Power consumption in KWH of motor

In summary, According to the information provided, a motor with a rating of 750 watts is needed to lift a 100 kg object 3 metres high in 5 seconds. The power required per cycle is 588 watts, and the total power used in a month is 162000 watts.
  • #1
Dave250526
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0

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
 
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  • #2
Dave250526 said:
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.
 
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  • #3
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 ?
 
  • #4
Dave250526 said:
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.
 
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  • #5
Dave250526 said:
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.
 
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  • #6
mfb said:
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 said:
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.
 
  • #7
Dave250526 said:
= 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
Dave250526 said:
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##
Dave250526 said:
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.
 
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  • #8

1. What is power consumption in KWH of a motor?

Power consumption in KWH (kilowatt-hour) of a motor refers to the amount of energy used by the motor over a period of time, measured in kilowatt-hours. This is an important measure of the motor's efficiency and can vary depending on factors such as the motor's size, speed, and load.

2. How is power consumption in KWH of a motor calculated?

To calculate power consumption in KWH of a motor, you multiply the motor's input power (in watts) by the amount of time it is in use (in hours), and then divide the result by 1000. This will give you the power consumption in kilowatt-hours. For example, if a 1000-watt motor is used for 5 hours, the calculation would be: (1000 watts x 5 hours) / 1000 = 5 KWH.

3. Why is it important to measure power consumption in KWH of a motor?

Measuring power consumption in KWH of a motor is important because it allows us to understand how much energy the motor is using and how efficiently it is operating. This information can help us make decisions about the motor's maintenance, usage, and potential upgrades to save energy and reduce costs.

4. How can power consumption in KWH of a motor be reduced?

There are several ways to reduce power consumption in KWH of a motor. One way is to use a more efficient motor or to upgrade the motor's components. Another way is to reduce the motor's load or run it at a lower speed, if possible. Regular maintenance and cleaning can also help improve the motor's efficiency and reduce power consumption.

5. Are there any external factors that can affect power consumption in KWH of a motor?

Yes, external factors such as temperature, humidity, and altitude can all affect power consumption in KWH of a motor. Higher temperatures can reduce the efficiency of the motor, while high humidity can cause corrosion and damage to the motor's components. Higher altitudes may also impact the motor's performance due to changes in air density.

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