Issue 64

F. Gugouch et alii, Frattura ed Integrità Strutturale, 64 (2023) 218-228; DOI: 10.3221/IGF-ESIS.64.14

The summation of the square of the residuals is:

2



  3 a

 

  

n

n

P

  i





  2 E

uri

(4)

1 2 3 , , f a a a

1 2 a a e

  

P





i

i

1

1

u

  3 ( ) a

i x e

We assume

 3 0 a is fixed and we put

, we use the classical least squares for the linear model we find:

P

p

      1 1 1 2 2 ( )( n n n uri uri i i i i u u n n i i x P P n x x      

n

x

)

i

 

 a a a 2 3 2



(5)





i

i

1

1

P

P

        2 1 1 1 1 2 2 ( )( ( ) n n n n uri uri i i i i i i u u n n i i x x P P n x x      

x

)

i

 

 a a a 1 1 3



(6)

i

i

1

1

Then we set:

         3 1 3 2 3 3 , , S a f a a a a a

(7)

This means:

n

n

n

n

n

n

n

P

P

P

P

        2 ( )( ) ( )( uri uri uri uri i i i i x x x n x 

x

)

i

n

P

Pu

Pu

Pu

Pu



2















uri

i

i

i

i

i

i

i

1

1

1

1

1

1

1







(8)

S a

x

3 ( )

(

(

))

i

n

n

n

n

Pu

   2 ( i

   2 ( i

2

2



i

1

n x

x

n x

x

)

)

i

i









i

i

i

i

1

1

1

1

Figure 7: Dimensionless pressure loss as a function of fraction of life

The optimization problem is now reduced to finding the minimum of S(a3). This can be done either using the Newton method [29], this gives a3=0.7713, we substitute the value of a3 in the value of a2 and a3 we get: a1=1.4838; a2= -0.4812; which allows to obtain the final form of the additional pressure loss function.

223