CHAPTER 3
EXACT ONE-DIMENSIONAL SOLUTIONS
3.1 Introduction
 Temperature solution depends on velocity
 Velocity is governed by non-linear Navier-Stokes eqs.
 Exact solution are based on simplifications governing
equations
3.2 Simplification of the Governing Equations
Simplifying assumptions:
(1) Laminar flow
(2) Parallel streamlines
(3.1)
v 0
1
(3.1) into continuity for 2-D, constant density fluid:

u
 0 , everywhere
x
 2u
0
x 2
(3) Negligible axial variation of temperature
T
 0 , everywhere
x
(3.4) is valid under certain conditions. It follows that
 2T
0
(3.2)
(3.3)
(3.4)
(3.5)
x 2
(4) Constant properties: velocity and temperature fields are
uncoupled (Table 2.1, white box)
2
TABLE2.1
B
asiclaw
No. of
E
quations
E
1
kk( p,T)
p


 k
T
3
Conductivityrelation
w
1
M
Viscosityrelation
  (p, T)
v
1
1
oS
TT u

w
uv w p  
p 
T
p
T

C
E
Unknowns
1
u v
p
n
o
q
k
Similar results are obtained for certain rotating flows.
3
Fig. 3.2:
 Shaft rotates inside sleeve
 Streamlines are concentric circles
 Axisymmetric conditions, no axial
variations
T
0


 2T
 2
(3.6)
0
(3.7)
3.3 Exact Solutions
3.3.1 Couette Flow
 Flow between parallel plate
 Motion due to pressure drop and/or moving plate
 Channel is infinitely long
4
Example 3.1: Couette Flow with Dissipation
 Very large parallel
plates
 Incompressible fluid
 Upper plate at To moves with velocity Uo
 Insulate lower plate
 Account for dissipation
 Laminar flow, no gravity, no pressure drop
 Determine temperature distribution
(1) Observations
 Plate sets fluid in motion
 No axial variation of flow
 Incompressible fluid
 Cartesian geometry
5
(2) Problem Definition.
Determine the velocity and temperature distribution
(3) Solution Plan
 Find flow field, apply continuity and Navier-Stokes
equations
 Apply the energy to determine the temperature distribution
(4) Plan Execution
(i) Assumptions
 Steady state
 Laminar flow
 Constant properties
 Infinite plates
 No end effects
 Uniform pressure
 No gravity
6
(ii) Analysis
Start with the energy equation
T
T
T 
 T
 c 
u
v
w

x
y
z 
 t
  2T  2T  2T 
k  2  2  2   
x
y
z 

 is dissipation
2
 u 2  v  2

w
  2          
y 
 z  
 x 



  u v  2  v w  2
w u  2 

     
  
 
z y 
 x z  
  y x 



2
2  u v w 

 

3  x y z 
(3.19b)
(3.17)
7
Need u, v and w. Apply continuity and the Navier-Stokes
equations
Continuity




  u v  w 
u
 v
w
 

0

t
x
y
z
 x y z 
Constant density
   



0
 t x y z
Infinite plates
(a) and (b) into (2.2b)
Integrate (c)
(2.2b)
(a)



w0
x z
(b)
v
0
y
(c)
v  f ( x)
(d)
8
f ( x ) is “constant” of integration
Apply the no-slip condition
v ( x ,0)  0
(d) and (e) give
Substitute into (d)
(e)
f ( x)  0
v 0
(f)
 Streamlines are parallel
To determine u we apply the Navier-Stokes eqs.
u
u
u 
 u
  u  v  w  
x
y
z 
 t
  2u  2u  2u 
p
g x     2  2  2 
x
y
z 
 x
(2.10x)
9
Simplify:
Steady state
u
0
t
(g)
No gravity
gx  0
(h)
Negligible axial pressure variation
p
0
x
(b) and (f)-(i) into (2.10x) gives
d 2u
Solution to (j) is
dy 2
0
u  C1 y  C 2
Boundary conditions
u(0)  0 and u( H )  U o
(i)
(j)
(k)
(l)
10
(k) and (I) give
(m) into (k)
C1  U o
and
C2  0
u y

