# Chapter 16 : Design for Physical Operations¶

## Example 1, Page 404¶

In [1]:
T=1000.;       #Operating temperature of calciner in degree celcius
deltaHr=1795.; #Heat of reaction in kJ/kg
M1=0.1;        #Molecular weight of Calcium carbonate in kg/mol
M2=0.056;      #Molecular weight of CaO in kg/mol
M3=0.044;      #Molecular weight of Carbon dioxide  in kg/mol
M4=0.029;      #Molecular weight of Air in kg/mol
M5=0.029;      #Molecular weight of Combustion gas in kg/mol
Cp1=1.13;      #Specific heat of Calcium carbonate in kJ/kg K
Cp2=0.88;      #Specific heat of CaO in kJ/kg K
Cp3=1.13;      #Specific heat of Carbon dioxide in kJ/kg K
Cp4=1.00;      #Specific heat of Air in kJ/kg K
Cp5=1.13;      #Specific heat of Calcium carbonate in kJ/kg K
Tf=20.;        #Temperature of feed in degree celcius
ma=15.;        #Air required per kg of fuel in kg
Hc=41800.;     #Net combustion heat of fuel in kJ/kg
Tpi=20.;       #Initial temperature of solids in degree C
Tgi=1000.;     #Initial temperature of gas in degree C

#CALCULATION
mc=1;#Based on 1 kg of Calcium carbonate
B=(1/(Hc-(ma+mc)*Cp5*(T-Tpi)))*(M3*Cp3*(T-Tf)+M2*Cp2*(T-Tf)+deltaHr)#Fuel consumption(kg fuel/kg calcium carbonate)
B1=B*M3/M2;#Fuel consumption(kg fuel/kg Cao)
H=Hc*B1;#Heat required for calcination
eta=deltaHr/(B*Hc);#Thermal efficiency

#OUTPUT
print 'Fuel consumption:%f kg fuel/kg Cao'%B1
print 'Heat requirement for calcination:%f kJ/kg Cao'%H
print 'Thermal efficiency:%f percentage'%(eta*100)

Fuel consumption:0.061731 kg fuel/kg Cao
Heat requirement for calcination:2580.366029 kJ/kg Cao
Thermal efficiency:54.657251 percentage


### Example 2, Page 405¶

In [1]:
from scipy.optimize import fsolve
import math

#INPUT
F = 400.        #Feed rate of Calcium carbonate in tons/day
T = 1000.       #Operating temperature of calciner in degree celcius
deltaHr = 1795.#Heat of reaction in kJ/kg
M1 = 0.1        #Molecular weight of Calcium carbonate in kg/mol
M2 = 0.056     #Molecular weight of CaO in kg/mol
M3 = 0.044     #Molecular weight of Carbon dioxide  in kg/mol
M4 = 0.029    #Molecular weight of Air in kg/mol
M5 = 0.029    #Molecular weight of Combustion gas in kg/mol
Cp1 = 1.13    #Specific heat of Calcium carbonate in kJ/kg K
Cp2 = 0.88    #Specific heat of CaO in kJ/kg K
Cp3 = 1.13   #Specific heat of Carbon dioxide in kJ/kg K
Cp4 = 1.00    #Specific heat of Air in kJ/kg K
Cp5 = 1.17    #Specific heat of Combustion gas in kJ/kg K
Tf = 20.     #Temperature of feed in degree celcius
ma = 15.    #Air required per kg of fuel in kg
uo = 0.8    #Superficial gas velocity in m/s
Hc = 41800.   #Net combustion heat of fuel in kJ/kg
Tpi = 20.     #Initial temperature of solids in degree C
Tgi = 1000.   #Initial temperature of gas in degree C
rhoa = 1.293    #Density of air in kg/m**3
pi = 3.14

