# Chapter 3: Mass-Transfer Coefficients¶

## Ex3.1:Page 53¶

In [8]:
# Illustration 3.1
# Page: 53

print'Illustration 3.1 - Page: 53\n\n'

# solution

#****Data*****#
# a = CO2 b = H2O
Ca0 = 0;#[kmol/cubic m]
Cai = 0.0336;#[kmol/cubic m]
Dab = 1.96*10**(-9);# [square m/s]
#*******#

density = 998.0;# [kg/cubic m]
viscosity = 8.94*10**(-4);#[kg/m.s]
rate = 0.05;#[kg/m.s] mass flow rate of liquid
L = 1;#[m]
g = 9.81;#[m/square s]
# From Eqn. 3.10
Del = ((3*viscosity*rate)/((density**2)*g))**(1.0/3);# [m]
Re = 4*rate/viscosity;
# Flow comes out to be laminar
# From Eqn. 3.19
Kl_avg = ((6*Dab*rate)/(3.141*density*Del*L))**(1.0/2);#[kmol/square m.s.(kmol/cubic m)]
bulk_avg_velocity = rate/(density*Del);#[m/s]
# At the top: Cai-Ca = Cai_Ca0 = Cai
#At the bottom: Cai-Cal
# From Eqn. 3.21 & 3.22
Cal = Cai*(1-(1.0/(exp(Kl_avg/(bulk_avg_velocity*Del)))));# [kmol/cubic m]
rate_absorption = bulk_avg_velocity*Del*(Cal-Ca0);# [kmol/s].(m of width)
print'The rate of absorption is ',round(rate_absorption,8),' kmol/sec.(m of width)'
# The actual value may be substantially larger.

Illustration 3.1 - Page: 53

The rate of absorption is  7.2e-07  kmol/sec.(m of width)


## Ex3.2: Page 56¶

In [17]:
# Illustration 3.2
# Page: 56

print'Illustration 3.2 - Page: 56\n\n'

# solution

#***Data****#
d = 0.025;# [m]
avg_velocity = 3;# [m/s]
viscosity = 8.937*10**(-4);# [kg/m.s]
density = 997;# [kg/m**3]
#*********#

kinematic_viscosity = viscosity/density;# [square m/s]
Re = d*avg_velocity*density/viscosity;
# Reynold's number comes out to be 83670
# At this Reynold's number fanning factor = 0.0047
f = 0.0047;
L = 1;# [m]
press_drop = 2*density*f*L*(avg_velocity**2)/(d);# [N/square m]
P = 3.141*(d**2)*avg_velocity*press_drop/4;# [N.m/s] for 1m pipe
m = 3.141*(d**2)*L*density/4;
# From Eqn. 3.24
Ld = ((kinematic_viscosity**3)*m/P)**(1.0/4);# [m]
# From Eqn. 3.25
Ud = (kinematic_viscosity*P/m)**(1.0/4);# [m/s]
print'Velocity of small eddies is',round(Ud,4),'m/s'
print'Length scale of small eddies is',round(Ld,7),'m'

Illustration 3.2 - Page: 56

Velocity of small eddies is 0.0549 m/s
Length scale of small eddies is 1.63e-05 m


## Ex3.3: Page 69¶

In [1]:
# Illustration 3.3
# Page: 69

print'Illustration 3.3 - Page: 69\n\n'

# solution

# Heat transfer analog to Eqn. 3.12
# The Eqn. remains the same with the dimensionless conc. ratio replaced by ((tl-to)/(ti-to))

# The dimensionless group:
# eta = 2*Dab*L/(3*del**2*velocity);
# eta = (2/3)*(Dab/(del*velocity))*(L/del);
# Ped = Peclet no. for mass transfer
# eta = (2/3)*(1/Ped)*(L/del);

# For heat transfer is replaced by
# Peh = Peclet no. for heat transfer
# eta = (2/3)*(1/Peh)*(L/del);
# eta = (2/3)*(alpha/(del*velocity))*(L/del);
# eta = (2*alpha*L)/(3*del**2*velocity);
print'Heat transfer analog to Eqn. 3.21 is eta = (2*alpha*L)/(3*del**2*velocity)'

Illustration 3.3 - Page: 69

Heat transfer analog to Eqn. 3.21 is eta = (2*alpha*L)/(3*del**2*velocity)


## Ex3.4: Page-69¶

In [21]:
# Illustration 3.4
# Page: 69

import math
print'Illustration 3.4 - Page: 69\n\n'

