ChE 126
SEPARATION PROCESSES
HEAT TRANSFER
Fourier Equation
𝒅𝑻
𝒒𝒙 = −𝑲𝑨 𝒅𝒙
(1-dimensional)
Where
qx= heat transfer rate (W or J/s)
K = thermal conductivity of material (W/m-°C or W/m-K)
A = area of a surface (𝑚2 )
°𝐶
dT/dx = temperature gradient ( 𝑚 or K/m)
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
𝑄 ∝ ∆𝑇 = (𝑇1 − 𝑇2 )
𝑇1
𝑇1
𝑄∝𝐴
1
∆𝑥
𝑄 ∝
𝑇2
𝑇2 < 𝑇1
𝑄∝(
𝑇1 −𝑇2
)𝐴
∆𝑥
Heat = energy of the molecules
𝒒𝒙 ⬚
0
X=0
∆𝑥
X=l
∆𝑇
∆𝑥
𝑄 = 𝐾 ( ) 𝐴; 𝑖𝑓 ∆𝑥 → 0
∆𝑇
∆𝑥
𝑄 = lim 𝐾 ( ) 𝐴
∆𝑥→0
𝑑𝑇
𝑑𝑥
𝑄 = −𝐾 ( ) 𝐴
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒,
𝑥 (𝑚)
𝑸
𝑨
𝒅𝑻
= −𝑲 ( )
𝒅𝒙
Conduction
-
Whenever heat transfer occurs by contact (requires contact)
environment (25 °C)
100°𝐶
air
Convection
-
Requires movement
Whenever heat (E) is transferred by the movement of a fluid (liquid/gas)
100°𝐶
air
Hot air rises
Cold air sinks
Example:
100 °C
Radiation
-
Typically, heat is transferred by infrared radiation (no contact at all)
100°𝐶
sun – emits infrared rays
- comes without anything that comes in between
heat
A
ΔT
𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 ∝ 𝑎𝑟𝑒𝑎
𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑖𝑛 + 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑜𝑢𝑡 + 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛
∆𝑄 𝐾𝐴∆𝑇
=
∆𝑡
𝑙
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 =
•
•
𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
A drivingforce is needed to overcome a resistance in order to transport a property
Similar to Ohm’s Law,
𝑉
𝐼=
𝑅
P (energy transferred per second)
𝑃=
𝐾↑ 𝑃↑
∆𝑡 ↑ 𝑃 ↑
𝐴↑ 𝑃↑
𝑙↑ 𝑃↓
𝐸 𝐽
( );
𝑡 𝑠
1𝑊 =1
𝐽
𝑠
What is the rate of heat flow through a glass window that is 2m x 3m and 14cm thick if the outside
temperature is 22°𝐶 and the inside temperature is 25°𝐶? 𝐾 = 0.84
𝐽
𝑠−𝑚−°𝐶
∆𝑄
∆𝑡
heat
=𝑃=
𝑃 = 0.84
25°𝐶
3m
𝐾𝐴∆𝑇
𝑙
𝐽
25°𝐶−22°𝐶
(
)6
𝑠−𝑚−°𝐶
0.014 𝑚
𝑚2
𝐽
𝑠
𝑃 = 1,080 (𝑊)
22°𝐶
If 1,080 𝐽 of heat E is transferred through the window every second, how much E is transferred in a
month?
𝐸
𝑡
𝑃 = ; 𝐸 = 𝑃𝑡
𝐽 3,600 𝑠
E = 1,080 (
𝑠 1 ℎ𝑜𝑢𝑟
)(
24 ℎ𝑜𝑢𝑟𝑠
1 𝑑𝑎𝑦
)(
30 𝑑𝑎𝑦𝑠
1 𝑚𝑜𝑛𝑡ℎ
) (1𝑚𝑜𝑛𝑡ℎ)
𝐸 = 2.799 × 109 𝐽
𝑅=
𝑙
𝐾
where
R describes the thermal resistance of an insulator
l = thickness of material (distance between hot and cold materials)
l
𝐾↑
𝑙↑
𝑅↓
𝑅↓
•
* K is an intrinsic value of materials
* R is not intrinsic
inc thickness of matl, inc insulation
Example:
Silver (excellent conductor of heat)
*metals conduct heat due to free-flowing E found in metals
𝐽
)
𝑠−𝑚−°𝐶
Material
K values (
Silver
Glass
Fiberglass
low R
420
0.84
0.048
high R
Conduction heat transfer through a Composite Solid/Slab
𝑇1 > 𝑇2 > 𝑇3
𝑇1
𝑇2
𝑇3
𝑏1
𝑏2
𝐾1
𝐾2
Assumptions:
1. One directional heat transfer
2. Constant thermal conductivity (𝐾 ≠ 𝑓(𝑡𝑖𝑚𝑒)
3. Steady-state heat transfer
→ 𝑡ℎ𝑒𝑟𝑚𝑜𝑝ℎ𝑦𝑠𝑖𝑐𝑎𝑙 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
Ohm’s Law
𝑉 = 𝐼𝑅
∆𝑇
∆𝑻∆T
𝑅𝑚
𝑉
=𝐼
𝑅
𝑄
T1
RTh1
𝑅𝑚1 =
T2
𝑏1
𝑘1 𝐴
𝑸=
VT1
𝑅𝑚2 =
R1
Gas
Molecular Collision
Molecular Diffusion
Liquid
-
Molecular Collision
Molecular Diffusion
Molecules do not move as freely as in gases
Solid
-
Lattice vibration
Flow of free 𝑒̅
T3
𝑏2
𝑘2 𝐴
𝑻𝟏 − 𝑻𝟑
𝒃𝟏
𝒃
+ 𝟐
𝒌𝟏 𝑨 𝒌𝟐 𝑨
Conduction Heat Transfer (H.T.)
