Homework Solution: The flow rate Q in m^3/s in an open channel of circular cross-section shown in the figure below is given by Q = 2^3/2…

    HW1_5 The flow rate Qin m/s in an open channel of circular cross-section shown in the figure below is given by whereiven 8m/s is the gravitational constant and D, = (1-cost) 232D5/2 V (θ _ 0.5 sin(20))3/2 De is given by 5/2 diameter 2m diameter 4m theta Q, m/s Q, m/s 10 0.0006 0.0031 15 0.0028 0.0158 20 0.0087 0.0493 25 0.0210 0.1186 30 0.0427 0.2415 35 0.0774 0.4380 40 0.1289 0.7291 45 0.2008 1.1358 50 0.2967 1.6783 55 0.4198 2.3746 60 0.5728 3.2400 De Note that the units for θ in these equations is radians. Create at least one function handle for this equation, which has has two input, diameter dand angle θ. Use the function handles to compute flow rate Q, and print the table shown above. Note that in the table, 0 is in degrees
    The flow rate Q in m^3/s in an open channel of circular cross-section shown in the figure below is given by Q = 2^3/2 D^5/2_c squareroot g (theta - 0.5 sin (2 theta))^3/2/8 squareroot sin theta (1 - cos theta)^5/2 where g = 9.8 m/s^2 is the gravitational constant and D_c is given by D_c = d/2 (1 - cos theta) Note that the units for theta in these equations is radians. Create at least one function handle for this equation, which has two input, diameter d and angle theta. Use the function handles to compute flow rate Q, and print the table shown above. Note that in the table, theta is in degrees.

    Expert Answer

    HW1_5 The course trounce Qin m/s in an known muniment of round cross-section shown in the shape adown is absorbed by whereiven 8m/s is the gravitational perpetual and D, = (1-cost) 232D5/2 V (θ _ 0.5 ungodliness(20))3/2 De is absorbed by 5/2 bisection 2m bisection 4m theta Q, m/s Q, m/s 10 0.0006 0.0031 15 0.0028 0.0158 20 0.0087 0.0493 25 0.0210 0.1186 30 0.0427 0.2415 35 0.0774 0.4380 40 0.1289 0.7291 45 0.2008 1.1358 50 0.2967 1.6783 55 0.4198 2.3746 60 0.5728 3.2400 De Note that the units ce θ in these equations is radians. Create at lowest undivided discharge feel ce this equation, which has has span input, bisection dand inclination θ. Use the discharge feels to value course trounce Q, and sculpture the board shown over. Note that in the board, 0 is in degrees

    The course trounce Q in m^3/s in an known muniment of round cross-section shown in the shape adown is absorbed by Q = 2^3/2 D^5/2_c squareroot g (theta – 0.5 ungodliness (2 theta))^3/2/8 squareroot ungodliness theta (1 – cos theta)^5/2 where g = 9.8 m/s^2 is the gravitational perpetual and D_c is absorbed by D_c = d/2 (1 – cos theta) Note that the units ce theta in these equations is radians. Create at lowest undivided discharge feel ce this equation, which has span input, bisection d and inclination theta. Use the discharge feels to value course trounce Q, and sculpture the board shown over. Note that in the board, theta is in degrees.

    Expert Counterpart

     

    Matlab Code clc;
    clear all;
    inclination = [10:5:60]; % Defining inclination

    g = 9.8;
    % Calculating ce d = 2
    D = @(d, theta) (d / 2)*(1 – cos(theta));
    ce i = 1:numel(angle)
    Dc1(i) = feval(D, 2, deg2rad(angle(i)));
    end
    % Calculating Q1 ce d = 2
    ce i = 1:numel(Dc1)
    Q1(i) = (2^(3/2)*Dc1(i)^(5/2)*sqrt(g).*(deg2rad(angle(i)) …
    – 0.5.*sin(2*deg2rad(angle(i)))).^(3/2) ) / …
    ( 8*sqrt(sin(deg2rad(angle(i)))).*((1 – cos(deg2rad(angle(i)))).^(5 / 2)) );
    end

    % Calculating ce d = 4
    ce i = 1:numel(angle)
    Dc2(i) = feval(D, 4, deg2rad(angle(i)));
    end
    % Calculating Q2 ce d = 4
    ce i = 1:numel(Dc2)
    Q2(i) = (2^(3/2)*Dc2(i)^(5/2)*sqrt(g).*(deg2rad(angle(i)) …
    – 0.5.*sin(2*deg2rad(angle(i)))).^(3/2) ) / …
    ( 8*sqrt(sin(deg2rad(angle(i)))).*((1 – cos(deg2rad(angle(i)))).^(5 / 2)) );
    end

    fprintf(‘ttbisection 2mttbisection 4mn’);
    fprintf(‘thetatQ m/stttQ m/snn’);
    ce j = 1:numel(Q1)

    fprintf(‘%dtt%.4fttt%.4fn’, inclination(j), Q1(j), Q2(j));
    end

    OUTPUT