ejercicios capa limite
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Ejercicios Capa Limite boundary layerTRANSCRIPT
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Worked Out Examples
(Thermal B.L.)
Example 1 (Convection Coefficient):
Air at a free stream temperature of T
= 20 C is in parallel flow over a flat plate of length L
= 5 m and temperature T s = 90 C . However, obstacles placed in the flow intensify mixing
with increasing distance x from the leading edge, and the spatial variation of temperatures
measured in the boundary layer is correlated by an expression of the form T( C) = 20 +
70 e (- 600 x y) , where x and y are in meters. Determine and plot the manner in which the local
convection h varies with x . Evaluate the average convection coefficient h for the plate.
1. Statement of the Problem
a)
Given Free stream air temperature T = 20 C
Plate length L = 5 m
Plate surface temperature T s = 90 C
Correlated measured temperature in the boundary layer: T( C) = 20 + 70e(- 600 x y),
where x and y are in meters
b) Find
Determine and plot the manner in which the local convection h varies with x.
Evaluate the average convection coefficient h for the plate.
2.
System Diagram
3. Assumptions
Steady state condition
Uniform free stream air temperature T = 20 C = constant
L = 5 m
T = 20 C
xye y xT
6007020,
x
y
T s = 90 C
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Uniform surface temperature T s = 90 C = constant
Constant thermal conductivity
4. Governing Equations
Newton's Law of Cooling
T T hq s s
One Dimensional Fourier's Law
y
T k q f
On the plate surface y = 0
0
y
f s y
T k q
Average Convection Coefficient Definition
s
A s
s
dAh
A
h 1
For the special case of flow over a flat plate, h varies with the distance x from the leading
edge. Thus,
L
dxh L
h0
1
5. Detailed Solution
Local Convection Coefficient, h
T T hq s s … Newton's law of cooling
0
y
f s y
T k q … One-dimensional Fourier's law on the plate surface
Thus,
0
y
f s s y
T k T T hq
Therefore,
160070
600700
7020
1
0
600
0
600
0
xT T
k
e xT T
k
e yT T
k
y
T k
T T h
s
f
y
xy
s
f
y
xy
s
f
y
f
s
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T T
xk xh
s
f 42000
Taking the average of the free stream and surface temperatures:
C T
55
2
9020
k f = 0.02837 W/m K
After plugging numbers into the expression obtained above, it becomes:
x xh 02.17 W/m2 K
Using MatLab, the variation of local convection coefficient can be plotted as:
Average Convection Coefficient, h
The average coefficient over the range 50 x m is
L
L
L
x
L
dx x L
dxh L
h
L
L
L
51.8
202.17
1
202.17
1
02.171
1
2
0
2
0
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
10
20
30
40
50
60
70
80
90Variation of Local Convection Coefficient
h(x)
x (m)
h(W/m2.K)
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6.42h W/m2 K
6.
Critical AssessmentBecause the local convection coefficient is a function of x, the average of the convection
coefficient must be obtained by integrating the function over the whole range of the flat plate.
Example 2 (Velocity and Temperature Profiles):
In flow over a surface, velocity and temperature profiles are of the forms
u(y) = Ay + By 2 - Cy 3 and
T(y) = D + Ey + Fy 2 - Gy 3
where the coefficients A through G are constants. Obtain expressions for the friction
coefficient C f and the convection coefficient h in terms of U
, T
, and appropriate profile
coefficients and fluid properties.
1. Statement of the Problema) Given
Velocity and temperature profiles
u(y) = Ay + By2 - Cy3 and
T(y) = D + Ey + Fy2 - Gy3
where the coefficients A through G are constants.
b) Find
Expression for the friction coefficient C f
Expression for the convection coefficient h
Both expressions must be in terms of U , T , and appropriate profile coefficients andfluid properties.
2. System Diagram
Velocity Profile, u(y)
Velocity B.L.
Thermal B.L.
