Power consumption in KWH of motor

[Dave250526]

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

 

[haruspex]

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

You are confusing energy with power. The power consumption is 750W during the active part of a cycle, but you want the energy consumed per cycle.

 

[Dave250526]

Thanks for the reply sir,

And yes Sir, I'm a bit confused. Please help me out by spending few minutes.

In the question

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 understand 750 watts is the power required.

Total work in a cycle = 100 kg * 9.8 * 3 metres = 2940 joules

Since i want it to be done in 5 seconds = 2940 / 5 second = 588 Watts

At efficiency loss of 22% = It works out to 750 Watts

And i thought 750 watts is the power consumed per cycle of 5 seconds. Am i wrong?

Is 750 watt the rated capacity of Motor? If so if it runs for 1 hour at full capacity will it consume 0.75 KWH ?

 

[mfb]

And i thought 750 watts is the power consumed per cycle of 5 seconds. Am i wrong?

750 Watts is needed for the whole duration of 5 seconds, it is not "750 W per cycle".
750 W * 5 s = 3750 J needed per cycle. You can multiply this number with your 216000 cycles per year and convert the resulting energy value to kWh, the result will be the same as with your correct first approach.

 

[haruspex]

i thought 750 watts is the power consumed per cycle of 5 seconds.

It makes no sense to say "power consumed per cycle of some number of seconds". It's like saying "a car went at speed 50km/h per one hour trip".
It consumes that power for a period of 5 seconds in each cycle, but what it consumes per cycle is energy, not power.
If it draws 750W for 5 seconds, how much energy is drawn?

Edit: I see mfb already answered my question for you.

 

[Dave250526]

750 Watts is needed for the whole duration of 5 seconds, it is not "750 W per cycle".
750 W * 5 s = 3750 J needed per cycle. You can multiply this number with your 216000 cycles per year and convert the resulting energy value to kWh, the result will be the same as with your correct first approach.

It makes no sense to say "power consumed per cycle of some number of seconds". It's like saying "a car went at speed 50km/h per one hour trip".
It consumes that power for a period of 5 seconds in each cycle, but what it consumes per cycle is energy, not power.
If it draws 750W for 5 seconds, how much energy is drawn?

Edit: I see mfb already answered my question for you.

Thanks a lot.
I think ( just think) i understand now. Please correct if I'm wrong...

1) So Total energy consumed(Kwh) will be the same IRRESPECTIVE of Power (Kw)..
2) What matters as per my question is the "work done" (Energy consumed in joules)
3) Power is derived from the rate of energy consumed.. ( Here, in how much seconds i want a cycle to be completed)

So total energy consumed = [(100 kg * 9.8 * 3 metres) / 78% efficiency] * 216000 cycles
= 810000000 Joules
= 810000000 watt

Since i want the power consumed in kwh = 810000000 watt /3600 seconds = 225000 Watts Hour= 225 Kwh = 225 Units of ENERGY CONSUMED

Hence to derive the power = Work done per cycle / time required per cycle
= 3750 joules / 5 sec
= 750 Watts of POWER Required

If i want it to be completed in 4 seconds = Power required will be 937.5 Watts (3750 Joules/ 4 seconds)
But the total ELECTRICITY CONSUMPTION to complete the said work will be 225 units, irrespective of whether the cycle is 4 or 5 seconds each... Right?

Kindly Reply, I think I've got a fair understanding between power and energy.

 

[haruspex]

= 810000000 Joules
= 810000000 watt

No, a Joule is a Watt-second (the energy delivered by 1 Watt of power flowing for 1 second).
810000000 Joules = 810000000 Watt-seconds

810000000 watt /3600 seconds = 225000 Watts Hour

(810000000 Watt-seconds) /(3600 seconds per hour) = 225000 Watt-hours.
You can manipulate the units as if algebraic variables:
$\frac{W \times s}{\frac sh}= W \times h$

the total ELECTRICITY CONSUMPTION to complete the said work will be 225 units, irrespective of whether the cycle is 4 or 5 seconds each... Right?

Yes.

 

[Dave250526]

@haruspex & @mfb

Ok, Understood.. Thanks a lot, Sir ...