电拖课后参考答案

彭鸿才《电机原理与拖动》习题解答

1 第一章 直流电机原理

电拖课后参考答案

解:3

141060.87()230

N N N P I A U ⨯=== 11416.37()0.855

N

N P P Kw η===

电拖课后参考答案

解:1110131430()N N P U I W ==⨯=

3

1

1.11076.92%1430N N P P η⨯=== 114301100330()N p P P W =-=-=∑

电拖课后参考答案

解:)(53.1150

230A R U I f N f === )(13.7153.16.69A I I I f N a =+=+= 23071.130.128239.1()a N a a E U I R V =+=+⨯=

)(6.647128.013.7122W R I p a a cu =⨯== )(1700713.711.239W I E P a a M =⨯==

)(5.18713855.010163

1W P P N N

=⨯==η

电拖课后参考答案

解:)(92.39683.06.40A I I I fN N a =-=-=

)(5.211213.092.39220V R I U E a a N a =⨯-=-= )(1.844392.395.211W I E P a a M =⨯==

)(4.339213.092.3922W R I p a a cu =⨯== )(26.150683.0220W U I p N fN f =⨯== )(8932

6.402201W I U P N N =⨯== %9

7.838932105.73

1

=⨯==P P N η

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