EPS 2018 2019 05

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Electric Power Systems Exam #5: 20/02/2020 1° Part Exercise 1 Consider the Elementary Electric Machine (EEM) illustrated in the Figure bellow: The EEM is supplied by a DC voltage V and the magnetic field B is constant in time. The characteristics of the EEM are: B d V R 2.0 T 100 cm 60 V 0.4 Ω The magnetic field B is perpendicular to the Figure’s plane and is exiting the Figure’s plane. Assuming the mobile iron bar, characterized by the electrical resistance R, can move along the fixed iron part without mechanical friction and at constant velocity: 1. draw, on the circuit of the EEM, the vectors of the mechanical and electrodynamic forces that act on the mobile part of the machine; 2. considering that at start-up, when the mobile part is still and breaker k1 is opened, determine the value of the start -up resistance ( Rsu) that will result in the reduction by half of the start-up electrodynamic force developed if breaker k1 would be closed; 3. considering that in normal operating conditions breaker k1 is closed, determine the analytical expression of the mechanical characteristic of the machine (𝐹𝑒𝑑 = 𝑓(𝑣)); 4. considering that breaker k1 is closed, that on the mobile iron bar a mechanical load is applied and that the characteristic of this load is gi ven by the equation 𝐹𝑚(𝑣) = 90 + 0.15 ∙ 𝑣 [𝑁], determine the velocity of the mobile part ( v), the electrodynamic force acting on the mobile part ( 𝐹𝑒𝑑), the emf induced by the magnetic field in the electric circuit (E), the real power absorbed from the grid (Pela) and the efficiency of the EEM (𝜂); Exercise 2 Consider the following electrical network Line L [km] xl [/km] Vn [kV] L1 40 0.4 400 L2.1 50 0.4 400 L2.2 80 0.4 400 Generator G An Vn Real power output: PG Operating voltage: VG [MVA] [kV] [MW] [kV] G 600 220 100…