Many problems regarding power etc

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In summary, the conversation discusses the required power and energy to run a system weighing 300 kg and moving at 25 km/h. The speaker plans to use a 1 hp motor, but is unsure if it will be enough. They also discuss the conversion between horsepower and watts, and the incorrect conversion between kilowatts and watts. The conversation concludes with a calculation of the time it would take for the motor to provide the required energy.
  • #1
shafeen
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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
 
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  • #2
Hello shafeen, welcome to PF.

shafeen said:
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.

shafeen said:
(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=

shafeen said:
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.

shafeen said:
2. 1h.p=746w

Right.

shafeen said:
1kw=100w

Wrong.

1 kW = 1000 W
(kilo = 103)

shafeen said:
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?
 
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  • #3


I would first suggest checking your calculations to ensure they are accurate. It is important to double check your work in order to avoid any errors that may lead to incorrect conclusions.

Next, I would recommend considering the efficiency of the motor. While a 1 hp motor may technically be able to provide the required power, it may not be able to do so efficiently. It is important to consider the efficiency of the motor in order to determine if it is suitable for your system.

Additionally, the weight and speed of the system also play a role in the power requirement. It is important to ensure that the motor is able to provide enough power to move the system at the desired speed while also considering the weight of the system.

In conclusion, it is important to accurately calculate the power requirement for your system and consider factors such as motor efficiency, weight, and speed in order to determine if a 1 hp motor will be sufficient. It may also be helpful to consult with a professional in the field for further assistance.
 

Related to Many problems regarding power etc

1. What is meant by "power" in the context of scientific research?

Power refers to the ability of a study or experiment to detect a true effect or difference, if one exists. It is influenced by factors such as sample size, effect size, and variability within the data.

2. Why is power important in scientific research?

Having sufficient power in a study is crucial because it increases the chances of detecting a true effect or difference. Low power can result in false negatives, where a real effect is not found, leading to incorrect conclusions.

3. How can power be increased in a study?

Power can be increased by increasing the sample size, using more sensitive measures or tests, reducing variability within the data, and controlling for confounding variables.

4. What are some common problems that can affect power in a study?

Some common problems that can affect power include small sample sizes, high variability within the data, and using insensitive measures or tests.

5. Can power be calculated or determined after a study has been conducted?

Yes, power can be calculated retrospectively after a study has been conducted. However, it is best to determine the desired level of power before the study begins in order to ensure appropriate sample sizes and methods are used.

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