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task_underseacable [2023/04/11 07:56] – angelegt mexleadmintask_underseacable [2023/04/11 23:02] (aktuell) – mexleadmin
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-Sure, here's the answer with the values and formulas in LaTeX notation:+Undersea cable capacity.
  
 +
 +You are working as an electrical engineer for a company that is planning to lay a power cable between two coastal cities that are $400 ~\rm km$ apart. The company wants to know what the maximum capacity of the cable should be to meet their power requirements. You have been tasked with calculating the cable's capacity.
 +
 +Assume that the cable's resistance is $0.1 ~\Omega$ per km, and that the voltage at the source end of the cable is $500 ~\rm kV$. The power factor of the cable is $0.8$, and the load at the destination end of the cable is $800~\rm MW$.
 +
 +Calculate the maximum capacity of the cable in $\rm MW$, assuming that the cable is operating at its maximum capacity.
 +
 +Provide your answer with a brief explanation of your calculations.
 +
 +Note: You may assume that the cable is a single-phase AC cable.
 +
 +#@HiddenBegin_HTML~1,Result~@#
 Given: Given:
  
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 \begin{align*} \begin{align*}
-R_{\rm total} &= R \times L \ +R_{\rm total} &= R \times L \\ 
-&= 0.1~{\rm \Omega/km} \times 400~{\rm km} \+&= 0.1~{\rm \Omega/km} \times 400~{\rm km} \\
 &= 40~{\rm \Omega} &= 40~{\rm \Omega}
 \end{align*} \end{align*}
Zeile 21: Zeile 33:
  
 \begin{align*} \begin{align*}
-I &= \frac{V_s}{R_{\rm total}} \ +I &= \frac{V_s}{R_{\rm total}} \\ 
-&= \frac{500~{\rm kV}}{40.01~{\rm \Omega}} \+&= \frac{500~{\rm kV}}{40~{\rm \Omega}} \\
 &= 12.5 ~{\rm kA} &= 12.5 ~{\rm kA}
 \end{align*} \end{align*}
Zeile 29: Zeile 41:
  
 \begin{align*} \begin{align*}
-P &= V_s \times I \times \cos \phi \ +P &= V_s \times I \times \cos \phi \\ 
-&= 500~{\rm kV} \times 12.5~{\rm kA} \times 0.8 \+&= 500~{\rm kV} \times 12.5~{\rm kA} \times 0.8 \\
 &= 5,000~{\rm MW} &= 5,000~{\rm MW}
 \end{align*} \end{align*}
Zeile 37: Zeile 49:
  
 \begin{align*} \begin{align*}
-Q &= V_s \times I \times \sqrt{1 - \cos^2 \phi} \ +Q &= V_s \times I \times \sqrt{1 - \cos^2 \phi} \\ 
-&= 500~{\rm kV} \times 12.5~{\rm kA} \times \sqrt{1 - 0.8^2} \+&= 500~{\rm kV} \times 12.5~{\rm kA} \times \sqrt{1 - 0.8^2} \\
 &= 2,500~{\rm MVAr} &= 2,500~{\rm MVAr}
 \end{align*} \end{align*}
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 \begin{align*} \begin{align*}
-S &= \sqrt{P^2 + Q^2} \ +S &= \sqrt{P^2 + Q^2} \\ 
-&= \sqrt{(5,000~{\rm MW})^2 + (2,500~{\rm MVAr})^2} \+&= \sqrt{(5,000~{\rm MW})^2 + (2,500~{\rm MVAr})^2} \\
 &= 5,590.17~{\rm MVA} &= 5,590.17~{\rm MVA}
 \end{align*} \end{align*}
  
 Therefore, the maximum capacity of the cable is $S = 5,590.17~{\rm MW}$, which is greater than the required power of $P = 800~{\rm MW}$. Therefore, the maximum capacity of the cable is $S = 5,590.17~{\rm MW}$, which is greater than the required power of $P = 800~{\rm MW}$.
 +#@HiddenEnd_HTML~1,Result~@#
 +