8.14 Discrete Solution (Step by Step Method)
343
That is:
P = R 1 L 1 Q
2
1 + · · · + R i L i Q
2
i + · · · + R n L n Q
2
n
And since, all the segments are equal of length (L i ) and resistance (R i ):
P = R i L i
Q
2
1 + · · · + Q
2
i + · · · + Q
2
n
Substituting in the expression for the airflow rate in the ith segment (Eq. 8.57):
P = R i L i Q
2
0
(1 − k)
2
+ (1 − k)
4
+ (1 − k)
6
+ · · · + (1 − k)
2n
The expression is a geometric progression
20 of ratio (1 − k)
2 , therefore, the sum
of the first n terms can be expressed as in Eq. 8.58:
P = R i L i Q
2
0
(1 − k)
2n+2
− (1 − k)
2
(1 − k)
2
− 1
(8.58)
Now, as the airflow rate at the gallery face (Q f ) is:
Q f = Q n = Q 0 (1 − k)
n
= Q v (1 − k)
n
Therefore, the characteristic equation of the system as a function of the required
airflow rate at the face is (Eq. 8.59):
P = R i L i
Q
2
f
(1 − k)
2n
(1 − k)
2n+2
− (1 − k)
2
(1 − k)
2
− 1
(8.59)
Exercise 8.10 Given Eq. 8.59, you are asked to:
(a) Express it for a segment located before the beginning of the duct (n = 0).
(b) Express it for a number of segments (n) when leakage is very small (k tends to
0).
Solution
(a)
The conditions where n = 0 are equivalent to those pertaining to a fan output without
any coupled segments, i.e. in free discharge. In this case, it is not possible to use the
derived formula since it is only valid for cases where the fan is connected to a duct
segment (n > 1).
20 The sum of n terms of a r-ratio geometric progression is calculated by the formula: S n =
an r −a1
r −1 .
Where a 1 and a n are the first and last term of the progression, respectively.