condensation and boiling

CHE 421: UNIT OPERATIONS II: HEAT TRANSFER

Steam saturated at 68.9 kPa is condensing on a vertical tube 0.305 m long having an OD of 0.0254 m and a surface temperature of 86.11OC. Calculate the average heat transfer coefficient using English and SI units.

Tsat = 89.44OC Tw = 86.11OC

Tf = Tw +Tsat = 89.44 + 86.11 = 87.8OC 2 2Latent Heat, λ = 2283.2 kJ/kg

ρl = 996.7 kg/m3 ρv = 0.391 kg/m3μl = 3.24 x 10-4 Pa.s kl = 0.675 W/m.KL = 0.305 m

ΔT = Tsat – Tw = 89.44 – 86.11 = 3.33 K

I. Assume Laminar flow

Nu = 1.13 ρl (ρl – ρv) gλL3 1/4

μlklΔT

=1.13 996.7 (996.7– 0.391)(9.8)(2.2832x106)(0.305)3 1/4

(3.24x10-4)(0.675)(3.33K)= 6129.3187

ho = 6129.3187(0.675) = 13564.8857W/m2.K 0.305A = ΠDL = Π(0.0254)(0.305) = 0.0243 m2q = hAΔT = ṁλ ṁ = 13564.8857 W/m2.K (0.0243 m2)(3.33 K) 2.2832x106 J/kg ṁ = 4.8075 x 10-4 kg/s

II. Check assumption:

Re = 4ṁ = 4 (4.81 x10 -4 kg/s) = 74.3791 Laminar! ΠDμl Π(0.0254m)(3.24x10-4kg/m.s)

1


Saturated steam condenses on the outside of a 5 cm-diameter vertical tube, 50 cm high. If the saturation temperature of the steam is 302 K, and cooling water maintains the wall temperature at 299 K, determine: (i) the average heat transfer coefficient, (ii) the total condensation rate, and (iii) the film thickness at the bottom of the tube.

Critical Temperature of water: 2 – 142 HBTc = 647.096 K

Calculate Tr : Tr=TsatTc

Tr=TsatTc

= 302K647.096K

Calculation of lambda , λλ = 2.4294 x 106 J/Kg

Calculate the film temperature

Tf=Tsat+Tw2

=302+2992

=300.5K

Properties @ Tf

ρL = 995.5116 kg/ m3

μL = 8.4583 x 10 -4

ρv = .0265 kg/ m3

KL = .6111 W/mK

ASSUME FLOW TO BE LAMINAR

Nu = 1.13 ρL (ρl – ρv) L3gλ 1/4 νlΔTkl

Nu=hLK

=1.13¿

∆T=Tsat−Twall=302−299=¿

h=9120.5414W

m2K

2

( ii ) The total condensation rate is:

{ m=Qh fg

=h AΔTh fg

=(9120 . 5414 )π (0 . 05 )( 0. 5 )(302−299 )(2 . 4294×106 )

=8 . 8457×10−4 kg/s¿

( iii ) The film thickness is

δ=(3ν l Γ

ρl g )1/3

ρv<< ρl

The mass flow rate per unit width of film Γ is:

Γ=mπD

=( 8.8457×10−4 )( π )(0 .05 )

=5 .6313×10−3 kg/ms

Hence, δ=(3(8 .4839×10-6 )(5. 6313×10−3 )(995 .5116 )(9 .8 ) )

1/3

=1 .1374×10−4 m


Check assumption Re = 4ṁ = 4 (8.8457 x 10-4kg/s) = 26.6311 < 1800 Laminar! ΠDμl Π(0.05m)( 8.4583 x 10 -4kg/m.s)

Saturated steam condenses on the outside of a 5 cm diameter vertical tube, 50 cm high. If the saturation temperature of the steam is 302 K, and cooling water maintains the wall temperature at 299 K, determine:

1. the average heat transfer coefficient, 2. the total condensation rate, and 3. the film thickness at the bottom of the tube.

