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
(3.2)
(3.3)
(3.4)
∂ 2T
(3.5)
=0
2
∂x
(4) Constant properties: velocity and temperature fields are
uncoupled (Table 2.1, white box)
2
TABLE 2.1
Basic law
No. of
Equations
Unknowns
p
ρ µ k
u
v
w
1
u
v
w
Momentum
3
u
v
w
Equation of State
1
T
p
ρ µ
ρ
Viscosity relation
µ = µ ( p, T )
1
T
p
µ
1
T
p
Energy
1
Continuity
Conductivity relation
k = k ( p, T )
TT
ρ
p
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
∂θ
∴
∂ 2T
∂θ 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
+u
+v
+w
ρ cΡ 
=
∂y
∂z 
∂x
 ∂t
 ∂ 2T ∂ 2T ∂ 2T 
k  2 + 2 + 2  + µΦ
∂x
∂z 
∂y

Φ is dissipation
2
 ∂u 2  ∂ v  2
∂
w
 
 
Φ = 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
∂ρ
∂ρ
∂ρ
∂ρ
 ∂ u ∂v ∂ w 
+u
+ +v
+w
+ρ +
+
=0

∂t
∂z
∂x
∂y
 ∂x ∂y ∂z 
(2.2b)
Constant density
Infinite plates
∂ρ ∂ ρ ∂ρ ∂ρ
=0
=
=
=
∂ t ∂x ∂y ∂z
(a) and (b) into (2.2b)
Integrate (c)
(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 
ρ + u + v + w  =
∂x
∂y
∂z 
 ∂t
 ∂ 2u ∂ 2u ∂ 2u 
∂p
ρg x − + µ  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 = Uo
and
C2 = 0
u y
=
Uo H
(m)
(3.8)
Dissipation: (b) and (f) into (2.17)
Use (3.8) into (n)
 ∂u 
Φ = 
 ∂y 
Φ=
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
+µ
dy 2
2
µU o
2
2kH
dT(0)
−k
=0
dy
U o2
H2
(p)
=0
y 2 + C3 y + C4
and
T ( H ) = To
(q)
(r)
B.C. and solution (q) give
C3 = 0 and C4 = To +
µUo2
(s) into (q)
T − To
µU o2
k
1 
y 2 
= 1 − 2 
2
H 
(s)
2k
(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 ) =
µU 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 Φ = 0, T(y) = To and q′′( H ) = 0
(iii) Inviscid fluid: no dissipation, uniform temperature To.
Set µ = 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 = τ ( H )U o
(t)
where
14
τ (H ) = shearing stress
(3.8) into (u)
du( H )
τ (H ) = µ
dy
(u)
Uo
τ (H ) = µ
H
(v)
(v) and (t)
W =
µU o2
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 =
µU 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 θ ∂T
∂T 
 ∂T
+ vr
+
+vz
=
∂r
r ∂θ
∂z 
 ∂t
 1 ∂  ∂T  1 ∂ 2T ∂ 2T 
k
+ 2  + µΦ
r
+ 2
2
∂z 
 r ∂r  ∂r  r ∂θ
ρ c P 
(2.24)
19
where
2
2
2
 ∂v z 
 1 ∂v θ v r 
 ∂v r 
Φ = 2
+  + 2
 +
 + 2
r 
 ∂r 
 r ∂θ
 ∂z 
2
2
 1 ∂v z ∂v 0 
 ∂v r ∂ v z 
 ∂v θ v θ 1 ∂v r 
−
+
+
+
 +


 +
r r ∂0 
∂z 
∂r 
 ∂r
 r ∂0
 ∂z
• Need v r, v θ and v z
• Flow field: use continuity and Navier-Stokes eqs.
1 ∂
∂
∂ρ 1 ∂
(ρ rv r ) +
(ρ v θ ) + (ρ v z ) = 0
+
∂ t r ∂r
r ∂θ
∂z
Constant ρ
∂ρ ∂ρ ∂ρ ∂ρ
=
=
=
=0
∂ t ∂r ∂θ ∂z
Axisymmetric flow
∂
vθ =
=0
∂θ
2
(2.25)
(2.4)
(a)
(b)
20
Long tube, no end effects
∂
=0
∂z
(c)
d
(r v r ) = 0
dr
(d)
r v 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 θ ∂ v z
+ vz
+
ρ v r
+
=
∂t 
∂r
r ∂θ
∂z