-
RTh2
R2
V2
Example.
The inner surface of a plane brick wall is at 60°𝐶 and the outer surface is at 35°𝐶. Calculate the rate of
heat transfer per 𝑚2 of the surface area of the wall, which is 220 m thick. 𝐾𝑏𝑟𝑖𝑐𝑘 = 0.51
𝐾 = 0.51
𝑡1 = 60°𝐶
𝑊
𝑚 − °𝐶
𝑡2 = 35°𝐶
𝐾(𝑡1 −𝑡2)
𝐿
𝑞=
𝑄
𝐴
𝑞=
0.51(60−35)
0.22
=
𝑊
𝑚−°𝐶
𝑞 = 57.95
𝑊
𝑚2
L=220mm=0.22m
Example.
An exterior wall of a house may be approximated by 0.1 m layer of common brick (𝐾 = 0.7
followed by a 0.04 m layer of gypsum plaster (𝐾 = 0.48
wool insulation (𝐾 = 0.065
𝑊
𝑚−°𝐶
𝑊
𝑚−°𝐶
) should be added to reduce the heat loss or (gain) through the wall by
Gypsum
Wool ins
Q
𝐾𝐴 = 0.7
𝑊
𝑚 − °𝐶
𝐿𝑎 =0.1 m
𝐾𝑏 = 0.48
𝑄 = 0.20𝑄𝐴−𝐵 = 20% 𝑄𝐴𝐵
Case 1:
𝑊
𝑚 − °𝐶
𝐿𝑏 =0.04 m
𝑄𝐴−𝐵 = 𝑥
)
). What thickness of loosely packed rock
80%?
Brick
𝑊
𝑚−°𝐶
𝐾𝑐 = 0.065
𝑊
𝑚 − °𝐶
𝐿𝑐
𝑄1 =
=
𝐾𝐴(𝑇1 −𝑇2 )
𝐿
𝐴(∆𝑇)
𝐿𝐴 𝐿𝐵
+
𝐾𝐴 𝐾𝐵
=
𝐴(∆𝑇)
𝐿
𝐾
𝐴(∆𝑇)
=
0.1 0.04
+
0.7 0.48
=
𝐴(∆𝑇)
0.2262
(1)
Case 2:
𝑄2 =
𝐴∆𝑇
𝐿𝐴 𝐿 𝐵 𝐿𝐶
+ +
𝐾𝐴 𝐾𝐵 𝐾𝐶
=
𝐴∆𝑇
0.1 0.04
𝑥
+
+
0.7 0.48 0.065
=
𝐴∆𝑇
(2)
𝑥
0.2262+0.065
𝑄2 = (1 − 0.8)𝑄1 = 0.2𝑄1
(3)
Substituting (1) and (2) in (3)
𝐴(∆𝑇)
𝑥
0.2262+0.065
=
1
0.2262+15.3846𝑥
0.2𝐴(∆𝑇)
0.2262
= 0.8842
𝑥 = 0.0428 𝑚 = 42.8 𝑚𝑚
Example:
A square plate heater (15cm x 15cm) is inserted between two slabs. Slab A is 2 cm thick (K=50 W/m-°C)
and slab B is 1 cm thick (K=0.2 W/m-°C). The outside heat transfer coefficients on side A and B are 200
W/m-°C and 50 W/m-°C respectively. 𝑇𝑠𝑢𝑟𝑟 = 25°𝐶. If the rating of heater is 1 kW, find:
a. maximum T in the system
b. outer surface T of 2 slabs. Draw an equivalent electrical circuit
Given:
A = 0.15 m x 0.15 m = 0.0225 𝑚2
