Rainwater Services Building – Draft

Technical Design Preview for Photovoltaic-Powered Rainwater Services Building

Abstract

This document outlines the technical specifications and calculations for a self-sustaining water services building powered by photovoltaic (PV) solar energy. The building will be located in the Philippines and designed to pump 500,000 liters of water per day from rainwater ponds, treat 20,000 liters for drinking, and provide energy for auxiliary electrical loads such as washing machines and lighting. The water is sourced from phytoremediation ponds, storing 1,000,000 liters over an area of 2,000 to 2,500 square meters. Updated pump sizing, energy calculations, and photovoltaic panel requirements are included to reflect adjustments such as reducing the head to 8 meters and increasing the washing machine operation to three machines running for six hours daily.

Table of Contents

  1. Introduction
  2. System Overview
    • 2.1. Hydraulic Requirements
    • 2.2. Energy Requirements
  3. Design Calculations
    • 3.1. Water Pumping System
      • 3.1.1. Power Requirement for Pumps
      • 3.1.2. Pump Sizing
    • 3.2. Photovoltaic System Sizing
      • 3.2.1. Daily Energy Demand
      • 3.2.2. Solar Panel Requirements
    • 3.3. Battery Storage Sizing
  4. Water Storage and Phytoremediation Ponds
  5. Conclusion

1. Introduction

This document provides the design for a self-contained water services building that operates independently using photovoltaic (PV) solar energy. The building, located in the Philippines, will pump 500,000 liters of water daily from rainwater ponds, treat 20,000 liters for drinking water through filtration and UV sterilization, and power auxiliary electrical appliances such as washing machines and lighting. The water will be sourced from nearby phytoremediation ponds with a total storage capacity of 1,000,000 liters. The design prioritizes sustainability and independence from the electrical grid.

2. System Overview

The system includes:

  • Pumping water from rainwater ponds into the water services building for filtration.
  • Treating 20,000 liters daily for drinking purposes through UV sterilization.
  • Supplying power to three washing machines and basic lighting.
  • Utilizing photovoltaic (solar) panels to generate electricity, storing energy in batteries for continuous operation.

2.1 Hydraulic Requirements

  • Daily water volume: 500,000 liters/day.
  • Water elevation (total head): Reduced to 8 meters.
  • Water density: 1000 kg/m³ (standard).

2.2 Energy Requirements

  • Electrical power for the pumps.
  • Electrical power for UV water treatment.
  • Additional power for operating washing machines and lighting.
  • Battery storage to ensure uninterrupted service, especially during non-sunlight hours.

3. Design Calculations

3.1 Water Pumping System

We need to calculate the power required to pump 500,000 liters of water daily, with a total head of 8 meters.

3.1.1 Power Requirement for Pumps

The hydraulic power required to pump water is calculated using the following equation:Ph=ρ⋅g⋅Q⋅HηP_h = \frac{\rho \cdot g \cdot Q \cdot H}{\eta}Ph​=ηρ⋅g⋅Q⋅H​

Where:

  • PhP_hPh​ = hydraulic power (W)
  • ρ\rhoρ = water density (1000 kg/m³)
  • ggg = acceleration due to gravity (9.81 m/s²)
  • QQQ = flow rate (0.0058 m³/s, corresponding to 500,000 liters/day)
  • HHH = total head (8 meters, updated)
  • η\etaη = pump efficiency (assumed 70%)

Substituting values:Ph=1000⋅9.81⋅0.0058⋅80.7=649.6 WP_h = \frac{1000 \cdot 9.81 \cdot 0.0058 \cdot 8}{0.7} = 649.6 \, WPh​=0.71000⋅9.81⋅0.0058⋅8​=649.6W

Adding a 20% safety margin for variations:Ppump=649.6 W×1.2=779.5 WP_{pump} = 649.6 \, W \times 1.2 = 779.5 \, WPpump​=649.6W×1.2=779.5W

Thus, the total power required for the pumping system is 779.5 W.

3.1.2 Pump Sizing

The work will be distributed across three pumps. The total power required for all pumps together remains 779.5 W. This means that each pump will handle approximately 260 W of load.

3.2 Photovoltaic System Sizing

The photovoltaic system must meet the total energy demand for pumping, UV filtration, washing machines, and lighting.

3.2.1 Daily Energy Demand

The total energy demand now considers the updated pump power, the increase in washing machine usage, and other auxiliary electrical loads:

  1. Energy demand for pumps:
    • The total power for the pumps is 779.5 W.
    • The pumps will operate for approximately 10 hours per day to pump 500,000 liters of water.