Uo H
(m)
(3.8)
Dissipation: (b) and (f) into (2.17)
Use (3.8) into (n)
 u 
  
 y 

2
2
Uo
2
(n)
(o)
H
Steady state: T / t  0
Infinite plates at uniform temperature:
T  2T
 2 0
x x
11
Use above, (b), (f) and (o) into energy (2.10b)
k
Integrate
T 
B.C.
d 2T
dy 2
2
U o
2
2kH
dT (0)
k
0
dy
B.C. and solution (q) give
(s) into (q)

U o2
H2
0
y 2  C 3 y  C4
and
T ( H )  To
U o2
C 3  0 and C 4  To 
2k
T  To
U o2
k
1 
y 2 
 1  2 
2
H 
(p)
(q)
(r)
(s)
(3.9)
12
Fourier’s law gives heat flux at y = H
(3.9) into the above
dT ( H )
q( H )   k
dx
q ( H ) 
U o2
H
(iii) Checking
Dimensional check: Each term in (3.8) and (3.9) is
dimensionless. Units of (3.10) is W/m2
(3.10)
Differential equation check: Velocity solution (3.8) satisfies
(j) and temperature solution (3.9) satisfies (p)
Boundary conditions check: Solution (3.8) satisfies B.C. (l),
temperature solution (3.9) satisfies B.C. (r)
13
Limiting check: (i) Stationary upper plate: no fluid motion.
Set Uo = 0 in (3.8) gives u(y) = 0
(ii) Stationary upper plate: no dissipation, uniform
temperature To, no surface flux. Set Uo = 0 in (o), (3.9) and
(3.10) gives   0, T(y) = To and q( H )  0
(iii) Inviscid fluid: no dissipation, uniform temperature To.
Set   0 in (3.9) gives T(y) = To
(iv) Global conservation of energy: Frictional energy is
conducted through moving plate:
W = Friction work by plate
q(H ) = Heat conducted through plate
W   ( H )U o
(t)
where
14
 (H ) = shearing stress
(3.8) into (u)
du( H )
 (H )  
dy
(u)
Uo
 (H )  
H
(v)
(v) and (t)
W 
2
U o
H
(w)
(w) agrees with (3.10)
(4) Comments
 Infinite plate is key assumption. This eliminates x as a
variable
 Maximum temperature: at y = 0 Set y = 0 in (3.9)
T (0)  To 
U o2
2k
15
3.3.2 Hagen-Poiseuille Flow
Problems associated with axial flow in channels
 Motion due to pressure drop
 Channel is infinitely long
Example 3.2: Flow in a Tube at Uniform
Surface Temperature
 Incompressible
fluid flows in a
long tube
 Motion is due to
pressure gradient p / z
 Surface temperature To
 Account for dissipation
 Assuming axisymmetric laminar flow
16
 Neglecting gravity and end effects
 Determine:
[a] Temperature distribution
[b] Surface heat flux
[c] Nusselt number based on [ T (0)  To]
(1) Observations
 Motion is due to pressure drop
 Long tube: No axial variation
 Incompressible fluid
 Heat generation due to dissipation
 Dissipated energy is removed by conduction at the
surface
 Heat flux and heat transfer coefficient depend on
temperature distribution
17
 Temperature distribution depends on the velocity
distribution
 Cylindrical geometry
(2) Problem Definition.
Determine the velocity and temperature distribution.
(3) Solution Plan
 Apply continuity and Navier-Stokes to determine flow field
 Apply energy equation to determine temperature
distribution
 Fourier’s law surface heat flux
 Equation (1.10) gives the heat transfer coefficient.
(4) Plan Execution
18
(i) Assumptions
 Steady state
 Laminar flow
 Axisymmetric flow
 Constant properties
 No end effects
 Uniform surface temperature
 Negligible gravitational effect
(ii) Analysis
[a] Start with energy equation (2.24)
T v  T
T 
 T
cP 
 vr

 vz

r
r 
z 
 t
 1   T  1  2T  2T 
k
 2   
r
 2
2
z 
 r r  r  r 
(2.24)
19
where
2
2
 v z 
 1 v  v r 
 v r 
  2
   2
 
