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Alekssandra [29.7K]
2 years ago
9

A 60 Hz three-phase 115kV transmission line has a series impedance of 30+j150 ohms/phase. The shunt susceptance of this line is

phasemhosBc/42. The line to neutral voltages at buses p and q are kVVp0.651.63and kVVq006.65, respectively. Compute lossP, the three phase real power losses in MW for this transmission circuit.

Engineering
1 answer:
bearhunter [10]2 years ago
5 0

The complete and the correct format for the question is as follows:

A 60 Hz three-phase 115kV transmission line has a series impedance of 30+j150 ohms/phase. The shunt susceptance of this line is \dfrac{B_c}{2} = 4 \ mhos /phase

The line to neutral voltages at buses p and q are and respectively. Compute , the three-phase real power losses in MW for this transmission circuit.

Answer:

0.183 MW

Explanation:

The diagrammatic illustration of the given information can be seen in the image attached below:

Thus: the current in the network can be computed as:

I =\dfrac{V_p-V_q}{Z +\dfrac{1}{Y}}

I =\dfrac{63.51 \times 10^3 \angle 6^0-656.06 \times 10^3 \angle 0^0 }{30 + j 150 -j0.125}

I  = 45.173 \angle 27.28^0 \ A

The P_{loss can be computed as:

P_{loss} = 3 \times I^2 \times  R

P_{loss} = 3 \begin {bmatrix} 45.173 \end {bmatrix} ^2  \times 30

Finally; the three-phase real power loss is:

P_{loss \ of  \ sphere} = 0.183  \ MW

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In[7] def pegasos(feature_matrix, labels, T, L):

   """

   .

   let learning rate = 1/sqrt(t),

   where t is a counter for the number of updates performed so far       (between 1   and nT inclusive).

Args:

       feature_matrix - A numpy matrix describing the given data. Each row

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       labels - A numpy array where the kth element of the array is the

           correct classification of the kth row of the feature matrix.

       T -  the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

       L - The lamba valueto update the pegasos

   Returns: Is defined as a  tuple in which the first element is the final value of θ and the second element is the value of θ0

   """

   (nsamples, nfeatures) = feature_matrix.shape

   theta = np.zeros(nfeatures)

   theta_0 = 0

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       for i in get_order(nsamples):

           count += 1

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           (theta, theta_0) = pegasos_single_step_update(

               feature_matrix[i], labels[i], L, eta, theta, theta_0)

   return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

   labels = np.array([1, 1])

   T = 1

   L = 1

   exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

   

   pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

Explanation:

In[7] def pegasos(feature_matrix, labels, T, L):

   """

   .

   let learning rate = 1/sqrt(t),

   where t is a counter for the number of updates performed so far       (between 1   and nT inclusive).

Args:

       feature_matrix - A numpy matrix describing the given data. Each row

           represents a single data point.

       labels - A numpy array where the kth element of the array is the

           correct classification of the kth row of the feature matrix.

       T -  the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

       L - The lamba valueto update the pegasos

   Returns: Is defined as a  tuple in which the first element is the final value of θ and the second element is the value of θ0

   """

   (nsamples, nfeatures) = feature_matrix.shape

   theta = np.zeros(nfeatures)

   theta_0 = 0

   count = 0

   for t in range(T):

       for i in get_order(nsamples):

           count += 1

           eta = 1.0 / np.sqrt(count)

           (theta, theta_0) = pegasos_single_step_update(

               feature_matrix[i], labels[i], L, eta, theta, theta_0)

   return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

   labels = np.array([1, 1])

   T = 1

   L = 1

   exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

   

   pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

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