Abstract
In this article, a mathematical model of working fluid inlet into chamber of rotary expansion machine from source of superheated steam. The main practical problem that can be solved on the basis of this model is to determine dependence of torque on the angular velocity, i.e. in the construction of mechanical characteristics of rotary expansion machine.
Highlights
Rotary vane and rotary machines have a number of advantages over engines of other designs, for example, connecting rod and piston: lower specific gravity, high specific liter power, can work stably at low speeds, less mechanical losses due to the absence of reciprocating movements, significantly less vibration due to design balance score
We perform following transformations of this equation. We divide both sides of Eq (3) by the internal energy MMccVVTT of the gas in the chamber and write it in the form dddd TT
Differentiating by angle of rotation of shaft αα, we obtain 1 dddd + 1 dddd = 1 dddd + 1 dddd, whence we find pp dddd VV dddd MM dddd TT dddd dddd TT
Summary
Rotary vane and rotary machines have a number of advantages over engines of other designs, for example, connecting rod and piston: lower specific gravity, high specific liter power, can work stably at low speeds, less mechanical losses due to the absence of reciprocating movements, significantly less vibration due to design balance score. Perminov A., et al Gas Dynamics of Working Fluid Inlet Into Chamber of Rotary Expansion Machine. The dependence of chamber’s volume VV(αα) on the angle, and angle of beginning of opening ααBBBB and angle of end of closing ααEEEE of inlet window are set In this process, three variables are unknown: the pressure pp of gas in chamber, its mass MM and temperature TT. Enthalpy flow entering chamber along with mass flow goes to change internal energy of gas and perform expansion work: ii0∗dddd − ddQQWW = dd(MMMM) + pp dddd,. We perform following transformations of this equation We divide both sides of Eq (3) by the internal energy MMccVVTT of the gas in the chamber and write it in the form dddd TT. Dividing (11) by differential dddd , we obtain the differential equation dddd dddd pp ccpp0MTMT0cc+VVυTυT2⁄2 dddd dddd dddd dddd ddQQWW dddd (12)
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