Group 3 workshop 11

REPORT
Amrish Mahadeo- 95606
Justin Young -92913
Kade De Gannes-95371
Introduction
The design problem being solved consists of 2 containers, one cone-shaped and the other is
cylindrical, containing its respective liquid and is being emptied into a cylindrical mixing tank
for mixing. The solution being pursued must result in the liquids in both containers pouring into
the mixing tank at a desirable rate over a certain period to properly produce the desired mixture.
To be more specific, the cone-shaped container must have a determined size and shape. Bernoulli
equations which are a form of the differential equation were used to determine pressure, speed,
and most importantly for this problem height.
Method
Firstly, each tank was taken into account as a separate entity and derived their theoretical
equations through the use of a Bernoulli equation. Once a differential equation was found for the
first 2 tanks through using the ode function and modeled in Mathlab for the plot graphs of height
vs time and observe which tank had a faster drainage time. These models were then tailored to
achieve the specific to deal with the problem. now with those equations obtained they were
verified by using Reynolds number and also plotting it against v to verify the shape of the graph
was correct. Bernoulli equation. Lastly, a third tank was then introduced so that a batch process
can be done so that all the graphs could be plotted into one that way better observation and
comparison could be taken.

Tank 2 plot task2.3
Task 2.4 Reynolds number

Task 3.1
Task 3.2
Minimum diameter=305
Discussion
The usual steady-state Bernoulli equation does not correctly describe the effect of the area
ratio a/A (where a is the hole area and A is the tank cross-sectional area) on the effluent
velocity. This is because the Bernoulli equation applies only to steady-state flow, and the
flow in this system is transient. Because the level of fluid is changing, the fluid velocity at
any constant elevation within the tank is varying with time. But the usual Bernoulli equation
does not take this part of the fluid acceleration into account. It includes only the advective
part of the acceleration. As the ratio of the area’s a/A gets higher, the error in the prediction
from the usual Bernoulli equation prediction gets worse. For the case where a/A = 1, the
Bernoulli equation fails to predict the required free fall. To determine the solution to this
problem correctly, a time-dependent modification to the Bernoulli equation must be used,
which properly includes the missing part of the acceleration. A possible limitation with
using this method though is the factor of having to use pressure and velocity which are
constantly changing as some might think but with using the differential equation we take
into account the rate of change.

Recommendations
Possible recommendations can be :
Neglecting the term V21V12 seemed reasonable during the derivation and allowed us to further
simplify Bernoulli’s Equation: V2^2−V1^2=2g(h1−h2) =2ghV22−V12=2g(h1−h2) =2gh.Instead of
neglecting this velocity of the top surface of the water, we could have chosen to relate it to the
other velocity of the water at the drain hole. Recall that our balance equation had led to the
following relationship: −a (velocity out) =Adhdt−a (velocity out) =Adhdt. Which turns out to be
V22−(aAV2)2=2ghV2^2−(aAV2) 2=2gh.In contrast to our previous simplified
result V2=2gh−−−√V2=2gh, we now arrive at V2=2gh1−(aA) that way accounts for all the
possibilities of v, and calculation can be done with simply substitution.
Appendix
Code for 1.3
clear all
[t,h]= ode45('one_tank',[0550],[5]);
plot(t,h);
xlabel('Time(s)');
ylabel('Height(m)');
title('Heightof fluid in Tank 1 vstime')

Code for 2.3
clear all
d2 = 0.028;
l2 = 0.5;
f2 = 0.00775;
g = 9.8;
rho = 1000
mu = 0.001
[t,h] = ode45 ('two_tanks',[0550],[3]);
plot(t,h);
xlabel('Time(s)');
ylabel('Tankheight(m)');
title('Heightof fluid in Tank 2 vstime')
v2 = sqrt((2*g*h)/(1+4*l2*f2/d2));
code for3.1
clear all
clc
H10 = 5;
H20 = 3;
H30 = 0;
[t,h] = ode45 ('batch_processing',[0600],[H10 H20 H30]);
figure(1)
plot(t,h);
xlabel('Time(s)');
ylabel('Tankheight(m)');
title('Heightsof fluid in Tank 1,Tank 2 and Tank 3 vs time');
legend('

code batch processing
functiondhdt = batch_processing(t,h)
d2 = 0.028;
l2 = 0.5;
f2 = 0.00775;
g = 9.8;
D2 = 1;
D3 = 2.5;
a = pi*(d2/2)^2;
A2 = pi*(D2/2)^2;
Cd1=0.9;
g= 9.8;
d1 = 0.09;
beta= 48.054*pi/180;
r = h(1)*tan(beta/2);
a1 = 0.0064;
A1 = pi*r^2;
A3 = pi*(D3/2)^2;
v1 = sqrt(2*g*h(1));
if h(1) < 1e-06 %Zero v1 if heightof fluid is negligible
v1 = 0;
end
v2 = sqrt(g*h(2)*inv(1/2+2*l2*f2/d2));
v3 = sqrt(2*g*h(3));
dhdt= [
-Cd1*a1*v1/A1
-a*v2/A2
(Cd1*a1*v1+a*v2)/A3
];

Group 3 workshop 11

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Group 3 workshop 11