Condensation

Chapter 10

Sections 10.6 through 10.11

General Considerations

General Considerations• Heat transfer to a surface occurs by condensation when the surface temperature is less than the saturation temperature of an adjoining vapor.

• Film Condensation

Entire surface is covered by the condensate, which flows continuously from the surface and provides a resistance to heat transfer between the vapor and the surface.

Thermal resistance is reduced through use of short vertical surfaces and horizontal cylinders.

Characteristic of clean, uncontaminated surfaces.

• Dropwise Condensation

Surface is covered by drops ranging from a few micrometers to agglomerations visible to the naked eye.

General Considerations (cont).

Thermal resistance is greatly reduced due to absence of a continuous film.

Surface coatings may be applied to inhibit wetting and stimulate dropwise condensation.

Film Condensation: Vertical Plates

Film Condensation on a Vertical Plate• Distinguishing Features

Generally, the vapor is superheated and may be part of a mixture that includes noncondensibles.

,v satT T

A shear stress at the liquid/vapor interface induces a velocity gradient in the vapor, as well as the liquid.

• Nusselt Analysis for Laminar Flow

Assumptions:

A pure vapor at .satT

Negligible shear stress at liquid/vapor interface.

0y

uy

Thickness and flow rate of condensate increase with increasing x

m

Vertical Plates (cont)

Negligible advection in the film. Hence, the steady-state x-momentum and energy equations for the film are

2

2

2

2

1

0

l l

pu Xy x

Ty

The boundary layer approximation, may be applied to the film.0/ ,p y Hence,

vp dp

gx dx

Solutions to momentum and energy equations

Film thickness:

1 44

/

l l sat s

l l v fg

k T T xx

g h

Vertical Plates (cont)

Flow rate per unit width:

3

3l l v

l

gmb

Average Nusselt Number:

1 43

0 943

/

.Ll l v fgL

l l l sat s

g h Lh LNuk k T T

1 0 68

Jakob number

.fg fg

p sat s

fg

h h Ja

c T TJa

h

Total heat transfer and condensation rates:

L sat s

fg

q h A T T

qm

h

Vertical Plates (cont)

• Effects of Turbulence:

Transition may occur in the film and three flow regimes may be identified and delineated in terms of a Reynolds number defined as

44 4Re l m

l l l

umb

Vertical Plates (cont)

Wave-free laminar region Re 30 :

1 32-1/31 47 Re

//

.L l

l

h g

k

3

2

4Re

3l l v

l

g

Wavy laminar region 30 Re 1800 :

(10.37)

1 32

1.22

Re

1.08 Re 5 2

//

.L l

l

h g

k

(10.38)

Turbulent region Re >1800 :

(10.39)

1 32

-0.5 0 75

Re

8750 +58 Pr Re 253

/

.

/L l

l

h g

k

Vertical Plates (cont)

Calculation procedure:

– Assume a particular flow regime and use the corresponding expression for (Eq. 10.37, 10.38 or 10.39) to determine Lh Re .

Re– If value of is consistent with assumption, proceed to determination of and .q m

– If value of is inconsistent with the assumption, recompute its value using a different expression for and proceed to determination of

Re

Lh and .q m

Film Condensation: Radial Systems

Film Condensation on Radial Systems

• A single tube or sphere:

1 43 /

l l l fgD

l sat s

g k hh C

T T D

Tube: C =0.729 Sphere: C=0.826

Film Condensation: Radial Systems (cont).

• A vertical tier of N tubes:

1 43

0 729

/

, . ll l fgD N

l sat s

g k hh

N T T D

Why does decrease with increasing N?,D Nh

How is heat transfer affected if the continuous sheets (c) breakdown and the condensate drips from tube to tube (d)?

What other effects influence heat transfer?

Film Condensation: Internal Flow

Film Condensation for a Vapor Flow in a Horizontal Tube• If vapor flow rate is small, condensate flow is circumferential and axial:

,iRe 35 000, , :m

i

u D

1 43

0 555

/

. l l l fgD

l sat s

g k hh

T T D

0 375.fg fg sat sh h T T

• For larger vapor velocities, flow is principally in the axial direction and characterized by two-phase annular conditions.

Dropwise Condensation

Dropwise Condensation

• Steam condensation on copper surfaces:

dc

51100 2044 22 C< 100 C

255 500 100 C

,

,

dc sat sat

sat

h T T

h T

dc sat sq h A T T

Problem: Condensation on a Vertical Plate

Problem 10.48 a,b: Condensation and heat rates per unit width for saturatedsteam at 1 atm on one side of a vertical plate at 54˚C if (a) the plate height is 2.5m and (b) the height is halved.

KNOWN: Vertical plate 2.5 m high at a surface temperature Ts = 54C exposed to steam at atmospheric pressure.

FIND: (a) Condensation and heat transfer rates per unit width, (b) Condensation and heat rates if the height were halved.

ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensables in steam.

SCHEMATIC:

Problem: Condensation on a Vertical Plate (cont)

PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100C, v = 0.596 kgm3, hfg = 2257 kJkg; Table A-6, Water, liquid (Tf = (100 54)C2 = 350 K): 973.7 kgm3, k 0.668

WmK, 365 10-6 Nsm2 , p,c = 4195 JkgK, Pr = 2.29.

ANALYSIS: (a) The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33, L sat s fgq h L T T m q h (1,2)

where, from Eq. 10.26, with Ja cp, (Tsat Ts)hfg ,

fg fg p, sat s fgh h 1 0.68 c T T h

fg 3

4195J kg K 100 54 KkJh 2257 1 0.68 2388kJ kg

kg 2257 10 J kg

.

Assuming turbulent flow conditions, Eq. 10.39 is the appropriate correlation,

1/ 32

0.5 0.75

hL g ReRe 1800

k 8750 58Pr Re 253

(3)

Problem: Condensation on a Vertical Plate (cont)

Not knowing Re or Lh , another relation is required. Combining Eqs. 10.33 and 10.35,

fg fgL

sat sat

mh hRe bh

A T T 4 A T T

. (4)

Substituting Eq. (4) for Lh into Eq. (3), with A bL,

fg1/ 30.5 0.75 2sat

Re bh Re k

4 bL T T 8750 58Pr Re 253 g

. (5)

Using appropriate properties with L = 2.5 m, find

6 2 3365 10 N s m 2388 10 J kg

4 2.5m 100 54 K

(6)

0.5 1/ 30.75 26 4 2 2

1 0.668W m K

8750 58 2.29 Re 253365 10 973.7 m s 9.8m s

Re 2979 .

Since Re 1800, the flow is turbulent, and using Eq. (4) or (3), find

2Lh 5645W m K .

Problem: Condensation on a Vertical Plate (cont)

From the rate equations (1) and (2), the heat transfer and condensation rates are

2q 5645W m K 2.5m 100 54 K 649k W m <

3 3m 649 10 W m 2388 10 J kg 0.272kg s m . <

(b) If the height of the plate were halved, L = 1.25 m, and turbulent flow was still assumed to exist, the LHS of Eq. (5) may be reevaluated and the equation solved to obtain Re 1280 .

Since 1800 Re , the flow is not turbulent, but wavy-laminar. The procedure now follows that of Example 10.3. For L = 1.25 m with wavy-laminar flow, Eq. 10.38 is the appropriate correlation. The calculation yields

Re 1372 2Lh 5199 W m K

q 299kW m m 0.125kg s m . <

COMMENT:

Note that the height was decreased by a factor of 2, while the rates decreased by a factor of 2.2. Would you have expected this result?