# Electricity consumed

• September 13th, 2013, 01:27 PM
rahulmohan
Electricity consumed
If i use 1Kw hydraulic pump to lift a 1000kg weighted mass to a height 25 m in 8 mins (ie work that should be done by pump =mgh=25000J,ie 52 J per sec=52 W)
How many unit of electricity is needed to drive the motor.ie in Kwhr.
• September 13th, 2013, 02:06 PM
Janus
First off, mgh will equal 250,000j, which gives 251 W. (if you use 10m/sec^2 for g)

So to find khw, you can either multiply the Watts by the time(in hrs) and divide by 1000, or you can just look up the conversion factor for joules to khw and multiply the total joules by that.
• September 13th, 2013, 03:24 PM
rahulmohan
ok but thats not the answer to my question
• September 13th, 2013, 03:34 PM
MacGyver1968
1.21 gigawatts
• September 13th, 2013, 04:16 PM
rahulmohan
How you got 1.21 gigawatts.can you please explain..
• September 13th, 2013, 04:45 PM
MacGyver1968
• September 13th, 2013, 04:55 PM
Strange
Quote:

Originally Posted by rahulmohan
ok but thats not the answer to my question

But it gives you enough information to work it out.
• September 13th, 2013, 07:07 PM
MacGyver1968
• September 13th, 2013, 09:54 PM
Harold14370
(9.8*1000*25)Joules/(8*60)seconds=510.4 watts.
• September 14th, 2013, 01:32 AM
rahulmohan
Its not homework.It is for a project.
• September 14th, 2013, 01:53 AM
rahulmohan
@ harold...weight is 1000 kg..not the mass..ie mg = 1000kg..so mass = 1000/9.81=101.9367 kg.so Work Done = (9.8*101.9367*25)Joules/(8*60)seconds=52.08 Joules/second
But i need the electricity consumed in units when i use 1 kw hydraulic pump.
• September 14th, 2013, 02:33 AM
Delta Flyer
Is your 1 kW hydraulic pump always going to be operating at 1 kW? If so, there's your answer.

If not, if that's simply an upper limit, and assuming negligible mass of hydraulic fluid, then the vertical distance times mass, will give you your kW, once you convert terms. For the -hr part, simply divide 8/60 and multiply times the kW portion of your answer.
• September 14th, 2013, 02:34 AM
Harold14370
Kilogram is not a unit of force. Something that "weighs" 1 kilogram has a mass of 1 kilogram.
http://en.wikipedia.org/wiki/Kilogram-force
Quote:

The kilogram-force (kgf or kgF), or kilopond (kp, from Latin pondus meaning weight), is a gravitational metric unit of force. It is equal to the magnitude of the force exerted by one kilogram of mass in a 9.80665 m/s2 gravitational field (standard gravity, a conventional value approximating the average magnitude of gravity on Earth).[1] Therefore one kilogram-force is by definition equal to 9.80665 N
The energy to lift the weight can be easily calculated by multiplying the power in watts as calculated above by the time, which is 8 minutes. However, the electricity consumed will be somewhat more than this due to inefficiencies in the motor and hydraulic pump. I don't think enough information was provided to calculate that.
• September 14th, 2013, 10:15 PM
jocular
Quote:

Originally Posted by Harold14370
Kilogram is not a unit of force. Something that "weighs" 1 kilogram has a mass of 1 kilogram.
Kilogram-force - Wikipedia, the free encyclopedia
Quote:

The kilogram-force (kgf or kgF), or kilopond (kp, from Latin pondus meaning weight), is a gravitational metric unit of force. It is equal to the magnitude of the force exerted by one kilogram of mass in a 9.80665 m/s2 gravitational field (standard gravity, a conventional value approximating the average magnitude of gravity on Earth).[1] Therefore one kilogram-force is by definition equal to 9.80665 N
The energy to lift the weight can be easily calculated by multiplying the power in watts as calculated above by the time, which is 8 minutes. However, the electricity consumed will be somewhat more than this due to inefficiencies in the motor and hydraulic pump. I don't think enough information was provided to calculate that.

Why, isn't efficiency in these kinds of problems always 100%, just as those problems advising to "neglect friction"? :) joc
• September 16th, 2013, 02:34 AM
rahulmohan
I m confused..so if we measure weight of a body using weight measuring equipment and got a value 100 kg then its mass is also 100 kg ?
• September 16th, 2013, 03:18 AM
Harold14370
Yes. In the MKS system, the unit of mass is the kilogram and the unit of force is the Newton. Weight is a force, so if you want to use the MKS system you should give the weight in Newtons. Alternatively, the scale can be calibrated to measure kilogram-force as explained in the Wikipedia article.
• September 16th, 2013, 12:52 PM
rahulmohan
Thanks i got it now..so considering my thread the pump should do 510 J/s work..so what will be the current consumed by pump in doing 245250 J in 8 mins (480 secs)
• September 16th, 2013, 04:48 PM
Harold14370
Well, if you take 510 watts times 8 minutes, that's 4080 watt-minutes. There are 60 minutes in an hour, so divide that by 60 and there's your answer in watt-hours, 68. If you use the rating of the pump instead of work done, that would be 1000*8/60=133.3 watt-hr. So, it would be somewhere between those two numbers.
• September 20th, 2013, 12:42 PM
rahulmohan
Ok so the value will be between 68 - 133 watts hr..