# Variables
P = 1.4 ; # [MPa]
T = 333. #[K]
T1 = 320. #[K]
T2 = 360. #[K]
P_low = 1. #[MPa]
P_high = 1.5
V_cap_T1_P1 = 0.2678
V_cap_T2_P1 = 0.2873
V_cap_T1_P1_5 = 0.1765
V_cap_T2_P1_5 = 0.1899
# Calculations
#At P = 1 MPa
V_cap_T333_P1 = V_cap_T1_P1 + (V_cap_T2_P1 - V_cap_T1_P1)*((T - T1)/(T2- T1))
#Similarly at P=1.5 MPa
V_cap_T333_P1_5 = V_cap_T1_P1_5 + (V_cap_T2_P1_5 - V_cap_T1_P1_5)*((T - T1)/(T2 - T1))
#At T=333*C
V_cap_P1_5 = V_cap_T333_P1_5 ;
V_cap_P1 = V_cap_T333_P1 ;
V_cap_P1_4 = V_cap_P1 + (V_cap_P1_5 - V_cap_P1)*((P - P_low)/(P_high - P_low)) ; #[m**3/kg]
# Results
print 'Required specific volume = %.5f m**3/kg'%(V_cap_P1_4);
# Variables
P = 1.4 ; #[MPa]
P_low = 1 ; #[MPa]
P_high = 1.5; #[MPa]
#At T=333*C from interpolation we have
v_cap_P1_5 = 0.18086 ; #[m**3/kg]
v_cap_P1 = 0.27414 ; #[m**3/kg]
# Calculations
#Molar volume is inversely proportional to pressure
v_cap_P1_4 = v_cap_P1 +(v_cap_P1_5 - v_cap_P1)*((1/P - 1/P_low)/(1/P_high - 1/P_low));
x=(0.19951-v_cap_P1_4)/v_cap_P1_4*100 ;
# Results
print 'Specific volume m**3/kg) = %g m**3/kg'%(v_cap_P1_4);
print 'Percentage difference = %.1f %%'%(x);
# variables
V1 = 1 # volume
m1 = 2.5 # water
Vv = 5.042 # m**3/kg
Vt = .001023 # m**3/kg
T_44632 = 145 # C
T_39278 = 150 # C
# Calculations
v1 = V1/m1
x1 = (v1 - Vt)/(Vv - Vt)
T2 = T_44632 + ( T_39278 - T_44632) * (( 0.4 - 0.44632)/(0.39278 - 0.44632))
# Restuls
print "Approximately %.f %% of the mass of water in the vapor"%(round(x1,2)*100)
print "Temperature T2 = %.1f C"%T2
# Note : Answer in book is wrong. Please calculate manually.