8.14 Discrete Solution (Step by Step Method)
341
…
i=0
i=1
i=2
i=n
Q1,P 1
Q2,P 2
Qn, P n
Q 0 ,P 0
Q f
∆Q0
∆Q 1
∆Q 2
P 0
ΔP 1
ΔP 2
∆Q n-1
ΔP n
ΔP 0 = 0
Fig. 8.15 Pressure, flow and friction losses in a segmented ventilation duct
At segment i = 1 (first duct segment):
Airflow rate : Q 1 = Q 0 − 0
where
ΔQ 0 is the leakage at the first joint between the fan and the duct.
At segment i = 2 (first duct segment):
Airflow rate : Q 2 = Q 1 − 1
Generalising for any segment i:
Airflow rate : Q i = Q i−1 − i−1
Assuming that each joint leaks the same fraction (k) of the airflow rate circulating
in the duct, then:
Segment i = 1:
Q 1 = Q 0 (1 − k)
Segment i = 2:
Q 2 = Q 0 (1 − k)(1 − k)
Segment i = 3:
Q 3 = Q 0 (1 − k)(1 − k)(1 − k)
Then, the flow rate (Q i ) for the ith segment of the geometric progression is
(Eq. 8.57):
Q i = Q 0 (1 − k)
i
(8.57)
341
…
i=0
i=1
i=2
i=n
Q1,P 1
Q2,P 2
Qn, P n
Q 0 ,P 0
Q f
∆Q0
∆Q 1
∆Q 2
P 0
ΔP 1
ΔP 2
∆Q n-1
ΔP n
ΔP 0 = 0
Fig. 8.15 Pressure, flow and friction losses in a segmented ventilation duct
At segment i = 1 (first duct segment):
Airflow rate : Q 1 = Q 0 − 0
where
ΔQ 0 is the leakage at the first joint between the fan and the duct.
At segment i = 2 (first duct segment):
Airflow rate : Q 2 = Q 1 − 1
Generalising for any segment i:
Airflow rate : Q i = Q i−1 − i−1
Assuming that each joint leaks the same fraction (k) of the airflow rate circulating
in the duct, then:
Segment i = 1:
Q 1 = Q 0 (1 − k)
Segment i = 2:
Q 2 = Q 0 (1 − k)(1 − k)
Segment i = 3:
Q 3 = Q 0 (1 − k)(1 − k)(1 − k)
Then, the flow rate (Q i ) for the ith segment of the geometric progression is
(Eq. 8.57):
Q i = Q 0 (1 − k)
i
(8.57)