= 2437 BTU/Hr-ft2
= 0.057= 5.7 %
Chapter 18
Show/Hide Problems
18.4
Solution
Bi =
hV 16 6/12
=
(
) = 0.058
kS 23 6
A lumped parameter problem
Fo =
αt
23
ft 2
t
=
∗
= 72t
2
(V/S)
460(0.1) hr (6/12)2
6
T − T∞
600 −(0.058)(72t)
=
e
To − T∞ 2000
t = 0.228 hr = 17.3 min
18.5
Solution
Lumped parameter solution applies if
Bi =
hV
< 0.1 or if h < 0.1
kS
kS
kπD2
= 3 = 0.47(6) = 2.82
V
πD /6
W
Therefore h must be < 0.1(2.82) = 0.282 m2K
But h = 15 ∴ Use Distributed parameter solution
(0.47)t
αt
=
= 5.26x10−5 t
ro 2 (940)(3800)(0.05)2
T − T∞
= 0.5
To − T∞
k
0.47
=
= 0.627
hro 15(0.05)
X ≅ 0.17 = 5.26x10−5 t
t = 3230 s = 53.9 min
18.7
Solution
T − T∞
500 − 1000
=
= 0.538
To − T∞
70 − 1000
V
D
1
=
=
A 4 + 2 D 17
L
v
hs
4
17
Bi =
=
k
k
a) Cu – Bi<0.1 Lumped
ρĈp V
1
t=
ln
= 27.9 min
hA
0.538
b) Al – Bi<0.1Lumped
t = 0.345 hr = 20.7 min
c) Zn - Bi<0.1Lumped
t = 0.381 hr = 22.9 min
d) Steel - Bi<0.1Lumped
t = 0.502 hr = 30.2 min
18.9
Solution
Bi =
v
hs
k
πD2
L 4⁄
15 [
πDL]
=
12.4
=
hD
D
4k (L + 2 )
=
85(0.6)(0.6)
= 0.0371
(229)(4)(0.9)
Lumped parameter case. Temperature may be considered uniform at any time.
Fo =
(9.16x10−5 )(3600)
αt
=
= 32.98
(V/S)2
(0.1)2
T − T∞
= e−Bi Fo = e−(0.0371)(32.98) = 0.294
To − T∞
T = 345 + 0.294(130) = 383.2 K
18.10
Solution
3
hV 40 πD ⁄6
Bi =
=
= 0.00575
kS
19.3πD2
Lumped parameter
0.8(15⁄3600)
αt
Fo =
=
= 432
(V/S)2
[0.2⁄2(6)]2
T − T∞
= e−Bi Fo = 0.0834
To − T∞
T = 115.9 ℉
Chapter 19
Show/Hide Problems
19.1
Solution
For a plane wall:
Variables
T
T0
T∞
x
L
Α
k
t
h
Dimension
T
T
T
L
L
L2 ⁄t
Q⁄LtT
t
Q⁄LtT
i=n−r=5
If temps are grouped as
T − T∞ ,
T0 − T∞ ,
i=n−r=4
π1 = ∆T a Lb k c αd (T − T∞ )
π2 = ∆T a Lb k c αd (x)
π3 = ∆T a Lb k c αd (t)
π4 = ∆T a Lb k c αd (h)
π1 =
T − T∞
,
T0 − T∞
x
π2 = ,
L
π3 =
αt
,
L2
π4 =
hL
k
19.5
Solution
For a single 4 ft long plate:
L
q = W ∫ (α + β sin (
0
πx
)) dx
L
βL
πx
q = W (αx + (cos ( )))
π
L
L
0
β
2(1500)
q = WL (α + 2 ) = 16 ft 2 (250 +
) = 19300 Btu⁄hr
π
π
For stack of plates:
q = 19300
(640)
= 772,000 Btu⁄hr
16
19.7
Solution
L
q
πx
πx
βL
15
3000
= α + β sin ( ) = πD ∫ [α + β sin ( )] dx = πD [αL + 2 ] = π ( ) [250 +
]
A
L
L
π
12
π
0
= 4730 Btu⁄Hr
TExit = Ti +
q
4730
= 60 +
= 60.3 ℉
ρvCp A
60(1)(0.067)(3600)
Tw = 60.3 +
250
≅ 60.6 ℉
976
19.11
Solution
For air @ Tf = 325 K:
ρ = 1.087 kg⁄m3
Cp =1.008 kJ⁄kg K
k = 2.816 W⁄m K
ν = 1.807x10−5 m2 ⁄s
Pr = 0.702
−1⁄
2
a) CfL = 1.328ReL
m
(1 m) (2.8 )
Lv
s
Re =
=
= 1.55x105
−5
2
⁄
ν
1.807x10 m s
−1
CfL = 1.328(1.55x105 ) ⁄2 = 0.00337
δv2
b) fd = CfL A 2 =
(0.00337)(0.25)(1)(1.087)(2.8)2
2
= 3.59x10−3 N
c) q = hA∆T Using Colburn analogy
St L =
(0.00337)
CfL
h
−2
−2
(Pr) ⁄3 =
(0.702) ⁄3 = 2.133x10−3 =
2
2
ρvCp
h = (2.133x10−3 )(1.087)(1008)(2.8) = 6.54 W⁄m2 K
q = 6.54(1)(0.25)(55) = 90.0 W
Chapter 20
Show/Hide Problems
20.1
Solution
q
750(3.413)
=
= 26100 Btu⁄hr ft 2
A π(3⁄ )(1⁄ )
48
2
Ends are neglected
2
1
k
k
h = NuL = 0.825 +
L
L
[
0.387ReL 6
8
9 27
0.492 16
{1 + ( Pr ) }
]
(a) For vertical orientation
By trial and error ∆𝑇 ≅ 103 ℉
HTR surface temp = 198 ℉
(b) Horizontal orientation
2
k
h = 0.6 +
D
[
1
0.387ReD 6
8
9 27
16
0.559
{1 + ( Pr ) }
]
Trial and error: ∆T = 99℉
HTR surface temp = 194 ℉
20.5
Solution
Ts = 140℃ T∞ = 25℃
Tf = 82.5℃
Bg
Air @ 355 K: ν2 = 0.625x108 (m3 K)−1
Re = (0.625x108 )(0.035)3 (115) = 3.08x105
Horizontal Cylinder:
2
1
Nu = 0.6 +
[
0.387ReD 6
8
9 27
0.559 16
{1 + ( Pr ) }
]
For Pr = 0.696 NuD = 10.47
q = hA∆T =
k
(πDL)(∆T) = (0.0304)(π)(0.8)(115)(10.47) = 92 W
D
Remainder of input goes to electrical & conduction losses & to illumination
20.9
Solution
k
For Tc to reach 320 K – use values calculated in problem 20.8 α = ρC = 2.1x10−7 m2 ⁄s
p
D, cm
h
αt⁄
x1 2
t
7.5
615
0.16
1.19 hr
5
710
0.16
31.7 min
1.5
1180
0.16
2.86 min
Tsurf ≅ 295 K at all times
20.13
Solution
Assuming each plate is independent
2
k
k
h = Nu = 0.825 +
D
D
[
1
0.387ReD 6
8
9 27
0.492 16
{1 + ( Pr ) }
]
Tplate = 200℉ T∞ = 80℉ Tf = 140℉
Pr = 3.08 Re = (540x106 )(3)3 (120)(3.08) = 5.39x1012
h = 287 Btu⁄hr ft 2 ℉
q = hA∆T = (278)(30 ∗ 1 ∗ 3 ∗ 2)(120) = 6.19x106 Btu⁄hr = 1.81 kW
Chapter 21
Show/Hide Problems
21.3
Solution
CL ∆T
= 0.0709
hfg
q
0.0709 3
Btu
Btu
) = 190 2 = 680000
=(
A
0.01235
s ft
hr ft 2
1 2
Btu
q = (680000)(π) ( ) = 178000
24
hr
h=
680000
Btu
= 10000
68
hr ft 2 ℉
21.5
Solution
Using English units:
A = πDL +
2πD2
π
= π(0.02)(0.15) + (2)(0.02)2 = 0.1082 ft 2
4
4
q 500(3.413)
Btu
=
= 15800
A
0.1082
hr ft 2
Assume nucleate boiling:
1
1 3
3.79x10−3 2
1∆T
15800
= 0.006 [
(
1.7
(1.8) (970)
(0.195x10−3 )(970)(3600)
∆T = 9.0℉
Surface temp = 221℉
h=
15800
Btu
= 1760
9
hr ft 2 ℉
60
) ]
21.7
Solution
h = 0.62 [
1
4
3 (∆ρ)g
k v ρv
(hfg + 0.4Cp ∆T)
v
Dμv (Ts − Tsat )
= 0.62 [
]
(0.0153)3 (0.0341 ∗
40
) (58.4)
14.7
1
(12) (0.914x10−5 )(933)
1
4
∗ 32.3(3600)(934 + 0.4 ∗ 0.481 ∗ 933)] = 26.9
q
0.2
Btu
= h∆T = 43.3π ( ) (1)(1960) = 25000
A
12
hr ft 2
Btu
hr ft 2 ℉
Chapter 22
Show/Hide Problems
22.1
Solution
q = ṁCp ∆T|H2O = UA∆TLM
UA = constant =
U=
q
∆TLM
1
1
∆i
r
∆i
)
( ) + (2πK ln ( r0 )) + (
hi
A
i
0 h0
≅ hi
hi Ai = constant
Using dittus-boelter correlation
0.8
k
k
k DQ
hi = (constant)Re0.8 Pr 0.4 = (Dv)0.8 = ( 2 )
D
D
D πD
4
hA = constant = A(constant)D−1.8
As diameter increases the required area increases as D−1.8
= (constant)D−1.8
22.2
Solution
Oil:
Tin = 400K Tout = 350K
ṁ = 2
kg
J
q = 1880
s
kgK
q = ṁCp ∆T = 2(1880)(50) = 188000 W
∆Tw =
q
188000
=
= 22.5 K
ṁCp 2(4187)
Tw in = 280K Tw out = 302.5 K
TLM =
A=
97.5 − 70
= 83K
97.5
ln ( 70 )
q
188000
=
= 9.85 m2
U∆TLM 230(83)
22.3
Solution
Dequiv =
4(0.1)(0.2)
= 0.0667 m
2(0.1 + 0.2)
(22.4 continued)
Tb avg = 295 K
RE = 0.667
Tf = 345 K
ρv
ṁ
= 0.0667
μ
Aμ
q = hA∆TLM
TLM =
105 − 95
= 99.9 K
105
ln (
)
95
Assuming turbulent flow:
2
ṁCp ∆T = ρvCp [0.023Re−0.2 Pr −3 ] As ∆TLM
2
ṁ
0.0667 ṁ −0.2
−
3 ] (2)(0.3)(2.5)(99.9)
(0.698)
)
ṁ(10) = [0.023 (
A
0.205x10−5 A
kg
Solving for ṁ: ṁ = 105 s
q = ṁCp ∆T = 105(1009)(10) = 1060 kW
22.16
Solution
NTU = 1.25
Cmin
= 0 ε ≅ 0.72
Cmax
q = εCmin (Tw in − TA in ) = 0.72(0.07)(4.18)(93) = 19.59 kW = Cw ∆Tw = 4.18(0.07)∆Tw
∆Tw =
0.72(0.07)(4.18)(93)
= 67 K
4.18(0.07)
Tw out = 280 + 67 = 347 K
Steam condensation rate:
ṁcond =
q
19.59
kg
=
= 8.68x10−3
hfg 2256
s
Chapter 23
Show/Hide Problems
23.1
Solutions
Radiant emission from sun = As Ebs
All passes through a spherical surface of radius, L
At the earth:
q πD2sEbs
D 2
=
= ( ) Ebs
A
4πL2
2L
Btu
Flux at earth= 360 + 90 = 450 hr ft2
2
8.6x105
450 = [
] σTs4
2(93x106 )
Ts = 10530 R
23.3
Solution
q
q
q
Ts 4
T∞ 4
)
) −(
) ]
( )| = ( )| − ( )| = 1000 − h(Ts − T∞ − ϵσ [(
A net
A in
A out
100
100
= 1000 − 12(30 − 20) − 5.676(0.3)(106) = 862
W
m2
23.8
Solution
Filament at 2910 K
q=100 W
λmax =
2897.6
= 0.999 μm
2910
Visible range: 0.38 < λ < 0.76
λ1 T1 = 0.38(2910) = 1102 F0−λT = 0.0009
λ2 T2 = 0.74(2910) = 2204 F0−λT = 0.1017
Fraction in V.R. = 0.1008
23.14
Solution
T=1500 K Peephole D=10 cm
τ = 0.78 for 0<λ<3.2μm
τ = 0.08 for 3.2<λ<∞
π
q max = 5.676(15)4 ( ) (0.01)2 = 22.57 W
4
For:
λ1 T1 = 0 F0−λT1 = 0
λ2 T2 = 4800 F0−λT2 = 0.6075
λ3 T3 = ∞ F0−λT3 = 1
Total heat loss
= 22.57[0.78(0.6075) + 0.8(0.3925)] = 11.40 W
23.15
Solution
q
1200 W
W
T 4
) − (2.8)4 ]
=
= 490 2 = ϵσ [(
2
)
A 5(0.49 m
m
100
T 4
) − (2.8)4 ]
490 = 0.7(5.676) [(
100
T = 369 K
23.16
Solution
q = 8 W through hole with D=0.0025 m2
Eb =
8
W
= 3200 2 = σT 4
0.0025
m
T = 4.87 K
24.5
Given:
P = 6.1 × 10−3 bar, T = 210 K
yCO2 = 0.9532, yN2 = 0.027, yAr = 0.016, yO2 = 0.0013, yCO = 0.0008
a. Partial pressure 𝑝𝐴 for CO2
A = CO2
 105 Pa 
p A = y A P = ( 0.9532 ) 6.110−3 bar 
 = 581 Pa
 1 bar 
b. Molar concentration 𝑐𝐴 for CO2
(
cA =
)
pA
581 Pa
gmole
=
= 0.333
3
RT
m3
8.314 m  Pa/gmole  K ( 210 K )
(
)
c. Total molar concentration c for the Martian atmosphere
c=
P
610 Pa
gmole
=
= 0.349
3
RT
m3
8.314 m  Pa/gmole  K ( 210 K )
(
)
d. Total mass concentration ρ for the Martian atmosphere
M avg = yCO2 M CO2 + yN2 M N2 + yAr M Ar + yO2 M O2 + yCO M CO



