Chapter 21 Simultaneous Heat and Mass Transfer

Example 21_1_2 pgno:600

In [1]:
#initialization of variables
from math import pi
Tdisc = 30. # Centigrade
T = 21. # Centigrade
T0 = 18. # Centigrade
R0 = 1.5 # cm
V = 1000. # cc
t = 3600. #seconds
Nu = 0.082 #cm**2/sec
omeg = 2*pi*10/60 #sec**-1
from math import log
#Calculations
k = -V*(log((Tdisc-T)/(Tdisc-T0)))/(pi*(R0**2)*t)# k = h/d*cp cm/sec
alpha = ((1/0.62)*(k)*(Nu**(1/6))*(omeg**(-0.5)))**1.5/2 # cm**2/sec
#Results
print"the value of thermal diffusivity is cm**2/sec",round(alpha,4)

the value of thermal diffusivity is cm**2/sec 0.0012

Example 21_3_1 pgno:606

In [2]:
#initialization of variables
d =1000. # kg/m^3
h = 30. # W/m^2-C-sec
Hvap = 2300*10**3 # J/kg
T = 75. # C
Ti = 31. # C
l = 0.04 # m
epsilon = 0.36
c = 3600 # sec/hr
t1 = (Hvap/h)*(1/(T-Ti))*(l*epsilon*d)# sec
t = (t1/c) # in hr
#Results
print"The time taken for drying is hr",round(t,3)# answer wrong in textbook

The time taken for drying is hr 6.97

Example 21_3_2 pgno:608

In [3]:
#intialization of variables
d = 100*10**-4 # cm
v = 10**-3# cm/sec
nu = 0.2 # cm**2/sec
DS = 0.3 # cm**2/sec
DG = 3*10**-7 # cm**2/sec
H = 4.3*10**-4 # at 60 degree centigrade
#Calculations
kG = (2+(0.6*((d*v/nu)**0.5)*((nu/DS)**(1/3))))*DS/d# cm/sec
k = kG*H 
t = 30*DG/k**2
#Results
print"The mass transfer coefficient is  cm/sec",round(k,3)
print"\nTHe time needed to dry the particle is  sec",round(t,3)
#Answer wrong in textbook starting from kG

The mass transfer coefficient is  cm/sec 0.026

THe time needed to dry the particle is  sec 0.013

Example 21_4_1 pgno:614

In [4]:
#initialization of variables
slope = 230. #J/g-mol-C
nair = 60. # gmol/cm^2-sec
CpH2O = 75. # J/gmol-C
f = 0.4 # Correction factor
F = 2150./(60*0.018)#gmol/m^2-sec
kc= 20./3.
a = 3 # m^2/m^3
k = 2.7 # integral of dH/Hi-H with limits Hout and Hin
#Calculations
nH2Omax = slope*nair/CpH2O#gmol/m^2-sec
nH2O = nH2Omax*(1-f) #gmol/m^2-sec
A = F/nH2O # m^2
l = (nair/(kc*a))*k # m
#Results
print"The flow rate of the water per tower cross section is  gmol H2O/m^2-sec",nH2O
print"\nThe area of tower cross section is  m^2",round(A,3)
print"\nThe length of the tower is  m",l

The flow rate of the water per tower cross section is  gmol H2O/m^2-sec 110.4

The area of tower cross section is  m^2 18.032

The length of the tower is  m 8.1

Example 21_5_1 pgno:619

In [5]:
#initialization of variables
A = 0.01 # cm**2
l = 1. # cm
VA = 3. # cc
VB = 3. # cc
alphagas = 0.29 
alphaliquid = -1.3
x1 = 0.5
x2 = 0.5 
deltaT = 50. # Kelvin Thot-Tcold = 50
Tavg = 298. # kelvin
Dgas = 0.3 # cm**2/sec
Dliquid = 10**-5 # cm**2/sec
#calculations
deltaY = alphagas*x1*x2*deltaT/Tavg # Y1hot-Y1cold = DeltaY
deltaX = alphaliquid*x1*x2*deltaT/Tavg# X1hot-X1cold = DeltaX
Beta = (A/l)*((1/VA)+(1/VB))#cm**-2
BetaDgasinverse = 1/(Beta*Dgas)# sec
BetaDliquidinverse = (1/(Beta*Dliquid))/(365*24*60*60)
#Results
print"The seperation achieved for gas is  ",round(deltaY,3)
print"\nThe seperation achieved for liquid is  ",round(deltaY,3)
print"\nThe time taken for seperation for gas will be seconds",BetaDgasinverse
print"\nThe time taken for seperation for liquid will be  year",round(BetaDliquidinverse,3)

The seperation achieved for gas is   0.012

The seperation achieved for liquid is   0.012

The time taken for seperation for gas will be seconds 500.0

The time taken for seperation for liquid will be  year 0.476