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 |
| 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
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