Energy and Lighting Lab - University of Michigan



ENERGY AND LIGHTINGUM Physics Demo Lab 07/2013Pre-Lab QuestionWhy do people say you should turn the lights off when you leave the room?EXPLORATION: Lighting MaterialsRegular (incandescent) 40 W light bulbRegular (incandescent) 60 W light bulb40W equivalent compact fluorescent bulbLED light bulbDigital voltmeter with alligator leadsSolar panelLamp base with cordInduction ammeter – clamp-onLight/lux meterCalculatorClear rulerMeter stickRag (for handling warm light bulbs)One third of all the energy used in the USA goes for lighting. In this experiment we will explore three different kinds of lighting and their efficiency (that is, how much light each produces for a certain amount of electrical power). We will also explore the use of solar energy panels which convert energy from light into electrical energy. Screw the 40W incandescent (regular old fashioned hot filament) bulb in the lamp base and turn it on. Bring your hand to within a few inches of the bulb. DO NOT TOUCH the bulb. What do you feel? Why?We can measure the power supplied to each bulb, by measuring the current and knowing the voltage (110 volts). Since the wall voltage (110 Volts) is dangerous, we need a non-contact way to measure the current. Do NOT TOUCH the bare contacts of the meter to the light bulb, or base or wall. We will instead measure the current using the induction ammeter provided. This meter clips around the wires attached to the lamp holder. Wrap one lead of the lamp cord through the induction meter 10 times. What is the relationship between the actual current flowing in the lamp cord and the current reported on the meter, given that the cord passes through the meter 10 times?Induction Meter Reading: __________ ampsActual current Supplied to the Incandescent bulb: ___________ ampsPower supplied to incandescent bulb:Power = Current (amps) x Voltage (Volts) = ____________ watts3. Does the power supplied to the bulb match the 40 Watts rating printed on the bulb? Why or why not?We will explain how the induction ammeter works based on coils and generators in a few weeks. Metal detectors at the airport work in very much the same way. 4. Unplug the lamp base, unscrew the incandescent bulb (careful it is hot) and insert the compact fluorescent bulb. Plug the lamp base back into the wall outlet and turn on the light. Again bring your hand within a few inches of the bulb. Is it giving off less or more heat than the incandescent bulb? 5. Measure the current for the compact fluorescent bulb and determine the power supplied to it: (also record the power in the table in question 7 below) Current = ____________ amps Power = Current x Voltage = ____________ watts 6. Repeat steps 4 and 5 for the LED bulb. Power supplied to LED bulb = Current x Voltage = ___________ watts Which of the three bulbs (incandescent, compact fluorescent, or LED) produces the most heat? And which appears to give the most light (is the brightest)?Most Heat ___________________________Most Light ___________________________7. Using the figures from parts 2, 5 and 6, calculate the cost per hour to operate each of the bulbs: (May 2013 Cost of power = $0.13/kWh) Show your calculations below the table.Type of bulbPower Supplied to Bulb (kW)Lifetime Hours(hr)Energy Consumed During Lifetime (kW-hr)Cost of Energy Consumed During lifetime ($)Purchase Cost Per Bulb($)Cost Per Hour of Operation ($)Incandescent1000$0.98Compact Fluorescent10,000$2.74LED50,000$12.86Assuming 6 hours per day of lighting, what is the cost for each light bulb over a 1 year period? Total hours used = 6 hours per day x 365 days = 2,190 hours. Show your calculations below the table. Remember to make sure you factor in the number of bulbs you need for a full year.Type of bulbCost per hour of Operation($)Cost to Run One Bulb for 1 year ($)Purchase Cost Per Bulb x#of Additional Bulbs Required for 1 Year($)Total Cost for One Year.