Engineering Essay

1377 Words Apr 1st, 2013 6 Pages
UNIVERSITY OF CALIFORNIA, DAVIS DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING COURSE: WATER RESOURCES SIMULATION (ECI 146)
CRN 29727; 4 units INSTRUCTOR: Fabián A. Bombardelli (fabombardelli@ucdavis.edu, fabianbombardelli2@gmail.com, bmbrdll@yahoo.com) OFFICE: 3105, Ghausi Hall Class: Tuesdays and Thursdays-12:10 PM to 1:30 PM (Olson 118) Computer lab: Fridays-1:10 PM to 2:00 PM (Academic Surge 1044) READER: Mr. Kaveh Zamani (kzamani@ucdavis.edu) TEACHING ASSISTANT: Ms. Kate Hewett (kmhewett@ucdavis.edu)

COMPUTER PROBLEM 1: Solution of the Colebrook-White equation via three different methods. Assigned on: Friday, January 13, 2012 Due on: Tuesday, January 24, 2012

Introduction The Moody diagram is the most reliable source of
…show more content…
Notice that the Reynolds number is given in logarithmic scale. Then, we exit the diagram horizontally, starting from the point defined above, towards the left of the diagram.

Page 1 of 4

The Colebrook-White formula was proposed in the first half of last century, and allows for the computation of f . It is expressed as follows:
1 = − 2 log10 f  ε 2.51  +  3 .7 D ℜ f     

(1)

where ε or ks indicates the pipe equivalent roughness, D is the pipe diameter, and ℜ is the Reynolds number. The main characteristic of Equation (1) is that it is implicit. It can be observed that the Darcy-Weisbach coefficient appears at both sides in (1), and there is no way of obtaining it solely as a function of known variables. Therefore, a numerical method needs to be applied to compute that coefficient. Equation (1) can be expressed as:  ε 1 2.51 + 0.869 ln  +  3.7 D ℜ f f   =0   (2)

where the logarithm on base 10 has been converted to the "natural" logarithm. Multiplying by f , Equation (2) yields:  ε 2.51 f 0.869 ln  +  3.7 D ℜ f   =0  

F =1+

(3)

which is still an implicit equation.

Problem 1
You are asked to: 1. Please, develop a code for the bisection method to obtain the value of f in the following 12 points, well distributed on the Moody diagram: ℜ =3 x 106ε D =0.0008; ℜ =3 x 106- ε D =0.00005; ℜ =3 x 107- ε D =0.00001; ℜ =3 x 107ε D =0.002; ℜ =3 x 107- ε D =0.015; ℜ =3 x 105- ε D =10-10; ℜ =3 x 105ε D =0.002; ℜ…

More about Engineering Essay