#CALCULATION
mc = 1.         #Based on 1 kg of Calcium carbonate
Bguess = 2.     #Guess value of B
def solver_func(B):         #Function defined for solving the system
phi = ((ma+mc)*Cp5*B+(M3*Cp3))/Cp1
T3 = (Tpi+(phi+phi**2+phi**3)*Tgi)/(1+phi+phi**2+phi**3)
phiplus = 30.6*B
Tr = (T+Tpi*phiplus)/(1+phiplus)
return Hc*B+Cp3*(T3-Tpi)+ma*B*Cp4*(Tr-20)-(ma+mc)*Cp5*(T-Tpi)-M3*Cp3*(T-Tpi)-M2*Cp2*(T-Tpi)-deltaHr
#fn = (1/20800)*(2470-T3-13.34*(Tr-20))

B = fsolve(solver_func,1E-6)#Using inbuilt function fsolve for solving Eqn.(23) for tou
phi = ((ma+mc)*Cp5*B+(M3*Cp3))/Cp1
#Temperature of various stages
T1 = (Tpi+(phi)*Tgi)/(1+phi)
T2 = (Tpi+(phi+phi**2)*Tgi)/(1+phi+phi**2)
T3 = (Tpi+(phi+phi**2+phi**3)*Tgi)/(1+phi+phi**2+phi**3)
phiplus = 30.6*B
Tr = (T+Tpi*phiplus)/(1+phiplus)
eta = deltaHr/(B*Hc)           #Thermal efficiency
H = B*Hc/M2                    #Heat requirement
#For lower heat recovery section
Ql = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(Tr+273)))#Volumetric flow rate of gas in the lower heat recovery section
dtl = math.sqrt(4/pi*Ql/uo)#Diameter of lower bed
#For calcination section
Qc = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T+273)))#Volumetric flow rate of gas in the calcination section
dtc = math.sqrt(4/pi*Qc/uo)#Diameter of calcination section
#For I stage
Q1 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T1+273)))#Volumetric flow rate of gas in the I stage
dt1 = math.sqrt(4/pi*Q1/uo)#Diameter of I stage
#For II stage
Q2 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T2+273)))#Volumetric flow rate of gas in the II stage
dt2 = math.sqrt(4/pi*Q2/uo)#Diameter of II stage
#For III stage
Q3 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T3+273)))#Volumetric flow rate of gas in the III stage
dt3 = math.sqrt(4/pi*Q3/uo)#Diameter of III stage

#OUTPUT
print '\nDiameter of lower bed:%fm'%(dtl)
print '\nDiameter of calcination section:%fm'%(dtc)
print '\nBed no.\t\t1\t2\t\t3'
print '\nDiameter(m)%f\t%f\t%f'%(dt1,dt2,dt3)

#The value of diameter of each section is largely deviating from the values in the textbook. This is because the fuel consumption B have not been included in the energy balance equation. And the value of molecular weight is wrong by one decimal point.

Diameter of lower bed:7.097814m

Diameter of calcination section:13.351483m

Bed no.		1	2		3

Diameter(m)12.728715	13.270865	13.340712


## Example 3, Page 413¶

In [2]:
T=20;         #Temeprature in degree C
M=0.018;      #Molecular weight of water in kg/mol
Q=10;         #Flow rate of dry air in m**3/s
R=82.06E-6;   #Universal gas constant
pi=0.0001;    #Initial moisture content in atm
pj=0.01;      #Final moisture content in atm

#CALCULATION
a=Q*(273+T)/273;  #Term At*uo
b=a*M/(R*(T+273));#Term C*At*uo
#The value of slope can be found only by graphical mehtod. Hence it has been taken directly from the book(Page no.414,Fig.E3)
m=10.2;
Fo=b/m;           #Flow rate of solids
Q3=(b/Fo)*(pj-pi);#Moisture content of leaving solids

#OUTPUT
print '\nMoisture content of leaving solids:%.3f kg H2O/kg dry solids'%Q3