Given: Film condensation of saturated steam

Required: (i) Average heat transfer coefficient, (ii) total condensation rate, (iii) and film thickness1. Effect of tube curvature negligible2. Effect of liquid sub cooling negligible3. Laminar

Evaluate hfg at the saturation temperature of 302 K

3

Condensate Film

g

y

y

x

Tsat

(x)

A

T

The Average heat transfer coefficent is given by:

h¿

=0 .943[h fg' g ( ρl−ρv )k l3

L(T sat−Tw )v l ]1/4

From Table of water properties:

hfg=2 .432×106

J /kg ρ v=0 .03kg /m3

Also, for water k l=0 . 611 W/mK

ρl=996 kg/m3

ν l=0 .87×10 -6 m2 /s

h=0 .943 [h fh g (ρl−ρv )k l3

L (T sat−Tw) ν l ]1/4

=0 . 943[(2. 432×10−6 )(9 . 81) (996−0. 03 )(0. 611)3

(0. 5 )(3)(0 . 87×10−6 ) ]1/4

=7570 W/m2K

( ii ) The total condensation rate is: { m

=Qhfg

= h AΔThfg

=(7570)(3 )π (0 . 05 )(0. 5 )(2. 432×106 )

=7 . 33×10−4 kg/s ¿

( iii ) The film thickness is

δ=(3ν l Γ

ρl g )1/3

ρv<< ρl

The mass flow rate per unit width of film Γ is:

Γ=mπD

=(7 .33×10−4 )( π )(0 .05 )

=4 .67×10−3 kg/ms

Hence, δ=(3(0 .87×10-6 )(4 .67×10−3 )(996)( 9.81 ) )

1/3

=1 .08×10−4 m


Air free saturated steam at 85OC condenses on the outer surfaces of 225 horizontal tubes of 1.27 cm OD, arranged in a 15 x 15 array. Tube surfaces are maintained at a uniform temperature of 75OC. Calculate the condensate rate per one m length of tube.

Tsat = 85OC Tw = 75OC

Tf = Tw +Tsat = 85 + 75 = 80OC 2 2

DO = 0.0127 m Latent Heat, λ = 2.3045 x 106 J/kgρl = 970.8918 kg/m3 ρv = .2973μl = 3.5533 x 10-4 Pa.s kl = 0.6698 W/m.K L = 1 m

ΔT = Tsat – Tw = 85 – 75 = 10 K

Nu = 0.725 ρl (ρl – ρv) gλD3 1/4

NμlklΔT hD = 0.725 970.8918 (970.8918-.2973)2 (9.81)(2.3045x106)(1)3 1/4 kl 15(3.5533 x 10-4)( 0.6698 W/m.K)(10 K) = 135.5272

hi = 135. 5272 (0.6698) = 7147.7295 W/m2.K 0.0127

A = ΠDL = Π(0.0127)(1) = 0.0399 m2

q = h n A ΔT = ṁλ

ṁ = 7147.7295 W/m2.K (Total N) (0.0399 m2)(10 K) = 7147.7295(225)(0.0399)(10) 2.3045 x 106 J/kg 2.3045 x 106 J/kg ṁ = .2784 kg/s

4

Also, for water k l=0 . 611 W/mK

ρl=996 kg/m3

ν l=0 .87×10 -6 m2 /s

h=0 .943 [h fh g (ρl−ρv )k l3

L (T sat−Tw) ν l ]1/4

=0 . 943[(2. 432×10−6 )(9 . 81) (996−0. 03 )(0. 611)3

(0. 5 )(3)(0 . 87×10−6 ) ]1/4

=7570 W/m2K

( ii ) The total condensation rate is: { m

=Qhfg

= h AΔThfg

=(7570)(3 )π (0 . 05 )(0. 5 )(2. 432×106 )

=7 . 33×10−4 kg/s ¿


Air free saturated steam at 90OC condenses on the outer surfaces of 2.5-cm OD 6 m long vertical tube, where outer surface temperature is maintained at a uniform temperature of 60OC. Calculate the total rate of condensation of steam at the total surface.

