Kilowatt Hour question

[RESmonkey]

The electric power usage for a household was 18 kW-h for a single day. Imagine that the power was supplied from batteries charged by a bicycle powered generator with 94% efficiency. For how many hours must someone pedal the bicycle at a peak power of 560 W in order to fully charge the batteries?

How many hours does it take?



No, this is *NOT* my hw. I've done this in the past; it's just to review for finals.

Thanks

 

[JohnCU]

34.2 hours?

 

[Fenixgoon]

560W*0.94 = actual output.

18 kW-h / actual output = time required.

go do your own homework

 

[Alone]

18/(560/1000)

 

[Rubycon]

That person will have cardiac arrest long before you top off your batteries. Human power is hardly constant. It's about as stable after the first minute as a transverse DPSS laser module!

 

[JohnCU]

Originally posted by: Alone
18/(560/1000)

you didn't take into account the efficiency

 

[RESmonkey]

Originally posted by: Fenixgoon
560W*0.94 = actual output.

18 kW-h / actual output = time required.

go do your own homework

Thanks

I dunno but i did this first:

(18E3*24)/(560*.94*60*60)

I thought 560W meant 560 joules per sec, and that i ...eh, fuck it.

 

[Alone]

Originally posted by: JohnCU

Originally posted by: Alone
18/(560/1000)

you didn't take into account the efficiency

Oops.

18/((560*0.94)/1000)

 

[ja1484]

The best part about this (and this is why I love hanging out with brilliant nerds, even though it makes me very tired some days) is that everyone jumped at trying to solve this rather than saying "Screw you, who cares? Pay your bill."

 

[silverpig]

Originally posted by: RESmonkey

Originally posted by: Fenixgoon
560W*0.94 = actual output.

18 kW-h / actual output = time required.

go do your own homework

Thanks

I dunno but i did this first:

(18E3*24)/(560*.94*60*60)

I thought 560W meant 560 joules per sec, and that i ...eh, fuck it.

It does.

kW-h is an energy. It's Joules/sec * hrs. The two time units cancel each other out up to a constant and you're left with an energy. It's just that we like to measure power in kW and hours are more convenient to describe how long we watch tv than seconds are.

You can divide by 3600 to do everything in Joules, but you'll have to multiply somewhere else by it too.

 

[RESmonkey]

Originally posted by: silverpig

Originally posted by: RESmonkey

Originally posted by: Fenixgoon
560W*0.94 = actual output.

18 kW-h / actual output = time required.

go do your own homework

Thanks

I dunno but i did this first:

(18E3*24)/(560*.94*60*60)

I thought 560W meant 560 joules per sec, and that i ...eh, fuck it.

It does.

kW-h is an energy. It's Joules/sec * hrs. The two time units cancel each other out up to a constant and you're left with an energy. It's just that we like to measure power in kW and hours are more convenient to describe how long we watch tv than seconds are.

You can divide by 3600 to do everything in Joules, but you'll have to multiply somewhere else by it too.

ah...understand now:

(18E3)(J/s)(Hr.) * 1/((560)(J/s)) = *

* x .94 = answer.

I never once did a problem like this where i wrote units down and canceled them. Always figured it out in the head and threw it in the calc. The whole Watt-hour and Kilowatt-hour had me confused.

I thought it was a rate of some sort, like 18 thousand joules used every hour or something.

 

[Jeff7]

Way over-complicating it by going to joules.

18 kWh = 18000 W * h

560W output * 0.94 efficiency= 526.4 W output

34.19 hrs - as JohnCU said.

 

[Demon-Xanth]

It's a trick question, as he can't power enough until the next day comes and drains the batteries again.

 

[randay]

Originally posted by: RESmonkey

The electric power usage for a household was 18 kW-h for a single day. Imagine that the power was supplied from batteries charged by a bicycle powered generator with 94% efficiency. For how many hours must someone pedal the bicycle at a peak power of 560 W in order to fully charge the batteries?

you do not give the capacity of the batteries.

How many hours does it take?

to do what? usually its over in 5-10 minutes.

 

[RESmonkey]

Well, I didn't write the question

 

[RESmonkey]

How about this one:

http://img354.imageshack.us/my...image=djfjdddddky4.jpg

 

[dighn]

average power = sum of energy / time interval
sum of energy here is integral of i(t) * v(t) * dt. v(t) is constant for t = [0, t1], 0 for other, so the energy is really 20V * (area of i(t) up to t1)

20 * (1.235 / 2 <area of triangle> + 6 * 1.235 <area of rectangle>) / 32 = 5.0171875 W (note that i'm using ms directly but both the top and the bottom are using ms so it cancels out)

 

[RESmonkey]

Originally posted by: dighn
average power = sum of energy / time interval
sum of energy here is integral of i(t) * v(t) * dt. v(t) is constant for t = [0, t1], 0 for other, so the energy is really 20V * (area of i(t) up to t1)

20 * (1.235 / 2 <area of triangle> + 6 * 1.235 <area of rectangle>) / 32 = 5.0171875 W (note that i'm using ms directly but both the top and the bottom are using ms so it cancels out)

So no use of VoltageRMS * CurrentRMS?

 

[dighn]

Originally posted by: RESmonkey

Originally posted by: dighn
average power = sum of energy / time interval
sum of energy here is integral of i(t) * v(t) * dt. v(t) is constant for t = [0, t1], 0 for other, so the energy is really 20V * (area of i(t) up to t1)

20 * (1.235 / 2 <area of triangle> + 6 * 1.235 <area of rectangle>) / 32 = 5.0171875 W (note that i'm using ms directly but both the top and the bottom are using ms so it cancels out)

So no use of VoltageRMS * CurrentRMS?

