KE

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Name
KE
Math 3150 Midterm 1
19 Feb, S2014
Problem 1. (Heat Conduction in a Rod, Ends at Different Temperatures)
Throughout, k is the mean thermal diffusivity, usually written as Fourier’s constant K
0 divided by specihe heat c and
moss density per unit voliune p.
j (a) [40%] Consider the heat coadiictioii problem in a laterally iiisulated rod of ieugth 1 with 000 011(1 at zero Celsius and
/—the other end at 1 Celsius. The initial temperature along the rod is given by function [(3;)
a;.
Ut
u(0.t)
u(1,t)
—
=
=
=
ku,.
.
2
,
0 <a; < 1,
0,
1,
.r,
0<x<1.
t> 0,
t>0,
t >0,
The answer u(x, t) to this problem is exactly the steady—state tenipero.ture. Find tile answer u(x, t) and display a
complete answer check.
(b) [60%] Consider tile heat conduction problem in a laterally insulated rod of length 1 with one end at zero Celsius
011(1 the other end at one Celsius. The initial temperature along the rod is given by function [(c) = 1 + z.
=
u(0,t)
=
u(1,t)
=
u(x,0)
kn.
0,
1,
1-!-x,
0 <x < 1,
t > 0,
t >0,
t >0,
0 <z <1.
Solve the rod problem for u(x. t). It is necessary to derive the product solutions. Provide Fourier coefficient formulas.
Evaluate all Fourier coefficients. Then display the final answer u(x, t).
o)
/
6,
t
C)
‘
a
I-i
L
o
—(3
()
-
<U
+
—0
—
0
)U
o
_
CI
c-
(
( a)
0
-fr
I-,
1’-’
C,
-
rL
r
-,
‘r
-1-
E-•
,
(
0
x
3
t
3
ii
0
F
::: w
ioblem 2. (Total Thermal Energy in a Rod)
Au
expussom ior
5’t ioiiah itito A is
the
total tilernial energy contained in a rod
tiuit—d’peiidtiit
J
,.
0 to x
=
L, with mutorni cross—
2
(‘
to obtain
‘ar(L, t)
At”
/
cpu(xJ)Adx.
=
satisf’ing the
is flit specific heat, p is the mOSS density per unit volume and u(x, t) is the rod temperature,
kit,... Assmniii ( 011(1 f) are coiistaiits.
11(111 iquitl 1011 l(
supply the saute total thermal
i’(x t) are two temperature distributions for the sonic rod which
and
1)
o(,r
Suppose
(lorgy br all I.
(a) j30%] Explain why J u,(x. t) dx
f’ v, t) dx.
..
1
1 = kv
usilig U)
k7trr and v
(ii) (i0%1 Differentiate on t across the equation of (a). Simplify the resulting ecluation
Xy’iihol
A
1
i:
—
‘ur(Df)
Lr(L,t)
=
Ur(0,t).
Low.
(c) [20%] Explain the meanilig of the equation in (b) in terms of heat flux and Fourierh
IdfrAl
1
T
becriuice
o
i
tJL
i
i;i1
ç
r?7y
b)
(,
A
/I
[L
jo
Jo
f
‘
/]
v&/
OYV1e)
an A
(:( dLrf
/e
i)Imaf
k
/
1
_;fA
(X
d)
1
CfM
)
1i&ra1
/1v()
4-
K
vW),)
/V()
v(i
L
,
L
/
/
\,
‘i
(, )
‘
_/c
-
4> (ô,
)
-
(±
,
Loi:
Thç
Ø’i
4-
bev’c
.i e
-
IlA
c Fc
17
I
]1
—(_,r,.,-l.,/)
L’...-c. ‘I.
Problem 2. (Total
rrheI1nal
Energy in a Rod)
total tlwriiial energy coiit,aiued in a rod c
An uxplessioli for the ti111c—dLpe11deHt
s’
1
n i titI
I
lou
01(0
=
0 to
.: =
L, with
uniform cross—
A is
J
hieiit
( is
f l1(
5l)(Clh( 1iett
(‘(flliltiofl IL(
t)
for all 1.
Snppose o(r,
cimergy
h0,pr.
[) IS tlW 1111155 (lensity
e(x. t) are
011(1
per unit volitnie and u(x, t) is the
two temperature distributions for
the same
(a) [30%] Explain wily J u,(:v. t) dx = j v(x, t) (Lc.
/ to(h) [60%] Differentiate on across the equation of (a). Simplify the
obt,aii
(0t)
1
r(L.t) —u
[20%] Explain the
(c)
(
mueanhllg
=
v(L,t)
rod winch
lr(0,t).
of the equation in (b) in terms of heat flux and Foi.irier’s Low.
i
c’.i%”t
1
Jc
L
11 A i
A
J)
.‘i
jf.j
(J
•lj
I
.
(i
J
/
\I/
I;,
4,.
I
Jo
ii
bL
/!
/4
supply the same total tliernial
resulting equatioli using
sJ)
j
rod teniperature, satisfying the
Asslilne ( 011(1 [1 1U0 coiistaiits.
i
-104+
,,
—J
\/ (I Li
Tkc*cr
°-
V Ot)
L,H
9
u,
ku audi m’
= kerr
;:;
÷
•
...
)
JiL
L4.
—
-Y
___\
--•
—
—
.c.
I
I
7-..
-.
)L
t—
IC
__1
Problem 3.
