# Chapter 27 : Ideal Processes Efficiency and the Mechanical Energy Balance¶

### Example 27.1 page no. 838¶

In [5]:
# Variables
V_w = 1. ;			#  Volume of given water -[L]
P_atm = 100. ;			# Atmospheric pressure - [kPa]

#W = -p*del_V
V_H2O = 0.001043 ;			# Specific volume of water from steam table according to book- [cubic metre]
V_vap = 1.694 ;			# Specific volume of vapour from steam table according to book- [cubic metre]
V1 = 0 ;			# Initial volume of H2O in bag-[cubic metre]

# Calculations
V2 = (V_w*V_vap)/(1000*V_H2O) ;			# Final volume of water vapour -[cubic metre]
W = -P_atm*(V2 -V1)* 1000 ;			# Work done by saturated liquid water -[J]

# Results
print ' Work done by saturated liquid water is %.3e J.'%W
Work done by saturated liquid water is -1.624e+05 J.

### Example 27.2 page no. 840¶

In [6]:
# Variables
m_N2 = 1. ;			# Moles of N2 taken -[kg mol]
p = 1000.;			# Pressure of cylinder-[kPa]
T = 20. + 273. ;			# Temperature of cylinder -[K]
a_pis = 6. ;			# Area of piston - [square centimetre]
m_pis = 2. ;			# Mass of pston - [kg]
R = 8.31 ;			# Ideal gas constant - [(kPa*cubic metre)/(K * kgmol)]

# Calculations
V = (R*T)/p ;			# Specific volue of gas at initial stage -[cubic metre/kg mol]
V1 = V * m_N2 ;			# Initial volume of gas - [cubic metre]
V2 = 2.*V1 ;			# Final volume of gas according to given condition -[cubic metre]

# Assumed surrounding pressure constant = 1 atm
p_atm = 101.3 ;			# Atmospheric pressure-[kPa]
del_Vsys = V2 -V1 ;			# Change in volume of system -[cubic metre]
del_Vsurr = - del_Vsys ;			# Change in volume of surrounding -[cubic metre]
W_surr = -p_atm*del_Vsurr ;			# Work done by surrounding - [kJ]
W_sys = -W_surr ;			# Work done by system - [kJ]

# Results
print ' Work done by gas(actually gas + piston system) is %.0f kJ.'%W_sys
Work done by gas(actually gas + piston system) is -247 kJ.

### Example 27.3 page no. 845¶

In [7]:
# Variables
p_plant = 20. ;			# Power generated by plant-[MW]
h = 25. ;			# Height of water level - [m]
V = 100. ;			# Flow rate of water -[cubic metre/s]
d_water = 1000. ;			# Density of water - [ 1000 kg / cubic metre]
g = 9.807 ;			# Acceleration due to gravity-[m/square second]

# Calculations
M_flow = V*d_water ;			# Mass flow rate of water -[kg/s]
del_PE = M_flow*g*h ;			# Potential energy change of water per second -[W]
eff = (p_plant*10**6) /(del_PE) ;			# Efficiency of plant

# Results
print ' Efficiency of plant is %.2f .'%eff
Efficiency of plant is 0.82 .

### Example 27.4 page no. 845¶

In [8]:
# Variables
LHV = 36654. ;			# LHV value of fuel - [kJ/ cubic metre]
Q1 = 16. ;			#- [kJ/ cubic metre]
Q2 = 0 ;			#- [kJ/ cubic metre]
Q3 = 2432. ;			#- [kJ/ cubic metre]
Q4 = 32114. ;			#- [kJ/ cubic metre]
Q41 = 6988. ;			#- [kJ/ cubic metre]
Q8 = 1948. ;			#- [kJ/ cubic metre]
Q9 = 2643. ;			#- [kJ/ cubic metre]
Q81 = 2352. - Q8 ;			# - [kJ/ cubic metre]
Q567 = 9092. ;			# Sum of Q5, Q6 and Q7- [kJ/ cubic metre]

