企業(yè)降壓變電所電氣初步設(shè)計【獨(dú)家課程畢業(yè)設(shè)計含4張CAD圖紙+帶任務(wù)書+開題報告+外文翻譯】
企業(yè)降壓變電所電氣初步設(shè)計
摘 要
本設(shè)計以變電站為研究對象。變電站不僅僅是電力系統(tǒng)的一個重要的組成部分,而且也是電廠和用戶之間的樞紐。主電路電氣配線的設(shè)計以及開發(fā),要對整個變電站電氣設(shè)備做出正確的選擇,其中包括配電設(shè)備裝置,繼電保護(hù)和自動裝置等設(shè)備。
在初步了解變電站的概念后,根據(jù)任務(wù)書上的要求,查閱相關(guān)的參考文獻(xiàn),進(jìn)一步了解所需要做的事情。由線路和負(fù)載的相關(guān)參數(shù),計算出相應(yīng)的負(fù)荷,進(jìn)行無功功率補(bǔ)償。然后根據(jù)變電站的位置及出線方向來考慮,并對負(fù)荷進(jìn)行數(shù)據(jù)分析,考慮安全性,經(jīng)濟(jì)性以及可靠性,確定10KV站以及電氣主布線,之后通過負(fù)載計算和供給的范圍,確定變壓器的容量和臺數(shù)以及型號,進(jìn)行相關(guān)的短路電流計算。再根據(jù)設(shè)計要求選擇斷路器,隔離開關(guān),變壓器和保護(hù)裝置的型號,最后還對相應(yīng)的繼電保護(hù)和防雷措施進(jìn)行了簡單的設(shè)計。
總而言之,一定要確保供電系統(tǒng)的安全和可靠性,在此基礎(chǔ)上再考慮其經(jīng)濟(jì)性,從而完成變電所的設(shè)計。
關(guān)鍵詞:主接線;數(shù)據(jù)分析;數(shù)據(jù)計算;變壓器;可靠性
Abstract
The design of transformer substation as the research object. Substation is not only power system an important component, and is also a hub between the power plants and users. Main circuit's electrical wiring design and development, it is necessary to make the right choice of the whole substation electrical equipment, including power distribution equipment, relay protection and automation device, equipment.
After a preliminary understanding of the concept of the substation, according to the requirements of the task book, refer to the relevant reference, to further understand the things that need to be done. The corresponding load is calculated by the parameters of the line and the load, and the reactive power compensation is carried out. After according to the substation location and outlet direction to consider, and the load for data analysis, consider the safety, economy and reliability, determine the 10kV station and main electrical wiring, after through the load calculation and supply determined transformer capacity and number of units, and types, related to the short-circuit current calculation.?According to the design requirements of the circuit breaker, isolating switch, transformer and the protection of the device model, and finally to the corresponding protection and lightning protection measures for a simple design.?
In short, we must ensure the safety and reliability of the power supply system, based on the consideration of its economy, so as to complete the design of substation.
