Calculation and application of the hottest dual fr

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Calculation and application of dual frequency induction quenching

Abstract: a new calculation method with better universality is proposed. This method can be applied to the actual production of induction heating devices. It can not only design the temperature distribution for special needs, but also analyze and simulate the temperature field. The test results show that the surface hardening quality of dual frequency heating quenching is better than that of single frequency heating quenching. Although the dual frequency heating quenching is usually more expensive than the traditional single frequency heating quenching, it has been proved to be economical in some cases and is the only process method in special cases

key words: dual frequency induction quenching; Temperature field; Calculation


the dual frequency quenching device is shown in Figure 1, which consists of two side-by-side inductors supplying currents of different frequencies. The upper sensor is connected with a lower frequency, and the workpiece moves to make the heating depth deeper (low frequency); The lower sensor is connected with a higher frequency, and the workpiece moves to make the heating depth shallow, which is equivalent to the required thickness of the hardened layer

1 workpiece; 2 and 3 sensors with frequencies of F1 and F2 respectively; 4, 5 magnetic core (guide magnet)

Figure 1 dual frequency quenching device should be loosened sketch

select the power of two inductors to make the upper inductor heat the workpiece to Curie temperature, and the lower inductor heat the workpiece to quenching temperature along the paramagnetic workpiece within the same time. In order to design various parameters of dual frequency quenching device, it is necessary to make systematic electromagnetic calculation, but there is no appropriate calculation method in the special literature

1 mathematical analysis of roll dual frequency induction quenching

1.1 simplified assumptions and calculation models for these problems

solving the electromagnetic problem of dual frequency induction quenching process is a complex mathematical problem. The difficulties lie in the finite size of system parts, the nonlinear changes of workpiece electromagnetic parameters and the influence of sensor moving speed on characteristic phenomena. If all these factors are fully considered, the solution is too complex to be practical. Therefore, on the premise that it has little impact on the accuracy of the solution, three types of simplified assumptions are introduced:

(1) assumptions about the geometric shape of the workpiece sensor system:

a. the heated workpiece with finite length is replaced by an infinite roll

b. replace the finite size sensor with a ring of finite height and small thickness

c. an infinite cylinder replaces the magnetic cores of two finite size sensors

(2) assumptions about electromagnetic parameters of workpiece:

a. the quenching and heating process can be divided into two stages. In the first stage, the ferromagnetic workpiece is only heated to Curie temperature in a certain thickness by the sensor of frequency f1 (this thickness is approximately equal to the penetration depth, i.e. the skin depth); In the second stage, the workpiece heated by the sensor with frequency f2 can be divided into two layers, and the outer layer is made of paramagnetic material( μ R=1), while the inner layer is a ferromagnetic material with high permeability, parameters μ and σ As a constant in the volume of each layer

b. the permeability of the two layers can be calculated based on the well-known method [2]. For simplicity, the permeability of the inner layer is taken as infinite

c. the thickness of the surface ferromagnetic layer heated by the frequency f1 sensor and the surface paramagnetic layer heated by the frequency f2 sensor are regarded as equal

(3) to avoid the problem of integral equation, assume that:

a. the moving speed of the sensor is ignored

b. separate the workpiece along the coordinate line z=0, so that each half of the roll is regarded as an infinite length, calculate separately according to the two cases in Figure 2, and then combine the results between the appropriate areas

Figure 2 shows the calculation model considering all these assumptions

Figure 2 calculation model after "separation" of workpiece

1.2 general solution of vector thermal differential equation

the differential equation of vector potential is derived from the calculation of Maxwell equation, which has the following forms in cylinder coordinates [3, 4]:


m2=j ωμσ;

ω= 2 π f current ripple number (i.e. angular frequency); μ—— Permeability of medium; σ—— The conductivity

electromagnetic field vector of the medium can be expressed in the following form:


e - electromotive force

b - magnetic induction intensity

j -- the special solution of eddy current density

equation (1) can be obtained by separating variables, which is

a (R, z) =[c (k) I1 (PR) +d (k) K1 (PR)]] × [f (k) coskz+g (k) sinkz] (4)

where: C (k), D (k), f (k), G (k) - integral constants

i1, K1, I0, K0 - modified Bessel function

k - separation variable constant

p= (k2+m2) 1/2 (5)

the general solution of equation (1) is the sum of all special solutions of equation (4), so:

in the special calculation interval of Figure 2, the solution of equation (6) has many forms. In fact, from the point of view of application, the most essential is the outer layer of the workpiece (region II). The result of only this region solution is as follows:

where: p1=k2+j ω one μ twenty-one σ 21

p2=k2+j ω two μ twenty-two σ 22

symbol II represents the calculated area II, and marks 1 and 2 represent the relevant quantities of frequency f1 [in the case of Fig. 2 (a)] and F2 [in the case of Fig. 2 (b)] sensors respectively

