free convection: cylinders, spheres, and enclosures

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1 Free Convection: Cylinders, Spheres, and Enclosures Chapter 9 Section 9.6.3 through 9.8 Cylinders The Long Horizontal Cylinder Boundary Layer Development and Variation of the Local Nusselt Number for a Heated Cylinder: The Average Nusselt Number: 2 1/6 12 8 / 27 9/16 0.387 0.60 10 1 0.559 / Pr D D D Ra Nu Ra See Example 9.4 in text How do conditions change for a cooled cylinder?

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Boundary Layer Development and Variation of the Local Nusselt Numberfor a Heated Cylinder:The Average Nusselt Number:Over such a large temperature range, the fluid properties are likely tovary significantly, particularly  and Pr. A more accurate solution could therefore beperformed if the temperature dependence of the properties were known. (2) Condensation ofthe steam is a significant process expense, which is linked to the equipment (capital) andenergy (operating) costs associated with steam production

TRANSCRIPT

Page 1: Free Convection: Cylinders, Spheres, and Enclosures

1

Free Convection:

Cylinders, Spheres, and Enclosures

Chapter 9

Section 9.6.3 through 9.8

Cylinders

The Long Horizontal Cylinder

• Boundary Layer Development and Variation of the Local Nusselt Number

for a Heated Cylinder:

• The Average Nusselt Number:

2

1/ 612

8 / 279 /16

0.3870.60 10

1 0.559/ Pr

DD D

RaNu Ra

•See Example 9.4 in text

•How do conditions change for a cooled cylinder?

Page 2: Free Convection: Cylinders, Spheres, and Enclosures

2

Spheres

Spheres

• The Average Nusselt Number:

1/ 4

4 / 99 /16

0.5892

1 0.469/ Pr

DD

RaNu

In the limit as how may conditions be characterized? Avg Nu=2 0,DRa

–tbl_09_02b

Page 3: Free Convection: Cylinders, Spheres, and Enclosures

3

Enclosures

Enclosures

• Rectangular Cavities

Characterized by opposing walls of different temperatures, with the

remaining walls well insulated.

3

1 2L

g T T LRa

1 2q h T T

Enclosures (cont)

Horizontal Cavities

Heating from Below 0

– , 1708:L L cRa Ra

Fluid layer is thermally stable. 1LhLNuk

– 41708 5 10 :LRa

Thermal instability yields a regular

convection pattern in the form of roll cells.

– 5 93 10 7 10 :LRa

Buoyancy driven flow is turbulent

1/ 3 0.0740.069 PrL LNu Ra

Horizontal Cavity 0, 180deg

Vertical Cavity 90 deg

For

For

For

Page 4: Free Convection: Cylinders, Spheres, and Enclosures

4

Enclosures (cont)

Heating from Above 180 deg

– Fluid layer is unconditionally stable.

1LNu

Vertical Cavities

310 :LRa

310 :LRa

– A primary cellular flow is established, as the core

becomes progressively more quiescent, and

secondary (corner) cells develop with increasing .LRa

Correlations for Eqs. (9.50) - (9.53).LNu

1LNu

Enclosures (cont)

Inclined Cavities

Relevant to flat plate solar collectors.

Heat transfer depends on the magnitude of relative to a critical angle ,

whose value depends on H/L (Table 9.4).

*

Heat transfer also depends on the magnitude of relative to a critical

Rayleigh number of LRa

, 1708/cos .L cRa

Heat transfer correlations Eqs. (9.54) – (9.57).

Page 5: Free Convection: Cylinders, Spheres, and Enclosures

5

Holland’s Correlation

1

6

1

3

1708(sin(1.8 ))17081 1.44 1 1

15830

Where tilt angle from HorizontalSee text

L

L L

L

NuRa Cos Ra Cos

Ra Cos

Holland’s Correlation

Page 6: Free Convection: Cylinders, Spheres, and Enclosures

6

Enclosures (cont)

Annular Cavities

• Concentric Cylinders

2

1n /

eff

i o

o i

kq T T

r r

1 4

1 40 3860 861

/

eff /

c

k Pr. Ra

k . Pr

or keff/k =1 if the value calculated above is less than unity.

Enclosures (cont)

The length scale in Rac is given by

• Concentric Spheres

4 3

5 33 5 3 5

2/

o i

c // /

i o

ln r / rL

r r

4

1 1

eff i o

i o

k T Tq

/ r / r

1 4

1 40 740 861

/

eff /

s

k Pr. Ra

k . Pr

or keff/k = 1 if the value calculated above is less than unity.