Key words: main connection; data analysis; calculation data; transformer; reliability
目 錄
摘 要 I
Abstract II
1 緒論 1
1.1 設(shè)計背景 1
1.2 變電所的總體分析 1
1.3 設(shè)計思路 1
2 負(fù)荷計算 3
2.1 負(fù)荷分級及計算過程 3
2.1.1負(fù)荷分級 3
2.1.2負(fù)荷計算過程 3
2.1.3 0.4kV側(cè)負(fù)荷的計算 4
3 無功功率補(bǔ)償 7
3.1 無功電源的組成及補(bǔ)償裝置的選擇 7
3.3 無功補(bǔ)償?shù)拇_定 7
4 變電所位置和主變壓器的選擇 9
4.1 變電所位置確定的一般原則 9
4.2變壓器實際容量的計算 9
4.3主變壓器臺數(shù)的確定以及型號的選擇 9
4.4主變壓器的運(yùn)行方式 10
5 主接線方式的選擇 11
5.1 電氣主接線設(shè)計的基本要求 11
5.2 主接線的設(shè)計 11
6 短路電流的計算 12
6.1 確定基準(zhǔn)值 12
6.2 計算各元件電抗標(biāo)幺值 12
6.3 k-1故障點(diǎn)的短路電流計算 13
6.4 k-2故障點(diǎn)的短路電流計算 13
7 變電所一次設(shè)備的選擇與校驗 15
7.1 變電所高壓一次設(shè)備的介紹 15
7.2 一次設(shè)備的選擇與校驗的條件和項目 15
7.3 變電所10kV側(cè)一次設(shè)備的選擇 17
7.4 變電所380V低壓一次設(shè)備的選擇 19
7.6 變電所進(jìn)出線的選擇 20
8 繼電保護(hù)及二次回路的選擇 23
8.1 變電所二次回路方案的選擇 23
8.2繼電保護(hù)裝置 23
8.2.1反時限過電流保護(hù) 23
8.2.2電流速斷保護(hù) 24
8.3 電力線路的繼電保護(hù)設(shè)計 25
9 防雷保護(hù)的設(shè)計 27
9.1 避雷針的配置原則 27
9.2 防雷設(shè)計 27
9.3 確定最終的防雷方案 28
結(jié)論與展望 29
致 謝 30
參考文獻(xiàn) 31
【詳情如下】【需要咨詢購買全套設(shè)計請加QQ1459919609】
企業(yè)降壓變電所電氣初步設(shè)計.doc
企業(yè)降壓變電所電氣初步設(shè)計任務(wù)書.doc
企業(yè)降壓變電所電氣初步設(shè)計開題報告.doc
外文翻譯原文.pdf
外文翻譯譯文.doc
平面布置圖.dwg
文件清單.txt
電氣主接線圖.dwg
繼電保護(hù)圖.dwg
防雷設(shè)計圖.dwg
A. W. A. J. A. of on 5 003/ 11 003/14 003003he of of on of is on on a to an in of of of on of of of is to 1], in is it is in 2]. In by to 3]4] 5, 6]to in At it of be to 7], of 8]. as to [5]), in of is of on is on of of on of a 9]. In a is of on of is by is to to of on of a MV)is 1). V) is of is m. A. &) ? W. A. J. A. 1,9000 +32+322004) 86: 181–190by s,V/LV be V 1a) or 1b). is in V is in to is 5h?3 is m;l;m;l;l;on of 1) V/MV V/an of V/MV of of is of is to V=||pu at of is pu at a of is or of is be is to of of s;4? e2TA is t=1. in to V:Zs?0:04815 5no 0:060890no 0:04626 0no ?? 0:02898 53T1 V a V at of b V in ?? 0:02070 55TZs?0:09300 5no ? 1Tpu to of a As is to be 0of is by or of 3) a or a of is .8 , to be of in a a of in A of 11]. A is [12]be on in of h=3, 9,15, ...) h ? 17) 9]of a by or of of a In as in 13]is of a of a is as MS of ???????????????????????????????????(h) in If of a a on of of 4) 3):=h>1=h>1of of in (5) In is in an of of a of I(h)/I(1)|1 ,h>25183 by of an V/Y):no =h>1=h>1=h>1, 6) (( In of is is is of be by V/LV by it Y of V in 1), of in CC be =h>1=h>1=h>1of to of V/MV , of in of in on 1. In of V/LV is ||0). V/LV be 5, a In of up to up to a of of of a pu at 1, is in to is V to be a as ) 0% of EC 14], By 4), is ???????????????????????????????????????????to in CC at of in ‘:is by in 1 is to be to of of on is by of . 2a. it 2 of 1 ( As HD in CC to of HD is in CC of of is in 13]. 1, 2, , of at of is to of in of is in at of of It in CC or of on is by 4–6. 2b. 4, of is 501. 2.3,of to a As of by as In a of HD at of 1 , 50of to by 6, It is to be of 1 )of 1 ) HD |(% |Zs,l| (% ( ) )1 0 5 1 V/V/V/MV 6· 16· 12 of a b c on HD in CC is it is HD in is is is by 11], of is 9]of on is of in it of of on is by 1,7, . 2c. 7 , ,as 1. on HD in CC at is is to s,k( be in of is by 1, 9, 0 . 3a. In 9, is Y c=0. It of HD in CC by 3.2)of as a of HD at of in C