Example 14:
Consider flow between two parallel plates separated by a distance 2 H with a uniform
heat flux imposed on both plates. The fluid is driven between the plates by an applied
pressure gradient in the x-direction. Assume the flow is laminar and fully developed
(du/dx = 0, v = 0, and w = 0). (a) Determine the fully developed velocity distribution of
the fluid as a function of the mean velocity. (b) Determine the fully developed
temperature distribution as a function of the surface and mean temperatures.
(c) Determine the Nusselt number for this flow.
Known: Pressure driven laminar flow between parallel plates with uniform surface heat
flux on both sides, H, du/dx = 0, v = 0, and w = 0
Assumptions:
1. steady flow
2. constant properties
3. Newtonian fluid
4. negligible radiation
5. negligible gravity effects
6. negligible viscous dissipation
6. laminar flow
7. fully developed
8. negligible end effects
9. conduction in y-direction much
greater than conduction in x-direction
y
Find: (a) u(y) and um, (b) T(y) and Tm, (c) Nu
Solution:
⎡ ⎛ y ⎞2⎤
H 2 dp
3
(a) u = um ⎢1− ⎜ ⎟ ⎥ , um = −
3 µ dx
2
⎣ ⎝H⎠ ⎦
€
⎡
⎛ y ⎞2 ⎛ y ⎞4 ⎤
17 um H 2 dTm
35
(b) T = Ts −
(Ts − Tm ) ⎢5 − 6 ⎜ ⎟ + ⎜ ⎟ ⎥, (Ts − Tm ) =
⎝H⎠ ⎝H⎠ ⎦
35 α
dx
136€
⎣
(c) Nu = 8.23
€
x
2H
Answer:
€
u(y)
(a) Derivation of velocity distribution, u(y), and mean velocity, um
Conservation of mass for constant property flow:
! ∂u ∂v ∂w
∇⋅V = + +
=0
∂x ∂y ∂z
For fully developed flow, v = w = 0, thus
€
∂u
= 0 and u( y )
∂x
Momentum equation for constant property flow of a Newtonian fluid:
!€
€
⎛ ∂V !
!⎞
!
!
ρ⎜
+ V ⋅ ∇V ⎟ = µ ∇ 2V − ∇p + ρ g
⎝ ∂t
⎠
x-component for 2-D steady flow in Cartesian coordinates:
€
⎛ ∂ 2 u ∂ 2 u ⎞ ∂p
⎛ ∂u
∂u ⎞
ρ ⎜ u + v ⎟ = µ ⎜ 2 + 2 ⎟ − + Fx
∂y ⎠
∂y ⎠ ∂x
⎝ ∂x
⎝ ∂x
µ
€
d 2 u ∂p
=
dy 2 ∂x
y-momentum reduces to ∂p ∂y = 0 , so p( x ) and
€
d 2 u 1 dp
=
dy 2 µ dx
€
€
Integrate twice to get:
€
u( y ) =
1 dp y 2
+ C1 y + C2
µ dx 2
Impose boundary conditions: u(−H ) = 0 and u( H ) = 0 (no-slip at wall):
€
u(−H ) =
u( H ) =
€
€
H 2 dp
− H C1 + C2 = 0
2 µ dx
€
€
H 2 dp
+ H C1 + C2 = 0
2 µ dx
Adding these equations we get:
H 2 dp
⇒ C2 = −
and ⇒ C1 = 0
2 µ dx
€
2
H 2 dp ⎡ ⎛ y ⎞ ⎤
u( y ) = −
€⎢1− ⎜ ⎟ ⎥
2 µ dx ⎣ ⎝ H ⎠ ⎦
Thus, the velocity profile is parabolic.
€ Recall definition for mean velocity, u , and mass flow rate, m˙ :
m
m˙ = ρ um A =
∫
A
ρ u dA
€
um =
€
1
ρA
∫
A
ρ u dA
Returning to flat plate example where W is the width of the plate perpendicular to flow:
€
um =
€
1
ρ (2H ) W
€
0
H
∫− H ρ u dy dz =
H 2 dp
3µ dx
dp
3µ
= − 2 um
dx
H
Substituting into velocity profile:
€
€
1
H
∫
H
0
⎞ H H 2 dp ⎛ 1 ⎞
H dp ⎛ y 3
um =
− y⎟ =
⎜ −1⎟
⎜
2 µ dx ⎝ 3H 2
⎠ 0 2 µ dx ⎝ 3 ⎠
um = −
€
∫
W
2
3 ⎡ ⎛y⎞ ⎤
u( y ) = um ⎢1− ⎜ ⎟ ⎥
2 ⎣ ⎝H⎠ ⎦
2
⎤
H 2 dp ⎡⎛ y ⎞
⎢⎜ ⎟ −1⎥ dy
2µ dx ⎣⎝ H ⎠
⎦
(b) Derivation of temperature distribution, T(x, y), and mean temperature, Tm
Energy equation for constant property flow of a Newtonian fluid:
⎛ ∂T !
