# Chapter 6 : Boiling and condensation¶

## Example 6.1 Page No : 177¶

In [3]:
from scipy.optimize import fsolve
import math
import warnings
warnings.filterwarnings('ignore', 'The iteration is not making good progress')
# Variables
#(a)
Tsat = 350            			#K, saturated temp.
Tl = Tsat+5           			#K, liquid temp.
#By antoine eqn.
T = Tl-273            			#C,

# Calculations and Results
pl = math.exp(4.22658-(1244.95/(T+217.88)))
ST = 26.29-0.1161*T   			#dyne/cm, Surface tension of liquid
ST_ = ST*10**-3        			#N/m  Surface tension of liquid
Lv = 33605            			#kj/kgmol, molar heat of vaporization
R = 0.08314           			#m**3  bar/kgmol K, gas math.cosmath.tant
r = (2*ST_*R*Tsat**2)/((Tl-Tsat)*pl*(Lv*10**3))
print "So a bubble nucleus that has been detached from a cavity will not collapse in \
the liquid if it is larger than %.2f micrometer "%(r*10**6)

#(b)
r1 = 10**-6       			#m
#pl1 = exp(4.22658-(1244.95/(Tl_-273+217.88)))   			#vapour pressure
#ST1 = 0.02629-1.161*10**-4(Tl_-273)              			#surface tension

def f(Tl):
return (Tl-Tsat)-2*(0.02629-1.161*10**-4*(Tl-273))*R*Tsat**2/(r1*Lv*10**3)
Tl = fsolve(f,0.1)
T_ = (Tl-273.5)-(Tsat-273)
print "The superheat of the liquid is %d C"%(T_)

# note : answers are slightly different because of rounding off error.

So a bubble nucleus that has been detached from a cavity will not collapse in the liquid if it is larger than 1.89 micrometer
The superheat of the liquid is 9 C


## Example 6.2 Page No : 180¶

In [2]:
import math

# Variables
d = 0.35                   			#m, diameter of pan
p = 1.013                  			#bar, pressure
T1 = 115.                   			#C, bottom temp.
T2 = 100.                   			#C, boiling temp.
Te = T1-T2                 			#C, excess temp.
#For Water
mu1 = 2.70*10**-4           			#Ns/m**2, vismath.cosity
cp1 = 4.22                 			#kj/kg C, specific heat
rho1 = 958.                 			#kg/m63. density
Lv1 = 2257.                 			#kj/kg, enthalpy of vaporization
s1 = 0.059                 			#N/m , surface tension
Pr1 = 1.76                 			#Prandtl no.
#For saturated steam
rho2 = 0.5955
#For the pan
Csf = 0.013               			#consmath.tant
n = 1.                     			#exponent
g = 9.8                   			#m/s**2, gravitational consmath.tant

# Calculations and Results
#from eq. 6.6  			#heat flux
Qs1 = mu1*Lv1*(g*(rho1-rho2)/s1)**(1./2)*(cp1*Te/(Csf*Lv1*(Pr1)**n))**3
Rate = Qs1/Lv1             			#kg/m**2 s. rate of boiling
Ap = math.pi/4*d**2            			#m**2, pan area
Trate = Rate*Ap           			#kg/s, Total rate of boiling
Trate_ = Trate*3600.5     			#kg/h. Total rate of boiling
print "total rate of boiling of water is %.0f kg/h "%(Trate_)

#umath.sing Lienhard's eq.,   			#critical heat flux
Qmax = 0.149*Lv1*rho2*(s1*g*(rho1-rho2)/(rho2)**2)**(1/4)
#by Mostinski eq.
Pc = 221.2                			#critical pressure
Pr = p/Pc                 			#reduced pressure
hb = 0.00341*(Pc)**(2.3)*Te**(2.33)*Pr**(0.566)     			#boiling heat transfer coefficient
hb_ = hb/1000              			#kW/m**2 C boiling heat transfer coefficient
Qs2 = hb_*(Te)
print "Qs2 compares reasonably well with the Qs1"

# note: rounding off error.

total rate of boiling of water is 69 kg/h
Qs2 compares reasonably well with the Qs1


## Example 6.3 Page No : 181¶

In [3]:
import math

# Variables
A = 12.5673
B = 4234.6
pv = 1.813
T1 = 200.                 			#C, tube wall temp.
#For methanol
Tc = 512.6               			#K, critical temp.
w = 0.556                			#acentric factor
Zra = 0.29056-0.08775*w
R = 0.08314             			#m**3bar/gmol K, universal gas consmath.tant
Pc = 80.9               			#bar, critical temp.
Mw = 32.                 			#g, molecular wt

