3

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

10 Comments: The comparison between Fig.5 and Fig.6 shows that the series resistance affects greatly the output power of the cell. This because of the voltage drop across it. For the effect of the parallel resistance on the P-V curves obtained, one can expect the power loss at reduced scale since the current flowing through it will cause Ohmic losses. However in order to clearly see the effect of the parallel resistance on the P-V curve, one should compare the obtained P-V curve (having parallel resistance) with the ide -V curve. Nevertheless, both series and parallel resistances reduces the output power of the cell. ELEC 437/ ELEC 6421 Renewable Energy Systems Homework No. (3) - Solution Problem 1 Find the optimal angle to orient a solar panel in your hometown to obtain the maximum power between the first day of spring and the first day of fall. Solution: You should choose the latitude of your hometown. You should choose the latitude of your hometown. Problem 2 Find the optimal angle to generate power at 1pm (solar time) in Montreal on the 30 th of May. Solution:

Question No. 3: Given the schematic diagram of an off-grid PV-connected house which is located in Montreal in the figure below. a. Use Homer simulation software (V.3.11.2) to find a technical feasible solution considering the following conditions: Location: Montreal Eastern Time (UTC-05:00 US & Canada) Location: Solar Global Horizontal Irradiance (GHI): download from internet Components: PV panel: - Selecting/adding the CanadianSolar MaxPower CS6U-330P as an example - The capital cost of 1kW PV installation is $ 6000 - PV capacity 0, 0.330, 0.660, 0.990, 1.320, 1.650, 1.980, 2.310, 2.640, 2.970, 3.300, 3.630, 3.960 and 4.290 kW with DC electrical bus - No tracking System and the use of default panel slope. Storage (battery): - From Discover Energy storage type, choose Discover 12VRE-3000TF-L (12 V, 1 kWh) battery as an example - choosing a capital cost of $ 410 - selecting between 0, 1, 2, 3, 4, up to 10 sets Daily load profile: - the daily DC load profile is specified in the charts given below for weekdays and weekends, - for Random Variability option, choose 10% Day-to-day and a 20% timestep Then, press the Calculate button to obtain the results The computed optimized solution shows a configuration of 2 panels and 3 batteries. For more details, you can double-click on each solution! b. Based on the electrical load, the daily energy consumption is 2.48 kWh/day, which corresponds to an annual load of 905.2 kWh/year. When looking at available possibilities, one can observe that the optimal configuration according to the electrical load is composed of 2 solar panels and three batteries. As expected, the optimal orientation for the panel is an azimuth of 0 and a slope of 45.5 , which corresponds to the latitude of the location. The less optimal configuration is composed of 3 panels and an array of 2 batteries. The

12 D) The annual electricity consumption is measured as 180m 2 *230 kWh/m 2 = 41,400 kWh. If we want to reduce the total annual fuel consumption by 25%, the energy saved by the panels must be equal to: As we can see, increasing the number of panels to 46, one can see that 25% savings of energy Fig. 6. PV System Parameters ELEC 437/ ELEC 6421 Renewable Energy Systems Homework No. (4) - Solution Submission format: an unzipped, unencrypted, readable and appropriate .pdf file ONLY. Problem 1 An industrial freezer is designed to operate with an internal air temperature of -20 0 C when the external air temperature is 25 0 C and the internal and external heat transfer coefficients are 12 W/m 2 K and 8 W/m 2 K, respectively. The walls of the freezer are composite construction, comprising of an inner layer of plastic (k = 1 W/m K, and thickness of 3 mm), and an outer layer of stainless steel (k = 16 W/m K, and thickness of 1 mm). Sandwiched between these two layers is a layer of insulation material with k = 0.07 W/m K. Find the width of the insulation that is required to reduce the convective heat loss to 15 W/m 2 . Solution: 2 /W. The total thermal resistance is due to the convective heat flow inside and outside the freezer and conductive heat flow in the plastic layer, insulation layer and the stainless steel layer. This total resistance is given by; Solving the above equation, we can obtain the thickness required of the insulator,