# Chapter 1 - General Introduction¶

## Example 1: pg 11¶

In :
#pg 11
#calculate the work done
print 'Example 1.1'

# Given values
P = 700.;   #pressure,[kN/m**2]
V1 = .28;   #initial volume,[m**3]
V2 = 1.68;   #final volume,[m**3]

#solution

W = P*(V2-V1);# # Formula for work done at constant pressure is, [kJ]

#results
print 'The Work done is (MJ) = ',W*10**-3

#End

Example 1.1
The Work done is (MJ) =  0.98


## Example 2: pg 13¶

In :
#pg 13
#calculate the new volume
print 'Example 1.2'

#Given values
P1 = 138.;  # initial pressure,[kN/m**2]
V1 = .112;  #initial volume,[m**3]
P2 = 690; #  final pressure,[kN/m**2]
Gama=1.4; #  heat capacity ratio

#  solution

#  since gas is following, PV**1.4=constant,hence

V2 =V1*(P1/P2)**(1/Gama); #   final volume, [m**3]

#results
print 'The new volume of the gas is (m^3) = ',round(V2,4)

#End

Example 1.2
The new volume of the gas is (m^3) =  0.0355


## Example 3: pg 15¶

In :
#pg 15
#calculate the work done by gas
print 'Example 1.3'

# Given values
P1 = 2070;  #  initial pressure,  [kN/m^2]
V1 = .014;  #  initial volume,  [m^3]
P2 = 207.;  #  final pressure, [kN/m^2]
n=1.35;  #  polytropic index

#  solution

#  since gas is following PV^n=constant
#  hence

V2 = V1*(P1/P2)**(1/n);  # final volume,  [m^3]

#  calculation of workdone

W=(P1*V1-P2*V2)/(1.35-1);  # using work done formula for polytropic process, [kJ]

#results
print 'Final Volume (m^3) = ',round(V2,3)
print 'The Work done by gas during expansion is (kJ) = ',round(W,1)

#End

Example 1.3
Final Volume (m^3) =  0.077
The Work done by gas during expansion is (kJ) =  37.2


## Example 4: pg 17¶

In :
#pg 17
#calculate the final pressure and work done
print 'Example 1.4'
import math

#  Given values
P1 = 100;  #  initial pressure, [kN/m^2]
V1 = .056;  #  initial volume,  [m^3]
V2 = .007;  #  final volume,  [m^3]

#  To know  P2
#  since process is hyperbolic so, PV=constant
#  hence

P2 = P1*V1/V2;  #  final pressure, [kN/m^2]

# calculation of workdone
W = P1*V1*math.log(V2/V1); #  formula for work done in this process, [kJ]

#results
print ' The final pressure (kN/m^2) = ',P2
print ' Work done on the gas (kJ) = ',round(W,2)

#End

Example 1.4
The final pressure (kN/m^2) =  800.0
Work done on the gas (kJ) =  -11.64


## Example 5: pg 21¶

In :
#pg 21
#calculate the heat required
print 'Example 1.5'

#  Given values
m = 5.; #  mass,  [kg]
t1 = 15.; #  inital temperature, [C]
t2 = 100.; #  final temperature, [C]
c = 450.; #  specific heat capacity, [J/kg K]

#  solution

#  using heat transfer equation,
Q = m*c*(t2-t1); #  [J]
#results
print 'The heat required (kJ) = ',round(Q*10**-3,2)

#End

Example 1.5
The heat required (kJ) =  191.25


## Example 6: pg 22¶

In :
#pg 22
print 'Example 1.6'

#Calculate the required heat transfer
#  Given values
m_cop = 2.; #  mass of copper vessel, [kg]
m_wat = 6.; #  mass of water, [kg]
c_wat = 4.19; #  specific heat capacity of water,  [kJ/kg K]

t1 = 20.; #  initial temperature, [C]
t2 = 90.; #  final temperature, [C]

# From the table of average specific heat capacities
c_cop = .390;  # specific heat capacity of copper,[kJ/kg k]

# solution
Q_cop = m_cop*c_cop*(t2-t1); #  heat required by copper vessel, [kJ]

Q_wat = m_wat*c_wat*(t2-t1); #  heat required by water, [kJ]

# since there is no heat loss,so total heat transfer is sum of both
Q_total = Q_cop+Q_wat ; #  [kJ]

#results
print 'Required heat transfer to accomplish the change (kJ) = ',Q_total

#End

Example 1.6
Required heat transfer to accomplish the change (kJ) =  1814.4


## Example 7: pg 22¶

In :
#pg 22
print('Example 1.7');
#calculate the final temperature

#  Given values
m = 10.; #  mass of iron casting, [kg]
t1 = 200.; #  initial temperature, [C]
Q = -715.5; #  [kJ], since heat is lost in this process

# From the table of average specific heat capacities
c = .50; #  specific heat capacity of casting iron, [kJ/kg K]

#  solution
#  using heat equation
#  Q = m*c*(t2-t1)

t2 = t1+Q/(m*c); # [C]

