Chapter four single phase metering meter is said to be self contained when the only path the load current can take from the line to the load is through the meter. Or put it another way the meter conducts the entire load current. The circuit is said to be to wire because there are only two wires to and from the load. The meter is described as being an a base meter because the connections are wired directly to the meter. Because it is important to maintain accuracy and consistency. standards have been developed for various types of meters and their connections.

This shows measurement, Canada's standard 112 one standard and there's more standards. We will talk about more And then later. However, there are other standards that are not measurement Canada standards, but they're very similar indeed and will standardize the connections to the various types of metering circuits and their components. This time the slide shows a socket or an SPS meter. Notice again how the current flows through the current coils of the meter. This shows measurement Canada standards 1202 and again, we'll talk more about the standards later.

Now let's look at a poly phase meter or poly phase metering. A poly phase meter has the same basic electric theory as a single phase meter. eddy currents are induced into this disk interacting with flux fields that cause them to rotate. There are more than one element acting on the same disk. But what they do is arithmetically sum up to the totals of both elements of polytheism polyphase meter could contain to two and a half or three elements, one potential coil and one current coil for each element. Notice the polyphase meter has one element on each the left and right side of the frame of this particular kilowatt hour meter.

Each element is made up of a potential element which is a half half the element and a current element which is the other half of the element. The disc is positioned so that it will rotate through the main air gap of both elements and therefore be affected by the flux of both elements. Since there is only one disc, there is only one set of retarding magnets and these are located between the two elements on the front of the frame. The register, as on a single phase meter is located above the braking magnet. Another difference in a polyphase meter is the design of the disk. In a meter where there are more than one element elements acting on a single disk, there'll be interference between the induced eddy currents of each.

To eliminate this interference Some manufacturers have use laminated disks while others use disks with a combination of triangular holes and or slots designed into it. The mechanical operation of a single phase or a polyphase meter are virtually the same, the register takes off take off gear is driven by the worm gear of a rotating disc, the term case of a still has the same meaning. Although some single phase meters use multipliers of less than 10 polyphase meters are more likely to have multipliers that are quite different. It could be 40 5090 or anything up to 200 depending on the manufacturer and the metering or the meter type. Remember that what meters and kilowatt hour meters are designed such that each element takes one voltage and one current and combines them in such a way that the dial register or readout is the product of the RMS voltage and rms current times the cosine of The angle between them.

That is to say, one element is fed with the one and is one voltage and current, and the other element is fed with the two, I two, voltage and current. The meter sums the cosine product of each of these energy consumptions of the poly phase load. We've got a head start into the understanding of how polyphase power meters operate by knowing and understanding single phase metering. The basics are the same potential coils and the current coils both produce flux fields. eddy currents are then introduced into the disk and react with these fluxes producing a discrete cation proportional to the load. To be more mathematically precise, than Meteor measuring the average power to the load takes the magnitude of the RMS voltage multiplied by the magnitude of the rms current times the cosine of the angle between them.

This is also the case for each element. It is the product of the magnitude of each of the RMS voltages and currents times the cosine of each of the angles between them for each element of the polyphase meter. When writing the formulas, the magnitude symbols are quite often left out of the formula as they add confusion to the variable symbols that are in the equation themselves. The term cosine phi or cosine theta, whatever variable you're using here is also left out of the form formula quite often. And it is assumed that we are taking the cosine of the angle between them the voltages and current when we're looking for the average power. This is sometimes referred to as a dot product.

But it's just left out out of the out of the formula as we progress through the equations just to make it a little bit simpler to view. So, anytime you see a voltage and a current product in a calculation for average power, it is assumed that we are multiplying it by the cosine of the angle between the two of them. Finally, when working with average power measurement, it is assumed that we are using RMS values for current and voltages. So, they to our left off the equation left out of the equation, making it much easier to visualize and deal with. Now, remember in a single phase electronic solid state meter there is a potential or voltage transducer plus an analog to digital converter and a current transducer plus an analog to digital converter feeding them directly into a digital processing unit that is used to generate several functions including the readout.

