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task_underseacable [2023/04/11 07:59] – mexleadmintask_underseacable [2023/04/11 23:02] (aktuell) – mexleadmin
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-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 km apart. The cable will be placed at a depth of 1000 meters below the ocean's surface. 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.+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 ohms per km, and that the voltage at the source end of the cable is 500 kV. The power factor of the cable is 0.8, and the load at the destination end of the cable is 800 MW.+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 MW, assuming that the cable is operating at its maximum capacity.+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. Provide your answer with a brief explanation of your calculations.
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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*}
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 \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*}
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 \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*}
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 \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*}