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LOAD SCHEDULE

These papers come up due to Meinhardt Philippines’s provided load schedule for our past projects. I have found out that the way we design our load schedule and Meinhardt’s way of making it is quite different. I started asking myself which is correct and why do they come up to that design based on principles and

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actual applications. Mixing up some ideas and observing closely every detail, this simple explanation was produced.

LOAD SCHEDULE:

Let’s recall first some topics

WYE AND DELTA CONNECTION

1. DELTA CONNECTION:

V line = V phase While, I line = √3 I phase

* On delta system, Phase Current and Line Current is 30° apart on Phasor diagram

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2. WYE (STAR) CONNECTION:

I line = I phase While, V line = √3 V phase

* On wye system, Phase Voltage and Line Voltage is 30° apart on Phasor diagram.

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General formula on 3ø system TOTAL POWER: Pt = 3 Pø

(Total power of a 3 phase system is the sum of the 3 PHASES POWER, usually equal on a balanced 3 phase system) So we can also say,

Pt = 3 VøIø general Question, is this formula applicable for both DELTA and WYE connection for 3 phase?Let us try to substitute the values above (Y and DELTA) to our general equation,

For delta:Pt = 3 VøIø ,Vø=Vline and Iø=Iline/√3 for delta Therefore;Pt = 3 x V L x I L x √3 √3 √3

Pt = √3 VL IL general

For wye:Pt = 3 VøIø ,Vø=Vline/√3 and Iø=Iline for wyeTherefore;Pt = 3 x V L x I L x √3 √3 √3

Pt = √3 VL IL general

Therefore, the above formula is applicable for both type of connection.

PHASE SEQUENCE

1. Positive Phase Sequence

ABC BCA CAB

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2. Negative Sequence

ACB BAC CBA

*IMPORTANT!!! If Phase Sequence is not given, assume a positive phase sequence. Three phase alternators are designed to operate with positive phase sequence voltages.Therefore, we will consider POSITIVE SEQUENCE for our computation.

RELATIONSHIP OF Vline to Vphase and Iline to Iphase ON +/- SEQUENCE

1. Vline leads the corresponding Vphase by 30° if the phase sequence is POSITIVE. 2. Vline lags the corresponding Vphase by 30° if the phase sequence is NEGATIVE. IN WYE3. Iline leads the corresponding Iphase by 30° if the phase sequence is NEGATIVE4. Iline lags the corresponding Iphase by 30° if the phase sequence is POSITIVE IN DELTA

*Since our computation is assumed to be @ POSITIVE SEQUENCE, we will consider item no. 1 and 4 w/c tells the relation of line and phase current/voltage. Observe that line voltage LEADS the phase voltage in + seq. and line current LAGS the phase current also in + seq. by 30 degrees.

WYE DELTA VL Iphase Vphase IL

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Phasor Diagram of Voltage @ WYE Phasor Diagram of Current @ DELTA Line leads Phase (take not of this) Line lags Phase (take note of this)

Using vector analysis, from the figure on the left, we can derive these formulas (w/c actually written on some books)

1. VAB = VAN- VBN

2. VBC= VBN-VCN

3. VCA=VCN-VAN

On Other figure(right)

1. IA = IAB- ICA

2. IB = IBC- IAB

3. IC = ICA- IBC

If you will not follow the phasor diagram above, the vector may have same resultant magnitude BUT will go to different direction.

Now that we have these explanation and formulas, let us apply it on actual. The purpose of this paper is to let us know and understand the current flowing on every conductor and bus of each panel. Most of the time, we just add algebraically some values of current which are actually vector quantities. Letting us know the relationship of line current and phase current on WYE and DELTA connection will help us to build a phasor diagram which we can use even unbalanced loads are encountered. Having that so many formulas and

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drawings are so exhausting to understand but if you will look on to it closely, you will see that its not that complicated as what you think. Examine the connection of loads on these different system configurations:

SINGLE PHASE LOADS CONNECTED ON 3WIRE + GRND

On Actual, this is how single phase loads are connected to a 3phasesupply. Please observe circuit nos. 1,3 and 5. 1 is connected to AB, 3 is on CA and 5 is on BC. Don’t you notice something? The loads are connected on AB,BC, and CA which is actually delta connection.

