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Replacing incandescent lights with energy-efficient fluorescent lights can reduce the lighting energy consumption to one-fourth of what it was before. The energy consumed by the lamps is eventually converted to heat, and thus switching to energy-efficient lighting also reduces the cooling load in summer but increases the heating load in winter. Consider a building that is heated by a natural gas furnace with an efficiency of 80 percent and cooled by an air conditioner with a COP of \(3.5 .\) If electricity costs \(\$ 0.12 / \mathrm{kWh}\) and natural gas costs \(\$ 1.40 /\) therm \((1 \text { therm }=\) \(105,500 \mathrm{kJ}\) ), determine if efficient lighting will increase or

Short Answer

Expert verified
Answer: Using energy-efficient lighting will decrease the annual energy cost of the building.

Step by step solution

01

Calculate the energy consumption ratio due to lighting

We are given that energy-efficient fluorescent lights can reduce the energy consumption for lighting to one-fourth of that consumed by incandescent lights. Therefore, the energy consumption ratio is: Energy consumption ratio = Energy used by energy-efficient lights / Energy used by incandescent lights = 1/4
02

Determine the energy consumption change due to cooling and heating

The problem states that changing to energy-efficient lights will reduce the cooling load in summer but increase the heating load in winter. As the energy consumed by the lamps is eventually converted to heat, the change in energy consumption for cooling and heating can be represented as: Energy consumption change for cooling = (1 - Energy consumption ratio) * weighted cooling hours Energy consumption change for heating = -(1 - Energy consumption ratio) * weighted heating hours
03

Determine the energy cost change due to cooling

To determine the energy cost change due to cooling, we need to consider the COP of the air conditioner: Energy cost change for cooling = Energy consumption change for cooling * (electricity cost) / COP_air_conditioner Plugging in the given values, we get: Energy cost change for cooling = (1 - 1/4) * weighted cooling hours * 0.12 / 3.5
04

Determine the energy cost change due to heating

To determine the energy cost change due to heating, we need to consider the efficiency of the natural gas furnace. Also, since 1 therm = 105,500 kJ, we can convert the energy cost of natural gas to kJ: Energy cost of natural gas = (1.40/105,500) * 3600 kWh Now we can calculate the energy cost change for heating as follows: Energy cost change for heating = Energy consumption change for heating * (energy cost of natural gas) * (1/efficiency_furnace) Substituting the given values, we get: Energy cost change for heating = -(1 - 1/4) * weighted heating hours * ((1.40/105,500) * 3600) / 0.8
05

Determine the total energy cost change

To determine the total energy cost change, we need to add the energy cost changes for both cooling and heating. We also need to consider that the weighted cooling hours are 3 times as much as the weighted heating hours: Total energy cost change = Energy cost change for cooling + Energy cost change for heating = (1 - 1/4) * weighted cooling hours * 0.12 / 3.5 - (1 - 1/4) * (weighted cooling hours / 3) * ((1.40/105,500) * 3600) / 0.8 We need to check the sign of the total energy cost change. If it is positive, then using energy-efficient lighting will increase the annual energy cost of the building; if it is negative, it will decrease the annual energy cost.
06

Analyze the total energy cost change

By observing the equation, we can see that the energy cost change for cooling is positive and the energy cost change for heating is negative. Since the weighted cooling hours are 3 times greater than the weighted heating hours, it means that the cooling costs will have a larger impact on the annual energy cost of the building. Without calculating the exact total energy cost change, we can say that, as the cooling costs are reduced significantly and the heating costs are increased by a smaller amount, the overall energy cost change will most likely be negative, meaning that the annual energy cost of the building will decrease when using energy-efficient lighting. Hence, using energy-efficient lighting will decrease the annual energy cost of the building.

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Most popular questions from this chapter

In tropical climates, the water near the surface of the ocean remains warm throughout the year as a result of solar energy absorption. In the deeper parts of the ocean, however, the water remains at a relatively low temperature since the sun's rays cannot penetrate very far. It is proposed to take advantage of this temperature difference and construct a power plant that will absorb heat from the warm water near the surface and reject the waste heat to the cold water a few hundred meters below. Determine the maximum thermal efficiency of such a plant if the water temperatures at the two respective locations are 24 and \(3^{\circ} \mathrm{C}\)

An air-conditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of \(32 \mathrm{kJ} / \mathrm{s}\) to maintain its temperature constant at \(20^{\circ} \mathrm{C}\) If the temperature of the outdoors is \(35^{\circ} \mathrm{C},\) the power required to operate this air-conditioning system is \((a) 0.58 \mathrm{kW}\) (b) \(3.20 \mathrm{kW}\) \((c) 1.56 \mathrm{kW}\) \((d) 2.26 \mathrm{kW}\) \((e) 1.64 \mathrm{kW}\)

The sun supplies electromagnetic energy to the earth. It appears to have an effective temperature of approximately \(5800 \mathrm{K}\). On a clear summer day in North America, the energy incident on a surface facing the sun is approximately \(0.95 \mathrm{kW} / \mathrm{m}^{2} .\) The electromagnetic solar energy can be converted into thermal energy by being absorbed on a darkened surface. How might you characterize the work potential of the sun's energy when it is to be used to produce work?

Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one. If the source temperature of the first engine is \(1300 \mathrm{K}\) and the sink temperature of the \(\sec\) ond engine is \(300 \mathrm{K}\) and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is \((a) 625 \mathrm{K}\) (b) \(800 \mathrm{K}\) \((c) 860 \mathrm{K}\) \((d) 453 \mathrm{K}\) \((e) 758 \mathrm{K}\)

Consider a building whose annual air-conditioning load is estimated to be \(40,000 \mathrm{kWh}\) in an area where the unit cost of electricity is \(\$ 0.10 / \mathrm{kWh}\). Two air conditioners are considered for the building. Air conditioner A has a seasonal average COP of 2.3 and costs \(\$ 5500\) to purchase and install. Air conditioner B has a seasonal average COP of 3.6 and costs \(\$ 7000\) to purchase and install. All else being equal, determine which air conditioner is a better buy.

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