4.9 Resistances in Parallel
121
R 1234567
Reversing the procedure used to find R eq , we can break this resistance down:
R 7
R 1
R 23456
And knowing that the total airflow rate through the network is Q T and that this
will be the same through all resistances in series, then:
1 = R 1 Q 2
T = 4
N s 2
m 8
12.16
m 3
s
2
= 591.46 Pa
23456 = R 23456 Q 2
T = 2.12
N s 2
m 8
12.16
m 3
s
2
= 313.48 Pa
7 = R 7 Q 2
T = 2
N s 2
m 8
12.16
m 3
s
2
= 295.73 Pa
R 23456 can be further broken down, thus:
R 1
R 234
R 56
R 7
Given that 234 = 56 , the respective airflow rates can be calculated:
Q 234 =
234
R 234
=
313.48 Pa
6.12
N s 2
m 8
= 7.16
m
3
s
Q 56 =
P 56
R 56
=
313.48 Pa
12.5 N s 2
m 8
= 5.0
m 3
s
Breaking down R 234 reveals a series arrangement with a common flow rate Q 234 .
Thus, the pressure losses at R 23 and R 4 can be calculated:
R 1
R 4
R 56
R 7
R 23