taller i de termodinamica

7/23/2019 Taller i de Termodinamica

1/13

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= +


2/13

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=


3/13

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


4/13

2

2

mUEc=

eem(#a.ano #os a#ores

2

2(1400 ) * (30 )

2

mkgsEc=

630000Ec J=

&aa en @A

630Ec kJ=

# tra"a


5/13

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


6/13

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=


7/13

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 = =


8/13

-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

= +


9/13

( )

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:


10/13

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

= =


11/13

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

&

= ==


12/13

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 $= +


13/13

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

= = =

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