Electrolysis power supply application:

Mainly electrolysis for non-ferrous metal such as aluminum, magnesium, zinc, lead, copper, manganese and dioxide manganese; gold, silver and other metal smelting; NdFeB rare earth smelting; carbide, diamond smelting; salt water, potassium electrolysis made caustic soda, potash, sodium; potassium chloride electrolysis potassium chlorate, potassium perchlorate; refractories electric heating, as well as other types of high-power electrolysis power supply.

Electrolysis power supply working in industry:

Electrolysis power supply is widely used in metallurgical industry, such as extracting metal from ore or compound (electrowinning) or extracting purified metal (electrolytic purification), and depositing metal (electroplating) from the solution. Metallic sodium and chlorine are produced by electrolysis of sodium chloride melt dissolved; electrolysis sodium chloride aqueous solution will generate sodium hydroxide and chlorine gas. Electrolysis water produce hydrogen and oxygen. Electrolysis of water is the external electric action to split water into H2 (g) and O2 (g). DC electrolysis power supply is a very powerful means of promoting the oxidation- reduction reaction, many difficult redox reaction is implemented by a DC electrolysis power supply. For example: molten fluoride can be oxidized at the anode to elemental fluorine, molten lithium salt is reduced to metallic lithium on the cathode.

Electrolysis power supply importance:

Electrolysis industry plays an important role in the national economy, many ferrous metals (such as sodium, potassium, magnesium, aluminum, etc.) and precious metals (such as zirconium, hafnium, etc.) smelting and metals refining (such as copper, zinc, lead), basic chemical products (such as hydrogen, oxygen, caustic soda, preparation of potassium chlorate, hydrogen peroxide, adiponectin, etc.), as well as electro plating, electrical polishing, anodizing, are achieved by a DC electrolysis power supply.

Electrolysis power supply specification and parameter:

Electrolysis power supply Input voltage3 phase 380V/415V/440V 50-60Hz
Output voltage6V/12V/15V/18V/24V/36V/48V optional or according to customers’ request
Output current0-8000A (Random optional)
Voltage stability≤1%
Current stability≤1%
Ripple factor1%-2%
Efficiency≥90%
 Stabilization modeCurrent stabilization/Voltage stabilization (Switchable easily)
Adjustment rangeVoltage/Current from 0-100% continuously adjustable within rated range
Cooling systemAir cooling/Water cooling/Oil cooling
Control methodManual/PLC
Display contentsVoltage meter/Current meter/Working/Over temp/Error, etc..
Protection methodInput over-voltage/Under-voltage/Over current/Output short/Over heating self protection etc..
Operating temperature-20-50℃
Ambient temperature-30-65℃
Ambient humidity≤90%
Working loadingFull capacity operation, more than 1,000 meters above sea level, reduced load operation

Electrolysis power supply Packing details:

2 replies
  1. andy
    andy says:

    1)A switched Power supply is known to be very efficient. Does anyone know have a circuit diagram this circuit or where I can buy a switched mode power supply for the conversion of 110 AC 12 amps, to 110 DC 12 amps

    2) if you do not provide power supply for a mains circuit, How do you select a capacitor when the ripple current appears to keep increasing? When I select a capacitor with lower ESR this causes the ripple current to increase? Again, how do you select the capacitor? Does anyone know where I can find the proper circuit diagram for or where I can purchase a linear power supply for the conversion of 110 AC 12 amps, to 110 DC 12 amps.

    We are calculating for three scenarios
    C= I load /Frequency *(voltage)
    Change in voltage = delta Volts = I load /Frequency * C
    I load = 12 amps, Frequency = 120 hertz, Voltage= 120 volts
    Solving for two cases
    C = 400 UF = .000400 Farad
    1 C = 10,000 UF = 0.01 Farad
    2 C = 20,000 UF = 0.02 Farad
    3 C = 20,000 UF = 0.1 Farad

    1) Calculating we get delta volts = 12 amps/(120 hertz) * (.0004 F) = 250 volts
    2) Calculating we get delta volts = 12 amps/(120 hertz) * (.01 F) = 10 volts
    3) Calculating we get delta volts = 12 amps/(120 hertz) * (.02 F) = 5 volts
    4) Calculating we get delta volts = 12 amps/(120 hertz) * (.1 F) = 1 volts …. This a very low ripple
    1) Change in voltage = ripple = 250 volts
    2) Change in voltage = ripple = 10 volts
    3) Change in voltage = ripple = 5 volts
    4) Change in voltage = ripple = 1 volts
    Using a higher capacitor reduces the ripple volts, reduces the ripple from 250 volts to 1 volt
    Calculating ripple Current

    I ripple = square root (P/ESR) = square root (1400watts/.062 ohm) = 150 amps and this value is larger than the stated value of 7 amps as stated in the table.

    As the capacitor increase from .0004 F to .01 F, the ESR reduces about 61 micro ohms. That is great, but now I am concerned with the rated increase in ripple Current where at .01 uF, the ripple current 7.7 amps, and that appears contrary to reduce the overall ripple.
    The chart below explains that the ESR for 2,700 uF capacitor is 62.6 milli ohms

    Name: Andy DuPont

    Reply

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