Temperature Gradients-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

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(SOLUTION OF PROBLEM # 3.3)

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PROBLEM # 3.

Two dimensional simplex element have been used for modeling a heated flat plate. the (x.y) coordinates of nodes i, j, k of an interior element are given by (5,4), (8,6) and (4,8)cm respectively. If the nodal temperatures are found to be Ti = 100 °C, Tj = 80 °C and Tk = 110 °C, find (i) temperature at point P located at (xp,yp)= 6.5 cm. (ii) the temperature gradients inside the element.

GIVEN DATA

xi = 5 cm

xj = 8 cm

xk = 4 cm

yi = 4 cm

yj = 6 cm

yk = 8 cm

Φi =Ti = 100 °C

Φj =Tj = 80 °C

Φk =Tk = 110 °C

TO FIND

(i) Temperature at point P located at ( xp , yp )= 6.5 cm. (ii) The temperature gradients inside the element.

SOLUTION

The interpolation model to be used for given coordinates is 2D linear interpolation model.

Φ ( x , y ) = α 1 + α 2 x+ α 3 y ----------------------------------------------- (1)

Boundary conditions are;

Φ ( x , y ) = Φi at (x = xi , y = yi)

Φ ( x , y ) = Φj at (x = xj , y = yj)

Φ ( x , y ) = Φk at (x = xk , y = yk)

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= (1/2A)/

= (1/2A)[ bi Φi+bj Φj + bk Φk]

Where,

bi = yj – yk -------------------------------------------------------------------------- (6)

bj = yk – yi -------------------------------------------------------------------------- (7)

bk = yi – yj -------------------------------------------------------------------------- (8)

In the similar fashion,

α 2 = (1/2A)[ ci Φi + cj Φj + ck Φk ]

Where,

ci = xk – xj -------------------------------------------------------------------------- (9)

cj = xi – xk ------------------------------------------------------------------------- (10)

ck = xj – xi ------------------------------------------------------------------------- (11)

Now put these coefficients into Equation 2, we get;

Φ ( x , y) = Ni ( x , y ) Φi + Ni ( x ,y ) Φj + Ni ( x , y ) Φk ------------------ (12)

Where,

Ni ( x , y ) = (1/2A)[ ai + bi x + ci y ] ------------------------------------------ (13)

Nj ( x , y ) = (1/2A)[ aj + bj x+ cj y ] -------------------------------------------(14)

Nk ( x , y ) = (1/2A)[ ak + bk x+ ck y ] ----------------------------------------- (15)

Now, put coordinates in respective equations,

First of all calculate area of triangle by putting given coordinates in Equation 2,

A = Area of triangle = (1/2)

= 7 cm^2

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Or,

2A = 14 cm^2

Now by using equations 3 to 11, and plug values in it, we get;

ai = (8)(8)- (6)(4) = 40

aj = (4)(4) - (8)(5) = -

ak = (5)(6) – (4)(8) = -

bi = 6 – 4 = 2

bj = 8 – 4 = 4

bk = 4 – 6 = -

ci = 4 – 8 = -

cj = 5 – 4 = 1

To find Temperature at (6,5), put all the above calculated values in Equations 13, 14 and 15 and replace x = 6 and y = 5,

We get;

Ni ( x, y ) = (1/14)[ 40 + (2)(6) + (-4)(5) ]

= 2.

Nj ( x , y ) = (1/14)[ -24 + (4)(6) + (1)(5) ]

= 0.

Nj ( x , y ) = (1/14)[ -2 + (-2)(6) + (3)(5) ]

= 0.

Now put these above calculated values in Equation 12, and also plug in values of corresponding temperatures, we get;

Φ ( x , y ) = (2.286)(100) + (0.357)(80) + (0.0714)(110)

= 265.014 C

Similarly Temperature gradients can be found out as under;

ð Φ/ ðx = (1/2A) [ bi Φi+ bj Φj + bk Φk ]