4.9 Resistances in Parallel
123
Summary table:
Resistance
Airflow rate (m 3 s −1 )
Pressure loss (Pa)
R 1
12.16
591.46
R 2
3.65
159.83
R 3
3.50
159.83
R 4
7.16
153.64
R 5
5.00
162.5
R 6
5.00
150.0
R 7
12.16
295.73
Question 4.4 Demonstrate that for a parabolic relationship between the airflow rate
and the pressure difference, the total equivalent orifice of a parallel circuit can be
obtained as the sum of the equivalent orifices of each of its branches. Use the following
expression for the equivalent orifice (A) of a ventilation network.
A =
1.2
√ P
Q
Answer
Let us consider a system with three resistances in parallel. In this case:
Q T = Q 1 + Q 2 + Q 3
The airflow rate through an equivalent orifice is:
Q =
A
√ P T
1.2
Thus:
A T
√
P T
1.2
=
A 1
√
P 1
1.2
+
A 2
√ P 2
1.2
+
A 3
√
P 3
1.2
Since we are dealing with a parallel circuit:
P T = P 1 = P 2 = P 3
Simplifying terms, we get that:
A T = A 1 + A 2 + A 3
123
Summary table:
Resistance
Airflow rate (m 3 s −1 )
Pressure loss (Pa)
R 1
12.16
591.46
R 2
3.65
159.83
R 3
3.50
159.83
R 4
7.16
153.64
R 5
5.00
162.5
R 6
5.00
150.0
R 7
12.16
295.73
Question 4.4 Demonstrate that for a parabolic relationship between the airflow rate
and the pressure difference, the total equivalent orifice of a parallel circuit can be
obtained as the sum of the equivalent orifices of each of its branches. Use the following
expression for the equivalent orifice (A) of a ventilation network.
A =
1.2
√ P
Q
Answer
Let us consider a system with three resistances in parallel. In this case:
Q T = Q 1 + Q 2 + Q 3
The airflow rate through an equivalent orifice is:
Q =
A
√ P T
1.2
Thus:
A T
√
P T
1.2
=
A 1
√
P 1
1.2
+
A 2
√ P 2
1.2
+
A 3
√
P 3
1.2
Since we are dealing with a parallel circuit:
P T = P 1 = P 2 = P 3
Simplifying terms, we get that:
A T = A 1 + A 2 + A 3