# CHAPTER 1.2: THERMAL STATIONS¶

## Example 2.1, Page number 25-26¶

In [1]:
#Variable declaration
M = 15000.0+10.0        #Water evaporated(kg)
C = 5000.0+5.0          #Coal consumption(kg)
time = 8.0              #Generation shift time(hours)

#Calculation
#Case(a)
M1 = M-15000.0
C1 = C-5000.0
M_C = M1/C1                                       #Limiting value of water evaporation(kg)
#Case(b)
kWh = 0                                           #Station output at no load
consumption_noload = 5000+5*kWh                   #Coal consumption at no load(kg)
consumption_noload_hr = consumption_noload/time   #Coal consumption per hour(kg)

#Result
print('Case(a): Limiting value of water evaporation , M/C = %.1f kg' %M_C)
print('Case(b): Coal per hour for running station at no load = %.f kg' %consumption_noload_hr)

Case(a): Limiting value of water evaporation , M/C = 2.0 kg
Case(b): Coal per hour for running station at no load = 625 kg


## Example 2.2, Page number 26¶

In [1]:
#Variable declaration
amount = 25.0*10**5          #Amount spent in 1 year(Rs)
value_heat = 5000.0          #Heating value(kcal/kg)
cost = 500.0                 #Cost of coal per ton(Rs)
n_ther = 0.35                #Thermal efficiency
n_elec = 0.9                 #Electrical efficiency

#Calculation
n = n_ther*n_elec                         #Overall efficiency
consumption = amount/cost*1000            #Coal consumption in 1 year(kg)
combustion = consumption*value_heat       #Heat of combustion(kcal)
output = n*combustion                     #Heat output(kcal)
unit_gen = output/860.0                   #Annual heat generated(kWh). 1 kWh = 860 kcal
hours_year = 365*24.0                     #Total time in a year(hour)
load_average = unit_gen/hours_year        #Average load on the power plant(kW)

#Result
print('Average load on power plant = %.2f kW' %load_average)
print('\nNOTE: ERROR: Calculation mistake in the final answer in textbook')

Average load on power plant = 1045.32 kW

NOTE: ERROR: Calculation mistake in the final answer in textbook


## Example 2.3, Page number 26¶

In [1]:
#Variable declaration
consumption = 0.5      #Coal consumption per kWh output(kg)
cal_value = 5000.0     #Calorific value(kcal/kg)
n_boiler = 0.8         #Boiler efficiency
n_elec = 0.9           #Electrical efficiency

#Calculation
input_heat = consumption*cal_value                 #Heat input(kcal)
input_elec = input_heat/860.0                      #Equivalent electrical energy(kWh). 1 kWh = 860 kcal
loss_boiler = input_elec*(1-n_boiler)              #Boiler loss(kWh)
input_steam = input_elec-loss_boiler               #Heat input to steam(kWh)
input_alter = 1/n_elec                             #Alternator input(kWh)
loss_alter = input_alter*(1-n_elec)                #Alternate loss(kWh)
loss_turbine = input_steam-input_alter             #Loss in turbine(kWh)
loss_total = loss_boiler+loss_alter+loss_turbine   #Total loss(kWh)
output = 1.0                                       #Output(kWh)
Input = output+loss_total                          #Input(kWh)

#Result
print('Heat Balance Sheet')
print('LOSSES:  Boiler loss      = %.3f kWh' %loss_boiler)
print('         Alternator loss  = %.2f kWh' %loss_alter)
print('         Turbine loss     = %.3f kWh' %loss_turbine)
print('         Total loss       = %.2f kWh' %loss_total)
print('OUTPUT:  %.1f kWh' %output)
print('INPUT:   %.2f kWh' %Input)

Heat Balance Sheet
LOSSES:  Boiler loss      = 0.581 kWh
Alternator loss  = 0.11 kWh
Turbine loss     = 1.214 kWh
Total loss       = 1.91 kWh
OUTPUT:  1.0 kWh
INPUT:   2.91 kWh