9c
(Steady-State Heat Conduction on a Rectangular Plate)
Consider Lapliwes equation
on the rectangle (I < ,t < 1, (1 < p < 1 subject to the boundary conditions
+
()y—
d,r—
iLr(.:, p) = 0 for :r = (I tuid ,r = 1, n(.r, y)
0 for y = 1, u.(,i, j) = [(a) for p = 0.
Ala) 70%) Find the product solutions a(.r, y) = X (a)Y(). Include a cheek that each product solution satisfies
required threc zero boundary conditions.
f\_
(b) (30%] Let
1\ u(m
f(:r)
be the smn of the first three eigciifiiiictioos which is the sum
of time first three X—answers. Find
y).
1
r
()
uj 0
L
0,
(,c(:
0
(
f( T
i
y:O
0
ch(: O
)1
w
Y
‘
/
4
1:-
X’ :Q
+
1
X’(o: X’(iO
(fl: 0
CCX
L C
1
(o (:Q
(:
6
.z
(O)
(
y
((: ce
( cv5
Y(
I
Corr!
/iv7i
7
o
3
L
OQr2/1_i
/
pi)
o
,
1
,nc1
O22
401
7
i:ec
(p.
,3ç)J
2
/
—
}
)
‘,)21i]7
(/2 ci
...(1
K
(i),I
2’)
P
I_JJ
;
9;’
joSJ 11
x2
o1
4l.,
77
(G()
LL
,)
Q -;
LF
A
<fL.
c7zj
)
(J,,.IJ/
(‘)
JI
iIL’);
XJ5’
:()(
/
.-2.
.?‘
‘)/.‘.1
‘
ç J
9
fr
’‘—
1
,Q
,k
:
-
(o
CQ’A)ul,
<2
(I
)1
ç)/
C2
0
1
(fl.r)7i
d
T
1
1t
J)
3’J ‘i
T1i—3(
IIo!4llIo
.10 TTTI
I
11T..)TiliJTl)’iT) )).l1 4H.ITJ )II} ) UhI1
‘T!’I’
q (i:)]
1
4J’
2-
(1)
1 0.T) •)J.II
1 1))1Il1))i
X110tfll)T10.) A.IUplIfl()1
4311p0.Id
(1i)(;)
i•i
Ii
HTI0i1)ITO.)
Ln])un()q iM14 0J 4)Jql0
I
(Ii
ioj (.i)f =
> 1 >
>
(wia
J1TL
.r)’ii
x
=
Ii
‘
.oj
(Ii ‘)
o
4 TIC)
1
J1
> 0
‘
1
411p01c
(Ii ‘.r) 1 =
_1i()
._.r()
-—----
Tfl3DOH U 110 11OT.)flpt10D
piiu
n
•:
P”!d [%OLI ()
o = (Ii .r)’’n
•oj
II03U1l1).) ))UT(IU’J .!J1)pI10)
s-pUos)
WOH
•g WOJJ
gc
//
Prob’em 3. (Steady-State Heat
Conduction
on a Rectangular Plate)
Conshkr Laplace’s equation
on flit rictongle () < i < 1, (1 < p < 1 sub ject to the boundary conditions
+
dc—
j—
1
d
ii..r(,i:, i) = 0 for :i
(I and .r = 1, u(, y)
(I for p = 1, u(.i’, y) = fCc) for p 0.
(a) [70%] Find the prodmt solutions u(,r, i) = X (.c)Y (y). Inelwhlk that each product solution satisfies the
required three zero boundary conditions.
(h) [30%] Let ICc) be lie suni of tlit first flirci’ cigeufunctiuns, winch is the suui of the first three X—answers. Find
u(.u, ti).
NI
-
y
-
e -e
1
C
D
-
çr c
c
1O
C
._
-
-.
-
C
NO
o
k
/‘(r’\L\)Hi
I
/
t
c_)
-)
me.’r-
,
—o
•.-l-( 2—
“
c(—’)
C-
F-o--
—
1-
—
CoLtQ.
0
i,-
,0
cCC
-‘
A0
11
tt()9’-
Co--
4—
.
5c co
1.
O1flt
l
0
L
0
0
-
C
c
q
\E-
7?
N
/
-
3
‘S
I
No
(IC3
4-
II
‘I
-,
t
I’
0
I
4
‘-J’
‘N
jc
I
c
\
Problem 4. (Steady-State Heat Conduction on a Disk)
(cushier till’ stead v—state heat conduction problem in polar coordinates
(r,8) + v.oo(r,8)
1
v
I
= 0,
0 <
i
<2,
0 <8 <2,
= ,f(8), 0<8< 2.
1(i0XJ Find the product solutions n = R(r)e(8), then identify tin
u(2,8)
orthogonal set and the interval, Stop at this stage:
(a)
omit superposition, omit the_series solution and do not develop formulas fur the Fourier coefficients.
8 < 2. Hint: Tue Poisson integral theorem
(b) [40Yi Calculate ‘u(O, 6) when f(8) = (1 on 0 < 8 < , 1(8) = 50 on
Theorem.
Value
Mcmi
the
mid
,/
a)
7/
(r, )
4 (r,&)
L
frU
ç
)
/
L
(e.);
C)
-f (p)- (:
M;
v’f
cf!ec.’
rO,
e
1
2
1
‘
(1
f/((
(5o)
4
‘-/9c 7 t
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