A 600 V DC motor drives a 60 kW at 900 rpm- The shunt field resistance is 100 - and the armature resistance is 0-16- If the efficiency is 85

P_in=600.85=70.59 kW I_L=70.59× 10000600=117.65 A, I_f=600100=6 A I_a=117.65 6=111.65 A nbsp; E_a=600 (111.65× 0.16)=582.14 V No load: nbsp; nbsp;E ...

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Standard X

Physics

Heating Effect of Current

A 600 V DC mo...

Question

A 600 V DC motor drives a 60 kW at 900 rpm. The shunt field resistance is 100 $Ω$ and the armature resistance is 0.16 $Ω$ . If the efficiency is 85% then the speed regulation of motor is _________%.
3.1

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Solution

The correct option is A 3.1
$P_{i n} = \frac{60}{0.85} = 70.59 k W$

$I_{L} = \frac{70.59 \times 10000}{600} = 117.65 A$ ,

$I_{f} = \frac{600}{100} = 6 A$

$I_{a} = 117.65 - 6 = 111.65 A$

$E_{a} = 600 - (111.65 \times 0.16) = 582.14 V$

$No load: E_{a} \approx V = 600$

$N_{0} = \frac{900 \times 600}{582.14} = 927.6 r p m$

$Speed regulation = \frac{927.6 - 900}{900} \times 100 % = 3.1 %$

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A 600 V DC motor drives a 60 kW at 900 rpm. The shunt field resistance is 100 Ω and the armature resistance is 0.16Ω. If the efficiency is 85% then the speed regulation of motor is _________%.3.1

The correct option is A 3.1Pin=600.85=70.59 kW IL=70.59×10000600=117.65A, If=600100=6 A Ia=117.65−6=111.65A Ea=600−(111.65×0.16)=582.14V No load: Ea≈V=600 N0=900×600582.14=927.6 rpm Speed regulation =927.6−900900×100%=3.1%

A 600 V DC motor drives a 60 kW at 900 rpm. The shunt field resistance is 100 $Ω$ and the armature resistance is 0.16 $Ω$ . If the efficiency is 85% then the speed regulation of motor is _________%.
3.1