05C-Chapter 5, Sec 5.9 Black
05C-Chapter 5, Sec 5.9 Black
05C-Chapter 5, Sec 5.9 Black
Finite-Difference Equations
and
Solutions
Chapter 5
Section 5.9
Finite-Difference Method
Procedure:
Represent the physical system by a nodal network, with an m, n notation used
to designate the location of discrete points in the network, and discretize the
problem in time by designating a time increment t and expressing the time
as t = pt, where p assumes integer values, (p = 0, 1, 2,).
where, according to convention, all heat flow is assumed to be into the region.
Tmp,n1 Fo Tmp1, n Tmp1, n Tmp, n 1 Tmp, n 1 1 4Fo Tmp, n (5.71)
t
Fo finite-difference form of Fourier number
x 2
Marching Solution
Transient temperature distribution is determined by a marching solution,
beginning with known initial conditions.
p t T1 T2 T3.. TN
Known
0 0 T1,i T2,i T3,i. TN,i
1 t -- -- -- --
2 2t -- -- -- --
3 3t -- -- -- --
. .
. .
. .
. .
. .
. .
Steady-state -- -- -- -- . --
Problem: Finite-Difference Equation
KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, T sur,
suddenly passes a current I; rod is in vacuum enclosure and has prescribed electrical
resistivity, e, and other thermophysical properties.
FIND: Transient, finite-difference equation for node m.
SCHEMATIC:
Problem: Finite-Difference Equation
4
p p p p
Tm-1 Tm Tm+1 Tm p 4 4 2 e x p+1
Tm Tm
p
kAc kAc Dx Tm Tsur I cAc x .
x x A c t
p+1
Dividing each term by cAc x/t and solving for Tm ,
p+1
Tm
k
c x 2 t p
Tm-1 p
Tm+1
2
k
t
c x 2
p
1 Tm
Ac c
Tm Tsur
P t p 4 4 I2 e t
.
Ac2 c
Problem: Finite-Difference Equation
2
or, with Fo = t/x ,
p 4
I2 e x 2
2
p Px
Tm Fo Tm-1 Tm+1 1 2 Fo Tm
p+1 p p 4
Fo Tm Tsur Fo.
kAc kAc2
p
Basing the stability criterion on the coefficient of the Tm term, it would follow that
Fo .
p 4
However, stability is also affected by the nonlinear term, Tm , and smaller values of Fo may be
needed to insure its existence.
Problem: Cold Plate
Features:
Cold plate is at a uniform temperature,
Ti=15C, when a uniform heat flux
of qo 10 5 W/m 2 is applied to its base
due to activation of chips.
During the transient process, heat
transfer into the cold plate q in increases
its thermal energy while providing for
heat transfer by convection to the
water qconv . Steady state is reached
when q conv q in.
Problem: Cold Plate (cont.)
ANALYSIS:
Nodes 1 and 5:
2t 2t p+1 2t p+1 2t p+1
1 T1 T2 T6 T1p
x 2 y2 x 2 y2
2t 2t p+1 2t p+1 2t p+1
1 T5 T4 T10 T5p
x 2 y2 x 2 y2
Nodes 2, 3, 4:
2t 2t p+1 t p+1 t p+1 2t p+1 p
1 Tm,n Tm-1,n Tm+1,n Tm,n-1 Tm,n
x 2 y2 x 2 x 2 y2
Node 11:
2t 2t 2ht p+1 t p+1 t p+1 t p+1 2ht p
1 2 2 T11 2 T8 2 2 T12 2 T16 T +T11
x y kx y x y kx
2t 2t p+1 t p+1 p+1 t p+1 p+1 p
1 Tm,n Tm,n+1 Tm,n-1 Tm-1,n Tm+1,n Tm,n
x 2 y2 y2 x 2
2t 2t p+1 t p+1 p+1 2t p+1 p
1 Tm,n Tm,n+1 Tm,n-1 Tm-1,n Tm,n
x 2 y2 y2 x 2
Node 19:
2t 2t p+1 t p+1 p+1 2t p+1 p
1 T19 T14 T24 T20 T19
x 2 y2 y2 x 2
2t 2t p+1 2t p+1 t p+1 p+1 2q t p+1
1 Tm,n Tm,n+1 Tm-1,n Tm+1,n o +Tm,n
x 2 2
y y 2 x 2 k y
q conv h x/2 T6 T x T7 T x y T8 T / 2 y T11 T x
y T16 T / 2 x T15 T x/2 T14 T q out.