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class="news_mes">文章来源Q电力工E技术官?作者:??点击Q?span id="HitNum"></span>?旉Q?span id="dTime">2020/6/16 14:18:10</span><input type="hidden" id="ID" value="100915165"/><input type="hidden" id="TypeID" value="2"/></div> </td> <td><div id="ckepop"></div></td></tr></table><div id="Content" style="display:none;">[甉|癄Q?20v늺可以拉多q,20公里外能否带?000瓦的电器Q]:늺늼只要传输用电׃有电压损耗,必定存在电压降,M电器都是存在电阻的,늺也不例外。电压降只要不低?%是正常Q现在是220v的单相电Q那?20*5%=11vQ即电压不能低于209vQ否则终...</div><div class="brand_news_con"><p style="text-indent:2em;"> 늺늼只要传输用电׃有电压损耗,必定存在电压降,M电器都是存在电阻的,늺也不例外。电压降只要不低?%是正常Q现在是220v的单相电Q那?20*5%=11vQ即电压不能低于209vQ否则终端电器不能正怋?</p> <p style="text-align:center;text-indent:2em;"> <img src="http://oss.cnelc.com/kindeditor/img/20200616/20200616141626_5759.png" alt="" /> </p> <p style="text-indent:2em;"> <br /> </p> <p style="text-indent:2em;"> 以一?.5斚wUؓ例,常温下(20℃)铜的电阻率ρ约0.0178Ω.mQ导U电d式R=ρ*l/sQ其中l为长度(单位为mQ,s为导U横截面U?单位为mm2Q。那?0公里长的导线Q其电阻为R=0.0178*20000/2.5=142.4QΩ)Q正负两根线Q电阻?Q共284.8Ω?000W的电器其工作甉|最约4.5AQ若串联142.4Ω的电阻,4.5A的电电M端的压降U?84.8Ω*4.5A=1281.6VQ远大于供电电压220VQ完全不可能Q必ȝ220V+1281.6?500V供电Q电器才能工作?</p> <p style="text-align:center;text-indent:2em;"> <img src="http://oss.cnelc.com/kindeditor/img/20200616/20200616141646_6233.png" alt="" /> </p> <p style="text-indent:2em;"> <br /> </p> <p style="text-indent:2em;"> 如果?20V的交电压如果不影响额定电压?20V的用电器L话,那么在线路中所损耗的压降不能低于额定电压?%Q也是说在U\中的压降不能低于11VQ我们可以根据线路压降的表达式△U=26ImL/S,q是以铜导线ZQIm表示U\中的相电、L表示U\的长度、S表示导线的长度,q样以来我们不难得出对于220V电压的供电距R?</p> <p style="text-align:center;text-indent:2em;"> <img src="http://oss.cnelc.com/kindeditor/img/20200616/20200616141706_4675.png" alt="" /> </p> <p style="text-indent:2em;"> <br /> </p> <p style="text-indent:2em;"> L=△U.S/26Im,假设U\的截面积?qx毫米的话Q输送额定电是4A的话Q我们把上述的数据代入就可以得出输电的最大距M。L=11X4/26X4=0.423千米。由此可见交?20V的电压供늚距离是和多大的线径和导线的材料以及所带多大的负荷都是有关pȝ。我q里l出的如果负荷额定电是4安培Q所用的?qx毫米铜线的话Q如果不影响额定电压220的用电器h常工作的话电U可以拉最q不过500c的距离?</p> <p style="text-indent:2em;"> <span>极简单的Ƨ姆定律q用Q次生定理,即电功率Q电热能Q与g截面U及电压、电之间的关系Q还需有材料因素)Q方E式为功率值等于电与电阻值^方的乘积。根据公式,如果不计成本Q拉个几万公里理Z也是可行的?/span> </p> <p style="text-indent:2em;"> 20公里之外Q还要保?000W电器能用。奢侈点用铜Uѝ?</p> <p style="text-indent:2em;"> <span>LQ?0000m。ΔU?1VQU?09V。ΔUQIRQIρL/SQ其中ρ=0.0175、IQ?.55。又因ؓ构成电\回\需要一根火U和一栚wUѝ因此距M变的情况下,每根U允许电压降不高?.5VQ即SQ?89.545mm2?/span> </p> <p style="text-indent:2em;"> <span>由此可见Q要20公里之外的电器能用,铜导U横截面U不能低?89.545mm?/span> </p> <p style="text-indent:2em;"> Ҏ题目说的问题Q带?0公里之外的电器肯定没问题。这是土豪之中的王中王,20公里处买U的qZ敢想Q?万米铜线Q淘宝卖的国标一?00mm2软线?65?mQ即660万元?</p> <p style="text-indent:2em;"> <span>存在理论上的可能性,比如采用极粗的电~,有些|友过要达?00qx毫米的线才可以,q还是小事,关键费用高的吓h。最l济的做法是用变压器升压Q比如升?0万伏Q这样对U缆的就不需要那么粗了?/span> </p></div><div class="meneame"><span class="disabled">< </span><span class="current">1</span> <span 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