Moisture content of leaving solids:0.101 kg H2O/kg dry solids


### Example 4, Page 422¶

In [2]:
from scipy.optimize import fsolve
import math

#INPUT
Qfi = 0.20          #Initial moisture fraction
Qfbar = 0.04        #Average final moisture fraction
rhos = 2000.        #Density of solid in kg/m**3
Cps = 0.84          #Specific heat of solids in kJ/kg K
Fo = 7.6E-4         #Flow rate of solids in kg/m**3
Tsi = 20.            #Inital temperature of solids in degree C
rhog = 1.            #Density of gas in kg/m**3
Cpg = 1             #Specific heat of gas in kJ/kg K
uo = 0.3            #Superficial gas velocity in m/s
Tgi = 200           #Initial temperature of gas in degee C
L = 2370            #Enthalpy of liquid in kJ/kg
Cpl = 4.2           #Specific heat of liquid in kJ/kg K
dt = 0.1            #Diameter of reactor in m
Lm = 0.1            #Length of fixed bed in m
ephsilonm = 0.45    #Void fraction of fixed bed
pi = 3.14
Fo1 = 1             #Feed rate for commercial-scale reactor in kg/s

#CALCULATION
#(a)Bed temperature
Teguess = 50#Guess value of Te
def solver_func(Te):#Function defined for solving the system
return (pi/4.)*dt**2*uo*rhog*Cpg*(Tgi-Te)-Fo*(Qfi-Qfbar)*(L+Cpl*(Te-Tsi))-Fo*Cps*(Te-Tsi)

Te = fsolve(solver_func,Teguess)

#(b)Drying time for a particle
xguess = 2#Guess value of x, ie term tou/tbar
def solver_func1(x):        #Function defined for solving the system
return 1-(Qfbar/Qfi)-(1-math.exp(-x))/x

x = fsolve(solver_func1,xguess)
W = (pi/4.)*dt**2*Lm*(1-ephsilonm)*rhos#Weight of soilds in bed
tbar = W/Fo#Mean residence time of solids from Eqn.(59)
tou = tbar*x#Time for complete drying of a particle

#(c)Commercial-scale dryer
W1 = Fo1*tbar
Atguess = 5#Guess value of area
def  solver_func3(At):          #Function defined for solving the system
return  At*uo*rhog*Cpg*(Tgi-Te)-Fo1*(Qfi-Qfbar)*(L+Cpl*(Te-Tsi))-Fo1*Cps*(Te-Tsi)

At = fsolve(solver_func3,Atguess)
dt1 = math.sqrt(4/pi*At)#Diameter of commercial-scale dryer
Q1 = At*uo*rhog#Flow rate necessary for the operation

#OUTPUT
print 'Bed temperature:%f degree C'%(Te)
print 'Time for complete drying of particle:%fs'%(tou)
print 'Flow rate of gas necessary for Commercial-scale dryer:%fkg/s'%(Q1)

Bed temperature:58.728126 degree C
Time for complete drying of particle:527.431202s
Flow rate of gas necessary for Commercial-scale dryer:3.098684kg/s


## Example 5, Page 425¶

In [3]:
import math

#Variable Declaration
rhos=1600.;     #Density of solid in kg/m**3
Cps=1.25;       #Specific heat of solids in kJ/kg K
Fo=0.5;         #Flow rate of solids in kg/s
Tsi=20.;        #Inital temperature of solids in degree C
Qwi=1.;         #Initial moisture fraction in water
Qwf=0.2;        #Final moisture fraction in water
Qhi=1.1;        #Initial moisture fraction in heptane
Qhf=0.1;        #Final moisture fraction in heptane
Tgi=240.;       #Initial temperature of gas in degee C
Te=110.;        #Bed temperature in degree C
ephsilonm=0.45; #Void fraction of fixed bed
ephsilonf=0.75; #Void fraction of fluidized bed
uo=0.6;         #Superficial gas velocity in m/s
di=0.08;        #Diameter of tubes in m
li=0.2;         #Pitch for square arrangement
hw=400.;        #Heat transfer coefficient in W/m**2 K
Tc=238.;        #Temperature at which steam condenses in degree C
#Specific heats in kJ/kg K
Cwl=4.18;       #Water liquid
Cwv=1.92;       #Water vapor
Chl=2.05;       #Heptane liquid
Chv=1.67;       #Heptane vapor
#Latent heat of vaporization in kJ/kg
Lw=2260.;       #Water
Lh=326.;        #Heptane
#Density of vapor in kg/m**3 at operating conditions
rhow=0.56;      #Water
rhoh=3.1;       #Heptane
Lf=1.5;         #Length of fixed bed in m
t=140.;         #Half-life of heptane in s
L=1.5;          #Length of tubes in heat exchanger
pi=3.14;

