Minimun Power needed

  • Thread starter arpsachin2
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  • #1
arpsachin2
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Homework Statement



A fan accelerates 2 m^3/min of air from rest to a velocity of 15 m/s. If the density of air is 1.25 kg/m^3, determine the minimum power that must be supplied to the fan?

Homework Equations





The Attempt at a Solution



Need some hints or something.
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi arpsachin2! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)

hint: momentum :wink:
 
  • #3
arpsachin2
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Thank you.

Okay a bit more I guess. :(
 
  • #4
chemisttree
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You can convert the volume of air to mass using density.
 
  • #5
tiny-tim
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(just got up :zzz: …)

force = rate of change of momentum

power = force times … ? :wink:
 
  • #6
arpsachin2
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Power = force * velocity

Am I heading the right way?


Fan accelerates at 2m^3/60 secs to 15 m/s

And density is 1.25kg/m^3.

So is it Power = 1.25kg/m^3 * Acceleration?

How do I get the acceleration?
 
  • #7
tiny-tim
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hi arpsachin2! :smile:
Power = force * velocity

Am I heading the right way?

yes :smile:
Fan accelerates at 2m^3/60 secs to 15 m/s

And density is 1.25kg/m^3.

So is it Power = 1.25kg/m^3 * Acceleration?

and now you've totally lost me! :confused:

you don't need the acceleration!

you only need the force, and you can get that from the change in momentum :smile:
 
  • #8
arpsachin2
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if possible could you show me an example or something?

Thanks
 
  • #9
arpsachin2
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hi arpsachin2! :smile:


yes :smile:


and now you've totally lost me! :confused:

you don't need the acceleration!

you only need the force, and you can get that from the change in momentum :smile:

Hey,

would it be possible to chat on Gmail or yahoo or something?
Please
 
  • #10
arpsachin2
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hi arpsachin2! :smile:


yes :smile:


and now you've totally lost me! :confused:

you don't need the acceleration!

you only need the force, and you can get that from the change in momentum :smile:

So change in momentum is

Final momentum - Initial Momentum right

So, would it be 15 m/s - xxxx?
 
  • #11
tiny-tim
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hi arpsachin2! :smile:

(just got up :zzz: …)
So change in momentum is

Final momentum - Initial Momentum right

So, would it be 15 m/s - xxxx?

15 m/s is a speed, you need a momentum (= mass times speed), in kg m/s,

and then you need a rate of change of momentum, in kg m/s2
 

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