kWh / KW ????

so if I had 331 Kwh on my electric bill last month..... how many kilowatts did I use?

Originally Posted by Unregistered

so if I had 331 Kwh on my electric bill last month..... how many kilowatts did I use?

Will someone please ansa this?

I need to know too - i have 665 KWh on my bill but need to know how many KW i consume so that I can decide upon a tariff. the actual KW value is not shown on my bill. (rather stupidly)

Or maybe they simplay are not comparable and therefore convertable.....

When you use your cattle, you consume around 1.5 KW (depends on how big it is). Electricity price in my region is around 0.25 TL / KWh. It means that if I use my cattle 10 hour per month, it will rise my bill by 3.75 TL per month ( 0.25 TL/kwh * 10 h * 1.5 kw ).

If you see kj / s ( kilo joule per second) and wonder what it is, it is exactly power again (kw).
We have power because we consume and produce joules ( energy ).

If you consume 8000 KJ per day, it means that you have an extra power that is exactly 8000 KJ / day which equals to 8000 / (30 * 24 * 3600 ) kj per second = KW

Actually this is in correct, because you need more info than this to calculate correctly, the water heater may be left on 24/7, but it would be cycling the element on and off as the thermostat would be turning the element on and off as the temp is met. The time it takes to heat the vessel in regards to how many litres, the insulation, the amount of hot water being used are all needed to determine the kwh. As a absolute high end value in KWH this is fine. I would probaly subtract 30% off the calculated value to get a realistic value

Think of it like this.

Kilowatts measure the intensity of the energy. If you thought of this in terms of a water pump, you would measure it in terms of gallons per hour (which is a rate of flow). A 5000 gal per hour pump, if run for one solid hour, pumps (or yields) 5000 gallons of water. A 4.5 kW heating element, if run for 1 hour, uses (or consumes) 4.5 kWh of electricity.

If I cut the time down to, say, 20 minutes, it is still a 4.5 kW heating element. But, I shut it down after 1/3 of an hour, or 20 minutes. 1/3 h * 4.5 kW = 1.5 kWh. Same thing with the water pump. If I shut the 5000 gal per hour pump off after 20 minutes, I will only have pupmed 1666.66~ gal of water, not 5000 gal.

What confuses people is the kWh vs kW. When you look at it compared to the water pump, kWh is the same type of measurement as gallons, which measure total volume, where as kW relates to gallons per minute, which measures a rate of flow.

Originally Posted by Layman

Think of it like this.

Kilowatts measure the intensity of the energy. If you thought of this in terms of a water pump, you would measure it in terms of gallons per hour (which is a rate of flow). A 5000 gal per hour pump, if run for one solid hour, pumps (or yields) 5000 gallons of water. A 4.5 kW heating element, if run for 1 hour, uses (or consumes) 4.5 kWh of electricity.

If I cut the time down to, say, 20 minutes, it is still a 4.5 kW heating element. But, I shut it down after 1/3 of an hour, or 20 minutes. 1/3 h * 4.5 kW = 1.5 kWh. Same thing with the water pump. If I shut the 5000 gal per hour pump off after 20 minutes, I will only have pupmed 1666.66~ gal of water, not 5000 gal.

What confuses people is the kWh vs kW. When you look at it compared to the water pump, kWh is the same type of measurement as gallons, which measure total volume, where as kW relates to gallons per minute, which measures a rate of flow.

I'm going to throw you a couple curveballs.

First, say I have a battery that's being measured in Wh (whether its kWh or Wh isn't important for my purpose). If I'm using this for a wireless sensor, and I want this sensor to last a year before I have to replace it, how would I convert Wh to W"year"? I believe I'd divide by the Wh rating by 8760 hours. However, this then confuses me with your gallon example; if you have a 5000 gal per hour pump, but you want this pump to last a year, so 0.57 gal per year, surely after one year the pump hasn't yielded only 0.57 gallons? What confuses me is sometimes I see the battery rating as an average, and sometimes a total. If you look at my W"year" example, it doesn't seem to make sense.

Second question, if i have a "pump function", i.e. I can calculate how much energy is used to execute a particular function (humor me), if i convert this energy to power through energy/time (call this a power cost), how will this power cost impact the remaining capacity of my Wh or W"year" battery? (if you just subtract it, I believe that remaining capacity isn't linear)

Third question, after a period of time, say 20 minutes, for a Wh battery rating, can I determine my current rate of consumption and compare it to battery rating to determine if I'm operating above or below average? Going back to your gallon example, if we have 5000 gal per hour, say if after 20 minutes we've only pumped 1000 gallons, then we're on pace to only use 3000 gallons per hour, which means we can pump up our consumption to 2000*(20/60) + x*(40/60) = 5000 where x = 6500 gallons per hour for the last 40 minutes.

I'm studying power consumption and the W vs. Wh is confusing me. You seem to have a firm grip on this so hopefully you can answer my questions quite easily!

Thanks!

