# Brookline College Oklahoma City MAE Project: Engineering Answers 2021

Brookline College Oklahoma City MAE Project: Engineering Answers 2021

## Brookline College Oklahoma City MAE Project: Engineering Answers 2021

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Brookline College Oklahoma City MAE Project

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MAE 316 Main Project
Fall 2020 Written Part
April 8, 2021
This is a problem about forced-convective heat transfer, as we might have in an airplane wing.
Airplane wings are generally not rectangular in shape, but this one will be rectangular to keep
things relatively simple. One end of the wing is near the engine, which exposes it to a constant
temperature, Tref . Outside, the ambient temperature is T0 , which is less than Tref . Heat is removed
from the wing via forced convection.
The flux is Fourier’s law, with a correction in the y-direction due to the changing thickness of
the wing:
q = −κ(1 + y)∇T
(1)
The source term is Newton’s law of cooling:
f = −η(T − T0 )
(2)
where η is the convection heat transfer coefficient and T0 is the temperature of the ambient air.
Forced convection is described by a dimensionless quantity called the Nusselt number, or Nu.
The definition of the Nusselt number is a ratio of convective to conductive heat transfer in the fluid:
ηy
Nu =
(3)
κf
where η is the convective heat transfer coefficient, κf is the heat transfer coefficient in the fluid,
and y is the length along the sheet in the direction of the fluid flow. For the particular situation of
1
forced convection over a planar surface, the Nusselt number can be approximated as a function of
the Reynolds number Re, which characterizes the flow, and the Prandtl number Pr, which depends
on the molecular properties of the fluid:
Nu = 0.332Re
1/2
1/3
Pr

= 0.332
ρvy
µ
1/2
Pr1/3
(4)
where ρ is the fluid density, v is the velocity of the flow and µ is the dynamic viscosity. Given
these two expressions for Nu along with the other flow and fluid properties, one can arrive at an
expression for the convection heat loss coefficient η in terms of y.
Derive the governing equation for the problem, including boundary conditions. Use finite differences to discretize, and write stencils for an arbitrary node in the center of the domain along with
stencils for all boundaries. Assume that the boundaries are insulated except for the left boundary,
which is fixed at Tref .
2
Finite differences in two spatial dimensions shares a lot in common with the onedimensional version. Let’s fix ideas right from the beginning by considering the following
flux and source expressions:
q = −κ∇T
f = 0;
(1)
(2)
−∇ · q + f = 0
h
i
⇒ ∇ · −κ∇T = 0
(3)
⇒ ∇2 T = 0
(5)
The continuity equation then yields:
(4)
Now imagine that this governing equation (which takes a famous form called Laplace’s equation) is valid on the following domain: If we want to use finite differences to solve this system
Figure 1: the domain for Laplace’s equation – the origin of coordinates is at the bottom left;
the rightmost edge is a fixed temperature T1 and half the leftmost edge is a fixed temperature
T2 , whereas every other boundary is insulated
it might be easier to write it like this:
∂ 2T
∂ 2T
+
=0
∂x2
∂y 2
(6)
We have, of course, finite difference approximations for second derivatives, but we’ll need to
introduce another set of indices for the second dimension. In the x-direction we can still use
the index i, whereas in the y-direction we could go for j, such that a generic node in the
center of the discretized domain and all of its neighbors looks like this: So when we discretize
(6), we get
Ti+1,j − 2Ti,j + Ti−1,j Ti,j+1 − 2Ti,j + Ti,j−1
+
=0
(7)
∆x2
∆y 2
If we consider that we have a regular mesh, which means that ∆x = ∆y, we get the stencil
Ti+1,j + Ti−1,j − 4Ti,j + Ti,j+1 + Ti,j−1 = 0
1
(8)
Figure 2: a single node on the domain, with its neighboring nodes
Implementation is a little bit different for the case of a multi-dimensional domain. The
main reason for this is the way the vector of temperatures is constructed. In the onedimensional case, we could simply create a vector out of the unknown temperatures, with
the positions of temperatures in the vector connected with the positions of nodes in the
domain:

T1
 .. 
 . 

Ti−1 

T =  Ti 
(9)

T
 i+1 
 . 
 .. 
Tn
Now however we’re faced with the problem of forming a vector containing temperatures at
nodes that are arrayed in a two-dimensional mesh. How do we accomplish that?
The most reasonable method seems to be: we stack them into the vector one row at a
time:

T1,1
 … 

T 
 i,1 
 . 
 .. 

Tn,1 

 T1,2 
T=
(10)

 .. 
 . 

 Ti,2 

 .. 
 . 

Tn,2 
..
.
and so on, until we have put all rows in the domain, from the row corresponding to j = 1 to
the row corresponding to j = m, into the vector.
What does that mean for the matrix? Consider how the terms that appear in the stencil
2
(8) are arranged in the temperature vector:

..
.

Ti,j−1 
 . 
 . 
 . 

Ti−1,j 

 Ti,j 

Ti+1,j 
 . 
 .. 

T

 i,j+1 
..
.
(11)
How are the coefficients that appear in the stencil arranged in the matrix as a result?

… … …

· · · 1 · · · 1 −4 1 · · · 1 · · ·

· · · 1 · · · 1 −4 1 · · · 1 · · ·
(12)

· · · 1 · · · 1 −4 1 · · · 1 · · ·

..
..
..
..
..
.
.
.
.
.
In other words, instead of the tri-diagonal matrix we obtained in the 1-D problem (where
every entry is zero except for the main diagonal and the two diagonals directly above &
below it), we now have two extra diagonals located n entries away – on either side – from
the main diagonal on every row. This is kind of an issue for Gaussian elimination, which
is O(n) complex for the tri-diagonal case but not for higher dimensions: when the nonzero
diagonal on the lower triangle is eliminated, one of the entries that was previously zero
becomes nonzero. This is the motivation for using iterative solvers that take advantage of
the matrix sparsity.
3

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