Chapter 10 : Properties of Pure Substance - Water

Example 10.1 Page No : 333

In [1]:
import math 
			
# Variables
m = 1.5 			#kg 			#mass of wet steam
mf = 0.08*m 			#kg 			#mass of liquid in wet steam
			
# Calculations and Results
x = (m-mf)/m 			#dryness fraction of mixture
print "Dryness fraction of mixture = %.2f"%(x)
Dryness fraction of mixture = 0.92

Example 10.2 Page No : 333

In [2]:
import math 

			
# Variables
x = 0.85 			#quality of steam
print "The ratio of mass of saturated liquid to saturated steam = %.5f"%((1-x)/x)
The ratio of mass of saturated liquid to saturated steam = 0.17647

Example 10.3 Page No : 350

In [3]:
import math 

# Variables
p1 = 200. 			#kPa 			#initial pressure
t1 = 100. 			#°C 			#initial temperature
ts = 120.23 			#°C 			#saturated steam temperature

			
# Calculations and Results
#From steam table
v1 = 0.001044 			#m**3/kg 			#volume per kilogram of water 
h1 = 419 			#kJ/kg 			#enthalpy per kilogram of water
s1 = 1.3068 			#kJ/kg 			#entropy per kilogram of water
u1 = h1-p1*v1 			#kJ/kg 			#internal energy per kilogram of water
print "Volume per kilogram of water = %.6f m**3/kg"%(v1)
print "Enthalpy per kilogram of water = %.1f kJ/kg"%(h1)
print "Entropy per kilogram of water = %.4f kJ/kgK"%(s1)
print "Internal energy per kilogram of water = %.1f kJ/kg"%(u1)
Volume per kilogram of water = 0.001044 m**3/kg
Enthalpy per kilogram of water = 419.0 kJ/kg
Entropy per kilogram of water = 1.3068 kJ/kgK
Internal energy per kilogram of water = 418.8 kJ/kg

Example 10.4 Page No : 350

In [3]:
import math
			
# Variables
p1 = 500. 			#kPa 			#initial pressure
s1 = 1.3625 			#initial entropy

			
# Calculations and Results
#Using Method 2:
Ts = 424.28 			#K 			#temperature at 500kPa
sf = 1.8606 			#kJ/kgK 			#entropy at 500kPa
Cwat = 4.189 			#kJ/kgK 			#specific heat of water
T1 = round((math.exp((sf-s1)/Cwat)/Ts)**-1) 			#K
print "Temperature = %.2f °C"%(T1-273)
v1 = 0.001 			#m**3/kg 			#volume per kg water
h1 = (640.21 - Cwat*(151.86-T1+273)) 			# kJ/kg 			#Enthalpy per kg water
u1 = h1 - p1*v1 			#kJ/kg 			#internal energy per kg water
print "Volume per kg water = %f m**3/kg"%(v1)
print "Enthalpy per kg water = %.1f kJ/kg"%(h1)
print "Internal energy per kg water = %.1f kJ/kg"%(u1)

# note : rounding off error. please check.
Temperature = 104.00 °C
Volume per kg water = 0.001000 m**3/kg
Enthalpy per kg water = 439.7 kJ/kg
Internal energy per kg water = 439.2 kJ/kg

Example 10.5 Page No : 353

In [5]:
			
# Variables
t = 50. 			#°C 			#temperature of water
h = 209.31 			#kJ/kg
			
# Calculations and Results
#From saturated property table
p = 12.35 			#kPa
v = 0.001012 			#m**3/kg
u = h - p*v 			#kJ/kg
s = 0.7037 			#kJ/kg
print "Pressure = %.2f kPa"%(p)
print "Volume per kg water = %.6f m**3/kg"%(v)
print "Internal energy per kg water = %.1f kJ/kg"%(u)
print "Entropy per kg water = %.4f kJ/kgK"%(s)
Pressure = 12.35 kPa
Volume per kg water = 0.001012 m**3/kg
Internal energy per kg water = 209.3 kJ/kg
Entropy per kg water = 0.7037 kJ/kgK

Example 10.6 Page No : 353

In [1]:
			
