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markotah
02-13-2006, 08:58 AM
I see on my bill I used 703KWH last month. My water heater has a 4500 watt (4.5KW) element in it. If I ran the unit continually for 1 hour how would I calculate the KWH?
Thanks,
Mark

Robert Fogt
02-14-2006, 10:51 PM
As its name implies, kilowatt hours is indeed kilowatts * hours.

4.5 kilowatts * 1 hour = 4.5 kWh

Unregistered
05-16-2007, 02:15 PM
thx for helping with my energy econ homework

want_a_pie
07-29-2007, 11:37 PM
Thanks for helping me with my homework too =D

Unregistered
10-11-2007, 11:48 AM
thanks for doing my homework

Unregistered
11-28-2007, 10:30 AM
that ******* is wrong! i failed my test because of him!!!

gubment_cheez
11-28-2007, 02:50 PM
that ******* is wrong! i failed my test because of him!!!no, Robert's answer is not wrong. kWh is KiloWatts Per 1 Hour. if a water heater had a 4.5 kW heating element, that ran for 1 hour (60 minutes, or 3600 seconds) --which is exactly what the question asked-- it would consume 4.5 kWh of power.

Pizza_Pi (Visitor)
12-04-2007, 07:13 PM
kilowatt-hour (kWh) is a unit of Power. 1 kilowatt-hour means that it will use (or transfer) one kilowatt every hour. One kilowatt is the power of one thousand joules (J) per second (s). the h in kWh basically signifies just the alternative time unit used. It is used because it often provides results that can be explained in whole numbers easily with out common household appliances.
If you really want to know what the conversion for kWh to kW...
1 kWh = 0.000277777...kW
1 kW = 3600 kWh

Unregistered
12-04-2007, 07:16 PM
The conversion:
1 kWh = 0.00027777... kW
1 kW = 3600 kWh
1 kW = 1000 J / s

So, the answer would be 4.5 kW = 16200 kWh, or 16.2 mWh (megawatt-hours)

Pizza_Pi
12-04-2007, 07:25 PM
1 kW = 1 kW consistent for 1 hour, or 1 kWh
1 kW = 1000 J / s = 1 kJ / s
1 kWh = 1000 J / h = 3.6 kJ / h
kW (kilowatts), kWh (kilowatt-hours), J (Joules), s (seconds), h (hours), kJ (kilojoules), mJ (megajoules)
But if you ant to find how much power was consumed in an hour, then you multiply 1 kWh times 3600. You can look above to find conversions to Joules per second, and, including the units, you find that the two time units cancel out each other:
kWh / kW = (1000 J / h) / (1000 J / s)
= (3600000 J / s) / (1000 J / s)
= (3600000 J / s) * (s / 1000 J)
= 3600

Just some prefixes for units if you ever want to know.
(f) fempto = 0.000000000000001
(p) pico = 0.000000000001
(n) nano = 0.000000001
(micro symbol) micro = 0.000001
(m) milli = 0.001
(k) kilo = 1,000
(M) mega = 1,000,000
(G) giga = 1,000,000,000
(G) tera = 1,000,000,000,000
(P) peta = 1,000,000,000,000,000

If anything is wrong, tell me now and prove it. I always want to be right.

Oh, and Mark, the heater used 4.5 kWh, and it will consume 108 kWh daily if you leave it on 24/7.

concept confusion
07-16-2008, 08:05 AM
passing by... Actually KWH is the unit of Energy or Work. Instead, KW is the unit of Power. 1 KWH= 1KW * 1Hour. (Work=Power * Duration)

Joule is also the unit of work or energy. 1 Joule = 1 Watt * 1 s. 1KWH=1000 * 3600 =3.6 MJ

Unregistered
07-27-2008, 12:16 PM
Wow...

Lets set this matter straight:

READ: Kilowatts can not be CONVERTED into Kilowatt-hours, PERIOD!

They are two different units of measurement!

A Kilowatt is a unit of POWER or the RATE OF ENERGY CONSUMPTION (or production).

A Kilowatt hour is a unit of ENERGY, much like the Joule, Calorie, etc.

YOU CANNOT CONVERT BETWEEN THE TWO

Simply put:
A Kilowatt hour IS the amount of energy you are using
A Kilowatt is how fast you are using it.

Unregistered
11-13-2008, 12:41 AM

As an ex plumber and now a Public health Engineer (a plumber with a shirt and tie on) we often use gas load calculations in our work and as far as I was concerned the calculation went something like this:-

1 kilowatt-hour = 3600 kilo-joules (KJ) - check this on any other web sites.

1 KJ/s (kilo-joule per second = 1 kw of energy used every second (remember your basic formula - 1KJ/s = 1Kw)

so 1 kWh = 3600 x 1KJ/s = 3600 x 1 kw

If you have a 1 kw heater it uses 1 kilowatt of energy for every second it is on otherwise it couldn't be rated. How could you rate a fire accurately as a 1 kw heater if this related to 1 hours use. If I had it turned on for 1 hour and you had it turned on for 20 minutes - in my eyes it would be a 1 kw heater and in your eyes it would be a 1/3 kw heater.

Unregistered
11-14-2008, 05:21 AM
1 kW = 1 kW consistent for 1 hour, or 1 kWh
1 kW = 1000 J / s = 1 kJ / s
1 kWh = 1000 J / h = 3.6 kJ / h
kW (kilowatts), kWh (kilowatt-hours), J (Joules), s (seconds), h (hours), kJ (kilojoules), mJ (megajoules)
But if you ant to find how much power was consumed in an hour, then you multiply 1 kWh times 3600. You can look above to find conversions to Joules per second, and, including the units, you find that the two time units cancel out each other:
kWh / kW = (1000 J / h) / (1000 J / s)
= (3600000 J / s) / (1000 J / s)
= (3600000 J / s) * (s / 1000 J)
= 3600

First of all, kW is a unit of power, kWh is a unit of energy, and a Joule (J) is a unit of energy. You cannot say that X kW = Y kWh = Z J/s, that's comparing apples to oranges.
1 kW = 1000 J/s (power)
1 kWh = 1000 Jh/s (energy)
1 h = 3600 s (time)
1 kWh = 1000 Jh/s = (3600*1000) Js/s = 3600000 J (energy)
To correct your final statement in the quote:
1kWh/1kW = (3600000 J)/(1000 J/s) = (3600000 J)*(s/1000 J) = 3600s = 1h
As you can see the Joules cancel out and you are left with a unit of time.
If you run your water heater which uses a steady 4.5kW of power for 1 hour you will have consumed a total of 4.5kWh of energy, if you used it for 2 hours straight it would consume 9kWh of energy.

gubment_cheez
11-16-2008, 12:17 PM
Wow...

