The energy required to break frozen rocks by means of a very simple combined electrothermomechanical instrument

Abstract

622.01.0131. At the Laboratory for Problems of Rock Breaking by Powerful High-Frequency Electromagnetic Fields (G. V. Plekhanov Mining Institute, Leningrad), we have developed a fundamentally new class of electrothermomechanical combined instruments which simultaneously combine the functions of rock-breaking tool and radiator of uhf electromagnetic energy. These instruments can minimize the loss of electromagnetic energy in the electrothermomechanical method of rock breaking.However, it is not clear whether in principle the power consumption is reduced by the use of combined ETM instruments in comparison with ordinary mechanical rock-breaking instruments. The answer to this question is far from obvious. The point is that intensification of the process of rock breaking by ETM ins=uments is due to additional energy consumption on softening up therock (the energy of the etectromagnetic field). A sharp rise in the cutting output (the technological efficiency) can be achieved at the expense of undue increase in power consumption.An estimate of the energy consumption of rock breaking by combined ETM instruments can hardly be obtained in general form, because it depends on the strength, moisture content, temperature, density, and thermal capacity of the rock, on the characteristics of the electromagnetic field-the frequency and the complex dielectric constant of the medium at this frequency -and on the characteristics of the rock-breaking tool-angle of cut, depth of cut, etc. Therefore in this article we have made a first attempt to study the process of cutting by the simplest model of a combined ETM instrument which emits an electromagnetic field in its direction of motion.For simplicity, we will assume that the electromagnetic field is uniform, with a Poynting vector directed atong the z axis.Then the density of the energy absorbed in unit time by the rock is known [1] to bewhere ~ ~ 2k0 (e) ''~ [ ( 1 -~-t g2 6) 's: 1 ] :" is twice the coefficient o f attenuation of the electromagnetic field, k 0 is the wave number of the electromagnetic field in air, ~ = e'-ie" is the complex dielectric constant of the medium, and tan 6 = e'/e' Let Pern be the electromagnetic power of the radiator, ~. its area, ~, a coefficient representing the fraction of the energy which is radiated by the emitter into a cylinder with generators parallel to the z axis and with a crosssectional area equal to r., Then obviously ~Pem: ~ .! qo exp (--az) dz, 0 whence qo= ~gerrlxlY. 9('2)