[JohnS]

Originally Posted by Unregistered

I'm going to throw you a couple curveballs.

First, say I have a battery that's being measured in Wh (whether its kWh or Wh isn't important for my purpose). If I'm using this for a wireless sensor, and I want this sensor to last a year before I have to replace it, how would I convert Wh to W"year"? I believe I'd divide by the Wh rating by 8760 hours. However, this then confuses me with your gallon example; if you have a 5000 gal per hour pump, but you want this pump to last a year, so 0.57 gal per year, surely after one year the pump hasn't yielded only 0.57 gallons? What confuses me is sometimes I see the battery rating as an average, and sometimes a total. If you look at my W"year" example, it doesn't seem to make sense.

Second question, if i have a "pump function", i.e. I can calculate how much energy is used to execute a particular function (humor me), if i convert this energy to power through energy/time (call this a power cost), how will this power cost impact the remaining capacity of my Wh or W"year" battery? (if you just subtract it, I believe that remaining capacity isn't linear)

Third question, after a period of time, say 20 minutes, for a Wh battery rating, can I determine my current rate of consumption and compare it to battery rating to determine if I'm operating above or below average? Going back to your gallon example, if we have 5000 gal per hour, say if after 20 minutes we've only pumped 1000 gallons, then we're on pace to only use 3000 gallons per hour, which means we can pump up our consumption to 2000*(20/60) + x*(40/60) = 5000 where x = 6500 gallons per hour for the last 40 minutes.

I'm studying power consumption and the W vs. Wh is confusing me. You seem to have a firm grip on this so hopefully you can answer my questions quite easily!

Thanks!

The pump analogy is flawed. The pump isn't used up at the end of an hour. It can pump as long as you provide it electricity. The apt analogy would be more like the resevoir of water from which it is pumping.

Batteries aren't very linear Amp-hour (or watt-hour ratings) do depend on the load. The figures are only good at a specific stated load. However, we pretend they are good over some useful range. It is very hard to character the caoacity curve over the whole useful range. Sometimes 2 or 3 points and interpolation suffice.

In the simple case of constant power, energy = power * time. When power varies, energy is the integral of power with respect to time.

[Ginu]

Originally Posted by JohnS

The pump analogy is flawed. The pump isn't used up at the end of an hour. It can pump as long as you provide it electricity. The apt analogy would be more like the resevoir of water from which it is pumping.

So my calculated average of say 0.57 gal year is accurate? In that, if the resevoir is meant to supply "water" (electricity) for a year, we can only pump at a rate of 0.57 gal such that it lasts a year? But then 0.57 gallons is an average rate, not total (like the 500 gallons that was provided in the previous example suggested). Can you clarify this please?

Originally Posted by JohnS

Batteries aren't very linear Amp-hour (or watt-hour ratings) do depend on the load. The figures are only good at a specific stated load. However, we pretend they are good over some useful range. It is very hard to character the caoacity curve over the whole useful range. Sometimes 2 or 3 points and interpolation suffice.

How does one calculate these 2 or 3 points? From what I gathered by analyzing this document, 3w dot sensorsmag.com/sensors/Feature+Articles/A-Practical-Guide-to-Battery-Technologies-for-Wire/ArticleStandard/Article/detail/533754, a non-linear equation exists or can be derived for each battery.

Originally Posted by JohnS

In the simple case of constant power, energy = power * time. When power varies, energy is the integral of power with respect to time.

Which equals the sum of the power over very small intervals (the definition of integration).

What I need to ascertain is if my 0.57 gal from my first question is correct. Then,

1) if 0.57 gal is correct, is this an average or a total for an entire year?

2) if you have a total or average rating in say mWh, how do you determine the fraction of an hour left that you have at any time?

3) if using the energy capacity of a battery in mWh or Wh, does plugging in the amount of energy consumed for a single function into a non-linear capacity equation suffice in updating the remaining battery powers? i.e., is there a 1 to 1 relationship between the energy drain that I calculate and the energy drain that will result in a drop in remaining battery capacity (based on the non-linear equation)? I hope I'm making sense - this is my biggest question.

I've ascertained that if you have a rating in mWh, to compute the capacity, you have to sum up all the energy used so far (or the integral of power over time) and use a non-linear equation to get the remaining capacity.

Am I on the right track?

[JohnS]

Originally Posted by Ginu

So my calculated average of say 0.57 gal year is accurate? In that, if the resevoir is meant to supply "water" (electricity) for a year, we can only pump at a rate of 0.57 gal such that it lasts a year? But then 0.57 gallons is an average rate, not total (like the 500 gallons that was provided in the previous example suggested). Can you clarify this please?



How does one calculate these 2 or 3 points? From what I gathered by analyzing this document, 3w dot sensorsmag.com/sensors/Feature+Articles/A-Practical-Guide-to-Battery-Technologies-for-Wire/ArticleStandard/Article/detail/533754, a non-linear equation exists or can be derived for each battery.