g 
g 
g 
M avg = ( 0.9532 )  44.01
 + ( 0.027 )  28.00
 + ( 0.0163)  9.95
+
gmole 
gmole 
gmole 





g 
g 
( 0.0013)  32.00
 + ( 0.0008 )  28.01

gmole 
gmole 


M avg = 43.41
g
gmole

 = M avg  c =  43.41

g 
gmole 
g
 0.349
 =15.2 3
3
gmole  
m 
m
24-1
24.12
A = SiH4, B = He
yA = 0.01, yB = 0.99
dpore = 10 × 10−4 cm, T = 900 K
g
g
MA = 32.12 gmole, MB = 4.00 gmole
κ = 1.38 × 10−16 ergs/K
a. Determine wA
g
0.01 ∙ 32.12
yA ∙ MA
gmole
wA =
=
= 0.750
yA ∙ MA + yB ∙ MB 0.01 ∙ 32.12 g + 0.99 ∙ 4.00 g
gmole
gmole
b. Estimate DAB at P = 1.0 atm and P = 100 Pa, T = 900 K
P = 1.0 atm
εA
κ
= 207.6 K,
εB
κ
= 10.22 K ,
εAB
κ
ε
εB
κ
κ
=√ A∙
= 46.06 K,
κT
εAB
= 19.54
Using Appendix K. 1, interpolate to get ΩD = 0.6676
σA = 4.08 Å,
σB = 2.576 Å, σAB =
σA +σB
2
= 3.328 Å
1
2
3
1
1
1
0.001858 ∙ 900 K 2 ∙ (
+
3
2
g
g )
1
1
32.12
4.00
0.001858 ∙ T 2 ∙ (M + M )
gmole
gmole
A
B
DAB =
=
2
P ∙ σAB ΩD
1.0 atm ∙ 3.328 Å2 ∙ 0.6676
= 3.60
cm2
s
𝑃 = 100 Pa ∙
1 atm
= 9.87 × 10−4 atm
101325 Pa
24-2
1
2
3
1
1
1
0.001858 ∙ 900 K 2 ∙ (
+
3
2
g
g )
1
1
32.12
4.00
0.001858 ∙ T 2 ∙ (M + M )
gmole
gmole
A
B
DAB =
=
2
P ∙ σAB ΩD
9.87 × 10−4 atm ∙ 3.328 Å2 ∙ 0.6676
= 3645
cm2
s
c. Assess importance of Knudsen Diffusion
Pressure = 1.0 atm
λ=
κT
√2πσA
Kn = d
λ
pore
2P
=
J
1.38 × 10−23 K ∙ 900 K
√2π ∙ (4.08 × 10−10 m)2 101325 Pa
= 1.66 × 10−7 cm
6.76×10−7 cm
= 10×10−4 cm = 1.66 × 10−4 < 1, Knudsen diffusion is not important.
Pressure = 100 Pa
λ=
κT
√2πσA 2 P
Kn =
λ
dpore
=
=
J
1.38 × 10−23 K ∙ 900 K
√2π ∙ (4.08 × 10−10 m)2 100 Pa
1.68 × 10−4 cm
= 0.168,
10 × 10−4 cm
= 1.68 × 10−4 cm
Knudsen diffusion plays moderate role
d. Gas velocity for Pe = 5.0 x 10-4
DKA = 4850 ∙ dpore ∙ √
T
900 K
cm2
= 4850 ∙ 10 × 10−4 cm ∙ √
=
25.67
g
MA
s
32.13
gmole
P = 1.0 atm
1
1
1
1
1
)=(
)
= (
+
+
2
cm
cm2
DAe
DAB DKA
3.60 s
25.67 s
DAe = 3.157
cm2
s
24-3
v∞∙dpore
= 5 × 10−4
DAe
cm
v∞ = 1.58
s
πd2
cm
cm3
V̇ = v∞
= 1.58
∙ π ∙ (5 × 10−4 cm)2 = 1.24 × 10−6
4
s
s
Pe =
P = 100 Pa
1
DAe
1
1
= (D
Pe =
AB
+ D ) , ∴ DAe = 25.49 cm2 /s
v∞∙dpore
DAe
KA
= 5 × 10−4
∴ v∞ = 12.75
V̇ = 12.75
cm
s
cm π
cm3
−4
2
−5
∙ ∙ (10 × 10 cm) = 1.00 × 10
s 4
s
24-4
24.17
A = CuCl2 (an ionic solute), B = H2O
T = 298 K, ℱ = 96,500 C/gmole
R = 8.316
J
gmole ∙ K
λ°+ (Cu2+ ) = 108
λ°− (Cl− ) = 76.3
A ∙ cm2
V ∙ gmole
A ∙ cm2
V ∙ gmole
n+ (valence of cation) = 2
n− (valence of anion) = 1
Nernst-Haskell Equation
1
1
( + + n− )RT
n
DAB =
1
1
( ° + ° )(ℱ)2
λ+ λ−
1 1
(2 + 1)(8.316
J
) ∙ 298K
cm2
gmole ∙ K
−5
DAB =
= 1.78 × 10
1
1
s
2
(
+
)(96,500
C/gmole)
A ∙ cm2
A ∙ cm2
108
76.3
V ∙ gmole
V ∙ gmole
24-5
24.23
a. CA
C A = y AC = y A
C A = ( 0.05 )
P
RT
(1.5 atm )

