# Chapter 2: Energy Conversion and General Energy Analysis¶

## Example 2-1 ,Page No.57¶

In :
#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 :
# 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 :
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 :
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 :
#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 :
#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 :
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 :
#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 :
#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 $504  ## Example 2-15 ,Page No.82¶ In : #Given values Ein=2.0;#Power of electric burner in kW n1=0.73;#Efficiency of open burners n2=0.38;#efficency of gas units CinH=0.09;#Unit cost of electricity in$
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 $0.123000/kWh rate of energy consumption by the burner 3.840000 kW the unit cost of utilized energy for burner 0.049000 kWh  ## Example 2-16 ,Page No.84¶ In : #there is a 0.204% error in the last part of the question due to rounding off the intermidiate steps in the solution #Constants used g=9.81;#acceleration due to gravity in m/s^2; #Given values h=50.0;#Depth of water in m m=5000.0;#mass flow rate of water in kg/sec Wout=1862.0;#generated electric power in kW ngen=0.95;#efficiency of turbine #calculation X=g*h/1000.0;# X stands for the differnce b/w change in mechanical energy per unit mass R=m*X;#rate at which mech. energy is supplied to turbine in kW nov=Wout/R;#overall efficiency i.e turbine and generator print'overall efficiency is %f'%round(nov,2); ntu=nov/ngen;#efficiency of turbine print'efficiency of turbine is %f'%round(ntu,2); Wsh=ntu*R;#shaft output work print'shaft power output %i kW'%round(Wsh,0)  overall efficiency is 0.760000 efficiency of turbine is 0.800000 shaft power output 1960 kW  ## Example 2-17 ,Page No.85¶ In : #Given values Pstd=4520.0; Phem=5160.0;#prices of std and high eff motor in USD R=60*0.7457;#rated power in kW from hp OpHrs=3500.0;#Operating hours Lf=1.0;#Load Factor nsh=0.89;#efficiency of shaft nhem=0.932;#efficiency of high eff. motor CU=0.08;#per unit cost in$

#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 :
#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 :
#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