For polyphase electronic meters, the construction is even easier. We just have to have three pairs of transducers feeding into three analog to digital converters. In this case there are three elements to the to this type of meter. The digital processing unit is programmed to handle three input pairs to generate several functions including digital readout and the EEPROM that retains the data when the power supply is switched off. Now, let's look at some practical applications or examples. A residential electrical supply comes from a single phase transformer.

That is, the primary is connected to one phase of a three phase distribution supply. And you can see a picture of one here you've seen many of them if you've traveled around the countryside or around the city you'll see these cans which are Transformers hanging from the pole, there may be one two or three of them depending on the type of supply you want. In the case of a residential house, there is usually only one transformer which again is a single phase transformer one phase feeding into the primary one phase of the three phase distribution supply feeding into the primary of the can. The secondary is a center tapped coil of a transformer providing 120 and 240 volt supply Sometimes referred to as two hot leads and one neutral lead. In order to understand how a residential electrical supply is metered, let's start by considering two independent power supplies v one and v2 supplying two loads.

The power to each of these loads can be measured by two meters. Notice the spot markings or the polarity markings on the white meters. With this arrangement, the power delivered in circuit one is given by W one, which is equal to v one times I one times cosine phi and five being the angle between the current and the voltage. And the power delivered to circuit two is given by W two which is equal to v two times I two times cosine of the angle between the voltage and the current. From here on out I'm going to not show the term cosine phi and assume that anytime you see a white reading given by the product of the voltage and the current, it will be the same as magnitude of the voltage times the magnitude of the current phasers multiplied by the cosine of the angle between them.

Now if we move watt meter number two like this, it would essentially be reversing the voltage to the watt meter, but keeping the current to the watt meter the same that is the current is still entering the spot mark. We would have to reverse the reading of W two in order to obtain the same reading as before. Hence w two is now equal to minus the two times I two connecting the circuits together at the what we'd call a neutral makes no electrical difference. To the loads, this arrangement is the same as connecting the loads to the center tap residential distribution transformer. v one will now equal the two will equal the T, the total voltage on the transformer divided by two. And this is a characteristic of the residential distribution transformer.

Since w t is equal to W one plus w two and we know that wt is v one times I one minus v two times I two, we can substitute the voltages vt over two or the voltage V t over two for the voltages v one and v two and we are left with this equation which can now be rewritten this way bringing the negative side in negative sign inside the brackets Where I too is and making eye to negative. Notice that the load is not included in the equation. And it makes no difference to the formulas calculating the metering as long as we connect the watt meters as shown here. This slide shows a three wire self contained meter. Now they call it three wire because they're three wires feeding a residential load. Notice how there is two current coils associated with one potential coil.

Since an element is made up of one current coil associated with one potential coil, each being a half element. These type of meters are often referred to as one and a half element meters. This is measurement Canada standard 1301 its connection to the load is governed by the formula that we've already developed here, which is shown here, the total power is given by the T over two times I one plus the quantity VP over two minus i two. So if we look at I one from line one current flows from the source through the meter through the outer current coil and to the load line two, or I two from line two flows like this through the inner core current coil, but notice this time that is flowing, essentially backwards through the inner current coil. And goes to the load. The potential element is connected between line one and line two.

So it is actually measuring the T. Remember that what meters and kilometers are designed such that each element takes one voltage and one current and combines them in such a way that the readout is the product of the RMS voltage and rms current times the cosine of the angle between them. So, let's look at the products of the RMS voltage and current as they are connected to the meter and see how they match our formula. vt is connected to the potential coil. It is voltage between line one and line two. The divide by two is taken into account because only half of the potential element reacts with one and the other half reacts with I two, or minus i two in this case. I one is the current through the outer current coil and I two is a current through the inner current coil, however, it's going backwards.

