What is the use of impedance rating in transformer?
It is used to calculate short circuit currents. .
Ideal Transformer Question?
An ac generator can be modeled as a voltage source in series with an impedance of (0.5 + j10) . It is connected to a transmission line having an equivalent series impedance of (5 + j12) via an ideal step-up transformer with an a ratio of 0.05. An ideal step-down transformer with a ratio of 25 connects the other end of the transmission line to a load impedance of (30 + j40) . If the load voltage is 240 V, determine (a) the power supplied to the load, (b) the power dissipated by the transmission line, (c) the generator voltage, and (d) the power supplied by the generator.
Electric transformers question?
With respect to single phase transformers:- For single phase transformers the full load primary current is Isp = 5MVAx1000/ 66 KV= 76 A (full load primary current) Iss = 5 MVAx1000/13.2 KV= 379 A (full load secondary current) Turn ratio = T=66/13.2 = 5 Transformer reactance (9%) wrt to primary Xsp = (% reactance) (Primary voltage) / ( Isp) = (9/100)(66 x1000)/76= 78 ohm Transformer resistance (0.65%) wrt to primary Rsp = (% resistance) (Primary voltage) / ( Isp) = (0.65/100)(66 x1000)/76= 5.645 ohm Zsp = (5.645 + j 78) ohm Now the transformers are connected in Δ Δ formation and put on load of 10 MW at 0.8 pf V(L)s = secondary Line to line voltage =13.2 KV I(L)s= secondary Line current =? cosØ= Power factor = 0.8 V(Δ)s= secondary Δ winding voltage = V(L)s I(Δ)s = secondary Δ winding current = I(L)s/√3 I(Δ)p = I(Δ)s/ T = I(Δ)s/ 5 For 3 phase transformer √3 V(L)s I(L)s cosØ = 10 MW I(L)s = (10 x1000) /√3 V(L)s cosØ = (10 x 1000) /{√3 (13.2) 0.8} = 547 A at pf 0.8 lag Therefore I(Δ)s = I(L)s/√3 = 547A /√3 =316 A at pf 0.8 lag I(Δ)p = I(Δ)s/ T = I(Δ)s/ 5 = 316/5 =63.2 A at pf 0.8 lag = (50.5 –j 38) A Output line to line voltage at secondary side is 13.2 KV at 0 angles WRT primary side this is 66 KV at 0 angles Therefore input voltage will be (66 + I(Δ)p Zsp/1000) KV = (66 KV + (50.5 –j 38) (5.645 + j 78)/1000) KV = (66+3.249+j3.725) KV = (69.249 + j3.725) KV Now with regard to books, the above problem encompasses almost the full background of Transformers. I refer “Advanced electrical technology” – H Cotton, as this one still with me . I also read Electrical Technology – B.L .Theraja during my college days (I do not remember the exact name of the book, is good for problem solving) and you may search the net. This one, sadly I lost.
For an impedance matching transformer, the ratio of (Rpri/RL) is equal to?
Rp/RL = (Np/Ns)² where, turns ratio: Np/Ns That is, the load impedance is reflected up to the primary and since this is a matching transformer then, Rpri = (Np/Ns)²RL, for maximum power transfer!
Transformers 250 kVA, 11/0.415 kV percentage impedance 4.75%.The rated current for LV side fuse should b?
The full load current of a 250 KVA, 11/0.415 v, 3-phase transformer secondary is 347.8 amperes from the formula I = 250000 / (415 * 1.732), which is I=(KVA * 1000) / (E * Sqrt(3)) The current for a single phase transformer of this size is 602.4 amperes from the formula I = 250000 / 415 , which is I=(KVA * 1000) / E There is a minimum and maximum factor that must be used to multiply the FLA values by to determine the current rating of the fuses, depending on the code of the country you live in. Check the latest version of the National Electrical Code for the USA. This rating is for the over current, not the maximum ground fault clearing rating. TexMav
How do you use a transformer for impedance matching ?
By presenting the inputs and outputs with a winding of the same impedance it maximizes power transfer. Maximum power transfer occurs when the input and output impedances are matched. For instance if you have an amplifier output with a 8 ohm output impedance and you want to drive another amplifier with a 10,000 ohm input impedance, you would have a transformer with one winding that presents an 8 ohm impedance connected to the output and a 10,000 ohm impedance winding connected to the input of the amplifier. The impedance generally follows the turns ratio. so such a transformer would have approximately a 1:1250 turns ratio.