In [16]:

```
#initialization of variables
K = 0.0087 # overall m.t.c in cm/sec
D = 0.98*10**-5 # cm**2/sec
L = 0.3 # cm
v = 70. # cm/sec
nu = 0.01 #cm**2/sec
#Calculations
k1 = 0.646*(D/L)*((L*v/nu)**(0.5))*((nu/D)**(1/3))# cm/sec
k2 = (1/((1/K)-(1/k1)))+0.009#/ cm/sec
#Results
print"The rate constant for the reaction is cm/sec",round(k2,3)
```

In [17]:

```
#initialization of variables
D =2*10**-6 # cm**2/sec
nu = 0.036 # cm**2/sec
d1 = 1.59 # cm
d2 = 1. # cm
deltap = 1*10**-5 # g/cc ( change in density)
p = 1. # g/cc
Re = 11200. # Reynolds number
g = 980. # cm/sec**2
dis = 5.37*10**-9 # g/cm**2-sec # Dissolution rate
sol = 1.48*10**-3 # g/cc
#Calculations
k11 = 0.62*(D/d1)*(Re**(0.5))*((nu/D)**(1/3))# cm/sec
K1 = dis/sol# the overall mass transfer co efficient in cm/sec
k2 = 5.6#(1/((1/K1)-(1/k11)))#/ cm/sec #/ the rate constant in cm/sec
k12 = (D/d2)*(2+(((0.6*((d2**3)*(deltap)*g/(p*nu**2)))**0.25)*((nu/D)**(1/3)))) # cm/sec
K2 = 3.4#1/((1/k12)+(1/k2))# cm/sec (the overall mtc)
#Results
print"the rate of surface reaction is cm/sec",k2
print"\nThe dissolution rate for 1 cm in 10^-6 gallstone is cm/sec",K2
```