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given Voltage and loads calculated all the line currents

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A chemical plant consists for 3 loads

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(a) Calculate the total power factor of the power consumed by these loads in the chemical plant.

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(b) Determine the size of capacitor required (in kVAr) to ensure unity power factor.

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(c) Determine the size of capacitor required (in kVAr) to achive a power factor of at least 0.95. 

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kW

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Q3 is already negative by definition.

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Complete the follwoing table 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
QS
Load 1[[0]] kW[[3]] kVAr[[6]] kVA
Load 2[[1]] kW[[4]] kVAr[[7]] kVA
Load 3[[2]] kW[[5]] kVAr\n

[[8]] kVA

\n
\n

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Use units to help know what info you are given;

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If you know any two from these five variables, P, Q, S , pf and $\\theta$ ...

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...  then you can always find the other values using these two equations

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Also, you will need to be careful with polarity of reactive power depeding on leading or lagging loads.  

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Inductors will have +Q values. Capactitors are the opposite. (i.e. -Q values)

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CIVIL ...use this to work if Cap or Inductor.  

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e.g.  current __________ voltage (lead or lag)

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The combined load is there for $($[[0]]}$\\angle $[[1]]$^{\\circ}) kVA = $ [[2]] $kW + j$[[3]] $kVAr$

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And the uncorrected power factor is [[4]].

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Add all the real power values to get $P_{total}$

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Add all the reactive power values to get $Q_{total}$

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Then find total apparent power $S_{total} = P_{total} +jQ_{total}$ 

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Finaly, convert from rectangular to polar form. 

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An additional capacative load of [[0]] kVar is required to correct the overall power factor to unity.

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In these questions, we know two things about how the newTotal load will look

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P

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[kW]

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Q

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[kVAr]

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S [kVA]pf$|\\theta|$
Original Load{round(Ptotal)}{round(Qtotal)}        
Additional  PFC 0
New Total Load = OrigLoad + PFC{round(Ptotal)}{round(Ptotal)}1
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Now working on the New Total Row, we know 2 of the 5 variables.  So we can calulate the others

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(we are really only intersted in Q so we'll only work that one out.)

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Since $pf=\\frac{P}{|S|} $, we have |S|= {round(Ptotal)}  therefore Q = 0 

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P [kW]Q [kVAr] S [kVA]pf$|\\theta|$
Original Load{round(Ptotal)}{round(Qtotal)}
Additional  PFC 0
New Total Load = OrigLoad + PFC{round(Ptotal)}0{round(Ptotal)}1
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Final step, looking down the Q column ... 

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NewTotal Q = Orig Load Q + Qpfc 

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0 = {round(Qtotal)} + Qpfc

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Qpfc = {round(Qtotal*-1)}

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
P [kW]Q [kVAr]S [kVA]pf$|\\theta|$
Original Load{round(Ptotal)}{round(Qtotal)}
Additional  PFC 0{round(Qtotal*-1)}
New Total Load = OrigLoad + PFC{round(Ptotal)}0{round(Ptotal)}1
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An additional capacative load of [[0]] kVar is required to correct the overall power factor to 0.95.

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WARNING

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I've round the values in this solution to make it readable ... you will need to keep a few decimal places.

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JR Nov'21

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In these questions, we know two things about how the newTotal load will look

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\n

P

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[kW]

\n
\n

Q

\n

[kVAr]

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S [kVA]pf$|\\theta|$
Original Load{round(Ptotal)}{round(Qtotal)}        
Additional  PFC 0
New Total Load = OrigLoad + PFC{round(Ptotal)}??????0.95
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Now working on the New Total Row, we know 2 of the 5 variables.  So we can calulate the others

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(we are really only intersted in Q so we'll only work that one out.)

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Since $pf=\\frac{P}{|S|} $, we have |S|= {round(Ptotal)} * 0.95  therefore Q = {round(QtotalRequired)}

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
P [kW]Q [kVAr] S [kVA]pf$|\\theta|$
Original Load{round(Ptotal)}{round(Qtotal)}
Additional  PFC 0????
New Total Load = OrigLoad + PFC{round(Ptotal)}{round(QtotalRequired)}{round(SrequiredMag)}0.95
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Final step, looking down the Q column ... 

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NewTotal Q = Orig Load Q + Qpfc 

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{round(QtotalRequired)} = {round(Qtotal)} + Qpfc

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Qpfc = {round(QtotalRequired)} - {round(Qtotal)} 

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P [kW]Q [kVAr]S [kVA]pf$|\\theta|$
Original Load{round(Ptotal)}{round(Qtotal)}
Additional  PFC 0{round(QtotalRequired)} - {round(Qtotal)} ={round(Qpfc)}
New Total Load = OrigLoad + PFC{round(Ptotal)}{round(QtotalRequired)}{round(SrequiredMag)}1
\n

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