  2
r 
 r 
 r 
 z 
2
2
2

v
v

v
1
 1  v z v 0 
  v r v z 
 


r




 


 
r r 0 
z 
r 
 r
 r 0
 z
2
(2.25)
 Need v r, v  and v z
 Flow field: use continuity and Navier-Stokes eqs.
 1 
1 

 r v r  
 v     v z   0

t r r
r 
z
Constant 
Axisymmetric flow
(2.4)
      



0
 t r   z
(a)

v 
0

(b)
20
Long tube, no end effects

0
z
(c)
d
r v r   0
dr
(d)
rv r  f (z )
(e)
(a)-(c) into (2.4)
Integrate
f (z ) is “constant” of integration. Use the no-slip B.C.
v ( ro , z )  0
(e) and (f) give
f (z)  0
Substitute into (e)
 Streamlines are parallel
vr  0
(f)
(g)
v z Determine : Navier-Stokes eq. in z-direction
21
v z v z 
 v z v  v z
 v r

vz


r
r 
z
t 

Simplify
 1   v z  1  2 v z  2 v z 
p
g z    

r
 2

2
2
z
z 
 r r  r  r 
Steady state:
No gravity:
(2.11z)

0
t
gr  g z  0
(b), (c) and (g)-(i) into (2.11z)
p
1 d  dv z 


r
0
z
r dr  dr 
 v z depends on r only, rewrite (3.11)
p
1 d  dv z 

r
  g( r )
z
r dr  dr 
(h)
(i)
(3.11)
(j)
22
Integrate
p  g( r ) z  C o
Apply Navier-Stokes in r-direction
(k)
 v r v  v r v  2
v r v r 

 vr


vz




r
r 
r
z
t 

2
2
  1 
p
 1  v r 2 v   v r 
g r     
( rv r )   2
 2

2
2 
r

r
r

r


 r 
r
z 
 
(2.11r)
(b), (g) and (i) into (2.11r)
Integrate
p
0
r
p  f (z )
(l)
(m)
f (z ) = “constant” of integration
 Equate two solutions for p: (k) and (m):
23

p  g( r ) z  C o  f ( z )
g(r) = C
C = constant. Use (o) into (j)
Integrate
integrate again
p
1 d  dv z 

r
C
z
r dr  dr 
(o)
(p)
dv z 1 d p 2
r

r  C1
dr 2  d z
vz 
Two B.C. on : v z
(n)
1 dp 2
r  C1 ln r  C 2
4 d z
dv z (0)
 0,
dr
(q) and (r) give C1and C2
v z ( ro )  0
(q)
(r)
24
C1  0
,
1 dp 2
C2 
ro
4 d z
Substitute into (q)
1 dp 2
vz 
( r  ro2 )
4 d z
For long tube at uniform temperature:
T  2 T
 2 0
z z
(b), (c), (g), (h) and (s) into energy (2.24)
1 d  dT 
k
r
    0
r dr  dr 
(b), (c) and (g) into (2.25)
 dv z 
 

 dr 
(3.12)
(s)
(t)
2
Substitute velocity solution (3.11) into the above
25
2
 1 d p 2
 
 r
2 d z 

(u) in (t) and rearrange
(u)
2
Integrate
d  dT 
1 d p 3

 r
r

dr  dr 
4 k  d z 
(3.13)
2
1 d p 4
T 

 r  C 3 ln r  C 4
64k  d z 
Need two B.C.
dT (0)
 0 and T ( ro )  To
dr
2
(v) and (w) give
1 d p 4
C 3  0 , C 4  To 

 ro
64k  d z 
Substitute into (v)
ro4  d p 
T  To 


64k  d z 
2
4

r
1  

ro4 

(v)
(w)
(3.14a)
26
In dimensionless form:
T  To
4

r
1  

2
4

4
r
ro  d p 

o 


64k  d z 
[b] Use Fourier’s
dT ( ro )
q ( ro )   k
dr
(3.14) into above
2
3
ro  d p 
q ( ro ) 


