Three Phase Power Measurment

Experiment : Two method watt meter of measurement of power

Apparatus   :

•  Two watt meters
• Volt meter ( 0- 600 V )
• Ammeter AC  (0-2 -10 A )
• Balanced Load ( 3ø)

Theory   :
In a three phase supply there should assentilly,by A,B and C and some times a fourth supply wire (N)

                   For a balanced system of Voltage the magnitued of voltages Van,Vbn,Vcn are equql and they are donate by Vp.
                  The magnitude of voltages AB,BG,CA are equal and they are Donated by V line or Vl from the diagram it can be that

                                       V_L   = Ґ3 V_{p .................(1)}

_{Fig 2 shows the vector diagram of the corcit in figure (1) since that the load used is a balanced one}

                     lE₁l = l E₂l =lE₃l  and lI₁l =lI₂l = l I₃l

Where E_{1 ,}E_2,s, E₃  are  the magnitudes of phasor representing the phase votage and I_1,I_2, I_3,are the magnitude phasor representing line current.

        The voltage applied to the pressure coils of wattmeter W_1,W2 are  pressure coils of wattmeter w1 and w2 are (E1 -E2 ) and ( E3 -E2 ) repectively. 

   The reading in the wattmeters W1 and W2 are related to the above quantities by,


W1  =I i1 l E1 - E2 l cos α l  ..........( a )
W2  =l i2 l E3 - E2 l cos β ............(b)
Let  lE₁l = l E₂l =lE₃l  and lI₁l =lI₂l = l I₃l
It can be seen that from the phasor diagram that 
W1  = i1 Ґ3 Vp cos ( Φ- 30) and l E1 -E2 l =Ґ3 Vp...........( c )
W2 = i1 Ґ3 Vp cos ( Φ + 30 )  .............(d)
W1 +W2 = i1 Ґ3 Vp cos ( Φ + 30 ) +cos ( Φ- 30) 
W1 + W2 = 3 Vp cos  Φ.......(E)
But Vp = VL / Ґ3
W1 + w2 =Ґ3VLIL cos  Φ,  cos  Φ = ( w1 +w2 ) / Ґ3 Vl IL
Similarly   W1 -W2 = Ґ3 IL Vp [ (  cos ( Φ- 30)  -cos ( Φ + 30 ) ]
W1 - W2 IL VL Sin Φ 


Tan Φ =Ґ3  W1-W2
                   W1+w2


Cos Φ = I / Ґ(1+3[w1-w2]2 / [ w1+w2 ] 2 ........(3)


Where total power = W1 + W2 ................. ( 4 )