Question 8.4: For a simple radial distribution system shown in Figure 8.13......
For a simple radial distribution system shown in Figure 8.13 and considered for Examples 8.1 through 8.3 (8.1 ,8.2, 8.3), determine the short-circuit at points E and F if the system specifications are
• Second Transformer: 200 \ kVA,480 \ Y/277-208 \ Y/120
• Z_{T,2}=5\%,\frac{X}{R}=1.5
• Subfeeder BE: 250 MCM THW (Cu), 50-ft length, steel conduit
• Impedance at point B: Z_{B,pu}=0.0565,\frac{X}{R}=2.4
Using Equation 8.8, the impedance at point E can be estimated:
Z_B=Z_A+Z_F (8.8)
Z_E(pu)=Z_{B,pu}+Z_{Feeder,BE}(pu)with
Z_B(pu) \text{given with}\frac{X}{R}=2.4Thus,
Z_B(pu)=0.0565[\cos 67^\circ +j\sin 67^\circ ]=0.0218+j0.0528For the feeder BE, the impedance can be obtained from Table 8.1:
\begin{matrix} R=\frac{0.054}{1000 \ ft}\rightarrow R=0.0027 \\ X=\frac{0.052}{1000 \ ft}\rightarrow R=0.0026 \end{matrix}The reference impedance, Z_{ref,1}, for the feeder BE is defined by Equation 8.9 using the first transformer rated at 1000~kVA and secondary phase voltage of E_{\phi ,S,1}=277~ V and phase current, I_{\phi ,S,1} = I_{L,S,1}:
Z_{ref}(\Omega )=\frac{E_{\Phi ,S}}{I_{\Phi ,S}} (8.9)
Z_{ref,1}=\frac{E_{\phi ,S,1}}{I_{\phi ,s,1}}=\frac{277}{1202}=0.23 \ \Omegawith the secondary current for the first transformer, IL,S,1, as calculated in Example 8.1:
I_{L,S,1}=\frac{1000 \ kVA}{\sqrt{3}(0.48 \ kV) }=1202 \ AThus, the per-unit impedance for feeder BE is
Z_{Feeder,BE}(pu)=\frac{0.0027}{0.23 }+j\frac{0.0026}{0.23}or
Z_{Feeder,BE}(pu)=0.0117+j00113Thus, the per-unit impedance at point E can be determined using Equation 8.8:
Z_B=Z_A+Z_F (8.8)
Z_E(pu)=(0.0218+0.0117)+j(0.0521+0.0113)=0.0335+j0.0634The magnitude of the per-unit impedance at point E, Z_E, is then
\left|Z_E(pu)\right|=\sqrt{0.0335^2+0.0634^2}=0.0718It should be noted that the X/R value of the impedance , Z_E can be calculated as X/R = 0.0634/0.0335 = 1.9, which corresponds to a phase angle of 62.1°.
The short-circuit current expressed in per-unit at point E is calculated to be
The short-circuit current in Amperes at point E is then
I_{SC,E}=13.94\times 1202=16,750 \ AThe per-unit impedance at point F (at the secondary side of the second transformer) is estimated using Equation 8.12:
Z_F=\frac{1}{a^2}Z_E+Z_{T,2} (8.12)
Z_F(pu)=Z_{T,2}(pu)+Z_{S,E}(pu)with
Z_F=\frac{1}{a^2}Z_ESince the equation here is only applicable when the impedances are expressed in Ohms, the impedance at point E is converted to be expressed in Ohms:
Z_E(\Omega )=Z_{ref,1}(\Omega )+Z_{E}(pu)or
Z_E(\Omega )=(0.23 \ \Omega )*(0.0718)=0.0165 \ \OmegaThe turn ratio of the second transformer can be calculated as
a=\frac{480}{208}=2.3Thus,
Z_{S,E}=\frac{1}{a^2}Z_E=0.00312 \ \OmegaThe impedance of the second transformer, Z_{T,2}, can be obtained from the values of X/Rand percent transformer impedance:
Z_{T,2}(pu)=\left\|Z_{T,2}(pu)\right\|\cdot \cos (\theta )+j\sin (\theta )with
\sin (\theta )\rightarrow \tan^{-1}(\frac{X}{R} )or