# solution

#***Data****#
# a = UF6 b = air
# The average heat transfer coefficient: Nu_avg = 0.43+0.532(Re^0.5)(Pr^0.31)
# The analogus expression for mass transfer coefficient: Sh_avg = 0.43+0.532(Re^0.5)(Sc^0.31)
d = 0.006;# [m]
velocity = 3.0;# [m/s]
surf_temp = 43.0;# [C]
bulk_temp = 60.0;# [C]
avg_temp = (surf_temp+bulk_temp)/2; #[C]
density = 4.10;# [kg/cubic m]
viscosity = 2.7*10**(-5);# [kg/m.s]
Dab = 9.04*10**(-6);# [square m/s]
press = 53.32;# [kN/square m]
tot_press = 101.33;# [kN/square m]
#******#

avg_press = press/2.0; # [kN/square m]
Xa = avg_press/tot_press;
Xb = 1-Xa;
Re = d*velocity*density/viscosity;
Sc = viscosity/(density*Dab);
Sh_avg = 0.43+(0.532*(2733**0.5)*(0.728**0.5));
c = 273.2/(22.41*(273.2+avg_temp));# [kmol/cubic m]
F_avg = Sh_avg*c*Dab/d;#[kmol/cubic m]
Nb = 0.0;
Ca1_by_C = press/tot_press;
Ca2_by_C = 0.0;
Flux_a = 1.0;
# Using Eqn. 3.1
Na = Flux_a*F_avg*math.log((Flux_a-Ca2_by_C)/(Flux_a-Ca1_by_C));#[kmol UF6/square m.s]
print'Rate of sublimation is',round(Na,8),' kmol UF6/square m.s'
# the answer is slightly different in textbook due to approximation

Illustration 3.4 - Page: 69

Rate of sublimation is 0.00102088  kmol UF6/square m.s


## Ex3.5: Page 73¶

In [27]:
# Illustration 3.5
# Page: 73

print'Illustration 3.5 - Page: 73\n\n'

# solution

#****Data****#
velocity = 15.0;# [m/s]
G = 21.3;# [kg/square m.s]
#******#

# Since the experimental data do not include the effects of changing Prandtl number.

# Jh = (h/(Cp*density*viscosity)) = (h/Cp*G)*(Pr^(2/3)) = Shi(Re);

# Shi(Re) must be compatible with 21.3*(G**0.6);
# Let Shi(Re) = b*(Re**n);
# Re = (l*G)/viscosity;

# h = (Cp*G/(Pr**(2/3)))*b*(Re**n);
# h = (Cp*G/(Pr**(2/3)))*b*((l*b/viscosity)**n) = 21.3*(G**0.6);

n = 0.6-1;
# b = 21.3*((Pr**(2/3))/Cp)*((l/viscosity)**(-n));

# Using data for air at 38 C & 1 std atm.
Cp1 = 1002;# [kJ/kg.K]
viscosity1 = 1.85*10**(-5);#[kg/m.s]
k1 = 0.0273;#[W/m.K]
Pr1 = (Cp1*viscosity1)/k1;
b_prime = 21.3*(Pr1**(2.0/3)/Cp1)*((1/viscosity1)**0.4);
# b = b_prime*l**(0.4);
# Jh = (h/(Cp*G))*Pr**(2/3) = b_prime*((l/Re)**(0.4)) = Shi(Re);

# The heat mass transfer analogy will be used to estimate the mass transfer coefficient. (Jd = Jh)

# Jd = (KG*Pbm*Mav*Sc**(2/3))/(density*viscosity) = Shi(Re) = b_prime*((l/Re)**0.4);

# KG*Pbm = F = (b_prime*density*viscosity)/(Re^0.4*Mav*Sc**(2/3)) = (b_prime*(density*velocity)**0.6*(viscosity^0.4))/(Mav*Sc**(2/3));

# For H2-H20, 38 C, 1std atm
viscosity2 = 9*10**(-6);# [kg/m.s]
density2 = 0.0794;# [kg/cubic m]
Dab = 7.75*10**(-5);# [square m/s]
Sc = viscosity2/(density2*Dab);

# Assuming desity, Molecular weight and viscosity of the gas are essentially those of H2

Mav = 2.02;# [kg/kmol]
F = (b_prime*(density2*velocity)**0.6*(viscosity2**0.4))/(Mav*Sc**(2.0/3));# [kmol/square m.s]
print'The required mass transfer: ',round(F,5),' kmol/square m.s'