# Variables
p = 12. 			#bar 			#pressure of steam leaving boiler
h = 2705. 			#kJ/kg 			#Enthalpy of steam
			
# Calculations and Results
#From pressure based saturated property table, at p = 12bar
hf = 798.64 			#kJ/kg
hg = 2784.8 			#kJ/kg
x = (h-hf)/(hg-hf) 			#Dryness fraction
v = (1-x)*0.001139 + (x)*0.1633 			#m**3/kg
s = (1-x)*2.2165 + (x)*6.5233 			#m**3/kgK
u = h - p*v*100 			#kJ/kg
print "Volume per kg water = %.4f m**3/kg"%(v)
print "Internal energy per kg water = %.0f kJ/kg"%(u)
print "Entropy per kg water = %.3f kJ/kgK"%(s)
Volume per kg water = 0.1568 m**3/kg
Internal energy per kg water = 2517 kJ/kg
Entropy per kg water = 6.350 kJ/kgK

Example 10.7 Page No : 354

In [2]:
			
# Variables
p = 15 			#bar
u = 2594.5 			#kJ/kg
			
# Calculations and Results
#From saturated steam table based on pressure at p = 15 bar
hf = 844.87 			#kJ/kg
hg = 2792.1 			#kJ/kg
vf = 0.001154 			#m**3/kg
vg = 0.13177 			#m**3/kg
uf = hf-100*p*vf 			#kJ/kg
ug = hg-100*p*vg 			#kJ/kg

if u-ug<0.1 :
    print "Temperature = %.2f °C"%(198.32)
    print "Volume per kg water = %.5f m**3/kg"%(vg)
    print "Enthalpy per kg water = %.1f kJ/kg"%(hg)
    print "Internal energy per kg water = %.1f kJ/kg"%(ug)
    print "Entropy per kg water = %.4f kJ/kgK"%(6.4448)
Temperature = 198.32 °C
Volume per kg water = 0.13177 m**3/kg
Enthalpy per kg water = 2792.1 kJ/kg
Internal energy per kg water = 2594.4 kJ/kg
Entropy per kg water = 6.4448 kJ/kgK

Example 10.8 Page No : 355

In [8]:
# Variables
p = 10e6 			#Pa
t = 550. 			#°C
			
# Calculations and Results
#From superheated property table
v_500 = 0.03279 			#m**3/kg
v_600 = 0.03837 			#m**3/kg
v_550 = v_500 + (v_500-v_600)/(500-600)*(550-500) 			#m**3/kg
h_500 = 3373.6 			#kJ/kg
h_600 = 3625.3 			#kJ/kg
h_550 = h_500 + (h_500-h_600)/(500-600)*(550-500) 			#kJ/kg
s_500 = 6.5965 			#kJ/kgK
s_600 = 6.9028 			#kJ/kgK
s_550 = s_500 + (s_500-s_600)/(500-600)*(550-500) 			#kJ/kgK
print "Volume per kg water = %.6f m**3/kg"%(v_550)
print "Enthalpy per kg water = %.1f kJ/kg"%(h_550)
print "Entropy per kg water = %.4f kJ/kgK"%(s_550)
Volume per kg water = 0.035580 m**3/kg
Enthalpy per kg water = 3499.4 kJ/kg
Entropy per kg water = 6.7496 kJ/kgK

Example 10.9 Page No : 355

In [3]:
			
# Variables
t = 250. 			#°C 
h = 2855.8 			#kJ/kg

#From superheated property table
p = 3e6 			#Pa
v = 0.07058 			#m**3/kg
s = 6.2871 			#kJ/kgK
u = h - p*v*.001 			#kJ/kg
print "Pressure = %.1f MPa"%(p*1e-6)
print "Volume per kilogram of water = %.5f m**3/kg"%(v)
print "Enthalpy per kilogram of water = %.1f kJ/kg"%(h)
print "Entropy per kilogram of water = %.4f kJ/kgK"%(s)
print "Internal energy per kilogram of water = %.0f kJ/kg"%(u)
Pressure = 3.0 MPa
Volume per kilogram of water = 0.07058 m**3/kg
Enthalpy per kilogram of water = 2855.8 kJ/kg
Entropy per kilogram of water = 6.2871 kJ/kgK
Internal energy per kilogram of water = 2644 kJ/kg

Example 10.10 Page No : 359

In [10]:
# variables
p = 2. 			#bar
m = 0.16 			#kg
V = 0.1 			#m**3
			
# Calculations and Results
#refereing to the saturation temperature corresponding to 2bar
v = V/m 			#m**3/kg
vf = 0.001061 			#m**3/kg
vg = 0.8857 			#m**3/kg
print 'Specific volume at saturated liquid phase vf) = %.6f m**3/kg '%(vf)
print 'Specific volume at saturated vapor phase vg) = %.6f m**3/kg '%(vg)
    
if v<vg and v>vf :
    print 'The temperature of the steam must be equal to saturation temperature corresponding to 2 bar'
    print "Temperature of steam = %.2f °C"%(120.23)
Specific volume at saturated liquid phase vf) = 0.001061 m**3/kg 
Specific volume at saturated vapor phase vg) = 0.885700 m**3/kg 
The temperature of the steam must be equal to saturation temperature corresponding to 2 bar
Temperature of steam = 120.23 °C

