Scribd
Upload
790 views6 pages

The - J-Factor of Chilton and Colburn: Unit Operation

The document discusses heat and mass transfer in tubes. It presents several equations for calculating the heat transfer coefficient (h), including the Chilton-Colburn analogy equation relating h to the Reynolds number. It also discusses analogous equations for mass transfer coefficients. An example problem is then presented to calculate the temperature rise of water flowing through a pipe using different methods, including the Chilton-Colburn equation. The temperatures calculated range from 313.9K to 324.1K depending on the specific method used.

Uploaded by

Ghazy alshyal
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
790 views6 pages

The - J-Factor of Chilton and Colburn: Unit Operation

The document discusses heat and mass transfer in tubes. It presents several equations for calculating the heat transfer coefficient (h), including the Chilton-Colburn analogy equation relating h to the Reynolds number. It also discusses analogous equations for mass transfer coefficients. An example problem is then presented to calculate the temperature rise of water flowing through a pipe using different methods, including the Chilton-Colburn equation. The temperatures calculated range from 313.9K to 324.1K depending on the specific method used.

Uploaded by

Ghazy alshyal
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 6

Unit OperatiOn

The -j- Factor of Chilton and Colburn


 For forced convection in tubes
. .
= 0.023
If both sides are divided by the product (Re.Pr)
. .
= = = 0.023 .
.

. .
. = = 0.023
i.e for heat transfer
.
= .

Chilton and Colburn found that a plot of (jh) against (Re) gave
approximately the same curve as the friction chart for flow
through tubes .
# By analogy with the derivation gives above for heat transfer
Chilton and Colburn have derived a factor for mass transfer
(jd) which they have expressed as :
.
= . ( )

Where :
CBm : The logarithmic mean of the conc. of B

CT : total concentration

: Schmidt number

Lecturer . Shymaa Ali Hameed 2013-2014


Unit OperatiOn
** Gillilond and Sherwood found the following relation for
mass transfer (hD) inside vertical tubes .
. .
. = 0.023 =

For a plane surface ≈


. .
. = .

So that :
.
ℎ = . ( )

** The general eq. for mass transfer in a wetted – wall column


is :
. . .
. = 0.023 . (*)

Where :
D v : vapor phase diffusivity

PB m : logarithmic mean of the press. of =

PT : total press.
d : column diameter
eq.(*) is frequently re arranged as :
. .
. = 0.023 = (a)

= . or ℎ =

KG : gas film transfer coefficient

Lecturer . Shymaa Ali Hameed 2013-2014


Unit OperatiOn
** For most gases , we may write

≈ ( the friction factor )

Gas phase

. ∗ . .
= ⁄ ⁄ +

Liquid phase
. ∗
= ⁄ ⁄

Ex : Calculate the rise in temperature of water which is passed at


3.5 m/s through a smooth 25 mm diameter pipe , 6 m long. The
water enters at 300 K and the tube wall may be assumed
constant at 330 K . The following methods may be used :
(a) the simple Reynolds analogy .
(b) the Taylor-Prandtl modification .
(c) the universal velocity profile .
. .
(d) = 0.023 .

Sol.
. ∗ . ∗
= = 1.25 ∗ 10
. ∗

. ∗ ∗ . ∗
= = 4.50
.

Lecturer . Shymaa Ali Hameed 2013-2014


Unit OperatiOn
(a) Reynolds analogy

= 0.032 Re-0.25

h = [4.18 x 1000 x 1000 x 3.5 x 0.032(1.25 x 105) -0.25 ]


= 24,902 W/m2 K or 24.9 kW/m2 K
Heat transferred per unit time in length dL of pipe =
ℎ 0.025 (330 − ) kW, where is the temperature at a
distance L m from the inlet .
Rate of increase of heat content of fluid

= ( (0.025) ∗ 3.5 ∗ 1000 ∗ 4.18) kw

The outlet temperature ′ is then given by:

∫ = 0.0109ℎ ∫
( )

Where : h is in kW/m2 K .
.
log (330 − ) = log 30 − = 1.477 − 0.0283ℎ
.

In this case :
h = 24.9 kW/m2 K
log (330 − ) = (1.477 − 0.705) = 0.772
and : = 324.1 k

(b) Taylor-Prattdtl equation

= 0.032 Re [1 + 2.0 Re (Pr − 1)]

. 2
ℎ=( = 9.53 kW/m k
. ∗ . ⁄ . )

Lecturer . Shymaa Ali Hameed 2013-2014


Unit OperatiOn
and :
log (330 − ) = 1.477 − (0.0283 ∗ 9.53) = 1.207
= 313.9 k

(c) Universal velocity profile equation

= 0.032 Re {1 + 0.82 Re [(Pr − 1) + ln(0.83Pr +


0.17)]}
.
=
( . / . )( . . )

= 12.98 kW/m2 K
log (330 − ) = 1.477 − (0.0283 ∗ 12.98) = 1.110
and : = 317.1 k

. .
d) = 0.023
. ∗ . . .
ℎ= (1.25 ∗ 10 ) (4.5)
.

= 0.596 x 1.195 x 104 x 1.64


= 1.168 x 104 W/m2 K or 11.68 kW/m2 K
and :
log (330 − ) = 1.477 − (0.0283 ∗ 11.68) = 1.147
= 316.0 k

Lecturer . Shymaa Ali Hameed 2013-2014


Unit OperatiOn

Lecturer . Shymaa Ali Hameed 2013-2014

You might also like