# Chapter 11 Mass Transfer¶

## Exa 11.1¶

In :
#Example Number 11.1
# diffusion coefficient for co2

# Variable declaration

T =298.0			# [K] temperature of air
Vco2 = 34.0 			# molecular volume of co2
Vair = 29.9 			# molecular volume of air
Mco2 = 44.0 			# molecular weight of co2
Mair = 28.9 			# molecular weight of air
P = 1.0132*10**(5) 		# [Pa] atmospheric pressure
# using equation (11-2)
#Calculation
D = 435.7*T**(3.0/2.0)*(((1/Mco2)+(1/Mair))**(1.0/2.0))/(P*(Vco2**(1.0/3.0)+Vair**(1.0/3.0))**(2))

#Result
print "value of diffusion coefficient for co2 in air is",round(D,3),"square centimeter/s"
print "From Table A-8,D=0.164 sq cm/s\n So,they are in fair agreement"

value of diffusion coefficient for co2 in air is 0.132 square centimeter/s
From Table A-8,D=0.164 sq cm/s
So,they are in fair agreement


## Exa 11.3¶

In :
#Example Number 11.3
# Wet-bulb temperature

#Variable declaration

Pg = 2107.0			# [Pa] from steam table at 18.3 degree celcius
Pw = Pg*18.0 			# [Pa]
Rw = 8315.0 			# [J/mol K] gas constant
Tw = 273+18.3	 		# [K]

RHOw = Pw/(Rw*Tw) 		# [kg/cubic meter]

Cw = RHOw 			# [kg/cubic meter]
RHOinf = 0.0 			# since the free stream is dry air
Cinf = 0.0
P = 1.01325*10**(5) 		# [Pa]
R = 287 			# [J /kg  K]
T = Tw 				# [K]
RHO = P/(R*T) 			# [kg/cubic meter]

Cp = 1004.0 			# [J/kg degree celsius]
Le = 0.845
Hfg = 2.456*10**(6) 		# [J/kg]
#Calculations
# now using equation(11-31)

Tinf = (((Cw-Cinf)*Hfg)/(RHO*Cp*(Le**(2.0/3.0))))+Tw 	# [K]
Tin = Tinf-273 					# [degree celsius]

print "Temperature of dry air is",round(Tin,2),"degree celsius"
print "  recalculate the density at the arithmetic-average temperature between wall and free-stream conditions"
print "With this adjustments these results are RHO = 1.143 kg/m**(3) and Tinf = 55.8 degree celcius"

Temperature of dry air is 53.66 degree celsius
recalculate the density at the arithmetic-average temperature between wall and free-stream conditions
With this adjustments these results are RHO = 1.143 kg/m**(3) and Tinf = 55.8 degree celcius


## Exa 11.4¶

In :
#Example Number 11.4
# relative humidity of air stream

#Variable declaration

# these data were taken from previous example
Rho = 1.212 					# [kg/cubic meter]
Cp = 1004 					# [J/kg]
Le = 0.845
Tw = 18.3 					# [degree celsius]
Tinf = 32.2 					# [degree celsius]
Rhow = 0.015666 				# [kg/cubic meter]
Cw = Rhow 					# [kg/cubic meter]

#calculation

Hfg = 2.456*10**(6) 				# [J/kg]
# we use eqn 11-31
Cinf = Cw-(Rho*Cp*Le**(2.0/3.0)*(Tinf-Tw)/Hfg) 	# [kg/cubic meter]
Rhoinf = Cinf 					# [kg/cubic meter]
Rhog = 0.0342 					# [kg/cubic meter]
RH = (Rhoinf/Rhog)*100

#Result

print "Relative humidity is therefore",round(RH,1),"percentage"

Relative humidity is therefore 27.8 percentage


## Exa 11.5¶

In :
#Example Number 11.5
# water evaporation rate

# Variable declaration

Ta = 38+273 			# [K] temperature of atmospheric air
RH = 0.30 			# relative humidity
u = 10.0			# [mi/h] mean wind speed
R = 0.287 			# universal gas constant
Dw = 0.256*10**(-4) 		# [square meter/s] from table A-8(page no.-610)
rho_w = 1000 			# [kg/cubic meter]
# for this calculation we can make use of equation(11-36). from thermodynamic 	steam tables
p_g = 6.545 			# [kPa] at 38 degree celsius
p_s = p_g 			# [kPa]
p_w = RH*p_s 			# [kPa]
p_s = 1.933 			# [in Hg]
p_w = 0.580 			# [in Hg]
# also
u_bar = u*24 			# [mi/day]
# equation(11-36) yields, with the application of the 0.7 factor

E_lp = 0.7*(0.37+0.0041*u_bar)*(p_s-p_w)**(0.88) 		# [in/day]
E_lp = E_lp*2.54/100 						# [m/day]

# noting that standard pan has the diameter of 1.2m, we can use the figure to 		calculate the mass evaporation rate per unit area as
m_dot_w_by_A = E_lp*rho_w/24 				# [kg/h square meter]

# as a matter of interest, we might calculate the molecular-diffusion rate of 		water vapour from equation(11-35), taking z1 as the 1.5m dimension above the 		standard pan.
z1 = 1.5 							# [m]

# since     rho = p/(R*T)
# equation(11-35) can be written as
m_dot_w_by_A1 = 0.622*Dw*p_g*3600/(R*Ta*z1) 			# [kg/h square meter]

#Result

print "Evaporation rate on the land under these conditions is",round(m_dot_w_by_A1,4),"kg/h square meter"

Evaporation rate on the land under these conditions is 0.0028 kg/h square meter