Question

For Problem 8.25, determine the complex power delivered to the load in terms of symmetrical components. Verify the answer by adding up the complex power of each of the three phases.

Short answer

The power deliver to load with using symmetrical component $2922.77\angle 74.40\circ$kVA and by adding up complex power of three phase is $2925.25\angle 74.40\circ$kVAwhich almost same.

Step-by-step solution

Write the given data from the question:

Write the line to neutral voltages.

$$\begin{gathered} V_{an}=200\angle 25°V \\ {V}_{bn}=100\angle -155°V \\ {V}_{cn}=80/100°V \end{gathered}$$

The series impedance is Z=$6-j24\Omega$

The mutual impedance is $Z_m=j3\Omega$

The data from problem 8.25.

$$\begin{gathered} Thesequencevoltages{V}_0,V_1,and{V}_{2}are47.78\angle 57.63°V,112.70\angle 359.97°Vand \\ 61.62\angle 45.88°Vrespectively. \end{gathered}$$

$$\begin{gathered} Thesequencecurrents{I}_0,I_{1}and{I}_{2}are1.56\angle -21.56°A,5.16\angle 285.92°Aand \\ 5.16\angle -28.16°Arespectively. \end{gathered}$$

$$\begin{gathered} Thephasecurrents{I}_a,I_{b}and{I}_{c}are10.89\angle 313.03°A,6.93\angle 122.27°Aand \\ 0.8\angle 32.69°Arespectively. \end{gathered}$$

Determine the equation to calculate the complex power delivered to the load.

The equation to calculate the complex power delivered to the load in terms of symmetrical component is given as follows.

$S=3\left[V_0{\left(I_0\right)}^x+V_1{\left(I_1\right)}^x+V_2{\left(I_2\right)}^x\right]$ …… (1)

The equation to calculate the complex power without symmetrical component is given as follows.

…… (2)

$S=V_a{\left(l_a\right)}^x+V_b{\left(l_b\right)}^x+V_c{\left(l_c\right)}^x$

Calculate the complex power delivered to load with using symmetrical component and by adding up complex power of three phase.

Calculate the complex power delivered to the load in terms of symmetrical component.

Substitute 47.78<57.63$°$V, for Vo , 112.70<359.97$°$V for V, 61.62/45.88$°$V

for V2 1.562-21.06" A, for 's, 5.164285.92" A for " and 5.162 -28.16" A for

l₂ into equation (1).

S = 3[(47.78257.630) (1.562-21.060) +(112.70_359.97.)(5.16_285.92)

+(61.62/45.88 V)(5.162-28.16)]

S = 3[(47.78257.63)(1.56_21.060)+(112.702359.97.)(5.164-285.929)]

+(61.62/45.88- V)(5.16_28.16)]

S =3 [74.54/78.69 +581.99/74.05+317.96274.059]

S =2922.77 74.40 KVA

Calculate the complex power by adding up complex power of three phase

Substitute 200/25" V for Va , 100/-155- V for V. 802100 V for 2

10.892313.03" A for ' 6.932122.27localid="1654851917637" $°$A for " and 0.8/32.69localid="1654851923291" $°$A for 12 into

equation(2).

S =[(200_25) (10.89<313.03) +(100<-155.)(6.932122.27)

+(802100- V)(0.8<32.690)]

S=[(200425.)(10.892 -313.03)+ (1002 -155)(6.932-122.27)+(804100- V)(0.82-32.69)]

S = 21782-288.03+693localid="1654851928302" $\angle$277.27localid="1654851932241" $°$+64.67.31

S =2925.25/74.40$°$KVA

Hence, the power deliver to load with using symmetrical component

2922.77 274.40localid="1654851945636" $°$KVA and by adding up complex power of three phase is

2925.25/74.40localid="1654851950509" $°$KVA which almost same.