please use example 1-5 concept and write matlab code for 3 pipes parallel flow network (please assume the data for pipe 3)

MATLAB code for Example 1-5 (this isn't answer)

clear;clc; %IW = 10; % Number of iterations NL = 3; % Number of loops NJ = 4; % Number of lines in Ith loop Km = 1; % minor loss constants e = .00085; v = .00001082; g = 32.2; % gravity constant i = [1;2;3];% loop numbers j = [16 10560;12 15840;14 10560;16 15840;16 15840;14 13200;12 10560;16 10560;12 15840;10 15840;12 15840;12 15840];% pipe diameter lengths D = j(:,1)./12;%pipe diameter in feet A = ((pi/4)*D.^2);%area of pipes Q = [3;1.5;-1.6;-4.0;3.8;1.7;.3;-3;1.8;.4;-1;-1.5]; % inital guessed flow rates V = Q(:,1)./A; ed = e./j(:,1);%flow rates Alpha = (8.*j(:,2))./(pi^2*g*D.^5);%alpha calculations Beta = (8.*Km)./(pi^2*g*D.^4); SBeta = sum(Beta); hm = (Q(:,1).^2)*SBeta; Re = ((abs(V).*D)./v); for IW =1:5; if Re>2000 f = (1.325./(log((.00085./((3.7).*D))+(5.74./(Re.^(.9)))).^2)) else f = 64./Re; end if Q(Q(:,1)>0) Hij = (Alpha.*(Q(:,1)>0).^2.*f)+hm elseif Q(Q(:,1)=2000 DfDq = ((13.69.*(((e./(3.7*D))+(5.74./(Re.^.9)))).^-1))./(Re.*Q(:,1).*(log((e./(3.7.*D)+(5.74./Re.^.9))).^3)) else DfDq = -64./Re.*Q(:,1) end if Q(Q(:,1)>0) DhDq =((2.*Alpha.*Q(:,1).*f)+(Alpha.*Q(:,1).*DfDq)+(2*SBeta.*Q(:,1))) elseif Q((Q(:,1)).0001 break else continue end syms('Q','Qnew',Q(:,1)) subs(Q,Q(:,1),Qnew) IW = IW+1 end

Please assume 12) Write MATLAB code for this problem. Thank you EXAMPLE 1-5 Consider the parallel flow network of Fig. 1-20 with the following specifications: L3,000 ft PA -80 psia L2-3,000 ft ZA 100 ft D28 in 1 0.001 ft 2 B 80ft 2 0.0001 ft 0.001 E0.00015 Find Q1.02,and Ps 0,00003 ft/ 121 FIGURE 1-20 Parallel flow network for Example 1-5 a, 5.3 ft3 /s Solution. This is a type 2 problem. Step 1 Assume Q-3 ft/s Apply the energy equation along line 1 from A to B to obtain assuming that V^ Va-Finding hh with Qi specified is a category 1 problem, and we have 01 vi- 3.82 ft/s 1.273 X 10 ofi0.022, and lbm Step 2 The loss for pipe 2 must then become -h-1497 t-lbf lbm Finding Q, for h, #14.97 ft-IbiIbm is a category Ll problem. We shall use vi as the it- eration variable"The results are given in Table 1-4 TABLE 1-4 Category l Problem for PIPE 2 Vi (fus) Rep (ft-lbt/lbm) 22,233 88,932 72,702 0.0265 1.859 21.47 14.94 3.27 0.0200 Step 3 From step 2,Q4-VA,'' 1.141 ft3/s and Z-1.141 + 3-4.141 ft% we I-1 find the corrected values by using a-mi 53m 3.84 ft/s 4.141 1.141 4.141 e-m 5.3-1.46 ft3/s Step 4 Using Q.-3.84 ft% and Q,-1.46 ft3/s, we compute V . v,' Rep', Reg, f. and /2, to find ft-lbf ibm 23.85 %-23.98 fulbf Ibm The head losses in the pipes then agree to 0.54 percent, which is sufficient With the flow rates in each line and the head loss for the parallel segments known, +23.91 ftlbr 32174 fi-lbm on - W, is computed from the first equation in step 1 32.174 Tbf-f ft-lbf Zs-hf., 80 ft-100ft + 23.91 bm 321743 ft W, Please assume 12) Write MATLAB code for this problem. Thank you EXAMPLE 1-5 Consider the parallel flow network of Fig. 1-20 with the following specifications: L3,000 ft PA -80 psia L2-3,000 ft ZA 100 ft D28 in 1 0.001 ft 2 B 80ft 2 0.0001 ft 0.001 E0.00015 Find Q1.02,and Ps 0,00003 ft/ 121 FIGURE 1-20 Parallel flow network for Example 1-5 a, 5.3 ft3 /s Solution. This is a type 2 problem. Step 1 Assume Q-3 ft/s Apply the energy equation along line 1 from A to B to obtain assuming that V^ Va-Finding hh with Qi specified is a category 1 problem, and we have 01 vi- 3.82 ft/s 1.273 X 10 ofi0.022, and lbm Step 2 The loss for pipe 2 must then become -h-1497 t-lbf lbm Finding Q, for h, #14.97 ft-IbiIbm is a category Ll problem. We shall use vi as the it- eration variable"The results are given in Table 1-4 TABLE 1-4 Category l Problem for PIPE 2 Vi (fus) Rep (ft-lbt/lbm) 22,233 88,932 72,702 0.0265 1.859 21.47 14.94 3.27 0.0200 Step 3 From step 2,Q4-VA,'' 1.141 ft3/s and Z-1.141 + 3-4.141 ft% we I-1 find the corrected values by using a-mi 53m 3.84 ft/s 4.141 1.141 4.141 e-m 5.3-1.46 ft3/s Step 4 Using Q.-3.84 ft% and Q,-1.46 ft3/s, we compute V . v,' Rep', Reg, f. and /2, to find ft-lbf ibm 23.85 %-23.98 fulbf Ibm The head losses in the pipes then agree to 0.54 percent, which is sufficient With the flow rates in each line and the head loss for the parallel segments known, +23.91 ftlbr 32174 fi-lbm on - W, is computed from the first equation in step 1 32.174 Tbf-f ft-lbf Zs-hf., 80 ft-100ft + 23.91 bm 321743 ft W