16   d z 
[c] Nusselt number:
hD 2hro
Nu 

k
k
(1.10) gives h
dT ( ro )
k
h
[T (0)  To ] dr
(3.14a) into (y)
(3.14b)
(3.15)
(x)
(y)
27
(z) into (x)
2k
h
ro
Nu  4
(z)
(3.16)
(iii) Checking
Dimensional check:
 Each term in (3.12) has units of velocity
 Each term in (3.14a) has units of temperature
 Each term in (3.15) has units of W/m2
Differential equation check: Velocity solution (3.12)
satisfies (p) and temperature solution (3.14) satisfies (3.13)
Boundary conditions check: Velocity solution (3.12) satisfies
B.C. (r) and temperature solution (3.14) satisfies B.C. (w)
Limiting check:
28
(i) Uniform pressure ( dp / dz  0 ): No fluid motion.
Set dp / dz  0 in (3.12) gives v z  0
(ii) Uniform pressure ( dp / dz  0): No fluid motion, no
dissipation, no surface flux. Set dp / dz  0 in (3.15) gives
q( ro )  0
(iii) Global conservation of energy:
H
l t

P
w
e
Pump work W for a tube of length L
(z-1)
p1 = upstream pressure
p2 = downstream pressure
= flow rate
29
(3.12) into the above, integrate
(z-2)
(z-1) and (z-2)
 ro4 dp
W 
( p1  p2 )
8 dz
Work per unit area W 
(z-3) into the above
(z-3)
W
W  
2 ro L
ro3 dp ( p1  p2 )
W   
16 dz
L
( p1  p2 )
dp
However

L
dz
Combine with (z-4)
ro3  dp  2
W  
 
16   dz 
(z-4)
30
This agrees with (3.15)
(5) Comments
 Key simplification: long tube with end effects. This is
same as assuming parallel streamlines
 According to (3.14), maximum temperature is at center r
=0
 The Nusselt number is constant independent of Reynolds
and Prandtl numbers
31
3.3.3 Rotating Flow
Example 3.3: Lubrication Oil Temperature
in Rotating Shaft
 Lubrication oil between shaft
and housing
 Angular velocity is 
 Assuming laminar flow
 Account for dissipation
 Determine the maximum temperature
rise in oil
(1) Observations
 Fluid motion is due to shaft rotation
 Housing is stationary
32
 No axial variation in velocity and temperature
 No variation with angular position
 Constant 
 Frictional heat is removed at housing
 No heat is conducted through shaft
 Maximum temperature at shaft
 Cylindrical geometry
(2) Problem Definition.
Determine the velocity and temperature distribution of oil
(3) Solution Plan
 Apply continuity and Navier-Stokes eqs. to determine flow
field
 Use energy equation to determine temperature field
 Fourier’s law at the housing gives frictional heat
33
(4) Plan Execution
(i) Assumptions
 Steady state
 Laminar flow
 Axisymmetric flow
 Constant properties
 No end effects
 Uniform surface temperature
 Negligible gravitational effect
(ii) Analysis
 Energy equation governs temperature
T v  T
T 
 T
cP 
 vr

 vz

r
r 
z 
 t
 1   T  1  2T  2T 
k
 2   
r
 2
2
z 
 r r  r  r 
(2.24)
34
where
2
2
2
 v z 
 1 v  v r 
 v r 
  2
   2
 
  2
r 
 r 
 r 
 z 
2
2

v
v

v
1
 1  v z v 0 
  v r v z 
 


r




 


 
r r 0 
z 
r 
 r
 r 0
 z
2
(2.25)
Need flow field v r , v  and v z
 Apply continuity and Navier-Stokes to determine flow
field
 1 
1 

 r v r  
 v     v z   0

(2.4)
t r r
r 
z
Constant 
      
(a)



0
 t r   z
Axisymmetric flow

(b)
0

35
Long shaft:
(a)-(c) into (2.4)
Integrate

vz 
0
z
d
r v r   0
dr
r vr  C
Apply B.C. to determine C
v r ( ro )  0
(e) and (f) give C = 0
Use (e)
vr  0
(c)
(d)
(e)
(f)
(g)
 Streamlines are concentric circles
Apply the Navier-Stokes to determine v 
36
v  v  
 v  v  v  v r v 
 v r


vz


r
r 
r
z
t 

2
  1 
1 p
2 v r  2 v  
 1  v
g 
  
( rv  )   2
 2


2
2
r 
 r 
r 
z 
 r  r r
(2.11  )
For steady state:

