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Problem session #1 – HT_Spring2024
Problem Session #1
Heat Transfer, MECE:315:001, Spring 2024
Name: ____________
(Jan. 17, Wednesday)
Ordinary Differential Equations (ODE) and Solutions:
1. First-Order and Second-Order Ordinary Differential Equations:
For a function,
y = −0.5 x 4 + 4 x 3 − 10 x 2 + 8.5 x + 1 ,
One has,
dy
= −2 x 3 + 12 x 2 − 20 x + 8.5
dx
d2y
dx
2
= −6 x 2 + 24 x − 20
(1)
- first-order ordinary differential equation
(2)
- second-order ordinary differential equation
(3)
2. Solution of an ODE by Integration – Integral Constants:
One can integrate Eq. (2) once to obtain the solution of this ODE:
y = −0.5 x 4 + 4 x 3 − 10 x 2 + 8.5 x + C
where C is the constant of integration and has to be determined based on a given condition (called initial
condition). For example, Eq. (1) become the solution of the ODE if one specifies that,
→
C = 1.
y (0) = 1
For the second-order ODE as Eq. (3), one has to integrate it twice. First integration gives,
dy
= −2 x 3 + 12 x 2 − 20 x + C1 ,
dx
and integrate one more time, one has
1
y = − x 4 + 4 x3 − 10 x 2 + C1x + C2
2
(4)
where we have two integral constants, C1 and C2.
3. Initial-Value and Boundary-Value ODEs:
We need to have two conditions to determine the two constants C1 and C2 in the solution (4).
If one specifies both values of function y and its first derivative y′ at x = 0,
y (0) = 1
y′(0) = 8.5
(5)
in this case, Eq. (4) together with conditions, Eq. (5), then make up to an initial-value problem (since for
transient problems, the two conditions are specified at the beginning of time).
We can also use the following tow conditions:
y (0) = 1
y (2) = 2
(6)
where the function values at two locations, x = 0 and x = 2, are specified. Then the problem is called
boundary-value ordinary differential equation. This is typical for problems involving changes in space.
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Problem session #1 – HT_Spring2024
4. Initial-Value ordinary differential equation of ingot cooling
The variation of the ingot temperature, T(t), with time, t, during transient cooling
of an metallic ingot quenched inside a liquid tank is governed by the following
initial-value, first order ordinary differential equation,
dT
(7)
τ
=
−(T − T∞ )
dt
where τ is the time constant and T∞ is the temperature of liquid far away from the
ingot surface. For the ingot with a given initial temperature Ti, the solution of Eq.
(7) has an initial condition,
(7a)
T (0) = Ti .
Equation (8) can be solved by the method of separation,
together with the initial condition, giving
dθ
dt
= −
θ
→
τ
θ (t )
∫
θ
i
dθ
θ
t
= −∫
dt
0
τ
%
%
t
τ
80
Ingot cooling in water
o
C)
70
60
50
40
30
50
100
150
200
250
300
t, (s)
Finally, one has,
−
MATLAB ode45
0
%
%
θ (t ) = θi e
Exact Solution for water
90
20
θ (t )
t
= −
θi
τ
→ ln
100
T, (
Let us define an excess temperature, θ=
(t ) T (t ) − T∞ ,
then Eq. (7) become a homogeneous first-order ODE
