146
4 Mine Ventilation Networks
Node N 1 : Q 14(n) − Q 12(n) = −150
Node N 2 : Q 12(n) − Q 23(n) − Q 24(n) = 0
Node N 3 : −Q 34(n) + Q 23(n) = 150
Mesh M 1 : 0.4
Q 14(n−1)
Q 14(n) + 0.3
Q 12(n−1)
Q 12(n) + 0.8
Q 24(n−1)
Q 24(n) = 0
Mesh M 2 : −0.8 ·
Q 24(n−1)
Q 24(n) + 0.2
Q 34(n−1)
Q 34(n) + 0.1
Q 23(n−1)
Q 23(n) = 0
Then, for iteration 1, represented in matrix form:
Branch (ij)
14
12
24
34
23
R ij
0.4
0.3
0.8
0.2
0.1
Q io
1
1
1
1
1
Node (N)/Matrix (M)
Q 14
Q 12
Q 24
Q 34
Q 23
Independent term
N 1
1
−1
0
0
0
=
−150
N 2
0
1
−1
0
−1
=
0
N 3
0
0
0
−1
1
=
150
M 1
0.4
0.3
0.8
0
0
=
0
M 2
0
0
−0.8
0.2
0.1
=
0
The system can be solved as:
A i j
[Q i ] =
B j
→ [Q i ] =
A i j
−1
B j
where
• [A ij ]: Square matrix of the coefficients. It has M + N − 1 rows and columns.
• [A ij ]
−1 : Inverse matrix of A ij .
• [Q j ]: Column matrix of the airflow rates in each branch (values to be calculated).
It has M + N − 1 rows.
• [B j ]: Colum matrix of the independent terms. It has M + N − 1 rows.
Therefore:
Inverse matrix
Solution
0.5643564
0.2376238
0.1584158 1.0891089
0.7920792 Q 14 = −60.89
−0.435644
0.2376238
0.1584158 1.0891089
0.7920792 Q 12 =
89.11
−0.118812
−0.207921
−0.138614
0.2970297 −0.693069
Q 24 =
−2.97
−0.316832
−0.554455
−0.70297
0.7920792
1.4851485 Q 34 = −57.92
−0.316832
−0.554455
0.2970297 0.7920792
1.4851485 Q 23 =
92.08
Iteration 2
The seed value for the airflow rate in the second iteration is Q 14(1) = −60.89. In
this way, for example, Q 14 , es Q 14(o) = −60.89. In order to avoid errors arising from