#Variable declaration
P=101.3e3
T=297
R=8314 #gas constant
RH=60 #Relative humidity
p_b1=12.2e3 #Vapor pressure at 297 K
p_b2=6e3 #Vapor pressure at 283 K
M_w=78 #molecular weight of benzene
M_a=28 #Mass of nitrogen
#Calculation
#From the definition of percentage relative humidity (RH)
P_w=(p_b1)*(RH/100.0)
#In the benzene -nitrogen mixture:
m_b=P_w*M_w/(R*T) #mass of benzene
m_n=(P-P_w)*M_a/(R*T) #mass of nitrogen
H=m_b/m_n #Humidity at 297 K
#In order to recover 80 per cent of the benzene, the humidity must be reduced to 20 per cent of the initial value
H_o=H*.20
#Thus in equation 13.2
P_r=p_b2+(p_b2/M_a*M_w)/H_o
#Result
print"\n The required pressure is =",round(P_r*1e-3,1),"kN/m**2"
#Variable declaration
P=101.3e3 #Given pressure
T=300 #Given Temperature
RH=25 #Percentage relative humidity of water
P_wo=3.6e3 #partial pressure of water vapour when air is saturated with vapour
M_w=18 #Molecular weight of water
M_a=29 #Molecular weight of air
R=8314 #gas constant
#Calculation
P_w=P_wo*(RH/100.0)
m_w=P_w*M_w/(R*T) #mass of water vapour
m_a=(P-P_w)*M_a/(R*T) #mass of water air
Vs_w=1/m_w #specific volume of water vapour at 0.9 kN/m**2
Vs_a=1/m_a #specific volume of air at 100.4 kN/m**2
H=m_w/m_a #Humidity
H_v=Vs_a #Humid volume
H_p=(P-P_wo)/(P-P_w)*RH #Percentage humidity
#Result
print"(a) The partial pressure of the water vapour in the vessel = ",P_w*1e-3,"kN/m**2"
print" b) Specific volume of water vapour =",round(Vs_w),"m**3/kg"
print" Specific volume of air =",round(Vs_a,3),"m**3/kg"
print"(c) Humidity of air =",round(H,4),"kg water/kg air"
print" Humid volume =",round(H_v,3),"m**3/kg"
print"(d) Percentage humidity =",round(H_p,1),"per cent"
#Variable declaration
T=310.0 #Temperature of moist air
T_w=300.0 #Wet bulb tempeature
L=2440e3 #Latent heat of vapourisation of water at 300 K
P=105e3 #Given total pressure
P_wo1=3.6e3 #Vapour pressure of water vapour at 300 K
P_wo2=6.33e3 #Vapour pressure of water vapour at 310 K
M_w=18.0 #Molecular weight of water
M_a=29.0 #Molecular weight of air
#Calculation
import sympy
H_w=(P_wo1/(P-P_wo1))*(M_w/M_a) #The humidity of air saturated at the wet-bulb temperature
#Therefore, taking (h/hD*rho*A) as 1.0 kJ/kg K, in equation 13.8:
H=H_w-(1e3/L)*(T-T_w)
#In equation 13.2:
x=sympy.Symbol('x')
P_w=sympy.solve(H*(P-x)*M_a-M_w*x)
RH=P_w[0]/P_wo2*100.0
#Result
print"\n The humidity of the air =",round(H,3),"kg/kg"
print"\n The percentage relative humidity (RH)=",round(RH,1),"per cent"
#Variable declaration
#Refer HUMIDITY ENTHALPY PLOT Figure 13.5 Page 748 as Humidity Chart
#According the given passes and situatuion
T = [325,301,308,312,315] #[K]
H = [.005,.015,.022,.027,.032] #[kg/kg]
#From Humidity Chart on humidifying to 60 percent humidity
Tw = [296,301,305,307] #[K]
#Calculation
Hin = H[4]-H[0] #[kg/kg] Increase in Humidity
#From Humitidy Chart at the obtained leaving conditions
v = .893 #[m**3/kg] Specific Volume of dry air
vs = .968 #[m**3/kg] Specific Volume of Saturated air
vh = .937 #[m**3/kg] Humid Volume of air of 60 per cent humidity by Interpolation of Curve in Humidity Chart
x = 5 #[m**3/s] Amount of moist air leaves the dryer in (b)
m = x/vh #[kg/s] Mass of air passing through the dryer
mw = m*Hin # [kg/s] Mass of water evaporated
Tb = 370.0 #[K] dry Bulb temperature corresponding to humidity of .005 kg/kg and wet-bulb temperature 307 K
#Result
print'\n\n (a) The temperature of the material on each tray (in Kelvin)'
print "\t",Tw,"K"
print' Thus the air leaving the system is at',T[4],' K and 60 per cent humidity.'
print'\n\n (b) If 5 m**3/s moist air leaves the dryer, The amount of water removed is',round(mw,3),' kg/s.'
print'\n\n (c) The Temperature to which the inlet air would have to be raised\n to carry out the drying in single stage is',Tb,'K.'
