Chapter 2: Energy Conversion and General Energy Analysis

Example 2-1 ,Page No.57

In [2]:
#there is a 0.00490% error as the answer in textbook is expressed in multiple of 10
#Constants used
Hu=6.73*10**10;#Energy liberated by 1 kg of uranium

# Given values
p=0.75;# assuming the avg density of gasoline in kg/L
V=5;# consumption per day of gasoline in L
Hv=44000; #heat value in kJ/kg
mu=0.1;# mass of uranium used

#Calculation
mgas=p*V;#mass of gasoline required per day
Egas=mgas*Hv;
Eu=mu*Hu;
d=Eu/Egas;
print'%i number of days the car can run with uranium' %round(d,0)
print'equivalent to %i years' %round(d/365,0)
40788 number of days the car can run with uranium
equivalent to 112 years

Example 2-2 ,Page No.59

In [2]:
# Given values
v=8.5;# wind speed in m/s
m=10;# given mass for part - b 
mf=1154;# given flowrate for part - c

#Calculations
e=(v**2)/2;
print'wind energy per unit mass %f J/kg' %round(e,1);
E=m*e;
print'wind energy for 10 kg mass %i J' %E;
E=mf*e/1000;
print'wind energy for mass flow rate of 1154kg/s %f kW'%round(E,1)
wind energy per unit mass 36.100000 J/kg
wind energy for 10 kg mass 361 J
wind energy for mass flow rate of 1154kg/s 41.700000 kW

Example 2-7 ,Page No.67

In [3]:
import math

# Given values
T=200;# applied torque in N
n=4000;# shaft rotation rate in revolutions per minute

#Calculation
Wsh=(2*math.pi*n*T)/1000/60;#factor of 1000 to convert to kW and 60 to convert to sec
print'Power transmitted %f kW'%round(Wsh,1)
Power transmitted 83.800000 kW

Example 2-8 ,Page No.69

In [4]:
import math

#Constants used
g=9.81;#acceleration due to gravity in m/s^2;

#Given values
m=1200;#mass of car in kg
V=90/3.6;#velocity ; converting km/h into m/s
d=30*math.pi/180;#angle of slope ; converting into radians

#Calculation
Vver=V*math.sin(d);#velocity in vertical direction
Wg=m*g*Vver/1000;
print'the addtional power %i kW'%Wg
the addtional power 147 kW

Example 2-9 ,Page No.69

In [21]:
#Given values
m=900;#mass of car in kg
v1=0;# intial velocity
v2=80/3.6;# final velocity; converting km/h into m/s
t=20;# time taken1

#Calculation
Wa=m*(v2**2-v1**2)/2/1000;
Wavg=Wa/t;
print'the average power %f kW'%round(Wavg,1)
the average power 11.100000 kW

Example 2-10 ,Page No.74

In [7]:
#Given values
Win=100;# work done in the process in kJ
Qout=500;# heat lost in kJ
U1=800;# internal energy of the fluid in kJ

#Calculations
# Win - Qout = U2- U1 i.e change in internal energy 
U2=U1-Qout+Win;
print'final internal of the system %i kJ'%U2
final internal of the system 400 kJ

Example 2-11 ,Page No.75

In [8]:
import math

#given values
Win=20;# power consumption in W
mair=0.25;# rate of air discharge in kg/sec

#calculation
v=math.sqrt(Win/2/mair)#Win = 1/2*m*v^2
if v >=8:
    print('True');
else:
    print('False')
False

Example 2-12 ,Page No.76

In [1]:
#Given values
Win=200.0;#Power of fan in W
U=6.0;#Overall heat transfer coefficient in W/m^2 C
A=30;#Surface area in m^2
To=25;#Outdoor temperature in C

#Calculations
Ti= (Win/U/A)+To;# Win = Qout = U*A*(Ti - To)
print'the indoor air temperature %f Celcius'%Ti
the indoor air temperature 26.111111 Celcius

Example 2-13 ,Page No.76

In [2]:
#Given values
Plamp=80.0;#Power of lamp in W
N=30;#no of lamps
t=12;#time period the light is in use in hours/day
y=250;#days in a year light is in function 
UC=0.07;#unit cost in $

#Calculation
LP=Plamp*N/1000;#Lighting power in kW
OpHrs=t*y;#Operating hours
LE=LP*OpHrs;#Lighting energy in kW
LC=LE*UC;#Lighting cost
print'the annual energy cost $%i'%LC
the annual energy cost $504

