Model an intermittent leak using active time
A resident tracks a leaking valve that flows at 0.5 L/min for 20 minutes every morning over 30 days. The logged active time is 600 minutes, or 5/12 of a day.
Open CalculatorInputs and calculation
| Tool input | Value |
|---|---|
| Currency | USD |
| Volume or billing unit | m³ |
| Flow (L/min) | 0.5 |
| Days or equivalent sessions | 0.416666666667 |
| Price per billing unit (USD) | 3 |
0.5 L/min × 1,440 min/day × (5/12) days = 300 L. Divide by 1,000 for m³, then multiply by 3 for USD 0.90.
Tool result · USD 0.90
| Tool input | Value |
|---|---|
| Currency | USD |
| Volume or billing unit | m³ |
| Flow (L/min) | 0.5 |
| Days or equivalent sessions | 30 |
| Price per billing unit (USD) | 3 |
0.5 L/min × 1,440 min/day × 30 days = 21600 L. Divide by 1,000 for m³, then multiply by 3 for USD 64.80.
Tool result · USD 64.80
What the comparison shows
Treating the active rate as continuous would multiply the estimated volume by 72. The intermittent schedule is the key input.
Assumptions and limits
Use active time with the active flow rate. If timing or flow varies, sum measured intervals before projecting.