2-D Conduction:Finite-Difference Methods

CH EN 3453 – Heat Transfer

Reminders…

• Homework #4 due Friday 4 pm

• Help session today at 4:30 pm in MEB 2325

• Exam #1 two weeks from today

• Homework available for pickup in ChE office

Graphical Method - Plotting Heat Flux

1. Consider lines of symmetry and choose sub-system if possible.

2. Symmetry lines adiabatic and count as heat flow lines.

3. Identify constant temperature lines at boundaries. Sketch isotherms between the boundaries.

4. Sketch heat flow lines perpendicular to isotherms, attempting to make each cell as square as possible.

Graphical Solution…

Conduction Shape Factor• Two- or three-dimensional heat transfer in

a medium bounded by two isothermal surfaces at T1 and T2 may be represented in terms of a conduction shape factor S

q = Sk(T1–T2)

• Corresponding two-dimensional conduction resistance:

Rcond,2D = (Sk)–1

Shape Factor S: q = Sk(T1–T2)

Shape Factors, Cont.

Example – Buried Hot PipeA pipeline, used for the transport of crude oil, is buried in the earth such that its centerline is a distance of 1.5 m below the surface. The pipe has an outer diameter of 0.5 m and is insulated with a layer of cellular glass 100 mm thick. What is the heat loss per unit length of pipe under conditions for which heated oil at 120°C flows through the pipe and the surface of the earth is at a temperature of 0°C?

SOLUTION – Buried Pipe

2-D Solid

Conduction between Nodes

Figure 1.5

Exterior Nodes

Example Problem 4.44Consider the square channel shown in the sketch operating under steady-state conditions. The inner surface is at 600 K while the outer surface is exposed to convection with a fluid at 300 K. A symmetrical element with a 2-dimensional grid is shown and temperatures for nodes 1,3,6, 8 and 9 are given.

(a) Derive finite-difference equations for nodes 2, 4 and 7 and determine the temperatures T2, T4 and T7.

(b) Calculate heat loss per unit length.

430 K 394 K

492 K

600 600

T∞ = 300 K

Problem 4.44 Figs