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— ee2:task_d9io924n0e3du21g_with_calculation [2024/07/16 00:17] (aktuell)
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 +{{tag>self:resonance impedance resonant_circuit exam_ee2_SS2024}}{{include_n>1100}}
 + 
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Magnetic Circuit \\
 +<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
  
 +A symmetric and balanced three-phase motor is driven with a $230 ~\rm V$ / $400 ~\rm V$ / $50 ~\rm Hz$ three-phase power net. 
 +Each single string has a resistor $R=5 ~\Omega$ and an inductance of $L=10 ~\rm mH$. 
 +
 +{{drawio>ee2:d9io924n0e3du21g_question1.svg}}
 +
 +1. Calculate the $\cos \varphi$, and the magnitude of the impedance $|Z|$ for a single string.
 +
 +#@HiddenBegin_HTML~d9io924n0e3du21g_11,Path~@#
 +
 +The phase $\varphi$ is given by:
 +\begin{align*}
 +\varphi &= \arctan \left( {{X_{L}}\over{R}} \right) \\
 +        &= \arctan \left( {{2\pi \cdot f \cdot L}\over{R}} \right) \\
 +        &= \arctan \left( {{2\pi \cdot 50 {~\rm Hz} \cdot 10 \cdot 10^{-3} {~\rm H}}\over{5 ~\Omega}} \right) \\
 +        &= 0.5609 ... \hat{=} +32° \\
 +\end{align*}
 +
 +With this, the $\cos \varphi$ becomes
 +\begin{align*}
 +\cos \varphi  &= \cos(0.5609 ...) \\
 +              &= 0.84673...\\
 +\end{align*}
 +
 +The impedance is given by:
 +\begin{align*}
 +|\underline{Z}_{RL}|  &= \sqrt{X_{L}^2 + R^2} \\
 +                      &= \sqrt{( 2\pi \cdot f \cdot L                                    )^2 + R^2} \\
 +                      &= \sqrt{( 2\pi \cdot 50 {~\rm Hz} \cdot 10 \cdot 10^{-3} {~\rm H})^2 +(5 ~\Omega)^2}  \\
 +                      &= 5.905... ~\Omega\\
 +\end{align*}
 +
 +#@HiddenEnd_HTML~d9io924n0e3du21g_11,Path~@#
 +
 +#@HiddenBegin_HTML~d9io924n0e3du21g_12,Result~@#
 +  * $|\underline{Z}_{RL}| = 5.90  ~\Omega$
 +  * $\cos\varphi = 0.84673$
 +#@HiddenEnd_HTML~d9io924n0e3du21g_12,Result~@#
 +
 +2. Calculate the true power, the apparent power, and the reactive power of the motor.
 +
 +#@HiddenBegin_HTML~d9io924n0e3du21g_21,Path~@#
 +The apparent power $S$ is given by 
 +\begin{align*}
 +S &= 3 \cdot U_s \cdot I_s \\
 +  &= 3 \cdot {{U_s^2}\over{Z_{RL}}} \\
 +  &= 3 \cdot {{(230~\rm V)^2}\over{ 5.90  ~\Omega }} \\
 +  &= 26.898... ~\rm kVA \\
 +\end{align*}
 +
 +The active power is
 +\begin{align*}
 +P &= S \cdot \cos \varphi \\
 +  &= 26.898... \cdot 0.84673 ~\rm kW \\
 +  &= 22.775... ~\rm kW \\
 +\end{align*}
 +
 +The reactive power is
 +\begin{align*}
 +Q &= \sqrt{S^2 - P^2} \\
 +  &= \sqrt{ (26.898... ~\rm kVA)^2 - (22.775... ~\rm kW)^2} \\
 +  &= 14.310...~\rm kVAr \\
 +\end{align*}
 +
 +#@HiddenEnd_HTML~d9io924n0e3du21g_21,Path~@#
 +
 +#@HiddenBegin_HTML~d9io924n0e3du21g_22,Result~@#
 +  * active power:  \\ $P=22.775~\rm kW$ \\ \\
 +  * reactive power:\\ $Q=14.310~\rm kVAr$ \\ \\
 +  * apparent power:\\ $S=26.898~\rm kVA$ \\ \\
 +#@HiddenEnd_HTML~d9io924n0e3du21g_22,Result~@#
 +
 +#@TaskEnd_HTML@#