Water is flowing in a trapezoidal channel at a rate of Q =20 m^3/s. The critical
ID: 1710339 • Letter: W
Question
Water is flowing in a trapezoidal channel at a rate of Q =20 m^3/s. The critical depth for such a channel must satisfy the equation: 0 = 1 Q^2/gA^3_c B where: g=981 m/s^2, A_c = the cross-sectional area (m^2), and B = the width of the channel at the surface (m). For this case, the width and the cross-sectional area can be related to depth y by B = 3 + y and A_c = 3y + y^2/2 Solve for the critical depth using the: bisection method (use initial guesses x_i= 0.5 and x_u= 2.5) false position method Do iterations until the relative approximate error falls below 1%.Explanation / Answer
SOLUTION:-
Given Data:- Critical depth is given by equation
1-Q/(g*Ac3)*b = 0
Q = 20 m3 /sec
g = 9.81 m/sec2
AC= 3y+y2/2
B = 3+y
STEP 1: let compute a equation using given data Q,g,AC,B
F=1-(20/(9.81)(3y+y2/2)3)*(3+y)
STEP 2: Using bisection method:
Start iteration using xl=.5 and Xu = 2.5
For xl=0.5,
Value of equation: F=1-(20/(9.81)(3(.5)+(.5)2/2)3)*(3+(.5))
=-0.66
F<0
For xu=2.5,
Value of equation: F=1-(20/(9.81)(3(2.5)+(2.5)2/2)3)*(3+(2.5))
=0.991
since F>0
So now we can say that root lie between Xl and Xu.
Now according to bisection method,
Xr=(Xl +Xu )/2
Xr=1.5
ITERATION 2:
Xl=0.5
F(.5)=-0.66
Xu=1.5
F(1.5)=1-(20/(9.81)(3(1.5)+(1.5)2/2)3)*(3+(1.5))
=0.95
Again one xl value is negative and xu value iss positive.
so,again
Xr =(Xl +Xu )/2
=1
ITERATION 3
For F(1)=.81
Again value lies between Xu and Xl.
Now relative approximate error can be calculated as :-
((Xu-Xl)/Xu)*100= rel. error
((1-0.5)/1)*100=50%
now we have to reduce this relative error by repeating above iterations .
follow above procedure to reduce error less than 1%
2. USING FALSE POSITION METHOD:
We have value of Xl=0.5 and Xu=2.5
So by computing value of Xl and Xu in euation = F=1-(20/(9.81)(3y+y2/2)3)*(3+y)
We get
ITERATION 1
F(0.5)=-0.66
F(2.5)=0.991
Now according to False position method
F(Xl)*F(xu)=-0.66*0.991<0
So, 1 root must lie between 0.5 and 2.5.
ITERATION 2
Xr= (Xu*F(Xl)-Xl*F(Xu))/(F(Xl)-F(Xu))
Xr= 1.3
now let Xr=Xu
Xl=0.5
So Xu=1.3
Again computing equation F=1-(20/(9.81)(3y+y2/2)3)*(3+y)
For Xu=1.3
we find
F(Xu)= 0.2
now again F(Xu)*F(xl)<0
so now we calculate relative error,
((Xu-Xl)/Xu)*100= rel. error
((1.3-.5)/1.3)*100= 61.53%
So, to reduce the error we have to repeat the above procedure
please follow the above ITERATION Similar to iteration 1 and 2 by using value Xr, XU and Xl = 0.5 (will remain fix) to reduce the rel. error to be fall below 1%.
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