-5 m  atm 
 8.206×10
 ( 353K )
gmole  K 

3
=2.58
gmole A
m3
'
b. DAe
at T = 80 oC (353 K) and 1.5 atm
P 
cm2  1.0 atm 
cm 2
DAB (T , P) = DAB (T , Pref )  ref  = 0.10
=
0.0.067


s  1.5 atm 
s
 P 
DKA = 4850 d pore
T
353
cm 2
−5
= 4850(1.5  10 )
= 0.202
MA
46
s
1
1
1
=
+
=
DAe DKA DAB
1
0.202
cm
s
2
+
1
0.067
cm 2
s
cm 2
DAe = 0.050
s

cm 2 
cm 2
'
DAe
=  2 DAe = (0.5) 2  0.05
=
0.0125

s 
s

24-6
24-7
25.7
𝑘1
𝐴 → 𝐵, 𝑅𝐴 = −𝑘1 𝑐𝐴 ,
𝑘𝑠
𝐴 → 2𝐶, 𝑟𝐴,𝑠 = −𝑘𝑠 𝑐𝐴𝑠 ,
𝑘1 [
1
]
𝑠
𝑘𝑠 [
𝑐𝑚
]
𝑠
4 species: A=1, B=2, C=3, D=4 (inert)
( 𝑅1 = −𝑘1 𝑐1, 𝑟1,𝑠 = −𝑘𝑠 𝑐1𝑠 )
a. Assumptions, Source and Sink for species 1 (A)
Assumptions:
1) Dilute process with respect to species 1
2) Constant source and sink for species 1 (steady state)
3) Right side and top side are impermeable barriers
4) 2-D flux of species 1 for catalyst II (source and sink antiparallel)
Source: Flowing fluid containing reactant 1, of constant concentration (𝑐1,∞)
Sinks: First-order homogeneous reaction of 1 in porous layer (catalyst I)
First-order heterogeneous surface reaction of 1 at nonporous boundary
surface (catalyst II)
b. Differential model for C1(x,y) (Shell balance on differential volume element for species 1)
IN – OUT + GEN = ACC = 0
𝑛1,𝑥 = 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑤 = 𝑚𝑎𝑠𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 1 𝑖𝑛 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
[𝑛1,𝑥 ∆𝑦𝑤|𝑥,𝑦̅ + 𝑛1,𝑦 ∆𝑥𝑤|𝑥̅ ,𝑦 ] − [𝑛1,𝑥 ∆𝑦𝑤|𝑥+∆𝑥,𝑦̅ + 𝑛1,𝑦 ∆𝑥𝑤|𝑥̅ ,𝑦+∆𝑦 ] + 𝑟1 ∆𝑥∆𝑦𝑤 = 0
÷ ∆𝑥, ∆𝑦; 𝑡𝑎𝑘𝑒 lim ∆𝑥 → 0, 𝑡𝑎𝑘𝑒 lim ∆𝑦 → 0 ; ÷ 𝑤
25-1
−
𝜕𝑛1𝑦
𝜕𝑛1,𝑥
+−
+ 𝑟1 = 0
𝜕𝑥
𝜕𝑥
General Flux Equation
𝜕𝐶
x-direction: 𝑛1,𝑥 = −𝐷1 𝜕𝑥1 (𝑜𝑡ℎ𝑒𝑟 𝑡𝑒𝑟𝑚𝑠 ≈ 0, 𝑑𝑖𝑙𝑢𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
𝜕𝐶
y-direction: 𝑛1,𝑦 = −𝐷1 𝜕𝑦1 (𝑜𝑡ℎ𝑒𝑟 𝑡𝑒𝑟𝑚𝑠 ≈ 0, 𝑑𝑖𝑙𝑢𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
Homogeneous reaction: 𝑟1 = −𝑘1 𝑐1
𝜕2𝐶
𝜕2𝐶
Combine: 𝐷1 [ 𝜕𝑥 21 + 𝜕𝑦 21] − 𝑘1 𝑐1 = 0, 𝑐1 (𝑥, 𝑦)
c. Boundary Conditions, 4 required: (2 for x, 2 for y)
𝑦 = 0, 0 ≤ 𝑥 ≤ 𝐿,
𝑐1 (𝑥, 0) = 𝑐1,∞
𝑦 = 𝐻, 0 ≤ 𝑥 ≤ 𝐿,
𝑛1 (𝑥, 𝐻) = 0,
𝑥 = 0, 0 < 𝑦 < 𝐻,
𝑟1,𝑠 = 𝑛1,𝑠 ,
𝑥 = 𝐿, 0 < 𝑦 < 𝐻,
𝑛1 (𝐿, 𝑦) = 0
∴
𝜕𝑐1 (𝑥, 𝐻)
=0
𝜕𝑥
−𝑘𝑠 𝑐1,𝑠 = −𝐷1
∴
𝜕𝑐1 (𝐿,𝑦)
𝜕𝑦
𝜕𝑐1 (0, 𝑦)
,
𝜕𝑥
∴
𝜕𝑐1 (0, 𝑦) 𝑘𝑠
=
𝑐 (0, 𝑦)
𝜕𝑥
𝐷1 1
=0
25-2
25.10
A = C6H6, B = H2O (liq)
T = 298 K, 𝑐𝐴,∞ 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
a. Differential model for cA(r,t)
Assumptions:
1) Unsteady state (control volume is also the sink)
2) No homogeneous reaction (𝑅𝐴 = 0)
3) Dilute with respect to A
4) 1-D flux along r (unimolecular diffusion)
General Differential Equation
′
𝑁𝐴,𝑟 ≅ −𝐷𝐴𝑒
𝜕𝑐𝐴
,
𝜕𝑟
(𝑑𝑖𝑙𝑢𝑡𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛)
Mass Conservation Equation
−∇𝑁𝐴 =
𝜕𝑐𝐴
, (𝑅𝐴 = 0),
𝜕𝑡
∴ ∇𝑁𝐴 =
1 𝜕 2
(𝑟 𝑁𝐴,𝑟 ),
𝑟 2 𝜕𝑟
𝑐𝐴 (𝑟, 𝑡)
Combine Equations
1 ′ 𝜕 2 𝜕𝑐𝐴
𝜕𝑐𝐴
)=
𝐷𝐴𝑒 (𝑟
,
2
𝑟
𝜕𝑟
𝜕𝑟
𝜕𝑡
𝜕 2 𝑐𝐴 2 𝜕𝑐𝐴
𝜕𝑐𝐴
′
𝑜𝑟 𝐷𝐴𝑒
[ 2 +
]=
𝜕𝑟
𝑟 𝜕𝑟
𝜕𝑡
b. Boundary and Initial Conditions
IC:
𝑡 = 0,
𝑐𝐴 (𝑟, 0) = 𝑐𝐴0 = 0
BC:
𝑟 = 0,
𝜕𝑐𝐴 (0, 𝑡)
=0
𝜕𝑟
𝑟 = 𝑅, 𝑐𝐴 (𝑅, 𝑡) = 𝑐𝐴𝑠
25-3
26.9
A = H2O vapor, B = O2 gas
𝑇 = 310 𝐾, 𝑃 = 1.2 𝑎𝑡𝑚, 𝑑𝑝𝑜𝑟𝑒 = 50 × 10−7 𝑐𝑚
𝜀 = 0.40, 𝑃𝐴 = 0.0618 𝑎𝑡𝑚, 𝑦𝐴∗ = 0.0515, 𝐻 = 800
𝐿 ∙ 𝑎𝑡𝑚
𝑔𝑚𝑜𝑙𝑒
a. Determine effective diffusion coefficient, DAe
Fuller-Schettler-Giddings Correlation for DAB
1
1 1/2
1
1 1/2
0.001𝑇 1.75 [𝑀 + 𝑀 ]
0.001 ∙ 310 1.75 [18.02 + 32.0]
𝑐𝑚2
𝐴
𝐴
𝐷𝐴𝐵 =
=
= 0.236
1
1
1/3
1/3 2
𝑠
2
3
3
𝑃[𝑉𝐴 + 𝑉𝐵 ]
1.2 ∙ [12.7 + 16.6 ]
Knudsen Diffusion Coefficient
𝐷𝐾𝐴 = 4850𝑑𝑝𝑜𝑟𝑒 √
𝑇
310
𝑐𝑚2
= 4850 ∙ 50 × 10−7 √
= 0.1006
𝑀𝐴
18.02
𝑠
Effective Diffusion Coefficient
1
1
1
1
1
𝑐𝑚2
=
+
=
+
∴
𝐷
=
0.0705
𝐴𝑒
𝑐𝑚2
𝑐𝑚2
𝐷𝐴𝑒 𝐷𝐴𝐵 𝐷𝐾𝐴
𝑠
0.236
0.1006
𝑠
𝑠
𝑐𝑚2
𝑐𝑚2
′
𝐷𝐴𝑒
= 𝜀 2 𝐷𝐴𝑒 = 0.42 ∙ 0.0705
= 0.0113
𝑠
𝑠
b. Determine L
Assume: 1) Steady state, 2) no homogenous reaction, 3) 1-D flux along z, 4) constant T and P
𝑐=
𝑃
=
𝑅𝑇
1.2 𝑎𝑡𝑚
𝑔𝑚𝑜𝑙𝑒
= 4.72 × 10−5
3
𝑐𝑚 ∙ 𝑎𝑡𝑚
𝑐𝑚3
82.06
∙ 310𝐾
𝑔𝑚𝑜𝑙𝑒 ∙ 𝐾
d
( N A,z ) = 0 , constant flux along z
dz
𝑑𝑐𝐴 𝑐𝐴
𝑑𝑦𝐴
𝑁𝐴 = −𝐷𝐴𝐵
+ (𝑁𝐴,𝑧 ) = −𝐷𝐴𝐵 𝑐
+ 𝑦𝐴 (𝑁𝐴,𝑧 )
𝑑𝑧
𝑐
𝑑𝑧
𝑁𝐴 = −
𝐷𝐴𝐵 𝑐 𝑑𝑦𝐴
1 − 𝑦𝐴 𝑑𝑧
Boundary Conditions:
𝑧 = 0, 𝑦𝐴𝑠 = 𝑦𝐴∗
26-1
𝑧 = 𝐿, 𝑦𝐴∞ = 0.2𝑦𝐴∗
𝐿
𝑦𝐴∞
𝑁𝐴 ∫ 𝑑𝑧 = −𝑐𝐷𝐴𝐵 ∫
0
𝑁𝐴 =
𝐿=
𝑦𝐴𝑠
𝑑𝑦𝐴
1 − 𝑦𝐴
𝑐𝐷𝐴𝐵
1 − 𝑦𝐴∞
)
𝑙𝑛 (
𝐿
1 − 𝑦𝐴𝑠
𝑐𝐷𝐴𝐵
1 − 𝑦𝐴∞
)
𝑙𝑛 (
𝑁𝐴
1 − 𝑦𝐴𝑠
𝑔𝑚𝑜𝑙𝑒
𝑐𝑚2
0.0113 𝑠 ∙ 4.72 × 10−5
𝑐𝑚3 ln [1 − 0.2 ∙ 0.0515] = 0.20 𝑐𝑚
𝐿=
𝑚𝑜𝑙
1 − 0.0515
1.15 × 10−7
𝑐𝑚2 ∙ 𝑠
∗
c. Determine 𝑐𝐵𝐿
∗
𝑐𝐵𝐿
=
𝑝𝐵 𝑃 − 𝑝𝐴 (1.2 − 0.062) 𝑎𝑡𝑚
𝑔𝑚𝑜𝑙𝑒
=
=
= 1.42 × 10−3
𝐿 ∙ 𝑎𝑡𝑚
𝐻
𝐻
𝐿
800
𝑔𝑚𝑜𝑙𝑒
26-2
26.17
Given:
A = O2, B = air, C = carbon (s), D = CO2
T = 1000 K, P = 2.0 atm
Electrode dimensions: L = 25 cm, d = 2.0 cm,  = 0.5 cm boundary layer
Electrode materials: C,solid = 2.25 g/cm3, MC = 12 g/gmole
a. Determine WA
Model for WA
Physical System: gas space in boundary layer between carbon electrode surface and bulk gas
(System I)
Source for A: bulk gas
Sink for A: consumption by carbon electrode to CO2 gas
Assumptions
1.
2.
3.
4.
PSS process for A
No homogenous reaction of A in gas space
1-D flux along r
Instantaneous consumption of O2 gas (A) at electrode surface so that cAs = 0 at r = R
(instantaneous surface reaction)
5. NA,r = − ND,r by reaction C ( s ) + O 2 ( g ) → CO 2 ( g )
CO2
solid carbon
electrode
bulk gas
cA∞
O2 (A)
25 cm
NA,r
r = R+ = 1.5 cm
r = R = 1.0 cm
cAs = 0
26-3
General Differential Equation for Mass Transfer (cylindrical system)
RA = 0,
c A
= 0 by assumptions 1 and 2
t
c
1 d
rN A,r ) + RA = A
(
r dr
t
1 d

