# Chapter 22 : Introduction to Energy Balances for Process without Reaction¶

### Example 22.1 page no. 651¶

In :
# Variables
#Assume that properties of water can be used to substitute properties of solution
# Given
V = 1.673 ;			# Volume of closed vessel-[cubic metre]
m = 1. ;			# mass of saturated liquid vaporized-[kg]
Pi = 1. ;			# Initial pressure -[atm]
Ti = 10. ;			# Initial temperature -[degree C]
Pf = 1. ;			# final pressure -[atm]
Tf = 100. ;			# final temperature -[degree C]

# Use steam table to obtain additional information at given condition
Ui = 35. ;			# Initial enthalpy-[kJ/kg]
Uf = 2506.0 ;			# Final enthalpy -[kJ/kg]

# Calculations
# Use eqn. 22.2 after modifiying it using given conditions(W = 0,del_KE = 0 and del_PE = 0 )
Q = m*(Uf - Ui) ;			# Heat transferred to  the vessel - [kJ]

# Results
print 'Heat transferred to  the vessel is %.1f kJ . '%Q

Heat transferred to  the vessel is 2471.0 kJ .


### Example 22.2 page no. 652¶

In :
# Variables
# Given
T1 = 80. ;			# Initial temperature -[degree F]
T1 = 40. ;			# final temperature -[degree F]

# Additional data obtained from steam table at given temperatures and corresponding vapour pressures
p1 = 0.5067 ;			# Initial saturation pressure-[psia]
p2 = 0.1217 ;			# Final saturation pressure-[psia]
V1 = 0.01607 ;			# Initial specific volume - [cubic feet/lb]
V2 = 0.01602 ;			# Final specific volume - [cubic feet/lb]
H1 = 48.02 ;			#Initial specific enthalpy -[Btu/lb]
H2 = 8.05 ;			# Final specific  enthalpy -[Btu/lb]

# Calculations
del_P = p2 - p1 ;			# Change in pressure -[psia]
del_V = V2 - V1 ;			# Change in specific volume -[cubic feet/lb]
del_H = H2 - H1 ;			# Change in specific enthalpy -[Btu/lb]
del_pV = p2*144*V2/778. - p1*144*V1/778. ;			# Change in pv-[Btu]
del_U = del_H - del_pV ;			# Change in specific internal energy - [Btu/lb]
del_E = del_U ;			# Change in specific total energy(since KE=0,PE=0 and W=0) -[Btu/lb]

# Results
print 'Change in pressure is %.3f psia . '%del_P
print 'Change in specific volume is %.5f cubic feet/lb (negligible value) . '%del_V
print 'Change in specific enthalpy is %.2f Btu/lb . '%del_H
print 'Change in specific internal energy is %.2f Btu/lb . '%del_U
print 'Change in specific total energy is %.2f Btu/lb . '%del_E

Change in pressure is -0.385 psia .
Change in specific volume is -0.00005 cubic feet/lb (negligible value) .
Change in specific enthalpy is -39.97 Btu/lb .
Change in specific internal energy is -39.97 Btu/lb .
Change in specific total energy is -39.97 Btu/lb .


### Example 22.3 page no. 662¶

In :
# Variables
#Lets take tank to be system
# Given
T = 600. ; 			# Temperature of steam  -[K]
P = 1000. ;			# Pressure of steam -[kPa]

# Calculations
# Additional data for steam obtained from CD database at T and P
U = 2837.73 ;			# Specific internal energy-[kJ/kg]
H = 3109.44 ;			# Specific enthalpy -[kJ/kg]
V = 0.271 ;			# Specific volume -[cubic metre/kg]
# By the reduced equation
Ut2 = H ;			# Internal energy at final temperature-[kJ/kg]

# Results
print 'The specific internal energy at final temperature is %.2f kJ/kg. Now use two properties\
of the steam (P = %i kPa and Ut2 = %.2f kJ/kg) to find final temperature (T) from steam table. \
From steam table we get T = 764 K.'%(Ut2,P,Ut2)

The specific internal energy at final temperature is 3109.44 kJ/kg. Now use two propertiesof the steam (P = 1000 kPa and Ut2 = 3109.44 kJ/kg) to find final temperature (T) from steam table. From steam table we get T = 764 K.


### Example 22.4 page no. 669¶

In :
# Variables
# Take milk plus water in tank to be system
# Given
T1_water = 70.  ;			# Temperature of entering water  -[degree C]
T2_water = 35. ;			# Temperature of exiting water  -[degree C]
T1_milk = 15. ;			#Temperature of entering milk  -[degree C]
T2_milk = 25. ;			#Temperature of exiting milk  -[degree C]

# Get additional data from steam table for water and milk,assuming milk to have same properties as that of water.
H_15 = 62.01 ;			#Change in specific internal energy-[kJ/kg]
H_25 = 103.86  ;			#Change in specific internal energy-[kJ/kg]
H_35 = 146.69  ;			#Change in specific internal energy-[kJ/kg]
H_70 = 293.10  ;			#Change in specific internal energy-[kJ/kg]

# Assumptions to simplify Equation 22.8 are:
print 'Assumptions to simplify Equation 22.8 are:'
print '1. Change in KE and PE of system = 0.'
print '2. Q = 0 ,because of way we picked the system,it is is well insulated.'
print '3. W = 0,work done by or on the system.'

# Calculations
#Basis m_milk = 1 kg/min , to directly get the answer  .
m_milk = 1 ;			# Mass flow rate of milk-[kg/min]
# By applying above assumtions eqn. 22.8 reduces to del_H = 0 .Using it get m_water-
m_water = (m_milk*(H_15 - H_25))/(H_35 - H_70) ; 			# Mass flow rate of water-[kg/min]
m_ratio = m_water/m_milk ;			# Mass flow rate of water per kg/min of milk-[kg/min]

# Results
print 'Mass flow rate of water per kg/min of milk is %.2f (kg water/min )/(kg milk/min).'%m_ratio

Assumptions to simplify Equation 22.8 are:
1. Change in KE and PE of system = 0.
2. Q = 0 ,because of way we picked the system,it is is well insulated.
3. W = 0,work done by or on the system.
Mass flow rate of water per kg/min of milk is 0.29 (kg water/min )/(kg milk/min).


### Example 22.5 page no. 670¶

In :
# Variables
# Take pipe between initial and final level of water
# Given
h_in = -20. ;			# Depth of water below ground-[ft]
h_out = 5. ;			# Height of water level above ground-[ft]
h = h_out - h_in ;			# Total height to which water is pumped-[ft]
V = 0.50 ;			# Volume flow rate of water - [cubic feet/s]
ef = 100.; 			# Efficiency of pump - [%]
g = 32.2; 			# Acceleration due to gravity -[ft/square second]
gc = 32.2 ;			#[(ft*lbm)/(second square*lbf)]

M = V * 62.4 ;			# mass flow rate - [lbm/s]
PE_in = 0 ;			# Treating initial water level to be reference level

# Calculations
PE_out = (M*g*h*1.055)/(gc*778.2) ;			# PE of discharged water -[lbm*(square feet/square second)]
W = PE_out - PE_in ;			#Work done on system = power delivered by pump, (since we are using mass flow rate and pump efficiency is 100 % , so W = Power) -[kW]

# Results
print 'The electric power required by the pump is %.2f kW. '%W

The electric power required by the pump is 1.06 kW.

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