Chapter 21 Simultaneous Heat and Mass Transfer¶

Example 21_1_2 pgno:600¶

In :
#initialization of variables
from math import pi
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 :
#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 :
#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 :
#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 :
#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