# Chapter 6 : Second Law of Thermodynamics¶

## Example 6.1 Page No : 138¶

In :
# Variables
# In first case     (T1-T2)/T1=1/6           or    T1= 1.2*T2                  (i)
# In seond case  (T1-(T2-62))/T1= 2/6  or   2*T1 -3*(T2-62)=0      (ii)
# From eq (i) and (ii)
T2= 186/0.6;			# in K

# Calculations
T1= 1.2*T2;			# in K

# Results
print "Temperature of the source in °C is : ",T2-273
print "Temperature of the math.sink in °C is : ",T1-273

Temperature of the source in °C is :  37.0
Temperature of the math.sink in °C is :  99.0


## Example 6.2 Page No : 138¶

In :
import math

# Variables
T1 = 25.;			# in °C
T2 = 1.; 			# in °C
T1 = T1 + 273.;			# in K
T2 = T2 + 273.;			# in K
HT= 2.;	    		# heat transfer across the wall and the roof in MJ/hr

# Calculations and Results
HT= HT*10**6;   			# in J/hr
Q = HT* (T1-T2);			# in J/hr
COP_heat = T1/(T1-T2);
W_net = Q/COP_heat;			# in J/hr
print "Power rquired for operating the pump in kW is : %.3f"%(W_net*10.**-3./3600)

# Part (b)
T2= 25.;			# in °C
T2=T2+273;			# in K
# COP= T2/(T1-T2)          (i)
# COP= HT*(T1-T2)/W_net        (ii)
# From (i) and (ii)
T1= math.sqrt(W_net*T2/HT)+T2;			# in K
T1= T1-273;			# in °C
print "The value of T1 in °C is : %.2f"%T1

# rounding off error

Power rquired for operating the pump in kW is : 1.074
The value of T1 in °C is : 49.00


## Example 6.3 Page No : 142¶

In :
# Variables
heatEngineEffi= 32./100;			# heat engine efficiency
COP= 5.                 			# COP of heat pump

# Calculations
# heat engine efficiency = Wnet/Q1 = (Q1-Q2)/Q1
Q1byWnet= 1/heatEngineEffi;
Q2byWnet= (1-heatEngineEffi)*Q1byWnet;
# COP = Q4/Wnet = Q4/(Q4-Q3)
Q4byWnet= COP;
ratio= (Q2byWnet+Q4byWnet)/Q1byWnet;			# ratio of heat transferred to the circulating water to heat trasferred to the engine

# Results
print "Ratio of heat trasferred to the circulating water to heat trasferred to the engine is : ",ratio

Ratio of heat trasferred to the circulating water to heat trasferred to the engine is :  2.28


## Example 6.4 Page No : 147¶

In :
# Variables
Q = 88;			# in MJ
Q=Q*10**3;			# in kJ
T1 = 190.;			# in °C

# Calculations
T1 = T1 + 273;			# in K
T3 = -15;			# in °C
T3 = T3 + 273;			# in K
Eta_carnot = (T1 - T3)/T1;
Wnet= Eta_carnot * Q;			# in kJ
CarnotPower= Wnet/3600.;			# in kWh

# Results
print "The value of Carnot Power in kWh is : %.3f"%CarnotPower
print ("As the actual power produced by the invented engine is more than the Carnot Power, ");
print ("so inverter claim is  not true")

The value of Carnot Power in kWh is : 10.823
As the actual power produced by the invented engine is more than the Carnot Power,
so inverter claim is  not true


## Example 6.5 Page No : 148¶

In :
# Variables
T1 = 24.;		    	# in °C
T1 = T1 + 273;			# in K
T2 = 2;		        	# in °C
T2 = T2 + 273;			# in K
Q = 100;    			#in MJ/h
Q = Q * 10**3;			#in kJ/h

# Calculations and Results
COP_heatPump = T1/(T1-T2);
W = Q/COP_heatPump;			#in kJ/h
W = W/3600.;			        # in kW
print "The theoretical minimum power required to drive the heat pump in kW is : %.3f"%W