Example 10.11 Page No : 360

In [9]:
			
# Variables
p1 = 2. 			#bar
v1 = 0.624 			#m**3/kg
t = 120.23 			#°C
m = 0.16 			#kg
			
# Calculations and Results
vf = 0.001061 			#m**3/kg
vg = 0.8857 			#m**3/kg
x1 = (v1-vf)/(vg-vf) 			#Dryness fraction
hf = 504.68 			#kJ/kg
hg = 2706.6 			#kJ/kg
h1 = (1-x1)*hf + x1*hg 			#kJ/kg
u1 = h1 - p1*v1*100			#kJ/kg

v2 = v1 			#m**3/kg
vf = 0.001044 			#m**3/kg
vg = 1.673 			#m**3/kg
x2 = (v1-vf)/(vg-vf) 			#Dryness fraction
hf = 419. 			#kJ/kg
hg = 2676. 			#kJ/kg
h2 = (1-x2)*hf + x2*hg 			#kJ/kg
p2 = 1.010325 			#bar
u2 = h2 - (p2*100)*v2 			#kJ/kg
print "Heat rejected from steam = %.1f kJ"%(m*(u2-u1))
Heat rejected from steam = -117.4 kJ

Example 10.12 Page No : 361

In [11]:
			
# Variables
m =0.1  			#kg
p1 = 10. 			#bar 
p2 = 1. 			#bar
			
# Calculations and Results
#From saturated steam table
v1 = 0.1944 			#m**3/kg
v2 = (p1/p2)**(1/1.3)*v1 			#m**3/kg
W = m*(p1*v1-p2*v2)*100/(1.3-1) 			#kJ
print "Work during expansion process = %.2f kJ"%(W)
h1 = 2778.1 			#kJ/kg
u1 = (h1 - p1*v1*100) 			#kJ/kg

vf = 0.001043 			#m**3/kg
vg = 1.694 			#m**3/kg
x2 = (v2-vf)/(vg-vf) 			#Dryness fraction
hf = 417.33 			#kJ/kg
hg = 2675.5 			#kJ/kg
h2 = (1-x2)*hf + x2*hg 			#kJ/kg
u2 = h2 - p2*v2*100 			#kJ/kg
print "Heat rejected from steam = %.2f kJ"%(W+m*(u2-u1))

# note : rounding off error
Work during expansion process = 26.71 kJ
Heat rejected from steam = -49.07 kJ

Example 10.13 Page No : 362

In [13]:
# Variables
p1 = 10. 			#bar
t1 = 300. 			#°C
V1 = 50. 			#m/s
p2 = 1. 			#bar
m = 1.2 			#kg/s

			
# Calculations and Results
			#From superheated steam table
h1 = 3051.2 			#kJ/kg
s1 = 7.1228 			#kJ/kgK
p2 = 1 			#bar
s2 = s1 			#kJ/kgK

sf = 1.3025 			#kJ/kgK
sg = 7.3593 			#kJ/kgK
x2 = (s2-sf)/(sg-sf) 			#Dryness fraction
hf = 417.44 			#kJ/kg
hg = 2675.5 			#kJ/kg
h2 = (1-x2)*hf + x2*hg 			#kJ/kg
vf = 0.001043 			#m**3/kg
vg = 1.694 			#m**3/kg
v2 = (1-x2)*vf + x2*vg 			#m**3/kg
V2 = (2*(1000*(h1-h2))+V1**2)**0.5 			#m/s
A2 = m*v2/V2*10000			#cm**3
print "The exit area of the nozzle = %.1f cm**2"%(A2)
The exit area of the nozzle = 20.3 cm**2

Example 10.14 Page No : 363

In [14]:
			
# Variables
m1 = 0.2 			#kg/s
p = 4. 			#bar
			
# Calculations and Results
#From superheated steam table
h1 = 2752.8 			#kJ/kg
h2 = 209.31 			#kJ/kg
h3 = 604.73 			#kJ/kg

m2 = (m1*h1-m1*h3)/(h3-h2) 			#kg/s
print "The flow rate of feed water into the heater = %.3f kg/s"%(m2)
The flow rate of feed water into the heater = 1.086 kg/s