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About LinkBacksIf 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, 01:27 PM
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.
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.
ok but thats not the answer to my question
1.21 gigawatts
How you got 1.21 gigawatts.can you please explain..
It's a joke:
But it gives you enough information to work it out.Originally Posted by rahulmohan
..and we don't answer homework questions...only help you with them.
(9.8*1000*25)Joules/(8*60)seconds=510.4 watts.
Its not homework.It is for a project.
@ 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.
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.
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
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.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
Why, isn't efficiency in these kinds of problems always 100%, just as those problems advising to "neglect friction"?joc
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 ?
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.
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)
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.
Ok so the value will be between 68 - 133 watts hr..
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