Lets set this matter straight:

READ: Kilowatts can not be CONVERTED into Kilowatt-hours, PERIOD!
incorrect. a kilowatt-hour is simply a second dimension on kilowatt (example, kilowatt is a 1 dimensional line, a kilowatt-hour would be a 2 dimensional graph) They are two different units of measurement!

A Kilowatt is a unit of POWER or the RATE OF ENERGY CONSUMPTION (or production).

A Kilowatt hour is a unit of ENERGY, much like the Joule, Calorie, etc. this is not correct. if I recall correctly,

a joule is the energy used by a force of one newton moving one meter.

a calorie is the amount of energy required to heat 1 gram of water 1 degree Celsius.

a kilowatt hour is the product of power in kilowatts, multiplied by hours
YOU CANNOT CONVERT BETWEEN THE TWO

Simply put:
A Kilowatt hour IS the amount of energy you are using
A Kilowatt is how fast you are using it.to clarify something,

a kilowatt is 1 000 watts

and,
a watt is the the rate at which work is done when an object is moving 1 meter a second against the force of 1 newton

JohnS
11-16-2008, 12:45 PM
incorrect. a kilowatt-hour is simply a second dimension on kilowatt (example, kilowatt is a 1 dimensional line, a kilowatt-hour would be a 2 dimensional graph) this is not correct. if I recall correctly,

a joule is the energy used by a force of one newton moving one meter.

a calorie is the amount of energy required to heat 1 gram of water 1 degree Celsius.

a kilowatt hour is the product of power in kilowatts, multiplied by hours
to clarify something,

a kilowatt is 1 000 watts

and,
a watt is the the rate at which work is done when an object is moving 1 meter a second against the force of 1 newton

A joule is also the energy of an electromotive force of 1 V moving a charge of 1 C. If it occurs in in 1 s, then it is 1 V moving 1 A, and the resulting power is 1 W. 1 W = 1 J/s, or 1 J = 1 W·s.

Clearly a watt-hour is 3600 W·s = 3600 J, and a kilowatt hour is 1000 W x 3600 s = 3.6 MJ.

The use of the accursed kilowatt-hour obscures the fact that electrical energy may be expressed in exactly the same units and form as mechanical or thermal energy. It should be deprecated, much like the calorie. Of course, the calorie is also deprecated because it comes in at least five flavors of slightly different size because the heat capacity of water is not constant enough to be an acceptable standard. (then, there is the whole big calorie, little calorie thing differing by 3 orders of magnitude)

JohnS
11-16-2008, 12:51 PM
kilowatt-hour (kWh) is a unit of Power.

Not so. The kilowatt is a unit of power. A kilowatt·hour is a unit of energy. If everything is constant load, it is the product of power in kilowatts and time in hours.

For flucuating loads, it is the integral of P(t)*dt

Layman
05-14-2009, 11:29 AM
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.

Jim
06-04-2009, 12:15 PM
I would be interested in how one would determine the cost to charge a 16kW battery using a standard 110 house system and how many kWh's would be needed to fully charge a 16kW battery.

Unregistered
06-05-2009, 09:51 AM
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!

Ginu
06-05-2009, 09:53 AM
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
06-05-2009, 10:54 AM
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
06-05-2009, 03:44 PM
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?

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.

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
06-05-2009, 06:39 PM
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
06-05-2009, 10:46 PM
To empty a 5000 gallon pond or resevoir in 1 year, you would pump 0.57 gallon per hour for 8760 hours.

Okay so it's an average of 0.57 gallons per hour for 8760 hours to last a year.

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.

What I was asking was, if the energy capacity equation is 0.001p^2 - 5p + 3080 for a battery, if I compute that the wireless sensor has a power drain of 5 watts due to a transmission (and consider that 5 watts is the only power drain to date), does the energy capacity decrease from 3080 to 3055? This would make the power drain a 1 to 1 relationship between what i calculate for one transmission and what decreases the capacity? (ignoring obvious sleep current draws, etc.) I'm just looking at the impact of the 5 watt power drain that I calculate and the effect that that single power drain has on the capacity. I would figure that if there was a second transmission that had a power drain of 2 watts, I would consider the total power drain of 7 watts in my calculation of the remaining energy capacity.

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.

I'm currently only looking at the simplest case with non-rechargable batteries.

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.[/QUOTE]

Maybe I need an example. Consider that I have a battery configuration with an internal 9V battery with 1500mAh rating and an external 9V battery with 8000mAh rating at the onset of my simulation, giving me 85.5Wh energy capacity.

Based on what you've suggested, say I want this battery to last a month (730 hours), to get the new rating I get 0.117Wmonths, which as you suggested allows me to draw 0.117 Watts per hour for one month.

Now say after 5 minutes, the device transmits. Say it is a device that has an energy consumption per bit of 3.5 uJ/bit and has a transmission rate 75 Mbps, and I calculate a power consumption of say 50 Watts (3.5 * 75 * another factor that deals with message sizes that we don't need to get into).

This means that I can transmit on average 85.5/50 = 1.71 or 730/(50/.117) = 1.708 times a month in order to satisfy my capacity restriction?

JohnS
06-06-2009, 03:20 AM
Okay so it's an average of 0.57 gallons per hour for 8760 hours to last a year.