equation,
dθ
(8)
τ
= −θ
dt
with an initial condition of
(8a)
θ (0)= θ=i Ti − T∞
or
T (t ) − T∞ = (Ti − T∞ )e
→ T (t ) = T∞ + (Ti − T∞ )e
−
−
t
τ
t
τ
%
%
%
%
%
%
Problem_session_1_1
Spring 2024
clc;clear;
Fe Ingot (Fe) cooling by convection
Given: (all SI units)
D = 0.05;
L = 0.08;
Ti = 80; % T, oC
rho_Fe = 7870; cp_Fe = 485;
% cp, J/kg-K
A = 2*(pi*D^2/4)+pi*D*L;
% surface area, m^2
Vol = pi*D^2/4*L;
% volume of ingot, m^3
Coolant: water
T_inf = 25;
h = 800;
% W/m^2-K
tao = Vol*rho_Fe*cp_Fe/h/A
% time constant,1/s
Analytical solution dT/dt = -(T-T_inf)/tao
T_e =@(t) T_inf + (Ti-T_inf)*exp(-t/tao);
t1 = linspace(0, 300, 20);
T_exact = T_e(t1);
plot(t1,T_exact,'k--','LineWidth',1)
axis([0 300 20 100])
xlabel('t, (s)','fontsize',15)
ylabel('T, (^oC)','fontsize',15)
grid on
hold on
Numerical solution by MATLAB built-in function, ode45
T0 = Ti;
% initial condition
f = @(t,T) -(T-T_inf)/tao;
[t2,T_n] = ode45(f, [0 300], T0);
plot(t2,T_n,'o','MarkerFaceColor','r')
text(120, 75, 'Ingot cooling in water', 'FontSize',15)
legend('Exact Solution for water','MATLAB ode45')
hold off
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Problem session #1 – HT_Spring2024
5. Numerical Methods to Solve a System of Initial-Value 1st-Order
Ordinary Differential Equations:
For example,
dy
= f (t , y ) = 4e 0.8t − 0.5 y
dt
y ( 0) = 2
The analytical solution,
4.1 MATLAB built-in function: ode45
[t,y] = ode45(f, [0 5], y0);
4 0.8t
(e
− e − 0.5t ) + 2e − 0.5t
1.3
70
Exact Solution
MATLAB ode45
60
50
40
y(t)
% Problem_session_1_2
%
%
MATLAB build-in ode solver
%
ode45 or ode23
%
clc;clear
f = @(t,y) 4*exp(0.8*t)-0.5*y;
y_e =@(t) 4/1.3*(exp(0.8*t)-exp(-0.5*t))+2*exp(-0.5*t);
%
t1 = linspace(0, 5, 100);
ye = y_e(t1);
%
plot(t1,ye,'k--','LineWidth',0.05)
axis([0 4 0 70])
xlabel('t','fontsize',15)
ylabel('y(t)','fontsize',15)
grid on
hold on
%
%
MATLAB ode45
y0 = 2;
[t,y] = ode45(f, [0 4], y0);
plot(t,y,'o','MarkerFaceColor','r')
legend('Exact Solution','MATLAB ode45')
hold off
y=
30
20
10
0
0
1
0.5
1.5
2
t
2.5
3
3.5
4
4.2 For a high-order ODE:
Solve the following second-order van der Pol equation by transforming it as a pair of 1st-order ODES:
− µ (1 − y12 )
dt 2
y1 (0) = 1
y1′ (0) = 1
dy1
+ y1 = 0
dt
y1 = y1
Let
dy1
→
dt
dy2 d 2 y1
= 2
dt
dt
y2 =
function Problem_session_1_3
% Problem_session_1_3
%
solve the {predator-prey models by MATLAB ode45
clc;clear
%
m = 100;
y1_0 = 1;
y2_0 = 1;
[t,y] = ode23(@dydt1, [0 1000], [y1_0 y2_0],[], m);
plot(t,y(:,1),'r','LineWidth',1.5)
hold on
axis([0 1000 -3 3])
xlabel('t','fontsize',15)
ylabel('y(t)','fontsize',15)
grid on
text(50,2.5,'mu=100','FontSize',15)
hold off
end
function [ dy] = dydt1(t,y,m)
%
Set up functions of system of ODEs
%
for van der Pol equation
f1 = y(2);
f2 = m*(1-y(1)^2)*y(2)-y(1);
dy = [f1; f2];
end
dy1
= y2
dt
dy2
= µ (1 − y12 ) y2 − y1
dt
y1 (0) = 1
y2 (0) = 1
3
mu=100
2
1
y(t)
d 2 y1
0
-1
-2
-3
0
100
200
300
400
500
t
600
700
800
900
1000