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{SECT 0 {PARA 256 "" 0 "" {TEXT 260 49 "Newton's Law for Cooling and C
ooling by Diffusion" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 96 " In Lectures 16 through 20, we model cooling of an \+
insulated rod with the diffusion equation" }}{PARA 0 "" 0 "" {TEXT -1
27 " " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"
tG" }{TEXT -1 5 " = k " }{XPPEDIT 18 0 "u[xx];" "6#&%\"uG6#%#xxG" }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 384 "The number k is determ
ined by the properties of the material and is identified as the rate o
f diffusion of heat within the material. In this worksheet, we take bo
undary conditions to be u(t, 0) =32 = u(t, 1). This model assumes that
the rod cools as heat redistributes itself because of the cold diffus
ing in from the ends. In Lecture 17, we saw that the solution for this
equation is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 25 " u(t, x) = 32 + " }{XPPEDIT 18 0 "sum(a[p]*sin(p*Pi*x)
*exp(-k*p^2*Pi^2*t),p = 1 .. infinity);" "6#-%$sumG6$*(&%\"aG6#%\"pG\"
\"\"-%$sinG6#*(F*F+%#PiGF+%\"xGF+F+-%$expG6#,$**%\"kGF+*$F*\"\"#F+F0F9
%\"tGF+!\"\"F+/F*;F+%)infinityG" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 1 " " }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 47 "Checking the sol
ution for the diffusion problem" }}{PARA 0 "" 0 "" {TEXT -1 60 " We
check the solution for the diffusion equation, again." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 61 "u:=(t,x)->32+sum(a[p]*sin(p*Pi*x)*exp(-k*p^2*Pi^2*t
),p=1..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "simplify(diff
(u(t,x),t)-k*diff(u(t,x),x,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 15 "u(t,0); u(t,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 92 " The coefficients are determined
by the initial value. This is a job for Fourier Series." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f(x)=u(0,x);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 18 "Radiation cool
ing." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 194 "
In that same Lecture 17, we introduced the notion of radiation co
oling along the sides of the rod. The equation we used to model this i
dea, in case the surrounding medium is 32 degrees, was" }}{PARA 0 ""
0 "" {TEXT -1 18 " " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG
6#%\"tG" }{TEXT -1 5 " = k " }{XPPEDIT 18 0 "u[xx];" "6#&%\"uG6#%#xxG
" }{TEXT -1 14 " - c (u - 32)." }}{PARA 0 "" 0 "" {TEXT -1 68 "Again, \+
we take the boundary conditions to be u(t, 0) = 32 = u(t,1). " }}
{PARA 0 "" 0 "" {TEXT -1 63 " Verify that you can derive the solut
ion for this equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 28 " u(t, x) = 32 + " }{XPPEDIT 18 0 "exp(-c
*t);" "6#-%$expG6#,$*&%\"cG\"\"\"%\"tGF)!\"\"" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "sum(a[p]*sin(p*Pi*x)*exp(-k*p^2*Pi^2*t),p = 1 .. infini
ty);" "6#-%$sumG6$*(&%\"aG6#%\"pG\"\"\"-%$sinG6#*(F*F+%#PiGF+%\"xGF+F+
-%$expG6#,$**%\"kGF+*$F*\"\"#F+F0F9%\"tGF+!\"\"F+/F*;F+%)infinityG" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 ""
0 "" {TEXT -1 71 "Checking the solution for the diffusion problem, wit
h radiation cooling" }}{PARA 0 "" 0 "" {TEXT -1 158 "We check the solu
tion for the diffusion equation with radiation cooling, again. Coeffic
ients are determined by the initial condition, which we call f(x) here
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "u:=(t,x)->32+exp(-c*t)*sum(a[p]*sin
(p*Pi*x)*exp(-k*p^2*Pi^2*t),p=1..5);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 62 "simplify(diff(u(t,x),t)-( k*diff(u(t,x),x,x)-c*(u(t,x
)-32) ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "u(t,0);u(t,1);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f(x)=u(0,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257
36 "The Main Question for this Worksheet" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 300 " Newton's Law of Cooling mode
ls how heat is lost from a point source to its environment. The model \+
assumes that the rate of change of the temperature of the body is prop
ortioned to the difference in the temperature of the object and the te
mperature of the surrounding medium. More specifically, \n" }}{PARA 0
"" 0 "" {TEXT -1 33 " T '(t) = " }{XPPEDIT 18
0 "alpha;" "6#%&alphaG" }{TEXT -1 9 " (T- 32)." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 442 "It is assumed in this mo
del that the temperature of the surrounding medium is constant. This c
ooling model takes the object as a point source. This model does not t
ake into account the shape of the object, or how heat diffuses within \+
the object. Usually, if the object is a pitcher of water, some stateme
nts are made about continuous stirring, so that the temperature of the
object is kept uniform. Else, where do you measure the temperature?"
