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Power check and help

  • #1
~christina~
Gold Member
715
0

Homework Statement


A 120 V motor has a mechanical power output of 2.50hp It is 90% effiecient in converting power that it takes in by electrical transmission into mechanical power.

a) find current in motor
b) find energy delivered to the motor by electrical transmission in 3.00 h of operation
c) If the electrical company charges $0.160kWh what does it cost to run the motor for 3.00h?

Homework Equations


The Attempt at a Solution



a) find current in motor
[tex]P=I \Delta V[/tex]

and I got

I= 17.08A

b) find energy delivered to the motor by electrical transmission in 3.00 h of operation

would I use this..
[tex]P \Delta t= I \Delta V ( \Delta t ) [/tex] ?

Thanks
 

Answers and Replies

  • #2
1,860
0
a) I got 17.26A. Check for rounding errors.

b) What have you really done by multiplying both sides by [itex]\Delta t[/itex]? Have you done anything?

You are on the right track, but executed wrong. Just remember how power and work are related.

[tex]P=\frac{dW}{dt}= \frac{\Delta W}{\Delta t}[/tex]

Since power has no explicit time dependence we can take derivatives to deltas. Then just remember the work-energy theorem.
 
  • #3
~christina~
Gold Member
715
0
a) I got 17.26A. Check for rounding errors.
I don't get it. It isn't rounding errors. I must have done something wrong in my calculations.
showing what I did, may help:
[tex]P= I \Delta V[/tex]
I think I'm screwing up here when I have to find the power.
I looked up 2.50 horsepower and it was equal to 1864.24968 Watts and
since they said it had 90% efficiency I did this:

[tex]P= \frac{1864.24968W + (1864.24968 W)(.10)} {120V}= 17.08A [/tex]
b) What have you really done by multiplying both sides by [itex]\Delta t[/itex]? Have you done anything?

You are on the right track, but executed wrong. Just remember how power and work are related.

[tex]P=\frac{dW}{dt}= \frac{\Delta W}{\Delta t}[/tex]

Since power has no explicit time dependence we can take derivatives to deltas. Then just remember the work-energy theorem.
If work incresaes then energy increases too. So if I'm not incorrect, I would just solve for W then and that would be the energy transfered?
thus...

[tex]P= \frac{\Delta W} {\Delta t} [/tex]

[tex]\Delta W= P \Delta t = (1864.24968 watts)(3hr)= 5592.74w*hr(3600 J/ 1W*hr) = 20133864 J = 20.133864 MJ [/tex]

I got that but the book has it as 22.4MJ so I'm not sure why I got the above answer.

Thanks
 
Last edited:
  • #4
1,860
0
a) So you know that [itex]P_e = IV[/itex] and that [itex]P_m = .9 P_e[/itex]. So then [itex]P_e = (10/9) P_m[/itex] or [itex](10/9) P_m = IV[/itex]. Solving for I shows that [itex]I= (10/9) P_m/V[/itex]. From here I just used the google calculator to do my conversions.

b) Careful. You forgot to account for the efficiency between the electric motor and the mechanical motor.
 

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