( r  N A, r ) = 0
r dr
d
( r  N A, r ) = 0
dr
−
By continuity N A,r r = R  R = N A,r  r NA,r r constant along r
Note WA = N A,r r = R  2 RL = N A,r  2 rL
Flux Equation
dc A c A
dc
c
+ ( N A,r + N B ,r + NC ,r + N D,r ) = − DAB A + A ( N A,r + 0 + 0 + − N A,r )
dr C
dr C
dc
N A,r = − DAB A
dr
dc 

Note: WA = N A,r r = R  2 RL = N A,r  2 rL = 2 rL  − DAB A 
dr 

WA is not a function of r; separate variables cA, r, pull WA outside of the integral, then integrate
with respect to limits shown below
N A,r = − DAB
r = R, cA = cAs = 0 (rapid consumption of O2 at surface)
r = R +  , cA = cA∞ (bulk gas composition beyond boundary layer)
R
0
dr
WA 
= −2 LDAB  dc A
r
R +
c A
WA =
2 LDAB c A
 R + 
ln 

 R 
Calculate WA
26-4
c A =
( 0.21)( 2.0 atm )
y A P
gmole
=
=4.65  10−6
3
RT 
cm3
cm atm 
82.06
1100
K
(
)


gmole  K 

 P  T 
DAB (T , P) = DAB (Tref , Pref )  ref  

 P   Tref 
3/2
3/2
= 0.136
cm 2  1.0 atm   1100 K 
cm 2
=
0.55



s  2.0 atm   273 K 
s

cm 2  
−6 gmole 
2π ( 25 cm )  0.55
  4.65  10

s 
cm3 
gmole

WA =
= 1.80  10−3
s
 2.5 cm 
ln 

 2.0 cm 
b. Time it takes for the rod to completely disappear (t at R = 0)
Material balance on carbon (PSS mass transfer) to model “shrinkage” of carbon electrode
(System II)
IN – OUT + GEN = ACC
0 − N A, r 
mC =
dm
1.0 mol C
S +0 = C
1.0 mol O 2
dt
C  R 2 L
MC

DAB c A
dR
−
2 RL = C 2 RL
MC
dt
 R + 
R  ln 

 R 
DAB c A

dR
−
= A R
 R +   M A dt
ln 

 R 
Separate variables R, t and integrate, then solve for time t at R = 0
t
− DAB c A dt =
0
R
 R + 
C 0  R +  
ln 
 RdR
M C R  R 
1 
 R + 
 1
 R + 


ln  R  RdR = 2 R  R  ln  R  +   − 2  ln  
0
2
26-5


 R + 
 R + 
2
R  R  ln 
 +   −  ln 

 R 
   C


t=
2 DAB c A
MC



2
 2.5 cm 
 2.5 cm   
g 
 2.0 cm  ( 2.0 cm ) ln 
 +0.5 cm  − ( 0.5 cm ) ln 
   2.25 3 
 2.0 cm 
 0.5 cm   
cm 


t=
= 5455 s
2


g 
cm  
−6 gmole 
2  0.55
  4.65  10
12.01 gmole 

s 
cm3 



26-6
26.19
A = acetone, B = air
𝑑 = 0.3 𝑐𝑚, 𝑇 = 293.9 𝐾
𝑃𝐴∗ = 0.254 𝑎𝑡𝑚, 𝑀𝐴 = 58
𝑔
𝑔
, 𝜌𝐴,𝑙𝑖𝑞 = 0.79 3
𝑔𝑚𝑜𝑙𝑒
𝑐𝑚
a. Determine DAB from Arnold Diffusion Cell data
Assume: 1) 1-D mass transfer along z, 2) No homogenous reaction, 3) PSS.
Material balance on acetone (PSS mass transfer)
IN – OUT + GEN = ACC
𝑑𝑚𝐴
0 − 𝑁𝐴,𝑧 𝑆 + 0 =
𝑑𝑡
𝑁𝐴 =
𝐷𝐴𝐵 𝑐
1 − 𝑦𝐴2
ln [
]
𝑍
1 − 𝑦𝐴1
𝜌𝐴,𝑙𝑖𝑞 𝑑𝑍
𝐷𝐴𝐵 𝑐
1 − 𝑦𝐴2
ln [
]=
∙
𝑍
1 − 𝑦𝐴1
𝑀𝐴 𝑑𝑡
𝑡
𝑍
0
𝑍𝑜
𝐷𝐴𝐵 𝑐𝐴𝑠 𝑀𝐴
∫ 𝑑𝑡 = ∫ 𝑍𝑑𝑍
𝜌𝐴,𝑙𝑖𝑞
Z2-Zo2 [cm2]
𝑍 2 − 𝑍0 2 =
2𝐷𝐴𝐵 𝑐𝑀𝐴
1 − 𝑦𝐴∞
ln [
] (𝑡 − 𝑡𝑜 )
𝜌𝐴,𝑙𝑖𝑞
1 − 𝑦𝐴𝑠
160
140
120
100
80
60
40
20
0
y = 0.6472x + 0.5537
0
50
100
150
200
250
time [hr]
26-7
Slope = 0.6472
𝑐=
𝑃
=
𝑅𝑇
𝑐𝑚2
𝑐𝑚2
= 1.80 × 10−4
ℎ𝑟
𝑠
1.0 𝑎𝑡𝑚
𝑔𝑚𝑜𝑙𝑒
= 4.15 × 10−5
3
𝑐𝑚 ∙ 𝑎𝑡𝑚
𝑐𝑚3
82.06
∙ 293.9 𝐾
𝐾 ∙ 𝑔𝑚𝑜𝑙𝑒
𝑃𝐴∗ 0.254 𝑎𝑡𝑚
𝑦𝐴𝑠 =
=
= 0.254
𝑃
1.0 𝑎𝑡𝑚
𝐷𝐴𝐵 =
𝜌𝐴 ∙ 𝑠𝑙𝑜𝑝𝑒
1−0
2 ∙ 𝑐 ∙ 𝑀𝐴 ln [1 − 𝑦 ]
𝐴𝑠
𝑔
𝑐𝑚2
∙ 1.80 × 10−4 𝑠
3
𝑐𝑚3
𝑐𝑚
𝐷𝐴𝐵 =
= 0.101
𝑔𝑚𝑜𝑙𝑒
𝑔
1−0
𝑠
2 ∙ 4.15 × 10−5
∙
58
∙
ln
[
]
𝑔𝑚𝑜𝑙𝑒
1 − 0.254
𝑐𝑚3
0.79
b. Determine DAB by correlation
Hirschfelder Correlation
𝐷𝐴𝐵 =
1
1 2
0.001858 ∙ 𝑇 3/2 [𝑀 + 𝑀 ]
𝐴
𝐴
2
𝑃𝜎𝐴𝐵
ΩD
508 𝐾
1
3
𝜎𝐴 = 2.44 (47.4 𝑎𝑡𝑚) = 5.380 Å, 𝜎𝐵 = 3.617 Å,, 𝜎𝐴𝐵 =
𝜀𝐴
𝜅
5.380+3.617
2
= 4.50 Å
𝜀
= 0.77(508 𝐾) = 391.16 𝐾, 𝜅𝐵 = 97 𝐾,𝜀𝐴𝐵 = √391.16 ∙ 97 𝐾 = 194.79 𝐾
𝜅𝑇
293.9 𝐾
=
= 1.509,
𝜀𝐴𝐵 194.79 𝐾
𝐷𝐴𝐵 =
∴ Ω𝐷 = 1.195
1
1 1/2
+
]
𝑐𝑚2
58.0 29.0
=
0.088
1.0 ∙ 4.50 2 ∙ 1.195
𝑠
0.001858 ∙ 293.93/2 [
26-8
26.23
nonporous
support
well-mixed
liquid phase
enzyme
layer
cA∞= 1.0 mmol/m3
z=0
z = L = 0.2 cm
cAL = 0.25 mmol/m3
a. Boundary Conditions
z = 0, CA = CA
z = L,
dC A
=0
dz
b. Rate constant k
at z = L

