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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 38 "Terminal Velocity of Fall
ing Particles" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 36 "The force balance on the particle is" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "forcebal := v[t]=sqrt(4*g*(r
ho[p]-rho)*D[p]/(3*C[D]*rho)):\nforcebal;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/&%\"vG6#%\"tG,$*&\"\"$#\"\"\"\"\"#*,%\"gGF,,&&%$rhoG6#
%\"pGF,F2!\"\"F,&%\"DGF3F,&%\"CG6#F7F5F2F5F+#F-F*" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 49 "The Reynolds number is defined for this system by
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "reeqn := re = D[p]*v[t]
*rho/mu: reeqn;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#reG**&%\"DG6#%\"
pG\"\"\"&%\"vG6#%\"tGF*%$rhoGF*%#muG!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 82 "The parameters in these equations have the following valu
es (in consistent units):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
68 "params := \{g=9.81,rho[p]=1800, D[p]=0.208e-3,rho=994.6,mu=8.931e-
4\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'paramsG<'/%$rhoG$\"%Y**!\"
\"/%#muG$\"%J*)!\"(/&%\"DG6#%\"pG$\"$3#!\"'/&F'F3\"%+=/%\"gG$\"$\")*!
\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "We substitute these parame
ters into the force balance to get" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "subs(params,forcebal);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/&%\"vG6#%\"tG,$*&\"\"$#\"\"\"\"\"#*$&%\"CG6#%\"DG!\"\"F+$\"+p,#
*4F!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 387 "We cannot solve this e
quation directly since the drag coefficient is a function of the termi
nal velocity through the Reynolds number. We continue by writing a Map
le procedure to compute the drag coefficient as a function of the Reyn
olds number. We have to take this approach since the drag coefficient \+
is computed from different expressions depending on the value of the R
eynolds number." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 276 "f := pr
oc(re)\nlocal CD;\nif not type(re,numeric) then RETURN(C[D](re)) fi;\n
if re < 0.1 then \n   CD := 24/re\nelif re >= 0.1 and re <= 1000 then \+
\n   CD := 24/re*(1+0.14*re^0.7)\nelif re > 1000 and re <= 350000 then
\n   CD := 0.44\nelif re > 350000 then\n   CD := 0.19-8e4/re\nfi;\nend
:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 "Note that we use re for the
 Reynolds number since Re is a reserved phrase in Maple for the real p
art of a complex variable. Let us test this proc to see what it does" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(10);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$\"+0\"*)R3%!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "f(100);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+TQ*R3\"
!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "f(2000);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"#W!\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "f(re);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-&%\"CG6#%\"
DG6#%#reG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "The next step is to
 create a Maple procedure to evaluate the force balance for different \+
values of the terminal velocity. There are many ways to do this; our e
xample follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 159 "eqn := \+
proc(vt)\nlocal C, re;\nglobal params;\n  re:= subs(params,D[p]*vt*rho
/mu);\n  C[D]:=f(re);\n  subs(params,vt-sqrt(4*g*(rho[p]-rho)*D[p]/(3*
C[D]*rho)));\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Finally, we
 invoke Maple's built in floating point solver" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 40 "vt1:=fsolve('eqn'(vt),vt=0.00001..0.05);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$vt1G$\"+#e='y:!#6" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 482 "This is the terminal velocity of the particle \+
under the given conditions. Note that eqn in the call to fsolve is con
tained within quote marks. If this were not done then Maple would atte
mpt to call the eqn procedure and pass that result to fsolve. Needless
 to say, that would be a disaster in this case. the quote marks force \+
Maple to re-evaluate the equation every time it tries a new value of t
he terminal velocity. The units here are m/s. The Reynolds number at t
his velocity is " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "subs(v[
t]=vt1,params,reeqn);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#reG$\"+ekp
cO!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "and the drag coefficien
t is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "f(rhs(\"));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+i5cS))!\"*" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 104 "For part (b) we are asked for the terminal velocity
 in a centifugal separator where the acceleration is " }{XPPEDIT 18 0 
"30*g" "*&\"#I\"\"\"%\"gGF$" }{TEXT -1 81 ". The only changes needed a
re to revise the set of parameters (in this case only " }{XPPEDIT 18 
0 "g" "I\"gG6\"" }{TEXT -1 13 " is changed)." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 72 "params := \{g=9.81*30,rho[p]=1800,  D[p]=0.208e-
3,rho=994.6,mu=8.931e-4\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'param
sG<'/%\"gG$\"&I%H!\"#/%$rhoG$\"%Y**!\"\"/%#muG$\"%J*)!\"(/&%\"DG6#%\"p
G$\"$3#!\"'/&F,F8\"%+=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 235 "
eqn := proc(vt)\nlocal C, re; \nglobal params;\n  params := \{g=9.81*3
0,rho[p]=1800,  D[p]=0.208e-3,rho=994.6,mu=8.931e-4\};\n  re:= subs(pa
rams,D[p]*vt*rho/mu);\n  C[D]:=f(re);\n  subs(params,vt-sqrt(4*g*(rho[
p]-rho)*D[p]/(3*C[D]*rho)));\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 20 "Again we call fsolve" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "vt2:=fsolve('eqn'(vt),vt);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$v
t2G$\"+PApg?!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "The units here
 are m/s. The Reynolds number at this velocity is " }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 28 "subs(v[t]=vt2,params,reeqn);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#/%#reG$\"+,rOtZ!\")" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 27 "and the drag coefficient is" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "f(rhs(\"));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+8(
Gkb\"!\"*" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }