Renewable Energy - Self-Built Solar Water Heater - Essay Example

RENEWABLE ENERGY PART ASSUMPTIONS A self-built solar water heater, was made by one of our recent and tested outdoors, with beam radiation nearly normal to the plane of the collector. The test results obtained are contained in the Excel spreadsheet (ENE800J1 Wk 1 c/w.xls). The following details will be necessary to complete this coursework. -A constant mass flow rate of 0.00755 kg/sec was used throughout the test. -The collector aperture area was 0.47m2. - An average value of Cp between the maximum and minimum fluid Temperatures may be used throughout, (4.1784 kJ/kg K) - linear regression analysis should be used to find the best-fit line, from which you can determine the following constants. First measurement, Qu = mCp (To - Ti). Second measurement= Gt & Ta. Qu = A FR [GT (t)e - UL(Ti - Ta)] Equation (2) A = Collector area (m2) FR = Heat removal factor (normally < 1) GT = Incident total global solar radiation (W/m2) = Transmission of the glazing (typically 0.9 for glass) = Absorptance of the absorber (typically 0.85 for black painted copper) (t) e = The transmission and absorption are angle dependent. As we are only considering a short period either side of solar noon we use the effective (t)e product. UL = Heat transfer (loss) coefficient from collector (depends in Levels of insulation behind back of absorber) Ta = Ambient temperature. Ti = Collector fluid inlet temperature ( rises with decreasing Ti) To = Collector fluid outlet temperature ( falls with increasing To) M=mass flow rate. And, = Qu / A GT The values of & (Ti-Ta)/ GT were calculated. And tabulated & charted as shown below: (Ti-Ta)/ GT 1 0.4719 0.0051 2 0.4293 0.0073 3 0.4273 0.009 4 0.4099 0.0099 5 0.4258 0.0121 6 0.4502 0.0162 7 0.3919 0.0156 8 0.4226 0.0156 9 0.4172 0.0171 10 0.4081 0.0177 11 0.4000 0.0194 12 0.3946 0.0206 13 0.3634 0.0236 14 0.3347 0.0275 15 0.3547 0.0251 16 0.3493 0.0247 17 0.2814 0.0354 18 0.3028 0.0277 19 0.3727 0.0329 20 0.3095 0.0487 21 0.3237 0.0400 22 0.3279 0.0550 The corresponding graph was drawn. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 (Ti-Ta)/GT From the chart, FR ()=intercept=0.4912. (49.12%) FR UL=Slope=(4.912/(-8.236))=0.596 2. Assess the suitability of the most appropriately oriented part of your roof for installing a solar water heater. How will this affect the performance of the proposed solar water heaters (If you live in an apartment with a flat roof I guess you'll have to pretend that you live on the top floor!) Would there be any practical limitations to the installation of a SWH system. On the roof, the best part would be dependent on two factors: first, it has to be near or at the center part of the roof so that more amount of sunlight falls on the SWH for a longer period of time (from dawn to dusk). Second, it has to be at a requisite to keep it out of reach from children. The limitations in any case would be the amount of open space that can be found on the roof and whether that open space is well lit by sunlight for most part of the day. Also the cost of maintaining and the safety measures adopted play a critical role in the successful and efficient role of the SWH system. 3. Using your own house as an example estimate how much hot water your family consumes each year. (As a rough estimate determine how many showers, baths, and basins of hot water are used per week and make a pro-rata calculation for the annual hot water consumption. Remember that most hot water used would be mixed with the cold water. It's not important to be 100% accurate but this will give you an idea of the magnitude of the hot water usage. Average European hot water usage is currently in the range of 15-35 liters of hot water per person per day depending on if you shower or bath!) The rough estimate per week is as follows: Showers: 21 Liters (average per shower): 45 Bath: 15 Liters (average per Bath): 25 Basins of Hot Water: 20 Liters (average per Basin): 20. Total consumption hot water per week = 945+375+400=1720 liters. Total consumption annually (assuming 52 weeks in a year)=89440 liters. 