I've got a prepaid meter and buy electricity online.
Last time I bought R500 I got 388 units. (R1.28 per unit) What exactly is one unit equal to? How do I find out the cost per kWh? Is it R1.28?
I've got a prepaid meter and buy electricity online.
Last time I bought R500 I got 388 units. (R1.28 per unit) What exactly is one unit equal to? How do I find out the cost per kWh? Is it R1.28?
Last edited by gregmcc; 26-11-2011 at 06:04 PM.
1 unit is 1000 watts = 1 kilowatt. If you switch on an appliance which is rated at 500 watts, it will take 2 hours to use up 1000 watts. R 1,28 is damn cheap by overseas standards, so when we get our electric mains rechargeable cars, we will be in the pound seats so to speak
Things won are done; joy's soul lies in the doing
Thanks - What I"m trying to work out is if for example if I plugged your 500 watt appliance in for 2 hours how much would it cost to run?
I know those terms are a bit inaccurate, professor, but I was just trying to help him out.
To answer his question, a 500 watt appliance run for 2 hours will consume R 1,28 worth of electricity
Things won are done; joy's soul lies in the doing
That's strange. I pay ± R0.70 per kWh - probably excl. VAT, but even with VAT it is only about R0.80/kWh. (In Johannesburg)
Edit: Maybe I'm greener than I thought.
Last edited by adrianx; 26-11-2011 at 06:46 PM.
ɸ Plus ça change, plus c’est la même chose. ɸ
Not sure what Cape Town's tariffs are, but here are Tshwane's.
It's worth noting that it's more expensive as you purchase more.
You need to purchase every month, and no more than you are going to need, well at least not more so that you go into the next 'step'.
A Joule is a quatity of energy. One joule is the energy requiered to accelerate a two kilogram object to one meter per second across a frictionless surface.
Power is d(E)/dt, thus a watt is the derivative of a joule with respect to time in seconds; in normal language the rate of change of energy transfered. One watt is the rate of energy transfer of one joule per second.
You are charged electiricty based on your energy usage. More powerful devices use energy quicker, and thus more expensive to run. To calculate how much energy a house used, rather than integrating the function of power use over time, you are charged in units of Power*time (since the units are the same as energy. E.g: one watt used for 3 seconds is 3 joules. Twenty watts for two hours is 72 kilojoules. A house using an average of 674 watts for a month = 674*60*60*24*30 = 1 747 008 000 Joules)
As you can see, charging a person per joule is complicated. So what Eskom does is charge us in units of kilowatt.hours. One kilowatt.hour is = 1000*60*60 = 3 600 000 joules. So the previous number of joules simplify to 485.28 kilowatt.hours. So now all you owe to Eskom are 485.28 kilowatt.hours, or as you know them: 485.28 units.
Last edited by agentrfr; 26-11-2011 at 06:22 PM.
I-Rah los Pruzah ahrk Mul
IMA CHEMICAL ENGINEER WEO
Agents are people too.
Just because I'm technically bloody and imperialist doesn't prove anything!
Thanks - is it that easy?!
Is my logic ok on this now?
I plug a device into the mains which draws 1A. So P = VI = 220V * 1A = 0.22kW.
If this is plugged in for 10 mins then it will cost R1.28 * 0.22kWh / 6 = 4.6cents
Thanks - good explanation.
Maybe gregmcc is on a higher tariff. Here in Cape Town, the council tried to make it as complicated as possible, with a daily service charge, a sliding scale which charged you more if you consumed more. Then in July they abandoned this idea, since very few council employees could understand it, let alone explain it to consumers. But anyway, there are now 3 tariffs, no service charge.
I went to a great deal of trouble installing a solar water heater, heat pump on a timer, led and cfl lamps throughout, switching off all charger transformers and managed to get my monthly useage down to about R 450,00. Very very strangely, my neighbour, who has a regular hwc and has not bothered much with energy saving lamps, also has 2 children of 11 and 13, manages to consume under R 300 a month. Admittedly their house is a bit smaller, but I can't figure it out.
Things won are done; joy's soul lies in the doing
Umm... no, its not that simple since we are dealing with alternating current. In Eskom three phase Delta is used in MV (medium voltage) networks and three phase Star is used in LV (low voltage). Depending on your line and phase voltage supplies to your house, there can be lots of factors introduced into the equation.
For example: If the RMS phase voltage to a plug is 220V (domestic supply, in our case at 50 Hertz), then the line RMS voltage is at sqrt(3)*phase = roughly 380V line supply. Chances are your house is hooked up to a Star 3 phase supply, meaning your phase currents are equal to your line current, but your line voltage is sqrt(3) times bigger than your phase voltage. Since the phasors are offset at 120 degrees per phase (3 phase = 360 degrees / 3), and 30 degrees from neutral referance (to satisfy phasor sums Vry + Vyn = Vrn etc.), (after some simple grade 10 trig, since Vry/sin(120) = Vrn/sin(30)); RMS line Voltage = sqrt(3) * RMS phase-to-neutral voltage.
