CollidingParticles.pdf

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CHEMICAL AND ENERGY ENGINEERING
SIMULATIONS OF MECHANICAL PROCESSES
PROBLEM 5.6: COLLIDING PARTICLES
Submitted by: Submitted to:
1. Angshuman Buragohain Prof. Berend van Wachem
Matriculation Nr. 221552 Universitätsplatz, 39106,
2. Devidas Khatri Otto-von-Guericke Universität,
Matriculation Nr. 221549 Magdeburg, G10-236
Chemical and Energy Engineering
Otto-von-Guericke Universität, Magdeburg
Authors: Angshuman Buragohain, Devidas Khatri
Date: 5th July 2019

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Problem 5.6: Colliding particles
Two particles at positions x1=(0,0,0)m and x2=(0,2,0)m have diameters=1m. Particle 1 is
stationary and fixed at the origin and particle 2 with a mass of 0.01kg moves at a velocity of (0,-
0.5,0)m/s.
Objectives:
1. Designing a function which takes positions of two particles and their diameters and returns
the overlap (scalar) and the normal (vector)
2. Using initial condition of the particles, move the particle 2 with a fixed time step=1e-4s using
Leapfrog algorithm (Using predictor -corrector)
3. Moving the particle with time step and evaluating the overlap. kn=1000 and computation of the
normal force based on the linear model given by Equation 1:
4. Verification of the working of the program using 40,000 time-steps
5. Plotting Kinetic energy of particle 2 as a function of time
6. Perform the objectives 3 and 4 using coefficient of restitution e=0.8
Objective 1: Function to take arbitrary position and diameter of two particles and return
the overlap and the normal
The two positions of the particles are x1 and x2. The distance between the particles and their
radii (r1 and r2) are used to calculated the overlap. Overlap = (r1+r2) - (Dist. betn. particles)
function [del, n] = Overlap(x1, x2, d1, d2)
% Radius of the particles
r1=d1/2;
r2=d2/2;
% Relative displacement between the two particles
x12 = x1-x2;
% Distance between the two particles
dist_part=norm(x12);
% Overlap
del = r1+r2-dist_part;
% Normal vector
n = x12/norm(x12)
end

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Objective 2: Function to move the particle 2 with a fixed time step using Leapfrog
mechanism.
The function described below is based on Leap frog mechanism to move particle 2 with the
fixed time step of dt = 10-4
. For leap frog mechanism the velocities for calculation of new
position are calculated at the mid-points of the time intervals (dt/2) which are assumed as vstar
values. Using vstar values x_new is calculated from x_old. The correction of velocity is done
by calculating the velocities. v_new (using v_old values) at the exact points of the iterations i.e
at (dt) intervals using vstar values and the average value of acceleration before and after
interval dt.
The prediction of velocity is given by the equation (1).
(1)
(2)
Replacing (1) in (2), we get
(3)
The correction of velocity after the interval dt is calculated using the vstar value and the
average of the acceleration before and after the interval dt.
(4)
The Leap-frog algorithm can be realised by the figure given below:
function [x_new,v_new] = Particle_move(x_old,v_old,F_new, F_old,m,dt)
% Accelaration from forces before and after time interval dt
a_new = F_new./m;
a_old = F_old./m;
% Prediction of velocity at mid interval using old velocity and old
acceleration
vstar = v_old + 0.5*dt.*a_old;

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% Position shifting calculation
x_new= x_old + vstar*dt;
% Correction of velocity after interval dt
v_new = vstar + 0.5*dt.*(a_old+a_new);
end
Objective 3&4: Moving the particle 2 with the time step given and calculating the force
and final velocities in case of overlap
Force{1}=[0 0 0];
for j=1:N
[del, n] = Overlap(x1, x2(j,:), d1, d2);
if (del<0)
Force{j+1}=[0 0 0];
F_new=Force{j+1};
F_old=Force{j};
[x2(j+1,:) , v2(j+1,:)] = Particle_move(x2(j,:),v2(j,:), F_new,
F_old, m2, dt);
elseif (del>0)
Force{j+1}=-kn_loading*del*n;
F_new=Force{j+1};
F_old=Force{j};
F_old, m2, dt);
end
KE(j+1) = 0.5*m2*dot(v2(j+1,:),v2(j+1,:));
end
The results of running for 40000 time steps is given as follows in the figure below

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The Kinetic energy should have been equal before after the overlap is over. Hence the time
step is not optimum.
Objective 5. Plotting Kinetic energy of particle 2 as a function of time
The Program was run for higher time steps with more precise time intervals and the results are
as follows:

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The best result was obtained for time step = 4E+6 and time interval = 1E-6 as shown below in
context to conservation of energy and momentum (Equal before and after overlap).
Objective 6: Performing the objectives 3 and 4 using coefficient of restitution e=0.8
Force{1} = [0 0 0];
for j=1:N
[del, n] = Overlap(x1, x2(j,:), d1, d2);
% Before and after overlapping
if (del<0)
Force{j+1}=[0 0 0];
F_new=Force{j+1};
F_old=Force{j};
F_old, m2, dt);
% During overlap
elseif (del>0)
v21 = v2(j+1,:)-v1;
% Loading condition
if (dot(v21,n)>0)
kn = kn_loading;
% Unloading condition (e=sqrt(k_unloading/k_loading))
elseif (dot(v21,n)<0)

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kn = kn_loading*(e_rest^2);
end
Force{j+1}=-kn*del*n;
F_new=Force{j+1};
F_old=Force{j};
F_old, m2, dt);
end
KEe(j+1) = 0.5*m2*dot(v2(j+1,:),v2(j+1,:));
end
The kinetic energy was calculated before and after the velocity of particle 2 changes direction
using the loading and the unloading conditions as given in the code above. The result is
plotted as shown below
Introducing a coefficient of restitution of 0.8 reduced the Kinetic energy after unloading to
80% of the initial value. This means the dissipation of Kinetic energy due to some degree of
inelasticity is 20% which is in accordance to concept of partially inelastic collision.

CollidingParticles.pdf

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