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Rainwater Harvesting Flow Requirements Estimator

This estimator converts a rainwater demand volume and operating window into the average flow rate a pump, filter, or treatment line must deliver. It also applies an optional peak factor to create a practical design flow for periods when demand is less uniform than the daily average. The tool is useful for preliminary pump and pipe sizing checks, treatment skid comparisons, or determining whether an existing system can meet a planned rainwater demand. It reports average liters per minute, peak design liters per minute, and equivalent cubic meters per hour. The result is a hydraulic screening value; final equipment selection should also account for pressure, elevation, friction losses, pump curves, simultaneous fixtures, control strategy, and minimum equipment turndown.

Inputs

L/day
h/day
×
Result
—
Peak design flow
Average flow—
Peak flow—
Peak flow—

1. Enter daily rainwater demand
Use the total volume the system must deliver during a typical design day.

2. Enter operating hours
Specify how many hours per day the pump or treatment train is expected to run.

3. Set the peak factor
Use 1.0 for a uniform flow or a larger project-specific factor for demand peaks.

4. Review the average flow
This is the continuous rate needed to deliver the full daily volume within the operating window.

5. Use peak flow for screening
Compare the peak design flow with pump, filter, and treatment capacities.

Average flow (L/min) = Daily volume ÷ (Operating hours × 60) Peak design flow (L/min) = Average flow × Peak factor Peak flow (m³/h) = Peak flow (L/min) × 0.06

The peak factor is an input rather than a fixed standard because actual demand patterns vary by application.

What the result means

Peak design flow is the screening flow rate equipment should be able to accommodate under the selected demand pattern.

This calculation does not determine required pressure or pipe diameter and does not replace a hydraulic analysis for final system design.

Given
Daily volume = 4,200 L; operating time = 8 h/day; peak factor = 1.5.

Calculation
Average flow = 4,200 ÷ (8 × 60) = 8.75 L/min.
Peak flow = 8.75 × 1.5 = 13.125 L/min.
Equivalent = 13.125 × 0.06 = 0.7875 m³/h.

Result
13.13 L/min

A preliminary pump or treatment unit should be checked against about 13.1 L/min, in addition to its pressure and operating-range requirements.

Why use operating hours instead of 24 hours?

Many pumps and treatment units do not run continuously. Using the actual daily operating window gives the average rate required while the equipment is available.

What does the peak factor represent?

It represents how much higher the short-term design flow is than the average operating flow. Choose a factor based on the demand pattern or design criteria for the specific project.

Can I use this to choose a pump?

It provides a target flow rate, but pump selection also requires head, pressure, efficiency, control method, and system-curve information.

What if the system operates continuously?

Enter 24 operating hours. The average flow will then spread the daily demand across the full day before the peak factor is applied.

How is this different from storage capacity?

Flow requirements describe how quickly water must be delivered. Storage capacity describes how much water the system should hold.