Temperature Profile, T(y)U , T
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3. Assumptions
Steady state condition
Constant air properties
Uniform U , T = constant 4. Governing Equations
Friction Coefficient Definition
2
2
1
U
C s
f
Shear Stress Definition
y
u
On the surface,
0
y
s
y
u
Newton's Law of Cooling
T T hq s s
One Dimensional Fourier's Law
y
T k q f
On the plate surface y = 0
0
y
f s y
T k q
5. Detailed Solution
Friction Coefficient, C f
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A
C B A
Cy By A
Cy By Ay y
y
u
y
y
y
s
2
02
0
32
0
0302
32
Therefore,
22
2
1
2
1
U
A
U
C s
f
Convection Coefficient, h
0
y
f s s y
T k T T hq
Thus,
0
1
y
f
s y
T k
T T h
Here, T s = T(y = 0) = D + E (0) + F (0)2 - G (0)3 = D
2
0
2
0
32
03020
320
1
G F E T D
k
Gy Fy E T D
k
Gy Fy Ey D yk T Dh
f
y
f
y f
Finally,
T D
E k h
f
6. Critical AssessmentIt is important to recognize (or know) that for both cases, the friction coefficient andconvection coefficient, an analysis must be done on the surface, which implies y = 0 m.
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Example 3 (Viscous dissipation and Heat Transfer Rate):
A shaft with a diameter of 100 mm rotates at 9000 rpm in a journal bearing that is 70 mm
long. A uniform lubricant gap of 1 mm separates the shaft and the bearing. The lubricant
properties are = 0.03 N s/m 2 and k = 0.15 W/m K , while the bearing material has a thermal
conductivity ofk b = 45 W/m K
.
(a)
Determine the viscous dissipation, (W/m 3 ), in the lubricant.
(b)
Determine the rate of heat transfer (W ) from the lubricant, assuming that no heat is lost
through the shaft.
(c)
If the bearing housing is water-cooled, such that the outer surface of the bearing is
maintained at 30 C , determine the temperature of the bearing and shaft, T b and T s .
1.
Statement of the Problem
a) Given
Di = 0.1 m
= 9000 rpm = 942.5 rad/s
L = 0.07 m
a = 0.001 m (gap)
= 0.03 N s/m2
k = 0.15 W/m K k b = 45 W/m K
T wc = 30 C
Do = 0.2 m
b) Find
Viscous dissipation in the lubricant, (W/m3)
Rate of heat transfer (W ) from the lubricant, assuming no heat loss through the shaft
Lubricant
Water-cooled surface,
T wc = 30 C
Bearing, k b
Shaft100 mm
diameter
200 mm
Lubricant
Shaft T s
Bearing, k b T b
x0
1
y (mm)
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Temperatures of the bearing and shaft, T b and T s
2. System Diagram
3. Assumptions
Steady state condition
Constant fluid properties ( , , and k 's)
Fully developed flow in the gap ( u/ x = 0)
Infinite width [ L/a = (0.07 m) / (0.001 m) = 70, so this is a reasonable assumption]
p/ x = 0 (flow is symmetric in the actual bearing at no load)
4. Governing Equations
2-D Dissipation Function
2222
3
22
y
v
x
u
y
v
x
u
x
v
y
u
Velocity Distribution in Couette Flow (flow in two infinite parallel plates, but one plate
moving with constant speed)
ay y x
p y
a
U yu
2
2
1)(
Heat Diffusion Equation in Cylindrical Coordinates
t
T cq
z
T k
z
T k
r r
T kr
r r p
2
11
Fourier's Law in Cylindrical Coordinates
z
T e
T
r e
r
T ek T k q z r
1
2-D Energy Equation
Shaft ( Di, )
Lubricant , k
Water-cooled sur ace T wc
Bearin k b
D
Lubricant
Sha t T s
BearinT b
x0
a1
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q
y
T k
y x
T k
x y
T v
x
T uc p
5. Detailed Solution
Viscous dissipation in the lubricant
Assume v 0 in the gap, and the fully developed flow (assumed) implies 0
x
u
. Thus,
2222
3
22
y
v
x
u
y
v
x
u
x
v
y
u
2
y
u
Because p/ x = 0 (assumed), the velocity distribution becomes:
ay y x
p y
a
U yu
2
2
1)(
y
a
U yu )(
Therefore, the viscous dissipation is
222
a
U y
a
U
y y
u
where U is the tangential velocity of the shaft, and it is
2
i
i
D RU
Finally, the viscous dissipation is
22
2
a
D
a
U i
After plugging in values into this expression above,
= 6.662 107 W/m3
Rate of heat transfer (W ) from the lubricant
The heat transfer rate from the lubricant volume through the bearing is
q L = = ( Di a L)
q L = (6.662 107 W/m3 )[( ) (0.1 m) (0.001 m) (0.07 m)] = 1465 W
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where L = 0.07 m is the length of the bearing normal to the page.
