(Vs)-1, q = 1.7 x 10-19 C, ni = 6.21 x 109 cm-3, and a desired 6.5 × 106 V s cm/

Question

(Vs)-1, q = 1.7 x 10-19 C, ni = 6.21 x 109 cm-3, and a desired 6.5 × 106 V s cm/C. Use (a) bisection and (b) the modified secant method. Mechanical and Aerospace Engineering 8.33 Beyond the Colebrook equation, other relationships, such as the Fanning friction factor f, are available to estimate friction in PROBLEMS 219 pipes. The Fanning friction factor is dependent on a number of pa- 8.36 Aerospace engineers sometimes compute the trajectories of rameters related to the size of the pipe and the fluid, which can all projectiles like rockets. A related problem deals with the trajectory be represented by another dimensionless quantity, the Reynolds of a thrown ball. The trajectory of a ball is defined by the (x, y) number Re. A formula that predicts fgiven Re is the von Karman coordinates, as displayed in Fig. P8.36. The trajectory can be equation, modeled as = 4 log10 (Rev/ f)-0.4 y = (tan@g)x-202cos2a-x" + yo Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to 0.01. Develop a function that uses bisection to solve for fgiven a user-supplied value of Re between 2,500 and 1,000,000. Design the function so that it ensures that the absolute error in the result is Ead0.000005 Find the appropriate initial angle o, if the initial velocity 10-20 m/s and the distance to the catcher xis 35 m. Note that the ball leaves the thrower's hand at an elevation of 2 m and the catcher receives it at 1 m. Express the final result in degrees. Use a value of 9.81 m/s for g and employ the graphical method to develop your initial guesses.

Explanation / Answer

problem asked that for a given value of Re find the f which should satisfy the given function or

which should satisfy 4*log ( Re*sqrt(f) )- 0.4- 1/sqrt(f) =0

%%%%%%%%%%% Matlab code

clc;
clear all;
close all;
format short
format compact;

R=input('Enter the raylond number in between 10,000 to 50,000 : ');
syms x  
y=4*log10(R*sqrt(x))-0.4-1/sqrt(x); %%%using symbol x instead of f
%%% Bisection menthod
disp('let plot function so that we can get initial guess')
t1=0.001:0.0001:0.01;
fx=subs(y,t1);
plot(t1,fx);
xlabel('f');
disp('mark 2 point on graph: 1 st one is positive value and next one is negative value');
[ m n ]=ginput(2);
a=m(1);
b=m(2);

%%% bisection method
% % % bisection method

tol=0.000005;
disp('Bisection method :');

fprintf('iteration a b (a+b)/2 ');
for n=1:100
l1=subs(y,a);
l2=subs(y,b);
  
c(n)=(a+b)/2;
fprintf(' %d %f %f %f ',n,a,b,c(n));
l3=subs(y,c(n));
if (n>2)
if (abs( c(n)-c(n-1))< tol)
break;
end
end
  
if ( l3 > 0 )
a=c(n);
else
b=c(n);
end

end

fprintf('value of Fanning fraction factor : %f ',c(end));

OUTPUT:

Enter the raylond number in between 10,000 to 50,000 : 30000
let plot function so that we can get initial guess
mark 2 point on graph: 1 st one is positive value and next one is negative value
Bisection method :
iteration a b (a+b)/2
1 0.009782 0.003437 0.006609
2 0.006609 0.003437 0.005023
3 0.006609 0.005023 0.005816
4 0.006609 0.005816 0.006213
5 0.006213 0.005816 0.006015
6 0.006015 0.005816 0.005915
7 0.005915 0.005816 0.005866
8 0.005915 0.005866 0.005891
9 0.005891 0.005866 0.005878
10 0.005878 0.005866 0.005872
11 0.005878 0.005872 0.005875
value of Fanning fraction factor : 0.005875
>>

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