k 
cosh  ( L − z )

DAe 
c A

c AL = c Ao
=
cosh( )
cosh( )
c 
 c AL 
 = cosh −1  A 
=L
k
DAe
2

-5 cm 
2.0×10

3
 
s 

D 
−1  c A 
-1  1.0 mmole/m
k = Ae
cosh
=
cosh




2
2 
3 
L 
( 0.2 cm ) 
 0.25 mmole/m  
 c AL  
k = 2.13  10−3 s −1
c. Total transfer rate of product B, WB
26-9
=L
k
= 0.2 cm
DAe
2.13×10-3s -1
= 2.063
2
-5 cm
2.0×10
s
NA =
DAec Ao
DAe y AoC

 tanh  =

 tanh 
2

mmole  1m 
−5 cm  
2.0

10

1.0


s 
m3  100 cm 
NA = 
(2.063)tanh(2.063)
( 0.2 cm )
mmole A
m2  s
1 mol B
 1 mol B 
-6 mmole A 
2
-6 mmole A
WB =
NA S = 
 2.0  10
 ( 2.0 m ) = 4.0  10
2
1 mol A
m s 
s
 1 mol A 
N A =2.0  10-6
d. At  = 2.063, process is between diffusion control and reaction control
26-10
26.27
A = CO, B = CO2
a. Determine total molar flow rate out n2
Material balance on reactor for CO
IN – OUT + GEN = ACC
y A,1n1 − y A,2 n2 + WA = 0
 d 2 DAe cA0
WA = S  N A =
 tan 
4

k
 =
= 1.0 cm
DAe
6.0 s -1
= 3.87
cm 2
0.4
s
yCO ,2 + yO2 ,2 + yCO2 ,2 = 1
 yCO2 ,2 = 1 − yCO,2 − yO2 ,2 = 1.0 − 0.05 − 0.025 = 0.925 = y A0
P
1.0 atm
gmole
= 0.925
= 1.051 x 10-5
3
RT
cm3

cm  atm 
82.06
1073
K
(
)


gmole  K 

 π (10.0 cm )2   0.40 cm 2 /s  
gmole 
WA = 
1.051 x 10-5
3.87 ) tanh ( 3.87 )


3 (

  1.0 cm  
4
cm



gmole
WA = 1.277 x 10-3
s
gmole CO 

(0.0)n1 − 0.05n2 + 1.277 x 10-3
=0
s


gmole CO  60 s 
gmole CO
n2 = 0.0255

 = 1.53
s
min
 1.0 min 
c A0 = y A0C = y A0
b. Mole fraction y A on backside of catalyst layer at z = 
yA ( z) =
(
y A0 cosh ( − z ) k / DAe
(
cosh  k / DAe
)
)=
y A0
0.925
=
= 0.0386
cosh ( ) cosh ( 3.87 )
26-11
27.5
Given:
A = Arsenic (As), B = solid silicon (Si)
T = 1050 C
Wafer dimensions: L = 1.0 mm, d = 10 cm
Solute A: cAs = cA* = 2.3 × 1021 atoms As/cm3 at z = 0
cAo = 2.3 × 1017 atoms As/cm3 initial As concentration in solid Si
DAB = 5.0 × 10−13 cm2/s
a. Time required and mA(t) to achieve CA(z,t) = 2.065 × 1020 atoms As/cm3 at z = 0.5 m
USS diffusion in semi-infinite medium—concentration profile
atoms
atoms
− 2.065  1020
3
cm
cm3
= erf ( ) =
atoms
atoms
c As − c Ao
2.3  1021
− 2.3  1017
3
cm
cm3
erf ( ) = 0.9103
2.3  1021
c As − c A ( z , t )
=
z
2 DAB t
From Appendix L,  = 1.2
z2
t= 2
=
4 DAB
−5
4 (1.2 )
2
2
2

−13 cm 
5.0

10


s 

= 868 s
( 5.0  10 cm /s ) (868 s ) = 2.08  10 cm < < L
−13
DAB  t =
Note
S=
( 5.0 10 cm )
2
−5
d2
4
USS diffusion in semi-infinite medium—total mass loaded in medium
mA ( t ) − mAo =
d2
4
4 DAB t

( cAs − cAo )
27-1
2

−13 cm 
4
5.0

10
2

 (868 s )
π (10 cm )
s 
atoms

mA ( t ) − mAo =
2.3  1021 − 2.3  1017
4

cm3
mA ( t ) − mAo = 4.25  1018 atoms As
(
d2
(
)
 (10cm )
)
2
L = 2.3  10 atoms As/cm
( 0.1cm ) = 1.8 1016 atoms As
4
4
mA ( t ) = 1.8  1016 atoms As + 4.25  1018 atoms As = 4.25  1018 atoms As
mAo = c Ao
17
3
b. Plot of z and vs. t at  = 1.2
1.20
1.00
0.80
z (m)
z1/2 (m)1/2
1.00
0.60
0.40
0.80
0.60
0.20
0.40
0.00
0
1000
2000
3000
Time, t (s)
20
4000
40
60
80
t1/2 (s)1/2
Note plot of z vs. t1/2 is linear
c. Transfer rate WA at t = 5.0 min, 10.0 min
 d 2 DAB
WA = S  N A z =0 =
( c − c ) for t > 0
4
 t As Ao
At t = 5.0 min = 300 s
WA =
 (10 cm )
2
4
( 5.0  10 cm /s ) 2.3  10
(
π ( 300 s )
−13
2
21
− 2.3  1017
) atoms
cm
− 2.3  1017
) atoms
cm
3
WA = 4.16  1015 atoms As / s
At t = 10.0 min = 600 s
WA =
 (10 cm )
4
2
( 5.0  10 cm /s ) 2.3  10
(
π ( 600 s )
−13
2
21
3
WA = 2.94 1015 atoms As / s
Note as t increases, WA decreases. At t = 0, the flux is not defined because no concentration
gradient for As in the solid Si initially exists.
27-2
27.16
A = solvent, B = polymer
x1 = L = 6.0 mm = 0.60 cm
𝐶𝐴∞ = 0
wA0 = 0.010
DAB = 2.0 x 10-6 cm2/sec
Determine time (t) required for wA(x,t) = 0.00035 at 1.2 mm (0.12 cm) from the exposed surface
USS Diffusion in a slab, use Concentration -Time charts
𝑚 = 0 (𝑛𝑜 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒)
𝑌=
𝐶𝐴∞ − 𝐶𝐴 (0, 𝑡) 𝐶𝐴𝑠 − 𝐶𝐴 (𝑥, 𝑡) 𝑤𝐴𝑆 − 𝑤𝐴 (𝑥, 𝑡) 0 − 0.00035
=
=
=
= 0.035
𝐶𝐴∞ − 𝐶𝐴0
𝐶𝐴𝑠 − 𝐶𝐴𝑜
𝑤𝐴𝑆 − 𝑤𝐴0
0 − 0.010
𝑛=
𝑥
0.60 𝑐𝑚 − 0.12 𝑐𝑚
=
= 0.80
𝑥1
0.60 𝑐𝑚
Figure F.1
𝑋𝐷 ≅ 1.0 =
𝑡 = 1.0
𝐷𝐴𝐵 𝑡
𝑥12
(0.60 𝑐𝑚)2
𝑥12
= 1.0
= 180,000 sec (50 hr)
𝐷𝐴𝐵
2.0 𝑥 10−6 𝑐𝑚2 /𝑠𝑒𝑐
27-3
28.1
Given:
T = 300 K, P = 1.0 atm


