Esercizio 9: soluzione
Dovendo calcolare la:
\(\Delta V = \sqrt{3},\Big(R_{2eq}\cos\varphi + X_{2eq}\sin\varphi\Big)\cdot I_2
\)
Sul lato BT si hanno: \(V_2 = 0{,}4\ \text{kV} = 400\ \text{V}\)
\(S_{2n} = \sqrt{3},V_{2n}I_{2n}
\quad \Longrightarrow \quad
I_{2n} = \frac{S_{2n}}{\sqrt{3},V_{2n}}
= \frac{150000}{\sqrt{3}\cdot 400}
= 216{,}5\ \text{A}
\) \(
R_{2eq} = \frac{P_{cc}}{3 I_{2n}^2}
= \frac{2000}{3\cdot 216{,}5^2}
= 0{,}0142\ \Omega
\) \(
Z_{2eq} = \frac{V_{cc}}{\sqrt{3} I_{2n}}
= \frac{\tfrac{V_{cc\%}}{100},V_{2n}}{\sqrt{3},I_{2n}}
= \frac{\tfrac{5}{100}\cdot 400}{\sqrt{3}\cdot 216{,}5}
= 0{,}053\ \Omega
\) \(
X_{2eq} = \sqrt{Z_{2eq}^2 - R_{2eq}^2}
= \sqrt{0{,}053^2 - 0{,}0142^2}
= 0{,}051\ \Omega
\)
Quindi:
\(\cos\varphi_2 = 0{,}9
\quad\Rightarrow\quad
\varphi_2 = 25{,}8^\circ
\quad\Rightarrow\quad
\sin\varphi_2 = 0{,}435
\) \(
\Delta V
= \sqrt{3},\Big(R_{2eq}\cos\varphi_2 + X_{2eq}\sin\varphi_2\Big)\cdot I_2
= \sqrt{3},(0{,}0142\cdot 0{,}9 + 0{,}051\cdot 0{,}435)\cdot 216{,}5
= 13{,}2\ \text{V}
\) \(
\Delta V\% = \frac{\Delta V}{V_2}\cdot 100
= \frac{13{,}2}{400}\cdot 100
= 3{,}3\%
\)