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I was trying to get a and b by verifying answers on LTspice. Iwas able to get a diagram and information going. How can I find aand b
A transformer whose nameplate reads 2300/230 V, 25 kVA operateswith primary and secondary voltages of 2300 V and 230 V rms,respectively, and can supply 25 kVA from its secondary winding. Ifthis transformer is supplied with 2300 V rms and is connected tosecondary loads requiring 8 kW at unity PF and 15 kVA at 0.8 PFlagging,
(a) what is the primary current?
(b) How many kilowatts can the transformer still supply to aload operating at 0.95 PF lagging?
(c) Verify your answers with PSpice.
It is given that: Vp:Vs = 2300:230 i.e turns ratio of 10:1 In LTSpice, there is no direct transformer model. Hence we need to use separate inductors for primary and secondary windings and couple them with K = 1 using the .op command (K L1 L2 1). Further, we know that the inductance is directly proportional to square of number of turns i.e L x №² Therefore, to get turns ratio of 10:1, the primary and secondary inductance ratio must be in the ratio of 100:1. Let primary inductance = 1000mH. Therefore, secondary inductance will be 10mH. Primary voltage = 2300Vrmss. Peak value = 2300 x 1.414 = 3252.2V. secondary voltage = 230Vrms, peak value = 230 x 1.414 = 325.22V To the secondary winding, there are two loads connected, 8kW @ UPF and 15kVA @ 0.8 pf lag. Load 1: Since the power factor is 1, the load is purely resistive. The value of R can be calculated using: R = V2/P=2302/8000 = 6.6125 ohm Load 2: Load 2 has a power factor of 0.8 lag i.e inductive load. theta cos-1(0.8) = 36.870 Active power = kVA x cos(theta) = 15k x 0.8 = 12kW. Corresponding value of resistive load can be calculated using: R = V2/P = 2302/12000 = 4.4083 ohm Reactive power = kVA x sin(theta) = 15 x 0.6 = 9kVar. Corresponding value of inductive reactance can be calculated using: XL = V2/Q = 2302/9000 = 5.8778 ohm But, we know that: XL = 2*pi*f*L. Assuming f = 50Hz, the value of inductance will be: L = XL/(2*pi*f) = 5.8778/(2 x 3.14 x 50) = 18.716 mH. With the above calculated values of primary and secondary inductance, load 1 & load 2 calculation, complete schematic is shown below. M Ci Vi-
With the above calculated values of primary and secondary inductance, load 1 & load 2 calculation, complete schematic is shown below. 16KW- 14KW- 12KW- 10KW- 8KW- 6KW- .tran 0 60m 0 4KW- ZKW- OKW- Oms V1 24KW 22KW- 20KW- 18KW- 16KW- 14KW- Vp 12KW- 10KW- 8KW- 6KW- 4KW 2KW- OKW L1 1000m SINE(0 3252.69 50) Power consumed in resistor 1 (ie load 1) is given below. Oms 6ms K L1 L2 1 6ms 12ms Vs 12ms L2 10m please mention if you have any doubts 18ms Power consumed in resistor 2 (ie resistive part of load 2) is given below. 24ms 18ms R1 6.6125 24ms V(vs)*(R1) 30ms V(vs)*1(R2) 30ms R2 4.4083 47 Interval Start: Interval End: Average: integral: 36ms 36ms Waveform: V(vs)*I(R1) Interval Start: Interval End: L3 18.7095m Average: integral: Os 60me 7.9976KW 479 85J Waveform: V(vs)*(R2) 42ms 05 60m 11.996KW 719.79J 42ms X 48ms 48ms