#CALCULATION
#(a) Dryer without Internals
xw=(Qwi-Qwf)/(Qhi-Qhf);                #Water-heptane weight ratio
xv=((Qwi-Qwf)/18.)/((Qhi-Qhf)/100.);   #Water-heptane volume ratio
T=(Qwi-Qwf)/18.+(Qhi-Qhf)/100.;        #Total volume
rhogbar=((Qwi-Qwf)/18.)/T*rhow+((Qhi-Qhf)/100.)/T*rhoh;           #Mean density of the vapor mixture
Cpgbar=(((Qwi-Qwf)/18.)/T)*rhow*Cwv+(((Qhi-Qhf)/100.)/T)*rhoh*Cwv;#Mean specific heat of vapor mixture
#Volumetric flow of recycle gas to the dryer in m**3/s from Eqn.(53)
x=(Cpgbar*(Tgi-Te))**-1*(Fo*(Qwi-Qwf)*(Lw+Cwl*(Te-Tsi))+Fo*(Qhi-Qhf)*(Lh+Chl*(Te-Tsi))+Fo*(Cps*(Te-Tsi)));
r=Fo*((Qwi-Qwf)/rhow+(Qhi-Qhf)/rhoh); #Rate of formation of vapor in bed
uo1=uo*(x/(x+r));                     #Superficial velocity just above the distributor
At=x/uo1;                             #Cross-sectional area of bed
dt=math.sqrt(4./pi*At);               #Diameter of bed
B=-math.log(Qwf/Qwi)/t;               #Bed height from Eqn.(63)
tbar=((Qhi/Qhf)-1)/B;                 #Mean residence time of solids
W=Fo*tbar;                            #Weight of bed
Lm=W/(At*(1-ephsilonm)*rhos);         #Static bed height
Lf=(Lm*(1-ephsilonm))/(1-ephsilonf);  #Height of fluidized bed

#(b) Dryer with internal heaters
f=1/8.0;                 #Flow rate is 1/8th the flow rate of recirculation gas as in part (a)
x1=f*x;                  #Volumetric flow of recycle gas to the dryer in m**3/s from Eqn.(53)
uo2=uo*(x1/float(x1+r)); #Superficial velocity just above the distributor
Abed=x1/uo2;             #Cross-sectional area of bed
q=(Fo*(Qwi-Qwf)*(Lw+Cwl*(Te-Tsi))+Fo*(Qhi-Qhf)*(Lh+Chl*(Te-Tsi))+Fo*(Cps*(Te-Tsi)))-Abed*uo2*Cpgbar*(Tgi-Te);#Heat to be added from energy balance of Eqn.(53)
Aw=q*10**3/(hw*(Tc-Te));#Total surface area of heat exchanger tubes
Lt=Aw/(pi*di);          #Total length of tubes
Nt=Lt/L;                #Total number of tubes
Atubes=Nt*(pi/4*di**2); #Total cross-sectional area of tubes
Atotal=Abed+Atubes;     #Total cross-sectional area of tube filled dryer
d=math.sqrt(Atotal*pi/4.);#Diameter of vessel
li=math.sqrt(Atotal/Nt);  #Pitch for square array of tubes

#OUTPUT
print '\t\t\tBed diameter(m)\tRecycle vapor flow(m**3/s)'
print 'Without internal heater\t%f\t%f'%(dt,x)
print 'With heating tubes\t%f\t%f'%(d,x1)

			Bed diameter(m)	Recycle vapor flow(m**3/s)
Without internal heater	3.630144	5.331235
With heating tubes	1.503916	0.666404

In [ ]: