# Chapter 1 : Measured thermodynamic Properties and Other Basic Concepts¶

## Example 1.3 Page No : 28¶

In [2]:
# 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);

Required specific volume = 0.19951 m**3/kg


## Example 1.4 Page No : 29¶

In [1]:
# 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);

Specific volume m**3/kg) = 0.194186 m**3/kg
Percentage difference = 2.7 %


## Example 1.5 page no : 29¶

In [8]:
# 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


Approximately 8 % of the mass of water in the vapor