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Chapter 1

Chapter 1#

Question 9: Steam and a Divider#

Consider \(m=10\:g\) of steam at \(T=250\:^{\circ}C\) stored in a container of volume \(V=100\:L\). For steam, calculate:

a) specific volume

b) density

c) pressure

d) specific internal energy

e) specific enthalpy

f) specific entropy

g) total internal energy

h) total enthalpy

i) total entropy

CH1-Q9.jpg

Solution Approach for a)#

for specific volume:

\(v=V/m\)

# define variables

m = 0.01   #mass of steam in kg
V = 0.1   #volume of steam in m3
T = 250 + 273.15   #tempearature of steam in K

v = V / m   #specific volume in m3/kg

print('specific volume of steam is:', f"{v:.1f}", 'm3/kg')

specific volume of steam is: 10.0 m3/kg

Solution Approach for b)#

for density:

\(D=1/v\)

D = 1 / v   #density in kg/m3

print('specific volume of steam is:', f"{D:.3f}", 'kg/m3')

specific volume of steam is: 0.100 kg/m3

Solution Approach for c)#

#importing the required library
import CoolProp.CoolProp as CP
fluid = 'water'

P = CP.PropsSI("P", "T", T, "D", D , fluid)   #pressure of steam in Pa

print('specific pressure of steam is:', f"{P/1000:.1f}", 'kPa')

specific pressure of steam is: 24.1 kPa

Solution Approach for d)#

\(u=U/m\)

where \(u\) is the specific internal energy and \(U\) is the total internal energy

u = CP.PropsSI("U", "T", T, "D", D , fluid)   #specific internal energy in J/kg

print('specific internal energy of steam is:', f"{u/1000:.1f}", 'kJ/kg')

specific internal energy of steam is: 2735.7 kJ/kg

Solution Approach for e)#

\(h=H/m\)

where \(h\) is the specific enthalpy and \(H\) is the total Enthalpy

h = CP.PropsSI("H", "T", T, "D", D , fluid)   #specific enthalpy in J/kg

print('specific enthalpy of steam is:', f"{h/1000:.1f}", 'kJ/kg')

specific enthalpy of steam is: 2977.0 kJ/kg

Solution Approach for f)#

\(s=S/m\)

where \(s\) is the specific entropy \(S\) is the total Entropy

s = CP.PropsSI("S", "T", T, "D", D , fluid)   #specific entropy in J/kgK

print('specific entropy of steam is:', f"{s/1000:.1f}", 'kJ/kgK')

specific entropy of steam is: 8.7 kJ/kgK

Solution Approach for g)#

\(u=U/m\)

so

\(U=m\times U\)

U = m * u   #total internal energy in J

print('total internal energy of steam is:', f"{U/1000:.1f}", 'kJ')

total internal energy of steam is: 27.4 kJ

Solution Approach for h)#

\(h=H/m\)

so

\(H=m\times h\)

H = m * h   #total enthalpy in J

print('total enthalpy of steam is:', f"{H/1000:.1f}", 'kJ')

total enthalpy of steam is: 29.8 kJ

Solution Approach for i)#

\(s=S/m\)

so

\(S=m\times s\)

S = m * s   #total entropy in J/K

print('total entropy of steam is:', f"{S/1000:.3f}", 'kJ/K')

total entropy of steam is: 0.087 kJ/K
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