第五章 变压器 5-10 解:(1)变压器的电压比为
'1L 2
L 6008.668
N
R k N R ==
== 回路电流
'
0L 1200.1(A)600600
E I R R ===++ 负载上获得的功率
2'
2L L
0.16006(W)P I R ==⨯= 信号源的输出功率
2'
20L
()0.1(600600)12(W)P I R R =+=⨯+= 效率
L 6
100%100%50%12
P P η=
⨯=⨯= (2)如果负载直接接至信号源,回路电流为 0L 120
0.197(A)6008
E I R R ==≈++
负载上获得的功率
22L L 0.19780.31(W)P I R ==⨯= 信号源的输出功率
220L ()0.197(6008)23.6(W)P I R R =+=⨯+≈ 效率
L 0.31
100%100% 1.
31%23.6
P P η=
⨯=⨯≈ 5-11 有一台单相变压器,已知1 2.19R =Ω,115.4X =Ω,20.15R =Ω,20.964X =Ω,
f 1250R =Ω,f 12600X =Ω,1876N =匝,2260N =匝,26000V U =,2180A I =,2cos 0.8
ϕ=(滞后),试用T 型等效电路和简化等效电路求1U 和1I 。
解:电压比12876 3.37260N k N ==≈
(1)T 型等效电路如下图所示
'
2222 3.370.15 1.70(Ω)R k R ==⨯≈
'2222 3.370.96410.95(Ω)X k X ==⨯≈
以2U 为参考向量,设260000V U =∠,则218036.87A I =∠-
'
2
2 3.3760000)202200(V)U kU ==⨯∠=∠ .
2'
218036.8753.4136.87(A )
k 3.37
I I ∠-=
=≈∠- ''''22220f f ()20220053.4136.87(1.710.95)
1.6396.89(A)+125012600
U I R jX j I R jX j --+-∠-∠-+=
=≈∠+ '
1021.6396.8953.4136.8754.1737.5854.17142.42
(A )I I I =-=
∠-∠-≈-∠-=∠
''
111122221110f f (j )(j )(j )(j )
54.1737.58(2.19j15.4) 1.6396.89(1250j12600)21243.66 2.77(V)
U I R X U I R X I R X I R X =+--+=
+++=-∠-⨯++∠⨯+≈-∠‘’ (2)简化等效电路如下图所示,将相关参数折算到高压侧来计算
'
sh 12
2.19 1.70
3.89()R R R =+=+=Ω '
sh 1215.410.9526.35()X X X =+=+=Ω
sh sh sh j 3.89j26.3526.6481.6()Z R X =+=+=∠︒Ω
以'2U 为参考向量,设'
2
6000 3.370202200(V)U =⨯∠︒=∠︒ '
2218036.8753.41-36.87(A)3.37
I I k ∠-︒
=
=≈∠︒ '
12
53.41-36.8753.41143.13(A)I I =-=-∠︒=∠︒
'
121sh
202200(-53.41-36.8726.6481.6)-20220-1422.8444.7221254.59 2.7(V)
U U I Z =-+=-∠︒+∠︒⨯∠︒=∠︒=-∠︒
5-13 解:(1)归算到高压侧的参数
1N 2N /43.3A/1082.6A I I = 1N 2N 10000
25400
U k U =
== 由空载试验数据,先求低压侧的励磁参数 '
2f 00400
3.85(Ω)360
U Z Z I ≈=
=≈⨯ '
Fe 0f 22200Δ38000.35(Ω)360p P R I I =
≈=≈⨯ '
'2'222f f f 3.850.35 3.83(Ω)X Z R =
-=-≈
折算到高压侧的励磁参数为