#Calculation
#Estimation of liquid and vapour properties
#from antoine eq.
T = B/(A-math.log(pv))       			#K, boiling point
Te = (T1+273)-T         			#K, excess temp.
Tm = ((T1+273)+T)/2     			#K, mean temp.
#Liquid properties
#(a)
Tr = T/Tc              			#K, reduced temp.
#from Rackett technique
Vm = R*Tc*(Zra)**(1+(1-Tr)**(2/7))/Pc      			#m**3/kg mol, molar volume
rhol = Mw/Vm                             			#kg/m**3, density of satorated liquid density
#(b)
#from Missenard technique
T2 = 348.               			#K,given data temp.
T3 = 373.               			#K,given data temp.
Cp2 = 107.5            			#j/g mol K specific heat at T2
Cp3 = 119.4            			#j/g mol K specific heat at T3
#By linear interpolation at T = 353.7 K
Cp = Cp2+(Cp3-Cp2)*((T-T2)/(T3-T2))    			#kj/kg mol C, specific heat at T = 353.7 K
Cp_ = Cp*0.03125                       			#kj/kg C
#(c)Surface tension at given temp.(K)
T4 = 313.
St4 = 20.96
T5 = 333.
St5 = 19.4
#By linear interpolation at T = 353.7 K
S = 17.8                               			#dyne/cm, surface temp.
#(d) liquid vismath.cosity
T6 = 298.
MUt6 = 0.55                           			#cP, liquid vismath.cosity at temp = 298
MU = ((MUt6)**-0.2661+((T-T6)/233))**(-1/0.2661)       			#cP
#(e)Prandtl no. a,b,c are consmath.tant
a = 0.3225
b = -4.785*10**-4
c = 1.168*10**-7
kl = a+b*T+c*T**2                     			#W/m C, thermal conductivity
Prl = Cp_*1000*MU*10**-3/kl           			#Prandtl no.
#(f)heat of vaporization at 337.5 K
Lv = 1100.                            			#kj/kg, enthalpy of vaporization

#Properties of methanol vapour at Tm
#(a)
Vm1 = R*Tm/pv                      			#m**3/kg mol, molar volume
rhov = Mw/Vm1                      			#kg/m**3, density of vapour
#(b) a1,b1,c1,d1 are math.cosmath.tants
a1 = -7.797*10**-3
b1 = 4.167*10**-5
c1 = 1.214*10**-7
d1 = -5.184*10**-11
#thermal conductivity of vapour
kv = a1+b1*Tm+c1*Tm**2+d1*Tm**3    			#W/m C
#(c)heat capacity of vapour,  a2,b2,c2,d2 are math.cosmath.tants
a2 = 21.15
b2 = 7.092*10**-2
c2 = 2.589*10**-5
d2 = -2.852*10**-8
#heat capacity of vapour,      in kj/kh mol K
Cpv = a2+b2*Tm+c2*Tm**2+d2*Tm**3

#(d)vismath.cosity of vapour
T7 = 67.
MUt7 = 112.
T8 = 127.
MUt8 = 132.
#from linear inter polation at Tm
MUv = 1.364*10**-5              			#kg/m s

#from Rohsenow's eq.
Csf = 0.027                   			#consmath.tant
n = 1.7                       			#exponent value
#from eq. 6.6
g = 9.8                       			#m/s**2, gravitational consmath.tant
#heat flux   			#kW/m**2
Q = MU*10**-3*Lv*(g*(rhol-rhov)/S*10**-3)**(1./2)*(Cp_*Te/(Csf*Lv*(Prl)**n))**3
#from eq. 6.11
#from eq 6.11,  critical heat flux
Qmax = 0.131*Lv*(rhov)**(1./2)*(S*10**-3*g*(rhol-rhov))**(1./4)
r = 0.016
r_ = r*(g*(rhol-rhov)/(S*10**-3))**(1./2)
#peak heat flux
Qmax1 = Qmax*(0.89+2.27*math.exp(-3.44*math.sqrt(r_)))
#from eq. 6.12
#heat transfer coefficient hb
d = 0.032                        			#m, tube diameter
hb = 0.62*((kv**3)*rhov*(694-rhov)*g*(Lv*10**3+0.4*Cpv*Te)/(d*MUv*Te))**(1./4)
Qb = hb*Te                      			#kw/m**2, heat flux
BR = Qb*10**-3/Lv                			#kg/m**2s, boilng rate

