Chapter 6 : Head Engines - The Second Law of Thermodynamics

Example 6.1 Page No : 148

			
# Variables
Q1 = 300. 			#kJ 			#heat supplied at the boiler
Wt = 100. 			#kJ 			#work output of turbine
Wp = 0.5 			#kJ 			#work input to pump

			
# Calculations and Results
Q2 = Q1 - (Wt - Wp) 			#kJ 			#heat rejected at the condensor
print "Heat rejected at the condensor = %.1f kJ"%(Q2);
efficiency = 1 - (Q2/Q1)
print "The thermal efficiency of plant = %.2f "%(efficiency)

Heat rejected at the condensor = 200.5 kJ
The thermal efficiency of plant = 0.33 

Example 6.2 Page No : 153

			
# Variables
COP_ref = 4 			#COP of refrigerator
Q1 = 0.5 			#kJ/s 			#rate of heat transfer at the condensor

			
# Calculations and Results
#Part(a)
print "Parta";
Wc = Q1/(COP_ref+1) 			#kJ/s 			#Power input to compressor
Q2 = COP_ref*Wc 			#kJ/s 			#Rate of heat transfer in evaporator
print "Rate of heat transfer in evaporator  = %.1f kJ/s"%(Q2)
print "Power input to compressor = %.1f kJ/s"%(Wc)

#Part(b)
print "Partb";
COP_hp = 1 + COP_ref 			#COP of heat pump
print "COP of heat pump = %.0f"%(COP_hp)

Parta
Rate of heat transfer in evaporator  = 0.4 kJ/s
Power input to compressor = 0.1 kJ/s
Partb
COP of heat pump = 5

Example 6.3 Page No : 154

#Part(a)
print "Part a";
I = 4.5 			#Amp 			#Current drawn
V = 220 			#V
Electricity_consumption = I*V 			#Watts
ElectricityUnitPerDay = Electricity_consumption/1000*8 			#kWh
MonthlyBill_part_a = ElectricityUnitPerDay * 5 * 30
print "The additional monthly electricity bill = Rs. %.2f"%(MonthlyBill_part_a);

#Part(b)
print "Part b";
Q1 = Electricity_consumption*.001 			#kW 			#Rate of heat transfer from heat pump
COP_hp = 4 			#COP of heat pump
W = Q1/COP_hp 			#kW 			#rate at which energy is consumed
ElectricityUnitPerDay = W*8
MonthlyBill_part_b = ElectricityUnitPerDay * 5 * 30
print "Saving in monthly electricity bill = Rs. %.2f"%(MonthlyBill_part_a - MonthlyBill_part_b)

Part a
The additional monthly electricity bill = Rs. 1188.00
Part b
Saving in monthly electricity bill = Rs. 891.00