Question

In fluid flow problems, the flow velocity in a long horizontal pipe depends on the pipe material, pipe geometry and fluid properties in addition to the pump power. For a horizontal pipe with a pump, the friction factor can be obtained from many correlations such as Colebrook-White Equation: 1.1998 5.0452 (6) log Re 28257 -)} (1) 3.7065 7--2 log +58506 (Re)-0.3981 In which, f is the friction factor and is the roughness ratio given by: given by: ρν Re Where is the roughness of the pipe material and D is the pipe diameter. Note that ris dimensionless and hence both and D should be expressed in the same units. Re is Reynolds number which can be calculated from: (2) Where is the fluid density, u is the fluid viscosity. A cast iron pipe that is 14.64 cm in diameter is 3.435 km long (3435 m). It is to convey octane. The estimated friction factor fis 0.023. a. Determine the velocity of the flow using secant method with an accuracy of 0.001%. By first solving for Re in (1) and then obtaining the required velocity in (2). b. In another part of the same code, Manipulate equation (1) to have f = F(Re) and plot f versus Re for the interval 3000 S Re s 500,000 For octane: p=701 kg/m =0.51x10 N.s/m? For cast iron: E=0.025 cm Note: in Python, In(2) is coded log (2) while the logarithm to the base of 10 like log(2) is coded log102) Make sure all units are consistent.

solve using paython

Answer 1

https://trinket.io/python3/88d3a020eb

import matplotlib.pyplot as plt
import numpy as np
import math
def f(x):
e=0.025;
p=701;
mu=0.51e-3;
D=14.64;
tf=e/D;
fp=0.023
return 1/np.sqrt(fp)+2*np.log(tf/3.7065-5.0452*np.log((tf**1.3098)/2.8257+5.8506*(x**(-0.8981)))/x);
def fun(x):
e=0.025;
p=701;
mu=0.51e-3;
D=14.64;
tf=e/D;
yy=-2*np.log(tf/3.7065-5.0452*np.log((tf**1.3098)/2.8257+5.8506*(x**(-0.8981)))/x);
return 1.0/(yy*yy);
def secant(x1, x2, E):
n = 0; xm = 0; x0 = 0; c = 0;
if (f(x1) * f(x2) < 0):
while True:
  
# calculate the intermediate value
x0 = ((x1 * f(x2) - x2 * f(x1)) /
(f(x2) - f(x1)));
  
# check if x0 is root of
# equation or not
c = f(x1) * f(x0);
  
# update the value of interval
x1 = x2;
x2 = x0;
  
# update number of iteration
n += 1;
  
# if x0 is the root of equation
# then break the loop
if (c == 0):
break;
xm = ((x1 * f(x2) - x2 * f(x1)) /
(f(x2) - f(x1)));
  
if(abs(xm - x0) < E):
break;
  
print("Reynolds number =",
round(x0, 6));
print("No. of iterations = ", n);
  
else:
print("Can not find a root in ",
"the given inteval");
return x0;
# Driver code
  
# initializing the values
x1 = 10;
x2 = 1000;
E = 0.001e-2;
Re=secant(x1, x2, E);
Re=np.linspace(3000,500000);
ff=fun(Re);
plt.plot(ff,Re);
plt.show();