final solution sp13 - weeklyjoys · 2014-04-21 · ! 3! node5!! q 25 =!k t 5 p!t 2 p "y...
Post on 14-Jul-2020
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1
Problem 1 (a)
!"
#"
$"
%"
&!%"
&#%"
&$%"
!"
#"
$"
%"
&!'"
&#'"
&$'"
'"&%'"
&()*+"
!"
#"$"
%!$"
%&'()"
%#$" %&'()"
2
Node 4 !x = !y
q14 = "kT4
p "T1p
!y#!x2#d
q54 = "kT4
p "T5p
!x# !y #d
q74 = "kT4
p "T7p
!y#!x2#d
!Egen = !q #V = !q # !x2# !y #d
!Estore =mCp$T$t
= !!x2# !y #d
%
&'
(
)*Cp
T4p+1 "T4
p
!t
!Est = !Ein " !Eout + !Egen + !Eout = 0 and !Ein = q14 + q54 + q74
!!x2# !y #1
%
&'
(
)*cp
T4P+1 "T4
P
!t%
&'
(
)*= k
!x2#1
%
&'
(
)*T1
P "T4P
!y%
&'
(
)*+ k
!x2#1
%
&'
(
)*T7
P "T4P
!y%
&'
(
)*+ k !y #1( ) T5
P "T4P
!y%
&'
(
)*+ !q
!x2# !y #1
%
&'
(
)*
Fo = !!t!x2
=k!t
"cp!x2 + T4
P+1 = T4P +Fo T1
P +T7P + 2T5
P " 4T4P{ }+ !q!t
!cp
T4P+1 = T4
P +Fo T1P +T7
P + 2T5P ! 4T4
P{ }+ !q"t!cp
3
Node 5
q25 = !kT5
p !T2p
"y# "x #d
q45 = !kT5
p !T4p
"x# "y #d
q65 = !kT5
p !T6p
"x#"y2#d
q85 = !kT5
p !T8p
"y#"x2#d
qconv = h T$ !T5p( ) # "x2 #d + h T$ !T5
p( ) # "y2 #d
!Egen = !q #V = !q # 3"x"y4
#d
!Estore =mCp%T%t
= !3"x"y4
#d&
'(
)
*+Cp
T5p+1 !T5
p
"t
!Est = !Ein ! !Eout + !Egen , !Eout = 0 and !Ein = q25 + q45 + q65 + q85 + qconv
!34"x # "y #1
&
'(
)
*+cp
T5P+1 !T5
P
"t&
'(
)
*+= k"y
T4P !T5
P
"x&
'(
)
*++ k"x
T2P !T5
P
"y&
'(
)
*++ k
"x2
T8P !T5
P
"y&
'(
)
*++ k
"y2
T6P !T5
P
"x&
'(
)
*+
+ h "x2+"y2
&
'(
)
*+ T$ !T5
P( )+ !q 34"x # "y #1
&
'(
)
*+
Fo = !"t"x2
=k"t
"cp"x2 , T5
P+1 = T5P +23Fo T8
P +T6P + 2T4
P + 2T2P ! 6T5
P{ }+ 43h"t!cP"x
T$ !T5P( )+ !q"t
!cp
T5P+1 = T5
P +23Fo T8
P +T6P + 2T4
P + 2T2P ! 6T5
P{ }+ 43h"t!cP"x
T# !T5P( )+ !q"t
!cp
4
Node 6
q36 = !kT6
p !T3p
"y#"x2#d
q56 = !kT6
p !T5p
"x#"y2#d
qconv = h T$ !T6p( ) # "x2 #d + h T$ !T6
p( ) # "y2 #d
!Egen = !q #V = !q # "x"y4
#d
!Estore =mCp%T%t
= !"x"y4
#d&
'(
)
*+Cp
T6p+1 !T6
p
"t
!Est = !Ein ! !Eout + !Egen , !Eout = 0 and !Ein = q36 + q56 + qconv
!"x # "y4
&
'(
)
*+cp
T6P+1 !T6
P
"t&
'(
)
*+= k
"y2
T5P !T6
P
"x&
'(
)
*++
k"x2
T3P !T6
P
"y&
'(
)
*++ h
"x2+"y2
&
'(
)
*+ T$ !T6
P( )+ !q "x # "y4
&
'(
)
*+
Fo = !"t"x2
=k"t
"cp"x2 , T6
P+1 = T6P +Fo 2T5
P + 2T3P ! 4T6
P{ }+ 4h"t!cP"x
T$ !T6P( )+ !q"t
!cp
T6P+1 = T6
P +Fo 2T5P + 2T3
P ! 4T6P{ }+ 4h"t
!cP"xT# !T6
P( )+ !q"t!cp
5
(b)
Stability limit for 2-D explicit method: !t "!x( )2
4!
