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A typical electric water heater has an efficiency of 95 percent and costs \(\$ 350\) a year to operate at a unit cost of electricity of \(\$ 0.11 / \mathrm{kWh}\). A typical heat pump-powered water heater has a COP of 3.3 but costs about \(\$ 800\) more to install. Determine how many years it will take for the heat pump water heater to pay for its cost differential from the energy it saves.

Short Answer

Expert verified
Answer: Approximately 3.36 years.

Step by step solution

01

Find the yearly energy consumed by the electric water heater

To find the yearly energy consumption, we will divide the yearly operating cost by the unit cost of electricity: Consumption = \(\frac{\$ 350}{\$0.11 / kWh} = 3181.82 kWh\) per year Now, we will find the input energy by dividing the consumption by the efficiency (as a decimal): Input Energy = \(\frac{3181.82}{0.95} = 3350.85 kWh\)
02

Calculate the energy consumed by the heat pump water heater

To find the energy consumption for the heat pump water heater, we will use the COP value given: Output Energy = Input Energy So, Input Energy = \(\frac{Output Energy}{COP}\) Input Energy = \(\frac{3350.85}{3.3} = 1015.41 kWh\)
03

Find the yearly cost of operating the heat pump water heater

Now, we will multiply the input energy by the unit cost of electricity to find the yearly cost of operating the heat pump water heater: Yearly Cost = 1015.41 kWh * \(\$ 0.11 / kWh = \$ 111.70\)
04

Calculate the yearly savings

To find the yearly savings by using the heat pump water heater, we will subtract its yearly cost from the yearly cost of the electric water heater: Yearly Savings = \(\$350 - \$111.70 = \$238.30\)
05

Calculate the number of years required for the cost differential to be paid off

Lastly, we will divide the cost differential between the two heaters by the yearly savings to find the number of years needed for the heat pump water heater to pay for its cost differential: Number of years = \(\frac{\$800}{\$238.30} = 3.36\) years Therefore, it will take approximately 3.36 years for the heat pump water heater to pay for its cost differential from the energy it saves.

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Most popular questions from this chapter

It is often stated that the refrigerator door should be opened as few times as possible for the shortest duration of time to save energy. Consider a household refrigerator whose interior volume is \(0.9 \mathrm{m}^{3}\) and average internal temperature is \(4^{\circ} \mathrm{C} .\) At any given time, one-third of the refrigerated space is occupied by food items, and the remaining \(0.6 \mathrm{m}^{3}\) is filled with air. The average temperature and pressure in the kitchen are \(20^{\circ} \mathrm{C}\) and \(95 \mathrm{kPa}\), respectively. Also, the moisture contents of the air in the kitchen and the refrigerator are 0.010 and \(0.004 \mathrm{kg}\) per \(\mathrm{kg}\) of air, respectively, and thus \(0.006 \mathrm{kg}\) of water vapor is condensed and removed for each kg of air that enters. The refrigerator door is opened an average of 20 times a day, and each time half of the air volume in the refrigerator is replaced by the warmer kitchen air. If the refrigerator has a coefficient of performance of 1.4 and the cost of electricity is 11.5 cents per \(\mathrm{kWh}\), determine the cost of the energy wasted per year as a result of opening the refrigerator door. What would your answer be if the kitchen air were very dry and thus a negligible amount of water vapor condensed in the refrigerator?

Using a thermometer, measure the temperature of the main food compartment of your refrigerator, and check if it is between 1 and \(4^{\circ} \mathrm{C}\). Also, measure the temperature of the freezer compartment, and check if it is at the recommended value of \(-18^{\circ} \mathrm{C}\)

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.

Why does a nonquasi-equilibrium expansion process deliver less work than the corresponding quasi-equilibrium one?

It is well known that the thermal efficiency of heat engines increases as the temperature of the energy source increases. In an attempt to improve the efficiency of a power plant, somebody suggests transferring heat from the available energy source to a higher-temperature medium by a heat pump before energy is supplied to the power plant. What do you think of this suggestion? Explain.

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