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TALLER I DE TERMODINAMICA
Pregunta 1
Datos:T: 25C h manmetro: 43,62cm g: 9,806 m/s2 Patmosferica: 101,45 kPa
erc!rio: 13,543 g/cm3 Pa"so#!ta: $%
Solucin
P gh=
Para poder trabajar en unidades de Pascal es necesario convertir la densidad aKg/m3
3
3 3 3
1 10013,534 * * 13534
1000 1
g kg cm kg
cm g m m
=
La altura en metros de la lectura del manmetro (43,62cm) euivale a!,4362m, una ve" igualadas las unidades se procede a operar#
( )3 213534 * 9, 806 * 0, 4362kg m
P mm s
=
57890,02P Pa=
$n %Pa es &', KPa
101, 45 57, 89Pabsoluta kPa kPa= +
159,34Pabsoluta kPa=
Pregunta 2
Datos:
&i'metro: 1,5ft masa (iston ) (esa: 300#" g: 32,158 ft/s2
Patmosferica: 29,84 *in +g
a. -e rea#i.a !n "a#ance e (resiones
Pgas Patm Ppistn= +
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Para ca#c!#ar #a f!er.a !samos #a ec!acin P
( )*Fgas Patm Ppistn A= +
( )*Fgas Patm A Fpistn= +
( )22 20, 75 1, 767A r A ft A ft = = =
( ) ( )( ) ( )2 229,84 * 1, 767 300 * 32,158ft
Fgas inHg ft lbs
= +
( )2 252, 727 9647, 4lbft
Fgas inHgfts
= +
Para (oer s!mar e"emos tener !niaes consistentes e #"f, (or #o tanto
efect!amos #as conersiones:
( )
( )23 2
2
22 2
30,483,386 10 1 1 152, 727 * * * * * *
1 1 1 * 4, 4482216 1100
cmkPa Pa N lbf minHgft
inHg kPa Pa m N ft cm
sto a como res!#tao:
3728,752lbf
hora e# 964,4 #"ft/s2 se iie entrecg
2 29647, 4 / 32,174
lbmft lbmft
s s lbf
# res!#tao es 299,851 #"f, con estos os a#ores en #i"ra7f!er.a a (oemos
hacer #a s!matoria:
3728,752 299,851Fgas lbf lbf= +
4028,603Fgas lbf=
. ti#i.ano #a ec!acin
FP
A=
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eem(#a.amos:
2 2
4028,6032279,911
1,767
lbf lbf P P
ft ft= =
ea#i.ano #as conersiones
Psi es (o!n (er s;!are inche *#i"ra (or (!#gaa c!araa entonces:
( )
2
22
12279,911 * 15,833
12
lbf ft psi
ft in=
15, 833 14, 7 3 0, 533Psia psi Psia= + =
c. # tra"a
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2
2
mUEc=
eem(#a.ano #os a#ores
2
2(1400 ) * (30 )
2
mkgsEc=
630000Ec J=
&aa en @A
630Ec kJ=
# tra"a
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40
0,35
20
4,18
*
m kg
! kW
" #
kJ#p
kg #
==
=
=
o
o
+
# cam"io e enta#(ia e# ag!a es e# tra"anices a*acero, E*ag!a t*tan;!e
Datos:
1
2
2
40
5
0,5*
0,5*
4,18*
500
2530
a
!
t
a
t
!
a
m kg
m kg
m kg
J#
kg #
J#
kg #
J#
kg #
" #
" #" #
=
=
=
=
=
=
=
==
o
o
o
o
o
o
?a tem(erat!ra 2 *T2 se s!(one (ara (oer ca#c!#ar #a rea#, ;!e a a ser !n
a#or entre 25 30 graos Cent>graosF
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Partieno e# "a#ance:
* * * 0a a ! ! t t m U m U m U + + = *1
G e #a ec!acin:
2
1
"
$
"
U # d" = *2
n este caso CC(C
-e reem(#a.a *2 en *1 e integrano:
*C *C *C 0a a a ! ! ! t t t m " m " m " + + =
eem(#a.ano
[ ] ( ) [ ]2* 0,5* (30 500) 40* 4,18 5*0, 5 0" " + + =
470 167, 2 2,5 0" " + + =
2
4702,77
169,7
25 2,77 27,77
"
" #
= =
= + = o
Pregunta %
Datos:
3
3
2
13600
1000
9,81
Hg
agua
kg
m
kg
m
mg
s
=
=
=
a. ?a (resin en Pasca#es (ara o"tener #a a#t!ra en metros
*
g
Hg
Ph
g=
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800000, 6
13600*9,81h hrs= =
.