0
t
Neglect gravity, use (b),(c), (g), (h) into (2.11  )
Integrate
B.C. are
d 1 d

(
r
v
)

 0
dr  r dr

C1
C2
v 
r
2
r
v  ( ri )   ri
v  ( ro )  0
(h)
(3.17)
(i)
(j)
37
(j) gives C1 and C 2
C1  
(k) into (i)
2 ri2
ro2
 ri2
C2 
 ri2 ro2
ro2
 ri2
v  ( r ) ( ro / ri )2 ( ri / r )  ( r / ri )

 ri
( ro / ri )2  1
(k)
(3.18)
Simplify energy equation (2.24) and dissipation function
(2.25). Use(b), (c), (g), (h)
1 d  dT 
k
r
    0
r dr  dr 
and
 dv  v  
 


r 
 dr
(l)
2
(3.18) into above
38
2

 1
 
2
4
1  ( ri / ro )  r
Combine (m) and (l)
2 ri2
d  dT 

r

dr  dr 
k
(m)
2
 2 ri
 1
1  ( r / r ) 2  r 3


i
o
2
Integrate(3.19) twice
2
2
 1
  2 ri
T (r )   
 2  C 3 ln r  C 4
2
4k 1  ( ri / ro )  r
Need two B.C.
dT ( ri )
and
T ( ro )  To
0
dr
(n) and (o) give C 3 and C 4
 
(n)
(o)
2
 1
C3  

 2
2
2k 1  ( ri / ro )  ri
2 ri2
(3.19)
39
 

C 4  To 


2
4k 1  ( ri / ro ) 
Substitute into (o)
2 ri2
2
or
2
1

2
 2  2 ln ro 
 ro ri




  2 ri
2
2
T ( r )  To 

 ( ri / ro )  ( ri / r )  2 ln(ro / r )
4k 1  ( r / r ) 2 
i o
(3.20a)
T ( r )  To
 
2
 ( ri / ro ) 2  ( ri / r ) 2  2 ln( ro / r )



2
4k 1  ( ri / ro ) 
Maximum temperature at r  ri
2 ri
 
2
(3.20b)



2
T ( ri )  To 
1

(
r
/
r
)


i o  2 ln( ro / ri )
2
4k 1  ( ri / ro ) 
(3.21)
2 ri
Use Fourier’s law to determine frictional energy per
unit length q( ro )
40
(3.20a) in above
(iii) Checking
dT ( ro )
q( ro )  2 ro k
dr
( ri )2
q( ro )  4 
1  ( ri / ro )2
(3.22)
 Each term in solutions (3.18) and (3.20b) is dimensionless
 Equation (p) has the correct units of W/m
Differential equation check:
 Velocity solution (3.18) satisfies (3.17) and temperature
solution (3.20) satisfies (3.19)
Boundary conditions check:
 Velocity solution (3.18) satisfies B.C. (j) and temperature
solution (3.20) satisfies B.C. (o)
41
Limiting check:
 Stationary shaft: No fluid motion. Set   0 in (3.18) gives
v  0
 Stationary shaft: No dissipation, no heat loss Set   0 in
(3.22) gives q( ro )  0
 Global conservation of energy:
H
lh

s
w
e
Shaft work per unit length
W   2 ri ( ri ) ri
 ( ri ) = shearing stress
 dv  v  
 ( ri )   
 
r  r  ri
 dr
(3.18) into the above
 ( ri )  2
( ro / ri ) 2  ri
( ro / ri )  1
2
(p)
(q)
(r)
42
Combining (p) and (r)
W   4 
( ri ) 2
1  ( ri / ro )
2
(s)
This is identical to surface heat transfer (3.22)
(5) Comments
 The key simplifying assumption is axisymmetry
 Temperature rise due to frictional heat increase as
the clearance s decreased
 Single governing parameter: ( ri / ro )
43