# Results
print "The boilins rate is %.0f kg/m**2 h"%(BR*3600)

# note : rounding off error.

The boilins rate is 63 kg/m**2 h


## Example 6.4 Page No : 188¶

In [4]:
import math

# Variables
W1 = 200.              			#kg/h, rate of entering toluene
muv = 10.**-5           			#kg/m s, vismath.cosity of toluene vapour
mul = 2.31*10**-4      			#kg/m s, vismath.cosity of  benzene
rhol = 753.            			#kg/m**3, density of benzene
rhov = 3.7            			#kg/m**3, density of toluene vapour
Cpl = 1968.            			#j/kg C, specific heat of benzene
kl = 0.112            			#W/m C, thermal conductivity of benzene
T1 = 160.              			#C tube wall temp.
T2 = 120.              			#C , saturated temp.
Te = T1-T2            			#C, excess temp.
Lv = 3.63*10**5        			#j/kg, enthalpy of vaporization
s = 1.66*10**-2        			#N/m, surface tension

#Calculation of hc & hb
w = 0.125             			#m, mean step size
d = 0.0211            			#, internal diameter of tube
G = W1/(3600*math.pi/4*(d**2))         			#kg/m**2 s, mass flow rate
Re1 = G*(1-w)*d/mul              			#Reynold no.
Prl = Cpl*mul/kl                 			#Prandtl no.
#from eq. 6.23
x = (w/(1-w))**(0.9)*(rhol/rhov)**(0.5)*(muv/mul)**0.1  			#let x = 1/succepsibility
#from eq. 6.22
F = 2.35*(x+0.231)**0.736        			#factor signifies 'liquid only reynold no.' to a two phase reynold no.
#from eq. 7.21
Re2 = 10**-4*Re1*F**1.25          			#Reynold no.
#from eq. 6.18
S = (1+0.12*Re2**1.14)**-1        			#boiling supression factor
#from eq. 6.15
hc = 0.023*Re1**(0.8)*Prl**(0.4)*(kl/d)*F  			#W/m**2 C, forced convection boiling part
#from eq. 6.16
mulv = (1/rhov)-(1/rhol)              			#m**3/kg, kinetic vismath.cosity of liquid vpaour
dpsat = Te*Lv/((T2+273)*mulv)         			#N/m**2, change in saturated presssure
#nucleate boiling part hb
hb = 1.218*10**-3*(kl**0.79*Cpl**0.45*rhol**0.49*Te**0.24*dpsat**0.75*S/(s**0.5*mul**0.29*Lv**0.24*rhov**0.24))
h = hc+hb                            			#W/m**2 C, total heat transfer coefficient

#calculation of required heat transfer area
a = 5.                              			#%, persentage change in rate of vaporization
W2 = W1*a/100                      			#kg/h, rate of vaporization
W2_ = W2/3600                      			#kg/s
A = Q/(h*Te)                       			#m**2, area of heat transfer
l = A/(math.pi*d)                      			#m, required length of tube
#from table 6.2
Tl = 0.393

# Results
print "The total tube length is %.3f m"%(Tl)

The total tube length is 0.393 m


## Example 6.5 Page No : 195¶

In [4]:
from scipy.optimize import fsolve
import math

# Variables
rhol = 483.                         			#kg/m**3, density of liquid propane
mul = 9.1*10**-5                    			#P ,vismath.cosity of liquid propane
kl = 0.09                          			#W/m K, thermal conductivity of liquid propane
Lv = 326.                           			#kj/kg. enthalpy of vaporization
Cpl = 2.61                         			#kj/kg K, specific heat of liquid propane
T1 = 32.
T2 = 25.                            			#C, surface temp.
p1 = 11.2
rhov = 24.7                       			#kg/m**3, density of vapour
g = 9.8
h = 0.3

#Calculation
Lv1 = Lv+0.68*Cpl*(T1-T2)
#h = 0.943*(g*Lv1*10**3*rhol*(rhol-rhov)*kl**3/(mul*L*(T1-T2)))**(1/4)
#Q = h*(L*1)*(T1-T2)
#m = Q/(Lv1*10**3) = 1.867*10**-2*L**(3/4)
Ref = 30.
#from the relation  4*m/mu = Re
L = (Ref*mul/(4*1.867*10**-2))**(4./3)
m = 1.867*10**-2*L**(3./4)        			#rate of condensation for laminar flow
#from eq. 6.32
#Nu1 = h_/kl*(mul**2/(rhol*(rhol-rhov)*g))**(1/3) = Ref/(1.08*(Ref)**(1.22)-5.2)
Lp = h-L        			#length of plate over which flow is wavy
A = Lp*1         			#m**2 area of condensation

def f(h1):
return h1/kl*(mul**2/(rhol*(rhol-rhov)*g))**(1./3)-(29.76+0.262*h1)/(1.08*(29.76+0.262*h1)**(1.22)-5.2)
h1 = fsolve(f,1000)
m2 = m+h1*A*(T1-T2)/(Lv1*10**3)
Ref1 = 4*m2/mul
m2 = m+h1*A*(T1-T2)/(Lv1*10**3)

