Electricity: Calculating power/energy/energy conversion efficency

[Femme_physics]

We were given holiday homework from our teacher, but I'm not sure if we were given the proper tools to solve it. Is it only a formulas game? If so, how close am I to not being embarrassingly wrong?

Homework Statement

An electrical engine that requires a flow of 5 A with the useful power of 1kW is connected to a voltage source of 220 V and has been operating for 3 hours. Calculate:
1) Invested power
2) Invested energy
3) The energy conversion efficency of the engine
4) The price of activating the engine, if the price of the electrical energy is 0.42 shekels for 1kWh.

Homework Equations

All the electronics formulas you know...

The Attempt at a Solution

Attached

 

[rock.freak667]

1) Correct. It can be more than 1kW since the engine is not 100% efficient, so not all the energy is converted into useful power.

2) You need to convert 3 hours to seconds, 1 hour = 60 minutes = 60*60 seconds = 3600 seconds.

3) Correct

4) The power it uses is 1100 W, so how many kW is this? It runs for 3 hours, so how much energy does it use in kWh? (kWh is nothing more than power in kW * time in h).
Then you are told the cost is 0.42 per kWh.

 

[I like Serena]

Hi FP,

You're very close to being embarrassingly right.

I think you answered your own question at (1), when you calculated the efficiency at (3).

You have to check your units though.
- In (2) an hour is not 60 seconds.
- In (4) the unit is kWh, that is, you only need 3 hours, not 180 hours.

 

[Femme_physics]

First off, phew! I'm relieved to know I was closer to the mark than I thought. I thought I was going out on a limb here.



So, I see, all my mistakes were in translating t to the correct figure.

In the following formulas, how do I know when t is hour and when it is seconds? (presumably it's never minutes)

 

[I like Serena]

First off, phew! I'm relieved to know I was closer to the mark than I thought. I thought I was going out on a limb here.
So, I see, all my mistakes were in translating t to the correct figure.

In the following formulas, how do I know when t is hour and when it is seconds? (presumably it's never minutes)

Normally you will work with basic SI-units, as in your formula 5) which says seconds.
So a Watt is a Joule per second.

However, in your problem 4) the unit is kWh, which is: kilo times Watt times hour.
This is an alternative unit for energy. The relationship with the Joule is given by:
1 kWh = 1 x 1000 x Watt x 3600 seconds = 3600000 Joule =3.6 MJ

[edit]Btw, time could also be minutes. When you start working with engines the speed of rotation is often given by "rpm", which is also shown on the dashboard of a car.
These are "revolutions per minute"[/edit]

 

[Femme_physics]

It can be more than 1kW since the engine is not 100% efficient, so not all the energy is converted into useful power.

Ah...I see. So in that case P can't be lower than 1000, or otherwise the engine wouldn't work, right?

Normally you will work with basic SI-units, as in your formula 5) which says seconds.
So a Watt is a Joule per second.

However, in your problem 4) the unit is kWh, which is: kilo times Watt times hour.
This is an alternative unit for energy. The relationship with the Joule is given by:
1 kWh = 1 x 1000 x Watt x 3600 seconds = 3600000 Joule =3.6 MJ

[edit]Btw, time could also be minutes. When you start working with engines the speed of rotation is often given by "rpm", which is also shown on the dashboard of a car.
These are "revolutions per minute"[/edit]

That's helpful. I can also write the units for question 2 in terms of [MJ x sec] right? I only have to plug in 10^-6 after I get the result in Joules, then I just add the [x sec] to whatever I make the Joules as [kilo, mega] etc. That's legit, yes?

 

[I like Serena]

Ah...I see. So in that case P can't be lower than 1000, or otherwise the engine wouldn't work, right?

Yep.

That's helpful. I can also write the units for question 2 in terms of [MJ x sec] right? I only have to plug in 10^-6 after I get the result in Joules, then I just add the [x sec] to whatever I make the Joules as [kilo, mega] etc. That's legit, yes?

Almost. You have the factor and the prefix right.
But the unit would be [MJ], and not [MJ x second].
That is because [J] = [W x second].

 

[Femme_physics]

Yep.

And if it were exactly 1000? Wouldn't then the energy conversion efficency of the engine be a 100%?

Almost. You have the factor and the prefix right.
But the unit would be [MJ], and not [MJ x second].
That is because [J] = [W x second].

Aha. I see :) Thanks a bunch.

 

[I like Serena]

And if it were exactly 1000? Wouldn't then the energy conversion efficency of the engine be a 100%?

Yes, but such engines do not exist. ;)

 

[Femme_physics]

Dooly noted. The stuff I learned this week is worth more than I learn in a typical study week, mainly thanks to you Ser! (we're on passover vacation, why is why I have so much time to overindulge myself with education and the scientific quest :) ).

I feel like I've gained a level and am now a level 2 science student, with a specialization in physics, lol. Thanks :D This is great.