4. Calculate the annual energy content of your conventionally heated water from the difference between the supply mains water temperature, (normally between 5C and 12C depending on location), and the final auxiliary heated temperature (normally in excess of 54C to avoid the risk of legionella bacteria forming in your hot water tank. Typically thermostats are set about 10C higher than this to meet health and safety requirements). Legionella grow in water temperatures between 20C and 50C, with optimal growth occurring between 35C and 46C. They will not multiply at temperatures below 20C, and cannot survive in water above 60C. Use an average specific heat capacity for water between these two temperatures of about 4.18 kJ/kg/C. C=specific heat capacity of water=4.18 kJ/ kg/C. T=Temperature Difference=64-((5+12)/2)=64-8.5C=55.5C. M=Mass of Water=volume of water in liters=89440 liters. Amount of heat consumed=M*C*T. =20749185.6 kJ. 5. For a typical 4m2 flat plate solar water heating system, and using typical annual insolations from the European solar radiation map from the solar thermal lecture in ENE805J1X last year, calculate how much energy you could hope to extract from both a self-built system and a flat plate system from an established manufacturer. I don't have the ENE805J1X lecture. This requires just the direct application of the formula. 6. What proportion of your hot water requirements can be met from such solar water heating systems You may be surprised at what you find! Calculate the ratio of answer of question 5 and question 4 and express it as a percentage. 7. Using the local cost of conventional auxiliary fuel for hot water heating calculate the financial contribution of both the DIY solar water heater and the flat plate solar water heating system from an established manufacturer. You may be even more surprised at what you find! Cost of conventional auxiliary fuel= $1.25. From question 5, we will obtain the energy that can be supplied by the DIY and flat panel heaters. We need to find out the energy that is supplied by this conventional fuel, and then apply the equation: Financial contribution=(energy from heater on an annual basis*1.25)/(energy supplied by unit amount of fuel). PART 2 A) Select at least two alternative PV systems from different manufacturers and obtain a quote from each manufacturer for a PV array suitable for domestic installation in your own home. (This will not always be easy. Many manufacturers are very reluctant to give even the most basic information but try your best and get in touch if/when you encounter any problems! Whatever you do, do not tell them that you are doing this as part of a piece of coursework!!) Alternatively you may select another suitable building so long as it is of a domestic scale i.e. no more than 200m2 floor area. The companies that provided us with quotes were: 1. PV systems Inc. 2. South West PV Systems. B) Select the most appropriate PV installation from the two quotes for your requirements detailing the reasons for your selection. The approximate electrical base load or total electrical load, depending on how much you intend to meet. This will determine the specification of your selection. Have any energy efficiency measures been considered to reduce this load still further How would you resolve the differences between the diurnal variations in supply and your current usage patterns Consider the following: - 1. Approximate monthly insolation for your location. (This will determine how much of the base load you decide to meet.) 2. Total area of panels required to meet the loads you decide to offset. (You may have to compare average monthly diurnal insulation with diurnal base load to establish PV area.) 3. The overall efficiency of the panels and the BOS to be installed calculated from the module area and max rated output of each module. 4. Total installed cost of panels, inverters and battery storage (if necessary). 5. Are there any site-specific maintenance issues We have chosen to go for the PV system offered by South West PV Systems. This is because of the load analysis and the cost offered by them. We offered both the companies the detailed load requirements of our home that were desired to be connected to the proposed PV system, a copy of which is shown below: Appliance Qty Volt AC/DC Run W Hrs/d WHrs/d Surge W Ph-L Fluorescent Lights 4 117 AC 60 5.0 1200 0 N 16cu ft Fridge 1 12 AC 141 10.0 1410 1300 N Blender 1 117 AC 350 0.1 10 1050 N Microwave Oven 1 117 AC 900 0.3 225 1200 Y Food Processor 1 117 AC 400 0.1 28.6 1200 N Espresso Maker 1 117 AC 1350 0.1 135 1350 N Coffee Grinder 1 117 AC 150 0.1 7.5 200 N 21" Color TV 1 117 AC 125 5 625 570 Y Stereo System 1 117 AC 30 8 240 60 Y Computer 1 117 AC 45 6 115.7 135 Y Power Tool 1 117 AC 750 0.5 160.7 2250 N Radio Telephone (receive) 1 12 AC 6 24 144 0 N Radio Telephone (transmit) 1 12 AC 20 1 20 0 N Phone answering machine 1 117 AC 6 24 144 0 N Washing machine 1 117 AC 800 0.5 228.6 100 Y Clothes dryer (motor only) 1 117 AC 500 1 285.7 1500 Y Sewing machine 1 117 AC 80 2 22.9 400 N Vacuum cleaner 1 117 AC 650 0.5 185.7 1950 N Hair dryer 1 117 AC 1000 0.2 200 1500 N Ni-Cd Battery Charger 1 117 AC 4 15 17.1 25 Y TOTAL 7367 The sole reason for choosing PV systems was the low cost of the equipment when compared to the other vendor. The other major reason was that the transporting cost was much cheaper, in the case of South PV systems. The various equipment that was offered by them were: PV array. Battery. Inverter. The total energy requirements were estimated to be around 3898.43900 watt-hours per day, which comes out to be 7367 watts. The average insolation values on a yearly basis are estimated to be around 5.5 hours per day. Therefore, the PV array size is given by (3894.8/5.5)=708.15 watts. The efficiency of the panels is around 20%. The cost per array=$5.57. The amp-hour of the battery (we assume a shallow-cycled discharged battery and estimated to be around 5 times the daily load and the voltage of the battery is 12 volts.)