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Cleo Madeline B. Lorque David Ian D. de Sagun BS Chemical Engineering III Principles of Transport Processes 1. Problem 2.B.3 Laminar flow in a narrow slit Fig. 2B.3 Flow through a slit, with B <<W<<L a. A Newtonian fluid is in laminar flow in a narrow slit formed by two parallel walls a distance 2B apart. It is understood that B<<W, so that “edge effects” are unimportant. Make a differential momentum balance, and obtain the following expressions for the momentum-flux and velocity distributions: τ xz = ( P o −P L L ) x v z = ( P o −P L ) B 2 2 μL [ 1− ( x B ) 2 ] In these expressions P=p+ρgh= p−ρgz b. What is the ratio of the average velocity to the maximum velocity for this flow? c. Obtain the slit analog of the Hagen- Poiseuille equation. d. Draw a meaningful sketch to show why the above analysis is inapplicable if B=W. e. How can the result in (b) be obtained from the results of chapter 2.5.
Transcript of Transpo.docx
Cleo Madeline B. LorqueDavid Ian D. de SagunBS Chemical Engineering IIIPrinciples of Transport Processes
1. Problem 2.B.3 Laminar flow in a narrow slit
Fig. 2B.3 Flow through a slit, with B <<W<<L
a. A Newtonian fluid is in laminar flow in a narrow slit formed by two parallel walls a distance 2B apart. It is understood that B<<W, so that “edge effects” are unimportant. Make a differential momentum balance, and obtain the following expressions for the momentum-flux and velocity distributions:
τ xz=( Po−PL
L )x
vz=( Po−PL ) B2
2μL [1−( xB )
2]In these expressions P=p+ ρgh=p− ρgz
b. What is the ratio of the average velocity to the maximum velocity for this flow?c. Obtain the slit analog of the Hagen- Poiseuille equation.d. Draw a meaningful sketch to show why the above analysis is inapplicable if B=W.e. How can the result in (b) be obtained from the results of chapter 2.5.
SOLUTION
Assume that:
density and viscosity are constant the flow of the fluid is in steady state the fluid flow is laminar Newton’s Law of Viscosity is applicable
The slit is very narrow (B<<W<<L);a.vx=0
v y=0
vz=v z(x)
Therefore,
MOMENTUM BALANCE IN z-DIRECTION
Convective momentum entering at z=0 is W ∆ xρ vz v z|z=0
Convective momentum leaving at z=L is W ∆ xρ vz v z|z=L
Momentum entering by viscous transport at x=x LW τ xz|x=x
Momentum leaving by viscous transport at x=x+∆x LW τ xz|x=x +∆ x
Gravity force LW ∆ xρ g
Pressure force at z=0 P0 W ∆ x
Pressure force at z=L -PL W ∆ x
W ∆ xρ vz v z|z=0−W ∆ xρ vz v z|z=L+LW τ xz|x=x−LW τ xz|x=x +∆ x+LW ∆ xρg−PL W ∆ x+P0 W ∆ x=0
LW τ xz|x=x−LW τ xz|x=x +∆ x+LW ∆ xρg−PL W ∆ x+P0 W ∆ x=0
LW τ xz|x= x−LW τ xz|x= x+∆x+LW ∆ xρg−PLW ∆ x+P0W ∆ x=0
LW ∆ x
τ xz=−μ¿)
τ xz=−μd v z
dx
τ zx=−μ(∂ v x
∂ z+
∂ v z
∂ x)
τ zx=−μd vz
dx
τ yx=0
τ yy=0
τ yz=0
τ zy=0
τ zz=0
τ xx=0
τ xy=0
τ xz=τ zx
τ xz|x= x−τ xz|x=x +∆ x
∆ x+
P0−¿ PL
L+ρg=0¿
τ xz|x= x−τ xz|x=x +∆ x
∆ x=
PL−¿ P0
L−ρg ¿
lim∆ x→ 0
τ xz|x= x−τ xz|x=x +∆ x
∆ x=
PL−¿ P0
L−ρg ¿
P=P+ ρ gh=P−ρgz
P0=P0−ρgz
PL=PL−ρgz
at z=0 atz=-L
P0=P0+ ρ g (0 )
P0=P0
d τ xz
dx−¿
d τ xz
dx−[ P0−PL− ρgL
L ]−ρg=0
d τ xz
dx=[ P0−PL− ρgL
L ]+ρg
d τ xz
dx=
P0−PL−ρgL+ρgL
L
d τ xz
dx=
P0−PL
L
∫ d τ xz
dx=∫ P0−PL
L
∫ d τ xz=∫P0−PL
Ldx
τ xz=P0−PL
Lx+C1
To solve for C1 set boundary conditions.
