The per unit impedance of s circuit element of 0.15. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element wii be ?
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The per unit impedance of s circuit element of 0.15. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element wii be ?
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- Consider three ideal single-phase transformers (with a voltage gain of ) put together as three-phase bank as shown in Figure 3.35. Assuming positive-sequence voltages for Va,Vb, and Vc find Va,Vb, and VC. in terms of Va,Vb, and Vc, respectively. (a) Would such relationships hold for the line voltages as well? (b) Looking into the current relationships, express IaIb and Ic in terms of IaIb and Ic respectively. (C) Let S and S be the per-phase complex power output and input. respectively. Find S in terms of S.Derive an expression of per unit impedance of a given base MVA and base KV in term of new base MVA and new base KVAThe p.u. impedance value of an alternator corresponding to base values of 13.2 kV and 30 MVA is 0.2 p.u. Then the p.u. impedance value of an alternator for the new base values of 13.8 kV and 50 MVA is......
- Consider a 3-phase, 50 HZ, 11 KV distribution system. Each of conductors is suspended by an insulator string having two identical porcelain insulators. The self-capacitance of the insulator is 5 times the shunt capacitance between link and ground. Find the voltages across two insulators.In a string of suspension insulator with three units, the line unit has 20kV and k=0.35. The voltage across tower unit is........ and across middle unit is.......No-load operation and short circuit tests were performed on three-phase, 50 Hz 480/220 delta / Y connected transformer. In these testsThe line voltage, line current and three-phase total power have been measured from the high voltage windings and the test results are below.pictures. 100 A DC current when 1.08 V DC voltage from non-transferable software is applied to the low voltage windingshas attracted.Find the single phase equivalent circuit parameters of this transformer.idle test:Voltage: 480V Power:114w current: 0.71Ashort circiut test:V: 10v P: 78w c: 18.3a
- Discuss the concept of reactive power and its role in grid stability.What effects are produced by change in voltage? 1. Iron los..........varies approximately as V². 2. Cu loss..........it also varies as V² but decreases with an increase in voltage if constant kVA output is assumed. 3. Efficiency...........for distribution transformers, efficiency at fractional loads decreases with in- crease in voltage while at full load or overload it increases with increase in voltage and vice- versa. 4. Regulation..........it varies as but decreases with increase in voltage if constant kVA output is assumed. 5. Heating.........for constant kVA output, iron temperatures increase whereas Cu temperatures decrease with increase in voltages and vice-versa.Select the base voltages for one section and calculate the base voltages for the rest of the sections. Remember that the voltages given are line-line. Calculate the current and impedance for all sections of the System. Assume a base complex power of 100 MVA.
- 7. TY Hat CHICL Ans. 1. Iron los.........varies approximately as V². 2. Cu loss..........it also varies as ² but decreases with an increase in voltage if constant kVA output is assumed. 3. Efficiency...........for distribution transformers, efficiency at fractional loads decreases with in- crease in voltage while at full load or overload it increases with increase in voltage and vice- versa. 4. Regulation..........it varies as but decreases with increase in voltage if constant kVA output is assumed. 5. Heating.........for constant kVA output, iron temperatures increase whereas Cu temperatures decrease with increase in voltages and vice-versa. quemsConsider a single phase transformer of rating 150 KVA, 3600/360, 50 Hz. Let the series resistance due to windings be 0.2 ohm and 0.002 ohm on the HV and LV sides, respectively. Let the series reactance due to leakage flux be j0.45 ohm and j0.0045 ohm on the HV and LV sides, respectively. While conducting Short Circuit test, the HV side should be excited with Volt Input power factor while conducting Short Circuit test is lagging6. A 5,000-kVA, 3-phase transformer, 6.6/33-kV, A/Y, has a no-load loss of 15 kW and a full-load loss of 50 kW. The impedance drop at full-load is 7%. Calculate the primary voltage when a load of 3,200 kW at 0.8 p.f. is delivered at 33 kV.