What Is A Unit Of Electricity

[RamdomChances]

Ok I've got my own bore hole and intend to try growing stuff throught the cold/hot season, using sprinklers, so I've been trying to work out how much it actually costs to pump the water.

I've got a 2 hp submersible pump that draws 7 amps while pumping (we put a meter on it) and a 3 hp pump for the sprinklers which although bigger only draws 5 amps.

I'm assuming the 1 Unit of electricity is 1 Kw/h which I think it is back home but I'm not sure and Watts= Volt x Amps so my subersible pump would use 7 x 220 Watts = 1540amps (I've got a feeling that the W = V x A is only for DC though)

1 Unit of electricity cost me just under 0.57 Bhat and it takes 2 hours to fill my tank which holds 15.5 thousand liters (2 meter wide circle x 5 meters tall). So be my reconing thats 2 hrs x 1540 Amps = 3.08 Kw/h (say 3 units) which only works out to about 1.7 bhat for 15.5 thousand liters (or 15.5 units) of water

The sprinkler pump will empty the tank in 1 hour so that only works out at just over 1 unit say 0.6 bhat, so I recon it will only cost about 2.3 bhat to put 1 tank on the land. (we actually run the sumbersible at the same time as the sprinkler pump, but it can"t keep up so its more like 20 thousand plus liters).

I'm a bit unsure on some of those figures, anyone got any ideas?

Oh it's just single phase electricity

RC

Edited November 28, 2005 by RamdomChances

[naka]

You've got the right idea, sort of.

7 amps x 220 volts = 1540 volt amperes

assume power factor = 0.75 so 1540/0.75 = 2053 watts

2053 watts = 2.053 kW, which is approx 2 units per hour.

So, 2 x 0.57 baht = 1.14 baht/hour or 2.28 baht for the two hour job.

similar for the other calc.

Note the bigger pump is taking less power cos it's obviously doing less actual work.

You could possibly get away with a smaller unit.

Basically actual work is flowrate x Backpressure ( height pumped ), sort of.

Naka.

Edited November 28, 2005 by naka

[chownah]

Your analysis seems mostly correct but I want to add a few things.

First, the lower the water level in your bore hole, the more energy it will take to pump it out so if you measured the amps that the submersible pump draws when the water was high then it will draw more when the water is lower.....during the dry season the water may alway be lower...I'm assuming that you actually measured the amperage that the two pumps used while working....using an ammeter.

In calculating the kilowatt hours used you got the numbers right but the units wrong. It should be 7 amps X 220 volts = 1540 watts.....and then later it should be 2 hours X 1540 watts = 3080 watt hours = 3.08 kilowatt hours...and notice that the abbreviation for kilowatt hours shoud be kWh...not kW/h which would be kilowatts per hour which is another thing entirely...but this detail is not important to the discussion since you did the calculations correctly.

When the pumps are running there will be some amount of voltage drop because of the load that they are drawing...this means that the voltage used in the equation above will not be 220 volts but could be as low as 200 volts I guess. This is a minor detail and it just means that you will use less electricity than you think. I'm just mentioning it here out of my neurotic need to be exact.

I think that W=VxA is correct for ac as well as dc....I know that even if its not perfectly true that it is close enough for these kinds of estimates...I'll ask my electrical engineer brother in law next time we chat. {Edit: after reading Naka's post and googling for 'power factor'...I think that I'm wrong here...so I'll leave it but dont' believe it!! Seems like a power factor adjustment is necessary here!!}

Are you sure about the cost of one kilowatt?....seems like it usually costs ten times as much...about 5 baht per kilowatt.

If you pump 20,000 litre per cycle that's 20 cubic metres per cycle...that will cover 400 square meters to a depth of 5 centimetres which is a reasonable ball park estimate of how much you want to apply at one time....so it will take 4 pumping cycles to irrigate one rai....ball park estimate. How much water you should apply at one time depends on the type of soil you have. Sandy soils will hold less water so you need to add less and do it more frequently...but you porbably know this already.

Also, are you running your sprinkler pump through your actual sprinkler system?...because the system you have will determine the actual amperage drawn when pumping through the system.

Edited November 28, 2005 by chownah

[chownah]

You've got the right idea, sort of.