What I was asking was, if the energy capacity equation is 0.001p^2 - 5p + 3080 for a battery, if I compute that the wireless sensor has a power drain of 5 watts due to a transmission (and consider that 5 watts is the only power drain to date), does the energy capacity decrease from 3080 to 3055? This would make the power drain a 1 to 1 relationship between what i calculate for one transmission and what decreases the capacity? (ignoring obvious sleep current draws, etc.) I'm just looking at the impact of the 5 watt power drain that I calculate and the effect that that single power drain has on the capacity. I would figure that if there was a second transmission that had a power drain of 2 watts, I would consider the total power drain of 7 watts in my calculation of the remaining energy capacity.

If you can really find such a model for your battery, it will tell you the energy capacity at different power draws. Dividing by power would tell how long you could continuously power a load at that power draw.

Your example of 7 watts of load is only valid if both loads are operating at the same time. You have to consider both the power and the time of operation; compute energy consumed by the load and subtract from the batteries total capacity.

I'm currently only looking at the simplest case with non-rechargable batteries.

Maybe I need an example. Consider that I have a battery configuration with an internal 9V battery with 1500mAh rating and an external 9V battery with 8000mAh rating at the onset of my simulation, giving me 85.5Wh energy capacity.

Based on what you've suggested, say I want this battery to last a month (730 hours), to get the new rating I get 0.117Wmonths, which as you suggested allows me to draw 0.117 Watts per hour for one month.

Now say after 5 minutes, the device transmits. Say it is a device that has an energy consumption per bit of 3.5 uJ/bit and has a transmission rate 75 Mbps, and I calculate a power consumption of say 50 Watts (3.5 * 75 * another factor that deals with message sizes that we don't need to get into).

This means that I can transmit on average 85.5/50 = 1.71 or 730/(50/.117) = 1.708 times a month in order to satisfy my capacity restriction?

You don't draw 0.117 watts per hour for a month. You draw 0.117 W continuous for a month. As to your transmission example you absolutely have to know the transmission length.

At 50 W, your battery set will NOT have 85.5 Wh capacity. You need to know the rated capacity at a 50 W load. Lets say it is 75 Wh, but I'm making that up. You could operate your transmitter for 1.5 h continuous or total of multiple intermittant transmissions. (Battery reality is that if 75 Wh is capacity for a continuous 50 W load, an intermittant 50 W load will be a little better, but probably not 85.5).

If 75 Wh is the right number, that is 270 kJ. At 3.5 µJ/bit, you can transmit 77 Gbits continuous, probably a little more intermittant.

I would suggest you still have confusion between watts and watt-hours. I would suggest you forget hours and think of time in seconds. Then power is still watts, but energy is joules (1 Wh = 3600 J). This may help clarity on whether you are dealing with energy or power. They are related, but the distinction is sometimes difficult for some to get their mind around.

Ginu
06-06-2009, 11:53 AM
If you can really find such a model for your battery, it will tell you the energy capacity at different power draws. Dividing by power would tell how long you could continuously power a load at that power draw.

Your example of 7 watts of load is only valid if both loads are operating at the same time. You have to consider both the power and the time of operation; compute energy consumed by the load and subtract from the batteries total capacity.

You don't draw 0.117 watts per hour for a month. You draw 0.117 W continuous for a month. As to your transmission example you absolutely have to know the transmission length.

At 50 W, your battery set will NOT have 85.5 Wh capacity. You need to know the rated capacity at a 50 W load. Lets say it is 75 Wh, but I'm making that up. You could operate your transmitter for 1.5 h continuous or total of multiple intermittant transmissions. (Battery reality is that if 75 Wh is capacity for a continuous 50 W load, an intermittant 50 W load will be a little better, but probably not 85.5).

I guess I still have confusion. There are two parts here for me to wrap my head around. First is what battery capacities I have to implement before my simulation begins. To find that I am considering the expected load.

The second part is how to reduce the energy capacity after a transmission. You suggest that I need to absolutely know how long the transmission is occurring, so let me go into more detail. Consider that I have a 256 byte packet with 48 byte payload. Energy consumption per bit (q) is given in relation to payload. The transmission time can be calculated as 256 bytes*8/transmission rate R in bits/sec. So from this information, we can calculate the power drain for one transmitter for a single transmission as:

Pd = total energy/total time = ((q/2) * 48 bytes * 8 bits/byte)/(256 bytes * 8 bits/byte / R) = (q/2)*(48/256)*R.

So for q = 7 uJ/bit and the average rate R of 75 Mbps, we have on average 49.22 Watts of power drain. So whenever this device transmits, the average power drain is 49.22 Watts.

You state that at 50 Watts, the battery won't have 85 Wh capacity. How would I go about assuring that I can guarantee one transmission a month by this device? To me it's obvious that the battery capacity has to be > 49.22 Watts. You say it may be around 75 Watts, this would be all I need, correct? How do I calculate the actual capacity ~ 75 Watts as you estimated?

If 75 Wh is the right number, that is 270 kJ. At 3.5 µJ/bit, you can transmit 77 Gbits continuous, probably a little more intermittant.

I get this calculation. If I'm only sending one 256 byte packet then, why is my consumption so high? There a HUGE difference between transmitting 77 Gbits continuously for one whole month, and sending one 256 byte packet. There's something that isn't being accounted for. Do you know what I mean? This 77 Gbits doesn't include the rate of transmission though.

I would suggest you still have confusion between watts and watt-hours. I would suggest you forget hours and think of time in seconds. Then power is still watts, but energy is joules (1 Wh = 3600 J). This may help clarity on whether you are dealing with energy or power. They are related, but the distinction is sometimes difficult for some to get their mind around.

I do want to work in terms of power because my simulation optimizes rates with respect to how much power drain occurs as a result. I know the expected power drain as a function of the rate, I use this as my power cost in an optimization function.

My devices aren't transmitting continuously. They are transmitting intermittently for 38 bytes/R secs and then stopping. I assumed I could only transmit once a month given my 85 Wh capacity. This is my problem - if you can use 85 Watts for 1 hour continuously, if I only use 49 or so Watts for milliseconds, surely I have a LOT of capacity left?