Which equals the sum of the power over very small intervals (the definition of integration).

What I need to ascertain is if my 0.57 gal from my first question is correct. Then,

1) if 0.57 gal is correct, is this an average or a total for an entire year?

2) if you have a total or average rating in say mWh, how do you determine the fraction of an hour left that you have at any time?

3) if using the energy capacity of a battery in mWh or Wh, does plugging in the amount of energy consumed for a single function into a non-linear capacity equation suffice in updating the remaining battery powers? i.e., is there a 1 to 1 relationship between the energy drain that I calculate and the energy drain that will result in a drop in remaining battery capacity (based on the non-linear equation)? I hope I'm making sense - this is my biggest question.

I've ascertained that if you have a rating in mWh, to compute the capacity, you have to sum up all the energy used so far (or the integral of power over time) and use a non-linear equation to get the remaining capacity.

Am I on the right track?

To empty a 5000 gallon pond or resevoir in 1 year, you would pump 0.57 gallon per hour for 8760 hours.

The nonlinearity of a battery depends on load. It is equivalent to saying the resevoir of water is a different size if you connect different size pumps to empty it. If you used a smaller pump, the pond might have 6000 gallons in it. If you used a bigger pump, it might only have 4000 gallons (of course, for a pond, this doesn't make sense, making batteries confusing)

The battery can only be characterized by measuring it at different loads (discharge rates), which are usually measured as constant current loads. The amp-hour capacity is normally less for higher loads (discharge time time decreases faster than reciprocal of load), but the profile is different for different battery types. Many rechargable batteries are commonly specified at a 10 hour discharge rate; this is usually the nominal capacity. If you double and half that current, you should get five and twenty hour loads. In reality, you will get less than five hours, and you will get more than twenty. In some batteries the discrepancy will be small, in others pretty noticable.

To further complicate things, rechargable batteries have internal self-discharge. They lose a significant state of charge in a month, even if no load at all is connected.
(the pond evaporates) Primary batteries are generally better for standby duty; alkalines will stand by for several years; once used, they are used, however.

If the load varies enough that you need to integrate, rather than just multiplying by time, you've already violated the constant current for which the amp-hours capacity have been measured. You are dealing with approximation. However, the only workable approximation is to subtract the amp-hours used from original capacity (measured at the average load) to get what's left. If the load is switched on/off and has a low duty cycle, the total "load on" hours of capacity will typically be greater than the same load driven continuously.

[Ginu]

Originally Posted by Layman

Think of it like this.

Kilowatts measure the intensity of the energy. If you thought of this in terms of a water pump, you would measure it in terms of gallons per hour (which is a rate of flow). A 5000 gal per hour pump, if run for one solid hour, pumps (or yields) 5000 gallons of water. A 4.5 kW heating element, if run for 1 hour, uses (or consumes) 4.5 kWh of electricity.

If I cut the time down to, say, 20 minutes, it is still a 4.5 kW heating element. But, I shut it down after 1/3 of an hour, or 20 minutes. 1/3 h * 4.5 kW = 1.5 kWh. Same thing with the water pump. If I shut the 5000 gal per hour pump off after 20 minutes, I will only have pupmed 1666.66~ gal of water, not 5000 gal.

What confuses people is the kWh vs kW. When you look at it compared to the water pump, kWh is the same type of measurement as gallons, which measure total volume, where as kW relates to gallons per minute, which measures a rate of flow.

I'm going to throw you a couple curveballs.

First, say I have a battery that's being measured in Wh (whether its kWh or Wh isn't important for my purpose). If I'm using this for a wireless sensor, and I want this sensor to last a year before I have to replace it, how would I convert Wh to W"year"? I believe I'd divide by the Wh rating by 8760 hours. However, this then confuses me with your gallon example; if you have a 5000 gal per hour pump, but you want this pump to last a year, so 0.57 gal per year, surely after one year the pump hasn't yielded only 0.57 gallons? What confuses me is sometimes I see the battery rating as an average, and sometimes a total. If you look at my W"year" example, it doesn't seem to make sense.

Second question, if i have a "pump function", i.e. I can calculate how much energy is used to execute a particular function (humor me), if i convert this energy to power through energy/time (call this a power cost), how will this power cost impact the remaining capacity of my Wh or W"year" battery? (if you just subtract it, I believe that remaining capacity isn't linear)

Third question, after a period of time, say 20 minutes, for a Wh battery rating, can I determine my current rate of consumption and compare it to battery rating to determine if I'm operating above or below average? Going back to your gallon example, if we have 5000 gal per hour, say if after 20 minutes we've only pumped 1000 gallons, then we're on pace to only use 3000 gallons per hour, which means we can pump up our consumption to 2000*(20/60) + x*(40/60) = 5000 where x = 6500 gallons per hour for the last 40 minutes.

I'm studying power consumption and the W vs. Wh is confusing me. You seem to have a firm grip on this so hopefully you can answer my questions quite easily!

Thanks!