#Variable declaration
G1=1 #flow rate of air at 350 K
PH1=10 #Percentage Humidity at 350 K
G2=5 #flow rate of air at 300 K
PH2=30 #Percentage Humidity at 300 K
#from fig 13.4
H1=0.043 #Humidity at 350 K and 10 percent humidity
H2=0.0065 #Humidity at 300 K and 30 percent humidity
#Calculation
import sympy
#Thus, in equation 13.23:
H=((G1*H1)+(G2*H2))/(G1+G2)
#from fig 13.5
H_1=192e3 #Entahlpy at 350 K and 10 percent humidity
H_2=42e3 #Enthalpy at 300 K and 30 percent humidity
x=sympy.Symbol('x')
H_=sympy.solve((G1*(x-H_1))-(G2*(H_2-x)))
#From Figure 13.5:
#at H_ (Enthalpy)= 67 kJ/kg and H(humidity) = 0.0125 kg/kg
T=309
#Result
print"\n Humidity of final stream =",round(H,4),"kg/kg"
print"\n Entahlpy of the resultant stream =",round(H_[0]*1e-3),"kJ/kg"
print"\n Temperature of the resultant stream =",T,"K"
#Variable declaration
G_s=0.15 #Mass flow rate of steam
T=400 #Temperature to which steam is superheated
T_a=320 #Tremperature of air
RH_a=20 #Percentage relative humidity of air
G_a=5 #Mass flow rate of air
L=2258e3 #latent heat of steam
Cp=2e3 #Specific heat of superheated steam
#Calculation
#Enthalpy of steam
H_3=4.18*(373-273)+L+Cp*(T-373)
#From Figure .13.5:
#at T=320 K and 20 percent Relative Humidity
H1=0.013 #Humidity
H_1=83e3 #Enthalpy
#By making required constructions we get
H=0.043
H_=165e3
T_s=324
#from chart and equation 13.28
G_case2=0.41
#Result
print"\n Relative humidity of stream=",round(H,3),"kg/kg"
print" Entahlpy of stream =",H_*1e-3," kJ/kg"
print" Temperature of stream =",T_s,"K"
print"\n\n When exit temperature = 330 K"
print" The required flow of steam = ",round(G_case2,2)," kg/s"
#Variable declartaion
theta1=300 #K
theta2=320 #K
H=0.2 #Percent humidity
#Calculation
#From standard charts(Fig 13.4):
H1=0.0045 #Humidity in kg/kg
H2=0.0140 #Humidity in kg/kg
#Heat the air from 300 to 318 K
theta=294.5 #K
#H=0.0140 kg/kg
#Result
print"Heat the saturated air at H=",H2,"kg/kg from",theta,"to",theta2,"K"
#Variable declaration
T_water_enter=301.0 #Water entering the tower in [K]
T_water_leave=294.0 #Water entering the tower in [K]
air_drybulb=287.0 #[K]
air_wetbulb=284.0 #[K]
m=4810.0 #[kg/s] Water flow rate
#Calculation
delta_T=T_water_enter-T_water_leave #Temperatue range of water
T_mean=0.5*(T_water_enter+T_water_leave) #MEan wate temperature in [K]
H_mean=92.6 #kJ/kg
H_drybulb=49.5 #kJ/kg
delTdash=T_mean-air_drybulb
delHdash=H_mean-H_drybulb
import math
Dt=m/(0.00369*(delHdash/delta_T)*math.sqrt(delTdash+(0.0752*delHdash)))
from scipy.optimize import fsolve
Ct=5.0
h=100.0 #height assumed in [m]
Ab=Dt/(19.50*math.sqrt(h)/(Ct**1.5))
Id=math.sqrt(round(Ab)*4/math.pi)
#Result
print "Thus the internal diameter of the column at sill level,",round(Id,1),"m"
print"Since this gives a height-dia ratio=3:2 the design is aceptable"
#Variable declaration
Ti=293 #Inlet temperature in [K]
L=2495 #Latent heat of water in kJ/kg
flow=0.68 #[m**3/m**2 s] flow of air
water_thru=0.26 #Water througput in [kg/m**2 s]
Cp_air=1.003 #Specific heat of air in kJ/kg K
Cp_wv=2.006 #Sp heat of water vapour
wv_in=0.003 #Water vapour in inlet air[kg/kg dry air]
wv_in_m=0.005 #kmol/kmol dry air
H=0.003
hd_a=0.2 #Resistance whole
