this is the derivation of th how heat transfer equation is derived. the book from which this presentation is made is written by m wakil

1. • 𝑡𝑚 − 𝑡𝑠 =
𝑞′′′𝑅2
4𝑘𝑓
• 𝑞(𝑟) = 𝜋𝑟2
𝐿𝑞′′′
• 𝑞𝑠 = 𝜋𝑅2𝐿𝑞′′′
• 𝑞𝑠 = 4𝜋𝑘𝑓𝐿(𝑡𝑚 − 𝑡𝑠)
• Therefore: 𝑞𝑠 = 2𝑘𝑓𝐴𝑠
𝑡𝑚−𝑡𝑠
𝑅
1

2. Syed Muhammad Haris
Chapter 5
Lecture 2

3. Heat Flow out of Solid-Plate-Type Fuel
Element
Assumptions:
• Thin, Bare and Plate-type fuel element
• Thermal Conductivity 𝑘𝑓
• Cross-Sectional Area = Constant
• y and z dimension → infinity
• 𝜙 𝑎𝑛𝑑 𝑞′′′
= Constant
3

4. Heat Flow out of Solid-Plate-Type Fuel
Element (Contd.)
1 (a) First Method to Get Poisson Equation:
– General Heat Conduction Equation:
𝛻2𝑡 +
𝑞′′′
𝑘
=
1
𝛼
𝜕𝑡
𝜕𝜃
– Poisson Equation:
𝛻2𝑡 +
𝑞′′′
𝑘
= 0
– For one-dimension i.e. x-dimension
𝑑2
𝑡
𝑑𝑥2
+
𝑞′′′
𝑘
= 0
4

5. Heat Flow out of Solid-Plate-Type Fuel
Element (Contd.)
1 (b) Second Method to Get Poisson
Equation:
– Applying Heat Balance:
𝐻𝑒𝑎𝑡 𝐶𝑟𝑜𝑠𝑠𝑖𝑛𝑔
𝑃𝑙𝑎𝑛𝑒 𝑥 + Δ𝑥
=
𝐻𝑒𝑎𝑡 𝐶𝑟𝑜𝑠𝑠𝑖𝑛𝑔
𝑃𝑙𝑎𝑛𝑒 𝑥
+
𝐻𝑒𝑎𝑡 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑
𝑖𝑛 𝐿𝑎𝑦𝑒𝑟
𝑞𝑥+Δ𝑥 = 𝑞𝑥 + 𝑞′′′
𝐴Δ𝑥
𝑞′′′𝐴Δ𝑥 = −𝑘𝑓𝐴
𝑑2𝑡
𝑑𝑥2
Δ𝑥
𝑑2𝑡
𝑑𝑥2
+
𝑞′′′
𝑘
= 0
5

6. Heat Flow out of Solid-Plate-Type Fuel
Element (Contd.)
2. Integration to get
temperature
distribution:
𝑑2𝑡
𝑑𝑥2
= −
𝑞′′′
𝑘
=
𝑞′′′𝑅2
𝑘𝑓
𝑡 = −
𝑞′′′
2𝑘𝑓
𝑥2
+ 𝐶1𝑥 + 𝐶2
6
3. Boundary Conditions:
𝑑𝑡
𝑑𝑥
= 0 𝑎𝑡 𝑥 = 0
𝑡 = 𝑡𝑚 𝑎𝑡 𝑥 = 0
𝑑𝑡(𝑥)
𝑑𝑥
= −
𝑞′′′
𝑘𝑓
𝑥
𝑡(𝑥) = −
𝑞′′′
2𝑘𝑓
𝑥2 + 𝑡𝑚
Surface Temperature:
𝑡 = 𝑡𝑠 𝑎𝑡 𝑥 = 𝑠
𝑡𝑠 = 𝑡𝑚 −
𝑞′′′
2𝑘𝑓
𝑠2

7. Heat Flow out of Solid-Plate-Type Fuel
Element (Contd.)
4. Heat Conducted out
from a plane
Between x=0 and x:
𝑞𝑠 = 𝐿𝑞′′′
𝑟2
Total Heat Generated from
one side (x=s):
𝑞𝑠 = 𝑞′′′
𝐴𝑠
As: 𝑡𝑠 = 𝑡𝑚 −
𝑞′′′
2𝑘𝑓
𝑠2
⇒ 𝑞′′′ = 2𝑘𝑓
𝑡𝑚 − 𝑡𝑠
𝑠2
7
Total Heat Generated
from both sides (As =
2𝐴):
𝑞𝑠 = 2𝑘𝑓𝐴𝑠
𝑡𝑚 − 𝑡𝑠
𝑠