 1 ∂  ∂v z  1 ∂ 2 v z ∂ 2 v z 
∂p
ρg z − + µ 
+
r
+ 2

2
2
∂z
∂z 
 r ∂r  ∂r  r ∂θ
Simplify
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 
−
+µ
=0
r
∂z
r dr  dr 
∴ v z depends on r only, rewrite (3.11)
∂p
1 d  dv z 
=µ
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 θ ∂v r v θ 2

∂
v
∂
v
r
r
=
−
+vz
+
+
ρ  v r

r
r
r
z
t
∂
∂
∂
∂
θ


2
2
 ∂ 1 ∂
∂p
 1 ∂ v r 2 ∂v θ ∂ v r 
( rv r )  + 2
− 2
+
ρg r − + µ  
2
2 
∂r
∂
r
r
∂
r
∂
θ
 r ∂θ
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 + Co = f ( z)
g(r) = C
C = constant. Use (o) into (j)
Integrate
integrate again
Two B.C. on : v z
∂p
1 d  dv z 
=µ
r
=C
∂z
r dr  dr 
(n)
(o)
(p)
dv z 1 d p 2
r
r + C1
=
dr 2µ d z
1 dp 2
vz =
r + C1 ln r + C2
4µ 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
4µ d z
Substitute into (q)
1 dp 2
vz =
( r − ro2 )
4µ 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
 + µΦ = 0
r dr  dr 
(b), (c) and (g) into (2.25)
 dv z 
Φ =

 dr 
(3.12)
(s)
(t)
2
Substitute velocity solution (3.11) into the above
25
2
 1 d p 2
Φ =
 r
2µ d z 

(u) in (t) and rearrange
(u)
2
Integrate
d  dT 
1 d p 3

 r
r
=−
dr  dr 
4kµ  d z 
(3.13)
2
1 d p 4
T =−
 r + C 3 ln r + C 4

64kµ  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

64kµ  d z 
Substitute into (v)
ro4
d p
T = To +


64kµ  d z 
2
4

r
1 − 
4

r

o 
(v)
(w)
(3.14a)
26
In dimensionless form:
T − To
4

r
1 − 
=
2
4

4
r
ro  d p 

o 


64kµ  d z 
[b] Use Fourier’s
dT ( ro )
q ′′( ro ) = − k
dr
(3.14) into above
2
3
ro  d p 
q ′′( ro ) =


16 µ  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:
Heat leaving tube = Pump work
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)
π ro4 dp
W =−
( p1 − p2 )
8µ dz
Work per unit area W ′′
(z-3) into the above
(z-3)
W
W ′′ =
2π ro L
ro3 dp ( p1 − p2 )
W ′′ = −
16 µ dz
L
( p1 − p2 )
dp
However
=−
L
dz
Combine with (z-4)
ro3  dp  2
W ′′ =
 
16 µ  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 ω
• 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 ρ
• 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 θ ∂T
∂T 
 ∂T
+ vr
+
ρcP 
+vz
=
∂r
r ∂θ
∂z 
 ∂t
 1 ∂  ∂T  1 ∂ 2T ∂ 2T 
k
+ 2  + µΦ
r
+ 2
2
∂z 
 r ∂r  ∂r  r ∂θ
(2.24)
34
where
2
2
2
 ∂v z 
 1 ∂v θ v r 
 ∂v r 
Φ = 2
+  + 2
 +
 + 2
r 
 ∂r 
 r ∂θ
 ∂z 
2
2
 1 ∂v z ∂v 0 
 ∂v r ∂ v z 
 ∂v θ v θ 1 ∂v r 
−
+
+
+
 +