#Calculation
Hg1=Cp_air*(Ti-273.0)+H*(L+Cp_wv*(Ti-273.0)) #;kJ/kg]
flow=(1.0-wv_in_m)*flow
rho=(29.0/22.4)*(273.0/Ti) #Density of air at 293 K[kg/m**3]
G_dash=rho*flow #[kg/m**s]
slope=water_thru*4.18/G_dash
theta_l1=293
Hg1=27.67 #kJ/kg
import numpy as np
Hg=np.array([27.7,30,40,50,60,70,76.5])
theta=np.array([293,294.5,302,309,316,323,328])
Hf=np.array([57.7,65,98,137,190,265,355])
Hf_Hg=Hf-Hg
one_Hf_Hg=1.0/(Hf_Hg)
%matplotlib inline
plot(Hg,one_Hf_Hg)
ylabel("1/(Hf-Hg)")
xlabel("Hg(kJ/kg)")
title("Evaluation of the integral of dHg/(Hf-Hg)")
from scipy.integrate import simps, trapz
area = simps(Hg,one_Hf_Hg,dx=10)
area=0.65
#At bottom of the column:
delta_H1=Hf[0]-Hg[0]
#At the top of the column:
delta_H2=Hf[6]-Hg[6]
m_water_temp=0.5*(theta[0]+theta[6])
Hgm=52
Hfm=145
del_Hm=93
f=0.79
intg=(Hg[6]-Hg[0])/(f*del_Hm)
z=intg*G_dash/(hd_a*rho)
show()
#Result
print "Height of packing,z=",round(z,1),"m"
#Variable declaration
L_dash=0.25 #n-butanol flow rate
Mw=74.0 #Molecular wt of butanol
Ma=29.0 #Molecular wt of air
hDa=0.1 #Mass transfer coefficient per unit volume
b=2.34 #Psychometric ratio
lamda=590 #kJ/kg
Cl=2.5 #kJ/kg K
s=1.05 #Humid heat of gas in [kJ/kg K]
Cp_dry=1.001 #Sp heat of dry air
G=0.7 #Vaour rate
T=[295,300,305,310,315,320,325,330,335,340,345,350]
Pwo=[0.59,0.86,1.27,1.75,2.48,3.32,4.49,5.99,7.89,10.36,14.97,17.5] #kN/m**2
Ho=[0]*12
Hf=[0]*12
Hf_dash=[0]*12
#Calculation
lamda_dash=lamda/b
print"______________________________"
print"T\tHo\tHf\tHf`"
print"_______________________________"
for i in range(0,12):
Ho[i]=Pwo[i]/(101.3-Pwo[i])*(74/29.0)
Hf[i]=(1/(1+Ho[i]))*Cp_dry*(T[i]-273.0)+Ho[i]*(Cl*(T[i]-273.0)+lamda)
Hf_dash[i]=Cp_dry*(T[i]-273.27)/(1+Ho[i])+Ho[i]*(Cl*(T[i]-273.0)+lamda_dash)
print T[i],"\t",round(Ho[i],4),"\t",round(Hf[i],2),"\t",round(Hf_dash[i],2)
%matplotlib inline
plot(T,Hf)
plot(T,Hf_dash)
xlim(285,340)
ylim(0,200)
theta_l1=T[0]
Hg1=s*(290.0-273.0)
Hg1=round(Hg1,1)
rho_air=(Ma/22.4)*(273/310.0)
G_dash=G*rho_air
slope=L_dash*Cl/G_dash
theta_l2=T[7]
Hg2=46 #kJ/kg
#y=mx+c
c=Hg2-slope*theta_l2
plot([theta_l1,theta_l2],[Hg1,Hg2])
xlabel("Temperature(K)")
ylabel("Enthalpy(kJ/kg)")
title("Graphical construction")
Hg1_dash=(Hg1+(b-1)*s*(290.0-273.0))/b #[kJ/kg]
slp=-3*s
theta_f1=293.0
import numpy as np
Hf_dash1=np.array([23.9,29.0,35.3,42.1,50.0,57.9,66.7,75.8])
Hg_dash=np.array([17.9,22.0,26.0,30.0,34.0,38.1,42.0,46.0])
Hfd_Hg=Hf_dash1-Hg_dash
HfHG=1/(Hfd_Hg)
mean=[0]*7
interval=[0]*7
value=[0]*7
summ=0
show()
print"____________________________________________________________________"
print"Mean value in interval\t\tInterval\tValue of integral"
print"____________________________________________________________________"
for i in range(0,7):
mean[i]=(HfHG[i]+HfHG[i+1])/2.0
interval[i]=Hg_dash[i+1]-Hg_dash[i]
value[i]=mean[i]*interval[i]
summ=summ+value[i]
print round(mean[i],3),"\t\t\t\t",interval[i],"\t\t\t",round(value[i],3),"\t"
z=G_dash*summ/(b*hDa)
print"Value of integral=",round(summ,3)
#Result
print "\n\n\nHeight=",round(z,1),"m"
print"The final point is given by theta=",308,"K"
print"NOTE:For 2nd part,calculation can't be done on python,it is drawn geometrically"