# Results
print "Total rate of condensation is %.2f kg/h"%(m2*3600)

Total rate of condensation is 33.08 kg/h


## Example 6.6 Page No : 199¶

In [2]:
import math

# Variables
#data fot TCE
T1 = 87.4                         			#C, normal boiling point
T2 = 25.                           			#C, surface temp.
Lv = 320.8                        			#kj/kg, heat of vaporization
cp = 1.105                        			#kj/kg C, specific heat
mu = 0.45*10**-3                   			#P. liquid vismath.cosity
k = 0.1064                        			#W/m C, thermal conductivity
rhol = 1375.                       			#kg/m**3, liquid density
rhov = 4.44                       			#kg/m**3, density of vapour
Tm = (T1+T2)/2.                    			#C, mean film temp.
d = 0.0254                        			#m, outside diameter of tube
l = 0.7                           			#m, length
g = 9.8                           			#m/s**2, gravitational consmath.tant

# Calculations and Results
#(a) from eq. 6.34
Lv1 = Lv+0.68*cp*(T1-T2)
h = 0.728*(g*Lv1*10**3*rhol*(rhol-rhov)*k**3/(mu*d*(T1-T2)))**(1./4)
A = math.pi*d*l                       			#m**2, area of tube
Q = h*A*(T1-T2)                   			#W, rate of heat transfer
m = (Q/Lv1)/1000                  			#kg/s rate of condensation
print "Rate of condensation is %.1f kg/h "%(m*3600)

#(b)   from eq. 6.35
N = 6.                             			#No. of tubes in vertical tire
h1 = 0.728*(g*Lv1*10**3*rhol*(rhol-rhov)*k**3/(N*mu*d*(T1-T2)))**(1./4)
TN = 36.                           			#total no. of tubes
TA = TN*math.pi*d*l                    			#m**2, total area
Q1 = h1*TA*(T1-T2)                   			#W, rate of heat transfer
m1 = (Q1/Lv1)/1000.                  			#kg/s rate of condensation
print "Rate of condensation is %.0f kg/h  "%(m1*3600)
#from chail's corelation
h2 = (1+0.2*cp*(T1-T2)*(N-1)/(Lv1))
print "thus there will be increase in the calculated rate of\
heat transfer and in rate of condensation as %.3f percent"%(h2)

# note : rounding off error.

Rate of condensation is 45.7 kg/h
Rate of condensation is 1052 kg/h
thus there will be increase in the calculated rate of heat transfer and in rate of condensation as 1.188 percent


## Example 6.7 Page No : 201¶

In [16]:
import math

# Variables
Gv = 20.                  			#kg/m**2 s, mass flow rate of benzene
di = 0.016               			#m, tube diameter
muv = 8.9*(10**-6)        			#P, vismath.cosity
Lv = 391.                 			#kj/kg., enthalpy of vaporization
cpl = 1.94               			#kj/kg C, specific heat
Tv = 80.                  			#C, normal boiling point of benzene
Tw = 55.                  			#C, wall temp.
g = 9.8                  			#m/s**2, gravitational consmath.tant
rhol = 815.               			#kg/m**3, density of benzene
rhov = 2.7               			#kg/m**3, density of benzene vapour
kl = 0.13                			#W/m C, thermal conductivity
mu = 3.81*10**-4          			#P, vismath.cosity of benzene
l = 0.5                  			#m, length  of tube

#calculation
Rev = di*Gv/muv          			#Reynold no. of vapour
#from eq. 6.38
Lv1 = Lv+(3./8)*cpl*(Tv-Tw)
#heat transfer corfficient , h
h = 0.555*(g*rhol*(rhol-rhov)*kl**3*Lv1*10**3/(di*mu*(Tv-Tw)))**(1./4)
Aavl = math.pi*di*l          			#m**2, available area
Q = Aavl*h*(Tv-Tw)       			#W, rate of heat transfer
m = Q/(Lv1*10**3)         			#kg/s, rate of condensation of benzene
Ratei = Gv*(math.pi/4)*di**2   			#kg/s rate of input of benzene vapour
n = m/Ratei

# Results
print "fraction of input vapour condensed is %.1f"%(n*100)

# note : rouding off error.

fraction of input vapour condensed is 52.7