taller de termodinamica

1) un recipiente rígido contiene 2 kg de refrigerante-134a a 800 kpa y 20c determinar el volumen del recipiente y la energía interna total.
2) un depósito rígido 5 ft3 contiene 5 lbm de agua a 20psia. determinar (a) la temperatura, (b) la entalpía total y (c) la masa de cada fase de agua
P = 20 psia 
T = 227.96 F ( ? ) 
vf = 0.016830 cu ft / lbm hf = 196.3 Btu / lbm 
vfg = 20.0732 cu ft / lbmhfg = 960.1 Btu / lbm 
vg = 20.09 cu ft / lbm hg = 1156.4 Btu / lbm 

v = V / m = ( 5.000 cu ft ) / 9 5.000 lbm ) = 1.000 cu ft / lbm 

v = vf + ( x ) ( vfg ) 

x = ( v - vf ) / ( vfg ) 

x = ( 1.000 -0. 016830 ) / ( 20.0732 ) 

x = 0.04898 = 4.898 % <--------- 

Since 1.0 > x > 0.0 , there are two phases present and T = Tsat at 20 psia 

T = Tsat at 20 psia = 227.96 F <-----[ a ]---------- 

h =hf + ( x ) ( hfg ) 

h = 196.3 + ( 0.04898 ) ( 960.1 ) = 243.3 Btu / lbm 

H = ( m ) ( h ) 

H = ( 5.000 lbm ) ( 243.3 Btu / lbm ) = 1216.6 Btu <-----[ b ]--------- 

mg = ( x ) ( m ) = ( 0.04898 ) (5.000 ) = 0.2449 lbm vapor <-----[ c ]--------------- 

mf = ( 1 - x ) ( m ) 

mf = ( 1.0 - 0.04898 ) ( 5.00 lbm ) = 4.755 lbm liquid <-----[ c ]---- 

Vg = ( mg ) ( vg ) 

Vg = ( 0.2449 lbm ) ( 20.09cu ft / lbm ) = 4.9200 cu ft vapor <----[ d ]---- 

Vf = ( mf ) ( vf ) = ( 4.755 lbm ) ( 0.016830 cu ft / lbm ) = 0.0800 cu ft liquid <---[ d ] 

Part d is some bonus information for you. 

You needsaturated R-134a Table Data : 
---------------------------------------... 

P = 50 psia 
T = 40.23 F 
vf = 0.01252 cu ft / lbm 
vfg = 0.93539 cu ft / lbm 
vg = 0.94791 cu ft / lbm 

Vf = ( 0.20 ) [5.000 cu ft ] = 1.000 cu ft 

mf = Vf / vf 

mf = ( 1.000 cu ft ) / ( 0.01252 cu ft / lbm ) = 79.87 lbm liquid 

mg = Vg / vg 

mg = ( 4.000 cu ft ) / ( 0.94791 cu ft / lbm ) = 3.792 lbm vapor 

mt = mf+ mg = 79.87 lbm + 3.792 lbm = 83.66 <------ 

x = mg / mt = ( 3.792 lbm ) / ( 83.66 lbm ) 

x = 004533 = 4.533 % <------------------------

3. calcular la variación de la energía de un sistema que...