Temperatures of the bearing and shaft, T b and T s
First, let us find out the bearing temperature (T b), which requires considering heat transfer between T b and Twc. See the diagram below.
Assume that the direction of heat transfer is in only r direction. Then Fourier's law becomes:
r
T k qr
or
r
T kAqr
In our case,
r
T rLk q br
2 … (1)
Heat diffusion equation is (assuming no heat generation in the bearing)
01
r
T kr r r
In our case, because k b = constant ,
01
r
T r k
r r b 0
1
r
T r
r r k b 0
r
T r
r
Boundary conditions for this differential equation are:
T = T b @ r = Di /2T = T wc @ r = Do /2
Let us solve the differential equation with the boundary conditions,
0
r
T r
r
1C
r
T r
r
C
r
T 1
21 )ln()( C r C r T
The first boundary condition: T b = C 1 ln(Di /2) + C 2
The second boundary condition: T wc = C 1 ln(Do /2) + C 2
T b T wc T s
Direction of Heat Transfer
r
Di /2
Do /2
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After rearranging, the temperature distribution becomes:
wc
ooi
wcb T D
r
D D
T T r T
2
lnln
)(
Substituting this temperature distribution into equation (1),
r D D
T T rLk T
D
r
D D
T T
r rLk q
oi
wcb
bwc
ooi
wcb
br
1
ln2
2ln
ln2
Therefore,
L
oi
wcbb
r q D D
T T Lk q
ln
2
C K mW m
mmW C
Lk
D DqT T
b
io L
wcb
3.81
/4507.02
1.02.0ln146530
2
ln
Finally, let us find out the shaft temperature (Ts).
The 2-D energy equation may be simplified for the prescribed conditions (see assumptions)
and further assuming v 0 and 0q , it follows that
2
y
u
y
T k
y x
T k
x x
T uc p
However, because the top and bottom plates are at uniform temperatures, the temperature
field must also be fully developed, in which case ( T/ x) = 0. For constant thermalconductivity the appropriate form of the energy equation is then
2
2
2
0
y
u
y
T k
T s
T b
U
x
k & a
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The desired temperature distribution may be obtained by solving this equation. Rearrangingand substituting for the velocity distribution,
22
2
2
a
U
dy
du
dy
T d k
Integrating twice, we obtain
21
2
2
2)( C yC y
a
U
k yT
The boundary conditions are, at y = 0, the surface is adiabatic
0
0
ydy
dT C3 = 0
and at y = a, the temperature is that of the bearing, T b
2
2
2
02
)( C aa
U
k T aT b
2
2
2U
k T C b
Hence, the temperature distribution is
2
22
2
22 1
221
2)(
a
y D
k T
a
yU
k T yT i
bb
and the temperature at the shaft, y = 0, is
222
/5.9422
1.0
/15.02
/03.03.81
22)0(
srad
m
K mW
m s N C
D
k T T T i
b s
C T s
4.303
6. Critical AssessmentWe have dealt with both heat conduction and convection situation on this problem. Make sure
you understand the difference between them and how to apply an appropriate equation for a particular case.
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Example 4 (Use of Similarity Rules and Correlation Parameters):
An industrial process involves evaporation of a thin water film from a contoured surface by
heating it from below and forcing air across it. Laboratory measurements for this surface
have provided the following heat transfer correlation:
4.058.0Pr Re43.0
L L Nu
The air flowing over the surface has a temperature of 290 K , a velocity of 10 m/s , and is
completely dry (
= 0 ). The surface has a length of 1 m and a surface area of 1 m 2 . Just
enough energy is supplied to maintain its steady-state temperature at 310 K .
