Many problems regarding power etc

[shafeen]

I am in a small project and i think i have got my self tangled in some equations...

1. I have a system and to run this system the required power is about 6kwh.(system weighs 300kgs and needs to go at 25kmph,the value is taken by calculating kinectic energy)i Plan to run this system by using a motor of 1 hp.Now i want to know two things
1.Will the 1 hp motor be enough to provide the power/energy
2. 1h.p=746w
1kw=100w
10 in² = 0.006451 m²
1w=1j/s
3. well i started off with attacking the horse power thing
1H.p=746W =746j/s=(746*60*60)=2685.6 *10^3 j/hr is the thing it would supply per hour

my requirement is 6kwh(assuming run for only one hour) =6kw=6000w=6000j/s=6000*60*60=2160*10^4

hence by this calculation the motor is not enough to power...but heck something in my mind says i am doing something wrong over here

 

[cepheid]

Hello shafeen, welcome to PF.

1. I have a system and to run this system the required power is about 6kwh.

Do you mean that the required energy is about 6kWh? Because you can use almost any motor to give an object a certain amount of kinetic energy. What the power of the motor determines is how much time it takes for the object to reach that energy. This is an important point, because your problem, as it stands right now, may not be very well-posed.

(system weighs 300kgs and needs to go at 25kmph,the value is taken by calculating kinectic energy)

First of all, try to use standard notation so that everyone knows what you are talking about. SI unit symbols don't get pluralized with an 's', it should be just 300 kg. Also, units that are named after people typically have capital letters as symbols (even though the unit names themselves are not capitalized). So, the unit named after Joule is the joule and it has symbol 'J'. The unit named after Watt is the watt and it has symbol W. EDIT: and also, we don't use the letter 'p' to mean 'per.' That speed should really be written as 25 km/h.

Second of all, I don't agree with your computation of the kinetic energy of the system (not when it is expressed in kilowatt-hours):

http://www.google.ca/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=gdM&q=(1%2F2)+*+300+kg+*+(25+km%2Fh)^2+in+kWh&btnG=Search&meta=&aq=f&oq=

Plan to run this system by using a motor of 1 hp.Now i want to know two things
1.Will the 1 hp motor be enough to provide the power/energy

As I said above, this is a very important distinction. Power and energy are not the same thing. The power of the motor is the rate at which it supplies energy. So the only really sensible question to ask is whether it can provide the required energy in a certain time interval.

2. 1h.p=746w

Right.

1kw=100w

Wrong.

1 kW = 1000 W
(kilo = 10³)

10 in² = 0.006451 m²

Can you explain what this has to do with anything?

Look, basically 746 W = 0.746 kW, meaning that the calculation goes like this (assuming the power is constant):

power = energy/time

time = energy/power

= (0.00200938786 kWh) / (0.746 kW)

= 0.00269354941 h

Which is about 9.7 seconds. Are you satisfied with that, or would prefer a more powerful motor to give the system that amount of kinetic energy faster than that?