7 amps x 220 volts = 1540 volt amperes

assume power factor = 0.75 so 1540/0.75 = 2053 watts

2053 watts = 2.053 kW, which is approx 2 units per hour.

So,  2 x 0.57 baht = 1.14 baht/hour or 2.28 baht for the two hour job.

similar for the other calc.

Note the bigger pump is taking less power cos it's obviously doing less actual work.

You could possibly get away with a smaller unit.

Basically actual work is flowrate x Backpressure ( height pumped ), sort of. 

Naka.

<{POST_SNAPBACK}>

Naka,

I made a post at the same time as you and I didn't include the 'power factor'. I've never heard of this. Can you explain it?

Chownah

[lopburi3]

How do you get electric at that price? Here is the city it averages over 3.70 baht per unit (total cost including VAT and adjustment charge and going up) for someone who uses 8,000 baht per month. Only those using a very, very small amount pay low costs.

[naka]

Chownah.

Apparent power (VA) = Volts x Amps. Real power (W) = Volts x Amps X Cos Phi

If a purely resistive load is connected to a power supply, current and voltage will change polarity in phase, the power factor will be unity (1), and the electrical energy flows in a single direction across the network in each cycle. Inductive loads such as transformers and motors (any type of wound coil) generate reactive power with current waveform lagging the voltage. Capacitive loads such as capacitor banks or buried cable generate reactive power with current phase leading the voltage. Both types of loads will absorb energy during part of the AC cycle, only to send this energy back to the source during the rest of the cycle.

For example, to get 1 kW of real power if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kVA = 1 kW × 1). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.5 power factor 2 kVA of apparent power needs to be transferred (1 kW = 2 kVA × 0.5).

Oops, just noticed in my calc I divided by PF (0.75) instead of multiplying.

RC take note.

Naka.

[RamdomChances]

Thanks guy's at least I'm on the right track, just got my bill, I't the house one not the farm, but we only use a bit more on the farm, unit price 0.5683, used 302.34 units last month, so I'm asuming that are Kwh as that works out to about 10 kwh/day ans I was away most of last month, air was very rarley on ect. I forgot to add the vat, so thats an extra 7%.

Yea we checked the amp used with an ammeter. Naka I thought it was about 75% and was jsut going to ask wether you should of multiplied rather than divided....but you beat me too it.

If you pump 20,000 litre per cycle that's 20 cubic metres per cycle...that will cover 400 square meters to a depth of 5 centimetres which is a reasonable ball park estimate of how much you want to apply at one time....so it will take 4 pumping cycles to irrigate one rai....ball park estimate. How much water you should apply at one time depends on the type of soil you have. Sandy soils will hold less water so you need to add less and do it more frequently...but you porbably know this already.

Thats good to know, as I was unsure. I'm only going to do about 5 rai to start and see how it goes. I'm limited a bit to the ammount of cycles I can get in in a day as we cant use it when we are doing the milk and it will take 3 hrs/cycle (2 hours filling 1 spraying).

I was really after a "ball park" figure, but it does seem really cheap, gov water here is 4 bhat a cube (1 unit)

Note the bigger pump is taking less power cos it's obviously doing less actual work.

You could possibly get away with a smaller unit.

Basically actual work is flowrate x Backpressure ( height pumped ), sort of.

First, the lower the water level in your bore hole, the more energy it will take to pump it out so if you measured the amps that the submersible pump draws when the water was high then it will draw more when the water is lower.....during the dry season the water may alway be lower

Yea I gathered that the submesible pump is using maor as it has to work harder, it pumps from 50m down. I would of thought that that would remain constant as the pump stays at the same level all the time regardless of the actuall water level and it pumps to the top of my tank so add an other 5 m.

Your right about the pump, but I deemed it better to get a 3 hp one and have spare capacity than risk buying a 2 hp and it not being enough. I've only actually got 2 sprinkler heads, at the monent, but they are the big ones, 3 will cover 1 rai at a time. I could probably use 4-5 with the pump I've got. I'm feeding it out in a 2 inch main (PVC) and then 1 inch flexible pipe to the sprinkler heads.

