I am struggling to get started on this MATLAB question. Any tips on how to use for loops in this? Or important things to take note of?

[Ned Williams]

Anything is helpful!

Answer 1

[Image Analyst]

Do you know how to program in the MATLAB language yet?

MATLAB Academy - Free 2 hour training

How do I get help on homework questions on MATLAB Answers? - MATLAB Answers - MATLAB Central

Here's a start

sa = 32760; % Surface area
wallHeight = 26;
data = readmatrix('data_river_discharge.dat');
inputRate = data(:, 1); % Column 1 is the river input charging rate (into the lake).
outputRate = data(:, 2); % Column 2 is the discharge rate from the dam (out of the lake).
numTimePeriods = size(data, 1); % Number of rows.  Each row is a different time point.
for k = 1 : numTimePeriods
    % Code goes here
    if waterLevel > 0.8 * wallHeight
        % Goes over spillway.
        Qout = 
    elseif waterLevel < 0.3 * wallHeight
        % Reduce outflow
        Qout = 
    else
        % Normal operating condition.
    end
    % Get water level at end of this time period
    waterLevel = something involving prior water level and Qout and Qin
end

[Ned Williams]

thanks for this reply!

however, i still do not understand how i would have the waterlevel at any day in the data. Does it use the intial level of 18.8m and build off that? I am sorry, I am new to this.

[Image Analyst]

At the beginning of the loop (start of the time period) you need to take the current volume and add in the input volume that flowed in during that time period and the output volume that flowed out during that time period. This will give you the new volume at the end of that time period. Then you compute the new water level = volume / (surface area) for the end of the time period. Then depending on the water level at the end of that time period, you need to adjust the output flow Qout -- this will be used on the next loop iteration to determine how much net volume change you will have.

[Ned Williams]

@Image Analyst, is there any chance you could show something like what that code would look like. with the loop that calculates new volume each day

[Image Analyst]

I explained it in words. I'm worried that if I fill in the remaining few steps you won't be able to turn it in. Aren't you afraid of getting caught "cheating" if you turn in my solution as your own? Post what you have so far (more than just my code) and we'll try to help if you can't figure it out.

Debugging in MATLAB | Doug's MATLAB Video Tutorials

[Ned Williams]

@Image Analyst

%% Data

clc;close;clear

data=xlsread('data_RiverDischarge.xls');

sa=352760;

wallHeight=26;

vol=sa*wallHeight;

Rin=data(:,1);

days=size(data,1);

mat=[];

y(1)=18.8;

Qin=data;

Qout(1)=5.33

p=[];

for i=2

y(i)=y(i-1)+((Qin(i-1))/sa)-(Qout(i-1)/sa)

p=y/wallHeight

for Qo=Qout

if 0.8>p>0.3

Qo=5.33

elseif 0.8<p

Qo=5.33+(2.2*Qo)

elseif p<0.3

Qo=(1/2)*Qin(i)

elseif p==1

fprintf('Dam is full!\n')

end

end

i=i+1

end

I am just playing around with it. I am obviouslty making many mistakes but all I am trying to achieve is using the formula for y(i) and incorporating the conditional Qout value using a nested if. If i make any sense, some tips in the right direction would be good. cheers

[Image Analyst]

When you do this:

y(i)=y(i-1)+((Qin(i-1))/sa)-(Qout(i-1)/sa)

p=y/wallHeight

you're setting the ith element of y in the first equation, so y is a vector. In the next line, you're using the entire y (all elements) so p will not be a single number, but a whole vector of numbers, as I'm sure you found out when you stepped through the code. So p < 0.3 will also be a logical vector, not a single true or false value but a whole array of them.

Also

if 0.8>p>0.3

is not allowed. You have to do it this way, with 2 comparisons

if p > 0.3 && p < 0.8

If you do this:

    for i=2

then the loop will only iterate once. Setting i to a new value at the end of the loop does not change the fact that it will still only go through once. The format is

    for k = startingValue : incrementValue : endingValue

Looks like you first need to spend 2 hours with this:

MATLAB Academy - Free 2 hour training

[Ned Williams]

@Image Analyst

clc;close;clear

format short g

data=xlsread('data_RiverDischarge.xls');

sa=352760;

H=26;

vol=sa*H;

Rin=data(:,1);

days=size(data,1);

y(1)=18.8;

Qin=data;

Qout=5.33;

Qr=2.2*Qout;

for i=2:1:days

I(i-1)=(Qin(i-1))/sa

O(i-1)=(Qout(i-1))/sa

R(i-1)=(Qr(i-1))/sa

y(i)=y(i-1)+I(i-1)-O(i-1)-R(i-1)

p=y(i)/H

for Q=Qout

if p>0.3 && p<0.8

Q=5.33; Qr=0

elseif p>0.8

Q=5.33;Qr=2.2*Qo

elseif p<0.3

Q=(1/1)*Qin(i)

else p>=1

disp('overflow')

end

end

y(i+1)=y(i)+I(i)-O(i)-R(i)

p=(y(i+1))/H

end

Does this make any more sense> I am still getting an error saying that the Index exceeds the number of array elements (1)