}}{PARA 0 "" 0 "" {TEXT -1 91 " An alternate choice is to take the
total heat. For this rod of length 1, we would have" }}{PARA 0 "" 0 "
" {TEXT -1 29 " T(t) = " }{XPPEDIT 18 0 "int(u(t,
x),x = 0 .. 1);" "6#-%$intG6$-%\"uG6$%\"tG%\"xG/F*;\"\"!\"\"\"" }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 255 " To match the boundary conditions of the diffusion p
roblem above, we assume that the environment is 32. We assume Newton's
Law for cooling as described above and as an ordinary differential eq
uation for T(t), with T(0) equal to the initial total heat." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161 " I wondere
d to what extent the rod would act similar to a point source. That is,
assume the rod has an initial distribution, f(x). Take the total heat
to be " }}{PARA 0 "" 0 "" {TEXT -1 21 " T(0) = " }
{XPPEDIT 18 0 "int(f(x),x = 0 .. 1);" "6#-%$intG6$-%\"fG6#%\"xG/F);\"
\"!\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 59 "Take k to be
something, why not k = 0.1? Is there a number " }{XPPEDIT 18 0 "alpha
;" "6#%&alphaG" }{TEXT -1 19 " so that if T(t) = " }{XPPEDIT 18 0 "int
(u(t,x),x = 0 .. 1);" "6#-%$intG6$-%\"uG6$%\"tG%\"xG/F*;\"\"!\"\"\"" }
{TEXT -1 105 ", and u satisfies the diffusion model with radiation coo
ling, then T satisfies the above Newton's model? " }}{PARA 0 "" 0 ""
{TEXT -1 26 " I will try to find a " }{XPPEDIT 18 0 "alpha;" "6#%&
alphaG" }{TEXT -1 20 " to make this work. " }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 46 "Determining Newton's Proportio
nality Constant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 196 " Take the initial temperature to be f(x) = 32 + 272 \+
x (1 - x). Take k = 1/10. Take c = 0.02 = 1/50. The value for c was ch
osen by an experiment. Ask me about the experiment if you are curious.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 331 "We h
ave recalled how to solve the diffusion model. I do this, check that I
have a solution, and draw the graph without further ado. I check the \+
boundary conditions by computing them. I check the initial condition b
y comparing graphs: I compare the graph of f(x) and the graph of u(0,x
)+0.007 in order to be able to tell them apart." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 31 "Generating the so
lution u(t, x)" }}{PARA 0 "" 0 "" {TEXT -1 52 "Here is the generation \+
for the solution and a check." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 40 "f:=x->32+272*x*(1-x); \nc:=1/50;\nk:=1/10;"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "for n from 1 to 5 do\na[n
]:=int((f(x)-32)*sin(n*Pi*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1):\nod:\n
n:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "u:=(t,x)->32+exp
(-c*t)*sum(a[n]*exp(-k*n^2*Pi^2*t)*sin(n*Pi*x),n=1..5);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u(0,x);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 32 "plot([f(x),u(0,x)+.007],x=0..1);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 76 "simplify(diff(u(t,x),t)-(k*diff(u(t,x),x,
x)-c*(u(t,x)-32)));\nu(t,0); u(t,1);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 62 "plot3d(u(t,x),x=0..1,t=0..3,axes=NORMAL,orientation=[
-30,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 30 "Computation of the to
tal heat." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
108 " Here, I compute the total heat of the system and show with a
graph how the total heat decays toward 32." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "T:=t->int(u(t,x),x=0..1);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 18 "plot(T(t),t=0..3);" }}}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 11 "L
ooking for" }{TEXT 263 1 " " }{XPPEDIT 264 0 "alpha;" "6#%&alphaG" }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 25 "We now seek the constan
t " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 18 " so that T '(t)
= " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 25 " (T(t) - 32) w
ith T(0) = " }{XPPEDIT 18 0 "int(u(0,x),x = 0 .. 1);" "6#-%$intG6$-%\"
uG6$\"\"!%\"xG/F*;F)\"\"\"" }{TEXT -1 24 ". The idea is to look at" }}
{PARA 0 "" 0 "" {TEXT -1 30 " " }
{XPPEDIT 18 0 "diff(T(t),t)/(T(t)-32);" "6#*&-%%diffG6$-%\"TG6#%\"tGF*
\"\"\",&-F(6#F*F+\"#K!\"\"F0" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "and see if this is nearly const
ant. If it is, we take that constant to be " }{XPPEDIT 18 0 "alpha;" "
6#%&alphaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
23 "diff(T(t),t)/(T(t)-32):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
19 "cons:=unapply(%,t):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "
plot(cons(t),t=0..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 64 "The la
st graph is nearly constant. Take the average value to be " }{XPPEDIT
18 0 "alpha;" "6#%&alphaG" }{TEXT -1 228 ". Don't extend the computati
on of the average value over too long an interval, for T(t)-32 -- the \+
bottom of the fraction -- approaches zero. Round off when a number alm
ost zero is divided by a number almost zero can clobber you!" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "alpha:=evalf(Int(cons(t),t=0
..3)/3,3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 265 23 "Verification of results" }}{PARA 0 "" 0 "" {TEXT -1
85 "How can we check our results? Why not solve the ordinary different
ial equation y ' = " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1
53 " (y(t) - 32) and compare its graph with that of T(t)?" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "y:='y'
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "y0:=evalf(T(0),3);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "ODE:=dsolve(\{diff(y(t),t)=
alpha*(y(t)-32),y(0)=y0\},y(t));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 23 "y:=unapply(rhs(ODE),t);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 43 "plot([T(t),y(t)],t=0..3,color=[BLACK,RED]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 11 "Conclusion:" }{TEXT -1 74 " Th
is worksheet provides connections between ODE cooling and PDE cooling.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 " \+
Another experiment needs to be done. Ask how " }{XPPEDIT 18 0 "alpha;
" "6#%&alphaG" }{TEXT -1 178 " changes when f, c, and k change. I wil
l resist doing that. A student might choose to tell us all the results
of another experiment: change only f and see what is the resulting "
}{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 54 ". In such an experi
ment, retain that f(0) = 32 = f(1)." }}}{MARK "0 0" 49 }{VIEWOPTS 1 1
0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }