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Autotransformer. Autotransformer connected for step- down operation N HS = # of turns on the High Side N LS = # of turns embraced by the Low Side.

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Presentation on theme: "Autotransformer. Autotransformer connected for step- down operation N HS = # of turns on the High Side N LS = # of turns embraced by the Low Side."— Presentation transcript:

1 Autotransformer

2 Autotransformer connected for step- down operation N HS = # of turns on the High Side N LS = # of turns embraced by the Low Side

3 Autotransformer Example Turns ratio = a = N HS / N Ls = N A / N B = 80 / 20 = 4 V LS = V HS / a = 120 V / 4 = 30 V I LS = V LS / Z LOAD = 30/0.5 = 60A >> I HS = I LS / a = 60/4 = 15A

4 Autotransformer Example continued How did the load current become 60A? 15A provided directly to the load by V HS 45A provided to the load by transformer action

5 Example 3.1 A 400-turn autotransformer, operating in the step-down mode with a 25% tap, supplies a 4.8-kVA, 0.85 F p lagging load. The input to the transformer is 2400-V, 60-Hz. Neglecting the small losses and leakage effects, determine –(a) the load current, –(b) the incoming line current, –(c) the transformed current, –(d) the apparent power conducted and the apparent power transformed.

6 Example 3.1 part a a = N HS / N LS = 400/(0.25)(400) = 4 V LS = V HS / a = 2400 / 4 = 600 V I LS = 4800 VA / 600 V = 8 A = I LOAD

7 Example 3.1 parts b, c, d (b)I LINE = I HS = I LS / a = 8 A / 4 = 2 A (c)I TR = I LS – I HS = (8 – 2) A = 6 A (d)S cond = I HS V LS = (2 A)(600 V) = 1200 VA S trans = I TR V LS = (6 A)(600 V) = 3600 VA

8 Two-Winding Transformer connected as an Autotransformer Two-Winding TransformerReconnected as Autotransformer

9

10 Example 3.2 A 10-kVA, 60-Hz, 2400240-V distribution transformer is reconnected for use as a step-up autotransformer with a 2640-V output and a 2400-V input. Determine –(a) the rated primary and secondary currents when connected as an autotransformer; –(b) the apparent-power rating when connected as an autotransformer.

11 Example 3.2 continued As a two-winding transformer

12 Example 3.2 continued As an autotransformer

13 Example 3.2 Simulation

14 Buck-Boost Transformer Buck>Subtract the low-voltage output from the line voltage Boost >>> Add the low-voltage output to the line voltage

15 Buck-Boost Transformer voltages 120 X 240 V primary 12 X 24 V or 16 X 32 V secondary

16 120/240 V operation For 120 V operation, connect H1 to H3 and H2 to H4 For 240 V operation, connect H2 to H3

17 12/24 V or 16/32V operation For 12 V or 16 V operation, connect X1 to X3 and X2 to X4 For 24 or 32 V operation, connect X2 to X3

18 Available Buck-Boost Voltage Ratios

19 Example 3.3 The rated voltage of an induction motor driving an air conditioner is 230-V. The utilization voltage is 212-V. –(a) Select a buck-boost transformer and indicate the appropriate connections that will closely approximate the required voltage. –(b) Repeat (a), assuming the utilization voltage is 246-V.

20 Example 3.3 continued The required step-up voltage ratio is a=V HS / V LS = 230/212 = 1.085 Choose the best available voltage ratio from Table 3.1 as a=1.100. Need a 240-V primary and 12-V secondaries –Connect the 120-V primaries in series –Connect the 12-v secondaries in series

21 Example 3.3 (a) Output Voltage = aV LS = (1.100)(212) = 233.2 V

22 Example 3.3 part (b) The required step-down voltage ratio is a = V HS / V LS = 246/230 = 1.070 Choose a = 1.0667 from Table 3.1 Need a 240-V primary and 16-V secondaries –Connect the 120-V primaries in series –Connect the 16-V secondaries in parallel

23 Example 3.3 (b) Check this connection Page 102 of the text by Hubert Output voltage = V HS / a = 246/1.0667 = 230.6 V


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