!x( )2
4!=
!x( )2
4 k!Cp
=0.01m( )2
4 #100W /m$K( )
2000 kg /m3( ) 300 J / kg$K( )
= 0.15 sec
%!t " 0.15 sec
(c) T1
P = T2P = T3
P = T4P = T5
P = T6P = 300 K and T7
P = T8P = 400 K
T! = 500 K
"x = "y = 0.01m and "t = 0.1s
Fo = k"t!cP"x
2 =100W /m#K( ) 0.1 s( )
2000 kg /m3( ) 300 J / kg#K( ) 0.01m( )2= 0.167
!q =106 W m3 and h =100W m2K
T4P+1 = 316.8 K
T5P+1 = 311.7 K
T6P+1 = 301.5 K
Δt = 0.1 sec is suitable because the stability limit for the given situation is Δt ≤ 0.15 sec.
6
Problem 2 (a) 1r!!r
k r !T!r
"
#$
%
&'+1r2
!!!
k !T!!
"
#$
%
&'+
!!z
k !T!z
"
#$
%
&'+ !q = !Cp
!T!t
Assuming 1-‐D conduction and steady state condition, 1r!!r
k r !T!r
"
#$
%
&'+ !q = 0 (
ddr
k r dTdr
"
#$
%
&'= ) !q r
k r dTdr
= )!q r2
2+C1 (
dTdr
= )!q r2k
+C1k r
BC 1: dTdr r=0
= 0 ( *C1 = 0
dTdr
= )!q r2k
( dT+ = )!q r2k
dr+
T = ) !q2k
r2
2+C2 ( T (r) = ) !q r
2
4k+C2
BC2 : Tr=0 = Tc =C2 ( *C2 = Tc
*T (r) = Tc )!q r2
4k
T1 = Tr=r0 = Tc )!q r0
2
4k
T1 = Tc !!q r0
2
4k
7
(b)
Assuming the length of the rod, L = 1m,
q = !q !V = !q ! !D12
4!L = 20000W /m3( ) ! ! (0.05m)
2
4! 1m( ) = 39.27W
Eb,1 "Eb,2
1"!1A1!1
+1
A1F12+1"!2A2!2
=" T1
4 "T24( )
1" 0.2! 0.05m( ) 1m( ) 0.2( )
+1
! 0.05m( ) 1m( ) 1( )+
1" 0.5! 0.06 m( ) 1m( ) 0.5( )
=5.67#10"8W /m2 "K 4( ) T14 " 5004( )K 4
1" 0.2! 0.05m( ) 1m( ) 0.2( )
+1
! 0.05m( ) 1m( ) 1( )+
1" 0.5! 0.06 m( ) 1m( ) 0.5( )
= 39.27W $ T1 = 544.99 K
T1 = Tc "!q r0
2
4k= Tc "
20000W /m3( )(0.025m)24 ! 15W /m"K( )
= 544.99 K $ Tc = 545.20 K
Tc = 545.20 K and T1 = 544.99 K
!"#$#%&&#'#("#$#&)%#
!*#$#+#'#(*#$#&)"#
!"#$%&% '$% '(% !"#(%
1!!1A1!1
1A1F12
1!!2A2!2
8
(c) !Ein ! !Eout + !Egen = !Estore
!Ein = !Estore = 0
" !Eout = !Egen
!Egen' = !q # !D1
2
4= 20000W /m3( ) # ! (0.05m)
2
4= 39.27W /m
" !Eout' = q1
' = !Egen' = 39.27W /m
!q1' = 39.27W /m
9
Problem 3 (a)
By neglecting the thickness of the container wall,
!!q A = qrad =! Tc
4 "Ts4( )
1"!c!cA
+1AFcs
+1"!s!sA
; Fcs =1 # !!q =" Tc
4 "Ts4( )
1"!c!c
+1+1"!s!s
Tc =!!q"1"!c!c
+1+1"!s!s
$
%&
'
()+Ts
4*
+,
-
./
14
= 372 K
!Tc = 372 K
!"#$#%#&#'"#$#()*#
+,-..#/#$#()0*#1#
!2344#
!2#$#50*#&#'2#$#()*#
67"331
#
89#$#:*(#;<10#
!=#$#5((#&#
!"#$%&%'%&()%
*$% *+% !"#+%
1!!cA!c
1AFcs
1!!sA!s
10
(b) !!qconv = h Ts "T#( )
Ra =g! Ts "T#( )H 3
!"= 3.01$107
h =Nu kH=kH0.825+ 0.387Ra
16
1+ 0.492Pr
%
&'
(
)*
916%
&''
(
)**
827
%
&
''''''
(
)
******
2
+ h = 4.54W m2 "K
!!qconv = 4.54W m2 "K( ) 325 K "300 K( ) =114W m2
! ""qconv =114W m2
(c) !!q =150W /m2 = !!qconv + !!qrad " !!qrad = !!q # !!qconv = !s" Ts
4 #Tsurr4( )
Tsurr = Ts4 #
!!q # !!qconv!s"
$
%&
'
()
14
= 315 K
!Tsurr = 315 K
11
Problem 4 (a)
(b)
Nu = 0.3+ 0.62Re1 2 Pr1 3
1+ 0.4 Pr( )2 3!"
