Chapter 9 : Second law of thermodynamics

Example 9.1 pageno : 308

In [1]:
# Variables
t2 = 300;			#temperature of the math.sink in K
n1 = 0.4;			#efficiency of the engine
n2 = 0.6;			#efficiency of the engine

# Calculations
t1 = t2/(1-n1);			#temperature of the source in K
t3 = t2/(1-n2);			#temperature of the source in K

# Result
print 'the temperature of the source when 0.4 efficiency is %3.2f K  \
\nthe temperature of the source when 0.6 efficiency is %3.2f K'%(t1,t3)

the temperature of the source when 0.4 efficiency is 500.00 K  
the temperature of the source when 0.6 efficiency is 750.00 K

Example 9.2 pageno : 308

In [3]:
# Variables
t2 = 273.;			#temperature of the math.sink in K
t1 = 373.;			#temperature of the source in K
q1 = 840.;			#heat supplied in joules
j = 4.2;			#joukes constant in erg/cal

# Calculations
w = (q1/t1)*(t1-t2);			#work done in joules
q2 = (q1/j)*(t2/t1);			#heat rejected in calories
n = 1-(t2/t1);			#efficiency of the engine

# Result
print 'work done is %3.f j  \
\nheat rejected is %3.f cal  \
\nthe efficiency of the engine is %3.1f %%'%(w,q2,n*100)

work done is 225 j  
heat rejected is 146 cal  
the efficiency of the engine is 26.8 %

Example 9.3 pageno : 309

In [4]:
# Variables
t1 = 90.;			#temperature of the oxygen boils in K
t2 = 20.;			#temperature of the liquid hydrogen in K
t3 = 300.;			#temperature of the sink in K

# Calculations
n = (t1-t2)/t1;			#efficiency of the engine
t4 = t3/(1-n);			#temperature of the source in K

# Result
print 'the efficiency of the engine is %3.2f  \
\nthe temperature of the source is %3.2f K'%(n,t4)

the efficiency of the engine is 0.78  
the temperature of the source is 1350.00 K

Example 9.4 pageno : 309

In [6]:
# Variables
t1 = 373.;	    	    	#temperature of the source in K
t2 = 273.;		    	    #temperature of the sink in K
w = 1200*10**5*980;			#work done in ergs
j = 4.18*10**7;		    	#joules constant in ergs/cal

# Calculations
q = (w/j)*(t1/(t1-t2));			#heat added in cal

# Result
print 'the heat added is %3.2f cal'%(round(q,-1))

the heat added is 10490.00 cal

Example 9.5 pageno : 309

In [12]:
# Variables
t1 = 273.;			#temperature of the source in K
t2 = 290.;			#temperature of the sink in K
l = 8*10.**11;			#latent of fusion in ergs/cal

# Calculations
n = (t2-t1)/t1;			#efficiency of the engine
w = n*l;			#energy to be supplied in ergs

# Result
print 'efficiency of the engine is %.2f %%  \
\nenergy to be supplied is %.3e ergs'%(n*100,w)
print "Note: answer in book are wrong please calculate manually."

efficiency of the engine is 6.23 %  
energy to be supplied is 4.982e+10 ergs
Note: answer in book are wrong please calculate manually.

Example 9.6 pageno : 309

In [14]:
# Variables
t1 = 373;			#temperature in K
t2 = 273;			#temperature of math.sink in K
q = 10**4;			#heat taken at higher temperature in cal
j = 4.2*10**7;			#joules consmath.tant in ergs/cal

# Calculations
w = q*j*(t1-t2)/t1;			#work done in ergs

# Result
print 'work done is %.1e ergs'%(w)

work done is 1.1e+11 ergs

Example 9.7 page no : 310

In [23]:
import math

# Variables
p = 100*746/4.2;			#power developed in cal/sec
t1 = 300.;			#temperature of the sink in K
t2 = 500.

# Calculations
te = 1 - (t1/t2)
Q1 = p * 100/40        # heat supplied
Q2 = Q1 * 0.6


# Result
print "Thermal efficiency = %.f %%"%(te*100)
print "Power developed by the engine %.2f calories/sec"%p
print "If Q1 heat supplied , Q1 = %.2e cal/sec"%Q1
print "Q2 = %.2e cal/sec"%Q2

Thermal efficiency = 40 %
Power developed by the engine 17761.90 calories/sec
If Q1 heat supplied , Q1 = 4.44e+04 cal/sec
Q2 = 2.66e+04 cal/sec

