#Given
P1 = 8.74;#Initial pressure in Kgf/sq cm
P2 = 2.41;#Final pressure in Kgf/sq cm
H1 = 327.13;#Enthalpy of inlet stream in Kcal/Kg
Hl = 26.8;#Enthalpy of liquid at the final condition in Kcal/Kg
H2 = H1#Enthalpy of exit stream in Kcal/Kg ,math.since throttling is isenthalpic
Hg = 340.3;#Enthalpy of gas at the final condition in Kcal/Kg
vl = 152*10**-5;#Specific volume of liquid at the final condition in cubic meter/Kg
vg = 0.509;#Specific volume of gas at the final condition in cubic meter/Kg
v1 = 0.1494;#Initial specific volume in cubic meter/Kg
#To Calculate the dryness fraction of exit stream and the ratio of upstream to downstream diameters
#(i)Calculation of the dryness fraction of exit stream
#From equation 3.13(a) (page no 82)
x = (H2- Hl)/(Hg-Hl);
print "i)The dryness fraction of the exit stream is %f"%(x);
#(ii)Calculation of the ratio of upstream to downstream pipe diameters
#From equation 3.13(b) (page no 82)
v2 = (vl*(1-x))+(x*vg);#Total specific volume at the final condition in cubic meter/Kg
#u1 = u2; math.since KE changes are negligible
#From continuity equation: A2/A1 = D2**2/D1**2 = v2/v1 ; let required ratio,r = D2/D1;
r = (v2/v1)**(1/2);
print " ii)The ratio of upstream to downstream diameters is %f"%(r);
#end
#Given
P1 = 1000*1.033*10**4;#Initial pressure in Kgf/sq m
P2 = 1*1.033*10**4;#Final pressure in Kgf/sq m
T1 = 300.0;#Inital temperature in K
Cp = 7.0;#Specific heat of the gas in Kcal/Kgmole K
#Gas obeys the relation: v = (R*T)/P+(b*(T**2))
b = 5.4392*10**-8;#in cubic meter/Kgmole K**2
#To Calculate the temperature of the throttled gas
#From equation (a) (page no 212);which we got after integration
T2 = 1/((1/T1)-((b/Cp)*((P2-P1)/427)));
print "The throttled gas is cooled to %f K"%(T2);
#Given
#From the figure 11.8 (page no 216) & from figure A.2.7
H3 = 0.0;
H7 = -47.0;#in Kcal/Kg
H6 = -93.0;#in Kcal/Kg
H8 = 7.0;#in Kcal/Kg
#To Calculate the fraction of air liquified at steady state and temperature of air before throttling
#(i)Calculation of fraction of air liquified
#From equation 11.3 (page no 215)
x = (H8-H3)/(H8-H6);
print "The fraction of air liquified is %f"%(x);
#(ii)Calculation of temperature
H4 = H3+(H7*(1-x))-(H8*(1-x));#enthalpy of the gas before throttling
#From figure A.2.7 temperature corresponds to pressure 160 atm and the enthalpy H4 is
T = -112;
print " The temperature of air before throttling is %d deg celsius"%(T);
#end