- #1
Dave250526
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Homework Statement
Hello Everybody,
When i was searching for power required to lift an object, i found that:-
For Example, Q :- 100 kgs to be lifted 3 metres in 5 seconds. (vertical)
A :- Mass * Gravity * (Distance / Time)
= 100 * 9.8 * (3/5)
= 588 Watts
Assuming a efficiency loss of 22% (588/78%) :- 750 watt required to lift a object weighing 100 kgs, 3 metre high in 5 seconds
I don't think there is any problem in above calculation (pls correct if I'm wrong)... But my question is , What will be the monthly electricity consumption (KWH) of 750 watt motor if run for the above said purpose 5 times in a minute (5 cycles) (i.e run for 25 seconds/min)
Homework Equations
.
The Attempt at a Solution
[/B]
A) Will the answer be :-
Total HOURS run in a DAY :- (25 seconds / 60 seconds) * 24 hrs a day
= 10 hrs
Total hrs in a month = 10 * 30 = 300 hrs in a month
Since 750 watt is 75% of 1 Kw = 300 * 75% = 225 KWH or 225 units ( I think this makes more sense)
Or
B)
Total cycles in a month = 5 times in a minute * 60 min * 24 hrs * 30 days
= 216000 cycles
Power consumption per cycle = 750 watt
Hence total power consumption is = 750 watt * 216000 cycles
= 162000000 watts
= 162000 Kwh ( I think this answer is absurd)
(I know this is a very basic doubt... but pardon me, since i don't belong to Physics field and i couldn't find a clear cut answer) May be i was looking in the wrong place..:) Sorry