EDIT: Ah so does that mean that if I have an 85 Wh capacity, and I use 49 Watts based on 256 byte packets and a 75 Mbps transmission rate, that I've only used 49*(256*8/75E6) / (85*3600) * 100% = 4.4x10^-7% of my battery capacity for one transmission?

JohnS
06-06-2009, 12:58 PM
I guess I still have confusion. There are two parts here for me to wrap my head around. First is what battery capacities I have to implement before my simulation begins. To find that I am considering the expected load.

The second part is how to reduce the energy capacity after a transmission. You suggest that I need to absolutely know how long the transmission is occurring, so let me go into more detail. Consider that I have a 256 byte packet with 48 byte payload. Energy consumption per bit (q) is given in relation to payload. The transmission time can be calculated as 256 bytes*8/transmission rate R in bits/sec. So from this information, we can calculate the power drain for one transmitter for a single transmission as:

Pd = total energy/total time = ((q/2) * 48 bytes * 8 bits/byte)/(256 bytes * 8 bits/byte / R) = (q/2)*(48/256)*R.

So for q = 7 uJ/bit and the average rate R of 75 Mbps, we have on average 49.22 Watts of power drain. So whenever this device transmits, the average power drain is 49.22 Watts.

You state that at 50 Watts, the battery won't have 85 Wh capacity. How would I go about assuring that I can guarantee one transmission a month by this device? To me it's obvious that the battery capacity has to be > 49.22 Watts.
Battery capacity CAN'T be watts (a unit of power) it HAS to be a unit of energy (joules or watts hours.) I said it might be around 75 Wh, but that's a SWAG. It needs to be measured You are using 49 W for a fraction of a second, What is your message length in units of time?
You say it may be around 75 Watts, this would be all I need, correct? How do I calculate the actual capacity ~ 75 Watts as you estimated?

I get this calculation. If I'm only sending one 256 byte packet then, why is my consumption so high? There a HUGE difference between transmitting 77 Gbits continuously for one whole month, and sending one 256 byte packet. There's something that isn't being accounted for. Do you know what I mean? This 77 Gbits doesn't include the rate of transmission though.

I do want to work in terms of power because my simulation optimizes rates with respect to how much power drain occurs as a result. I know the expected power drain as a function of the rate, I use this as my power cost in an optimization function.

My devices aren't transmitting continuously. They are transmitting intermittently for 38 bytes/R secs and then stopping. I assumed I could only transmit once a month given my 85 Wh capacity. This is my problem - if you can use 85 Watts for 1 hour continuously, if I only use 49 or so Watts for milliseconds, surely I have a LOT of capacity left?

EDIT: Ah so does that mean that if I have an 85 Wh capacity, and I use 49 Watts based on 256 byte packets and a 75 Mbps transmission rate, that I've only used 49*(256*8/75E6) / (85*3600) * 100% of my battery capacity for one transmission?

The numerator is apparently correct for the total energy of one transmission, in joules. The denominator is the total energy of your 85 Wh battery in joules (although at 50 W load, I believe its capacity will be less. However, my 75 Ah figure is just a guess, please don't take it as better than that). It looks like a message is about 27 µs at 50 W. You should be getting somewhere around an hour and half total out of your battery, or 200 million messages. How many are you hoping to send in a month?

Ginu
06-06-2009, 02:44 PM
Battery capacity CAN'T be watts (a unit of power) it HAS to be a unit of energy (joules or watts hours.) I said it might be around 75 Wh, but that's a SWAG. It needs to be measured You are using 49 W for a fraction of a second, What is your message length in units of time?

Since I've simplified my problem, let me provide some more details. I have 3 types of devices:

1) first set of devices (low power sensors) has 9V batteries with 1500 mAh ratings. This gives my sensors 13.5 Wh energy capacity. The expected load on these sensors is 4.04 Watts based on the same calculation we used earlier to determine the 50W load.

2) second set of devices (also low power sensors of a diff type) also has 9V batteries with 1500 mAh rating. Thus, same 13.5 Wh capacity, but they only have a load of 0.105W.

3) mesh nodes that are given more resources and have 9V batteries with 2800 mAh rating. Thus, the capacity is 25.2 Wh, but they have a load of 49.22*2 = 98.5 W load 11% of the time (in acting as both a receiver and a transmitter in relaying the data, thus 2*50 we used before), and 49.22W the remaining 89% of the time.

So given these expected loads, how do I compute the actual capacity? If I send the, say, 200 million messages, how will I update the capacity after one message is sent? (just for completeness - please confirm that this is done using the non-linear equation I provided previously? Thanks) Is it based on elapsed time?

The numerator is apparently correct for the total energy of one transmission, in joules. The denominator is the total energy of your 85 Wh battery in joules (although at 50 W load, I believe its capacity will be less. However, my 75 Ah figure is just a guess, please don't take it as better than that). It looks like a message is about 27 µs at 50 W. You should be getting somewhere around an hour and half total out of your battery, or 200 million messages. How many are you hoping to send in a month?

The relay mesh node with 98.5 Watt load consumes 98.5 * (256 * 8 * 4)/(75E6) = 0.0108 Joules in sending 4 packets one after the other. Using the same logic as before, if we use our 25.2 Wh capacity, we can send 1/(0.0108/(25.2*3600)) = 8.4 million messages.

For the sensors, the first group of sensors uses 4.04 * (38 * 8 * 4)/(480E6) 10.2 uJ per transmission. Given their 13.5 Wh capacity, they can send 1/(1.0235e-005/(13.5*3600)) = 4.7 billion messages.

For the second group of sensors sensors, we have 0.105 * (38 * 8 * 1)/(250E3) = 127 uJ per transmission. Given their 13.5 Wh capacity, they can send 1/(1.2768e-004/(13.5*3600)) = 380 million messages.

So with the bottleneck of 8.4 million messages, I would like the battery to last a year. I should be able to do this if I transmit roughly every 4 seconds.

JohnS
06-06-2009, 03:25 PM
Since I've simplified my problem, let me provide some more details. I have 3 types of devices:

1) first set of devices (low power sensors) has 9V batteries with 1500 mAh ratings. This gives my sensors 13.5 Wh energy capacity. The expected load on these sensors is 4.04 Watts based on the same calculation we used earlier to determine the 50W load.

