Induction Furnace Melt Rate Calculator
Calculate production capacity and melting time based on power and efficiency.
The rated power output of the induction power supply.
Typical induction efficiency is 60-75% (coil + converter losses).
Custom
Steel / Iron
Aluminum
Brass
Copper
Zinc
Energy required to raise 1 ton to pouring temp (Enthalpy).
Total weight of metal to be melted in one heat.
Please enter valid numerical values for Power and Capacity.
Hourly Melt Rate (kg/hr):
0
Hourly Melt Rate (tons/hr):
0
Time to Melt Full Charge:
0 min
Actual Power Used for Melting:
0 kW
How to Calculate Melt Rate in an Induction Furnace
Determining the melt rate of an induction furnace is critical for production planning, sizing power supplies, and ensuring operational efficiency. The melt rate basically describes how much metal your furnace can convert from solid to liquid at the desired pouring temperature within a specific timeframe.
The Core Physics of Induction Melting
The calculation is based on the principle of energy conservation. You supply electrical energy (kW), which is converted into thermal energy within the metal. However, not all electrical energy drawn from the grid ends up in the metal due to losses in the converter, the induction coil, and thermal radiation from the crucible.
Key Formula
The simplified industrial formula to calculate melt rate is:
Melt Rate (tons/hr) = (Power Supply (kW) × System Efficiency) / Theoretical Energy Required (kWh/ton)
Understanding the Variables
- Power Supply Rating (kW): The maximum power output your inverter can deliver.
- System Efficiency (%): For most coreless induction furnaces, the total electrical efficiency (Converter Efficiency × Coil Efficiency) ranges between 55% and 75%. A safe estimate for calculations is often 60-65% for steel melting.
- Theoretical Energy (kWh/ton): This is the enthalpy required to heat the specific metal from room temperature to pouring temperature (including latent heat of fusion).
- Charge Weight (kg): The total mass of metal loaded into the crucible.
Typical Energy Requirements (to Pouring Temp)
Below are standard theoretical energy values used in the industry. Note that "Practical Consumption" will be higher due to efficiency losses.
| Metal Type | Pouring Temp (°C) | Theoretical Energy (kWh/ton) | Typical Practical Consumption (kWh/ton)* |
|---|---|---|---|
| Steel / Iron | 1650 | 380 – 400 | 550 – 650 |
| Aluminum | 760 | 260 – 280 | 500 – 550 |
| Copper | 1150 | 190 – 210 | 350 – 400 |
| Brass | 1050 | 170 – 180 | 300 – 350 |
*Practical consumption assumes typical efficiency losses.
Example Calculation
Let's say you have a 1000 kW induction furnace melting Steel. You want to know how fast it can melt.
- Power: 1000 kW
- Efficiency: Assume 65% (0.65)
- Useful Power: 1000 × 0.65 = 650 kW (Power actually going into the metal)
- Energy Needed: Steel requires approx 385 kWh/ton (Theoretical).
- Calculation: 650 kW / 385 kWh/ton = 1.69 tons/hour.
If your crucible holds 3 tons, the time to melt would be: 3 tons / 1.69 tons/hr = 1.77 hours (approx 1 hour 46 minutes).
Factors Reducing Melt Rate
In real-world operations, several factors can slow down your calculated rate:
- Lid Management: Leaving the furnace lid open causes massive radiant heat loss.
- Lining Thickness: A lining that is too thick reduces coil efficiency; too thin is dangerous.
- Charge Density: Loose scrap couples poorly with the magnetic field, drawing less power initially.
- Slag Build-up: Excessive slag acts as an insulator and absorbs energy without melting the metal.