### Mine ventilation case calculation

Let consider the following ventilation network in a mine. The network is made of five branches and four nodes. The airflow which gets in the mine is 47 m3/s.

The Hardy Cross method will be used to compute the natural distribution of air flow in the different parts of the mine.

Branch R Q
2 0.4 30
3 0.6 17
4 0.1 3
5 0.5 27
6 0.7 20

R is the airway resistance in kg/m7

1st case

We will start by considering an arbitrary distribution which respects the nodes law (the sum of incoming flows is equal to the sum of outgoing flows in a node)

If the distribution is real, the sum of head losses along any loop is zero.

The number of independent loops is given by b-n+1 where b is the number of branches and n the number of nodes.

In the present case the number of loops is 5-4+1=2

The chosen loops are 243 and  564. We walk along the loops clockwise. If the flow is in the same direction as the walk, the flow is considered positive. Otherwise it is negative.

Loop Σ R|Q|Q 2Σ R|Q| Correction Δ
243 0.4 30^2 + 0.1 3^2 - 0.6 17^2=187.5 2(0.4 30 + 0.1 3 + 0.6 17)=45 -4.167
564 0.5 27^2 - 0.7 20^2 - 0.1 3^2=83.6 2(0.5 27 + 0.7 20 + 0.1 3)=55.6 -1.504

The correction to be made to the assumed flow is (Σ R|Q|Q)/(2Σ R|Q|).

The new flows are:

Branch R Q
2 0.4 30-4.167=25.833
3 0.6 17+4.167=21.167
4 0.1 3-4.167+1.504=0,337
5 0.5 27-1.504=25.496
6 0.7 20+1.504=21.504

The branch 4 is part of the two loops therefore the two corrections should be considered when calculating the new flow. The correction is to be added when the flow is in the same direction as the walk and subtracted otherwise.

The same process is repeated again by calculating the corrections to be made to the new flows.

Loop Σ R|Q|Q 2Σ R|Q| Correction Δ
243 0.4 25.833^2 + 0.1 0,337^2 - 0.6 21.167^2=-1,861 2(0.4 25.833 + 0.1 0,337 + 0.6 21.167)= 46,134 0.04
564 0.5 25.496^2 - 0.7 21.504^2 - 0.1 0,337^2=1,339 2(0.5 25.496 + 0.7 21.504 + 0.1 0,337)=55,669 -0,024

Note that the new corrections are much more smaller.

The new flows are:

Branch R Q
2 0.4 25.833+0.04=25.874
3 0.6 21.167-0.04=21,126
4 0.1 0,337+0.04+0.024=0,401
5 0.5 25.496-0.024=25.472
6 0.7 21.504+0.024=21.528

Loop Σ R|Q|Q 2Σ R|Q| Correction Δ
243 0.4 25.8742 + 0.1 0,4012 - 0.6 21.1262=0,002 2(0.4 25.874 + 0.1 0,401 + 0.6 21.126)= 46,131 -0,00004
564 0.5 25.8742 - 0.7 21.5282 - 0.1 0,4012=-0,003 2(0.5 25.874 + 0.7 21.528 + 0.1 0,401)=55,691 0,0001

We are satisfied that the process has converged and the definitive flows are:

Branch R Q
2 0.4 25.874
3 0.6 21,126
4 0.1 0,401
5 0.5 25.472
6 0.7 21.528

Summary of the method :

Branch R Q it 1 it 2 it 3 it 4 it 5
2 0.4 30 25,833 25,874 25,874 25,874 25,874
3 0.6 17 21,167 21,126 21,126 21,126 21,126
4 0.1 3 0,337 0,401 0,401 0,401 0,401
5 0.5 27 25,496 25,472 25,472 25,472 25,472
6 0.7 20 21,504 21,528 21,528 21,528 21,528
Loop 1 Σ R|Q|Q 187,500 -1,861 0,002 0,000 0,000 0,000
2Σ R |Q| 45,000 46,134 46,131 46,131 46,131 46,131
Δ -4,167 0,040 -0,00004 0,000 0,000 0,000
Loop 2 Σ R|Q|Q 83,600 1,339 -0,003 0,000 0,000 0,000
2Σ R |Q| 55,600 55,669 55,691 55,691 55,691 55,691
Δ -1,504 -0,024 0,0001 0,000 0,000 0,000

2nd case

What happens if the assumed initial flows are unlikely where all the incoming flow transits by branch 2 and 5 and goes out.

Branch R Q it 1 it 2 it 3 it 4 it 5
2 0.4 47 23,500 25,850 25,874 25,874 25,874
3 0.6 0 23,500 21,150 21,126 21,126 21,126
4 0.1 0 0,000 0,392 0,401 0,401 0,401
5 0.5 47 23,500 25,458 25,472 25,472 25,472
6 0.7 0 23,500 21,542 21,528 21,528 21,528
Loop 1 Σ R|Q|Q 883,600 -110,450 -1,089 -0,001 0,000 0,000
2Σ R |Q| 37,600 47,000 46,138 46,131 46,131 46,131
Δ -23,500 2,350 0,024 0,000 0,000 0,000
Loop 2 Σ R|Q|Q 1104,500 -110,450 -0,782 -0,002 0,000 0,000
2Σ R |Q| 47,000 56,400 55,695 55,691 55,691 55,691
Δ -23,500 1,958 0,014 0,000 0,000 0,000

We can see that within few iterations we converge towards the same solution.

3rd case

In this case, we will consider an illogical scenario. A flow of 47 m3/s arrives to the mine, in branch 2 we consider a bigger flow (of 57 m3/s). Therefore a flow of 10 m3/s passes in the opposite direction in branch 3. We assume also that the flow in branch 4 is zero.

Branch R Q it 1 it 2 it 3 it 4 it 5
2 0.4 57 33,396 25,620 25,873 25,874 25,874
3 0.6 10 -13,604 -21,380 -21,127 -21,126 -21,126
4 0.1 0 0,262 0,364 0,401 0,401 0,401
5 0.5 57 33,134 25,257 25,472 25,472 25,472
6 0.7 10 -13,866 -21,743 -21,528 -21,528 -21,528
Loop 1 Σ R|Q|Q 1359,600 335,076 -11,678 -0,028 0,000 0,000
2Σ R |Q| 57,600 43,094 46,225 46,131 46,131 46,131
Δ -23,604 0,253 0,001 0,000 0,000 0,000
Loop 2 Σ R|Q|Q 1694,500 414,328 -12,003 -0,028 0,000 0,000
2Σ R |Q| 71,000 52,599 55,770 55,691 55,691 55,691
Δ -23,866 -7,877 0,215 0,000 0,000 0,000

Note that the flows in branches 3 and 6 are negatives. This means that the assumed flow direction considered initially is wrong and that the flow in these branches is in the opposite direction.

This gives the real distribution