有一台三相异步电动机,额定功率为7.5千瓦,额定电压380v,功率因数0.85,效率0.87,求 一台三相感应电动机,其额定值如下:功率因数0.8滞后,22K...
\u4e09\u76f8\u5f02\u6b65\u7535\u52a8\u673a,\u989d\u5b9a\u529f\u7387\u4e3a7.5 kW,\u989d\u5b9a\u7535\u538b\u4e3a380 V,\u529f\u7387\u56e0\u6570\u4e3a0.85,\u6548\u7387\u4e3a0.87,\u8bd5\u8ba1\u7b97\u5176\u989d\u5b9a\u7535\u6d41.\u89c6\u5728\u529f\u7387S=P/cosø=7.5/0.85=8.823KVA
\u989d\u5b9a\u7535\u6d41In=S/\u221a3*U=13.4A
\u7194\u65ad\u5668\u7684\u9009\u578b\u4e00\u822c\u4e3a\u989d\u5b9a\u7535\u6d41\u76841.5-1.6\u500d\uff0c\u6211\u4eec\u6309\u71671.5\u500d\u9009\u578b\uff0c\u5373\uff1a13.4*1.5=20.1A
\u70ed\u7ee7\u7535\u5668\u4e00\u822c\u4e3a\u8fc7\u8f7d\u4fdd\u62a4\uff0c\u8fc7\u8f7d\u4fdd\u62a4\u6574\u5b9a\u503c\u4e3a\u989d\u5b9a\u503c\u76841.2\u500d\uff0c\u5373\uff1a13.4*1.2=16.08A
\u5b9e\u9645\u9009\u578b\u6309\u7167\u8ba1\u7b97\u6807\u51c6\u4f5c\u4e3a\u53c2\u8003
\u7b54
\u3000\u3000\u4e09\u76f8\u5e94\u7535\u673a\u8f6c\u901f\u4e0e\u989d\u5b9a\u529f\u7387\u5173\u4e0e\u989d\u5b9a\u7535\u538b\u5173\u4e0e\u63a5\u4ecd\u5173
\u3000\u3000\u5e94\u7535\u673a\uff08\u540c\u6b65\uff09\u8f6c\u901f\u53d6\u51b3\u4e8e\u7535\u6e90\u9891\u7387\u4e0e\u7535\u673a\u78c1\u6781\u6570\u4f8b50Hz\u7535\u6e904\u6781\uff082\u78c1\u6781\uff09\u7535\u673a\u540c\u6b65\u8f6c\u901f1500\u8f6c/\u5b9e\u9645\u8f6c\u901f\u7565\u4f4e\u4e8e\u540c\u6b65\u8f6c\u901f1450\u8f6c/\u5de6\u53f3
电压380v
功率因数0.85
效率0.87
P= 1.732 X U X I X COSφ X η
电机的电流 I = 7500 /(1.732 X 380 X 0.85 X 0.87) =15.41 A
看你求什么，三楼给的是额定线电流。线电流与相电流的关系跟你的接法是三角形还是星形接法有关。采用三角形接法，线电压与相电压相等，均是380V，线电流是相电流的√3倍，采用星形接法，相电流等于线电流，线电压等于相电压的√3倍，上诉380V是线电压，所以其相电压是220V
求什么呀
绛旓細鍚屾杞焠1=60f/p=3000/3=1000r/min锛岃浆宸巼s=(n1-n)/n1=(1000-950)/1000=0.05锛涙晥鐜囄=Pn/(Pn+螖P)脳100锛=28/锕28+2.2+1.1+0.186)脳100锛=89锛呫涓夌浉浜ゆ祦寮傛鐢靛姩鏈鐨勬晥鐜囪秺楂橈紝鍏惰浆宸巼瓒婂皬銆備篃鍗冲湪棰濆畾棰戠巼涓庣數鍘嬩笅锛岀浉鍚岃礋杞界殑杞熻秺楂樸傚浜庢煇浜鍔熺巼涓庤浆閫熸垚姝ｆ瘮锛岀敋鑷...