Sc =
=
 DAB DAB
 P  T 
DAB ,T2 , P 2 = DAB ,T1 , P 1  1  2 
 P2  T1 
3/2

 D |T1
 D |T2
Neglect temperature dependency on  D
 P  T 
DAB ,T2 , P 2 = DAB ,T1 , P 1  1   2 
 P2   T1 
3/2
Appendix I:  H2O,liquid (300K ) = 8.788  10−7
m2
m2
,  air (300 K) = 1.5689  10−5
s
s
O2 gas (A) in Air (B)
Appendix J: DAB (273 K,1.0 atm) = 0.175
cm 2
s
3/2
cm2  1 atm  300 K 
cm2
DAB ,T2 , P 1 = 0.175
=
0.202



s  1 atm  273 K 
s
2
m
1.5689  10−5
vB
s = 0.786
Sc =
=
2
m
DAB
0.202  10−4
s
O2 (A) dissolved in liquid H2O (B)
 L ,H O (300 K) = 8.76  10−4 Pa s = 0.876 cP , VA = 25.6
2
cm3
gmole
Determine DAB using Hayduk and Laudie correlation
DAB = 13.26  10 −5  L −1.14VA−0.589
(
)
DAB = 13.26×10 −5 ( 0.876 )
v
Sc = B =
DAB
8.788  10−3
2.28  10−5
−1.14
( 25.6 )
−0.589
= 2.28  10 −5
cm 2
s
2
cm
s = 385
2
cm
s
CO2 gas (A) in air (B)
28-1
Appendix J: DAB (273 K,1 atm) = 0.136
DAB ,T2 , P 1 = 0.136
cm2  1 atm  300 K 



s  1 atm  273 K 
cm 2
s
3/2
= 0.157
cm2
s
CO2 (A) dissolved in liquid H2O (B)
Determine DAB using Hayduk and Laudie correlation
cm3
VCO2 = 34.0
gmole
DAB = 13.26 10−5  L −1.14VA−0.589
(
DAB = 13.26  10
v
Sc = B =
DAB
−5
)( 0.876
8.788  10−7
1.93  10−9
−1.14
) ( 34.0 )
−0.589
=1.93  10
−5
cm 2
s
2
m
s =455
2
m
s
Schmidt numbers are higher in liquids relative to gases
28-2
28.4
gas distributor
Drying Chamber
cA, ≈ 0
60 m3/min air H = 1.0 m
27°C, 1.0 atm (W = 1.5 m)
L = 1.5 m
painted
steel plate
heated surface (27°C)
Given:
A = solvent, B = air
T = 300 K, P = 1.0 atm
g
,
cm3
Plate dimensions: L = 150 cm, H = 100 cm, W = 150 cm
m3
1.0
3
3
v
m
m
m
s
v = o =
= 0.667
vo = 60
= 1.0
W  L 1.0 m 1.5 m
s
min
s
Paint coating:
= 0.10 cm , wA = 0.30 ,  A,liq = 1.5
Physical parameters:
2
cm2
g
*
−5 m
p
=
0.138
atm
,
,
,
 air (300K) = 1.5689 10
DAB = 0.097
M A = 78
A
s
s
gmole
Physical System: boundary layer between surface and bulk gas
Source for A: solvent coating surface
Sink for A: bulk flowing gas
a. ShL over entire plate
Re and Sc:
m

0.667  (1.5 m )

v L
s 
Re =  = 
= 63, 771 laminar flow
2


−5 m 
1.5689  10

s 

28-3
m2
1.5689  10
v
s = 1.617
Sc = air =
2
m
DAB
0.097  10−4
s
−5
Laminar flow over a flat plate:
1/3
ShL = 0.664 Re1/2
= ( 0.664 )( 63, 771)
L Sc
1/2
(1.617 )
1/3
= 197
b. Emissions rate WA
Transfer rate across boundary layer
WA = N A S = kc (cAs − cA, )S
cA,  0 in bulk air flow
S = L W = (150 cm)(150 cm) = (150 cm)2
cm 2
197  0.097
Sh D
s = 0.127 cm
kc = L AB =
L
150 cm
s
( 0.138 atm )
p*A
mol
cAs =
=
=5.61  10−6 3
3
RT 
cm
cm  atm 
 82.06
 ( 300 K )
mol  K 

cm 
g  60 s 
g
2

−6 mol 
WA =  0.127
150 cm )  78
 5.61  10

 =75.0
3 (
s 
cm 
min

 mol  min 
c. Time required for complete drying of paint
Material balance on solvent in paint layer
IN – OUT + GEN = ACC
0 − WA + 0 =
0
t
mAo
0
dmA
dt
 dmA =  WAdt
t=
mAo
WA
28-4
g 

mA,0 = V    x A =  L2    wA = ( 0.1 cm ) (150 cm) 2 1.5 3  (0.30) = 1012.5 g
 cm 
m
1012.5g
t = A,0 =
= 13.5 min
WA 75 g/min
d. Boundary layer thickness at x = L = 1.5 m
( 5)(1.5 m ) = 2.97 cm
5 x
=
Re x
63771

2.97 cm
 c = 1/3 =
=2.53 cm
1/3
Sc
(1.617 )
=
These are both much smaller than the height of the drying chamber.
e. Volumetric flowrate of air needed for WA = 150 g A/min, all other variables unchanged
1/2
v

=
=   ,new 
WA,old
kc ,old  v,old 
WA,new
kc ,new
2
1/2
v

=  o ,new 
v

 o,old 
W
 
m3  150 g/min 
m3
vo ,new = vo ,old  A,new  =  60
=
240


W

min
 A,old   min   75 g/min 
2
28-5
29.3
Given:
A = O2 (solute), B = H2O (solvent)
T = 293 K, P = 2.0 atm
Gas: y A = 0.21, yN2 = 0.78, yAr = 0.010
Solvent: B,liq = 1000 kg/m3, MB = 18 kg/kgmole
Solute: H = 40,000 atm with p A,i = HxA,i
a. Determine xA*
x*A =
p A y A P ( 0.21)( 2.0 atm )
=
=
= 1.05  10−5
H
H
( 40,100 atm )
b. Determine cAL*
c* AL = x*AcL
kg 

1000 3 
kgmole
m 
xA 2
= xA
= 1.05  10−5 
=5.83  10 −4
M H2O
MB
m3

kg 
18 kgmol 


 H O,liq
*
 B ,liq
*
c. Determine cAL* if P = 4.0 atm
xA * =
p A y A P ( 0.21)( 4.0atm )
=
=
= 2.10  10−5
H
H
( 40,100 atm )
kg 

1000 3 

 H O,liq
kgmole
m 
c* AL = x*AcL = x*A 2
= 2.10  10−5 
= 1.17  10 −3
M H2O
m3

kg 
18.0 kgmole 


(
)
29-1
29.4
Given:
A = ClO2 (solute), B = H2O (solvent)
T = 293 K, P 1.5 atm
Operating point: yA = 0.040, xA = 0.00040
Solute: MA = 67.5 kg/gmole
Solvent: B,liq = 992.3 kg/m3, MB = 18 kg/kgmole
Film Mass Transfer Coefficients:
k x =1.0
a.
kgmole
kgmole
(liquid film), kG = 0.010 2
(gas film)
2
m s
m  s  atm
pA vs. CAL equilibrium line and operating point
Operating point:
c AL = x AcL
xA
 H O ,liq
2
M H2O
= xA
 B ,liq
MB
( 992.3 kg/m ) = 0.022 kgmole
= ( 0.00040 )
3
(18 kg/kgmole )
m3
p A = y A P = (0.06)(1.5 atm) = 0.090 atm
0.12
cAL , pA
0.10
pA (atm)
0.08
0.06
0.04
cAL,i , pA,i
0.02
pA*
cAL*
0.00
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
cAL (kgmole/m 3)
Gas Absorption process since the operating point is above the equilibrium line
29-2
b. Determine m
Get H from slope of PA vs. CAL data, H = 0.772 atm∙m3/kgmole
(
)(
)
0.772 atm  m3 /kgmole 55.1 kgmole/m3
HcL
=
= 28.4
P
(1.5 atm )
c. Determine kL

kg 
18 kgmole 
k
MB 
gmole 
m


kL = x  kx
= 1.0
= 1.81  10−5

2
CL
 B ,liq 
m s 
s
kg   1000 gmole 
 992.3 3 

m   kgmole 

∗
d. Determine 𝑐𝐴𝐿
m=
c*AL =
( 0.06 atm )
pA
kgmole
=
= 0.078
3
H
m3
0.772 atm  m /kgmole
(
)
e. Compositions at the gas-liquid interface, pA,i and cAL,i
Recall −
k L ( p A − p A ,i )
=
kG ( c AL − c AL ,i )
For linear equilibrium line, pA,i = HcAL ,i


atm  m3
m

0.090
atm
−
0.772
c AL ,i 
− 1.81  10−5 

kgmole
( pA − HcAL,i )  
k
s 

− L =
=
kgmole
kG



( cAL − cAL,i ) 1.0  10−5 kgmole
− c AL ,i 

 0.022
2
3
m  s  atm 
m



kgmole
m3

atm  m3  
kgmole 
p A,i = HcAL ,i =  0.772
 0.0503
 = 0.039 atm
kgmole  
m3 

c AL ,i = 0.0503
f. Determine Ky by two approaches
Path 1: Convert kG to ky, get Ky from kx, ky, m
kgmole 

−5 kgmole
k y = kG P = 1.0  10−5 2
 (1.5 atm ) =1.5  10
m  s  atm 
m 2s

29-3
-1


 1 m


1
28.4
−5 kgmole
Ky =  +  = 
+
 = 1.05  10
kgmole
kgmole
m 2s
 k y k x 
1.5  10−5 2

