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: 1.1090 7--2109 C 5.0452 3.7065 Re log +5.8506 (Re) -6.8901 (1) 2.8257 In which, f is the friction factor and r is the roughness ratio given by: given by: Re = PDV Where is the roughness of the pipe material and D is the pipe diameter. Note that is dimensionless and hence both & and should be expressed in the same units. Re is Reynolds number which can be calculated from: (2) Where p is the fluid density, 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 log10(2) • Make sure all units are consistent. • Save your file as s123456.py where 123456 is your student ID.
use python program

Answer 1

The code is given below.

import numpy as np
import matplotlib.pyplot as plt

# Given data
rho = 701       # [kg/m^3]
mu = 0.51e-3    # [Ns/m^2]
eps = 0.025e-2  # [m]
D = 14.64e-2    # [cm]
L = 3435        # [m]



def f(R):
    rf = eps/D
    f = (-2*np.log10(rf/3.7065 - 5.0452/R*np.log10(rf**1.1098/2.8257 + 5.8056*R**-0.8981)))**-2 - 0.023
    return f


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("Root of the given equation =",  round(x0, 6))  
        #print("No. of iterations = ", n)  
          
    else: 
        print("Can not find a root in ", 
                   "the given inteval")
  
    return x0

# initializing the values  
x1 = 1e5  
x2 = 5e5
E = 0.001e-2  
Re = secant(x1, x2, E)  


# Velocity
V = mu*Re/rho/D

print("Reynolds Number =",  round(Re, 2))  
print("Velocity = ", round(V,4)," m/s") 

R = np.linspace(3e3,5e5,1000)
plt.plot(R,f(R)+0.023)
plt.xlabel("Reynolds Number")
plt.ylabel("Friction Factor")
plt.grid(True)

plt.show()

Output

Reynolds Number 395076.47 Velocity 1.9633 m/s =

0.045 0.040 0.035 Friction Factor 0.030 0.025 100 O 400 500000 200000 300000 Reynolds Number

Please rate the solution if found satisfactory.