# Help a newb out please.

1. Sep 8, 2010

### sooperdrave

basic electrical math question.

1. The problem statement, all variables and given/known data

A certain power supply provides a continuous 371W to a load. It is opperating at 0.13% efficiency. In a 129 day period, how much does it cost (in dollars) to run the device if electricity costs 0.34 $per kWh? 2. Relevant equations im not sure..... everything has to be to 3 significant digits after the decimal. 3. The attempt at a solution i have tried this a few different ways.... im not sure where the efficiency comes in. this is what i did: 371W=.371kW 129days x 24= 3096hours .371kW x 3096= 1.4932008kWh 1.4932008kWh x .34 = .507688272 answer is:$0.508

Last edited: Sep 9, 2010
2. Sep 8, 2010

### cepheid

Staff Emeritus
Welcome to PF! In the future, try to use a thread title that is more descriptive of the problem you want help with.

The efficiency is simply the ratio of the power you get out of the supply to the power you have to put into it. Does that make sense?

The basic method here is good. You multiplied the power by the time in order to get the total energy. HOWEVER, you're not calculating the right thing. You want to know how much it's going to cost, which means you need to know how much electrical energy from the grid is consumed by the supply during operation. This requires knowledge of the power INTO the supply, whereas you have used the power OUT of the supply. Hopefully now it is obvious where the efficiency must be used.

Right. This part is obvious: (cost per unit energy) * (total energy) = total cost. Once you correct the step that came before this one, this step will give you the right answer.

3. Sep 8, 2010

### sooperdrave

sorry, double post.

Last edited: Sep 9, 2010
4. Sep 9, 2010

### sooperdrave

THANKS! ill try to post better titles in the future.

yes it does make sense.

i guess im still not understanding how to get the P-input.

5. Sep 9, 2010

### cepheid

Staff Emeritus
If, as you have claimed, you understand that the efficiency is the ratio of the power you get out of the supply to the power you have to put into it, then you should also understand that because you know the efficiency, you KNOW what that ratio is. Let me emphasize that: you know the ratio of output power to input power. You also know what the output power is. Therefore, you can calculate the input power.

6. Sep 9, 2010

### sooperdrave

thanks for your time. im just not getting it. my last answer was 256682.215384
this just doesnt look right to me.

i do get the ratio thing, im just not sure what to do..... 371/.13 = x/100 ?????

oh well, its only 1 question out of 35 on this assignment. it is difficult for me to solve equations without the instuctor in my class going over them first. ill have to go in early to get some help from him, not that your help is unappreciated, im just kinda lost on this one.
(its not you, its me...lol)

7. Sep 9, 2010

### sooperdrave

think i got it...how does this answer look?

$391.037 8. Sep 9, 2010 ### cepheid Staff Emeritus 0.13% = 0.0013 This is the ratio of output power to input power: P_out / P_in = 0.0013 Hence, P_in = P_out / 0.0013 This result makes sense since it means the input power is much greater than the output power (due to the inefficiency in the supply). Once you have P_in, you can just do the same calculations as before. I was hoping that you would write these equations down and work it out for yourself. They are nothing more than the mathematical version of what I said in post #5 using words. 9. Sep 9, 2010 ### sooperdrave i actually did write them out...but i used multiplication instead of division. this is only my second week of class, and i have been out of school for a decade. i know this question probably isnt a hard one, but its 1:43am here and my brain hurts from 4 hours of homework. i really do appreciate the help! 10. Sep 9, 2010 ### cepheid Staff Emeritus Alright, no problem. Well I get that the input power is about 285 kW, what do you get? What result does that lead to for the total energy used in kWh? 11. Sep 9, 2010 ### sooperdrave (285.384615385kW)(3096hrs)=883550.76923196kWh (883550.76923196kWh)(.34)=300407.261539 this still looks wrong to me..... 12. Sep 9, 2010 ### cepheid Staff Emeritus Those are the numbers I get too.$300,000 is certainly a LOT higher than four month's worth of my electricity bills. However, there are a couple of things to keep in mind:

1) This supply is REALLY inefficient. We can't tell whether that's realistic or not without more information about the application. For instance, maybe the supply needs to convert AC mains electricity to some much lower DC voltage for the device in question. Typically it would do that by stepping down the voltage using a transformer and then converting that to DC using a rectifier. There would be losses at every stage. But this is speculative of course. I've only given ONE example of how a power supply might work, and my main point in doing so has been to point out that the problem might not have been formulated with realism in mind, esp. since making it realistic would have required the problem maker to look into details that are irrelevant to the problem.

2) I did a quick internet search for residential electricity prices in my area and found that they were a cheaper than the electricity cost given in your problem by a factor of 5 or so. So that number may not be too realistic either.

3) 371 W is a fairly substantial load. Again, some quick internet searching reveals that that's sort of like running a garage door mechanism or certain types of washing machines or certain types of vacuum cleaners -- continuously for four months.

Bottom line: I don't think that we made a mistake with the the math, but it's late for me as well (I'm in the same time zone), and I'm multi-tasking. My advice to you would be to get some sleep and check it over again tomorrow morning.

Last edited: Sep 9, 2010
13. Sep 9, 2010

### sooperdrave

thanks for all the help! i think the math is right too. i guess ill find out for sure in class today. ill let you know how it turns out.