Example
The Hardy Cross method can be used to calculate the flow distribution in a pipe network. Consider the example of a simple pipe flow network shown at the right. For this example, the in and out flows will be 10 liters per second. We will consider n to be 2, and the head loss per unit flow r, and initial flow guess for each pipe as follows:
| Pipe | Q12 | Q13 | Q23 | Q24 | Q34 |
|---|---|---|---|---|---|
| r | 1 | 5 | 1 | 5 | 1 |
| Q guess (L/s) | 5 | 5 | 0 | 5 | 5 |
We solve the network by method of balancing heads, following the steps outlined in method process above.
1. The initial guesses are set up so that continuity of flow is maintained at each junction in the network.
2. The loops of the system are identified as loop 1-2-3 and loop 2-3-4.
3. The head losses in each pipe are determined.
| Loop 1-2-3 | Q12 | Q13 | Q23 |
|---|---|---|---|
| Head loss = | 25 | 125 | 0 |
| Direction | Clockwise | Counter-clockwise | Clockwise |
For loop 1-2-3, the sum of the clockwise head losses is 25 and the sum of the counter-clockwise head losses is 125.
| Loop 2-3-4 | Q23 | Q24 | Q34 |
|---|---|---|---|
| Head loss = | 0 | 125 | 25 |
| Direction | Counter-clockwise | Clockwise | Counter-clockwise |
For loop 2-3-4, the sum of the clockwise head losses is 125 and the sum of the counter-clockwise head losses is 25.
4. The total clockwise head loss in loop 1-2-3 is . The total clockwise head loss in loop 2-3-4 is .
5. The value of is determined for each loop. It is found to be 60 in both loops (due to symmetry), as shown in the figure.
6. The change in flow is found for each loop using the equation . For loop 1-2-3, the change in flow is equal to and for loop 2-3-4 the change in flow is equal to .
7. The change in flow is applied across the loops. For loop 1-2-3, the change in flow is negative so its absolute value is applied in the clockwise direction. For loop 2-3-4, the change in flow is positive so its absolute value is applied in the counter-clockwise direction. For pipe 2-4, which is in both loops, the changes in flow are cumulative.
| Pipe | Q12 | Q13 | Q23 | Q24 | Q34 |
|---|---|---|---|---|---|
| Q (L/s) | 6.66 | 3.33 | 3.33 | 3.33 | 6.66 |
The process then repeats from step 3 until the change in flow becomes sufficiently small or goes to zero.
3. The total lead loss in Loop 1-2-3 is
| Loop 1-2-3 | Q12 | Q13 | Q23 |
|---|---|---|---|
| Head loss = | 44.4 | 55.5 | 11.1 |
| Direction | Clockwise | Counter-clockwise | Clockwise |
Notice that the clockwise head loss is equal to the counter-clockwise head loss. This means that the flow in this loop is balanced and the flow rates are correct. The total head loss in loop 2-3-4 will also be balanced (again due to symmetry).
| Loop 2-3-4 | Q23 | Q24 | Q34 |
|---|---|---|---|
| Head loss = | 11.1 | 55.5 | 44.4 |
| Direction | Counter-clockwise | Clockwise | Counter-clockwise |
In this case, the method found the correct solution in one iteration. For other networks, it may take multiple iterations until the flows in the pipes are correct or approximately correct.
Read more about this topic: Hardy Cross Method
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