Use MATLAB code to solve the following problem:

I am very lost. I have had no formal MATLAB training and barely understand the problem. I really want to understand. Please be clear with all steps and code. Thank you in advance.

I believe I am to use Bernoulli's equation. The below is sample code we did as a class. I just not sure how to make this problem work with that code.

clear clc close all %% Variable Initialization K_en=0.5; K_elbow=1.5; K_exit=1.0; LA=100;% meter LB=100;% meter DA=0.075;% meter DB=0.05;% meter Kine_Vis=1*10^-6;% m^2/s Rough_A=0.002; Rough_B=0.002; HL_A=10.5;% meter HL_B=10.5;% meter g=9.8;% m/s^2 %% Iteration N=100; e=1; Re_A_1=10^8; figure(1) hold on for i=1:N V_A=(HL_A/(friction_coeff(Re_A_1,Rough_A)*LA/DA+K_en+K_elbow*2+K_exit)*2*g)^.5; Re_A_2=V_A*DA/Kine_Vis; if(abs(Re_A_2-Re_A_1)>e) Re_A_1=(Re_A_2+Re_A_1)/2; plot(i,abs(Re_A_2-Re_A_1),'*r'); xlim([0 100]) else break end end Re_B_1=10^8; for i=1:N V_B=(HL_B/(friction_coeff(Re_B_1,Rough_B)*LB/DB+K_en+K_elbow*2+K_exit)*2*g)^.5; Re_B_2=V_B*DB/Kine_Vis; if(abs(Re_B_2-Re_B_1)>e) Re_B_1=(Re_B_2+Re_B_1)/2; plot(i,abs(Re_B_2-Re_B_1),'sg'); xlim([0 100]) else break end end hold off function [ f ] = friction_coeff( Re,Rough ) %FRICTION_COEFF Summary of this function goes here if (Re Two galvanized iron pipes of diameter D are connected to a large water reservoir, as shown. Pipe A has length L and pipe B has length 2L. Both pipes discharge to atmosphere. Which pipe will pass the larger flow rate? Justify without calculating the flow rate in each pipe. Compute the flow rates if H = 10m,D = 50mm, and L-10m. 2L Pipe A Pipe B Two galvanized iron pipes of diameter D are connected to a large water reservoir, as shown. Pipe A has length L and pipe B has length 2L. Both pipes discharge to atmosphere. Which pipe will pass the larger flow rate? Justify without calculating the flow rate in each pipe. Compute the flow rates if H = 10m,D = 50mm, and L-10m. 2L Pipe A Pipe B