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Induction Coil Design Conditions

2025.09.05

MF furnace induction

MF furnace induction coil is the load of MF power supply, usually with a wall thickness of 3 to 5 mm square copper tube wound.The common faults are as follows:

1. The induction coil is leaking the water, which may cause ignition between the turns of the coil. Repair welding must be carried out in time before operation.

2.The molten steel sticks to the induction coil, the steel slag will get hot and red, which will cause the copper pipe to burn through. It must be cleaned in time.

3. The induction coil turns short circuit, this kind of fault in small induction heating furnace is particularly prone to occur. Because the furnace is small,and it will be deform in the work of the thermal stress, which resulting in the turn short circuit. The fault performance is large current and working frequency is higher than usual.

induction coil


For through-heating, we need to understand:

  1. Production capacity per ton-hour, kilogram-hour, or blank-hour;
  2. Workpiece size;
  3. Temperature rise range;
  4. Material.
    Based on these conditions, we must determine the power density and frequency for induction coil design. We must also calculate the temperature distribution and heat losses described above.
    Care should be taken to maintain the power supply impedance. If the power used is the maximum available power from the power supply, the load must be matched to the impedance. For example, if the power supply has a 400V, 250A or 200V, 500A output, the rated power of the motor-inverter unit can be 100kW. Based on the induction coil design, assume that 5000 ampere-turns are required. If 50 turns are selected, even at the highest voltage (400V), only 100A current can be obtained, and the maximum power can only reach 40kW. For this purpose, a 10-turn induction coil with a 200V output or a 20-turn induction coil with a 400V output should be used, as these values ​​provide the appropriate current (500A and 250A, respectively) and maximum effective power. If the induction coil only has a single turn (for example, in a hardening application), the coil current is 5000A, and a transformer is required. For a variable-frequency unit with a 200V output, the turns ratio would be 5000:500 or 10:1.
    Two examples (including comments) for heating aluminum and steel, respectively, will be used to illustrate the calculation of the data required for induction coil design. Obviously, this approach can also be applied to heating other profiles, plates, etc.

As with most through-heating applications, the workpiece is enclosed within the induction coil.
As a rule of thumb, radiation losses are reduced to 75% of the free-space heating losses. In this chapter, it is shown that the average radiated power can be calculated as the difference between the workpiece length and the workpiece's external dimensions. The length of the induction coil winding depends largely on the selected processing method and the metal being heated. In most steel heating systems, when production rates are high, the induction coils are often split. The resulting cold zone has a negligible effect on the temperature distribution. Large quantities of billet are fed through the induction coil path, eliminating unheated areas. Therefore, the length of each induction coil is primarily determined by the power supply used and the maximum allowable length of the copper tube per coil. In most induction systems for non-ferrous metal heating, the billet size is larger than the steel billet size, and each heating unit has only one induction coil, with no gaps along its length. For electrical calculations, it is sufficient to use the minimum workpiece length plus 25-75 mm as the induction coil length.

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