($)IncandescentCompact FluorescentLEDWhich bulb is the cheapest to operate over a year? 8. Assuming they perform equally well for their task (lighting), which is the most efficient in terms of energy usage? (Check with your instructors, but move on for now if they are busy)9. Now we will quantify this a bit by measuring the brightness. We do that using a photocell. This is a device that converts light energy into electrical energy. Set the digital multimeter on DC volts and clip the multimeter leads across the photocell output wires (located on the back of the photocell).For each of the three bulbs, hold the bulb 30 cm above the photocell, using the lamp base and cord to power the bulbs. For each bulb, measure the photocell voltage using the digital multimeter. Repeat the measurements using the light meter. Record all your data in the table below. Note: The light meter measures light output in lux or lumens/m2 (equivalent to power/area or w/m2). Compare each measured photocell voltage with the light meter reading. The light meter is just a photocell with its output voltage calibrated in different units (the photocell voltage is calibrated to correspond to w/m2 of light output). Type of bulbPhotocell Voltage (V)Light Meter Output (Lumens/m2 ≡ lux ≡ w/m2)IncandescentCompact FluorescentLED10. Are the photocell and light meter measurements consistent? Are these results what you expected? Everyday Applications - Street lighting (some streets in Ann Arbor are lit by LED’s) - LED traffic lights (long lifetime gives more reliability) - LED automotive tail lights - Home lighting. Compact fluorescents are now common and LEDs will eventually become the lighting of choice - Adjustable color lighting (mix red, green, and blue LEDs) - Efficient flashlights powered by energy stored in capacitors and generated by mechanical work done when shaking the flashlight (Faraday induction)-no batteries required.Solar Power Everyday Applications- Power for watches and calculatorsSolar powered traffic signs on highways (needs a battery)Solar powered residential outdoor lighting (needs a battery)Providing power in remote areas far from the electrical supply grid Powering electric fences on farmsRecharging batteries for cell phones and lap-top computers APPLICATION: SOLAR ENERGY CONVERSION1. The photocell you are using for this experiment can also be used to convert solar energy to electrical power. Scaled up and placed on the roofs of buildings, such panels can generate enough electrical power to supply your daily needs for heating, lighting and cooking, etc. At the present time they are quite expensive, but costs are coming down as more are produced. To approximate exposure to bright sunlight, shine the 60W light bulb on the solar panel from a distance of 10 cm.Measure the voltage Vsun Vsun = ___________volts Now, with the multimeter on the current setting (mA) measure the current I = ___________ milliamps = ______________ amps Psun = I x Vsun = ____________watts In Ann Arbor, the sun provides about 700 W of power to an area of 1 square meter at the brightest part of the day in midsummer. The area of your solar cell is only about 7 cm x 4 cm or 0.0028 m2, so you would only expect to collect 700 (w/m2) x 0.0028 (m2) = 1.96 watts with this small photocell in bright sunlight. How does this compare with what you actually measured above (Psun)? The efficiency of solar cells is an important factor in their usefulness as an alternate source of energy. Calculate the efficiency of your solar cell as: Efficiency = Psun/ 1.96 watts =____________ x 100 = ___________% Manufacturers strive to make solar cells that are as efficient as possible. The cell you are using, based on Polycrystalline Silicon, is one of the first versions available. It is still far from 100% efficient, as you can see. There will be a very big effort worldwide to improve solar energy cells over the next few years. Typical commercial solar cells are currently 18 – 20% efficient. Challenge Work:What would be the area in square meters of an array of solar cells which generates 5,000 Megawatts of power (typical for conventional power generating plant burning coal)? Remember that 1 Megawatt (Mw) is one million watts = 1,000,000 w = 106 w. Use the power per unit area you determined for the actual solar cells we used in class today = Psun/(area of solar cell). Assuming this area corresponds to a square array of cells calculate the length of one side of the array by taking the square root of the area you calculated. Finally, divide this length by 1000m to express it in kilometers (1 km = 1000 m). Summary Engineering trade studies are a tool for making informed technical decisions.The Kilowatt-hour is a unit of energy. 1kw-hr = 3.6 x 106 Joules.Lighting energy efficiency can be greatly improved by replacing incandescent bulbs with new technologies such as compact fluorescent tubes and LEDs.Photovoltaic cells produce electricity directly from sunlight and can be used for practical power generation. The major impediment to their more widespread use is cost, not technical feasibility. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download