Fluid Mechanics, SG2214, HT2013

September 30, 2013

Exercise 8: Exact solutions to energy equation

Example 1: Energy Dissipation in Poiseuille Flow

a) Calculate the dissipation function for the plane Poiseuille flow where
P 2
u= (h − y 2 ), v = w = 0,
2µ
or in terms of the bulk velocity U
3U 2
u= (h − y 2 ), v = w = 0.
2h2
The mass-flow rate through the channel is
Z h
Q= udy = 2U h .
−h

The dissipation function is defined as (dissipation to heat due to viscous stresses)

∂ui
Φ = τij .
∂xj
For incompressible flows, it can be re-written as
∂ui
Φ = τij = 2µeij (eij + ξij ) = 2µeij eij ,
∂xj
where we used the fact that eij ξij = 0.
 ∂u 
0 ∂y 0
The deformation tensor for the Poiseuille flow becomes eij = 1/2  ∂u
∂y 0 0 and therefore
0 0 0
" 2  2 #  2
1 ∂u 1 ∂u ∂u
Φ = 2µ + =µ .
2 ∂y 2 ∂y ∂y

b) Calculate the total dissipation for unit area

h h 2
6µU 2
Z Z 
3U
φ= Φ dy = µ − 2y dy = .
−h −h h h

c) Write the mechanical energy equation for this flow. Integrate over the channel width and relate the
total dissipation φ to the pressure gradient and the mass flux.
The mechanical energy equation is obtained by multiplying the Navier-Stokes equations by ui (the
energy is ρ(1/2)ui ui ). One gets
 
D 1 ∂p ∂τij
ρ ui ui = ρFi ui − ui + ui .
Dt 2 ∂xi ∂xj

Considering the Poiseuille flow and re-writing the last term as

∂τij ∂ui τij
ui = − Φ,
∂xj ∂xj

1
 the energy equation reduces to
∂
0 = uP + (uτxy ) − Φ.
∂y
Integrating across the channel each term in the expression above, one obtains for the first term
Z h Z h
uP dy = P udy = QP,
−h −h

where Q is the flow rate. This term represents the work rate by pressure forces.
The second term Z h
∂ h
(uτxy ) dy = [(uτxy )]−h = 0
−h ∂y
due to the no-slip boundary conditions.
Rh
The third term is the total dissipation φ = −h
Φ dy defined above. Summarising
Z h
0 = QP − Φ dy.
−h

One can check the results, using the expression for φ obtained in b). Just recall that
Z h
Q= udy = 2U h,
−h

3µU
and the pressure gradient can be expressed in terms of U as P = h2 . Therefore QP = 6µU 2 /h = φ.

Example 2: Exact solution for energy equation

Consider plane Poiseuille flow in a straight channel with walls at y = ±h. The temperature at the lower wall
is T (−h) = TW + ∆T , whereas the upper wall is at T (h) = TW . The velocity field is
3U 2
u= (h − y 2 ), v = w = 0.
2
a) Derive and plot the temperature distribution.

Let us consider the energy equation for incompressible fluid.

D
ρ 0 cp T = ∇ · (K ∇T ) + Φ
Dt
In this case, the equation of state is simply ρ = ρ0 , and cp = cv . We also assume K to be independent of
the temperature T .
∂ ∂
We can therefore assume a steady solution and a fully developed field: ∂t T = 0; ∂x T = 0. The material
derivative term is then
D ∂T ∂T ∂T ∂T
T = +u +v =0+u·0+0· = 0,
Dt ∂t ∂x ∂y ∂y
2
while the diffusion term reduces to K ∇2 T = K ∂∂yT2 . Finally, as shown in recitation 5, the dissipation function
Φ reduces for this case to
" 2  2 #  2
1 ∂u 1 ∂u ∂u
Φ = 2µeij eij = 2µ + =µ .
2 ∂y 2 ∂y ∂y

Summarising the energy equations to be solved is

2
∂2T

µ ∂u
0= + , (1)
∂y 2 K ∂y

2
 Y

−h
Figure 1: Coordinate system for plane Poiseuille flow.

with boundary conditions 

T (y = −h) = TW + ∆T
T (y = h) = TW
Let us introduce dimensionless variables y ∗ = hy , T ∗ = T −T
∆T
W
and u∗ = u
U. Re-writing equation (1), one
obtains  ∗ 2
∂2T ∗ µU 2 ∂u
0= ∗2
+ . (2)
∂y K∆T ∂y ∗
The following adimensional numbers can then be introduced

µU 2 µcp U 2
= .
K∆T K cp ∆T

The first term can be written as

µcp ν ν
= = = P r,
K K/(ρcp ) κ
where P r denotes the Prandtl number, the ratio between the kinematic viscosity ν and the thermal diffusivity
2
κ. cpU∆T = E is the Eckert number defined mainly for fluids. Indeed, the same expression can be rewritten
for gases in terms of the Mach number M and the adiabatic constant γ as

U2 M 2 a2w M 2 γRTW M 2 TW
= = = (γ − 1) .
cp ∆T cp ∆T cp ∆T ∆T

Summarising the energy equations in adimensional form is

2
∂2T ∗ ∂u∗

0= + Pr E , (3)
∂y ∗2 ∂y ∗

with boundary conditions

T (y ∗ = −1) = 1

T (y ∗ = 1) = 0
∂u∗
Recalling the expression for the velocity field in adimensional variables u∗ = 3
1 − y ∗2 , = −3y ∗


2 ∂y ∗
and Φ = 9y ∗2 . Therefore equation (3) can be integrated to yield

∂2T ∗
+ P r E 9y ∗2 = 0,
∂y ∗2
∂T ∗ 9
+ P r E y ∗3 = C,
∂y ∗ 3
3
T ∗ + P r E y ∗4 = Cy ∗ + D.
4

3
C and D are determined imposing the boundary conditions at y ∗ = ±1.
 
1 3 1
D= 1 + P rE ; C = − .
2 2 2

Finally the temperature field can be written as

1 3 1 3
T∗ = + P rE − y ∗ − P rE y ∗4


2 4 2 4
1 3
T = (1 − y ) + P rE 1 − y ∗4 ,
∗ ∗


2 4
and in dimensional form
U2
  y 4 
1 ∆T  y 3
T =1+ 1− + Pr 1− .
2 TW h 4 cp TW h

The solution is composed of two parts, the first, linear in y, is the temperature distribution one would obtain
in the presence of a temperature difference between the two walls. The second, fourth order contribution, is
the heating due to dissipation in the fluid.

0.5 Pr E
y*

−0.5

−1
0 0.5 1 1.5 2
T*

Figure 2: Temperature distribution for increasing P rE.

b) Write the heat flux at the wall. Determine the value of P rE for which the heat flux at the lower wall is
zero.

∂T ∗
 
1 3 ∗3

1
|y∗ =−1 = − + P rE −4y = (−1 + 6P rE).
∂y ∗ 2 4 y ∗ =−1 2
The wall heat flux is
∂T K ∂T ∗ K 1
qy = −K |y=−h = − ∆T ∗ |y∗ =−1 = − ∆T (−1 + 6P rE).
∂y h ∂y h 2
Normalising
qy K∆T 1
=− (−1 + 6P rE) =
ρcp U ∆T hρcp U ∆T 2
K/(ρcp ) κ 1 1
− (−1/2 + 3P rE) = − (−1/2 + 3P rE) = − (−1/2 + 3P rE).
hU Uh Re P r
Thus qy = 0 if P rE = 1/6.