Engineering Efficiency Calculation of a DC Series Motor | Homework Solution

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A DC series motor operates at a full load current of 10 A from a 12 V source, with losses due to iron, friction, and windage accounting for 9% of input power. The calculations show that the input power is 120 W, while total losses, including windage, core, and brush losses, amount to 55.8 W. This results in a power output of 64.2 W and an overall efficiency of 53.5%. The voltage drop of 2 V is confirmed to be for both brushes combined. The calculations appear to be correct based on the provided data.
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Homework Statement

A dc series motor takes a full load current of 10 A from a 12 V voltage source .The iron,friction and windage losses account to 9% of the input power .the field winding resistance is 0.1 ohms and the armature resistance is 0.15 ohms and the voltage drop is 2 V .Calculate the power output and the overall efficiency ?

Homework Equations


The Attempt at a Solution


Okay so first the I armature =I load=I field =10Amp hence power in= Ia?*Vt =10*12=120 W
The I calculated all the losses p losses (windage) = 0.09*120 = 10.8 Watts ,p losses in the core = Ia^2*r =100*(0.25)=25 Watts
Also the brush losses =Vb*Ia= 2*10 =20 Watts ,therefore p output= 120-10.8-25-20 =64.2 watts and the efficiency is p output/p input =64.2/120 =53.5% .

Okay I just need someone to check if I have done the right thing here please ? Thank u
 
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The brush voltage drop of 2 volts is the total for both brushes, is it?
 
yes the voltage drop is for both brushes
 

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