Rating = 1 kW = 1, 000 W
1 kW
ℎ1 = 200
ℎ2 = 50
𝑡max
𝑡1
𝑇𝑎 = 25°𝐶
A
𝑡2
B
𝑇𝑎 = 25°𝐶
K= 50
𝐿𝐴 = 0.02 𝑚
K=0.2 𝑊/𝑚 − °𝐶
𝐿𝐵 = 0.01 𝑚
𝑡𝑎
𝑡𝑎
1
ℎ1 𝐴
a. 𝑄 =
𝑡max − 𝑡𝑎
𝐿𝐴
1
+
𝐾𝐴 𝐴 ℎ1 𝐴
+
𝑡max − 𝑡𝑎
𝐿𝐵
1
+
𝐾𝐵 𝐴 ℎ2𝐴
𝑡1
𝑡𝑎
𝐿𝐵
𝐾𝐵 𝐴
𝐿𝐴
𝐾𝐴 𝐴
=
1
ℎ2 𝐴
𝑡max − 𝑡𝑎
𝑡
𝐿
𝐿
1
+ 𝐴 + 𝐵 +
ℎ1 𝐴 𝐾𝐴𝐴 𝐾𝐵 𝐴 ℎ2 𝐴
1
𝑡
max − 𝑎
= 𝑅1+𝑅2+𝑅3+𝑅4
𝑡2
𝑡𝑤
A
B
𝐾𝐴
𝐾𝐵
𝐿𝐴
𝐿𝐵
ℎ1
ℎ2
Q = 𝐴(𝑡𝑚𝑎𝑥
− 𝑡𝑎 ) [ 𝐿𝐴
1
1
𝐾𝐴 ℎ1
+
+ 𝐿𝐵
1
1
𝐾𝐵 ℎ2
+
]
1
+ 0.011 1 ]
1
+
+
0.2 50
50 200
1000 = 0.0225 (𝑡𝑚𝑎𝑥 − 25) [0.02
𝑡𝑚𝑎𝑥 = 247.8 °𝐶
b. 𝑄𝐴 =
𝐾𝐴 𝐴(𝑡𝑚𝑎𝑥−𝑡1)
ℎ1 𝐴(𝑡1−𝑡𝑎)
𝐾𝐴 𝐴(𝑡𝑚𝑎𝑥 − 𝑡1 )
= ℎ1 𝐴(𝑡1 − 𝑡𝑎 )
𝐿𝐴
50(247.8 − 𝑡1 )
= 200(𝑡1 − 25)
0.02
𝑡1 = 231.3°𝐶
𝑄𝐵 =
𝐾𝐵 𝐵(𝑡𝑚𝑎𝑥−𝑡2)
𝐿𝐵
𝐾𝐵 𝐵(𝑡𝑚𝑎𝑥−𝑡2)
𝐿𝐵
0.2(247.8−𝑡2)
0.01
= ℎ2 𝐴(𝑡2 − 𝑡𝑎 )
= ℎ2 𝐴(𝑡2 − 𝑡𝑎 )
= 50(𝑡2 − 25)
𝑡2 = 88.6°𝐶
Thermal Conductivity (K)
Gases – Low K
Liquids – intermediate
Solid metals – High K
For liquids:
K = a+bT
Where: a and b – empirical constants
K for L independent of P
Convective Heat-Transfer Coefficient
-rate of heat transfer from the S → fluid (V.V.-vice versa)
𝒒 = 𝒉𝑨(𝑻𝒘 − 𝑻𝒇 )
Where:
𝐽
𝑞= heat-transfer coefficient rate (𝑊 𝑜𝑟 )
𝑠
A = area (𝑚2 )
𝑇𝑤 = temperature of solid surface (K)
𝑇𝑓 = average of bulk T of the fluid flowing past (K)
h = heat transfer coefficient (
𝑊
𝑚2 −𝐾
function of : system geometry
: fluid properties
𝑜𝑟
𝐵𝑇𝑈
)
ℎ−𝑓𝑡 2 −°𝐹
: flow velocity
: temperature difference
(empirical correlations are available)
-
known as film coefficient – when a fluid flows past a surface there is a thin almost stationary
layer of film of fluid adjacent to the wall which presents most of the resistance to heat transfer.
See Table 4.1-2 for h values
1 BTU/hr-ft2-oF = 5.6783 W/m2-K