RMS works only for sinusoidal waveforms, where the average works out to be 1/2 the peak v * peak i, so that individual components are 1/sqrt(2) of the peak

well, not only sinusoids, anything that averages to 1/2 the power of the peak * peak, which I don't think is the case here.

 

[RESmonkey]

Originally posted by: dighn

Originally posted by: RESmonkey

Originally posted by: dighn
average power = sum of energy / time interval
sum of energy here is integral of i(t) * v(t) * dt. v(t) is constant for t = [0, t1], 0 for other, so the energy is really 20V * (area of i(t) up to t1)

20 * (1.235 / 2 <area of triangle> + 6 * 1.235 <area of rectangle>) / 32 = 5.0171875 W (note that i'm using ms directly but both the top and the bottom are using ms so it cancels out)

So no use of VoltageRMS * CurrentRMS?

RMS works only for sinusoidal waveforms, where the average works out to be 1/2 the peak v * peak i, so that individual components are 1/sqrt(2) of the peak

well, not only sinusoids, anything that averages to 1/2 the power of the peak * peak, which I don't think is the case here.

Hmm...

my poor attempt:

Vrms = sqrtroot( (20^2 Volts * 7E-3 ) / 32E-3 )

 

[dighn]

doesn't make sense to apply Vrms here. the v/i are not sinusoidal and the v/i relationsihps are not linear. gotta integrate v*i to get total energy within the period then divide by time.

 

[RESmonkey]

integrate? but i dont have the function(s) of I(t) nor V(t)

i think i know what u mean tho, 1/T integratal of P(t) respect from 0 to T.

 

[dighn]

Originally posted by: RESmonkey
integrate? but i dont have the function(s) of I(t) nor V(t)

i think i know what u mean tho, 1/T integratal of P(t) respect from 0 to T.

but you do, v(i) = 20 for t <= t2, 0 for others, v(t) is a line before t1, constant for t1 to t2, and a line for t2 to t3, then 0.

but you don't have to do it purely analytically. v is 0 after t2 which means P(t) is 0 after t2, so you only need to integrate before t2, where V(t) is constant i.e.

energy = integrate(20*i(t)*dt) for 0 to t2= 20 * integrate (i(t)*dt) for 0 to t2

you can use the geometric intepretation of integration which is "area under". it just becomes 20 * area under the i(t) graph for 0 to t2. or you can integrate it analytically if you so wish, you just gotta dividie it into two integrations, first form 0 to t1, then t1 to t2.

 

[Born2bwire]

Originally posted by: dighn

Originally posted by: RESmonkey

Originally posted by: dighn
average power = sum of energy / time interval
sum of energy here is integral of i(t) * v(t) * dt. v(t) is constant for t = [0, t1], 0 for other, so the energy is really 20V * (area of i(t) up to t1)

20 * (1.235 / 2 <area of triangle> + 6 * 1.235 <area of rectangle>) / 32 = 5.0171875 W (note that i'm using ms directly but both the top and the bottom are using ms so it cancels out)

So no use of VoltageRMS * CurrentRMS?

RMS works only for sinusoidal waveforms, where the average works out to be 1/2 the peak v * peak i, so that individual components are 1/sqrt(2) of the peak

well, not only sinusoids, anything that averages to 1/2 the power of the peak * peak, which I don't think is the case here.

Just to clarify, RMS is not relegated to only sinusoidal signals. People associate RMS with sinusoidal signals for two reasons. The first is that the average value of a sine wave is always zero, so RMS is a way to get a non-trivial average for the voltage and currents. The second is that it is easy to do the conversion to the peak power due to the root 2 and such. But RMS is simply the square root of the mean of the square of the quantity, so it is not only for sine waves. So in the given problem, the RMS power is 11.132 W. The main reason not to use it here is that the question did not ask for RMS power, but the average power.

EDIT: Meant peak, not average.

 

[RESmonkey]

Originally posted by: Born2bwire

Originally posted by: dighn

Originally posted by: RESmonkey

Originally posted by: dighn
average power = sum of energy / time interval
sum of energy here is integral of i(t) * v(t) * dt. v(t) is constant for t = [0, t1], 0 for other, so the energy is really 20V * (area of i(t) up to t1)

20 * (1.235 / 2 <area of triangle> + 6 * 1.235 <area of rectangle>) / 32 = 5.0171875 W (note that i'm using ms directly but both the top and the bottom are using ms so it cancels out)

So no use of VoltageRMS * CurrentRMS?

RMS works only for sinusoidal waveforms, where the average works out to be 1/2 the peak v * peak i, so that individual components are 1/sqrt(2) of the peak

well, not only sinusoids, anything that averages to 1/2 the power of the peak * peak, which I don't think is the case here.

Just to clarify, RMS is not relegated to only sinusoidal signals. People associate RMS with sinusoidal signals for two reasons. The first is that the average value of a sine wave is always zero, so RMS is a way to get a non-trivial average for the voltage and currents. The second is that it is easy to do the conversion to the average power due to the root 2 and such. But RMS is simply the square root of the mean of the square of the quantity, so it is not only for sine waves. So in the given problem, the RMS power is 11.132 W. The main reason not to use it here is that the question did not ask for RMS power, but the average power.

Wait, average power doesn't = Vrms * Irms? I thought it did. RMS power is something completely diff, isn't it?