2) second set of devices (also low power sensors of a diff type) also has 9V batteries with 1500 mAh rating. Thus, same 13.5 Wh capacity, but they only have a load of 0.105W.

3) mesh nodes that are given more resources and have 9V batteries with 2800 mAh rating. Thus, the capacity is 25.2 Wh, but they have a load of 49.22*2 = 98.5 W load 11% of the time (in acting as both a receiver and a transmitter in relaying the data, thus 2*50 we used before), and 49.22W the remaining 89% of the time.

So given these expected loads, how do I compute the actual capacity? If I send the, say, 200 million messages, how will I update the capacity after one message is sent? (just for completeness - please confirm that this is done using the non-linear equation I provided previously? Thanks) Is it based on elapsed time?

The relay mesh node with 98.5 Watt load consumes 98.5 * (256 * 8 * 4)/(75E6) = 0.0108 Joules in sending 4 packets one after the other. Using the same logic as before, if we use our 25.2 Wh capacity, we can send 1/(0.0108/(25.2*3600)) = 8.4 million messages.

For the sensors, the first group of sensors uses 4.04 * (38 * 8 * 4)/(480E6) 10.2 uJ per transmission. Given their 13.5 Wh capacity, they can send 1/(1.0235e-005/(13.5*3600)) = 4.7 billion messages.

For the second group of sensors sensors, we have 0.105 * (38 * 8 * 1)/(250E3) = 127 uJ per transmission. Given their 13.5 Wh capacity, they can send 1/(1.2768e-004/(13.5*3600)) = 380 million messages.

So with the bottleneck of 8.4 million messages, I would like the battery to last a year. I should be able to do this if I transmit roughly every 4 seconds.

I don't understand where the the new 98.5 W load came from. If the device functions as a receiver then transmitter, I would expect it to be on twice as long (to receive, then retransmit the sequence). I would also expect it to operate a much lower power as a receiver than transmitter.

Calculate the total energy available from your battery, whether it is 85 Wh or some other number based on the relatively heavy load. (It is probably better to convert to kilojoules, so 306 kJ in this case). If there is any nonlinear equation, it is only in battery capacity at different loads. Once you decide that capacity, it is a linear problem.

Calculate the energy of one transmission (power multiplied by length of transmission). Count the number of transmissions. Multiply these two figures for energy consumed by cumulative number of transmissions. Subtract from battery's total capacity for what is left.

Don't expect 3 and 4 digit precision. Batteries are not precision devices. Capacity varies unit to unit, and with load. Allow a substantial safety factor, and test when you get it built. I'd want my best calculations to say I could send 50 - 100% more messages than I need to send, as a realm for the safety factor.

Ginu
06-07-2009, 10:41 AM
I don't understand where the the new 98.5 W load came from. If the device functions as a receiver then transmitter, I would expect it to be on twice as long (to receive, then retransmit the sequence). I would also expect it to operate a much lower power as a receiver than transmitter.

The energy consumption/bit value 'q' is from the perspective of the link entirely, both transmitter and receiver, hence my use of q/2 previously when just talking about the transmitter. In receiving packets, the receiver takes (q/2) joules/bit to receive the packets, and then (q/2) joules/bit to re-transmit the packets. Hence a total load of q*l*n*8/(8*m*n/R) = q*(l/m)*R as total energy/total time, where m is the packet size in 256 bytes, n is the number of packets in 4, and R is the transmission rate in 75E6. The receiver must decode the packets to determine the destination before re-transmission, hence a load of 49 Watts x 2.

Calculate the total energy available from your battery, whether it is 85 Wh or some other number based on the relatively heavy load. (It is probably better to convert to kilojoules, so 306 kJ in this case). If there is any nonlinear equation, it is only in battery capacity at different loads. Once you decide that capacity, it is a linear problem.

The part in bold, is this what you meant by the 75 Wh capacity that you guessed before? If there isn't a non-linear equation (I only found one source that had it and its not widely available), is there any other way to find the capacity at different loads?

Calculate the energy of one transmission (power multiplied by length of transmission). Count the number of transmissions. Multiply these two figures for energy consumed by cumulative number of transmissions. Subtract from battery's total capacity for what is left.

Thanks so much.

Don't expect 3 and 4 digit precision. Batteries are not precision devices. Capacity varies unit to unit, and with load. Allow a substantial safety factor, and test when you get it built. I'd want my best calculations to say I could send 50 - 100% more messages than I need to send, as a realm for the safety factor.

With this in mind, if I consider my bottleneck device which could send 8 million messages, I should calculate my arrival of transmissions based on say, 4-6 million. I'll do that. Thank you.

You've really helped me. I really appreciate this.

Unregistered
06-17-2009, 02:49 AM
Don't forget that there is probably a thermostat in the system so it won't really work all the time

Unregistered
09-07-2009, 03:48 AM
you people have just a huge assesment peice for me
thnx

Unregistered
09-15-2009, 06:08 PM
I read with interest the conversion rules. So if I install one of those solar power generators quoted at 1KW. Does that mean it will generate 3600kW (per hour) or what?

JohnS
09-16-2009, 02:45 AM
I read with interest the conversion rules. So if I install one of those solar power generators quoted at 1KW. Does that mean it will generate 3600kW (per hour) or what?

No, that means it will generate 1 kW at noon, if it is sunny. In the hour around noon, it will generate 1 kWh.

It produces NO power at night, and reduced power in early morning and late afternoon. It will probably generate 4 to 6 kWh each day, but that is an estimate.

Unregistered
10-20-2009, 11:37 AM
so if I had 331 Kwh on my electric bill last month..... how many kilowatts did I use?

TR Power
10-26-2009, 02:45 AM
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

Viyanasl
12-14-2009, 04:25 AM
As its name implies, kilowatt hours is indeed kilowatts * hours.

4.5 kilowatts * 1 hour = 4.5 kWh

Unregistered
01-12-2010, 09:42 AM
OK...all this is very confusing...heres what i think...KWH Kilo=1000 watts=watts...kwH=1000 watts...a 60 WATT light that runs for 24 hrs is 60x24= 1440...which is 1.44 KWH's...if it only runs for 12 hrs its 60x12= 720w= .720KWH??? No?