\tan ^{-1}(1.5)=56.3^\circThus, the per-unit transformer impedance is
Z_{T,2}(pu)=0.05\times \cos (56.3^\circ )+j\sin (56.3^\circ )or
Z_{T,2}(pu)=0.0278+j0.0416It should be noted that for points F and G, the reference impedance is related to second transformer characteristics and is defined as
Z_{ref,2}=\frac{E_{\phi ,s,2}}{I_{\phi ,S,2}}=\frac{120 \ V}{555 \ A}=0.216 \ \Omegasince the reference voltage is the secondary phase voltage for the second transformer, that is, E_{\phi ,s,2}= 120 \ V, and the reference current is
I_{L,S,2}=\frac{200 \ kVA}{\sqrt{3}(0.208 \ kV) }=555 \ AThus,
Z_{E,S}(pu)=\frac{Z_E(\Omega )}{Z_{ref,2}} Z_{E,S}(pu)=\frac{0.00312 }{0.216}[\cos 62.1^\circ +j\sin 62.1^\circ ]=0.00675+j0.01276Therefore, the per-unit impedance at point F can be calculated:
Z_F(pu)=0.0345+j0.0544The magnitude of the per-unit impedance at point F is then
\left\|Z_F(pu)\right\|=\sqrt{0.0345^2+0.0544^2}=0.06438Finally, the per-unit and the actual short-circuit current at point F can be estimated:
I_{SC,F}(pu)=\frac{1}{0.06438}=15.53Therefore and since the reference current is 555 \ A,
I_{SC,F}(A)=15.53\times 555 \ A=8600 \ ATABLE 8.1
Resistance and Reactance of Feeders Made Up of Three Copper Conductors in a Conduit for Three-Phase Systems
| Reactance (Ω/1000 ft [Ω /1000 m]) | Resistance (Ω/1000 ft [Ω /1000 m]) | Wire Size in AWG or MCM |
|||
| Steel Conduit | PVC, Aluminum Conduit | Steel Conduit | Aluminum Conduit | PVC Conduit | |
| 0.073 (0.240) | 0.058 (0.190) | 3.1 (10.2) | 3.1 (10.2) | 3.1 (10.2) | 14 |
| 0.068 (0.223) | 0.054 (0.177) | 2.0 (6.6) | 2.0 (6.6) | 2.0 (6.6) | 12 |
| 0.063 (0.207) | 0.050 (0.164) | 1.2 (3.9) | 1.2 (3.9) | 1.2 (3.9) | 10 |
| 0.065 (0.213) | 0.052 (0.171) | 0.78 (2.56) | 0.78 (2.56) | 0.78 (2.56) | 8 |
| 0.064 (0.210) | 0.051 (0.167) | 0.49 (1.61) | 0.49 (1.61) | 0.49 (1.61) | 6 |
| 0.060 (0.197) | 0.048 (0.157) | 0.31 (1.02) | 0.31 (1.02) | 0.31 (1.02) | 4 |
| 0.059 (0.194) | 0.047 (0.154) | 0.25 (0.82) | 0.25 (0.82) | 0.25 (0.82) | 3 |
| 0.057 (0.187) | 0.045 (0.148) | 0.19 (0.62) | 0.19 (0.62) | 0.19 (0.62) | 2 |
| 0.057 (0.187) | 0.046 (0.151) | 0.15 (0.49) | 0.15 (0.49) | 0.15 (0.49) | 1 |
| 0.055 (0.180) | 0.044 (0.144) | 0.12 (0.39) | 0.12 (0.39) | 0.12 (0.39) | 1/0 |
| 0.054 (0.177) | 0.043 (0.141) | 0.10 (0.33) | 0.10 (0.33) | 0.10 (0.33) | 2/0 |
| 0.052 (0.171) | 0.042 (0.138) | 0.079 (0.259) | 0.082 (0.269) | 0.077 (0.253) | 3/0 |
| 0.051 (0.167) | 0.041 (0.135) | 0.063 (0.207) | 0.067 (0.220) | 0.062 (0.203) | 4/0 |
| 0.052 (0.171) | 0.041 (0.135) | 0.054 (0.177) | 0.057 (0.187) | 0.052 (0.171) | 250 MCM |
| 0.051 (0.167) | 0.041 (0.135) | 0.045 (0.148) | 0.049 (0.161) | 0.044 (0.144) | 300 MCM |
| 0.050 (0.164) | 0.040 (0.131) | 0.039 (0.128) | 0.043 (0.141) | 0.038 (0.125) | 350 MCM |
| 0.049 (0.161) | 0.040 (0.131) | 0.035 (0.115) | 0.038 (0.125) | 0.033 (0.108) | 400 MCM |
| 0.048 (0.157) | 0.039 (0.128) | 0.029 (0.095) | 0.032 (0.105) | 0.027 (0.089) | 500 MCM |
| Note: This table is adapted from NEC-2014 Table 9, Chapter 9 (NEC, 2014). | |||||