绛旓細锛3锛鐢靛姩鏈洪瀹杩愯鏃惰緭鍏鍔熺巼涓哄灏戯紵---杈撳叆鍔熺巼Pr=Pc/畏=10/0.9=11.11KW銆
绛旓細绌哄帇閲囩敤瑗块棬瀛3UT绯诲垪锛20-35A 鍙傛暟鍙互璋冨湪27鍒28A灏卞ソ浜
绛旓細鏁堢巼0.87 P= 1.732 X U X I X COS蠁 X 畏 鐢垫満鐨鐢垫祦 I = 7500 /(1.732 X 380 X 0.85 X 0.87) =15.41 A
绛旓細1銆佸悓姝ヨ浆閫焠1=60f/p=3000/3=1000r/min 杞樊鐜噑=(n1-n)/n1=(1000-950)/1000=0.05 2銆佹晥鐜囄=Pn/(Pn+螖P)脳100锛=28/锕28+2.2+1.1+0.186)脳100锛=89锛 3銆佸畾瀛愮數娴両n=锕28+2.2+1.1+0.186)/1.732/0.38/0.85/0.89=63.2A 璁鍔熺巼鍥犳暟涓0.85 ...
绛旓細涓鍙颁笁鐩稿紓姝ョ數鍔ㄦ満鐨勯瀹氭暟鎹负棰濆畾鍔熺巼PN=15KW锛岄瀹鐢靛帇UN=380V锛岃绠楅瀹氱數娴佷负锛氳冭檻鍔熺巼鍥犳暟0.79锛屾晥鐜88%锛屽洜姝ら瀹氱數娴佷负32.78A銆傛棰樼殑鍔熺巼鍥犳暟鍜屾晥鐜囬兘鏄亣瀹氬硷紝鎵浠ョ數娴佽绠椾篃鏄繎浼煎笺
绛旓細棣栧厛锛屾垜浠潵瀹氫箟涓浜涘彉閲忥細P_n = 棰濆畾鍔熺巼 = 28kWU_n = 棰濆畾鐢靛帇 = 380Vf_n = 棰濆畾棰戠巼 = 50Hzn_n = 棰濆畾杞 = 950r/minP_Cu + P_Fe = 閾滆楀強閾佽 = 2.2kWP_m = 鏈烘鎹熻 = 1.1kWP_add = 闄勫姞鎹熻 = 186W 杞樊鐜囩殑璁＄畻锛氳浆宸巼 s 鍙互鐢变互涓嬪叕寮忚绠楋細s = (n_n...
绛旓細涓夌浉鐢垫満鍔熺巼璁＄畻鍏紡锛歅=鈭3脳U脳I脳COS桅脳畏P:鐢垫満杞寸锛22KW锛塙锛氱數鏈虹嚎鐢靛帇I锛氱數娴侊紙42.2A锛塁OS桅锛氬姛鐜囧洜鏁帮紙0.89锛壩凤細鏁堢巼1銆佷笁鐩镐氦娴寮傛鐢靛姩鏈鐨勬晥鐜囷細畏=P/锛堚垰3*U*I*COS蠁锛夊叾涓紝P鈥旀槸鐢靛姩鏈鸿酱杈撳嚭鍔熺巼U鈥旀槸鐢靛姩鏈虹數婧愯緭鍏ョ殑绾跨數鍘婭鈥旀槸鐢靛姩鏈虹數婧愯緭鍏ョ殑绾跨數娴丆OS蠁鈥旀槸鐢靛姩...
绛旓細鐢靛姩鏈鐨勯瀹氱數娴=棰濆畾鍔熺巼涓22kW/棰濆畾鐢靛帇涓380V/1.732/鍔熺巼鍥犵礌涓0.86=38.9A
绛旓細棰濆畾鐢垫祦I=P/鈭3*U*c0S =0.75/鈭3*380鈭*0.8=1.4A
