# Chapter 4 : The Dense Bed¶

## Example 1, Page 106¶

In [1]:
#Variable declaration
dt=4;               #Vessel diameter in m
Lmf=2;              #Length of the bed in m
ephsilonmf=0.48;    #Void fraction of bed
rhos=1500;          #Density of solid in kg/m**3
rhog=3.6;           #Density of gas in kg/m**3
myu=2E-5;           #Viscosity of gas in kg/m s
po=3;               #Pressure of inlet gas in bar
uo=0.4;             #Superficial velocity of gas in m/s
uorm=40;            #Maximum allowable jet velocity from holes in m/s
g=9.80;             #Acceleration due to gravity in m/s**2
gc=1;
pi=3.1428;

#CALCULATION
#Computation of minimum allowable pressure drop through the distributor
deltapb=((1-ephsilonmf)*(rhos-rhog)*g*Lmf)/gc;  #Calculation of pressure drop in bed using Eqn.(3.17)
deltapd=0.3*deltapb;                            #Calculation of pressure drop in distributor using Eqn.(3)

#Computation of orifice coefficient
Ret=(dt*uo*rhog)/myu;
if Ret>=3000:
Cd=0.60;
elif Ret>=2000:
Cd=0.61;
elif Ret>=1000:
Cd=0.64;
elif Ret>=500:
Cd=0.68;
elif Ret>=300:
Cd=0.70;
elif Ret>=100:
Cd=0.68;

#Computation of gas velocity through orifice
uor=Cd*((2*deltapd)/rhog)**0.5;   #Calculation of gas velocity through orifice by using Eqn.(12)
f=(uo/uor)*100;                   #Calculation of fraction of open area in the perforated plate

#Computation of number of orifices per unit area of distributor
dor=[0.001,0.002,0.004];          #Different orifice diameters in m
n=len(dor);
i=0;
Nor = [0.,0.,0.]
while i<n:
Nor[i]=(uo*4)/(pi*uor*(dor[i])**2);#Calculation of number of orifices by using Eqn.(13)
i=i+1;

#OUTPUT
print 'The pressure drop in bed:%fPa'%deltapb
print 'The minimum allowable pressure drop in distributor:%fPa'%deltapd
if uor<uorm:
print 'The gas veleocity of %fm/s is satisfactory'%uor
else:
print 'The gas veleocity of %fm/s is not satisfactory'%uor

if f<10:
print 'The fraction of open area of %f percent is allowable'%f
else:
print 'The fraction of open area of %f percent is not allowable'%f

print 'Diameter of orifice(m)',
print '\tNumber of orifices per unit area(per sq.m)'

j=0;
while j<n:
print '%f'%dor[j],
print '\t\t%f'%Nor[j]
j=j+1;

print 'This number can be rounded off.'
print 'Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter %fm'%dor[2]

The pressure drop in bed:15251.308800Pa
The minimum allowable pressure drop in distributor:4575.392640Pa
The gas veleocity of 30.250265m/s is satisfactory
The fraction of open area of 1.322302 percent is allowable
Diameter of orifice(m) 	Number of orifices per unit area(per sq.m)
0.001000 		16829.610145
0.002000 		4207.402536
0.004000 		1051.850634
This number can be rounded off.
Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter 0.004000m


## Example 2, Page 108¶

In [5]:
lor=0.1;      #Minimum allowable tuyere spacing in m
uorm=30;      #Maximum allowable jet velocity from the tuyere in m/s
uo=0.4;       #Superficial velocity of gas in m/s
uor=30.2;     #Gas velocity through orifice,from Exa 1, in m/s
Cd=0.6;       #Dicharge coefficient from Exa 1
rhog=3.6      #Density of gas in kg/m**3
pi=3.1428;

#CALCULATION
Nor=1/(lor**2);                     #Calculation of number of orifices per unit area by assuming minimum spacing for tuyeres
dor=((4/pi)*(uo/uor)*(1/Nor))**0.5; #Calculation of diameter of inlet orifiec by using Eqn.(13)

#Computation of diameter of hole for different number of holes per tuyere
q=(lor**2)*uo;                      #Volumetric flow rate in m**3/s
Nh=[8.,6.,4.];                         #Different number of holes per tuyere
n=len(Nh);
i=0;
dh = [0.,0.,0.]
while i<n:
dh[i]=((((q/Nh[i])*(4./pi))/uorm)**0.5);#Calculation of diameter of holes
i=i+1;

deltaph=(rhog/2.)*((uor/Cd)**2)