x=0 τ xz=0 v z=0 Solid-liquid interface
0=P0−PL
L(0)+C1
PL=PL−ρgz
PL=PL−ρ g (−L)
0=C1
Substitute C1 in the original equation
τ xz=P0−PL
Lx+C1
τ xz=P0−PL
Lx+0
τ xz=P0−PL
Lx
τ xz=−μd v z
dx
−μd vz
dx=(Po−PL
L ) x+C1
d vz
dx=−[( Po−PL
μL )x+C1
μ]
∫ d vz=∫ [−( Po−PL
μL )x ]−C1
μdx
vz=−( Po−PL
μL ) x2
2+
C1
μx+C2
Set boundary conditions to solve for C2 x=B v z=0
0=−( Po−PL
μL ) B2
2+0+c2
C2=( Po−PL
2μL )B2
vz=−( Po−PL
μL ) x2
2+( Po−PL
2μL )B2
vz=( Po−PL
2 μL )B2[1−( xB )
2]b. At x=0, we can obtain the maximum velocity
vz=( Po−PL
2 μL )B2[1−( xB )
2]
vmax=(Po−PL
2 μL )B2[1−( 0B )
2]vmax=(Po−PL
2 μL )B2
vz=vmax [1−( xB )
2]The average velocity= ¿ vz>¿
¿ vz≥∫0
w
∫−B
B
v z ( x )dxdy
∫0
w
∫−B
B
dxdy
=
∫0
w
∫−B
B
[1−( xB )
2]dxdy
∫0
w
∫−B
B
dxdy
=1B∫
0
B
v z dx
=1B∫
0
B
vmax [1−( xB )
2]dx
vz=vmax
B∫0
B
[1−( xB )
2]dx
vz=vmax
B [B−( B )3
3 B2 ]= vmax
B [B−B3 ]
vz=vmax
B [ 2B3 ]
¿23
vmax
v z
vmax
=23
c. Hagen-Poiseuille
Mass Flowrate = ρ∗¿v z>¿ A
A = 2BW
= ρ∗¿v z>¿2BW
= ρ∗2
3 ( Po−PL
2 μL )B2∗2 BW
= ρ∗23 ( Po−PL
μL )B3W
=23¿W]ρ
d. The analysis is inapplicable if B = W. In the case of B <<< W, the velocity at z-direction ( vz ) exists as function of x, while if B = W the presence of wall at y=0 and y=B makes ( vz ) a function of x and y.
vz = vz (x,y)
e. Using Flow of Two Adjacent Immiscible Fluids we can get the ratio of average velocity < vz > to maximum velocity ¿¿).
vzI=1
b∫−b
0
v zI dx
¿( p¿¿o−pL)b
2
12 μI L ( 7 μ I+μII
μ I+μII )¿
vzII=1
b∫
0
b
v zII dx
= ( p¿¿o−pL)b2
12 μ II L¿ ( μI+7 μ II
μI+μ II )let b=B
since, it is a single phase μ I=μ II=μ
Fluid in
Fluid out
L
W
B
vz=(p¿¿o−pL)B2
12 μL ( 7 μ+μμ+μ )¿
¿( p¿¿o−pL)B2
12 μL ( 8 μ2 μ )¿
¿( p¿¿o−pL)B2
12 μL ( 82 )¿
¿( p¿¿o−pL)B2
3 μL¿
¿( p¿¿o−pL)B2
3 μL ( 22 )¿
¿( 23 )( p¿¿o−pL)B2
2 μL¿
since , vmax is equal ¿ ( p¿¿o−pL)B2
2 μL¿
vz=( 23 )vmax
v z
vmax
=23
2.a. Obtain Equation 2.6-9 from Equation 2.6-7b. Obtain Equation 2.6-12 from Equation 2.6-10
SOLUTIONa.
Equation 2.6-7
F(n)=∫0
2 π
∫0
π
(−( p+τ rr )¿r=R cosθ) R2 sinθ dθdϕ
Equation 2.6-9
F(n)= 43
π R3 ρg+2 πμ Rv∞
In the equation 2.6-5, the normal stress τ rr is equal to zero at r=R so,
τ rr¿r=R=0
Equation 2.6-4, the pressure distribution at the surface sphere is
p¿r=R=p0−ρgRcosθ−32
μv ∞
Rcosθ
The term p0 will be zero if we integrate Equation 2.6-7
p¿r=R=− ρgRcosθ−32
μ v∞
Rcosθ
∫0
2 π
∫0
π
cosθ¿¿¿
(ρgR+ 32
μv∞
R )R2∫0
2 π
∫0
π
cos2θ sin θ dθdϕ
∫0
2 π
(¿ ρgR+ 32
μv ∞
R)R2dϕ ¿
`23
(2π )(ρgR+ 32
μ v∞
R )R2
F(n)= 43
π R3 ρg+2πμ Rv∞
b.
Equation 2.6-10
F(t )=∫0
2 π
∫0
π
(τ rθ ¿r=R sinθ ) R2 sinθ dθdϕ
Equation 2.6-12
F(t )=4 π μ R v∞
since, τ rθ¿r=R=32
μ v ∞
Rsinθ
F(t )=¿ ∫0
2 π
∫0
π32
μ v∞
Rsinθ ( sinθ ) R2 sinθdθdϕ
F(t )=∫0
2 π
∫0
π32
μ v∞
R¿¿¿
F(t )=32
μ v∞ R∫0
2 π
∫0
π
¿¿¿
F(t )=43∫0
2 π32
μv∞ R dϕ
F(t )=2 μ v∞ Rϕ|2 π0
F(t )=4 πμR v∞
.
