AC/DC Solar Water Pump 200m Head Height

AC/DC Solar Borehole Pump

4SP3-18

MOTOR 3 HP

Q.max.100 l/min H.max.200m

90-360VDC,1X90-240VAC,50/60Hz Motor-(4”*4”)-BSP 1.25”

Maximum OD 4”

APPLICATION AREA:

This project products are mainly used in dry region for irrigation of agriculture, It can be used for drinking water and living water. The living condition could be much improved. It also can be used for fountains.

MATERIAL OF PARTS:

Outlet: stainless steel
Pump body: stainless steel
Motor body: stainless steel
Bearing: C&U

ADVANCED TECHNOLOGY:

1.Application innovation
Compared with the traditional alternating current machine, the efficiency is improved 25% by the permanent magnetism, direct current, brushless, non-sensor motor.
2.Technics innovation
Adopt double plastic package for rotor and stator, motor insulation≥300MΩ, the motor security was much improved.
3.Structure innovation
Oil filling, convenient installation and environmental protection

HIGHLIGHTS

Energy-saving and environment-protected green products
High technique products adopting MPPT and DSP chip technique.
100% copper wire, cold-rolled silicon steel sheet
CE certificate
Advanced three phase brushless DC motor
Stainless steel 316 screws
3 years warranty
PRINCIPLE OF OPERATION:

Solar panel collects sunlight→DC electricity energy → solar controller(rectification, stabilization, amplification, filtering)→available DC electricity→(charge the batteries)→pumping water

ADVANTAGES OF SOLAR PUMP SYSTEM:

It is easier and more widely used than any other dynamoelectric driven pumps.
It is more economical and more environmentally friendly.
MODEL SELECTION:

The power of solar panel = power of pump ×1.3
The voltage of solar panel = the voltage of pump
The controller should be matched
Select the batteries according to the following formulas:
The use hour of battery =
The battery capacity ÷(the machine power÷the battery voltage)×0.6 For example, the machine power is 200W, the battery
capacity is 100AH,the voltage is 12V,and the battery is fully charged, then the use hour is:100÷(200÷12)×0.6=3.6hours
The battery capacity=
the use hour ÷0.6×(the machine power÷the battery voltage) For example, the machine power is 200W,the battery voltage
is 12V,and the battery need to be used for 3.6hours,then the battery capacity is:3.6÷0.6×（200÷12）=100AH