VM-LSDYNA-SOLVE-012

VM-LSDYNA-SOLVE-012
Steady State Thermal Analysis of 2D Axisymmetric Cylindrical Shell

Overview

Reference: Kreith, F. (1959). Principles of Heat Transfer (2nd ed.). International Textbook Co.
Analysis Type(s): Steady State Thermal 2D Axisymmetric
Element Type(s): Shell
Input Files:Link to Input Files Download Page

Test Case

A cylindrical shell has inner radius temperature Ta = 0°C. The length of the cylinder is 2 m from ra = 1 m to rb = 2 m. The length of the cylinder from rb = 2m to rd = 4 m is 3 m. The outer surface has a flux of -20 W/m2. Find the steady state temperature of the outer surface Td. Heat transfer conductance at the contact surface rb = 2 m and rc = 3 m is set as 1000 W/°C

Figure 44: Problem Sketch

Problem Sketch

Material PropertiesGeometric PropertiesLoadingThermal Contact
ka = kb = 1 W/(m°C)La = 2.0 mq" = -20W/m2 Ccontactb = 1000 W/°C
Lb = 3.0 m Ta = 0°CCcontactc = 1000 W/°C
ra = 1.0 m Td = ?
rb = 2.0 m
rd = 4.0 m

Analysis Assumptions and Modeling Notes

LS-DYNA Thermal Solver 1 is used. Three separate parts of equal material make up the model. Section shell with ELFORM 15 and thickness 0.001 m is used to create the 2D axisymmetric elements. *BOUNDARY_TEMPERATURE is used to set the boundary temperature on ra. A flux boundary condition is set on the outer surface, rd, using *BOUNDARY_FLUX. Heat transfer conductance is specified between parts using *CONTACT_2D_AUTOMATIC_SURFACE_TO_SURFACE_THERMAL.

Figure 45: Simplified Problem Schematic

Simplified Problem Schematic

At steady state, the rate of heat dissipation Q is

where

And the conduction resistances Rab, Rbc, and Rcd, and for cylindrical shell can be expressed as below

From the above equations, the analytical solution of Td is obtained as 141.644°C.

Figure 46: Fringe plot of model temperature, °C

Fringe plot of model temperature, °C

Results Comparison

ResultTargetLS-DYNAError (%)

Temperature Td

141.644°C

143.5° C

(average of 4 nodes on rd)

1.3%