Anyway thanks all, if anyone else can pick some holes in it feel free

Cheers RC

[RamdomChances]

Just re-checked my bill, and I think the 0.5683 is the wrong price/unit, looks like it might be a "surcharge" I'll check it out when I can find someone to read the Thai on it, but it looks like 532 units used for 1399 bhat (2.63 bhat/unit) then a surcharg of 0.5683 (302 baht), then Vat bringing it to 1820 for 532 units or 3.42 bhat/unit. Which actually makes more sence to me.

RC

[Thetyim]

Ramdom,

Electricity is costing me 3.3 baht per unit.

I also have a figure of 0.5683 on my bill but that is not the price per unit.

[Ijustwannateach]

If you suspect that your power factor is less than ideal, you can add a small capacitor in series with your motor to improve performance and reduce costs- though most motor-based equipment I have examined in Thailand already includes one (it's worth opening the pump case to see if there's a capacitor- look for something with mC or micro (Greek letter) C in a separate casing- but don't try to touch it if it's there!!!

"Steven"

[Ijustwannateach]

The power factor is related to the fact that coils cause the phase of the voltage to anticipate the current while capacitors cause the voltage to drag behind the current.... to get maximum average voltage and thus power you need this fighting effect (between the capacitance and the inductance (coils)) to be balanced out. A motor is made up of coils of wire and is a natural inductor; thus, it is less efficient without an accompanying capacitor.

[chownah]

Naka, I guess this means that apparant power (volts x amps) is an overestimate of the electricity required and the actual amount you will pay for is less. Is this right?

Ijustwannateach, If you put in the right sized capacitor would this cancel out the effect totally? This seems like magic and too good to be true.

Random Chances, When measuring how high a pump must lift water you measure from the surface of the water from which the pump is pumping to the surface of the water to where the pump is pumping to. How far the pump is below the surface of the water from where it is pumping doesn't matter....except for the fact that the pipe you need is longer and this adds a tiny bit more flow friction. I'm not sure if my explanation is clear or not...if not I can reword it maybe. Also, when I said that 5 centimetres is a ball park estimate of how much water to apply at one time I was using a very large ball park....it could be that 10 centimetres could be applied at one time.....depends on the soil....and the crop...how much water is in the soil when you start...etc. but 10 cm would be alot!

Chownah

[RamdomChances]

Chownah, I understand what you are saying....not sure I agree with it though, but I wont argue

Bases on the lates fig's (taking into account the right price for electricity, power factor and voltage drop) for anyone who still has an interest.

Straight pumping water into the tank, works out at 46satang/cubic meter (1 unit)

I cycle of irrigation( about 20,000L) works out to around 13.4 baht

RC

[solo siam]

2 pounds

[chownah]

Chownah, I understand what you are saying....not sure I agree with it though, but I wont argue

............

RC

<{POST_SNAPBACK}>

Want to bet? What odds would you give me? (Sinister Smile)

Edited November 28, 2005 by chownah

[Thetyim]

Ramdom,

I have always wanted to know how much my water is costing per month as I also have a deep bore.

Presumeably cost is proportional to the pumping height and my bore is 85 metres deep and with the water tower the total lift must be about 100 metre. What is your total lift ?

[Thetyim]

Sorry Ramdom, I see you have already said a head of 50 metres.

I have tried working it out by calculating the energy required to lift 1000 litres up 50 metres and then converting horse power into kilowatts. Guess what, I got nearly the same figure as you 44satang per M3. I have not made an allowance for pump efficiency so I reckon the cost will be slightly higher.

[h90]

W=VxA is correct. The 220 volt are called Veffective. The peak of voltage is higher than 220. But the 220 are somehow the DC which would have the same effect than the AC which you use (sorry for my english), so it is already a corrected value.

the power factor (and here I am not 100 % sure): you measure the Amper (or maybe not), but the box which measures how many kWh you needed does not measure it, so for this you don't need more, you need less. Big companies also must measure that "waste" and get somehow charged for it.

Your analysis seems mostly correct but I want to add a few things.

First, the lower the water level in your bore hole, the more energy it will take to  pump it out so if you measured the amps that the submersible pump draws when the water was high then it will draw more when the water is lower.....during the dry season the water may alway be lower...I'm assuming that you actually measured the amperage that the two pumps used while working....using an ammeter.