Now comes the fun bit. The phasor quantities for Voltage and Current are linked by the constant of Impedence. Vphase = Iphase*Zphase (where V = voltage, I current and Z = Impedence), but the current phasors can be out of "sync" with the angle of the voltage, but related by Impedence. Luckily, for both a star or delta supply, the equation holds true for Vphase/Iphase = Z.... thank goodness! That makes life much easier, since now we dont have to worry about the phasor sums to relate everything depending on the type of network.
For the rest, we will assume your house has a star 3 phase supply.
Since we know from the phasor diagrams that for a star network RMS line Voltage = sqrt(3) * RMS phase-to-neutral voltage = RMS Vphase and that Iphase = I line = I phase-to-neutral, Power = dE/dt where W = work in watts and t is time in seconds.
Since Voltage = dE/dC (E is energy in joules, C is coulombs of elctrons) and Current = dC/dt , V*I = dE/dC * dC/dt = dE/dt = Power. Awesome.
Assuming your house's star supply is for three identical loads (for the sake of simplicity, we really don't need to do it phase by phase) i.e. one phase per load: P = 3*Vphase*Iphase. Luckily, we can integrate this function to get it in terms of joules, since a joule is equal to one amp thourgh one ohm for one second. (<------ This is why SI units are king)
BUT since you use a star supply, Ptotal = sqrt(3)*Vline*Iline*cos(angle of impedence), since we work with only real units of joules expended, not lost though inductance and impedence to the ether.
Eskom gets real iffy about that angle of impedence (a.k.a the Real Power factor), since their peak supply of power does not equal the real usage of their customers, since the sin(angle of impedence) is "lost", and is not measured by their watt-meters. This is where it gets complicated with serious phasor diagrams, but I will leave it for you to look on wiki if you want to know how they calculate total actual power supplied.
The real question is how Eskom charges you. They have different rates for the real and ether power supplied by their network that you consume. For you to calculate what each thing in your house uses, you need to calculate everything of their impedence values, and offset of current against voltage. Each device usually has the angle of its impedence on the small writing on the bottom or in the manual. You need to use that in your calculations, not the pure current and voltage.
Why is this? Because as a domestic customer, Eskom will only charge you for your real power usage, not your etherithic as well since it is too expensive for them to calculate it for every house hold.
So what you pay for in usage is actually equal to, per each device, P = Vphase*Iphase*Real Power Factor, where real power factor = cos(angle of impedence), or P=(V^2)/Z
For example. Say you have an induction motor to run (simple expample to have a simple inductive load), and for argument sake it has an impedence angle of 60 degrees. Its power factor is therefore 0.5 (cos(60) = 0.5).
Say we run this bad boy on a plug in your house. Let's say it has an impedence magnatude of 110 ohms. The current running through the motor is thus 220V RMS/110 Ohms angle 60 degree = lagging current 2 Amps at 60 degrees.
So the power consumed and measured by the eskom wattmeter on the side of your house = 220*2*0.5 = 220 watts, BUT the actual true power usage is 440 watts.
Run that motor for an hour and Eskom will be able to charge you 0.22 kWh like the meter on the side of your house says, but they will sit angrily scratching their heads because their supply meter says they sent 0.22 kWh more to your neighbouhood and have no idea where the other 0.22kWh went in your neighbourhood. Since they cant proove who used what (or watt hehehe) since they are not measuring the phase line phasor angles to each house, they can't change anyone fairly for it.
You technically get 0.22 kWh free. Cool huh
I-Rah los Pruzah ahrk Mul
IMA CHEMICAL ENGINEER WEO
Agents are people too.
Just because I'm technically bloody and imperialist doesn't prove anything!
Check and see if they are measureing your current angle too. People that run lots and lots of incandecent lightbulbs and those tube bulb spotlight things can get whacked for having a "heavy" inductive current load. You are charged for peak etherithic power used, not quatity. I immagine you turn on all your lights at the same time :/.... that is if they are even measuring your current angle.
What can I say, I'm a conspirisy nut.
I-Rah los Pruzah ahrk Mul
IMA CHEMICAL ENGINEER WEO
Agents are people too.
Just because I'm technically bloody and imperialist doesn't prove anything!
OMG! Leaves scratching my head.
Great info there - I had to read it a few times
I"ll check out the wiki and start doing some more reading up calculating cost taking the PF into account.
Sounds like its not going to be too simple to work out what the running costs of certain items are
Gees!!! I pay R1.4 per unit what a rip!
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