Tsat = 90OC Tw = 60OC

Tf = Tw +Tsat = 90 + 60 = 75OC 2 2

DO = 0.025 m Latent Heat, λ = 2592.8468 kJ/kgρl = 973.9543kg/m3 ρv = .2442 kg/m3

μl = 3.7861 x 10-4 Pa.s kl = 0.6666 W/m.K L = 1 mΔT = Tsat – Tw = 90 – 60 = 30 K

Assume Laminar flow Nu = 1.13 ρl (ρl – ρv) gλL3 1/4 μlklΔT Nu = 1.13 (973.9543) (973.9543 - .2442) 9.8(2.283x106)(6)3 1/4

(3.7861 x 10-4)(0.6666)(30 K)= 32 338.3122

hi = 32 338.3122 (0.6666) = 3517.3455 W/m2.K 6A = ΠDL = Π(0.025)(6) = 0.471239 m2

q = hAΔT = ṁλ ṁ = 3517.3455 W/m2.K (0.471239 m2)(30 K) 2.5928 x 106 J/kg ṁ = .0201 kg/s

Check assumption:Re = 4ṁ = 4 (0.0201 kg/s) ΠDμl Π(0.025m)(3.7861 x 10-4kg/m.s)Re = 2703.7970Re = 2703.7970 ≠ assumed 2500

Nu = 0.0077 gρl(ρl – ρv) L3 1/3 Re0.4

μl2

Assume Re = 2500hD = 42441.5272 kl Nu = 42441.5272hi = 42441.5272 (0.6666) = 4715.2357W/m2.K 6A = ΠDL = Π(0.025)(6) = 0.4712 m2

q = hAΔT = ṁλ ṁ = 0.0264 kg/s

Check assumption:Re = 4ṁ = 4 (0.0264 kg/s) ΠDμl Π(0.025m)(3.7861 x 10-4kg/m.s)Re = 3551.2558Re = 3551.2558 ≠ 2500

5


Assume Re = 4500hD = 53690.8161 kl Nu = 53690.8161hi = 5390.8161 (0.6666) = 5965.0497 W/m2.K 6A = ΠDL = Π(0.025)(6) = 0.4712 m2

q = hAΔT = ṁλ ṁ = .0333kg/s

Check assumption:Re = 4ṁ = 4 (0.0333 kg/s) ΠDμl Π(0.025m)(3.7861 x 10-4kg/m.s)Re = 4479.4249Re = 4479.4249 ≈ assumed 4500

Saturated steam at atmospheric pressure condenses on a 2-cm tube whose surface is maintained at 340 K. Determine the rate of condensation and the heat transfer coefficient for the case of 1.5-m long tube oriented vertically.

6


Boiling

A: Free convection; B: Nucleate boiling; C: Transition Boiling;D: Film boiling

Modes of Pool Boiling Excess Temperature (deg. C)Free Convection: < 5Nucleate Boiling: 5 – 25Transition Boiling: 25 – 120Film Boiling: > 120

The excess temperature is the surface temperature above the boiling point of the liquid.

Nucleate Boiling

Bubbles form at surface and separate, mixing the fluid. As temperature increases, the bubbles join to form bubble columns or jets. These eventually merge to form slugs. This type of boiling is the most desirable for a chemical process due to the high heat rates

Transition Boiling

Film begins to form at surface. Surface changes from nucleate to film boiling.

7


Film Boiling

Surface completely covered with film blanket.

Nucleate Boiling Correlations

8


Film Boiling Correlations

Water is being boiled at 1 atm abs pressure in a jacketed kettle with steam condensing in the jacket at 115.6⁰C. The inside diameter of the kettle is 0.656 m and the height is 0.984 m. the bottom is slightly curved but it will be assumed to be flat. Both the bottom and the sides up to the height of 0.656 m are jacketed. The kettle surface for heat transfer is 3.2 mm stainless steel with a k of 16.27 W/m.K. the condensing steam coefficient hi inside the jacket has been estimated as 10200 W/m.K. predict the boiling heat transfer coefficient ho for the bottom surface of the kettle.

