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ELEC 437/ ELEC 6421 Renewable Energy Systems Homework No. (1) - Solution Problem 1 (a) When the capacitor is not connected = cos -1 0.75 = 41.41 o S = 4500 VA P = S cos o = 4500(0.75) = 3 375 W = S sin o = 4500(0.66) = 2 976.49 var lagging When the capacitor is connected in parallel with the impedance Z , the = 0.9 lagging = 25.84 o , and P = Therefore, 3 375 W = cos 3 375 W = (0.9) = 3 750 VA, = sin (cos -1 0.9) = 1 634.47 Var lagging Capacitor var ( ) = - = 2 976.49 1634.47 = 1342.02 var leading Since = V 2 /X c = V 2 /(1/ C) = V 2 C C = /( V 2 ) = 1342.02/(2 x 60 x 240 2 ) = 61.3 µ F (b) Desired PF is 0.9 leading. = sin (cos -1 0.9) = 1 634.47 Var leading Capacitor var ( ) = - = 2 976.49 (-1634.47) = 4 610.96 var leading Since = V 2 /X c = V 2 /(1/ C) = V 2 C C = /( V 2 ) = 4 610.96/(2 x 60 x 240 2 ) = 212.34 µ F Problem 2 Branch 1: I 1 = V/Z 1 = (20 o )/(4 o ) = 5 o A S 1 = V I 1 *= (20 o )*(5 o ) = 100 o VA P 1 = S 1 cos o = 100(0.87) = 86.6W Q 1 = S 1 sin o = 100(0.5) = 50 Var lagging PF 1 = P 1 /S 1 = 0.866 lagging Branch 2: I 2 = V/Z 2 = (20 o )/(5 o ) = 4 o A S 1 = V I 1 *= (20 o )*(4 o ) = 80 o VA P 1 = S 1 cos o = 80(0.5) = 40 W Q 1 = S 1 sin o = 80(0.866) = 69.2 Var lagging PF 1 = P 1 /S 1 = 0.5 lagging Then, the complete power triangle can be found as follows: P T = P 1 +P 2 = 86.6 + 40 = 126.6W Q T = Q 1 +Q 2 = 50 +69.2 = 119.2 Var lagging S T = P T +jQ T = 126.6 +j119.2 = 174 o VA PF T = 126.6/174 = 0.728 lagging Here is the total power triangle: Problem 3 Average Power Problem 4 1 ELEC 437/ ELEC 6421 Renewable Energy Systems Homework No. (2) - Solution Problem A Solar cell has following specifications: Area: 1.6 Is: Ideality factor: 1.0 Current density: 41 Temperature: 300 K ( 27 ° ) 1- is neglected. a- Calculate the characteristic resistance and new filling factor FF To find the characteristic resistance R CH and the fill factor FF , four values are needed which are the open circuit voltage short circuit current , voltage corresponding to the maximum power , and current corresponding to the maximum power I mp . To find , the load current I should be set to zero in Eq. (1). (1) Therefore, = 0.692V To find , the load voltage V should be set to zero in Eq. (1). Therefore.

2 By iteration, To find and , the maximum power should be found. The power of the cell can be calculated as follows: Hint : to draw P-V curve of the cell, just use P = V * I Fig. 1. P-V curve with only series resistance From the figure above, The value of can be calculated as follows: = 0.1671 A Now, the characteristic resistance R CH will be found by using the following equation: Which leads to a value of R CH = 2.09 . The fill factor FF can be calculated according to the following equation: Which leads to a value of FF = 0.251 3 b- Calculate the maximum power Pmax. The maximum power can be directly determined from the P-V curve. P max = 0.0585 W. c- Find the slope of the I-V curve near VOC. To find the slop, two arbitrary points near V OC on the I-V curve can be selected. The two points are: a(0.6511, 0.02), b(0.6715,0.01) By using the following equation: The slope ( m ) is found to be (-0.49). d- How does the slope compare with the value of the series resistance? Increasing the series resistance would decrease the slope near Voc. 2- Repeat the questions 1-a. and 1-b. by assuming a parallel resistance of the cell is 10 series resistance can be neglected. 1-a Calculate the characteristic resistance and new filling factor FF To find the characteristic resistance R CH and the fill factor FF , four values are needed which are the open circuit voltage V oc , short circuit current I sc , voltage corresponding to the maximum power V mp , and current corresponding to the maximum power I mp . To find V oc , the load current should be set to zero in Eq. (2). (2) We have: (Negative term can be neglected) By iteration, the value of the open circuit voltage is found to be To find , the load voltage V should be set to zero in Eq. (2), so, 4 By substituting the given values, will be found to equal 0.64 A To find and , the maximum power should be found. The power of the cell can be calculated as follows: Fig. 2. P-V curve with only parallel resistance and the corresponding voltage The value of can be calculated as follows: = 0.6247 A Now, the characteristic resistance R CH will be found by using the following equation: Which leads to a value of R CH = 0.972 . The fill factor FF can be calculated according to the following equation: 5 Which leads to a value of FF = 0.8366 1-b Calculate the maximum power Pmax. The maximum power can be determined from the P-V curve of Fig.2 which shows a value of P max = 0 a. Find the slope of the I V curve near I sc To find the slop, two arbitrary points on the I-V curve can be selected. The two points are: a(0.0105,0.6559), b(0.0863, 0.6551) By using the following equation: The slope m is found to be (-0.010). b. How does it compare with the value of the parallel resistance? Decreasing parallel resistance would increase the slope near I sc . 3- Assuming a parallel resistance of 10 series resistance of . a. Find the To find V oc , the load current should be set to zero in Eq. (3). (3) We have: (-1 term can be neglected) By iteration, the value of the open circuit voltage is: Voc= 0.6916 V b. Find the To find I sc , the load voltage should be set to zero in Eq. (3), so

6 By iteration, Isc = 0.3364 A 4- Draw the I-V curves of the cell as described in question 1and 2 in the same graph. Comment on the effect of series resist The equation describing the P-V characteristic of the cell is described as follows: By plugging different values of , the corresponding values of V can be obtained through iteration. Hint: For the I-V curve of question 1, use : For the I-V curve of question 2, use: 7 Fig. 3. I-V curve with only series resistance Fig. 4. I-V curve with only shunt resistance 8 Comments: The series resistance greatly affects the output voltage of the cell. It reduces the short circuit current of the cell. However it does not affect the open circuit voltage. The shunt resistance has a negligible effects on the output current of the cell compared to the effect of the series resistance output voltage. The shunt resistance slightly reduces the short circuit current of the cell. However, it does not affect the open circuit voltage. 5- Draw the P-V curves of the cell as described in question 1 and 2 in the same graph. Comment on the effect of series resistance and shunt resistance on the P-V curves obtained. The equation describing the P-V characteristic of the cell is described as follows: By plugging different values of , the corresponding values of I can be obtained via iteration. Then to obtain the values of the corresponding power, one can multiply the obtained current by the voltage. Hint: For the P-V curve of question 1, use: ) For the P-V curve of question 2, use: 9 Fig. 5. P-V curve with only series resistance Fig. 6. P-V curve with only shunt resistance

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