8. Heat Flow out of Solid-Plate-Type Fuel
Element With Clad
5. Effect of Cladding:
• Cladding thickness = c
• Thermal Conductivity = 𝑘𝑐
• Cladding act as extra resistance to
heat transfer
• No heat generated in cladding
• Heat leaving out of any surface in
the direction of heat transfer is
constant at steady state
• If 𝑘𝑐 is constant
𝑑𝑡
𝑑𝑥
is constant
𝑞𝑠 = 𝑞′′′𝐴𝑠 = 2𝑘𝑓𝐴
𝑡𝑚 − 𝑡𝑠
𝑠
= −𝑘𝑐𝐴
𝑑𝑡
𝑑𝑥
= 𝑘𝑐𝐴
𝑡𝑠 − 𝑡𝑐
𝑐
8

9. Heat Flow out of Solid-Plate-Type Fuel
Element With Clad (Contd.)
6. Temperature Differences:
𝑡𝑚 − 𝑡𝑠 =
𝑞𝑠𝑠
2𝑘𝑓𝐴
=
𝑞′′′
𝑠2
2𝑘𝑓
𝑡𝑠 − 𝑡𝑐 =
𝑞𝑠𝑐
𝑘𝑐𝐴
=
𝑞′′′
𝑠𝑐
𝑘𝑐
Temperature Difference:
𝑡𝑚 − 𝑡𝑐 =
𝑞𝑠
𝐴
𝑠
2𝑘𝑓
+
𝑐
𝑘𝑐
=
𝑞′′′𝑠2
2𝑘𝑓
+
𝑞′′′𝑠𝑐
𝑘𝑐
Total Heat Generated from one side:
𝑞𝑠 =
𝐴 𝑡𝑚 − 𝑡𝑐
𝑠
2𝑘𝑓
+
𝑐
𝑘𝑐
Total Heat Generated from both sides:
𝑞2𝑠 =
𝐴𝑠 𝑡𝑚 − 𝑡𝑐
𝑠
2𝑘𝑓
+
𝑐
𝑘𝑐
9

10. Heat Flow out of Solid-Plate-Type Fuel
Element With Clad and Coolant
7. Heat transfer form Fuel Element to
Coolant:
• Bulk Coolant Temperature = 𝑡𝑓
• Heat Transfer Coefficient= h
• No heat generated in cladding and
coolant
• Heat leaving out of any surface in
the direction of heat transfer is
constant at steady state
• If 𝑘𝑐 is constant
𝑑𝑡
𝑑𝑥
is constant
𝑞𝑠 = 𝑞′′′𝐴𝑠 = 2𝑘𝑓𝐴
𝑡𝑚 − 𝑡𝑠
𝑠
= −𝑘𝑐𝐴
𝑑𝑡
𝑑𝑥
= 𝑘𝑐𝐴
𝑡𝑠 − 𝑡𝑐
𝑐
= ℎ𝐴(𝑡𝑐 − 𝑡𝑓)
10

11. Heat Flow out of Solid-Plate-Type Fuel
Element With Clad (Contd.)
6. Temperature Differences:
𝑡𝑚 − 𝑡𝑠 =
𝑞𝑠𝑠
2𝑘𝑓𝐴
=
𝑞′′′
𝑠2
2𝑘𝑓
𝑡𝑠 − 𝑡𝑐 =
𝑞𝑠𝑐
𝑘𝑐𝐴
=
𝑞′′′
𝑠𝑐
𝑘𝑐
𝑡𝑐 − 𝑡𝑓 =
𝑞𝑠
ℎ𝐴
=
𝑞′′′
𝑠
ℎ
Temperature Difference:
𝑡𝑚 − 𝑡𝑐 =
𝑞𝑠
𝐴
𝑠
2𝑘𝑓
+
𝑐
𝑘𝑐
+
1
ℎ
=
𝑞′′′𝑠2
2𝑘𝑓
+𝑞′′′
𝑠
𝑐
𝑘𝑐
+
1
ℎ
Total Heat Generated from one side:
𝑞𝑠 =
𝑡𝑚 − 𝑡𝑐
𝑠
2𝑘𝑓𝐴
+
𝑐
𝑘𝑐𝐴
+
1
ℎ𝐴
11