Droplet size distribution of a spray image

Hi I'm currently trying to identify the spray characteristics of the attached image.

I want to obtain the DROPLET SIZE DISTRIBUTION, I have attemped to do so but I am getting an exponential distribution (attached) as opposed to the expected log-normal distribution.

I therefore assume i am using the wrong method and was wondering if someone could help me figure out how to do this.

function DROPLET_SIZE_DISTRIBUTION
clc;
close all;
workspace;
format longg;
format compact;
fontSize = 18;

%Convert to grayscale.
grayImage = imread('image1.tif');
[rows, ~, numberOfColorBands] = size(grayImage);
if numberOfColorBands > 1
    grayImage = grayImage(:,:,2); 
end

subplot(2, 3, 1);%Display the original image.
imshow(grayImage);
title('Original Image', 'FontSize', fontSize);
set(gcf, 'Units', 'Normalized', 'Outerposition', [0, 0, 1, 1]);

% Threshold the image
thresholdValue = 150;
binaryImage = grayImage > thresholdValue; 
subplot(2, 3, 2);% Display the image.
imshow(binaryImage);
title('Binary Image', 'FontSize', fontSize);

cc = bwconncomp(binaryImage); %Separate droplets    
stats = regionprops(cc, 'Area');% calculate area of each separate region
area = cat(1,stats.Area);
area(area>50) = []; % filter bigger blobs/regions

%Plot the Histogram
subplot(2, 3, 3);
histogram(area,100)
xlabel('relative size')
ylabel('droplets')
title('Droplet size distribution','FontSize',fontSize);

Answer 1

[chansheng liu]

It's hard to identify the particle size only by the gray image. Because there are always dense areas where droplets get together, it's impossible to discriminate these droplets in 3D from the 2D image. One of the feasible solutions is apply the laser diagnostic technology, such as Phase Doppler Anemometry (PDA) or Particle/Droplet Image Analysis (PDPA) systems.

Answer 2

[Image Analyst]

I've done several spray angle programs here in Answers. Just search for spray.

However if you want individual particle information, that would be difficult or impossible to obtain directly from images and you may have to either measure other things (like weight of sprayed liquid as a function of time) or measure it in other ways, like @chansheng liu mentioned.