## Finding minimum Cushion Pressure needed to break an Ice for it's different Thickness by Newton-Raphson Method in Python

import matplotlib.pyplot as plt
import numpy as np

# Functions

def f(p):
return pow(p,3)*(1-pow(beta,2))+(0.4*h*pow(beta,2)-sigma*pow(h,2)/pow(r,2))*pow(p,2)+(pow(sigma,2)*pow(h,4)/(3*pow(r,4)))*p-pow((sigma/3)*(pow(h,2)/pow(r,2)),3)

def fprime(p):
return (1-pow(beta,2))*3*pow(p,2)+(0.4*h*pow(beta,2)-sigma*pow(h,2)/pow(r,2))*2*p+pow(sigma,2)*pow(h,4)/(3*pow(r,4))

# Given Inputs

beta = 0.5
r = 40*12                                          # inch
sigma = 150                                        # psi

thickness_ft = np.linspace(0.6,4.2,7)              # feet
thickness_in = thickness_ft*12                     # inch

# Initialization

p = []
Iterations = []

p_guess = 90                                       # psi
alpha = 0.5
tol = 1e-6
iter = 1

# Newton-Raphson with Loop Operation

for h in thickness_in:

while (abs(f(p_guess)) > tol):

p_guess = p_guess-alpha*(f(p_guess))/(fprime(p_guess))

iter = iter+1

p.append(p_guess)

# Table Construction

print ('%29s%44s' % ('Thickness of Ice (feet)','Minimum Cushion Pressure (psi)'))

for k in range(0,len(p)):
print ('%19.1f%45.10f' % (thickness_ft[k],p[k]))

# Plotting the Graph

plt.plot(thickness_ft,p,'r')
plt.title( 'Minimum Cushion Pressure variation A/C to Thickness of Ice')
plt.xlabel('Thickness of Ice (feet)')
plt.ylabel('Minimum Cushion Pressure (psi)')
plt.xlim([0.5,4.5])
plt.ylim([0,0.14])
plt.grid()
plt.savefig('hp')
plt.show()


Table  Obtained:

Graph  Obtained:

Conclusion:

Minimum Cushion Pressure needed to break an Ice always increases with increasing Ice Thickness.

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