The integral constant appearing in

expression (7) can be calculated by vector potential and in accordance with the boundary conditions in Figure 2. There is

where: μ I - permeability in region I

where: I1, I2 - induced current of frequencies F1 and F2

n1, N2 -- the number of inductor turns of frequency f1 and F2

h, h -- the length of the workpiece corresponding to inductor 1 and 2 respectively (see Figure 2)

calculate the integral constant based on the boundary condition (9), and then obtain the vector potential of the workpiece

where: n1=k1 (kr3) I0 (kr4) +i1 (kr3) K0 (kr4)

n2=k0 (kr2) I0 (kr4) -i0 (kr2) K0 (kr4)

n3=k1 (kr2) I0 (kr4) +i1 (kr2) K0 (kr4)

n4=i1 (p1r2) K0 (p1r1) +k1 (p1r2) I0 (p1r1)

n6=i0 (p1r1) K0 (p1r2) -k0 (p1r1) I0 (p1r2)

m4=i1 (p2r2) K0 (p2r1) +k1 (p2r2) I0 (p2r1)

m6=i0 (p2r2) K0 (p2r1) -k0 (p2r2) I0 (p2r1)

x1=p1/k; x2=p2/k (12)

μ r1, μ R2 -- relative permeability in Fig. 2 (a) and (b)

1.3 magnetic induction intensity, eddy current density and effective power density in the workpiece

region II consistent with formula (3) has two components, brii and bzii. According to formula (11), the following formula can be obtained:

since the differential of vector potential (11) has no practical significance, the expression formula of magnetic induction intensity is not given, However, the magnetic induction intensity component (BZ) of the roll surface (r=r2) is calculated by means of the computer through equation (13), and the results are shown in Figure 3

the eddy current density on the workpiece can be obtained according to formula (3):

the eddy current density on the roll surface (r=r2) is calculated according to the above formula, and the results are shown in Figure 4

the effective power density of the workpiece surface can be calculated according to the following formula:

obtained from (14):

the effective power density calculated according to formula (16) is shown in Figure 5

2 roll dual frequency quenching test

according to the above calculation, 50 Hz is selected for the upper inductor and 1000 Hz is maintained for the lower inductor. During the double frequency quenching test of the roll, the temperature distribution of the roll was measured, and the results are shown in figure 6

1, 2, 3, 4, 5, 6 temperature distribution at different distances from the surface

Fig. 6A temperature distribution during dual frequency induction quenching

(distance from the roll surface)

a temperature at the beginning of quenching heating; B temperature after passing through 50 Hz sensor

c temperature after passing through 1000 Hz sensor; D instantaneous temperature of spray

Fig. 6B temperature distribution of roll section I

it can be seen from Fig. 6A that the temperature distribution during quenching has three highest points, the first highest point (t=3 min) corresponds to the temperature after preheating, the second highest point (t=21 min) corresponds to the temperature after passing through the 50 Hz sensor, and the third highest point (t=25 min) corresponds to the temperature after passing through the 1000 Hz sensor. The figure also shows that the appropriate quenching temperature starts at the 16th minute after the workpiece temperature is 250 ℃, and accurately speaking, it is only heated for 10 minutes. After the 27th minute, the treatment is completed by spray. The average speed of temperature rise of roll during quenching is about 1.2 ℃/sec. See Fig. 6B for the temperature distribution of roll body section Compared with the temperature distribution obtained by single frequency quenching, it can be seen that the temperature difference of the roll surface within the depth of 30 mm is about 210 ℃, while the former is only 120 ℃, thus ensuring a thicker hardened layer and a smaller hardness gradient (Fig. 7)

(distance from roll surface)

1 single frequency induction quenching with one preheating

2 double frequency induction quenching with one preheating

Fig. 7 hydraulic testing machine for roll has high power 1 weight ratio section hardness distribution

3 conclusion

1 The above calculation method of dual frequency induction quenching has strong universality and practicability

2. Compared with single frequency quenching, dual frequency quenching can make the workpiece obtain better properties (surface hardness, section hardness gradient, hardening depth), improve work efficiency and reduce energy consumption


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