A heat transfer engineer is working on the better understanding of a chimney with a square-shaped cross-section.
engineering
Description
A heat transfer engineer is working on the better understanding of a chimney
with a square-shaped cross-section.
a) Fundamentals.
We consider a steady heat conduction problem in 2D. These can be solved analytically,
numerically or through a conductance shape factor look-up table.
The latter are particularly popular while easy to use. Take ‘Shape factor #5’
from table 3.2 in ‘Mills’: conduction through concentric square cylinders.
We investigate how good it is…
·
Write a computer program to calculate this shape
factor; based on my numerical code as attached: Adapt the grid (to a square
and/or #cells), and impose the appropriate BC for outside and for the inner
square (by simply overwriting the intended inner block of cells after each
Jacobi- step with the intended initial condition). A hint how to do that is
already added.
Apply ratios a/b = 2, 3/2, 4/3, 5/4 and 8/7. Take care
for yourself that the grid is correctly fitting the values, and also for each
be critical on the number of grid points to use.
·
Compare your numerical result with the outcome of Shape
factor #5. (directly)
·
Construct the same geometry from ‘flat plates’ and
‘edges’ (Shape factor #10).
Compare the three methods tabular and in a graph.
Conclusion?
b) Application. Assuming the models in a) are appropriate, design
a reasonable chimney for a 1 GWe coal-fired power plant of 40%
efficiency.
The chimney will be 100 m high, and of (constant over height) inner
dimension 12 m.
It will also need to withstand windforce 12.
- What is a realistic wall thickness regarding wind loading?
Brick material has been used as a building material.
Hint: determine compressive and tensional stresses at the foot of the pipe.
- What will be the exhaust temperature of the Chimney, assuming we have an
inlet temperature to the Chimney of 100oC, at a throughput of 100 kg
of exhaust gas/s?
Will condensation occur?
A small
report ~10 pages, containing some relevant pictures and the code for a/b
= 1.5.
Good luck!