# Calculations and Results
#(a)
G_ef = (LHV+ Q1 +Q2 + Q3 - Q9)/(LHV) ;			# Gross efficiency
print '(a) Gross efficiency is %.3f .'%G_ef

#(b)
T_ef = (Q567+Q8)/(LHV+ Q1 +Q2 + Q3) ;			#Thermal efficiency
print ' (b) Thermal efficiency is %.3f .'%T_ef

#(c)
C_ef = Q4/(Q4 + Q41) ;			# Combustion efficiency
print ' (c) Combustion efficiency is %.3f .'%C_ef
(a) Gross efficiency is 0.995 .
(b) Thermal efficiency is 0.282 .
(c) Combustion efficiency is 0.821 .

### Example 27.5 page no. 850¶

In [1]:

# Variables
V1 = 5. ;			# Volume of gas initially - [cubic feet]
P1 = 1. ;			# Initial pressure - [atm]
P2 = 10. ;			# Final pressure - [atm]
T1 = 100. + 460 ;			# initial temperature - [degree Rankine]
R = 0.7302 ;			# Ideal gas constant -[(cubic feet*atm)/(lb mol)*(R)]
#Equation of state pV**1.4 = constant

# Calculations and Results
V2 = V1*(P1/P2)**(1/1.4) ;			# Final volume - [cubic feet]

def f(V):
return -(P1)*(V1/V)**(1.4)

W1_rev = quad(f,V1,V2)[0] ;			# Reversible work done in compresion in a horizontal cylinder with piston -[cubic feet *atm]
W1 = W1_rev *1.987/.7302 ;			# Conversion to Btu -[Btu]
print '(a)Reversible work done in compression in a horizontal cylinder with piston is %.1f Btu . '%W1

#(b)
n1 = (P1*V1)/(R*T1) ;			# Number of moles of gas

def f1(P):
return (V1)*(P1/P)**(1/1.4)
W2_rev = quad(f1,P1,P2)[0]		# Reversible work done in compresion in a rotary compressor -[cubic feet *atm]
W2 = W2_rev *1.987/.7302 ;			# Conversion to Btu -[Btu]

print '(b)Reversible work done in a rotary compressor is %.1f Btu . '%W2
(a)Reversible work done in compression in a horizontal cylinder with piston is 31.7 Btu .
(b)Reversible work done in a rotary compressor is 44.3 Btu .

### Example 27.6 page no. 853¶

In [1]:
# Variables
m_water = 1. ;			# Mass flow rate of water -[lb/min]
P1 = 100. ;			# Initial pressure - [psia]
P2 = 1000. ;			# Final pressure - [psia]
T1 = 80. + 460 ;			# initial temperature - [degree Rankine]
T2 = 100. + 460 ;			# final temperature - [degree Rankine]
h = 10. ;			# Difference in water level between entry and exit of stream-[ft]
g = 32.2 ;			# Accleration due to gravity - [ft/ square second]
gc = 32.2 ;			#[(ft*lbm)/(lbf*square second)]

v1 = .01607 ;			# specific volume of liquid water at 80 degree F -[cubic feet/lbm]
v2 = .01613 ;			# specific volume of liquid water at 100 degree F -[cubic feet/lbm]
v= 0.0161 ;			# -[cubic feet/lbm]

# Calculations
del_PE = (h*g)/(gc*778) ;			# Change in potential energy - [Btu/lbm]

def f(P):
return (v)*(12**2/778.)

PV_work = quad(f,P1,P2)[0]			# PV work done  -[Btu/lbm]
#From eqn. (A)
W = PV_work + del_PE ;			# Work per minute required to pump 1 lb water per minute - [Btu/lbm]

# Results
print ' Work per minute required to pump 1 lb water per minute is %.2f Btu/lbm . '%W
Work per minute required to pump 1 lb water per minute is 2.69 Btu/lbm .
In [2]: