In [1]:

    

%pylab


    

    




Using matplotlib backend: MacOSX
Populating the interactive namespace from numpy and matplotlib

In [16]:

    

def S(mu, Di, Do):
    return (mu/4.0)*(1-np.square(Di*1.0/Do))+1


    

In [29]:

    

S(20000,67,70)
S(20000,49,52)


    

    
Out[29]:





561.28106508875737

In [12]:

    

def B_Attenuation(mu, Di, Do):
    return 4.0*np.square(Do)*mu/(np.square(Do*(1+mu))-np.square(Di*(mu-1)))


    

In [27]:

    

B_Attenuation(20000,70,73)
B_Attenuation(20000,52,55)


    

    
Out[27]:





0.0018813774372473384

In [14]:

    

1/0.0024784726519025803


    

    
Out[14]:





403.47429261822106

In [35]:

    

def S2(mu, D1i, D1o, D2i, D2o): #D1 is inner can
    S_1 = (mu/4.0)*(1-np.square(D1i*1.0/D1o))
    S_2 = (mu/4.0)*(1-np.square(D2i*1.0/D2o))
    S12 = (1-np.square(D1o*1.0/D2i))
    Stotal = 1.0+S_1+S_2+S_1*S_2*S12
    print('Stotal is {0}, S_1 is {1}, S_2 is {2}, S12 is {3}'.format(Stotal, S_1,S_2,S12))
    
    return Stotal


    

In [33]:

    

S2(20000, 49, 52, 67,70)


    

    




S_1 is 560.281065089, S_2 is 419.387755102, S12 is 0.448163265306






    
Out[33]:





106287.84020341607

In [36]:

    

#for my can, I used thickness is 1mm. Use inscribed hex circle diameter
S2(60000,2.796-2*0.039 , 2.796, 3.2747- 2*0.039,3.2747)


    

    




Stotal is 138449.326498, S_1 is 825.236235702, S_2 is 706.05911941, S12 is 0.234983905382






    
Out[36]:





138449.3264980063

In [38]:

    

%matplotlib?


    

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