Illustration 3.5 - Page: 73

The required mass transfer:  0.00525  kmol/square m.s


## Ex3.6:Page 77¶

In [7]:
# Illustration 3.6
# Page: 77

print'Illustration 3.6 - Page: 77\n\n'

# solution
from scipy import integrate
import math
#***Data***#
Dp = 0.0125;# [m]
viscosity = 2.4*10**(-5);# [kg/m.s]
Sc = 2.0;
E = 0.3;
Go = (2*10**(-3))/0.1;# molar superficial mass velocity [kmol/square m.s]
#********#

# a = CO b = Ni(CO)4
# Nb = -(Na/4);
Flux_a = 4.0/3;
Ca2_by_C = 0;# At the metal interface
# Ca1_by_C = Ya #mole fraction of CO in the bulk

# Eqn. 3.1 becomes: Na = (4/3)*F*log((4/3)/((4/3)-Ya));

# Let G = kmol gas/(square m bed cross section).s
# a = specific metal surface
# z = depth
# Therefore, Na = -(diff(Ya*G))/(a*diff(z));# [kmol/((square m metal surface).s)];
# For each kmol of CO consumed, (1/4)kmol Ni(CO)4 forms, representing a loss of (3/4) kmol per kmol of CO consumed.
# The CO consumed through bed depth dz is therefore (Go-G)(4/3) kmol;
# Ya = (Go-(Go-G)*(4/3))/G;
# G = Go/(4-(3*Ya));
# diff(YaG) = ((4*Go)/(4-3*Ya)**2)*diff(Ya);

# Substituting in Eqn. 3.64
# -(4*Go/((4-3*Ya)**2*a))*(diff(Ya)/diff(z)) = (4/3)*F*log(4/(4-3*Ya));

# At depth z:
# Mass velocity of CO = (Go-(Go-G)/(4/3))*28;
# Mass velocity of Ni(CO)4 = ((Go-G)*(1/3))*170.7;
# G_prime = 47.6*Go-19.6G; # total mass velocity [kg/square m.s]
# G_prime = Go*(47.6-19.6*(4-3*Ya));# [kg/m.s]
# Re = (Dp*G')/viscosity

# With Go = 0.002 kmol/square m.s & Ya in the range 1-0.005, the range of Re is 292-444;
# From table 3.3:
# Jd = (F/G)*(Sc**(2/3)) = (2.06/E)*Re**(-0.575);
# F = (2.06/E*(Sc)**(2/3))*(Go/(4-3*Ya))*Re**(-0.575);

a = 6*(1-E)/Dp;

# Result after arrangement:

X2=lambda Ya:-((4*Go)/((4-(3*Ya))**2.0*a))*(3.0/4)*(E*(Sc**(2.0/3))*(4-(3*Ya))/(2.06*Go)*(1/math.log(4.0/(4-(3*Ya)))))*(((Dp/viscosity)*(Go*(47.6-(19.6/(4.0-(3*Ya))))))**0.575);# [m]
print'The bed depth required to reduce the CO content to 0.005 is',round(Z[0],3),'m'
#the answers are slightly different in textbook due to approximation while here answers are precise

Illustration 3.6 - Page: 77

The bed depth required to reduce the CO content to 0.005 is 0.132 m


## Ex3.7: Page 80¶

In [36]:
# Illustration 3.7
# Page: 80

print'Illustration 3.7 - Page: 80\n\n'

# solution

#****Data*****#
# a = water b = air
out_dia = 0.0254;# [m]
wall_thick = 0.00165;# [m]
avg_velocity = 4.6;# [m/s]
T1 = 66.0;# [C]
P = 1.0;# [atm]
Pa1 = 0.24;# [atm]
k1 = 11400.0;# [W/(square m.K)]
T2 = 24.0;# [C]
k2 = 570.0;# [W/square m.K]
k_Cu = 381.0;# [w/square m.K]
#******#

# For the metal tube
int_dia = out_dia-(2*wall_thick);# [m]
avg_dia = (out_dia+int_dia)/2;# [mm]
Nb = 0;
Flux_a = 1;
Ya1 = 0.24;
Yb1 = 1-Ya1;
Mav = (Ya1*18.02)+(Yb1*29);# [kg/kmol]
density = (Mav/22.41)*(273/(273+T1));# [kg/cubic m]
viscosity = 1.75*10**(-5);# [kg/m.s]
Cpa = 1880.0;# [J/kg.K]
Cpmix = 1145.0;# [J/kg.K]
Sc = 0.6;
Pr = 0.75;
G_prime = avg_velocity*density;# [kg/square m.s]
G = G_prime/Mav;# [kmol/square m.s]
Re = avg_dia*G_prime/viscosity;
# From Table 3.3:
# Jd = Std*Sc**(2/3) = (F/G)*Sc**(2/3) = 0.023*Re**(-0.17);
Jd = 0.023*Re**(-0.17);
F = (0.023*G)*(Re**(-0.17)/Sc**(2.0/3));