JohnS
01-12-2010, 03:21 PM
OK...all this is very confusing...heres what i think...KWH Kilo=1000 watts=watts...kwH=1000 watts...a 60 WATT light that runs for 24 hrs is 60x24= 1440...which is 1.44 KWH's...if it only runs for 12 hrs its 60x12= 720w= .720KWH??? No?

I would like to correct "kwH=1000 watts" A kilowatt-hour is 1 kilowatt (1000 W) used for one hour.

Kilowatt is a measure of power, a kilowatt hour is a unit of energy, which is the product of power and time.

Unregistered
01-26-2010, 04:12 AM
so if I had 331 Kwh on my electric bill last month..... how many kilowatts did I use?

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.....

JohnS
01-26-2010, 04:59 AM

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.....

Each electrical appliance needs a certain amount of power to run properly, expressed in watts or kilowatts. The kilowatt-hour is the product of the power and the number of hours it is run. Your electric meter sums that up over all the appliances. Assuming 30 days x 24 h/day = 720 h/month, your average power use was 665 kWh/month x 1 month/720 h = 0.924 kW.

However, there is not way to tell whether you really used 0.924 kW continuously, or 1 kW for 665 h or 5 kW for 133 h, or some other combination (such as 500 W for 720 h PLUS 5 kW for 61 h) that gives the same total. Reality is that between you turning switches on and off, and thermostats in loads(like the refrigerator) turning them on and off automatically, your power consumption varies continuously, and your electric bill in kilowatt-hours is the integral of P(t)*dt over the month, where P(t) is the instantaneous power vs time.

You need to consider the power drain of each load, and the number of hours per month it runs. Some loads like TVs don't turn all the way off. They draw a small standby current even when "off" but a lot more when they are on. Both have to be considered.

Unregistered
01-27-2010, 08:28 AM
I use 3180 kwh per month. (acreage in the winter with geothermal, two kids..bla bla)

Average wind speed is 12-12.5mph

I am interested in a 5kw system, but worried it wont be enough to make up much of my bill.

According to the chart for 12mph it will product 2200watts(2.2kw)

So do I figure its working 2.2kw per hr per second or what?

This is were I am confused. Can anyone help explain this?

JohnS
01-27-2010, 10:50 AM
I use 3180 kwh per month. (acreage in the winter with geothermal, two kids..bla bla)

Average wind speed is 12-12.5mph

I am interested in a 5kw system, but worried it wont be enough to make up much of my bill.

According to the chart for 12mph it will product 2200watts(2.2kw)

So do I figure its working 2.2kw per hr per second or what?

This is were I am confused. Can anyone help explain this?

1 kWh is literally 1 kW for an hour, perhaps this is clearer if you force a dot between the W and the h, ie 1 kW·h. There are 30 days x 24 h/day = 720 h in a month, so 1 kW for a month is 720 kWh. The wind won't be constant, your average may be higher or lower, but assuming 2.2 kW constant output, it will produce 1584 kWh per month.

That is around half the total. Any larger and you will need to think about energy storage. Your 3180 kWh usage represents an average power draw of 4.4 kW, BUT this will have peaks and valleys.

If you are making 2.2 kW but using less than 2.2 kW, it has nowhere to go unless you can feed it back to the grid or store it in a battery. When you draw more than 2.2 kW, it comes from the electric company.

Unregistered
03-09-2010, 09:22 AM
The conversion:
1 kWh = 0.00027777... kW
1 kW = 3600 kWh
1 kW = 1000 J / s

So, the answer would be 4.5 kW = 16200 kWh, or 16.2 mWh (megawatt-hours)

I gotta question;

If you are saying that 1 kWh = 0.00027777... kW what that really mean is that 1 kWh = 0.00027777... kW/s mean per second, ain't I'm Right??? it's not that you can convert kWh in kW 'cause one is a measure of energy, other is power. don't you think so??

Unregistered
04-07-2010, 08:27 AM
1kw=1000j/s
1kwh=1000j/s*h
1kwh=1000j/s*3600s
so??/
1kwh=3.6mj??????

JohnS
04-07-2010, 09:10 AM
1kw=1000j/s
1kwh=1000j/s*h
1kwh=1000j/s*3600s
so??/
1kwh=3.6mj??????

Case is important in metric symbols. As a prefix, "m" = 10^-3, "M" = 10^6

1 kWh = 3.6 MJ (megajoules), not millijoules.

While being picky, note that "W" is the proper symbol for watt, and "J" for joule.

Rocky
08-17-2010, 06:04 PM
How much coal is required at a power plant with a heat rate of 10000 BTU/kWh to run a 6kW electric resistance heater constantly for 1 week (168 hours). Assume 1 ton of coal has an energy content of 25,000,000 BTU.

Thanks, Rocky

Chris2380
08-18-2010, 05:07 PM
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

Unregistered
09-12-2010, 10:37 PM
Thanks for this, helped so much with an assignment on the use and production of energy :)

Unregistered
09-27-2010, 04:38 AM
1 kW = 1 kW consistent for 1 hour, or 1 kWh
1 kW = 1000 J / s = 1 kJ / s
1 kWh = 1000 J / h = 3.6 kJ / h

surely 1 kwh = 3600 J/h

JohnS
09-27-2010, 07:48 AM
Quote:
Originally Posted by Pizza_Pi
1 kW = 1 kW consistent for 1 hour, or 1 kWh
1 kW = 1000 J / s = 1 kJ / s
1 kWh = 1000 J / h = 3.6 kJ / h
=============================

surely 1 kwh = 3600 J/h

1 kW·h x 3600 s/h = 3.6 MJ since 1 J = 1 W * 1 s = 1 W·s

Unregistered
01-19-2011, 03:21 AM
Hi there,
Pls help me out with my question
If we have a consumption of 281 Kwhr between previous and current reading of the meter(monthly), how can we convert it into Kw and is it possible that this conversion leads to calculate the I, Amps. As per the answers above:
1 kw = 3600 Kwhr
means 281 kwhr = 0.08 Kw = 80 w
as we know P=1.732 U x I x Pf (for 3 phase circuit)
assume U = 400V, Pf = 0.9
Then I = 80/692.8x0.9 = 80/623.52 = 0.13 A