1.0  10−3
2
m s  atm
ms 

−1
Path 2: KG from kL kG, H, then convert KG to Ky
-1

atm  m3 
−1
0.772

 1 H
1
kgmole
kgmole 
KG =  +  = 
+
= 7.10  10−6 2
m  s  atm
1.0  10−5 kgmole 1.81  10−5 m 
 kG k L 
2

ms
s 

kgmole 

−5 kgmole
K y = K G P =  7.10  10−6 2
 (1.5 atm) = 1.05  10
m  s  atm 
m 2s

g. Flux NA
Base on overall gas phase mole fraction driving force as an example

atm  m3  
kgmole 
p A * = Hc AL =  0.772
 0.022
 = 0.017 atm
kgmole  
m3 

p * 0.017 atm
yA* = A =
= 0.0113
P
1.5 atm
kgmole 

N A = K y ( y A − y* A ) = 1.05  10−5
 ( 0.040 − 0.0113)
m 2s 

kgmole
N A = 3.0  10−5
m 2s
29-4
29.14
gmole
gmole
m3
c AL ,o = 0.5
, vo = 2.0
, c AL = 0.35
, L = 10.0 m, p A  0 , c*AL  0
3
3
m
m
s
T = 293 K, H = 0.50
kc = 2.67 10−4
m3  atm
m3  atm
= 500
, P = 1.0 atm
gmole
kgmole
m
m
, k L = 5.5 10−3
s
s
a. Determine K L
m
k
gmole
s
kG = c =
= 0.0111 2
3
m  atm
RT
m  s  atm
(8.206 x 10−5
)(293 K)
gmole  K
2.67 x 10−4
1
1
1
= +
K L k L H kG
−1




1
1
m
 = 0.00276
KL = 
+
3
m  atm
gmole 
s
 5.5 x 10−3 m
(0.50
)(0.0111 2
)

s
gmole
m  s  atm 

b. Develop material balance model and determine surface area S
Assume: 1) steady state, 2) no reaction, 3) constant T and P, 4) liquid stripping process.
A = Species A, B = Air, C = Water
IN – OUT – FLUX OUT = 0
cAL ,o vo − cAL vo − N A S = 0
cAL ,o vo − cAL vo − SK L (c AL − c*AL ) = 0
Determine S , the surface area between the two phases
29-5
m3
gmole
gmole
2.0 (0.5
− 0.35
)
3
vo (cAL ,o − cAL )
s
m
m3
S=
=
= 311 m 2
*
m
gmole
gmole
K L (cAL − cAL ) 0.00276 (0.35
−0
)
s
m3
m3
c. Determine v
Laminar flow over a flat plate:
B =
B
=
B
kg
2
m  s = 1.503 x 10−5 m
kg
s
1.206 3
m
1.813 x 10−5
B 
13
kc L  DAB 

v =


L  0.664 DAB   B 




2
2


m2 
cm 2
−4 m
−4 m
(2.67
x
10
)(10
m)
(0.08
)(1
x
10
)


2 
s 
s
s
cm 

2
2
cm 2
m
cm

−4


1.503 x 10 −5

 0.664(0.08 s )(1 x 10 cm 2 ) 
s


13
v =
1.503 x 10
−5
10 m
2


 = 0.249 m
s



d. Determine pA,i
p A − p A ,i
kL
m3  atm
− =
= −0.495
kG c AL − c AL ,i
gmole
0 atm − p A,i
m3  atm
=
gmole 0.35 gmole − c
AL ,i
m3
m3  atm
p A,i = HcAL ,i = 0.50
cAL ,i
gmole
−0.495
Two equations, two unknowns: p A,i = 0.087 atm, cAL ,i = 0.174
gmole
m3
29-6
29-7
30.1
Given:
A = solvent, B = air
T = 298 K, P = 1.0 atm
Sphere: D = 1.0 cm, 0.12 g A/cm2 liquid solvent on sphere at t = 0
Physical parameters:
MA = 78 g/gmole, pA* = 1.17 × 104 Pa (298 K), DAB = 0.0962 cm2/s, B = 0.156 cm2/s
Physical System: boundary layer between surface of sphere and bulk gas
Source for A: solvent coating surface of sphere
Sink for A: bulk flowing gas
a. Time to dry for still air
Mass balance on solvent evaporating from paint-coated sphere
Initial mass of solvent on sphere
mAo =
 D 2 0.12 g
cm 2
MA
=
π(1.0 cm) 2  0.12 g 
= 4.83 x 10-3 gmole

2 
( 78 g/gmole )  cm 
IN – OUT + GEN = ACCUMULATION (moles A/time)
0 − N AS + 0 =
t
mA
0
m Ao
dmA
dt
−WA  dt =  dmA
WA  t = mAo − mA
WA = kc ( c*A − c A )  D 2
Determine kc, WA, then t
For still air, Sh = 2.0
D
0.0962 cm 2 /s
kc = Sh AB = 2.0
=0.192 cm/s
D
1.0 cm
30-1
1.17 x 104 Pa )
(
pA
c =
=
= 4.72 gmole/m3 = 4.72x10-6 gmole/cm3
3
R T ( 8.314 m  Pa/gmole  K ) (298 K)
*
A
WA = D 2 kc (c*A − c A ) = π (1.0 cm ) ( 0.192 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.85 x 10-6 gmole/s
2
Final mass of solvent on sphere − mA = 0 for dried paint coating on sphere
t=
mAo
4.83 x 10-3 gmole
=
=1697 s
WA 2.85 x 10-6 gmole/s
b. Time to dry for flowing air with v∞ = 1.0 m/s
Determine kc, WA, then t
For air at 298 K and 1.0 atm, air = 1.56 × 10−5 m2/s (Appendix I)
Re =
Sc =
v D
 air
 air
DAB
=
=
(1.0 m/s )( 0.01m ) = 641
1.56 x 10-5 m 2 /s
0.156 cm2 /s
=1.62
0.0962 cm 2 /s
Froessling equation for gas flow around a single sphere
Sh =
kc D
= 2 + 0.552 Re1/2 Sc1/3 =2.0+0.552(641)1/2 (1.62)1/3 =18.4
DAB
kc = Sh
DAB
0.0962 cm/s
= 18.4
= 1.77 cm/s
D
1.0 cm
WA = D 2 kc (c*A − c A ) = π (1.0 cm ) (1.77 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.63 x 10-5 gmole/s
2
t=
mAo
4.83 x 10-3 gmole
=
= 184 s
WA 2.63 x 10-5 gmole/s
30-2
30.5
Let A = O2, B = H2O (liquid)
a. Material balance
In − Out + Generation = Accumulation
[ S  N A + v LWC Alo ] − [v LWC Al ] + 0 = 0
*
[( N t DL)(k L (C AL
− C AL )) + v LWC Alo ] − [v LWC Al ] = 0, combineterms to solve for C AL
C AL =
*
(v LW )C Alo + ( N t DLk L )C AL
(v LW ) + ( N t DLk L )
b. Mass transfer coefficient, kL
ShD =
kL D
= 0.281( Re ') 0.6 ( Sc) 0.44
DAB
v =
0.44
  
 DAB 


 D 


 DAB 
( 50 mol/s )( 0.018 kg/mol )
v D
k L = 0.281  
  
0.6
nM w
=
= 1.80×10-3 m/s
LW 
(1 m )( 0.50 m ) ( 998.2 kg/m3 )
 (1.80×10-3 m/s)(0.02 m) 
k L = 0.281

(0.995×10-6 m 2 /s)


0.6
 (0.995×10-6 m 2 /s) 


-9
2
 (2×10 m /s) 
0.44
 (2×10-9 m 2 /s) 
−6

 = 3.72  10 m/s
(0.02
m)


c. Outlet concentration, CAL
For CAL,o = 0
C AL =
*
( N t DLk L )C AL
(10)π(0.02 m)(1 m)(3.72×10-6 m/s)(5 mol O 2 /m3 )
=
(v LW ) + ( N t DLk L ) (1.80×10-3 m/s)(0.02 m)(1 m)+(10)π(0.02 m)(1 m)(3.72×10-6 m/s)
C AL = 0.0129 mol/m3
Increase CAL* by increasing pressure on the oxygen side to increase CAL.
30-3
30.13
Let A = CO2, B = H2O (liquid)
a. Determine kL
Gr =
db3  L g 
 L2

=
(2.0 x 10-3 m)3 (998.2 kg/m 3 )(9.81m/sec 2 )( ( 998.2-1.7967 ) kg/m3 )
(993 x 10-6 kg/m  sec) 2
= 79161
0.995 x10−6 m2 / sec
= 562
DAB 1.77 x10−9 m2 / sec
db = 2.0 mm
Sc =
Sh =
=
k L db
= 0.31Gr1/3 Sc1/3
DAB
D 
k L = 0.31Gr1/3 Sc1/3  AB 
 db 
 1.77 x10−9 m2 / sec 
−5
k L = (0.31)(79161) (562) 
 = 9.721x10 m / sec
−3
2 x10 m