 +
r r ∂0 
∂z 
∂r 
 ∂r
 r ∂0
 ∂z
2
(2.25)
Need flow field v r , v θ and v z
• Apply continuity and Navier-Stokes to determine flow
field
∂ρ 1 ∂
1 ∂
∂
(ρ rv r ) +
(ρ v θ ) + (ρ v z ) = 0
+
(2.4)
∂ t r ∂r
r ∂θ
∂z
Constant ρ
∂ρ ∂ρ ∂ρ ∂ρ
(a)
=
=
=
=0
∂ t ∂r ∂θ ∂z
Axisymmetric flow
∂
(b)
=0
∂θ
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 θ
36
∂v θ ∂v θ 
 ∂v θ v θ ∂v θ v r v θ
−
+vz
+
ρ v r
+
=
∂t 
∂r
r ∂θ
r
∂z

2
 ∂ 1 ∂
1 ∂p
2 ∂v r ∂ 2 v θ 
 1 ∂ vθ
ρgθ −
+ µ 
( rv θ )  + 2
+ 2
+

2
2
r ∂θ
 r ∂θ
r ∂θ
∂z 
 ∂ r  r ∂r
(2.11 θ )
For steady state:
∂
=0
∂t
Neglect gravity, use (b),(c), (g), (h) into (2.11 θ )
d 1 d

(
r
v
)

θ =0
dr  r dr

Integrate
B.C. are
C1
C2
vθ =
r+
2
r
v θ ( ri ) = ω ri
v θ ( ro ) = 0
(h)
(3.17)
(i)
(j)
37
(j) gives C1 and C 2
C1 = −
(k) into (i)
2ω ri2
ro2
− ri2
C2 =
ω ri2 ro2
ro2
− ri2
v θ ( r ) ( ro / ri ) 2 ( ri / r ) − ( r / ri )
=
ω 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
 + µΦ = 0
r dr  dr 
and
 dv 0 v θ 
−
Φ =

r 
 dr
(l)
2
(3.18) into above
38
2

 1
Φ =
2
4
1
−
(
r
/
r
)
r


i
o
Combine (m) and (l)
2ω ri2
d  dT 
µ
r
=−
dr  dr 
k
(m)
2
 2ω ri
 1
1 − ( r / r ) 2  r 3


i
o
2
Integrate(3.19) twice
2
2
 1
µ  2ω ri
T (r ) = −

 2 + C 3 ln r + C 4
2
4k 1 − ( ri / ro )  r
Need two B.C.
dT(ri )
T (ro ) = To and
=0
dr
(n) and (o) give C 3 and C 4
µ 
(n)
(o)
2
 1
C3 = −

 2
2
2k 1 − ( ri / ro )  ri
2ω ri2
(3.19)
39
µ 

C 4 = To +


2
4k 1 − ( ri / ro ) 
Substitute into (o)
2ω ri2
2
or
2
1

2
 2 + 2 ln ro 
 ro ri


µ  2ω 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
µ 
2
[
]
= ( ri / ro ) 2 − ( ri / r ) 2 + 2 ln( ro / r )



2
4k 1 − ( ri / ro ) 
Maximum temperature at r = ri
2ω ri
µ 
2
(3.20b)

2
T ( ri ) − To =
1
+
(
r
/
r
)


i o + 2 ln( ro / ri )
2
4k 1 − ( ri / ro ) 
(3.21)
2ω 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 ) = −2π ro k
dr
(ω ri ) 2
q′( ro ) = 4π µ
2
1 − ( ri / ro )
(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 ω = 0 in (3.18) gives
vθ = 0
• Stationary shaft: No dissipation, no heat loss Set ω = 0 in
(3.22) gives q′( ro ) = 0
• Global conservation of energy:
Heat leaving housing = shaft work
Shaft work per unit length
W ′ = −2π riτ ( ri )ω ri
τ ( ri ) = shearing stress
 dv 0 v θ 
τ ( ri ) = µ 
− 
r  r = ri
 dr
(3.18) into the above
τ ( ri ) = −2
( ro / ri ) 2 µω ri
2
( ro / ri ) − 1
(p)
(q)
(r)
42
Combining (p) and (r)
W ′ = 4π µ
(ω 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