2'2f f 25 3.852406.25(Ω)Z k Z ==⨯= 2'2f f 250.35218.75(Ω)R k R ==⨯= 2'2f f 25 3.832393.75(Ω)X k X ==⨯= 由短路试验数据,计算高压侧室温下的短路参数 sh sh sh 440 5.87(Ω)343.3U Z I =
=≈⨯ Cu sh sh 22
2
sh 1N Δ10900
1.94(Ω)343.3p P R I I =
≈=≈⨯ 2222sh sh sh 5.87 1.94 5.54(Ω)X Z R =
-=-≈
换算到基准工作温度o
75C 时的数值 o sh sh75C 2287522875
1.94
2.37(Ω)22822820
R R θ++==⨯≈++
o o 2222sh sh75C sh75C
2.37 5.54 6.03(Ω)Z R X =+=+≈ 额定短路损耗为
o o 2
2
1N shN75C sh75C 3343.3 2.3713330.5(W)P I R ==⨯⨯= 短路电压(阻抗电压)为
o o 1N shN75C sh75C 43.3 6.03261.1(V)U I Z =⨯=⨯≈
o shN75C sh 1N
261.1
100%100% 4.52%10000/3
U u U =
⨯=
⨯≈
(2)满载(1β=)及2cos 0.8ϕ=(滞后)时
o 1N
2sh 2sh75C 1N
Δ%(cos sin )100%43.3
(2.370.8 5.540.6)100%10000/33.91%I U R X U β
ϕϕ=+⨯=
⨯⨯+⨯⨯≈
22N (1Δ)(1 3.91%)400384.4(V)U U U =-=-⨯≈
o o
20shN75C
2N 20shN75C
3
(1)100%
cos 380013330.5
(1)100%750100.8380013330.597.2%P P S P P βηβϕβ+=-
⨯+++=-
⨯⨯⨯++≈ (3)当o 0m shN75C
3800
0.5313330.5
P P ββ==
=
≈时
max N 20
3
2(1)100%
cos 223800
(1)100%0.53750100.82380097.7%P S P ηηβϕ==-
⨯+⨯=-
⨯⨯⨯⨯+⨯≈ 5-14 解:(1)归算到高压侧的短路参数 变比1N 2N 6000
26.1230
U k U =
=≈ '
22sh 1212 4.3226.10.006318.62(Ω)R R R R k R =+=+=+⨯≈ '22sh 12128.926.10.01317.76(Ω)X X X X k X =+=+=+⨯≈
22
22sh sh sh 8.6217.7619.74(Ω)Z R X =
+=+≈
(2)满载时,1β=。
另外,3
N 1N
1N 1001016.67(A)6000
S I U ⨯==≈ 当2cos 1ϕ=时,2sin 0ϕ=,
1N
sh 2sh 21N
Δ%(cos sin )100%16.67
8.621100%60002.39%I U R X U β
ϕϕ=+⨯=
⨯⨯⨯≈
当2cos 0.8ϕ=(滞后)时,2sin 0.6ϕ=,
1N
sh 2sh 21N
Δ%(cos sin )100%16.67
(8.620.817.760.6)100%60004.88%I U R X U β
ϕϕ=+⨯=
⨯⨯+⨯⨯≈ 当2cos 0.8ϕ=(超前)时,2sin 0.6ϕ=-,
1N
sh 2sh 21N
Δ%(cos sin )100%16.67
(8.620.817.760.6)100%60001.04%I U R X U β
ϕϕ=+⨯=
⨯⨯-⨯⨯≈- 由于在电力变压器中,一般sh R 和sh X 都比较小,因此在纯电阻负载,即2cos 1ϕ=时,%U ∆很小,说明负载变化时2U 下降得很小;在感性负载,即20ϕ>时,2cos ϕ和2sin ϕ均
为正值,%U ∆也为正值且较大,说明2U 随负载电流2I 的增大而下降,而且在相同负载电流2I 下,感性负载时2U 的下降比纯电阻负载时2U 的下降来得大;在容性负载,即20ϕ<时,
2cos 0ϕ>,而2sin 0ϕ<,若1s h 21s h 2c o s s
in I R I X ϕϕ<,则%U ∆为负值,这表明负载时二次侧端电压比空载时高,即2U 随2I 的增大而升高。