Chapter 7 : The joule thomson cooling efect

Example 7.1 page no : 239

In [1]:
# Variables
t = 33.18;			#critical temperature in K
pc = 12.80*76*981*13.6;			#critical pressure in dynes/sq.cm
r = 83.15;			#universal gas constant in kj/kg.K
d = 0.08987;			#density of hydrogen in gm/lit
v = 2000/0.08987;			#gram molecular volune of hydrogen in cc

# Calculations
b = r*10**6*t/(8*pc);			#vanderwaal constant in cm**3/mol
to = 2*27*t*(1-(b/v))/8;			#inversion temperature of the hydrogen in K

# Result
print 'the inversion temperature of hydrogen is %3.2f K'%(to)
the inversion temperature of hydrogen is 223.70 K

Example 7.2 pageno : 240

In [2]:
# Variables
b = 0.00136;			#vanderwaal constant in suv/gm
a = 0.011;			#vanderwaal constant in atm(suv)**2/gm**2
r = 0.003696;			#universal gas constant in atm(suv)/gm.deg
t = 423;			#temperature of steam in K
cp = -0.674/0.024205;			#specific heat at 423K in atm(cc)gm(deg)

# Calculations
dt = (-b+(2*a/(r*t)))/cp;			#change of temperature per atm drop of pressure in deg/atm

# Result
print 'the change of temperature per atmosphere drop of pressure is %3.7f deg/atm'%(dt)
the change of temperature per atmosphere drop of pressure is -0.0004565 deg/atm

Example 7.3 pageno : 241

In [2]:
# Variables
r = 8.3*10**7;			#universal gas constant in ergs/deg.C
a = 1.36*10**6*76*13.6*981;			#vanderwaal constant in atm.(suv**2)/(gm**2)
b = 32;			#vanderwaal constant in cc
cp = 7.03;			#specific heat at constant pressure in cal
j = 4.18*10**7;			#joules constant in ergs/cal
t = 273;			#temperature of the gas in K

# Calculations
dt = ((2*a/(r*t))-b)*10**6/(cp*j);			#change of temperature in atmosphere drop of pressure in deg/atm/cm**3

# Result
print 'the change of temperature in atmosphere drop of pressure is %3.2f deg C/atm/cm**2'%(dt)
the change of temperature in atmosphere drop of pressure is 0.31 deg C/atm/cm**2

Example 7.4 pageno : 241

In [6]:
# Variables
u = 1.08;
cp = 8.6;	    	        	#specific heat in kj/kg.K
j = 4.2;		    	        #joules constant in j/cal
p1 = 1*1.013*10**6;			    #pressure at intial in N/sq.m
p2 = 20*1.013*10**6;			#pressure at final in N/sq.m

# Calculations
dh = -u*cp*j*(p1-p2);			#change in enthalpy in joules

# Result
print 'the change in enthalpy is %3.3e joules'%(dh)
the change in enthalpy is 7.508e+08 joules

Example 7.5 pageno : 241

In [7]:
# Variables
tc = 5.26;			#critical temperature of the helium in K

# Calculations
ti = 27*tc/4;			#inversion temperature of the helium in K

# Result
print 'the inversion temperature of the helium is %3.2f K'%(ti)
the inversion temperature of the helium is 35.50 K

Example 7.6 pageno : 241

In [9]:
# Variables
a = 0.245*10**6*10**6;			#vanderwaal constant in cm**4.dyne/mole**2
b = 2.67*10;			#vanderwaal constant in cc/mole
r = 2*4.2*10**7;			#universal gas constant in ergs/mole.K

# Calculations
ti = 2*a/(b*r);			#inversion temperature in K

# Result
print 'inversion temperature of hydrogen is %.f K'%(round(ti,-1))
inversion temperature of hydrogen is 220 K

Example 7.7 pageno : 242

In [11]:
# Variables
dp = 50*10**6;			#change in pressure in dynes/sq.cm
cp = 7*4.2*10**7;			#specific heat constant pressure in ergs/mole.K
a = 1.32*10**12;			#vanderwaal constant in cm**4.dyne/mole**2
b = 31.2;			#vanderwaal constant in cm**2/mole
t = 300;			#inital temperature in K
r = 2*4.2*10**7;			#ergs/mole.K

# Calculations
dt = ((2*a/(r*t))-b)*dp/cp;			#change in temperature in K

# Result
print 'the change in temperature is %3.1f K'%(dt)
the change in temperature is 12.5 K

Example 7.8 pageno : 242

In [16]:
# Variables
p1 = 1.;			    #inital pressure in atm
p2 = 51.;   			#final pressure in atm
t1 = 300.;	    		#inital temperature in K
y = 1.4;		    	#coefficient of expansion

# Calculations
t2 = t1*(p2/p1)**((1-y)/y);			#final temperature in K
dt = t1-t2;	                		#drop in temperature in K

# Results
print 'the drop in temperature is %3.2f K'%(dt)
print "Note : answer is slightly different because of rounding error"
the drop in temperature is 202.45 K
Note : answer is slightly different because of rounding error