8.13 Calculating Losses Along a Duct Using the Integral Method
335
I (z) =
1
3
ln
(z − 1)
2
z 2 + z + 1
−
2
√
3
arctan
2z + 1
√
3
(8.54)
Moreover:
I (z 1 ) + I (z 2 ) = r
2θ
r
2
3
x + C 2
Therefore, we get Eq. 8.55:
I (z) = r
2θ
r
2
3
x + C 2
(8.55)
Returning to our true values
19 :
z =
1 −
C 1 r
2θ
P
−1.5
1
3 =
1 −
C 1 r
2θ
1
P 1.5
1
3
If we now substitute the value of P
1.5 , given by Eq. 8.50, we have:
z =
1 −
C 1 r
2θ
2θ
r
Q 3 + C 1
1
3
=
1 −
C 1
Q 3 + C 1
1
3
=
Q 3
Q 3 + C 1
1
3
=
Q
3
Q 3 + C 1
Therefore, we get Eq. 8.56:
z =
3
1 −
C 1 r
2θ
1
P 1.5 =
Q
3
Q 3 + C 1
(8.56)
Which has the alternative forms shown in Eqs. 8.56a and 8.56b:
P =
C 1 r
2θ
1 − z 3
2
3
(8.56a)
Q =
C 1 z
3
1 − z 3
1
3
(8.56b)
19 The changes made can be reverted with the expressions:
(t+b)
t
= z 3 ; z =
3
t+b
t ; t = P
2
3 ; therefore: z =
3
P
3
2 +b
P
3
2
= 3
1 +
b
P
3
2
; b = −
C1·r
2θ ;
z =
3
1 −
C1r
2θ P
−
3
2 .