# Chapter 9 : Agitation¶

### Example 9.3 - Page No :389¶

In :
import math

# Variables
d = 9/12.;  			 #[ft] - diameter of the impeller
dt = 30/12.;  			 #[ft] - diameter of the math.tank
Nbaffles = 4.  			 #  no. of baffles
h = 30.;  	    		 # [inch]  - height of unit
mu = 10.;    			 #[cP] - vismath.cosity of fluid_
sg = 1.1;  		    	 # specific gravity of fluid_
s = 300.      			 #[rpm] - speed of agitator
CbyT = 0.3;

# Calculations
V = (math.pi*dt**3)/4; 	 #volume of math.tank in ft**3
V1 = V*7.48;  			 #[gal] - volume of math.tank in gallons
mu = mu*(6.72*10**-4);   #[lb/ft*sec]
p = sg*62.4;  			 #[lb/ft**3] - density of fluid_
N = s/60.;  			 #[rps] - impeller speed in revolutions per second
Nre = ((d**2)*N*p)/mu;

# Results
print "Nre = %.2e"%Nre
print " Therefore the agitator operates in the turbulent region"
Npo = 1.62;
gc = 32.174;
P = (Npo*(p*(N**3)*(d**5)))/(gc*550);
Cf = 63025.;
Tq = (P/s)*Cf;
PbyV = P/V;
PbyV1 = P/V1;
TqbyV = Tq/V;
TqbyV1 = Tq/V1;
print " The power per unit volume and the torque per unit volume is  \nP/V = %.2ef hp/ft**3 = %.2e \
hp/gal \nTq/V = %.2f in*lb/ft**3 = %.3f in*lb/gal"%(PbyV,PbyV1,TqbyV,TqbyV1);

Nre = 2.87e+04
Therefore the agitator operates in the turbulent region
The power per unit volume and the torque per unit volume is
P/V = 1.52e-02f hp/ft**3 = 2.03e-03 hp/gal
Tq/V = 3.19 in*lb/ft**3 = 0.427 in*lb/gal


### Example 9.4 - Page No :391¶

In :
# Variables
Tpilot = 30.;
Tlab = 10.;
N1 = 690.;
N2 = 271.;
D2 = 3.;
D1 = 1.;

# Calculations
n = (math.log(N1/N2))/(math.log(D2/D1));
V = 12000/7.48;  			 #[ft**3]
T = ((4.*V)/math.pi)**(1./3);  			 #[ft]
R = 12.69/(30/12.);
N3 = N2*(1./R)**n;  			 #[rpm] - impeller speed in the reactor

# Results
print "impeller speed in rpm = %f"%round(N3,4)
D3 = 0.75*R;  			 #[ft] - reactor impeller diameter
print "reactor impeller diameter in ft = %.3f"%D3
P = 0.1374*((N3/N2)**3)*(R**5);
print "power in hp = %.3f"%P
Cf = 63025.;
Tq = (P/N3)*Cf;  			 #[inch*lb]
print "torque in inch*lb = %.0f"%Tq
print "At this point, the design is complete. \nA sarc ard size impeller would be chosen as \
well as a tan ard size motor7.5 hp or 10 hp"

impeller speed in rpm = 68.044500
reactor impeller diameter in ft = 3.807
power in hp = 7.329
torque in inch*lb = 6789
At this point, the design is complete.
A sarc ard size impeller would be chosen as  well as a tan ard size motor7.5 hp or 10 hp


### Example 9.5 - Page No : 393¶

In :
from numpy import *

# Variables
# given
n = array([0.5, 0.6, 0.7, 0.8, 0.9, 1.0]);
D2 = 3.806;
D1 = 0.25;
R = D2/D1;
N1 = 690.;

# Calculations
N2 = N1*((D1/D2)**n);
P1 = 9.33*10**-3;  			 #[hp]
P2 = P1*R**(5.-3*n);

# Results
print " n                   N,rpm         P,hp"
for i in range(6):
print " %f            %4.0f         %4.0f"%(n[i],N2[i],P2[i]);

# Answers may be differ because of rounding error.

 n                   N,rpm         P,hp
0.500000             177          128
0.600000             135           57
0.700000             103           25
0.800000              78           11
0.900000              60            5
1.000000              45            2