340
8 Secondary Ventilation
I (z) = −1.9224
Solved using Excel Solver gives z = 3.0565
Solving by means of Table 8.6, gives: z = 3.00
Step 5. Calculating the fan pressure and airflow rate by means of Eq. 8.56a:
P =
C 1 r
2θ
1 − z 3
2
3
Substituting we have:
P v =
−4.913 · 0.2
2 · 0.5 × 10 −5 ·
1 − 3.0565 3
2
3
= 1.72
m
3
s
and Eq. 8.56b:
Q =
C 1 z
3
1 − z 3
1
3
Substituting in all known values:
Q v =
−4.913 · 3.0565
3
1 − 3.0565 3
1
3
= 233.4 mmwc (2289.7 Pa).
8.14 Discrete Solution (Step by Step Method)
One calculation method that is frequently used assumes that a duct is composed of n
segments each with the same aerodynamic characteristics. It is further assumed that
leaks will occur at the joints between segments, and that these are always the same
fraction (k) of the airflow rate at each joint (Flügge 1953).
In this way, the airflow rate provided by the fan (Q v ), must be equal to the one
that arrives at the duct outlet (face) (Q f ), plus the losses (ΔQ) (Fig. 8.15):
Q v = Q f + Q
For segment i = 0 (a special segment which contains the fan and to which the
remaining segments of the duct are connected):
Airflow rate : Q v = Q 0