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...

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.... 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? (i.e) to lift 100 kgs (3 metres/5 sec)  five cycles in a minute (i.e run for 25 seconds in a minute at full capacity of 750 watt)

A) Will the answer be :-

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)

or

B) 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)

Kindly Reply Whether the Answer is A / B / or someother Answer

(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

Asked on by rrgpro2000

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t-nez's profile pic

t-nez | High School Teacher | (Level 3) Associate Educator

Posted on

Your calculations look correct, but you need to consider the peak power capacity of the turbine (KW) rather than the energy consumed (KWH). If it can generate 1 kW that means that at any one time it can power appliances totaling 1KW in power ratings. The KWH per month is kind of irrelevant in this case. What matters is if its peak capacity is enough for the maximum power you need at any one time. You could possibly use power in excess of what you need to charge a deep cycle battery to store energy for later use.

I hope this helps with your project. 

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t-nez's profile pic

t-nez | High School Teacher | (Level 3) Associate Educator

Posted on

Your second calculation, B, is correct. Kilowatts is used to express power, which is the rate of doing work. Thus power is work/time, which you correctly calculated as 750 W (expressed to 2 significant digits.) 

Electrical energy consumed is power x time. Watts x hours gives electrical energy in the units of kilowatt-hours, KWH. You were correct to find the number of hours that the motor operated in a month and multiply its power usage by the hours, giving 225 KWH (or 230 KWH to be consistent with 2 significant digits.)

The reason the method you used to come up with your answer in A doesn't work is because it doesn't factor in time. 750 Watts is the rate at which power is consumed. You would need to multiply the number of cycles by the amount of energy consumed by each cycle, not by the rate of power consumption.

The amount of energy consumed by each cycle is:

(750W)(1 KW/1000 W)(5 sec)(1hour/3600 sec) = 1.04 x10^-3 KWH

Notice that multiplying this by the number of cycles in a month, 216,000, gives the same answer as B: 225 KWH

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rrgpro2000's profile pic

rrgpro2000 | eNotes Newbie

Posted on

Wow thanks a lot t-nez.. Hope i had a teacher like you in High school.. Honest.. I might have took science as my first subject :-)  [Late realization, bcoz I do find physics quite interesting)

So,I do want to clarify few points. Please correct if i'm wrong in any of the following conclusions..

1) Basically 750 watts is the rated capacity of motor, like in appliances. i.e an iron box of 1 Kw if used for an hour will consume 1 unit (1 Kwh)

2) Actually i was looking for how much motor and appliances i could run if i installed a pico turbine , to help in my granny's country side farm.

The stream has a Water flow of 16 litres / second (water bucket method) proper head of 8 metres.. So assuming a efficiency lvl of 80% :- Turbine's RATED capacity will be 

                        = Flow rate (M^3)*gravity* Head height* Efficiency(%)

                        = 0.016 * 9.8 * 8 * 80%

                        = 1 Kw (approx)

(Here 1 Kw is as same as 750 watt of the earlier Question, Right?)

3) That means assuming the turbine runs 24 hours a day for 30 days at full capacity, it could produce 

                         = 1Kw * 24 hrs * 30 days

                         =  720 Kwh

4) So, does it mean after consuming 230 KWH for motor , i would have 490 Units (720-225) for other household appliances like Fan, light etc .. Right?

Thanks in advance T-nez.. I much appreciate your help.

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