# Chapter 26 : Fluid-Particle Reactors: Design¶

## Example 26.1 pageno : 592¶

In [1]:
#Lets say F(Ri)/F = F_ri

# Variables
F_50 = 0.3
F_100 = 0.4
F_200 = 0.3;
t_50 = 5.
t_100 = 10.
t_200 = 20.
tp = 8.

# Calculations
a = ((1-(tp/t_50))**3)*F_50
b = ((1-(tp/t_100))**3)*F_100
c = ((1-(tp/t_200))**3)*F_200;
g = [a,b,c];
sum1 = 0;
for p in range(3):
if g[p]>0:
sum1 = sum1+g[p];
f_converted = 1-sum1;

# Results
print " The fraction of solid converted equals %.1f %%"%(f_converted*100)

 The fraction of solid converted equals 93.2 %


## Example 26.2 pageno : 597¶

In [3]:
# Variables
t_avg = 60.         # min
t = 20.             #min

# Calculations
unconverted = ((1./4)*(t/t_avg))-((1./20)*(t/t_avg)**2)+((1./120)*(t/t_avg)**3);
unconverted1 = ((1./5)*(t/t_avg))-((19./420)*(t/t_avg)**2)+((41./4620)*(t/t_avg)**3);
c_avg = (unconverted+unconverted1)/2;

# Results
print "Fraction of original sulfide ore remain unconverted is %.2f"%(c_avg)

Fraction of original sulfide ore remain unconverted is 0.07


## Example 26.3 page no : 600¶

In [5]:
# Variables
F = 1000.               #gm/min
W = 10000.              #gm

# Calculations
t_avg = W/F;
F_50 = 300.
F_100 = 400.
F_200 = 300.             #gm/min
t_50 = 5.
t_100 = 10.
t_200 = 20.             #min

unconverted = ((((1./4)*(t_50/t_avg))-((1./20)*(t_50/t_avg)**2)+ ((1./120)*(t_50/t_avg)**3))*(F_50/F))+((((1./4)*(t_100/t_avg))-((1./20)*(t_100/t_avg)**2)+((1./120)*(t_50/t_avg)**3))*(F_100/F))+((((1./4)*(t_200/t_avg))-((1./20)*(t_200/t_avg)**2)+((1./120)   *(t_50/t_avg)**3))*(F_200/F))
converted = 1-unconverted;

# Results
print "The mean conversion of soild is %f"%(converted)
print " The answer slightly differs from those given in book as we have considered \
only significant terms in infinite series"

The mean conversion of soild is 0.795208
The answer slightly differs from those given in book as we have considered only significant terms in infinite series


## Example 26.4 pageno : 601¶

In [7]:
# Variables
t1 = 1.             #hr
t2 = t1/0.1
a = 0.
r = 1               #ton/hr
# Calculations
while a<=1:
x = (1./4)*(a)-((1./20)*(a)**2)+((1./120)*(a)**3);
if x >0.099 and x<0.1005:
r = a;
a += .0001

FBo = 1.            #tons/hr
t_avg = t2/r;
W = t_avg*FBo;

# Results
print " The needed weight of bed is %.f"%(W),
print "tons"

 The needed weight of bed is 23 tons

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