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Furnace Melting Energy Calculator

Work out furnace melting energy instantly with clear inputs, formula shown and shareable results.

About Furnace Melting Energy Calculator

Melting energy has two parts: sensible heat to raise the charge to melting and then to pouring temperature, m x c x ΔT, and latent heat of fusion to change state, m x L. Steel needs roughly 0.4 kWh per kilogram theoretically; real furnaces use 1.5 to 2 times that because of radiation, wall and off-gas losses.

Formulas

Melting energy

kJ = mass x (specific heat x ΔT + latent heat of fusion)

Power rating

kW = kWh / melt hours

Worked example

With the default values below, Furnace Melting Energy Calculator returns Theoretical melting energy of 130.4. Change any input above and every figure here is recalculated in your browser.

Inputs used

Charge massminimum 0.1 kg
500 kg
Specific heat capacityminimum 0.01 kJ/kg·K
0.46 kJ/kg·K
Temperature rise to pouringminimum 1 K
1,450 K
Latent heat of fusionminimum 0 kJ/kg
272 kJ/kg

Results

Theoretical melting energy
130.4 kWh
Energy in megajoules
469.5 MJ
Energy per kilogram
0.261 kWh/kg
Furnace rating for a 2 hour melt
65.2 kW

Step by step

  1. 1Values used
  2. 2Melting energy
  3. 3Power rating
  4. 4Theoretical melting energy= 130.4 kWh
  5. 5Energy in megajoules= 469.5 MJ
  6. 6Energy per kilogram= 0.261 kWh/kg
  7. 7Furnace rating for a 2 hour melt= 65.2 kW

Assumptions and limits

  • The result is derived only from the 4 inputs listed above. Anything not among them is not part of the calculation.
  • In particular, this calculator does not take inflation, tax or compounding frequency into account.
  • Figures are rounded for display to 3 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.

How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.

Frequently Asked Questions

How is the Furnace Melting Energy figure worked out?

kJ = mass x (specific heat x ΔT + latent heat of fusion). Melting energy has two parts: sensible heat to raise the charge to melting and then to pouring temperature, m x c x ΔT, and latent heat of fusion to change state, m x L.

Why is real consumption so much higher?

Furnace shell losses, door openings, electrode or burner inefficiency and holding time all add energy that never reaches the metal. Induction furnaces are closest to theoretical, cupolas the furthest.

What values do I need to enter?

This calculator takes 4 inputs: Charge mass, Specific heat capacity, Temperature rise to pouring, Latent heat of fusion. The pre-filled defaults are a realistic worked example — replace them with your own site or project figures.

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