So it does take into account the minus sign. The meter sum the product of the two RMS voltages and rms current times the cosine of the angle between each of them. In actual fact, the meter elements work more like this. The potential coil is split in half, where the meter sums the power measured by both elements number one and number two as in this equation here where the total power is given by the one RMS times I one RMS times a cosine of the angle between the one and I one plus v two RMS times I two RMS and the cosine of the angle between the two and I two. So, the left element is the first part of the equation. And the right element is the other part of the equation.

Where v one is the T over two and The two is also vt, all over two. And I one is I one, and I two in this case is minus, because it's going through the coil backwards. It requires people fuel and equipment to operate in an electrical utility system. The number of people in the amount of fuel and equipment are directly related to the collective requirements of the utility customers. If the utility is to be financially self sustaining, it must charge enough to recover the costs of generating and distributing power. Recovering cost would be easy if all customers used a steady amount of power all the time.

Unfortunately, this steady load situation doesn't exist in practice, commercial and Industrial and sometimes residential areas like those in Freeman Ville have characteristic load peaks during which excessive power is demanded. The demand may last for a split second or for many hours. Naturally, a utility would prefer to have a customer who uses load evenly over a period of time to one who uses the same amount of total energy in a fraction of the time. As reality reflects the latter case is more often. The utility must be equipped not only to supply the constant level requirements of the system, but also the generating disk and distribution equipment must have sufficient reserve capacity to meet excessive power demands. Reserve capacity in items like generators and Transformers mean that means that a utility must Spend money to purchase large components and lots of fuel, if only what our billing is used, it could financially constrict the utility.

So, quite often, the demand charge is associated with the purchase of these large ticket items and the capital investment of the utility. The energy charge is usually related to such things as fuel consumption and water flow and routine maintenance items. And if if the if there is a higher demand for power, then there's more of these large ticket items have to be purchased such that the customer should be charged for these or share the cost of these large ticket items using a demand charge in order to achieve fairness in this that Demand energy combination meter was developed without going into too much detail this meter while and while measuring energy is also used to measure peak demand over a set period. And that period is usually a month. It does this by means of measuring and displaying the average demand over a period of say 15 minutes, and it is measured in watts or kilowatts or megawatts.

And it holds that reading until it's surpassed by a larger demand reading or it is re manually reset at the end of the one month period. The customer is then billed with a combination of the amount of energy used in kilowatt hours plus the peak demand that it was recorded in kilowatts. This used to be accomplished and still is in most place in some places with the use of thermal elements in the meter that are heated proportionally to the kilowatt demand and it drives a Push your pointer into an idle pointer. How that works is the pusher pointer would move with the demand pushing the idle pointer upscale. When the pusher pointer drops with a falling demand, the idle pointer would hold the highest reading until it is manually reset. Of course, the more modern digital meters do this with a built in algorithm that the digital processing unit has and simply records it for future reference.

As the display cycles through its various read readings. One of them will be say a PD or power demand reading and in this case it's showing 6.87 which is 6.87 kilowatts. Let's walk through a couple of Classroom problems and solutions just to reinforce some of the theory that we've just discovered. In this power measurement working example, we have measurements Canada standard connection for 1301, which you see here. In this figure we've placed a load of 24 ohms across line one and two. We see that the 24 ohm resistor connected line to line causes a current flow of 10 apps, which we can find just simply by applying ohms law of 24 ohms being fed by 240 volts is equal to 10 amps.

Our task will be to find out what the amount of power is being used by the low What is the amount of power measured by each element? And subsequently, what is the amount of power measured by the watt meter? In answer to the first question, the amount of power used by the load is simply a calculation of the power formula voltage times the current. And since it's a resistive load, we have the current in phase with the voltage so it's 240 volts times 10 amps, which should be drawing 2400 watts or 2.4 kilowatts from the source. Now, let's look at what each element of the meter is reading the current coils each share Half an element of a potential coil therefore only half the flux is reacting with each coil. The meter is measuring in this case, element number one is 240 volts times the inner current coil amps which is 10 amps times half the flux which is 1200 watts plus element number two which is 240 volts times the outer current coil 10 amps times half the flux which is also equal to 1200 watts.

