Colección de problemas de Sistemas Cerrados (Gases Reales)

↑
ARCIAL 2017-2018

B
m(x2) 120g =
=
0,12k


A
*, W= 840 =0,084m

T1 5732 530,15K
Estado 1




i ......
=
=




x=128KN/m


iami.
F -

Acm
.
Embolo: Aceroi (E = 508 /kg-k
one
#




94c 5
0,085.ps
=




Too =
298,15k

100 98.183Pa
=




I, I I I I



FT




0.72.259,8725.330,15.P1 122.538,77*a
mRgTx
=




1: Ecacion termica de estado:pp - =




8,084



Rg =

Rn 8,31Us.moltk= 259,8125-157at
=




* PIA


Equilibrio mecanico del embolo: Fm (P1 Px)A
=
- -


mEg
,BoA
amegotm
V 0.01m. 8,12m
=
=
0.0012m
=




f.V =830.0.012 -
ME
9,4kg
=

mE
=
=




128.183(86 (1) (122.538.1 98.103).0,12 - = - -
9,4.9,8 + 10 11 0.024m
-
=




2.-12 V1 A.X =
-
0,884m3 0,12m=.0.0 + m=12 0.8756m
=
- =




* PeA

Equilibrio mecanico del embolo: P2 = P00 + MEg 98.103Pa 9.4kg.9,8N/kg
= + ;P2 =
98, 8KPa
0.72 m2
I ME9 -PoxA
98.8.103.0.0556; 12:239,5
Ecacion termica de estado: 12 -
12V2 =


259,8125.8,12


3. Balance
energetico: Sistema complete


DEg 0Ug: = m(v(Tz Ts) - m(x(1z- +1) +
mgy + mz((+z +1)
-
-



zk(t
- (1)500
-
-
A. + + 0g




UEE NEp +Nug mgx + mECE (Tz T1)
=
=
-




van 0Ep
=
-
k)110-85(10 1,1) - =

zk(ee)-

0.12.653.(239,5+ 298.15) + 9.4.9,87.(-0.0+) + 9,4.500(239,5+ 298.15)
- -
-



21.120.703(0,024)2 = 98.183.0.12.0.0 + Gg

0g -
=
280,78k5 0g=-433,115 No me sale el mismo valor,



4.Balance
energetics: Sistema gas



Ng Δug m(x(+z T1)
= =
-
+ m(x)Tz T1) pg + WE,g
-
=




WE,g 0,12.653,1851239,5
=
-
298,15) + 433,7.103



WE,g 428,5k)
=

, 3. us
047+
=




mpIn-Rgincameceino -




0.129731239.55. 259,87251n98.8.103 433,7.10
9.4.5081n239.5.
-



+

298,15
-
98.163 239,5



1 225,244kj
r 55,55/k
=


+
=