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Presentation regarding Parabolic Partial Differential Equation including theory and sample problem
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Parabolic Partial Differential Eq’n
Obillo, Alvin
Gonzales, Patricia Shaine
INTRODUCTION
GENERAL FORM OF SECOND ORDER LINEAR PDE WITH TWO INDEPENDENT VARIABLES AND ONE DEPENDENT VARIABLE:
Where u = u(x, y) and a, b, c, d, e, f, and g are functions of x and y ONLY. If g is zero, then the equation is
homogeneous. The PDE is said to be elliptic if The PDE is said to be hyperbolic if The PDE is said to be parabolic if
HEAT-CONDUCTION EQUATION
Where a = k, b = 0, c = 0, and d = -1
Then:
THEREFORE, HEAT-CONDUCTION EQUATION IS A PARABOLIC DIFFERENTIAL EQUATION
THE EXPLICIT METHOD
where:
T is the temperature
x is the location
j is the time step
i is the interval of location
THE EXPLICIT METHODSubstituting the approximations from the last slide to the general equation…
Solving for the temperature at time interval j + 1 would give….
Where:
If we know the temperature at the first time step and the boundary temperatures, it is possible to get the temperature of each location at the next time step.
THE EXPLICIT METHODEXAMPLE 1: Consider a steel rod that is subjected to a temperature of 100 degrees Celsius on the left end and 25 degrees Celsius on the right end. If the rod is of length 0.05 m, use the explicit method to find the temperature distribution in the rod from t = 0 and t = 9 seconds. Use ∆x = 0.01 m, and t = 3 seconds. = 1.4129 x / s. The initial temperature of the rod is 20 degrees Celsius.
0 1 2 3 4 5 6
25 deg C
100 deg C
EXAMPLE 1SOLUTION:
When t = 0 seconds, j = 0:
= 100 degrees C
= 20 degrees C
= 20 degrees C
= 20 degrees C
= 20 degrees C
= 25 degrees C
When t = 3 seconds, j = 1
= 100 degrees C
= 53.912 degrees C
= 20 degrees C
= 20 degrees C
= 22.120 degrees C
= 25 degrees C
EXAMPLE 1SOLUTION:
When t = 6 seconds, j = 2:
= 100 degrees C
= 59.073 degrees C
= 34.375 degrees C
= 20.889 degrees C
= 22.442 degrees C
= 25 degrees C
When t = 9 seconds, j = 3
= 100 degrees C
= 65.953 degrees C
= 39.132 degrees C
= 27.266 degrees C
= 22.872 degrees C
= 25 degrees C
EXAMPLE 1If the temperature vs the location is to be plotted in a graph…
THE EXPLICIT METHOD Based on the graph, as the time goes on, the graph
seems to become linear. This means that as time approaches infinity, the temperature becomes linear – which corresponds to a steady-state system.
CRANK – NICHOLSON METHOD
where:
T is the temperature
x is the location
j is the time step
i is the interval of location
CRANK – NICHOLSON METHODSubstituting the approximations from the last slide to the general equation…
Solving for the temperature at time interval j + 1 would give….
Where:
In order to solve for the temperature of every location at any time, this method makes use of simultaneous linear equations.
CRANK – NICHOLSON METHOD
EXAMPLE 1: Consider a steel rod that is subjected to a temperature of 100 degrees Celsius on the left end and 25 degrees Celsius on the right end. If the rod is of length 0.05 m, use the explicit method to find the temperature distribution in the rod from t = 0 and t = 9 seconds. Use ∆x = 0.01 m, and t = 3 seconds. = 1.4129 x / s. The initial temperature of the rod is 20 degrees Celsius.
0 1 2 3 4 5 6
25 deg C
100 deg C
EXAMPLE 1SOLUTION:
For time = 3 seconds;
i = 1, j = 0:
(-0.4239)(100) + (2)(1+0.4239) - (0.4239) (0.4329)(100) + (2)(1-0.4329)(20) + (0.4329)(20)
Which would then give:
2.8478 - 0.4239 = 116.30
EXAMPLE 1SOLUTION:
By further solving i = 2, 3, and 4, this matrix can be created:
EXAMPLE 1SOLUTION:
At time = 3 seconds;
EXAMPLE 1SOLUTION:
At time = 6 seconds;
EXAMPLE 1SOLUTION:
At time = 6 seconds;
EXAMPLE 1SOLUTION:
At time = 9 seconds;
EXAMPLE 1SOLUTION:
At time = 9 seconds;
COMPARISON OF RESULTSLocation Explicit Implicit Crank-
Nicholson
Analytical
65.953 59.043 62.604 62.510
39.132 36.292 37.613 37.084
27.266 26.809 26.562 25.844
22.872 24.243 24.042 23.610
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