⎞
ρ c p ⎜ + V ⋅ ∇T ⎟ = k ∇ 2T + µ Φ + q˙
⎝ ∂t
⎠
For 2-D steady flow in Cartesian coordinates:
€
€
⎛ ∂T
∂T ⎞ ⎛ ∂ 2T ∂ 2T ⎞
ρ c p ⎜ u + v ⎟ = k⎜ 2 + 2 ⎟ + µΦ + q˙
∂y ⎠ ⎝ ∂x
∂y ⎠
⎝ ∂x
u ∂T ∂ 2T
=
α ∂x ∂ y 2
where we assume
2
3 ⎡ ⎛y⎞ ⎤
Substitute in u( y ) = um ⎢1− ⎜ ⎟ ⎥ :
2 ⎣ ⎝H⎠ ⎦
€
∂ 2T
∂ 2T
>>
∂y 2
∂x 2
€
2
∂ 2T 3 um dTm ⎡ ⎛ y ⎞ ⎤
⎢1 − ⎜ ⎟ ⎥
2 =
2 α dx ⎣ ⎝ H ⎠ ⎦
€ ∂y
For constant surface heat flux recall that:
€
∂T dTm
=
= constant
∂x
dx
Integrate twice to obtain:
€
3 um dTm ⎛ y 2
y4 ⎞
T=
⎜ −
⎟ + C1 y + C2
2 α dx ⎝ 2 12 H 2 ⎠
Impose boundary conditions T (−H ) = T (+H ) = Ts (note that Ts is unknown):
€
€
T (H ) =
3 um dTm ⎛ H 2 H 2 ⎞
−
⎜
⎟ + C1 H + C2 = Ts
2 α €
dx ⎝ 2 12 ⎠
T (H ) =
3 um dTm ⎛ H 2 H 2 ⎞
−
⎜
⎟ − C1 H + C2 = Ts
2 α dx ⎝ 2
12 ⎠
⇒ C2 = Ts −
€
€
3 um H 2 dTm ⎛ 5 ⎞
⎜ ⎟ and ⇒ C1 = 0
2 α
dx ⎝12 ⎠
€
2
4
3 um H 2 dTm ⎡ 5 1 ⎛ y ⎞
1 ⎛y⎞ ⎤
Ts − T =
⎢ − ⎜ ⎟ + ⎜ ⎟ ⎥
2 α
dx ⎣12 2 ⎝ H ⎠ 12 ⎝ H ⎠ ⎦
Recall definition for mean temperature:
€
€
Tm =
Tm =
1
ρ um A c v
1
um H
∫
H
0
∫
A
ρ u c v T dA =
1
2 um H
∫
H
−H
u T dy
4
2
⎧⎪ 3 ⎡ ⎛ y ⎞ 2 ⎤⎫⎪ ⎧⎪
3 um H 2 dTm ⎡ 1 ⎛ y ⎞ 1 ⎛ y ⎞
5 ⎤⎫⎪
⎨ um ⎢1− ⎜ ⎟ ⎥⎬ ⎨Ts −
⎢ ⎜ ⎟ − ⎜ ⎟ + ⎥⎬ dy
⎪⎩ 2 ⎣ ⎝ H ⎠ ⎦⎪⎭ ⎪⎩
2 α
dx ⎣12 ⎝ H ⎠ 2 ⎝ H ⎠ 12 ⎦⎪⎭
After many steps:
€
(Ts − Tm ) =
€
17 um H 2 dTm
35 α
dx
(Ts − T ) = 35 ⎡⎢5 − 6 ⎛ y ⎞2 + ⎛ y ⎞4 ⎤⎥
⎜ ⎟ ⎜ ⎟
(Ts − Tm ) 136 ⎣ ⎝ H ⎠ ⎝ H ⎠ ⎦
(c) Derivation of Nusselt number, Nu
€ Recall definition of heat transfer coefficient:
h=
kf
∂T
(Ts − Tm ) ∂y
Nu =
€
€
€
€
hDh
Dh
∂T
=
kf
(Ts − Tm ) ∂y
where
y= H
Dh =
4 As 4 (2H )W
=
= 4H
P
2W
⎡⎛ y ⎞ 4
⎛ y ⎞2 ⎤⎪⎫
4H
∂ ⎪⎧
35
⎨T −
Nu =
(Ts − Tm )⎢⎜⎝ H ⎟⎠ − 6 ⎜⎝ H ⎟⎠ + 5⎥⎬⎪
(Ts − Tm ) ∂y ⎪⎩ s 136 €
⎣
⎦⎭ y=H
Nu =
€
y= H
⎡ ⎛ y3 ⎞
⎛ y ⎞⎤
4H ⎛ 35 ⎞
−
T
−
T
⎜
⎟( s m ) ⎢4 ⎜ 4 ⎟ −12 ⎜ 2 ⎟ ⎥
⎝ H ⎠ ⎦ y=H
(Ts − Tm ) ⎝ 136 ⎠
⎣ ⎝H ⎠
⎛ 35 ⎞⎛ −8 ⎞ 140
Nu = 4H ⎜ −
= 8.2353 (Same as in Table 8.1)
⎟⎜ ⎟ =
⎝ 136 ⎠⎝ H ⎠ 17