Thus, expressed in per unit, the combined motor current is obtained by using the equation: At unity power factor, the load is given as three-quarters or 0.75 p.u. If the motors are operating at 12 kV, this represents 12 kV/13.8 kV = 0.87 per-unit voltage. Calculate Operating Conditions of the Motors The completed reactance diagram is shown in Figure 5:įigure 5 – Single line reactance circuit diagram (reactances shown on a per-unit basis)ħ. Use the correcting equation from Step 3, above. Xper unit = (65) (25,000)/(1000)(72.1)2 = 0.313 p.u.Ĭorrections need to be made in the nameplate ratings of both motors because of differences of ratings in kVA and kV as compared with those selected for calculations in this problem.Xper unit = (ohms reactance)(base kVA)/(1000)(base kV)2 =.Calculate the Transmission-Line Reactance (nameplate per-unit reactance) (base kVA/nameplate kVA) (nameplate kV/base kV)2 = Use the equation for correction: per-unit reactance: In this problem, 25,000 kVA is chosen as the base S, and simultaneously, at the generator end 13.8 kV is selected as a base voltage Vbase. It should be of the general magnitude of the components, and the choice is arbitrary. Establish Base Voltage through the Systemīy observation of the magnitude of the components in the system, a base value of apparent power S is chosen. Specifications are given in above table.ġ. X” = 15 percent XL = 15 percent X” = 15 percent X” = 15 percent X = 65 Ωįigure 4 – Single line diagram of electric-power system supplying motor loads. Per-unit method of solving of 3-phase problemsįor the system shown in Figure 4, draw the electric circuit or reactance diagram, with all reactances marked in per-unit (p.u.) values, and find the generator terminal voltage assuming both motors operating at 12 kV, three-quarters load, and unity power factor.ġ3.8kVĒ5,000 kVAđ5,000 kVAđ0,000 kVA –Ģ5,000 kVA 3-phaseđ3.2/69 kVđ3.0 kVđ3.0 kV – It is necessary to make a correction when the transformer nameplate reactance is used because the calculated operation is at a different voltage, 13.8 kV / 72.136 kV instead of 13.2 kV / 69 kV.ĮEP – Nov1, 2017 Corrections by Ovide Brudo, Nov 18th If a different S base were used in this problem, then a correction would be necessary as shown for the transmission line, electric motors, and power transformers. No calculation is necessary for correcting the value of the generator reactance because it is given as 0.15 p.u. The selected base S value remains constant throughout the system, but the base voltage is 13.8 kV at the generator and at the motors, and 72.136 kV on the transmission line. The base voltage of the motors is determined likewise but with the 72.136 kV value, thus: Since Gullstrand powers and focal lengths are measured with respect to a hypothetical plane, it is often more useful to deal with front and back vertex focal lengths.The base voltage of the transmission line is then determined by the turns ratio of the connecting transformer: For a thick lens with surface powers P 1 = and P 2 = diopters The powers in the equation are then the surface powers. The equivalent power given by Gullstrand's equation isīy putting in the index of refraction of the glass, Gullstrand's equation can be applied to a thick lens. For lenses with powers P 1 = diopters and P 2= diopters It is often useful to calculate the front and back vertex powers or focal lengths. Gullstrand's equation can be used to calculate the effective focal length of a thick lens or two separated lenses with respect to the second principal plane. Gullstrand's Equation Gullstrand's Equation Calculations
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