#OUTPUT
print 'Number of holes(number of holes/tuyeres)',
print '\tDiameter of hole(m)'
j=0;
while j<n:
print '%f'%Nh[j],
print '\t\t\t\t\t%f'%dh[j]
j=j+1;

print 'The design chosen is as follows'
print '\tTuyeres are as shown in Fig.2(b),page 97'
print '\tNumber of holes = %f(Since rectangular pitch is chosen for tuyeres)'%Nh[1]
print '\tDiameter of hole = %fm'%dh[1]
print '\tDiameter of incoming high-pressure-drop orifice = %fm ID'%dor
print 'Checking the pressure drop in tuyeres'
print 'Since pressure drop of %.1f Pa gives sufficiently high \
distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.'%deltaph

Number of holes(number of holes/tuyeres) 	Diameter of hole(m)
8.000000 					0.004606
6.000000 					0.005318
4.000000 					0.006513
The design chosen is as follows
Tuyeres are as shown in Fig.2(b),page 97
Number of holes = 6.000000(Since rectangular pitch is chosen for tuyeres)
Diameter of hole = 0.005318m
Diameter of incoming high-pressure-drop orifice = 0.012984m ID
Checking the pressure drop in tuyeres
Since pressure drop of 4560.2 Pa gives sufficiently high distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.


## Example 3, Page 110¶

In [3]:
deltapd=[3,10]    #Distributor pressure drop in kPa
deltapd2=10.0;    #Distributor pressure drop in kPa
po=101.0;         #Entering air pressure in kPa
To=20.0;          #Entering air temperature in degree C
y=1.4;            #Fugacity of air
deltapb=10;       #Pressure drop in bed in kPa
p3=103;           #Pressure at the bed exit in kPa
F=8;              #Feed rate of coal in tons/hr
H=25;             #Gross heatig value of coal in MJ/kg
Fa=10;            #Air required at standard condition in nm**3/kg
etac=0.75;        #Efficiency of compressor
etap=36;          #Efficiency of plant in %

#CALCULATION
#Calculation of volumetric flow rate of air
vo=((F*1000)*Fa*((To+273)/273.0))/3600.0;

#Case(a) Distributor Pressure drop = 3kPa and Case(b) Distributor Pressure drop = 10kPa
n=len(deltapd);
i=0;
p1= [0,0]
p2 = [0,0]
ws = [0.,0.]
while i<n:
p2[i]=p3+deltapb;      #Calculation of pressure at the entrance of the bed
p1[i]=p2[i]+deltapd[i];#Calculation of pressure before entering the bed
ws[i]=(y/(y-1))*po*vo*((p1[i]/po)**((y-1)/y)-1)*(1.0/etac);#Calculation of power required for the compressor by Eqn.(18) & Eqn.(20)
i=i+1;

#Case(c) 50% of the required bypassed to burn the volatile gases. Distributor Pressure drop = 3kPa
#No change in pressure drop from case(a)
v1=vo/2.0;   #New volumetric flow rate of air
ws1=455/2.0; #Power required for blower for primary air
ws2=(y/(y-1))*po*v1*((p3/po)**((y-1)/y)-1)*(1/etac);#Power required for blower for bypassed air
wst=ws1+ws2;                #Total power required for the two blowers
p=((ws[1]-wst)/ws[1])*100;  #Saving in power when compared to case(a)

#OUTPUT
print 'Case(a)'
print '\tVolumetric flow rate of air = %.2f m**3/hr'%vo
print '\tPower required for compressor = %.0f kW'%ws[0]
print 'Case(b)'
print '\tVolumetric flow rate of air = %.2f m**3/hr'%vo
print '\tPower required for compressor = %.0f kW'%ws[1]
print 'Case(c)'
print '\tVolumetric flow rate of air = %.3f m**3/hr'%v1
print '\tPower required for compressor for primary air = %.1f kW'%ws1
print '\tPower required for blower for bypassed air = %.1f kW'%ws2
print '\tTotal power required for the two blowers = %.0f kW'%wst

Case(a)
Volumetric flow rate of air = 23.85 m**3/hr
Power required for compressor = 454 kW
Case(b)
Volumetric flow rate of air = 23.85 m**3/hr
Power required for compressor = 651 kW
Case(c)
Volumetric flow rate of air = 11.925 m**3/hr
Power required for compressor for primary air = 227.5 kW
Power required for blower for bypassed air = 31.6 kW
Total power required for the two blowers = 259 kW

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