COP_refrigerator = T2/(T1-T2);
W = Q/COP_refrigerator;			# in kJ/h
W = W/3600.;         			# in kW
print "The theoretical power required to drive the heat pump when it is used as a refrigerator in kW is : %.2f"%W

The theoretical minimum power required to drive the heat pump in kW is : 2.058
The theoretical power required to drive the heat pump when it is used as a refrigerator in kW is : 2.22


## Example 6.6 Page No : 150¶

In :
# Variables
Q1= 278.;			# in kJ/s
T1= 283.+273;			# in K
T2= 50.+273;			# in K
print ("Part (a)")
Q2= 208.;			# in kJ/s

# Calculations and Results
# By Clausius inequality
V= Q1/T1-Q2/T2;
if V<0:
print ("The cycle is irreversible")
elif  V>0:
print ("Reversible or irreversible cycle is not possible and the result is impossible")
else:
print ("The cycle is reversible")

print ("Part (b)")
Q2= 139.;			# in kJ/s
V= Q1/T1-Q2/T2;
if V<0:
print ("The cycle is irreversible")
elif V>0:
print ("Reversible or irreversible cycle is not possible and the result is impossible")
else:
print ("The cycle is reversible")

print ("Part (c)")
Q2= 161.5;			# in kJ/s
V= Q1/T1-Q2/T2;
if V<0:
print ("The cycle is irreversible")
elif V>0:
print ("Reversible or irreversible cycle is not possible and the result is impossible")
else:
print ("The cycle is reversible")

Part (a)
The cycle is irreversible
Part (b)
Reversible or irreversible cycle is not possible and the result is impossible
Part (c)
The cycle is reversible


## Example 6.16 Page No : 155¶

In :
# Variables
Wnet_compresser= 3.;	                    		# in kW
Wnet_compresser=Wnet_compresser*3600.;			# in kJ/h
Qabsorbed= 50.;	    		                    # in MJ/h
Qabsorbed=Qabsorbed*10**3;			            # in kJ/h

# Calculations and Results
T1 = 46+273;			# in K
T2 = 1+273; 			# in K
Qrejected= Wnet_compresser+Qabsorbed;			# in kJ/h
print "The heat rejected in MJ/h is : ",Qrejected*10**-3

I= -(-Qrejected/T1+Qabsorbed/T2);			# in kJ/h
print "Irreversibility in kJ/h is : %.3f"%I

The heat rejected in MJ/h is :  60.8
Irreversibility in kJ/h is : 8.114


## Example 6.17 Page No : 156¶

In :
#Exa 6.17
import math

# Variables
T1 = 12.;    			# in °C
T2 = 92.;	    		# in °C
T1 = T1 + 273.;			# in K
T2 = T2 + 273.;			# in K
del_T = T2 - T1;			# in K
m = 20;			        # in kg
C_v = 4.187;
s= 1;

# Calculations and Results
Q = m * s * del_T;			# in cal
Q = Q * 4.18;	    		# in J
H = 2;	        	    	# heat given by the heater in kw
H = H * 10**3;		    	# in J/sec
t = Q/H;			        #time taken by the heater to raise the temp. in sec
print "Time taken by the heater to raise the temperature in sec is : %.2f"%t

del_phi = m * C_v * math.log(T2/T1);			# in kJ/K
print "Entrophy generated during the process in kJ/K is : %.2f"%del_phi

Time taken by the heater to raise the temperature in sec is : 3.34
Entrophy generated during the process in kJ/K is : 20.72


## Example 6.18 Page No : 156¶

In :
# Exa 6.18
import math

# Variables
Q1 = 1000.;			# in kW
Q2 = 492.;			# in kW
T1 = 285.;			# in °C
T1 = T1 + 273;			# in K
T2 = 5.;			    # in °C
T2 = T2 + 273;			# in K

# Calculations and Results
Eta_carnot = (T1-T2)/T1*100;			# in percentage
print "Carnot efficiency in %% is : %.2f"%Eta_carnot

Eta_heat = (Q1 - Q2)/Q1*100;			# in percentage
print "Efficiency of the heat engine in % is : ",Eta_heat
if Eta_heat>Eta_carnot:
print ("As the efficiency of heat engine cannot be more than Carnot efficiency, Hence\
engine cannot execute irreversible cycle.")

