The interior lighting of refrigerators is usually provided...

The interior lighting of refrigerators is usually provided by incandescent lamps whose switches are actuated by the opening of the refrigerator door. Consider a refrigerator whose 40 -W lightbulb remains on about 60 h per year. It is proposed to replace the lightbulb by an energy-efficient bulb that consumes only 18 W but costs $\$ 25$ to purchase and install. If the refrigerator has a coefficient of performance of 1.3 and the cost of electricity is 8 cents per $\mathrm{kWh}$, determine if the energy savings of the proposed lightbulb justify its cost.

Key Concepts

Energy Consumption Calculation

This concept involves determining the total energy used by a device over a period by multiplying its power rating (in watts) by the length of time it is operated (converted to hours). Energy is commonly expressed in kilowatt-hours (kWh), which is a standard unit for billing and comparing electrical energy use.

Electricity Cost Calculation

Understanding how to calculate the cost of electricity involves multiplying the energy consumption (in kWh) by the unit price of electricity. This provides a direct measure of the monetary expense associated with operating an electrical device over a given period.

Energy Efficiency and Savings

This concept compares the energy consumption of two devices performing the same function but with different power requirements. By evaluating the difference in energy use and associated costs, one can determine the potential savings from using a more efficient device. This is fundamental when assessing upgrades or replacement of equipment in terms of both environmental and economic benefits.

Break-even Analysis

Break-even analysis determines the point at which the savings achieved from reduced energy consumption equal the initial cost of investing in a more efficient device. This concept helps in deciding whether the investment in energy-efficient technology is financially justified over a given time horizon.

Coefficient of Performance

The coefficient of performance (COP) measures the efficiency of a system, such as in refrigeration, by comparing the useful output (e.g., cooling effect) to the work input. It is a crucial concept when assessing the overall energy impact of a device, as lower wattage components can also reduce the load on energy-intensive systems like refrigeration.

Transcript

00:01 So here in this problem we are given that there is a refrigerator in which a light bulb is installed and which consumes 40 watt of light energy and it is remained open for 60 hour in a year and we are proposed to replace it with a light bulb which consumes only 18 watt and it also is being open for 60 hour per year and we are given that the cost of of this the second light bulb it is $25 and the cop of the refrigerator we are given as 1 .3 and the cost of electricity is given as 8 cents per kilowatt hour or we could say that is equals to 0 .08 dollar per kilowatt hour and we have to determine the energy saving of the light bulb justify its cost or not.

01:05 So for that we have to calculate the total energy saved by the this light bulb the second light bulb but for that before that we have to calculate the light energy saved by using the second light bulb so this is a light energy saved so it will be equals to the lighting power saved and lighting power saved would be like this is 40 watt it consumed and it consumes 18 watt so power it the light power saved is 14 minus 18 watts and it runs for 60 hour in a year that means times 60 hour in a year so this much of light energy is saved in one year if we use this light bulb and it comes out to be as nearly 1 ,320 kilowatt hour or we could say 1 .32 kilowatt hour per year this much of light energy we are able to save if we are using using the second light bulb next we'll be calculating the refrigeration energy saved refrigeration energy so it will be equals to the light energy saved divide by cob so we have all the values here this one is 1 .3 to kilowatt hour per year divide by 1 .3 so it could it will come out to be as nearly 1 .0 to kilowatt hour per year so this much of energy the refrigeration energy we are able to save so total energy saved it will be equals to some some the submission of this one and this one that is 1 .32 kilowatt hour per year plus 1 .0 2 kilowatt hour per year so it will come out to be nearly 2 .34 kilowatt hour per year so this much of total energy we are able to save next will be cal calculating the money saved by saving this much of energy so the money saved will be equals to the total energy saved this much of energy saved 2 .34 kilowatt hour and divide by a kilo watt hour per year that much of energy is saved in a year times the cost we are given as 0 .08 per kilowatt hour so it would become out to be nearly 0 .19 dollar per year so we are only we are able to save only point to one nine dollar in a year so if we see the total year that would be needed to recover twenty five dollar it will be equals to twenty five dollar divide by zero point one nine dollar so it would be nearly one thirty two year so this much of year is needed to cover this twenty five dollar of cost so i would recommend that this is not justified.

04:46 This cost is not justified...

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