Example 9.8 page no : 310

In [34]:
import math 

# Variables
l = 964.8;			#latent heat of steam in B.Th.U per lb
q = 4*15*l*778;			#heat developed in ft lbs
w = 30000*60;			#work done is ft lbs
pv = 12*1.013*10**6*10**3 
T = 600                # K

# Calculations
n = (w/q)*100;			#efficiency of the engine
p = 100-n;			#percentage of heat wasted
T2 = 600./(6**.4)
R = pv/T
W = R * (T - T2) * 2.303 * math.log10(6)
e = 1 - (T2/T)
# Result
print "Lowest temperature T2 = %.f K"%T2
print "Work done W = %.2e ergs"%W
print "Efficiency = %.1f %%"%(e*100)

Lowest temperature T2 = 293 K
Work done W = 1.11e+10 ergs
Efficiency = 51.2 %

Example 9.9 page no : 311

In [49]:
# Variables
l=964.8;            #latent heat of steam in B.Th.U per lb
q=4*15*l*778;       #heat developed in ft lbs
w=33000*60;         #work done is ft lbs

#CALCULATIONS
n=(w/q)*100;       #efficiency of the engine
p=100-n;          #percentage of heat wasted

# Results
print ('the percentage of the heat wasted is %3.2f'%p)

the percentage of the heat wasted is 95.60

Example 9.10 page no : 311

In [4]:
# Variables
ip = 16.3*500*778/33000;			# Variables power of the engine in HP
me = 0.72;			#mechanical efficiency of the engine
bhp = 31;			#brake horse power in b.h.p
ihp = bhp/me;			#indicated horse power in HP

# Calculations
i = ihp/ip;			#indicated thermal efficiency

# Result
print 'the indicted thermal efficiency is %3.2f %%'%(i*100)

the indicted thermal efficiency is 22.41 %

Example 9.11 pageno : 312

In [5]:
# Variables
p = 200.;			#horse power of steam engine in lbs coal per hour
j = 770.;			#joules constant in ft lbs per B.Th.U

# Calculations
w = 12500*p*j;			#equivalent work in ft.lb.per.hr
hp = w/(60*33000);			#horse power

# Result
print 'horse power of the engine is %3.2f'%(hp)
print "Note : answer in book is wrong. Please check manually."

horse power of the engine is 972.22
Note : answer in book is wrong. Please check manually.

Example 9.12 pageno : 312

In [43]:
# Variables
t1 = 340.;			#temperature of the atmosphere in K
t2 = 612.;			#temperature of the compression stroke in K
y = 1.39;			#adiabatic expansion 
t3 = 2040.;			#temperature after consmtant volume ignition in K

# Calculations
d = (t2/t1)**(1/(y-1))			#density in gm/cc
n = 1-(1/d)**(y-1);		    	#efficiency of the engine
p = ((d)**(y))*(t3/t2);			#maximum temperature of the temperature in atm

# Result
print 'the maximum pressure of the engine is %3.f atm'%(p)

the maximum pressure of the engine is  27 atm

Example 9.13 pageno : 313

In [40]:
# Variables
t1 = 915;			#temperature at the beggining in K
t2 = 2040;			#temperature at the end in K
d = 12.6;			#adiabatic expansion ratio
y = 1.39;			#coefficent of expansion

# Calculations
x = t2/t1                               			#ratio temparatures
n = 1-(1/d)**(y-1)*((x**y)-1)/(y*(x-1));			#efficiency of the engine

# Result
print 'the efficiency of the engine is %3.3f'%(n)
print "Note : answer slighty different because of rounding error"

the efficiency of the engine is 0.566
Note : answer slighty different because of rounding error

Example 9.14 pageno : 313

In [6]:
# Variables
p1 = 15.;			#intial pressure in lb/sq.inch
dv = 15.;			#ratio of intial to final volume
t1 = 520.;			#temperature at intial in K
y = 1.4;			#coefficient of expansion

# Calculations
p2 = p1*(dv)**(y);			#final pressure in lb/sq.inch
t2 = t1*(dv)**(y-1);			#final temperatire in K

# Result
print 'the final pressure is %3.2f lb/sq.inch  \
\nthe final temperature is %.f K'%(p2,t2)

the final pressure is 664.69 lb/sq.inch  
the final temperature is 1536 K