Description
Electric Machinery and Power System fundamentals, you can see the details of each questions the attachment.
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Explanation & Answer
Attached.
𝟏)
𝑆𝑙𝑜𝑎𝑑 =
100
∠53.13 + 61.46∠ − 11.47 = 200.976∠37𝐾𝑉𝐴
0.6
𝑆𝑙𝑜𝑎𝑑 = 160.5 + 120.95𝑗 𝐾𝑉𝐴
𝐓𝐡𝐞 𝐠𝐞𝐧𝐞𝐫𝐚𝐭𝐢𝐨𝐧 𝐬𝐢𝐝𝐞 𝐨𝐟 𝐓𝐚 its VLine = 13.2KV and VLn = 7.6kv ≈ 8kv
𝐓𝐡𝐞 𝐥𝐨𝐚𝐝 𝐬𝐢𝐝𝐞 𝐨𝐟 𝐓𝐚 its Vline = 4.16 V and VLN = 2.4kv
𝐓𝐨 𝐜𝐨𝐯𝐞𝐫 𝐭𝐡𝐞 𝐥𝐨𝐚𝐝 𝐭𝐡𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦𝐞𝐫 𝐫𝐚𝐭𝐢𝐧𝐠 𝐦𝐮𝐬𝐭 𝐛𝐫 𝐦𝐨𝐫𝐞 𝐭𝐡𝐚𝐧 𝟐𝟎𝟎𝐊𝐕𝐀
𝒂) 𝑰 𝒘𝒊𝒍𝒍 𝒄𝒉𝒐𝒔𝒆 𝑻𝒂 𝟕𝟓𝑲𝑽𝑨
𝟖
𝑲𝒗
𝟐. 𝟒
𝐙 =. 𝟎𝟒 + 𝟎. 𝟏𝐣 𝐩𝐮 𝐚𝐧𝐝 𝐜𝐨𝐧𝐧𝐞𝐜𝐭𝐞𝐝 𝐢𝐧 𝐬𝐭𝐚𝐫 − 𝐬𝐭𝐚𝐫
𝑡ℎ𝑒 3 𝑝ℎ𝑎𝑠𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑟𝑎𝑡𝑖𝑛𝑔 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑠 = 3 ∗ 75 = 225𝑘𝑉𝐴
𝐻𝑉 𝑠𝑖𝑑𝑒: 𝑉𝐿. 𝐿 = 8 ∗ √3 = 14𝑘𝑣
𝐿𝑉𝑠𝑖𝑑𝑒: 𝑉𝐿. 𝐿 = 2.4 ∗ √3 = 4.16𝑘𝑣
200
𝑇𝑎: 𝑍 = (0.04 + 0.1𝑗)1 ∗ (
) = 0.0355 + 0.0889𝑗 𝑝𝑢
225
𝑎𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑡 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑟𝑒𝑓𝑟𝑒𝑒𝑛𝑐𝑒 𝑉𝑙𝑜𝑎𝑑 = 1∠0 𝑝𝑢
𝑠𝑙1 =
166.67
61.46
= 0.833𝑝𝑢 𝑠𝑙2 =
= 0.3073𝑝𝑢
200
200
𝑽𝒃𝟐 = 𝟏𝟑. 𝟒𝑲𝑽 𝒂𝒕 𝑯𝑽 𝒔𝒊𝒅𝒆 𝒐𝒇 𝑻𝑨
200 13.4 2
𝑮: 𝑿 = 0.15 ∗
∗(
) = 0.1333 𝑝𝑢
225 13.4
𝐼𝑙1 =
𝐼𝑙2 =
𝑠𝑙1
√3 ∗ 𝑉𝐿
𝑠𝑙2
√3 ∗ 𝑉𝐿
=
=
0.833
√3 ∗ 1
0.3073
√3 ∗ 1
= 0.48
→
= 0.17742
𝐼𝑙1 = 0.48∠ − 53.13
→
𝐼𝑙2 = 0.17742∠11.48 𝑝𝑢
𝑍𝑡 = 𝑋𝑔 + 𝑍𝑡 = 0.1333𝑗 + 0.0355 + 0.0889𝑗 = 0.0355 + 0.2222𝑗 𝑝𝑢
𝐼𝑡 = 𝐼𝑙1 + 𝐼𝑙2 = 0.5787∠ − 37.05 𝑝𝑢
𝒃) 𝑽𝒈 = 𝑉 + 𝐼𝑡 ∗ 𝑍𝑡 = 1∠0 + (0.5787∠ − 37.05) ∗ 0.0355 + 0.2222𝑗
= 1.0976∠4.72 𝑝𝑢
𝑉𝑔 = 1.0976∠4.72 ∗ 𝑉𝑏𝑎𝑠𝑒 = 1.0976∠4.72 ∗ 13.4 = 14.71∠4.72 𝑘𝑣
𝒄)
𝑺𝒈 = 𝐼𝑡 ∗ ∗ 𝑉𝑔 = √3 ∗ 0.5787∠37.05 ∗ 1.0976∠4.72 = 0.82 + 0.733𝑗 𝑝𝑢
𝑆𝑔 = 0.82 + 0.733𝑗 ∗ 𝑆𝑏𝑎𝑠𝑒 = (0.82 + 0.733𝑗) ∗ 200 = 164.1 + 146.57𝑗
𝑃𝑔 = 164.1𝑘𝑊
𝒅) 𝜼 =
𝑽𝑹 =
𝑄𝑔 = 146.57𝐾𝑉𝐴𝑅
𝑝𝑜𝑢𝑡
159.7
∗ 100% =
∗ 100% = 97.335 %
𝑝𝑖𝑛
164.1
𝑉𝑛𝑙 − 𝑉𝑓𝑙 1.0976 − 1
=
∗ 100% = 9.76%
𝑉𝑓𝑙
1
2)
The base voltage VBG1on the LV side of T1 is 2 kV. Hence the base on its HV
side is
15
𝑉𝐵1 = 2 ∗
= 12.5 𝑘𝑣
2.4
This fixes the base on the HV side of T2 at VB2 = 12.5 kV, and on its LV side at
220
𝑉𝐵𝑚 = 12500 ∗
= 208.33𝑣
13200
𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑎𝑛𝑑 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒𝑠 𝑖𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑜𝑛 𝑎 250 𝑀𝑉𝐴 𝑏𝑎𝑠𝑒,
250
𝐺: 𝑋 = 0.1 ∗ (
) = 0.1 𝑝𝑢
250
250
𝑇𝑎: 𝑍 = (0.01 + 0.05𝑗)1 ∗ (
) = 0.0083 + 0.04167𝑗 𝑝𝑢
300
𝑇𝑏: 𝑍 = (0.01 + 0.05𝑗) ∗ (
𝑍(𝑏𝑎𝑠𝑒 𝑙𝑜𝑎𝑑) =
𝐿𝑜𝑎𝑑: 𝑍 =
𝑉𝑏𝑎𝑠𝑒 2 208.332
=
= 0.1736𝑝𝑢
𝑆𝑏𝑎𝑠𝑒
250𝑘
(0.14 + 0.087𝑗)
= 0.806 + 0.51𝑗 𝑝𝑢
0.1736
𝑍(𝑏𝑇𝑙𝑖𝑛𝑒) =
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