ejercicios capa limite

7/21/2019 Ejercicios Capa Limite

http://slidepdf.com/reader/full/ejercicios-capa-limite 1/16

Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

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)



Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

3. Assumptions


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


Friction Coefficient, C f



Ghosh - 550 1/5/2016

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.



Ghosh - 550 1/5/2016

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)



Ghosh - 550 1/5/2016

Temperatures of the bearing and shaft, T b and T s

2. System Diagram

3. Assumptions


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)


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



Ghosh - 550 1/5/2016

q

y

T k

y x

T k

x y

T v

x

T uc p


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



Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

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.



Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

3. Assumptions


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


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



Ghosh - 550 1/5/2016

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



Ghosh - 550 1/5/2016

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

http://people.rit.edu/angeme/550cd/Introduction.doc

http://people.rit.edu/angeme/550cd/Introduction.doc

ejercicios capa limite

Documents