Example 2-15 ,Page No.82

In [3]:
#Given values
Ein=2.0;#Power of electric burner in kW
n1=0.73;#Efficiency of open burners
n2=0.38;#efficency of gas units
CinH=0.09;#Unit cost of electricity in $
CinB=0.55;#Unit cost of natural gas in $

#Calculations
QutH= Ein * n1;
print'rate of energy consumption by the heater %f kW'%round(QutH,2);
CutH= CinH / n1;
print'the unit cost of utilized energy for heater $%f/kWh'%round(CutH,3);
QutB= QutH / n2 ;
print'rate of energy consumption by the burner %f kW'%round(QutB,2);
CutB= CinB / n2 / 29.3; # 1 therm = 29.3 kWh
print'the unit cost of utilized energy for burner %f kWh'%round(CutB,3);
rate of energy consumption by the heater 1.460000 kW
the unit cost of utilized energy for heater $0.123000/kWh
rate of energy consumption by the burner 3.840000 kW
the unit cost of utilized energy for burner 0.049000 kWh

Example 2-16 ,Page No.84

In [12]:
#there is a 0.204% error in the last part of the question due to rounding off the intermidiate steps in the solution

#Constants used
g=9.81;#acceleration due to gravity in m/s^2;

#Given values
h=50.0;#Depth of water in m
m=5000.0;#mass flow rate of water in kg/sec
Wout=1862.0;#generated electric power in kW
ngen=0.95;#efficiency of turbine

#calculation
X=g*h/1000.0;# X stands for the differnce b/w change in mechanical energy per unit mass
R=m*X;#rate at which mech. energy is supplied to turbine in kW
nov=Wout/R;#overall efficiency i.e turbine and generator
print'overall efficiency is %f'%round(nov,2);
ntu=nov/ngen;#efficiency of turbine
print'efficiency of turbine is %f'%round(ntu,2);
Wsh=ntu*R;#shaft output work
print'shaft power output %i kW'%round(Wsh,0)
overall efficiency is 0.760000
efficiency of turbine is 0.800000
shaft power output 1960 kW

Example 2-17 ,Page No.85

In [5]:
#Given values
Pstd=4520.0;
Phem=5160.0;#prices of std and high eff motor in USD
R=60*0.7457;#rated power in kW from hp
OpHrs=3500.0;#Operating hours
Lf=1.0;#Load Factor
nsh=0.89;#efficiency of shaft
nhem=0.932;#efficiency of high eff. motor
CU=0.08;#per unit cost in $

#calculation
PS=R*Lf*(1/nsh-1/nhem);#Power savings = W electric in,standard -   W electric in,efficient
ES=PS*OpHrs;#Energy savings = Power savings * Operating hours
print'Energy savings %i kWh/year'%ES;
CS=ES*CU;
print'Cost savings per year $%i'%CS;
EIC=Phem-Pstd;#excess intial cost
Y=EIC/CS;
print'simple payback period %f years'%round(Y,1)
Energy savings 7929 kWh/year
Cost savings per year $634
simple payback period 1.000000 years

Example 2-18 ,Page No.91

In [6]:
#Given values
#NOx details
m1=0.0047;#emissions of gas furnaces of NOx in kg/therm
N1=18*10**6;#no. of therms per year 
#CO2 details
m2=6.4;#emissions of gas furnaces of CO2 in kg/therm
N2=18*10**6;#no. of therms per year 

#Calculation
NOxSav=m1*N1;
print'NOx savings %f kg/year'%round(NOxSav,1);
CO2Sav=m2*N2;
print'CO2 savings %f kg/year'%round(CO2Sav,1)
NOx savings 84600.000000 kg/year
CO2 savings 115200000.000000 kg/year

Example 2-19 ,Page No.95

In [7]:
#Constants used
e=0.95;#Emissivity
tc=5.67*10**-8;#thermal conductivity in W/m^2 K^4

#Given values
h=6;#convection heat transfer coefficient in W/m^2 C
A=1.6;#cross-sectional area in m^2
Ts=29;#average surface temperature in C
Tf=20;#room temperature in C

#Calculation
#convection rate
Q1=h*A*(Ts-Tf);
#radiation rate
Q2=e*tc*A*((Ts+273)**4-(Tf+273)**4);
Qt=Q1+Q2;
print'the total rate of heat transfer %f W'%round(Qt,1)
the total rate of heat transfer 168.100000 W