#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"
#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"
#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"
#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"