#$1 4 % 1+
Re282000&
'(
)
*+5 8!
",,
#
$--
4 5
Re = !uDµ
=0.94 kg m3( ) 5 m s( ) 0.2 m( )
218.10/7 N % s m2 = 4311.93
Nu = 0.3+0.62 4311.93( )1 2 0.7( )3
1+ 0.4 0.7( )2 3!"
#$1 4 % 1+ 4311.93
282000&
'(
)
*+5 8!
",,
#
$--
4 5
= 33.858 = h %Dk
=h % 0.02 m( )0.031W mK( )
0h = 52.48W m2K
qconv = h T1 /Th( )Ah; Ah = !DhL
qconv = h T1 /Th( )Ah = 52.48W m2K( ) 375 K /350 K( )! 0.02 m( ) 0.1m( )= 8.24353W
!qconv = 8.24 W (into the hotdog)
!"#$%
&'(!"#'%
&"% !"#"%
1!!hAh!h
1AbFbh
1!!bAb!b
1AhFhR
1AbFbR
12
(c) !Ein ! !Eout + !Egen = !Estore
!Egen = !Eout = 0
!Estore =mCpdTdt
= 0.03 kg( ) 2500 J / kg!K( ) 0.25 K / s( ) =18.75W
!Ein = qconv + qrad
" !Estore = !Ein = qconv + qrad # qrad = !Estore ! qconv =Eb,b !Eb,h
Rtot=! Tb
4 !Th4( )
Rtot
Rtot =1!!b!bAb
+1!!h!hAh
+ AbFbh +1
AbFbR+
1AhFhR
$
%&
'
()
!1*
+,,
-
.//
!1
FbR =1!Fbh = 0.93
Fhb =AbAhFbh = 0.23
FhR =1!Fhb = 0.77
"Rtot = 558.21m2
Tb =!Estore ! qconv( ) 0Rtot
!+Th
4*
+,,
-
.//
14
= 586.6 K
!Tb = 586.6 K
13
Problem 5 (a)
x fdD
!
"#
$
%&thermal
= 0.05RePr =0.05 20( ) 1.0( ) =1 Pr =1.0( )0.05 20( ) 10( ) =10 Pr =10( )
'()
*)
x fd,thermal =D Pr =1.0( )10D Pr =10( )
'()
*)
Tm = Tm,x=0 +++q P!mcp
x = Tm,x=0 + constant( ) , x - Linear Line
!"#
$#%#&#
'(#)#*+#
'(#)#*,+#-,./#
01#
$#%#
'(#)#*,+#2#*+#013$)+#
013$)%#
14
(b)
Tm =!ucpT dAcAc
!!mcp
=!ucp T dAcAc
!!uAccp
Ac = y "d and dAcdy
= d # dAc = d "dy
Tm =T "d dy
$H
H!2H "d
=
%%q2k f H
y2 $H 2( )+Tw dy$H
H!
2H=12H
%%q2k f H
13y3 $H 2y
&
'(
)
*++Twy
,
-..
/
011$H
H
=12H
%%q2k f H
13H 3 $H 3&
'(
)
*++TwH
,
-..
/
011$
%%q2k f H
$13H 3 +H 3&
'(
)
*+$TwH
,
-..
/
011
234
54
674
84
=12H
%%q2k f H
$23H 3&
'(
)
*++TwH $
%%q2k f H
23H 3&
'(
)
*++TwH
,
-..
/
011
=12H
2TwH $%%q
2k f H43H 3&
'(
)
*+
,
-..
/
011=12H
2TwH $%%qkf
23H 2&
'(
)
*+
,
-..
/
011=12H
2TwH $2 %%q3k f
H 2,
-..
/
011
= Tw $%%q
3k fH # Tm = Tw $
%%q3k f
H
%%q = h Tw $Tm( ) # h = %%qTw $Tm
=%%q
Tw $ Tw $%%q
3k fH
&
'((
)
*++
=%%q%%q
3k fH=3k fH
!h =3k fH
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