1/3
1/3
b. Determine if the inlet flow rate of CO2 gas is sufficient
PA = 2.0 atm
PA
2.0 atm
=
=0.06757 kmol/m3
3
H 29.6 atm  m /kmol
6g
A
*
WA = N A i V = k L
V ( C AL
− C AL ,o ) =
V
db
*
C AL
=
(9.721 x 10-5 m/sec)
6(0.05)
2.0 m 3 ) (0.06757 kmol /m 3 ) = 1.97 x 10-3 kmol CO 2 /sec
(
-3
2.0 x 10 m
Convert WA in kmol/sec to volumetric flowrate of STD m3 CO2/ min
VCO 2 = WA M A /  A = (1.97 x 10-3 kmol CO 2 /sec)(44kg/kmol)(1m3 /1.7967 kg)(60 s/min)
= 2.89 m3 CO 2 /min
The inlet flow rate of CO2 from the problem statement (4.0m3/min) is larger than 2.89 m3/min,
thus the inlet flow rate of CO2 gas is sufficient to ensure that the CO2 dissolution is mass transfer
limited.
c. Determine CAL,out
Assumptions : 1 ) no chemical reaction, 2) steady state, 3) dilute system
30-4
In – Out +Generation = Accumulation (liquid phase – mole A/time)
N A  Ai − C ALV0 + 0 = 0
−CALV0 + kL a V  (C*AL − C AL ) + 0 = 0
6(0.05)
(0.06757 kmolCO 2 /m 3 )
-3
k L a  V  (C )
2.0 x10 m
C AL ,out =
=
(0.45m3 /min)(1min/60 sec)
6(0.05)
V0
+(9.721 x 10-5 m/sec)
+ kL a
3
2.0 m
2.0 x 10-3 m
V
C AL ,out = 0.0537 kmol/m3
*
AL
(9.721 x 10-5 m/sec)
30-5
30.18
Let A = CO2, and B = H2O (liquid)
DAB = 1.77x10-9 m2/sec
w
36 g / sec
V0 =
=
= 3.61x10−5 m3 / sec
 L 998.2 x103 g / m3
a. Determine CAL*
P
2.54 atm
*
C AL
= A =
=0.10 kgmol/m3
3
H
atm  m
25.4
kgmol
b. Determine kL
4w
4(36 g / sec)
Re =
=
= 769
 D L  (6.0cm)(9.93x10−3 g/ cm − sec)
Sc =
L
 L DAB
=
993x10−6 kg / m2 sec
= 562
(998.2 kg/ m3 )(1.77 x10−9 m2 / sec)
1/6
  2 g  z3 
k z
Sh = L = 0.433( Sc)1/2  L 2  (Re L ) 0.4
DAB
 L 
1/6
  2 g  z3 
D 
k L = 0.433( Sc)  L 2  (Re L ) 0.4  AB 
 z 
 L 
1/2
1/6
-9
2
 (998.2 kg/m 3 ) 2 (9.81 m/sec 2 )(2m)3 
0.4  1.77 x 10 m /sec 
k L = 0.433(562) 

 (769) 
2.0 m
(993 x 10-6 kg/m-sec) 2




-5
=2.69 x 10 m/sec
1/2
c. Determine CAL,out
Assumption: 1) no reaction, 2) Steady state, 3) dilute system
Material balance on differential volume element (A = CO2)
In – Out +Generation = Accumulation (liquid phase – mole A/time)
V0C AL + N A   D  z − V0C AL
z
z +z
+0=0
÷ Δz, rearrangement, z → 0
dC AL
V0
+ N A  D = 0
dz
30-6
dC AL  D
*
+
k L (C AL
− C AL ) = 0
dz
V0
C AL ,out
L
dC AL
k
−
= L  D  dz
C AL ,o C * − C
0
V0
AL
AL
*
 C AL
− C AL ,o  k L
ln  *
 =  DL
C −C
V0
AL ,out 
 AL


 −k L

 -π(0.06 m)(2.69x10-5 m/sec)(2 m)  
*
3
C AL ,out = C AL
1
−
exp

DL
=
0.10
kgmol/m
1-exp
) 


 (

3.61x10-5 m3 /sec



 V0
 

3
C AL ,out = 0.0245 kgmol/m
30-7
31.4
a. Determine AGs,min
(1 − R ) xA2 AL2 = xA1 AL1
(1 − 0.8)( 0.01)(100 ) = xA1 AL1 = 0.2
lbmol
hr
ALs = AL2 (1 − x A 2 ) = (100 )(1 − 0.01) = 99
lbmol
hr
 x

x A1 AL1 = ALs  A1 
 1 − x A1 
 x

0.20 = ( 99 )  A1 
 1 − x A1 
x A1 = 2.02  10−4
AL1 =
AL s
99
lbmol
=
= 99.02
−4
1 − x A1 1 − 2.02  10
hr
y A1 AG1,min + x A 2 AL2 = x A1 AL1 + y A 2,min AG2,min
y*A 2,min =
H
 515 atm 
xA2 = 
 ( 0.01) = 0.412
PT
 12.5 atm 
0 + ( 0.01)(100 ) = ( 0.20 ) + ( 0.412 ) ( AG2,min )
AG2,min = 1.94
lbmol
hr
AGs ,min = (1 − y A2,min ) AG2,min = (1 − 0.412 )(1.94 ) = 1.14
lbmol
hr
b. Determine yA2,min
From part (a) above,
y*A 2,min = 0.412
31-1
31.9
a. Characterize molar flowate and mole fraction composition of all terminal streams
kgmol
kgmol
AGs = AG2 (1 − y A1 ) = 2.0
(1 − 0.10) = 1.8
sec
sec
AGS 1.8 kgmol/sec
kgmol
AG2 =
=
=1.836
1 − y A2
1-0.02
sec
AG1 + AL2 = AG2 + AL1
kgmol
AL1 = AG1 + AL2 − AG2 = 2.0 - 1.837 + 3.0 = 3.163
sec
AG1 y A1 + AL2 x A 2 = AG2 y A 2 + AL1 x A1
(2.0)(0.10) + (3.0)(0.01) = (1.837)(0.020) + (3.163) xA1
xA1 = 0.061
Material Balance Summary:
AG1 (kgmol/sec) 2.0
AG2 (kgmol/sec) 1.84
AL1 (kgmol/sec) 3.163
AL2 (kgmol/sec) 3.0
yA1
0.1
yA2
0.02
xA1
0.061
xA2
0.01
b. Tower diameter (D) at 50% of flooding condition
Cf = 98 for 1.0 inch ceramic Intalox saddles (Table 31.2)
AL '1 = AL1 ( x A1 M A + (1 − x A1 ) M B ) = ( 0.061(17)+(1-0.061)(18) )( 3.163) =56.75 kg/sec
1/2
x − axis =
AL1'  G 


AG1   L − G 
1/2
kg 

kg


56.75
2.8




sec 

m3
=

 = 0.056
kgmol  
kg  1000 kg − 2.8 kg 


 2.0
  26.9
m3
m3 
sec  
kgmol  

y − axis = Y = 0.25
1/2

   (  −  ) g 
( 2.8)(1000 − 2.8 )(1.0 ) 
kg
 = 3.77 2
G = Y  G L 0.1 G c   = 0.25
0.1

C f L J
m sec

 
( 98)( 0.001) (1) 

kg
kg
53.8
53.8
AG1'
sec =14.27 m 2 at flooding, A =
sec = 28.54 m 2 at 50% of flooding
A= ' =
kg
kg
Gf
3.77 2
1.885 2
m sec
m sec
1/2
'
f
31-2
1/2
 4A 
D=

  
1/2
 4  14.27 m 2 
=
 = 4.25 m at flooding
π


1/2
 4  28.54 m 2 
D=
 = 6.03 m at 50% of flooding
π


c. Log-mean mass transfer driving force (yA – yA*)lm
y A1 − y*A1 ) − ( y A 2 − y*A 2 )
(
*
( y A − y A )lm =
 ( y A1 − y*A1 ) 

ln 
 ( y A 2 − y*A 2 ) 
x A1 = 0.061, y*A1 = 0.035
x A 2 = 0.01, y*A 2 = 0.005
( y − y ) = ( 0.10 − 0.035) = 0.065
( y − y ) = ( 0.02 − 0.005) = 0.015
A1
*
A1
A2
*
A2
(y − y ) =
A
*
A lm
( 0.065 − 0.015)
 0.065 
ln 

 0.015 
= 0.034
0.12
XA1, YA1
0.10
0.08
YA0.06
0.04
XA2, YA2
0.02
0.00
0.00
0.02
0.04
0.06
0.08
XA (mole dissolved NH 3 / mole water)
31-3
31.13
a. Determine ALs,min
y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1,min AL1,min
H 0.46 atm
=
=0.023
P
2.0 atm
y
0.03
x A1,min = A1 =
= 0.130
m
0.23
AG1 (1 − y A1 ) 2.0(1-0.03)
lbmol
AG2 =
=
= 1.95
1-0.005
hr
(1 − y A2 )
m=
( 0.03)( 2 ) + 0 = ( 0.005 )(1.95 ) + ( 0.130 ) AL1,min
AL1,min = 0.387
lbmol
hr
ALs ,min = AL1,min (1 − x A1,min ) = ( 0.387 )(1-0.130 ) =0.336
lbmol
hr
b. Estimate required K’ya for z = 6.0 ft
z = H OG N OG
H OG =
Z
N OG
AL2 + AG1 = AL1 + AG2
1.0 + 2.0 = AL1 + 1.95
lbmol
hr
x A 2 AL2 + y A1 AG1 = x A1 AL1 + y A 2 AG2
AL1 = 1.05
0 + 0.03(2.0) = x A1 (1.05) + 0.005(1.95)
x A1 = 0.048
( AG1 + AG2 )
lbmol
2
hr
( AL1 + AL2 )
lbmol
AL =
= 1.025
2
hr
*
y A1 = mx A1 = 0.23(0.048) = 0.011
AG =
= 1.975
y*A 2 = mx A 2 = 0.23(0) = 0.0
31-4
N OG =
( y A1 − y A2 )
(y − y )−(y − y )
(y − y )

ln 
 ( y − y ) 
*
A1
A1
H OG =
G=
A2
A1
*
A1
A2
*
A2
=
( 0.03 − 0.005)
= 2.39
( 0.03 − 0.011) − ( 0.005 − 0 )
 ( 0.030 − 0.011) 
ln 

 ( 0.005 − 0 ) 
6.0 ft
G
=2.51ft = '
2.39
Kya
( AG1 + AG2 )
K y' a =
*
A2
2A
G
=
H OG
= 10.07
lbmol
ft 2 hr
lbmol
ft 2 hr = 4.0 lbmol
2.51 ft
ft 3 hr
10.07
c. New size of packing at 45% of flooding
 G1' 
 '  = 0.26
 G f old
G 'f ,old =
G1'
0.26
 G1' 
 '  = 0.45
 G f new
G 'f ,new =
G1'
0.45
 G1' 


 0.45 
1/2
 380 
=
=


G 'f ,old  G1'   C f ,new 


 0.26 
C f ,new = 1138
G 'f ,new
Therefore 3/8 inch packing size is suitable
31-5