#results
print 'The final temperature is  (C) = ',t2

# End

Example 1.7
The final temperature is  (C) =  56.9


## Example 8: pg 23¶

In :
#pg 23
#calculate the specific heat capacity
print('Example 1.8');
# Given values
m = 4.; #  mass of the liquid, [kg]
t1 = 15.; #  initial temperature, [C]
t2 = 100.; #  final temperature, [C]
Q = 714.; #  [kJ],required heat to accomplish this change

#  solution
#  using heat equation
#  Q=m*c*(t2-t1)

#  calculation of c
c=Q/(m*(t2-t1)); #  heat capacity, [kJ/kg K]

#results
print 'The specific heat capacity of the liquid is  (kJ/kg K) = ',c

#End

Example 1.8
The specific heat capacity of the liquid is  (kJ/kg K) =  2.1


## Example 9: pg 27¶

In :
#pg 27
#calculate the energy rejected by the engine
print('Example 1.9');

#  Given values
m_dot = 20.4; #  mass flowrate of petrol, [kg/h]
c = 43.; #  calorific  value of petrol, [MJ/kg]
n = .2; #  Thermal efficiency of engine

#  solution
m_dot = 20.4/3600; # [kg/s]
c = 43*10**6; #  [J/kg]

#  power output
P_out = n*m_dot*c; #  [W]

#  power rejected

P_rej = m_dot*c*(1-n); #  [W]
P_rej = P_rej*60*10**-6; #  [MJ/min]

#results
print 'The power output of the engine is  (kJ) = ',round(P_out*10**-3,1)
print 'The energy rejected by the engine is (MJ/min) = ',round(P_rej,1)

#End

Example 1.9
The power output of the engine is  (kJ) =  48.7
The energy rejected by the engine is (MJ/min) =  11.7


## Example 10: pg 28¶

In :
#pg 28
print('Example 1.10');
#calculate the thermal efficiency

#  Given values
m_dot = 3.045; #  use of coal, [tonne/h]
c = 28; # calorific value of the coal, [MJ/kg]
P_out = 4.1; #  output of turbine, [MW]

#  solution
m_dot = m_dot*10**3/3600; # [kg/s]

P_in = m_dot*c; #  power input by coal, [MW]

n = P_out/P_in; #  thermal efficiency formula

#results
print 'Thermal efficiency of the plant = ',round(n,3)

#End

Example 1.10
Thermal efficiency of the plant =  0.173


## Example 11: pg 29¶

In :
#pg 29
#calculate the power output of the engine
print('Example 1.11');

#  Given values
v = 50.; #  speed, [km/h]
F = 900.; #  Resistance to the motion of a car

#  solution
v = v*10**3/3600; #  [m/s]
Power = F*v; #  Power formula, [W]

print 'The power output of the engine (kW) = ',Power*10**-3

#  End

Example 1.11
The power output of the engine (kW) =  12.5


## Example 12: pg 31¶

In :
#pg 31
#calculate the power output from the engine

print('Example 1.12');

#  Given values
V = 230.; #  volatage, [volts]
I = 60.; #  current, [amps]
n_gen = .95; #  efficiency of generator
n_eng = .92; #  efficiency of engine

#  solution

P_gen = V*I; #  Power delivered by generator, [W]
P_gen=P_gen*10**-3; #  [kW]

P_in_eng=P_gen/n_gen;#Power input from engine,[kW]

P_out_eng=P_in_eng/n_eng;#Power output from engine,[kW]

#results
print 'The power output from the engine (kW) = ',round(P_out_eng,2)

#  End

Example 1.12
The power output from the engine (kW) =  15.79


## Example 13: pg 32¶

In :
#pg 32
#calculate the current taken by heater
print('Example 1.13');

#  Given values
V = 230.; # Voltage, [volts]
W = 4.; #  Power of heater, [kW]

#  solution

#  using equation P=VI
I = W/V; #  current, [K amps]
#results
print 'The current taken by heater (amps) = ',round(I*10**3,1)

#  End

Example 1.13
The current taken by heater (amps) =  17.4


## Example 14: pg 32¶

In :
#pg 32
#calculate the mass of coal burnt
print('Example 1.14');

#  Given values
P_out = 500.; #  output of power station, [MW]
c = 29.5; #  calorific value of coal, [MJ/kg]
r=.28;

#  solution

#  since P represents only 28 percent of energy available from coal
P_coal = P_out/r; #  [MW]

m_coal = P_coal/c; #  Mass of coal used, [kg/s]
m_coal = m_coal*3600; #  [kg/h]

#After one hour
m_coal = m_coal*1*10**-3; #  [tonne]
#results
print 'Mass of coal burnt by the power station in 1 hour (tonne) = ',round(m_coal,0)

#  End

Example 1.14
Mass of coal burnt by the power station in 1 hour (tonne) =  218.0