## 4 Case Study
### Calculation of Load Flows  
The load flow is calculated for four lines (1-3-7, 1-5-9, 1-4-8, and 1-2-6). We will demonstrate the application of the forward/backward sweep technique for line 1-3-7 as an example.  
Step 1: Initialization of voltages in eight nodes, \(V_{2}^{0}=V_{3}^{0}=V_{4}^{0}=V_{5}^{0}=V_{6}^{0}=V_{7}^{0}=V_{8}^{0}=V_{9}^{0 }=V_{line}=33\) Kv.  
Step 2: Iteration count initialization, \(k=1,\epsilon=10^{-5}\).  
Step 3: Load Current computation,  
\(I_{7}^{1}=\frac{S_{7}^{0}}{v_{7}^{1}}=\frac{9.2(MW)}{33(KV)}=278.78\)  
\(I_{3}^{1}=\frac{S_{7}^{0}}{v_{7}^{1}}=\frac{8(MW)}{33(KV)}=242.42\).  
Step 4: Update current starting from the end nodes (backward sweep),  
\(I(3-7)^{1}=I_{7}^{1}+\sum_{branch,downstream}=278.78+0=278.78\)  
\(I(1-3)^{1}=I_{3}^{1}+\sum_{branch,downstream}=I_{3}^{1}+I_{(3-7)}^{1}=242.42+2 78.78=524.21\).  
Step 5: Forward sweep,  
\(V_{3}^{1}=V_{line}^{1}-Z_{1-3}*I_{1-3}^{1}=33,000-2.051*524.21=31,930.56\)  
\(V_{7}^{1}=V_{3}^{1}-Z_{3-7}*I_{3-7}^{1}=31,930.56-6.16*278.78=30,212.48\).  
Step 6: Error,  
\(E_{3}^{1}=\frac{|V_{3}^{1}-V_{4}^{2}|}{Vline}<=\)_tolerance_  
\(E_{3}^{1}=\frac{|31,930.56-33.000|}{33000}=0.0324>10^{-5}\)  
\begin{table}
\begin{tabular}{c c} \hline
**Parameters** & **Values** \\ \hline \(P_{v}\) (Kw) & 15, 50 \\ ES (Kwh) & 50, 100 \\ \(t_{disp}\) (hours/year) & 200 \\ \(Pr_{p}(PeakPrice)\) USD kWh & 0.5 \\ \(C_{ac}\) USD/year for each vehicle & 304 \\ Car availability & 100\% \\ SOC levels & 0.3, 0.45, 0.7 \\ Number of Vehicles per SOC & 0.25, 0.25, 0.5 \\ Number of parking lots & 3 \\ \(Pr_{off}\) USD/kWh & 0.05 \\ \(\mu_{conv}\) & 0.85 \\ \(C_{d}\) & 0.225 \\ Number of load levels & 3 \\ \(V_{line}\) & 33 kv \\ \(V_{min}\) & 0.9 pu \\ \(V_{max}\) & 1.1 pu \\ Price (per load level) USD/kWh & Light load & 0.035 \\  & Medium load & 0.049 \\  & Peak load & 0.07 \\ \(t_{k}\) (Time duration) h/year & Light load & 2190 \\  & Medium load & 4745 \\  & Peak load & 1825 \\ \hline \end{tabular}
\end{table}
Table 4: Technical/cost parameters.  
\(E_{7}^{1}=\frac{|V_{7}^{1}-7\eta_{7}^{0}|}{Vline}<=tolerance\)  
\(E_{7}^{1}=\frac{|30.2122.48-33,000|}{33000}=0.0844>10^{-5}\).  
Step 7: Maximum error,  
\(e_{max}=max(E_{3}^{1},E_{7}^{1})=0.0844>10^{-5}\).  
Step 8: If \(e_{max}\) is more than tolerance (\(\epsilon\)), then update the iteration count to k = k + 1 and go to Step 3 and repeat the steps. K = 2.  
Iteration 2  
Step 3: Load Current computation (line 1-3-7),  
\(I_{7}^{2}=(S_{7}^{1})/(v_{7}^{1})=\frac{9.2(MW)}{30(212.48}=304.51\)  
\(I_{3}^{2}=(S_{3}^{2})/(v_{3}^{2})=\frac{8.4(MW)}{31903.56}=250.54\),  
Step 4: Update the current starting from the end nodes (backward sweep),  
\(I_{3-7}^{2}=I_{7}^{2}+\sum_{branch,downstream}=304.051+0=304.051\)  
\(I_{1-3}^{2}=I_{3}^{2}+\sum_{branch,downstream}=I_{3}^{2}+I_{3-7}^{2}=250.54+ 304.51=555.05\).  
Step 5: Forward sweep,  
\(V_{3}^{2}=V_{line}^{2}-Z_{1-3}*I_{1-3}^{2}=33,000-2.051*555.05=31,861.12\)  
\(V_{7}^{2}=V_{3}^{2}-Z_{3-7}*I_{3-7}^{2}=31,861.12-6.16*304.052=29,984.53\).  
Step 6: Error,  
\(E_{3}^{2}=|V_{3}^{2}-V_{3}^{1}|/V_{line}\leq tolerance\)  
\(E_{3}^{2}=|31,861.12-31,930.56|/33,000=0.021>10^{-5}\)  
\(E_{7}^{2}=|V_{7}^{2}-7_{7}^{1}|/V_{line}\leq tolerance\)  
\(E_{7}^{2}=|29,984.53-30,212.48|/33000=0.0069>10^{-5}\).  
Step 7: Maximum error,  
\(e_{max}=max(E_{3}^{2},E_{7}^{2})=0.0069>10^{-5}\).  
Step 8: \(e_{max}\) is greater than tolerance (\(\epsilon\)), so update the iteration count to \(k=k+1\) and go to Step 3. These steps are repeated, and in K = 6, the \(e_{max}\) becomes less than tolerance (\(\epsilon\)), so the load flow is converged at iteration 6.  
\(e_{max}=max(E_{3}^{6},E_{7}^{6})=4.28*10^{-7}>10^{-5}\)