Download PROBLEM 3.3 Incompressible fluid is set in motion between two

PROBLEM 3.3
Incompressible fluid is set in motion between two
large parallel plates by moving the upper plate with
constant velocity U o and holding the lower plate
stationary. The clearance between the plates is H.
The lower plate is insulated while the upper plate
exchanges heat with the ambient by convection. The heat transfer coefficient is h and the
ambient temperature is Tf . Taking into consideration dissipation determine the temperature of
the insulated plate and the heat flux at the moving plate. Assume laminar flow and neglect
gravity effect.
(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are
infinite the flow field does not vary in the axial direction x. (iii) The fluid is incompressible
(constant density). (iv) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field.
Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity
and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible
gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat
transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for
constant properties is given by (2.19b)
§ wT
wT
wT
wT ·
¸
u
v
w
wx
wy
wz ¸¹
© wt
U c 5 ¨¨
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wy
wz ¹
© wx
(2.19b)
where the dissipation function ) is given by (2.17)
)
ª
2 § wv · 2
2 º ª§ wu wv · 2 § wv ww · 2
2º
§ ww ·
§ ww wu · »
§ wu ·
¸¸ ¨
¸ 2«¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸¸
¨¨ 3 © wx wy wz ¹
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution
requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in
Cartesian coordinates
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wy wz ¼
For constant density
(2.2b)
PROBLEM 3.3 (continued)
wU
wt
wU
wx
wU
wy
wU
wz
0
(a)
Since plates are infinite
w
wz
w 0
(b)
wv
wy
0
(c)
f (x)
(d)
w
wx
Substituting (a) and (b) into (2.2b), gives
Integrating (c)
v
To determine the “constant” of integration f (x ) we apply the no-slip boundary condition at the
lower plate
v ( x,0) 0
(e)
Equations (d) and (e) give
f ( x) 0
Substituting into (d)
v 0
(f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel.
To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,
(2.10x)
§ wu
wu ·
wu
wu
w ¸¸
v
u
wz ¹
wy
wx
© wt
U ¨¨
Ug x § w 2u w 2u w 2u ·
wp
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¸¹
©
(2.10x)
This equation is simplified as follows: Steady state
wu
wt
0
(g)
gx
0
(h)
wp
wx
0
(i)
0
(j)
Negligible gravity effect
Negligible axial pressure variation
Substituting (b) and (f)-(i) into (2.10x) gives
d 2u
dy 2
The solution to (j) is
u
C1 y C 2
(k)
where C1 and C 2 are constants of integration. The two boundary conditions on u are:
u (0)
These conditions give
0 and u ( H ) U o
(l)
PROBLEM 3.3 (continued)
Uo
and C 2
H
C1
(m)
0
Substituting (m) into (k)
u
Uo
y
H
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy
equation. Substituting (b) and (f) into (2.17) gives
§ wu ·
¨¨ ¸¸
© wy ¹
)
2
(n)
Using solution (3.8) into (n) gives
U o2
)
Noting that for steady state wT / wt
simplifies to
(o)
H2
0 and using (b), (f) and (o), the energy equation (2.10b)
d 2T
k
P
dy 2
U o2
H2
0
(p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at
uniform surface temperature. Equation (p) is solved by direct integration
T
PU o2
2kH
2
y 2 C3 y C 4
(q)
where C3 and C 4 are constants of integration. The two boundary conditions on (q) are
(1)
dT (0)
dy
(2) k
0
dT ( H )
dy
h>T ( y ) Tf @
These boundary conditions and solution (q) give
C3
C4
Tf (r)
0
PU o2
khH
PU o2
2k
(s)
Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give
T Tf
PU o2
k
1 k
1 y2
2 hH 2 H 2
(t)
The dimensionless parameter hH / k is known as the Biot number, Bi . It is associated with
convection boundary conditions.
PROBLEM 3.3 (continued)
The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.
T (0) Tf 1 k
2 hH
PU o2
k
The heat flux at the moving surface is determined by applying Fourier’s law at y
qcc( H )
k
(u)
H
dT ( H )
dy
Substituting (t) into the above
PU o2
qcc( H )
H
(v)
(iii) Checking. Dimensional check: Each term in (3.8), (t) and (u) is dimensionless. Each term
in solution (v) has units of W/m 2 .
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature
solution (t) satisfies (p).
Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and
temperature solution (t) satisfies boundary conditions (r).
Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting U o
in (3.8) gives u ( y ) 0.
0
(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be
uniform equal to the ambient temperature Tf . Setting U o 0 in (u) gives T ( y ) Tf . Similarly
the heat flux qcc( H ) vanishes. Substituting U o 0 (v) gives qcc( H ) 0.
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to
Tf . Setting P 0 in (u) gives T ( y ) Tf . Similarly the heat flux qcc( H ) vanishes. Substituting
P 0 (v) gives qcc( H ) 0.
Global energy balance: Energy leaving the channel must equal to the work done to move the
plate. Consider the work done by the plate on the fluid
W cc W oU o
(w)
where
W cc
Wo
work done per unit surface area by the plate on the fluid
shearing stress at the moving plate
However, shearing stress is given by
Wo
P
wu( H )
wy
(x)
Uo
H
(y)
(3.8) into (x)
Wo
(y) into (w)
P
PROBLEM 3.3 (continued)
W cc
PU o2
(z)
H
This is identical to the heat removed from the upper plate given in equation (v).
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This
eliminates the x-coordinate as a variable and results in governing equations that are ordinary.
Alternatively, one could state that the streamline are parallel. This means that v wv / wy 0. (ii)
The solutions is characterized by a single dimensionless parameter hH / k , which is the Biot
number. (iii) The Nusselt number at moving plate, Nu(H ), is defined as
hH
k
Nu( H )
(w)
The heat transfer coefficient h is based on the overall temperature drop, defined as
h
qcc( H )
T (0) T ( H )
(x)
Centerline temperature and moving plate temperature are obtained by evaluating (t) at y = 0 and
y=H
PU o2 ª 1 k º
T (0) Tf
k «¬ 2 hH »¼
and
1 PU o2
T ( H ) Tf
2 Hh
The above two equations give
T ( 0) T ( H )
1 PU o2
2 k
(y)
(v) and (y) into (x)
h
2
k
H
Substituting into (w) gives the Nusselt number
Nu( H )
2
(z)