(a)
Determine the heat transfer coefficient and the rate at which the surface loses heat by
convection.
(b) Determine the mass transfer coefficient and the evaporation rate (kg/h ) of the water on
the surface.
(c)
Determine the rate at which heat must be supplied to the surface for these conditions.
1. Statement of the Problema) Given
Heat transfer correlation equation:4.058.0
Pr Re43.0 L L Nu
Forcing air properties:
T = 290 K (temperature)
U = 10 m/s (velocity)
= 0 (completely dry)
Surface dimensions and property:
L = 1 m (length) A s = 1 m2 (area)
T s = 310 K (temperature) b) Find
Heat transfer coefficient
Rate at which the surface loses heat by convection
Mass transfer coefficient
Evaporation rate (kg/h) of the water on the surface
Rate at which heat must be supplied to the surface for these conditions
2. System Diagram
Air
T
U
Thin water film
Surface
T s L
A s
Heat transfer correlation:4.058.0
Pr Re43.0 L L Nu
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3. Assumptions
Steady state condition
Constant properties
Heat-mass analogy applies:
Heat Transfer Mass Transfer
Pr ,Re1 L f Nu Sc f Sh L ,Re
2
Correlation requires properties evaluated at K T T
T s
mean 3002
4. Governing Equations
Reynolds Number:
LV
L
Re
Prandtl Number:
Pr
Schmidt Number: AB
DSc
Average Nusselt Number: f k
Lh Nu
Average Sherwood Number: AB
m
D
LhSh
Newton's Law of Cooling:
T T Ahq s sconv
Convection Mass Transfer Equation: ,, A s A sm Ahm
First Law of Thermodynamics (for steady flow process): 0out in
E E
5.
Detailed Solution
Properties:
Air (at T mean = 300 K, 1 atm)
= 15.89 10-6 m2 /s
k f = 0.0263 W/m K Pr = 0.707
Air-water mixture (at T mean = 300K, 1 atm)
D AB = 0.26 10-4 m2 /s
Saturated water (at T s = 310 K )
A, sat = 1/ v g = 1/22.93 m3 /kg = 0.04361 kg/m3
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h fg = 2414 kJ/kg
Heat transfer coefficient
First of all, evaluate Re L at T mean to characterize the flow
5
26 10293.6
/1089.15
1/10Re
sm
m sm LU
L
and substituting into the prescribed correlation for this surface, find
f
L Lk
Lh Nu 1.864707.010293.643.0Pr Re43.0
4.058.054.053.0
72.221
/0263.01.864
m
K mW
L
k Nuh
f LW/m2 K
Rate at which the surface loses heat by convection
W K K m K mW T T Ahq s sconv 2.4542903101/71.22 22
Mass transfer coefficient
Using the heat-mass analogy,
Heat:4.058.0
Pr Re43.0 L L Nu
Mass:4.058.0
Re43.0 ScSh L L
where
6112.0/1026.0
/1089.15
24
26
sm
sm
DSc
AB
Substituting numerical values, and find
AB
m
L L
D
LhScSh 2.8156112.010293.643.0Re43.0
4.058.054.053.0
smm
sm
L
DSh
h AB L
m /1012.21
/1026.02.815 2
24
Evaporation rate (kg/h) of the water on the surface
The evaporation rate, with A,s = A,sat (Ts), is
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3322
,, /0/04361.01/1012.2 mkg mkg m sm Ahm A s A sm
hkg skg m /327.3/10243.9 4
Rate at which heat must be supplied to the surface for these conditions
Applying the first law of thermodynamics,
0out in
E E
0 evapconvin qqq
where qin is the heat supplied to sustain the losses by convention and evaporation.
W q
W W q
kg J skg W q
hmT T Ahq
qqq
in
in
in
fg s sin
evapconvin
2685
3.22312.454
/102414/10243.92.454 34
6. Critical Assessment
Heat-mass analogy has been applied in this problem. Note that convection mass transfer
can be analyzed like convection heat transfer. Equations are very similar to each other.
Notice that the heat loss from the surface by evaporation is nearly 5 times that due toconvection.
The End
Air qconv qevap
qin