Pls advise if it is correct

JohnS
01-19-2011, 07:20 AM
Hi there,
Pls help me out with my question
If we have a consumption of 281 Kwhr between previous and current reading of the meter(monthly), how can we convert it into Kw and is it possible that this conversion leads to calculate the I, Amps. As per the answers above:
1 kw = 3600 Kwhr
means 281 kwhr = 0.08 Kw = 80 w
as we know P=1.732 U x I x Pf (for 3 phase circuit)
assume U = 400V, Pf = 0.9
Then I = 80/692.8x0.9 = 80/623.52 = 0.13 A

Pls advise if it is correct

It is not correct. For average power, you need to know the number of hours between meter readings. For a monthly reading, 30 days x 24 h/day = 720 h

281 kWh/ 720 h = 0.39 kW or 390 W average

As things are turned on and off during the day, actual power will vary above and below this average.

Unregistered
02-23-2011, 05:50 PM
lol this is on my AP environmental science ws lmao

Unregistered
04-08-2011, 11:30 AM
To all the people who keep posting:
1 kWh = 3600 kW..... YOU ARE WRONG

1 kWh = 3600 kJ not kW

Unregistered
05-16-2011, 06:12 AM
no, Robert's answer is not wrong. kWh is KiloWatts Per 1 Hour. if a water heater had a 4.5 kW heating element, that ran for 1 hour (60 minutes, or 3600 seconds) --which is exactly what the question asked-- it would consume 4.5 kWh of power.

Do not mix kWh with kW per hour, different, although you explained most of it correctly but check Pizza Pi explanation is more accurate.

Unregistered
05-31-2011, 02:32 AM
Hi,
Correct me if im wrong,but i think most of the confusion here is because of the household items. On most electrical appliances their power consumptions is rated in watts (W) or kilowatts (kW). In reality they are meaning Wh (watts per hour) or kWh (kilowatts per hour). If the heater is rated 4500W (they mean 4500Wh or 4.5kWh) and is working 1 hour, your electrometer will show 4.5 kWh used.
From engineering point of view W =\= Wh , kW=\=kWh , but in everyday household they are the same.

JohnS
05-31-2011, 02:39 AM
Hi,
Correct me if im wrong,but i think most of the confusion here is because of the household items. On most electrical appliances their power consumptions is rated in watts (W) or kilowatts (kW). In reality they are meaning Wh (watts per hour) or kWh (kilowatts per hour). If the heater is rated 4500W (they mean 4500Wh or 4.5kWh) and is working 1 hour, your electrometer will show 4.5 kWh used.
From engineering point of view W =\= Wh , kW=\=kWh , but in everyday household they are the same.

kilowatts are a unit of power, kilowatt-hours are a unit of energy. If your electric heater is rated 4.5 kW, and the element is on, it will use 4.5 kWh in 1 hour, 9 kWh in 2 hours, etc. The energy is the product of power and running time.

For a hot water heater, the element is controlled by a thermostat. It is only on when the water needs to be heated, it does NOT run continuously.

Freddy Avila
07-04-2011, 11:55 AM
kW and kWh are not units to be interchanged just like you change miles to kilometers, because they refer to different types of measurements.
kW is "power" and kWh is "energy"

The definition of power is:
Power = Energy / Time
or
Energy = Power * Time
So...
1 kWh = 1 kW * 1 h
1 kWh = 1 kW * 3600 s
1 kWh = 1 kJ/s * 3600 s
1 kWh = 3600 kJ

kJ are energy units just the same like kWh, that is why they can be interchanged. But kWh cannot be interchanged with kW without knowing exactly another measurement: Time.

HerrWarum
07-04-2011, 04:21 PM
Hi,
Correct me if im wrong,but i think most of the confusion here is because of the household items. On most electrical appliances their power consumptions is rated in watts (W) or kilowatts (kW). In reality they are meaning Wh (watts per hour) or kWh (kilowatts per hour).
No.
Consider yourself corrected.

Why is this so difficult?

seshu
09-23-2011, 10:20 PM
Co2 emission 0.47kg/kWh,
how much emission for 1 KW & 1 MW

JohnS
09-24-2011, 03:15 AM
Co2 emission 0.47kg/kWh,
how much emission for 1 KW & 1 MW

For 1 kW, 0.47 kg/h
For 1 MW, 470 kg/h

Unregistered
10-26-2011, 11:09 AM
so if i average 61kWh per day, how many kilowatts did i consume that day?

JohnS
10-26-2011, 02:06 PM
so if i average 61kWh per day, how many kilowatts did i consume that day?

Divide by 24 h/day. Your average power consumption is 2.54 kW.

Unregistered
11-26-2011, 05:18 AM
ok but my bill said I used 29,757 kwh last year if i Divide by 365 then I get day useage but how can I figure hourly or Minute useage

JohnS
11-26-2011, 05:33 AM
ok but my bill said I used 29,757 kwh last year if i Divide by 365 then I get day useage but how can I figure hourly or Minute useage

After you divide by 365 (366 in leap year) to get daily usage, divide by 24 hours to get average power usage in kilowatts. That will be an average. At times during the day, you are undoubtedly using more at some times, less at others.

Unregistered
11-26-2011, 05:52 AM
Ok but if I put up solar pannels I will need too know about how much power to expect ... so I guess over 3400 watts according to the math

ok but my bill said I used 29,757 kwh last year if i Divide by 365 then I get day useage but how can I figure hourly or Minute useage

29757/ 365 / 24 = 3.39691... Yes I know this is a guess not real data...

JohnS
11-26-2011, 06:12 AM
Ok but if I put up solar pannels I will need too know about how much power to expect ... so I guess over 3400 watts according to the math

ok but my bill said I used 29,757 kwh last year if i Divide by 365 then I get day useage but how can I figure hourly or Minute useage

29757/ 365 / 24 = 3.39691... Yes I know this is a guess not real data...

