How to calculate the melting rate and productivity of induction melting furnace?

It should be pointed out that the melting capacity data of the electric furnace provided by the general индукциялық балқыту пеші manufacturer in the sample or technical specification is the melting rate. The melting rate of the electric furnace is the characteristic of the electric furnace itself, it is related to the power of the electric furnace and the type of power source, and has nothing to do with the production operation system. The productivity of the electric furnace is not only related to the melting rate performance of the electric furnace itself, but also related to the melting operation system. Usually, there is a certain no-load auxiliary time in the melting operation cycle, such as: feeding, skimming, sampling and testing, waiting for the test results (related to the test means), waiting for pouring, etc. The existence of these no-load auxiliary times reduces the power input of the power supply, that is, reduces the melting capacity of the electric furnace.

For clarity of description, we introduce the concepts of electric furnace power utilization factor K1 and operating power utilization factor K2.

The electric furnace power utilization factor K1 refers to the ratio of the output power of the power supply to its rated power during the entire melting cycle, and it is related to the type of power supply. The K1 value of an intermediate frequency induction furnace equipped with a silicon controlled (SCR) full-bridge parallel inverter solid power supply is usually around 0.8. Xi’an Institute of Mechanical and Electrical Technology has added inverter control to this type of power supply (usually this type of power supply only has rectifier control ), the value can be close to 0.9 or so. The K1 value of the intermediate frequency induction furnace equipped with (IGBT) or (SCR) half-bridge series inverter power sharing solid power supply can theoretically reach 1.0.

The size of the operating power utilization coefficient K2 is related to factors such as the process design and management level of the melting workshop, and the configuration scheme of the electric furnace power supply. Its value is equal to the ratio of the actual output power of the power supply to the rated output power during the entire operating cycle. Generally, the power utilization coefficient K2 is selected between 0.7 and 0.85. The shorter the no-load auxiliary operation time of the electric furnace (such as: feeding, sampling, waiting for testing, waiting for pouring, etc.), the larger the K2 value. Using Table 4 Scheme 4 (dual power supply with two furnace system), the K2 value can theoretically reach 1.0, in fact, it can reach more than 0.9 when the no-load auxiliary operation time of the electric furnace is very low.

Therefore, the productivity N of the electric furnace can be calculated by the following formula:

N = P·K1·K2 / p (t/h)…………………………………………………………(1)

Қайда:

P — rated power of electric furnace (kW)

K1 — Electric furnace power utilization factor, usually in the range of 0.8 ~ 0.95

K2 — Operating power utilization factor, 0.7 ~ 0.85

p — electric furnace melting unit consumption (kWh/t)

Take a 10t intermediate frequency induction melting furnace equipped with a 2500kW silicon controlled (SCR) full-bridge parallel inverter solid power supply produced by the Institute of Mechanical and Electrical Engineering as an example. The unit melting consumption p indicated in the technical specifications is 520 kWh/t, and the electric furnace power utilization factor The value of K1 can reach 0.9, and the value of operating power utilization factor K2 is taken as 0.85. The productivity of the electric furnace can be obtained as:

N = P·K1·K2 / p = 2500·0.9·0.85 / 520 = 3.68 (t/h)

It should be pointed out that some users confuse the meaning of melting rate and productivity, and regard them as the same meaning. They did not consider the electric furnace power utilization coefficient K1 and the operating power utilization coefficient K2. The result of this calculation would be N = 2500/520 = 4.8 (t /h). The electric furnace selected in this way cannot achieve the designed productivity.