- #1
Master1022
- 611
- 117
- Homework Statement
- A 230 V DC shunt motor has an armature resistance of ## 0.2 \Omega ## and a field resistance of ## 216 \Omega ##. At full rated load, the output power is ## 18642 ## W and the armature current is 90 A. Find the rotational losses at this full load condition and therefore find the motor efficiency in this case.
- Relevant Equations
- ## V = E + I_a R_a ##
## P = I V ##
Hi,
My main question is: should we be subtracting the rotational power losses from the motor output power in our efficiency calculation?
My approach:
Will skip extra work as I have done the above parts of the question correctly.
I used power conservation to state that:
$$ P_{input} = P_{motor} + P_{field} + P_{armature} + P_{losses} $$
We end up with: ## P_{motor} = 18642 ## W, ## P_{field} = \frac{230^2}{216} = 244.907 ## W, ## P_{input} = 230(90 + \frac{230}{216}) = 20944.907 ## W, and therefore ## P_{losses} = 437.5 ## W.
When it comes to calculating the efficiency, I was always under the impression that we did:
$$ \eta = \frac{want}{pay for} = \frac{P_{motor}}{P_{input}} = \frac{18642}{20944.907} = 0.89 $$
However, in a previous question the solution takes into account the 'frictional losses (rotational losses)' such that we do:
$$ \eta = \frac{P_{motor} - P_{losses}}{P_{input}} $$
in that solution we subtract the 'windage losses'... and ## P_{motor} = I_a \cdot (V - I_a R_a) ##
Potential reasons for the calculation method:
1. Negligible effect by excluding the ## P_{losses} ##: perhaps, but the inclusion of the value does change the efficiency by 3% which doesn't seem that insignificant to me
I am struggling to see why the solution methods differ and would really appreciate any help.
My main question is: should we be subtracting the rotational power losses from the motor output power in our efficiency calculation?
My approach:
Will skip extra work as I have done the above parts of the question correctly.
I used power conservation to state that:
$$ P_{input} = P_{motor} + P_{field} + P_{armature} + P_{losses} $$
We end up with: ## P_{motor} = 18642 ## W, ## P_{field} = \frac{230^2}{216} = 244.907 ## W, ## P_{input} = 230(90 + \frac{230}{216}) = 20944.907 ## W, and therefore ## P_{losses} = 437.5 ## W.
When it comes to calculating the efficiency, I was always under the impression that we did:
$$ \eta = \frac{want}{pay for} = \frac{P_{motor}}{P_{input}} = \frac{18642}{20944.907} = 0.89 $$
However, in a previous question the solution takes into account the 'frictional losses (rotational losses)' such that we do:
$$ \eta = \frac{P_{motor} - P_{losses}}{P_{input}} $$
in that solution we subtract the 'windage losses'... and ## P_{motor} = I_a \cdot (V - I_a R_a) ##
Potential reasons for the calculation method:
1. Negligible effect by excluding the ## P_{losses} ##: perhaps, but the inclusion of the value does change the efficiency by 3% which doesn't seem that insignificant to me
I am struggling to see why the solution methods differ and would really appreciate any help.