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Autotransformer
Autotransformer connected for step-down operation
NHS = # of turns on the High Side
NLS = # of turns “embraced by the Low Side
Autotransformer Example
Turns ratio = a = NHS / NLs = NA / NB = 80 / 20 = 4
VLS = VHS / a = 120 V / 4 = 30 V
ILS = VLS / ZLOAD = 30/0.5 = 60A >> IHS = ILS / a = 60/4 = 15A
Autotransformer Example continued
How did the load current become 60A?
15A provided directly to the load by VHS
45A provided to the load by “transformer action”
Example 3.1• A 400-turn autotransformer, operating in the
step-down mode with a 25% tap, supplies a 4.8-kVA, 0.85 Fp 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.
Example 3.1 part a
a = NHS / NLS = 400/(0.25)(400) = 4
VLS = VHS / a = 2400 / 4 = 600 V
ILS = 4800 VA / 600 V = 8 A = ILOAD
Example 3.1 parts b, c, d
• (b) ILINE = IHS = ILS / a = 8 A / 4 = 2 A
• (c) ITR = ILS – IHS = (8 – 2) A = 6 A
• (d) Scond = IHSVLS = (2 A)(600 V) = 1200 VA Strans = ITRVLS = (6 A)(600 V) = 3600 VA
Two-Winding Transformer connected as an Autotransformer
Two-Winding Transformer Reconnected as Autotransformer
1 2 2
2 2 2
2
( )
( 1)
at
w
at w
S V V IS V I
S a S
Example 3.2
• A 10-kVA, 60-Hz, 2400—240-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.
Example 3.2 continued
10 41.6724041.67 4.16710
LS
HS
kVAI AVAI A
As a two-winding transformer
Example 3.2 continuedAs an autotransformer
2
2400( 1) ( 1) 10 110240at wS a S kVA
Example 3.2 Simulation
V1
2400 V 60 Hz 0Deg
LOAD63.35 Ohm
U1AC 1MOhm 2.640k V
+
-
T110
ItransformedAC 1e-009Ohm 4.357 A
+
-IconductedAC 1e-009Ohm 41.672 A
+
-
XWM1
V IIinput
AC 1e-009Ohm
45.859 A+ -
Buck-Boost Transformer
“Buck”>Subtract the low-voltage output from the line voltage
“Boost” >>> Add the low-voltage output to the line voltage
Buck-Boost Transformer voltages
120 X 240 V primary
12 X 24 V or 16 X 32 V secondary
120/240 V operation
For 120 V operation, connect H1 to H3 and H2 to H4
For 240 V operation, connect H2 to H3
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
Available Buck-Boost Voltage Ratios
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.
Example 3.3 continued
• The required step-up voltage ratio is a’=VHS / VLS = 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
Example 3.3 (a)
Output Voltage = a’VLS = (1.100)(212) = 233.2 V
Example 3.3 part (b)
• The required step-down voltage ratio is a’ = VHS / VLS = 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
Example 3.3 (b)
Check this connection
Page 102 of the text by
Hubert
Output voltage = VHS / a’ = 246/1.0667 = 230.6 V