# The heat transfer coeffecient in the absence of mass transfer will be estimated through Jd = Jh
# Jh = Sth*Pr^(2/3) = (h/Cp*G_prime)*(Pr^(2/3)) = Jd
h = Jd*Cpmix*G_prime/(Pr**(2.0/3));

U = 1/((1/k1)+((wall_thick/k_Cu)*(int_dia/avg_dia))+((1/k2)*(int_dia/out_dia)));# W/square m.K

# Using Eqn. 3.70 & 3.71 with Nb = 0
# Qt = (Na*18.02*Cpa/1-exp(-(Na*18.02*Cpa/h)))*(T1-Ti)+(Lambda_a*Na);
# Qt = 618*(Ti-T2);
# Using Eqn. 3.67, with Nb = 0, Cai/C = pai, Ca1/C = Ya1 = 0.24;
# Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1));

# Solving above three Eqn. simultaneously:
Ti = 42.2;# [C]
pai = 0.0806;# [atm]
Lambda_a = 43.4*10**6;# [J/kmol]
Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1)));# [kmol/square m.s]
Qt1 = 618*(Ti-T2);# [W/square m]
Qt2 = ((Na*18.02*Cpa/(1-exp(-(Na*18.02*Cpa/h))))*(T1-Ti))+(Lambda_a*Na);# [W/square m]

# since the value of Qt1 & Qt2 are relatively close
print'The local rate of condensation of water is ',round(Na,6),' kmol/square m.s'

Illustration 3.7 - Page: 80

The local rate of condensation of water is  0.000232  kmol/square m.s


## Ex3.8: Page 81¶

In [42]:
# Illustration 3.8
# Page: 81

import math
print'Illustration 3.8 - Page: 81\n\n'
print'Illustration 3.8 (a)\n\n'

# Solution (a)

#***Data****#
# a = water b = air
Nb = 0;
h = 1100.0;# [W/square m]
#*****#

Ma = 18.02;# [kg/kmol]
Cpa = 2090;# [J/kg.K]
T1 = 600.0;# [C]
Ti = 260;# [C]
# The positive dirn. is taken to be from the bulk gas to the surface.
Has = 2.684*(10**6);# enthapy of saturated steam at 1.2 std atm, rel. to the liquid at 0 C in [J/kg]
Hai = 2.994*(10**6);# enthalpy of steam at 1 std atm, 260 C in [J/kg]

# Radiation contributions to the heat transfer from  the gas to the surface are negligible. Eqn. 3.70 reduces to
Na = -((h/(Ma*Cpa))*log(1-((Cpa*(T1-Ti))/(Has-Hai))));# [kmol/square m.s]
print'The rate of steam flow reqd. is',round(Na,4),' kmol/square m.s\n\n'
# negative sign indicates that the mass flux is into the gas

print'Illustration 3.8 (b)\n\n'

# Solution (b)

#***Data****#
# a  =  water b  =  air
h  =  572.0;# [W/square m]
T1  =  25.0;# [C]
#******#

Ti  =  260.0;# [C]
# The positive dirn. is taken to be from the bulk gas to the surface.
Has  =  1.047*10**(5);# enthapy of saturated steam at 1.2 std atm, rel. to the liquid at 0 C in [J/kg]
Hai  =  2.994*(10**6);# enthalpy of steam at 1 std atm, 260 C in [J/kg]

# Radiation contributions to the heat transfer from  the gas to the surface are negligible. Eqn. 3.70 reduces to
Na  =  -((h/(Ma*Cpa))*math.log(1-((Cpa*(T1-Ti))/(Has-Hai))));# [kmol/square m.s]
print'The rate of steam flow reqd. is',round(Na,4),' kmol/square m.s'
# the answer of part B in textbook is incorrect

Illustration 3.8 - Page: 81

Illustration 3.8 (a)

The rate of steam flow reqd. is -0.0348  kmol/square m.s

Illustration 3.8 (b)

The rate of steam flow reqd. is 0.0028  kmol/square m.s