The length scale in Ras is given by

4 3

5 31 3 7 5 7 5

1 1

2

/

i o

s // / /

i o

/ r / rL

r r

Page 7: Free Convection: Cylinders, Spheres, and Enclosures

7

–p_09_100

–tbl_09_03

Page 8: Free Convection: Cylinders, Spheres, and Enclosures

8

–tbl_09_04

Problem: Batch Reactor

Problem 9.74: Use of saturated steam to heat a pharmaceutical in a batch reactor.

KNOWN: Volume, thermophysical properties, and initial and final temperatures of a

pharmaceutical. Diameter and length of submerged tubing. Pressure of saturated steam

flowing through the tubing.

FIND: (a) Initial rate of heat transfer to the pharmaceutical, (b) Time required to heat the

pharmaceutical to 70C and the amount of steam condensed during the process.

Page 9: Free Convection: Cylinders, Spheres, and Enclosures

9

Problem: Batch Reactor (cont)

SCHEMATIC:

Pharmaceutical = 1100 kg/m3

c = 2000 J/kg-K

= 4.0x10 /s

Pr = 10, = 0.002 K-1

k = 0.250 W/m-K-6 m2

D = 15 mm, L = 15 mTubing

T(t)

T Cs o= 127

psat = 2.455 bars

T C, V = 200 L

i o= 25 T = 70 C

f

o

Saturated steam

ASSUMPTIONS: (1) Pharmaceutical may be approximated as an infinite, quiescent fluid of

uniform, but time-varying temperature, (2) Free convection heat transfer from the coil may be

approximated as that from a heated, horizontal cylinder, (3) Negligible thermal resistance of

condensing steam and tube wall, (4) Negligible heat transfer from tank to surroundings, (5)

Constant properties.

PROPERTIES: Table A-4, Saturated water (2.455 bars): Tsat = 400K = 127C, hfg = 2.183

106 J/kg. Pharmaceutical: See schematic.

ANALYSIS: (a) The initial rate of heat transfer is s s iq hA T T , where As = DL = 0.707

m2 and h is obtained from Eq. 9.34.

Problem: Batch Reactor (cont)

With = /Pr = 4.0 10-7

m2/s and RaD = g (Ts – Ti) D

3/ = 9.8 m/s

2 (0.002 K

-1) (102K)

(0.015m)3/16 10

-13 m

4/s

2 = 4.22 10

6,

D

2 21/ 6

61/ 6D

8/ 27 8/ 279/16 9/16

0.387 4.22 100.387 RaNu 0.60 0.60 27.7

1 0.559 / Pr 1 0.559 /10

Hence, 2

Dh Nu k / D 27.7 0.250W / m K / 0.015m 462W / m K

and 2 2s s iq hA T T 462W / m K 0.707m 102 C 33,300W <

(b) Performing an energy balance at an instant of time for a control surface about the liquid,

s sd cT

q t h t A T T tdt

where the Rayleigh number, and hence h, changes with time due to the change in the

temperature of the liquid.

Page 10: Free Convection: Cylinders, Spheres, and Enclosures

10

Problem: Batch Reactor (cont)

Integrating the foregoing equation numerically, the following results are obtained for the

variation of T and h with t.

0 100 200 300 400 500 600 700 800 900

Time, t(s)

25

35

45

55

65

75

Te

mp

era

ture

, (C

)

0 100 200 300 400 500 600 700 800 900

Time, t(s)

370

390

410

430

450

470

Co

nve

ctio

n c

oe

ffic

ien

t, h

ba

r (W

/m^2

.K)

The time at which the liquid reaches 70C is

ft 855s <

The rate at which T increases decreases with increasing time due to the corresponding

reduction in (Ts – T), and hence reductions in DRa , h and q.

The Rayleigh number decreases from 4.22 106 to 2.16 10

6, while the heat rate decreases

from 33,300 to 14,000 W.

The convection coefficient decreases approximately as (Ts – T)1/3

, while q ~ (Ts – T)4/3

.

Problem: Batch Reactor (cont)

The latent energy released by the condensed steam corresponds to the increase in thermal

energy of the pharmaceutical. Hence, c fgm h f ic T T ,

and

3 3

f ic 6

fg

c T T 1100kg / m 0.2m 2000J / kg K 45 Cm 9.07kg

h 2.183 10 J / kg

<

COMMENTS: (1) Over such a large temperature range, the fluid properties are likely to

vary significantly, particularly and Pr. A more accurate solution could therefore be

performed if the temperature dependence of the properties were known. (2) Condensation of

the steam is a significant process expense, which is linked to the equipment (capital) and

energy (operating) costs associated with steam production.