Assume Tw = 110⁰CΔT = Tw – Tsat = 110 – 100 = 10⁰C

Using the simplified equation for horizontal surfaces: ho = 5.56 ΔT3 = 5.56 (10)3 = 5560 W/m2.Kq/A = h ΔT = 5560(10) = 55600 =55.6 kW/m2

Check assumed Tw:Ri = 1/hiA = 1/(10200)(1) = 9.8 x 10-5

Rw = Δx = 3.2/1000 = 19.66 x 10-5

kA 16.27 (1)

Ro = 1/hoA = 1/(5560)(1) = 17.98 x 10-5

ΣR = 47.44 x 10-5

ΔT = ΔTo ; ΔTo = Ro(ΔT) = 17.98 x 10-5 (115.6 – 100)ΣR Ro ΣR 47.44 x 10-5

ΔTo = 5.9 = (Tw – 100) ; Tw = 105.9 ⁰C ≠ 110⁰C

Reassume Tw and Recompute Tw…

Assume Tw = 108⁰CΔT = Tw – Tsat = 108 – 100 = 8⁰C

Using the simplified equation for horizontal surfaces: ho = 5.56 ΔT3 = 5.56 (8)3 = 2846.72 W/m2.Kq/A = h ΔT = 2846.72(8) = 28467.2 = 28.4672 kW/m2

Check assumed Tw:Ri = 1/hiA = 1/(10200)(1) = 9.8 x 10-5

Rw = Δx = 3.2/1000 = 19.66 x 10-5

kA 16.27 (1) Ro = 1/hoA = 1/(2846.72)(1) = 3.51 x 10-4

ΣR = 6.4588 x 10-4

ΔT = ΔTo ; ΔTo = Ro(ΔT) = 3.51 x 10-4 (115.6 – 100)ΣR Ro ΣR 6.4588 x 10-4

ΔTo = 8.48 = (Tw – 100) ; Tw = 108.48 ⁰C ≈ 108⁰C

9

1/ 43( )

( )

where 0.62 for horizontal cylinders

0.67 for spheres

is the density

is the enthalpy change from liq. to vapor

l v fgNu

v v v s sat

fg

g h DhDN C

k v k T T

C

C

h

is the diameter of the pipe

is the specific volume

is the thermal conductivity

is the surface temperature

is the boiling point temperature s

sat

D

v

k

T

T of the liquid

Subscript refers to the vapor

Subscript refers to the liquid.

v

l

Film boiling not desirable! Better to have nucleated!


A vessel with a flat bottom and 0.1 m2 in area is used for boiling water at atmospheric pressure. Find the temperature at which the vessel must be maintained if a boiling rate of 80 kg/h is desired. Assume that the vessel is made of copper and the boiling is nucleate. Take ρV = 0.60 kg/m3

Latent Heat, λ = 2257 kJ/kgρl = 960.6 kg/m3 ρv = 0.60 kg/m3 μl = 2.824 x 10-4 Pa.s

q = hAΔT = ṁλ q/A = ṁλ /A

q/A = 501.556 kW/m2 ho = 5.56 ΔT3

501555.5556 W/m2 = h (Tw – Tsat) 501555.5556 W/m2 = 5.56 (Tw – Tsat)4 501555.5556 W/m2 = 5.56 (Tw – 100)4

Tw = 117.33 OC

Calculate the heat transfer coefficient and the mass flow rate during stable film boiling of water from a 0.9 cm diameter horizontal carbon tube. The water is saturated and at 100OC. The tube surface is at 1000OC.

Tsat = 100OC Tw = 1000OCTf = Tw +Tsat = 100 + 1000 = 550OC

2 2

DO = 0.009 m

Latent Heat, λ = 2257 kJ/kgρl = 960.63 kg/m3 ρv = 0.4005 kg/m3 νv = 4.70 x 10-5 m2/s kv = 0.0379 W/m.K

ΔT = Tw – Tsat = 1000 – 100 = 900 K

FILM BOILING IN HORIZONTAL TUBES

Nu = 0.62 g (ρl – ρv) λD3 1/4 νvkvΔT

Nu = 0.62 9.81(960.63 – 0.4005)(2.257x106)(0.009)3 1/4 (4.70 x 10-5)(0.0379)(900 K)

= 34.5718

hi = 34.5718(0.0379) = 145.5859 W/m2.K 0.009A = ΠDL = Π(0.009)(1) = 0.0283 m2

q = hAΔT = ṁλ ṁ = 145.5859 W/m2.K (0.0283 m2)(900 K) 2.283 x 106 J/kg ṁ = 1.6414 x 10 -3 kg/s

10

q/A = 80 kg/h (1h/3600 s) 2.257 x 106 J/kg 0.1 m2

condensation and boiling

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