In calculating the kilowatt hours used you got the numbers right but the units wrong.  It should be 7 amps X 220 volts = 1540 watts.....and then later it should be 2 hours X 1540 watts = 3080 watt hours = 3.08 kilowatt hours...and notice that the abbreviation for kilowatt hours shoud be kWh...not kW/h which would be kilowatts per hour which is another thing entirely...but this detail is not important to the discussion since you did the calculations correctly.

When the pumps are running there will be some amount of voltage drop because of the load that they are drawing...this means that the voltage used in the equation above will not be 220 volts but could be as low as 200 volts I guess.  This is a minor detail and it just means that you will use less electricity than you think. I'm just mentioning it here out of my neurotic need to be exact.

I think that W=VxA is correct for ac as well as dc....I know that even if its not perfectly true that it is close enough for these kinds of estimates...I'll ask my electrical engineer brother in law next time we chat. {Edit: after reading Naka's post and googling for 'power factor'...I think that I'm wrong here...so I'll leave it but dont' believe it!!  Seems like a power factor adjustment is necessary here!!}

Are you sure about the cost of one kilowatt?....seems like it usually costs ten times as much...about 5 baht per kilowatt.

If you pump 20,000 litre per cycle that's 20 cubic metres per cycle...that will cover 400 square meters to a depth of 5 centimetres which is a reasonable ball park estimate of how much you want to apply at one time....so it will take 4 pumping cycles to irrigate one rai....ball park estimate.  How much water you should apply at one time depends on the type of soil you have.  Sandy soils will hold less water so you need to add less and do it more frequently...but you porbably know this already.

Also, are you running your sprinkler pump through your actual sprinkler system?...because the system you have will determine the actual amperage drawn when pumping through the system.

<{POST_SNAPBACK}>

[Chang_paarp]

All these calculations work on assumption that the voltage remains at the is the claimed 220V.

If as happens in this part of the world (Perth West Oz) the voltage varies and usually down, the current required to do the job will change, usually up. (The power utility denies this but ask any insurance person on the relative number of surge / fusion claims.)

The magic little meter box is a current measuring device, it ignores the fluctuations in voltage while it just measures current. So when things are not up to spec, you are paying more per hour of running when the supply volage is low.

Long dodgey cords or lines to the pump can give similar results but this is usually a lesser problem if the cabling does not get warm at night.

[naka]

Naka, I guess this means that apparant power (volts x amps) is an overestimate of the electricity required and the actual amount you will pay for is less.  Is this right?

Ijustwannateach, If you put in the right sized capacitor would this cancel out the effect totally?  This seems like magic and too good to be true.

Random Chances,  When measuring how high a pump must lift water you measure from the surface of the water from which the pump is pumping to the surface of the water to where the pump is pumping to.  How far the pump is below the surface of the water from where it is pumping doesn't matter....except for the fact that the pipe you need is longer and this adds a tiny bit more flow friction.  I'm not sure if my explanation is clear or not...if not I can reword it maybe.  Also, when I said that 5 centimetres is a ball park estimate of how much water to apply at one time I was using a very large ball park....it could be that 10 centimetres could be applied at one time.....depends on the soil....and the crop...how much water is in the soil when you start...etc. but 10 cm would be alot!

Chownah

<{POST_SNAPBACK}>

1) Chownah, difficult to explain but look on it as apparent power (as measured) as opposed to Usefull or Useable power.

2) Capacitors ... forget it for such a small installation ... not cost effective !

3) Chownah, you are correct, load is determined by 'pumped elevation',

so nothing to do with the depth of the pump below the water surface.

i.e If your borehole is 100m deep and has 99m of water in it, then it doesnt

matter whether the pump is down 1m or 98m.

[Ijustwannateach]

Exactly... if there were a direct way to measure the inductance of your motor, you would need a capacitor that was 1/[(4x(pi)^2x(frequency=50)^2xinductance], but a good starting point would probably be to add about 50 microfarads of capacitance at a time and see where you stop getting improvements, and then fine tune it.

This is how the old dial radio/TV sets worked! The dial tuner was a variable capacitor which in combination with an inductor amplified different frequencies depending on the capacitance.

Most fluorescent lights are inefficient electrically- they are too inductive; a small capacitor in series with each one would save a lot in terms of the overall power bill.

It's not magic, it's just that the use of current (which you are paying for) is inefficient unless the AC circuit is properly tuned.