*
g
agua
Ph
g=
800008,155
1000*9,81h hrs= =
Pregunta &
1 aceiteP gh=
1 720* 9,81*0,75 5297, 4P Pa= =
2 agua aireP gh P= +
2 1000*9,81*0,3 80000 82943P Pa= + =
82943 5297, 4 77645,6
77645,6
0, 58 5813600*9,81
P Pa
h m cm
= =
= = =
Pregunta '
* *con%& h A " =
eem(#a.ano #os atos en #a ec!acin
2
26 *1, 6 *9con%W
& m #m #
= oo
86,4con%& W=
Pregunta 1(
maletaE Ep mg z = =
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-e (asa e# (eso e #a ma#eta *90#" a kg
190 * 41
2, 2
kglb kg
lb=
eem(#a.amos en #a ec!acin
241 * 9,81 * 35 m 14077,35 J
mEp kg
s = =
14,07Ep J =
Pregunta 11
airem %=
( ) ( )3 31,18 / 4 / 4, 72 /airem kg m m s kg s= =
2
2
mUEc=
( ) ( )2
4, 72 / 10 /236 / s
2
kg s m sEc J= =
Pregunta 12
Datos:
1
2
1
2
3 /
200 /
334, 9 /
2726, 5 /
u m s
u m s
H J g
H J g
=
=
=
=
&aos (or #a sig!iente ec!acin:2 2
2 12 1
2
u u& H H
= +
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( )
2 2
200 31
2726,5 / 334,9 / *2 1000
m m
Js s& J g J g
J
= +
2411,6J
&g
=
Pregunta 1!
a. Datos:
1
2
25
400
10
20,8*
" #
" #
n molesJ
#%mol #
=
=
==
o
Partieno e #a ec!acin, si e# reci(iente tiene ca(acia ca#or>fica
insignificante:
1 2 1
1
1
1
* ( )
10*20,8* (400 25)
78000
78
& n #% " "
&
& J
& J
=
=
=
=
Para !na ca(acia ca#or>fica e 0,5A@g71C71, entonces:
150
0,5*
%m kg
J#%
g #
=
=o
[ ]2 2 1 1* C ( )% %& m " " &= + +
( )2
2
150*0,5*375 78000
106125 106,125
&
& J J
= +
= =
. Datos:
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1
2
475 273,15 201.85
72 22, 22
29,1*
" #
" F #
J#p
mol #
= =
= =
=
o
o o
o
&e #a ec!acin:
( )2 1* *& n #p " " =
Tenemos 2 incognitas, e# ca#or #os mo#es, entonces ha##amos e# ca#or en
tDrminos e #os mo#es ;!e e"er>an e es(ecificar#osF
( )29,1 * 201,85 22, 22*
J& #
mol # = o
o
5227, 238J
&mol
=
# ca#or ;!e e"e eHtraerse es e 522,238A (or caa mo# e nitrgeno
Pregunta 1"
1
2 1
3
20
100000 1
80, 33 300
3
*/ 83,14
*
airem kg
P Pa #te 'ar
" F
$ $
bar cmP$ "
mol
=
= = =
= =
=
=
o o
# (eso mo#ec!#ar e# aire es 28,9g/gmo#, entonces ca#c!#amos #as mo#es
20000692,04
28,9
gn mol
g
gmol
= =
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3
11
3
1
3
1
*83,14 *
*
* 30083,14 *1*
24942
bar cm "$
Pmol
bar cm $barmol
cm$ mol
=
=
=
2
1
1 2
1 1
1
(V )
( 3 )
* * 2
$
$
W n Pd$
W nP $
W nP $ $
W n P $
=
=
=
=
3
(692, 04 ) * (100000 ) * (2 * 0, 024942 )mW mol Pa mol=
3252172,336 3452,172W J kJ = =
2 12 1 1 1
1 1
3* * 3$ $
" " " " $ $
= = =
2 13" "=
29*J#p
mol =
2 1
1 1
1
( )
(3 )
(2 )
29 (2 *300 )*
17400 17, 4
H #p " "
H #p " "
H #p "
JH
mol
J JH
mol mol
=
=
=
=
= =
*
(692,04 )* (17, 4 KJ/ mol)
12041,496KJ
& n H
& mol
&
= ==
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124041, 496 3452,172
692,04
12, 41 /
& WU
n
J JU
mol
U J mol
+ =
=
=
Pregunta 1#
a. Datos:
( )
1
3 3 33
1 3
1
356 180
10, 023 1002300 1002, 3
1 10001,128 * * 1,128 10
1100
762 /
" F #
P bar Pa Pa
cm m g m$ x
g g gcm
U J g
= == = =
= =
=
o o
&e #a ec!acin:
1 1 1 1*H U P $= +
33
1
1
762 / (1002.3 *1,128 10 )
763,131 /
mH J g pa x
g
H J g
= +
=
. Datos:
2 2784, 4 /U J g =
2 14, 804 1500P atm pa= =
( )
3 3 3
2 3
1 1000169, 7 169, 7 * * 0,1697
1100
ml cm m g m$
g g g gcm= = =
2 2 2 2*H U P $= +
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3
2 2784, 4 / (1500 * 0,1697 )
mH J g pa
g= +
2 3038,95 /H J g=
2 1
3038, 95 / 763,131 /
2275,819 /
H H H
H J g J g
H J g
= = =
2 1
2784, 4 / Kg 762 /
2022, 4 /
U U U
U J J g
U J g
= = =