Yes. Keep in mind the 3.4 kW average is a 24/7 figure. The sun is not out at night, and some days are cloudy.

Unless you can sell excess to the grid or you arrange battery storage, it does not pay to put in a system much bigger than the average use. However, you may want to arrange major usage to occur during the day, while sun is powering your array and minimize night usage. If you can do that 10-15 kW solar array may make sense. A lot depends on how you can fit your power usage schedule to the available solar power.

However, if you want to go "off grid" and put in battery storage, then to get you through nights and cloudy days, your solar array should produce at least 5 X that number (say 17 kW) at midday.

coltoe1
11-26-2011, 06:23 AM
thank you John s I finally got on as a real person not a guest. and yes I do use about 3.4 kW average per day. But I did not know how to figure usage needed if I used a generator or solar power ... So how big do I need for backup generator?

coltoe1
11-26-2011, 06:43 AM
according to my best guess I need 4000w generator using the math we discussed earlier.. 29,757 / 356 /24 /1000? = 3.4kwh average

coltoe1
11-26-2011, 07:02 AM
I am guessing that 4000w generator is a good answer to my own question if I do not here from some one who know more on this subject...

Unregistered
11-26-2011, 07:07 AM
thank you John s I finally got on as a real person not a guest. and yes I do use about 3.4 kW average per day. But I did not know how to figure usage needed if I used a generator or solar power ... So how big do I need for backup generator?

I think you will kneed a 5000w generator just to be safe.

JohnS
11-26-2011, 07:29 AM
I am guessing that 4000w generator is a good answer to my own question if I do not here from some one who know more on this subject...

Generators are rated for their maximum power not average. Your 3.4 kW is an average, more at some times, less at others.

You need to assess what things would you run at once, and what are ther wattages. Also any heavy load you can't run at all on backup (I can't run my stove or my dryer). My average load is about 1 kW, and I use a 4 kW generator.

That lets me pretty well run what I need to (except the stove and dryer)

coltoe1
11-26-2011, 07:38 AM
Generators are rated for their maximum power not average. Your 3.4 kW is an average, more at some times, less at others.

You need to assess what things would you run at once, and what are ther wattages. Also any heavy load you can't run at all on backup (I can't run my stove or my dryer). My average load is about 1 kW, and I use a 4 kW generator.

That lets me pretty well run what I need to (except the stove and dryer)

OK but if I buy a 100 kw generator it will cost more than my home.. so if I buy a 8 kw generator will I be close to what I need.

JohnS
11-26-2011, 07:57 AM
OK but if I buy a 100 kw generator it will cost more than my home.. so if I buy a 8 kw generator will I be close to what I need.

100 kW would be way too much.

With some load management, 8 kW would probably work, 12 - 20 kW would mean you could run just about anything and not worry. More than 20 kW would likely be wasteful and unnecessary.

Just noticed you are in Texas. You might not be able to run a/c, but that is probably a big part of your load. As a minimum, I would look at your monthly bills and see how load varies throughout the year.

coltoe1
11-26-2011, 08:18 AM
Yes A/C is my big ticket .. 3,561KWH is the largest draw for the last 36 mo.
about 5000w x 2 = 10 kw or close

thanks for the math...

HerrWarum
11-26-2011, 08:33 AM
Yes A/C is my big ticket .. 3,561KWH is the largest draw for the last 36 mo.
about 5000w x 2 = 10 kw or close

thanks for the math...
Don't you use an evap. cooler?

coltoe1
11-26-2011, 08:41 AM
Don't you use an evap. cooler?

Well one day everything is dry
and wet the next..

a evap. cooler uses water that makes mold inside a Mobile home unless you vent with large fans.. talk about Noisy

HerrWarum
11-26-2011, 08:59 AM
Well one day everything is dry
and wet the next..

In SA we had two months of bad weather and 10 months of dry heat.

Unregistered
02-15-2012, 05:42 PM
how can you tell what the kw a electric heater package is when looking at a hvac unit

11-08-2012, 04:24 AM
u mean 4.5 kw heater = 4.5 x 24 hours x 7 days = 756 kwh/day/week

thanks,

1 kwh means killo watt consumption for time equal one hour
so if i have machine with 0.4 kw, and i use it for 6 hours

will be costing me electircity : 0.4 x 6 = 2.4 kwh

:)

Unregistered
01-11-2013, 10:16 AM
If my demand rate is \$.10/ kwh what is my rate in \$/kw?

israel
03-13-2013, 02:12 PM
Boa noite.

tenho 2 cargas e gostaria de converter como faço?

318 KWM para KW
2.196 KWM para KW

Unregistered
07-10-2013, 02:24 AM
Hello fella.

Please note. Kilowatts (kW) is unit of the power while kilowatts-hour (kWh) is the amount of power used in an hour. To simplify, if you have 5000 watts (5kW) device that you plug in for 1 hour, then it means you consume 5kWh. If you plug it 24hours, then in 1day you consume (5kW x 24hrs) 120kWh.

To make it complicated, if you plug the device for only half an hour, then you consume (5kW x 1/2hr) 2.5kWh. Or, if you plug it 25minutes, then you consume (5kW x 25min/60min) 2.0833...kWh.

Thanks,

Unregistered
10-01-2013, 09:09 AM
There seems to be a lot of confusion about kw and kwh. Simple answer is they don't convert. Asking how many kwh are in a kw is like asking how many inches are in a pound.

Unregistered
12-17-2013, 08:51 AM
hari om

hello every one i think i have simplified the conversion.
if u go he theoretical way i is very confusing.
so I did This

just every electrical appliance which uses current and just the heater for say 10 minutes .

and directly check the meter

for me
I power On the room heater with 1000 W for 12 minutes

so the KWh reading is 0.2 KWh

so the eqaution is simple

1000 W * 12 =12 KW
12 KW / 60 = 0.2 KWh

so equation is

(heater Wattage * time used) / 60 = meter reading in KWh

I hope it helped

Hari om