Energy for pumps=779.5 W×10 h=7,795 Wh/day\text{Energy for pumps} = 779.5 \, W \times 10 \, h = 7,795 \, Wh/dayEnergy for pumps=779.5W×10h=7,795Wh/day

  1. Energy demand for UV filtration:
    • The UV system consumes 60 W and operates for 8 hours daily:

Energy for UV filtration=60 W×8 h=480 Wh/day\text{Energy for UV filtration} = 60 \, W \times 8 \, h = 480 \, Wh/dayEnergy for UV filtration=60W×8h=480Wh/day

  1. Energy demand for washing machines:
    • Three washing machines, each consuming 500 W, operating for 6 hours daily:

Energy for washing machines=3×500 W×6 h=9,000 Wh/day\text{Energy for washing machines} = 3 \times 500 \, W \times 6 \, h = 9,000 \, Wh/dayEnergy for washing machines=3×500W×6h=9,000Wh/day

  1. Energy demand for lighting and small appliances:
    • Estimate for lighting: 200 W for 4 hours daily.

Energy for lighting=200 W×4 h=800 Wh/day\text{Energy for lighting} = 200 \, W \times 4 \, h = 800 \, Wh/dayEnergy for lighting=200W×4h=800Wh/day

3.2.2 Solar Panel Requirements

The total daily energy demand is now:Total energy demand=7,795 Wh+480 Wh+9,000 Wh+800 Wh=18,075 Wh/day\text{Total energy demand} = 7,795 \, Wh + 480 \, Wh + 9,000 \, Wh + 800 \, Wh = 18,075 \, Wh/dayTotal energy demand=7,795Wh+480Wh+9,000Wh+800Wh=18,075Wh/day

Assuming an average of 5.5 peak sun hours per day in the Philippines, the energy generated by a single 400 W solar panel is:Energy per panel=400 W×5.5 h=2.2 kWh/day\text{Energy per panel} = 400 \, W \times 5.5 \, h = 2.2 \, kWh/dayEnergy per panel=400W×5.5h=2.2kWh/day

The number of panels required to meet the demand:Number of panels=18.075 kWh/day2.2 kWh/panel/day=8.21 panels\text{Number of panels} = \frac{18.075 \, kWh/day}{2.2 \, kWh/panel/day} = 8.21 \, \text{panels}Number of panels=2.2kWh/panel/day18.075kWh/day​=8.21panels

Rounding up, the system will require 9 solar panels (400 W each) to meet the daily energy requirements.

3.3 Battery Storage Sizing

The battery system must provide power during periods of no sunlight (e.g., night), especially for critical loads.

  • Assume 1 day of autonomy.
  • Critical loads include UV filtration, washing machines, and lighting:

Critical load=480 Wh+9,000 Wh+800 Wh=10,280 Wh/day\text{Critical load} = 480 \, Wh + 9,000 \, Wh + 800 \, Wh = 10,280 \, Wh/dayCritical load=480Wh+9,000Wh+800Wh=10,280Wh/day

To ensure longevity, the battery should not discharge beyond 50%, so the battery capacity is:Battery capacity=10,280 Wh0.5=20,560 Wh=20.56 kWh\text{Battery capacity} = \frac{10,280 \, Wh}{0.5} = 20,560 \, Wh = 20.56 \, kWhBattery capacity=0.510,280Wh​=20,560Wh=20.56kWh

Thus, the required battery capacity is 20.56 kWh.

4. Water Storage and Phytoremediation Ponds

The rainwater will be collected in phytoremediation ponds near the building, with the following characteristics:

  • Storage volume: 1,000,000 liters.
  • Surface area: 2,000 to 2,500 square meters.
  • Average depth: 0.4 to 0.5 meters.

The ponds will provide a natural filtration process (phytoremediation), aiding in the pre-treatment of the water before it enters the system for pumping, filtration, and UV sterilization.

To express the battery capacity in ampere-hours (Ah), we first need to use the relationship between energy (in watt-hours), voltage, and ampere-hours.

The formula to convert watt-hours (Wh) to ampere-hours (Ah) is:

[
\text{Capacity (Ah)} = \frac{\text{Energy (Wh)}}{\text{Voltage (V)}}
]

For solar systems, battery banks are commonly designed at voltages of 12V, 24V, or 48V. Let’s calculate the ampere-hour capacity at each of these voltage levels.

Battery Capacity Calculation for 20.56 kWh

The total energy requirement for the battery is 20,560 Wh. Now, let’s calculate the required capacity in Ah for common system voltages.

1. At 12V System Voltage:

{Capacity (Ah)} = \frac{20,560 \, \text{Wh}}{12 \, \text{V}} = 1,713.33 \, \text{Ah}

2. At 24V System Voltage:

{Capacity (Ah)} = \frac{20,560 \, \text{Wh}}{24 \, \text{V}} = 856.67 \, \text{Ah}

3. At 48V System Voltage:

{Capacity (Ah)} = \frac{20,560 \, \text{Wh}}{48 \, \text{V}} = 428.33 \, \text{Ah}

Conclusion:

  • For a 12V system, the required battery capacity is 1,713.33 Ah.
  • For a 24V system, the required battery capacity is 856.67 Ah.
  • For a 48V system, the required battery capacity is 428.33 Ah.

Given the substantial energy requirements, a 24V or 48V system would be more practical, as higher voltage systems reduce the ampere-hour (Ah) capacity needed, making it more efficient and feasible to store energy without needing excessively large batteries.

5. Conclusion

This essay provides a technical preview for a solar-powered “water services building, based on rainwater ponds”. The system pumps 500,000 liters of water daily, with a total head of 8 meters, and treats 20,000 liters for drinking water through UV filtration. The daily energy demand for the entire system has been calculated to 18.075 kWh/day, and the system will require 9 photovoltaic panels (400 W each) to meet this demand. A battery capacity of 20.56 kWh will be necessary to ensure the system’s autonomy during periods without sunlight. The water will be sourced from phytoremediation ponds storing 1,000,000 liters, ensuring a sustainable and eco-friendly water supply for the community.

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