Let us redraw our panel and see how it looks

…. see it clearly? DELTA.Now let us examine the loads of eachConductors. For example: ckt 1 is a1000 VA load and Vline@ 230 V.If we will divide 1000 by 230,we’ll get 4.34øAB current, not line current. After doing this to some number of loads,its time to determine the line current w/cwill determine the size of our feeder.1000 VA load of ckt1 is actually PHASEPOWER. So according to what have Mentioned before, Pt = 3 Pø, we cansay the other 2 phase power are 1000VA Assuming it’s a balanced 3phase.let ckt

3 and 5 have 1000 VA each phase, computing for total power we will have 3000VA as Pt. Using the formula Pt = √3 VL IL we will be able to determine the overall line current which we needed to get the size of the main feeder. IL=7.53 A. Now we can say that line currents @ bus A,B and C are all equal to 7.53.Letting IA=7.53, let us use this formula, IA = IAB- ICA to verify our statement.

By just seeing the diagram,we can easily imagine theresultant current.

=4.34cis -210 =4.34 cis 30 Iabx =cos30 (4.34) = 3.76 Iaby =sin30 (4.34) = 2.17 Iabx =cos-30 (4.34) = 3.76 =7.53 Iabx =sin-30 (4.34) = -2.17 =4.34 cis -30 ∑x= 3.76 + 3.76 = 7.52

∑y= 2.17 – 2.17 = 0

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IA = √ 7.52² + 0² = 7.52≈7.53 AMP

Ø = arctan (∑y/∑x) = 0 degress

SINGLE PHASE LOADS CONNECTED ON 3WIRE + NEUTRAL + GROUND (line to neutral)

For Single Phase loads which are connected on line to neutralConnection, we can simply say that the objective of this connection Is to use the line to neutral voltage. Just like we have done before, lets Observe the connection of these loads. We can see that each otherPole of the load is connected or terminated at NEUTRAL BUS.This is usual in WYE connection. Let us redraw the figure using loads 1, 3 and 5

Now we formed a WYE connection.Let us set each load to have 1000 VA.Power total =3000VA and Line voltageOf 230 V. Thru computations we’ll get A Line current of 7.53 A. remember that Line current at WYE is equal to its Phase current. Each phase has 1000 VA.This time, we will get phase voltage,Which is actually the line to neutral Voltage. Getting Neutral Voltage usingThe formula V line = √3 V phase, We will a get Phase voltage of 132.79 VNow getting the phase current by This equation 1000/132.79, PhaseCurrent will be 7.32 A, equal to line Current

Verify : VAB = VAN- VBN

Vabx =cos30 (132.79) = 115 Vaby =sin30 (132.79) = 66.40 =132.79 cis 30 Vabx =cos-30 (132.79) = 115

Vabx =sin-30 (132.79) = -66.40 =230 cis 0 ∑x= 115 + 115 = 230

∑y= 66.40 – 66.40 = 0 =132.79 cis 30

IA = √ 230² + 0² = 230V

Ø = arctan (∑y/∑x) = 0 degress

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EXAMPLE 1:

3-Wire + Neutral + Ground (BALANCED)

I Ø = I FL

Size of Feeder:

Feeder Amp = 125% (IFL * D.F) FOR NORMAL LOADS = 1.25 * (30.1* .85) =31.98 A USE 40 AT MAIN

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EXAMPLE 2:

3-Wire + Ground (BALANCED)

I Ø = I FL/√3

Size of Feeder:

Feeder Amp = 125% (IFL * D.F) FOR NORMAL LOADS = 1.25 * (30.12* .85) =32.0 A USE 40 AT MAIN

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EXAMPLE 3:

3-Wire +Neutral + Ground (UNBALANCED)

I Ø = I FL (Consider HIGHEST PHASE)

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EXAMPLE 4:

3-Wire + Ground (UNBALANCED)

I Ø ≠ I FL/√3( USE PHASOR DIAGRAM + SEQ)

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