"Steven"

[Crossy]

The magic little meter box is a current measuring device, it ignores the fluctuations in voltage while it just measures current. So when things are not up to spec, you are paying more per hour of running when the supply volage is low.

Not true actually. Your electric meter really is a Watt Hour meter, take one apart and you'll find coils for sensing current AND voltage, with some clever magnetic stuff that makes the disc go around at a speed proportional to the power consumption (it's actually a calibrated induction motor). Within reason even if the voltage decreases the meter still measures your actual power consumption, so you ain't getting ripped off (at least not on your bill).

As others have said, for the purposes of these calculations Volts x Amps = Watts (as opposed to Voltamperes or VA).

Adding a capacitor won't reduce your bill as it doesn't change the actual power, it just brings the figures for VA and Watts closer together (making power factor nearer to unity [1]).

Edited November 29, 2005 by Crossy

[Crossy]

On the subject of capacitors.

Power Factor correction capacitors are always in PARALLEL with the load, you often find them in strip light fittings to cancel some of the inductance of the choke. These capacitors reduce the current requirements by bringing the PF back to near unity. They do NOT reduce the amount of power consumed and they will not reduce the meter reading.

The capacitors you find hung on the outside of motors are starter capacitors, they are in SERIES with a small winding. Single phase induction motors are not self starting. They will happily sit and buzz unless they get a spin in one direction or the other and will then happily run in that direction. The cap gives enough phase shift in conjunction with the auxiliary winding to provide an automatic spin to start the motor. There is usually a centifugal switch to disconnect the cap and aux winding once the motor starts so things don't get too hot.

[chownah]

The magic little meter box is a current measuring device, it ignores the fluctuations in voltage while it just measures current. So when things are not up to spec, you are paying more per hour of running when the supply volage is low.

<{POST_SNAPBACK}>

Not true actually. Your electric meter really is a Watt Hour meter, take one apart and you'll find coils for sensing current AND voltage, with some clever magnetic stuff that makes the disc go around at a speed proportional to the power consumption (it's actually a calibrated induction motor). Within reason even if the voltage decreases the meter still measures your actual power consumption, so you ain't getting ripped off (at least not on your bill).

As others have said, for the purposes of these calculations Volts x Amps = Watts (as opposed to Voltamperes or VA).

Adding a capacitor won't reduce your bill as it doesn't change the actual power, it just brings the figures for VA and Watts closer together (making power factor nearer to unity [1]).

<{POST_SNAPBACK}>

Is there any advantage to adding the capacitor so that the power factor is nearer to unity?

Also, I think I've figured out about VxA=W. Actually this is always true...BUT...you have to use the real voltage and the real amerage at every instant of time....this means the you can not use the rms voltage nor the rms amperage because they are averages over time and in general this will not work...it will work in some instances...like when there is no capacitance and no inductance in the circuit. The 220v figure for domestic electricity is the rms voltage not the actual voltage and ammeters measure rms amperage (I think) not the actual amperage....that's why it doesn't work in examples using motors when you use the rms values.

Random Chances, Sorry but I have to withdraw my offer to make a bet because I have found out that betting is not allowed on TV.

Edited November 30, 2005 by chownah

[Crossy]

Chownah. You're pretty near in your analysis.

The RMS values for voltage and current relate to the power delivered into a resistive load (eg a heater). They are actually equivalent to the values obtained by using DC.

Now with a resistive load the voltage and current waveforms are in phase with one another. An inductive or capacitive load causes a phase difference of up to 90 degrees. A pure inductor or capacitor will draw a current determined by it's impedence so you will get lots of amps but there will be no heating effect because of the phase difference.

The phase angle between voltage and current is the power factor (actually it's the sine of the phase angle), power factor is always < 1.

Motors tend to be inductive. the result is that the apparent power (Volts x Amps) is greater than the true power (V x A x Power Factor). Adding a capacitor across the motor improves the power factor by cancelling out some of the inductance.

The advantage of using the capacitor is that it reduces the CURRENT so you can use smaller wires, transformers, fuses. without changing the actual power consumed. BUT in small installations (less than several kilowatts) the difference is too small to be of any practical use.

[chownah]

Chownah. You're pretty near in your analysis.

The RMS values for voltage and current relate to the power delivered into a resistive load (eg a heater). They are actually equivalent to the values obtained by using DC.