Carnot efficiency in % is : 50.18
Efficiency of the heat engine in % is :  50.8
As the efficiency of heat engine cannot be more than Carnot efficiency, Hence    engine cannot execute irreversible cycle.


## Example 6.19 Page No : 156¶

In :
# Variables
n = 1080.;			# in cycle/min
Q_s = 57.;			# in J/cycle
T1 = 12.;			# in °C
T1 = T1 + 273;			# in K
T2 = 2.;			# in °C
T2 = T2 + 273;			# in K

# Calculations
# 1-(Q_r/Q_s) = 1- (T2/T1)
Q_r = (T2/T1)*Q_s;			# in J/cycle
W = Q_s - Q_r;			# in J/cycle
P_o = W * n;			# in J/min
P_o = P_o/60;			# in W

# Results
print "The output of the engine in watt is",P_o

The output of the engine in watt is 36.0


## Example 6.20 Page No : 157¶

In :
# Variables
Q2 = 10.**5;			# in kJ/hr
T1 = -3.;			# in °C
T1 = T1 + 273;			# in K
T2 = 22.;			# in °C
T2 = T2 + 273;			# in K

# Calculations
COP_heat = 1./(1-T1/T2);
W = Q2/COP_heat;			# in kJ/hr
W = W/3600.;    			# in kW

# Results
print "Minimum power required in kW is : %.3f"%W

Minimum power required in kW is : 2.354


## Example 6.21 Page No : 158¶

In :
# Variables
T_A= 927.+273;			# in K
T_B= 127.+273;			# in K
T_C= T_B;			# in K

# Calculations
# Q_A= Q_B+Q_C+W = 2*Q_B+W (math.since Q_B=Q_C)     (i)
# Q_A/T_A= Q_B/T_B+Q_C/T_C or
# Q_A= 2*Q_B*T_A/T_B           (ii)
# From eq (i) and (ii)W= 2*Q_B*(T_A/T_B-1)     (iii)
# Dividing (iii) by (ii)
WbyQ_A= (T_A/T_B-1)/(T_A/T_B);

# Results
print "engine efficiency in %% is : %.1f"%(WbyQ_A*100)

engine efficiency in % is : 66.7


## Example 6.22 Page No : 158¶

In :
# Variables
T_A= 700.;			# in K
T_B= 600.;			# in K
T_C= 500.;			# in K
Q_A= 1200.;			# in kJ

# Calculations
# Q_B+Q_C= Q_A-200                             (i)
# Q_A/T_A = Q_B/T_B+Q_C/T_C    (ii)
# From eq(i) and (ii)
Q_B= (Q_A*(1/T_B-1/T_A)-200/T_B)/(1/T_B-1/T_C);			# in kJ
Q_C= Q_A-Q_B-200;			# in kJ

# Results
print "The heat rejected at B in kJ is : %.2f"%Q_B
print "The heat rejected at C in kJ is : %.2f"%Q_C

The heat rejected at B in kJ is : 142.86
The heat rejected at C in kJ is : 857.14


## Example 6.23 Page No : 159¶

In :
# Variables
T1= 180+273;			# in K
T2= 20+273;			# in K

# Calculations
# W_A/Q1= 1-T3/T1      (i)
# W_B/QB= 1-T2/T3      (ii)
# W_A= W_B                 (iii)
# Q1= Q_B+W_A          (iv)
# From eq(i),(ii),(iii) and (iv)
T3= (T1+T2)/2;			# in K

# Results
print  "intermediate temperature in °C is : ",T3-273

intermediate temperature in °C is :  100


## Example 6.24 Page No : 160¶

In :
# Variables
Q2 = 1.75;			# in kJ/sec
T1 = -15;			# in °C
T1 = T1 + 273;			# in K
T2 = 30;			# in °C
T2 = T2 + 273;			# in K

# Calculations
del_T = T2 - T1;			# in K
# Q2/W_net = T2/(del_T)
W_net = Q2 * del_T/T1;			# in kW

# Results
print "Least power required in kW is : %.3f"%W_net

Least power required in kW is : 0.305