=(7367*5)/12=3069 amp-hours. Cost of using battery= 3069*0.65 = $2915.55 Cost of using inverter power=(size of PV array * $0.95 per watt)=708.15* 0.95=$672. The total cost of PV array =5.57*708.15=$3944. Therefore, the total cost=3944+672+2916 =$7532 Balance of System cost=(0.18 * total cost)= $1355 Therefore, the total cost= $8887. The only consideration on-site must be that there must be good insulation of all the wiring along with efficient energy designs. Also, the orientation and proper shading of the PV array must be appropriate enough. Calculation of cost of electricity produced for 20, 30 & 60 years A) Calculate the cost/kWh of the electricity produced by the PV system for lifetimes of 20, 30 and 60 years. The average consumption in terms of kWh for the household is: 3.9 kWh. For a year, it would be 1423 kWh. Money invested at the time of installation would be $8887. Total energy requirements for 20 years would be= 28460 kWh. Total energy requirements for 30 years would be=42690 kWh. Total energy requirements for 60 years would be=85380 kWh. Therefore, power cost per KWH energy is= money invested/total power = $0.310 (for a 20-year lifetime). =$0208 (for a 30-year lifetime). =$0.104 (for a 60-year lifetime). B) Evaluate the cost effectiveness of your selected installation. - Is it worth it Thus, it can be seen that the cost has drastically reduced from $0.95 to $ 0.310 for a period of 20 years, and it is reduced as the lifetime of the PV system is increased. Hence, it is definitely worth it. - What are the most expensive components The most expensive components are the PV array and the inverter. - Can you be stand-alone No, we cannot be standalone, as the PV system can satisfy our needs, only when there is adequate sunshine, or when the battery and inverters are working. In the absence of these, we would have to depend on the usual power from the grid. PART 3 1. Use an atlas to estimate the hydro-potential of your country or province/state/region, as follows: (a) Call the place in question X. (b) What is the lowest altitude in X (c) What area of X lies more than 300 meters above the lowest level (d) How much rain falls per year on this part of X What would be the potential energy per year given up by this mass of water if it all ran down to the lowest level Express this in Megawatts. a) The country that I choose is INDIA. At present the hydroelectric project under consideration is the bhakra nangal dam in the northern part of India. b) The lowest altitude is 252 meters above sea level. c) Around 53% of the area lies above 300 meters than the lowest level. d) The total annual rainfall on this area equals 5 million kiloliters of water. The water falls through a height of 310 meters. So total energy generated would be 155+10^7 kJ per annum. So power generated would be 49.15 megawatts. 2. Refine this power estimate by allowing for the following: (a) Not all the rain that falls appears as surface runoff; (b) Not all the runoff appears in streams that are worth damming; (c) If the descent is at too shallow a slope, piping difficulties limit available head. a) It has been estimated that around 34% of the rainfall is wasted and uncollected as surface runoff. b) Out of the remaining 66%, around 12% of the water is not worth damming. Therefore, a net amount of 54% of the annual rainfall is estimated to be utilized for the production of hydroelectricity. So the total annual rainfall, which is termed as utilized, would be (0.56* 5 million=2.8 million) kiloliters. Therefore, the actual annual production would be 27.52 megawatts. 3. If hydroelectric power has in fact been installed in X, compare your answer with the installed capacity of X, and comment on any large differences. Upon consultation with some people who are known to be connected with the hydroelectric project, it has been found that the actual annual production of electricity is around 175 megawatts. This is a huge difference from the estimated value. The reason for this disparity has been attributed to the fact that the Bhakra Nangal dam is fed by a perennial river and as such there is a much larger inflow of water than what can be obtained from rainfall alone. Read More