Now with a resistive load the voltage and current waveforms are in phase with one another. An inductive or capacitive load causes a phase difference of up to 90 degrees. A pure inductor or capacitor will draw a current determined by it's impedence so you will get lots of amps but there will be no heating effect because of the phase difference. I disagree with the statement that there will be lots of amps but no heating.  I believe that electrons flowing in a conductor will heat the conductor and the more electrons flowing the more heat they will generate...and this is independent of the voltage.  This would mean that high amperage (high current) would creat alot of heating regardless of the voltage (regardless of phase angle).

The phase angle between voltage and current is the power factor (actually it's the sine of the phase angle), power factor is always < 1.  Shouldn't the power factor be the cosine of the phase angle?  If the phase angle is zero then sine of zero equals zero and the power factor would be zero but with a phase angle of zero you should have a power factor of 1.

Motors tend to be inductive. the result is that the apparent power (Volts x Amps) is greater than the true power (V x A x Power Factor). Adding a capacitor across the motor improves the power factor by cancelling out some of the inductance.  In this paragraph I think that all references to volts and amps are rms....correct?

The advantage of using the capacitor is that it reduces the CURRENT so you can use smaller wires, transformers, fuses. without changing the actual power consumed. BUT in small installations (less than several kilowatts) the difference is too small to be of any practical use.

<{POST_SNAPBACK}>

[astral]

The cost depends on the amount of electricity used.

The more you use, the higher the cost per unit!!

It might be worth looking at wind energy as an alternative, with electricity when the wind stops.

[Crossy]

Chownah. You're pretty near in your analysis.

The RMS values for voltage and current relate to the power delivered into a resistive load (eg a heater). They are actually equivalent to the values obtained by using DC.

Now with a resistive load the voltage and current waveforms are in phase with one another. An inductive or capacitive load causes a phase difference of up to 90 degrees. A pure inductor or capacitor will draw a current determined by it's impedence so you will get lots of amps but there will be no heating effect because of the phase difference. I disagree with the statement that there will be lots of amps but no heating.  I believe that electrons flowing in a conductor will heat the conductor and the more electrons flowing the more heat they will generate...and this is independent of the voltage.  This would mean that high amperage (high current) would creat alot of heating regardless of the voltage (regardless of phase angle). Correct. I am implying PERFECT inductors and capacitors (no parasitic resistances). In practice it is not possible to achieve a power factor of zero.

The phase angle between voltage and current is the power factor (actually it's the sine of the phase angle), power factor is always < 1.  Shouldn't the power factor be the cosine of the phase angle?  If the phase angle is zero then sine of zero equals zero and the power factor would be zero but with a phase angle of zero you should have a power factor of 1.Yeah, cosine, it's been a l-o-n-g time

Motors tend to be inductive. the result is that the apparent power (Volts x Amps) is greater than the true power (V x A x Power Factor). Adding a capacitor across the motor improves the power factor by cancelling out some of the inductance.  In this paragraph I think that all references to volts and amps are rms....correct? Correct. Unless stated otherwise always assume RMS values.

The advantage of using the capacitor is that it reduces the CURRENT so you can use smaller wires, transformers, fuses. without changing the actual power consumed. BUT in small installations (less than several kilowatts) the difference is too small to be of any practical use.

Edited November 30, 2005 by Crossy

[johna]

If the cost of power is critical to your farm operation an alternative is drip feed irrigation, your water tank is 5 meters high, all you need is the correct type of plastic hose and punch the appropriate number of holes of the correct diameter,

The initial cost of these pipes is high but I have been using them for more than 5 years with no problem.

[RamdomChances]

johna.

It's not critical, just wanted a "ballpark" fig for the cost's as eventuallt it will all come out of any prifit I make. I did think about drip feed irrigation, but could'nt really find anyone who knows much about it, or anywhere that sld the stuff. I've seen sort of "micro" sprinklers that look good but not sure that it would work with the ammount of land I want irrigated.

Probably the main reason I went for the system I've got is the ease of moving it around, I've only got 2 big sprinkler heads, but they will do about 1 rai at a time, hence the need for a big pump. Actually most of the cost involved come from actually pumping the water up, so I probably would'nt of saved much anyway.

RC