Notice that the inner current coil has the power flowing backwards through the core through the current coil and the actual current coming from the load is is actually in reverse of the current flowing through the outer current coil from line one. So in actual fact, you've kind of got to double negatives which add to a positive. So the element number two is indeed going to read 240 times 10 times one half, which is 1200 watts. And the two meter total will be simply the arithmetic sum of the two elements, which is 2400 watts, which checks out against our original calculation of what the load should be dissipating for the same standard meter connection. This time we're going to connect a 12 volt load across line one to neutral We see that a 12 ohm resistor connected line to neutral causes a current flow of 10 amps.

This is simply a matter of calculating ohms law, which is i is equal to V over R, which is 120 divided by 12, which is 10 or 10 amps. Again, we're going to try and find what the amount of power used by the load, we're going to find out the power measured by each element and what is the amount of power measured by the wire. This time we've placed the 12 ohms. Across line one to neutral and the amps is 10. As we've calculated, of course, the power to that load is simply the voltage times the current which is 120 times 10 or 1200 watts we'll see which each element of the meter is reading element number one is 240 volts, times the inner current coil of zero amps, times half the flux, while because the current is zero in the inner coil, then the meter is or that that element of the meter is reading zero.

Plus we're going to add to it element number two, which is 240 volts times the outer current coil 10 amps times half the flux which is 1200 watts. And if we add up element one into we still have 1200 watts which agrees with our original calculation of Power of the connected load. In this example problem, again, we're going to use the same standard one and a half element meters metering a single phase load as we have in the past two examples, but this time, we're going to connect the loads like this. We've placed a 10 out of 10 ohm load across line one in line two. And we're going to be placing 50 ohm load from line one to neutral and a 30 ohm load from line two to neutral. We're going to calculate the amount of power used by the load and then we're going to find out what is the amount of power measured by each element and what is the power unit Power used by or the total amount of power measured by the watt meter.

Moving on to the solution will answer the first question and that is, what is the power consumed by the load regardless of how the kilowatt hour meter is connected to the load itself. The current through load r1 is simply ohms law again 240 volts over 10 ohms which will draw 240 amps power used by that load will be 240 volts times 24 amps which is equal to 5760 watts for our load are two using ohms law to calculate the current Is 120 volts is time because it's lined to neutral voltage divided by 50 ohms which is equal to 2.4 amps. Therefore, the power used by or dissipated by that load is 120 volts times 2.4 amps or 288 watts. Finally, the current through R three will be 120 volts over 30 ohms which will give us four amps, giving us a power draw of 120 volts times four amps which is equal to 480 watts.

Summing those three power draws together we will get a total load dissipation of 6528 watts. Now, looking at the meter connected to the load the meter will read for r1 It will read 240 volts times 24 amps in the outer current coil times half the flux is equal to 2880 watts plus 240 volts times 24 amps for the inner current coil times half the flux again is equal to 2880 watts, making a total power measurement for that particular load of 5760 watts. Now, this is measured by the meter the meter for load are two will read 240 volts times 2.4 amps for the inner current coil times one half because of Sona using half the flux amounts to 288 watts for the The outer current coil the current is zero, so there's no contribution to the from the from the outer coil for r two. However, the meter will read for r three 240 volts times four amps for the going through the inner current coil times half the flux will give us 480 watts, there will be no current flowing through the outer current coil due to our three, so the contribution would be zero.

So if we add up those power draws, we see that we have 5760 plus 288 plus 480, which totals to 6528 watts, which agrees with our load calculations activity. Getting so this ends chapter four