. A text-book of electrical engineering;. Fig. 412From equation (i8i) on page 401, we know that {i + Ti)KFA\Zj^ — -fa so that CS- ^1 ■ We get, therefore, for the copper loss in the rotor 3V-^2 = 3^1. CS. Subtracting this from the power transmitted to the rotor, we have P = = 4i6 Electrical Engineering This is the power which is transformed into mechanical power in the rotor. It includes the no-load losses, which, however, can be allowed for by drawing a horizontal Une at a suitable distance above the base hne. This line, which is shown dotted, is then taken as the foot of the
. A text-book of electrical engineering;. Fig. 412From equation (i8i) on page 401, we know that {i + Ti)KFA\Zj^ — -fa so that CS- ^1 ■ We get, therefore, for the copper loss in the rotor 3V-^2 = 3^1. CS. Subtracting this from the power transmitted to the rotor, we have P = = 4i6 Electrical Engineering This is the power which is transformed into mechanical power in the rotor. It includes the no-load losses, which, however, can be allowed for by drawing a horizontal Une at a suitable distance above the base hne. This line, which is shown dotted, is then taken as the foot of the output ordinates. For the sake of simphcity we add the stator hysteresis loss to the friction loss, and remember that we have still a part of the stator copper loss, viz. lo^.Ri, to allow for. The sum of these losses is practically equal to the power taken to run the motor light at the normal voltage. Calling this Pg, we have „ P„ = ,.TT. The ordinate RT is then a measure of the nett mechanical output of Fig- 413 We must now construct a scale of slip in our diagram. For this purposethe line GJ is drawn at right angles to GM^ (Fig. 413). This line cuts theoriginal circle in the point /. Since GJ is a tangent to the lowest circle, itdoes not intersect it; in other words, the point R in Fig. 412 coincides withthe point G, and the mechanical output is nil. The motor is therefore station-ary, and the point / corresponds to a slip of 100 per cent. The vector OJrepresents the primary starting current. To find the slip for any primary current I^ we join GA, and drop ajser-pendicular from any point D in GJ produced, on to GMj^. Since the sidesof any triangle are proportional to the sines of the opposite angle, we have XYGY sin^ sin;8 sin [90° — (ai + y8)] cos (aj + ^) 129. The most Convenient Form of the Circle-diagram 417 Similarly, since the angle at D is equal to y^ — a^, we have GY _ sin (yi — «^) _ sin (yj — «i)DY ~ sin (90° - 7]) cosy^ By multiply
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