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Arctic explorers are unsure if they can use a \(5-\mathrm{kW}\) motor-driven heat pump to stay warm. It should keep their shelter at \(60 \mathrm{~F}\); the shelter loses energy at a rate of 0.3 Btu/s per degree difference from the colder ambient. The heat pump has a COP that is \(50 \%\) that of a Carnot heat pump. If the ambient temperature can fall to \(-10 \mathrm{~F}\) at night, would you recommend this heat pump to the explorers?

Short Answer

Expert verified
No, the heat pump would not be sufficient since its heating capacity is less than the shelter's heat loss.

Step by step solution

01

Understand the Problem

We need to determine if the 5-kW motor-driven heat pump is sufficient to keep a shelter at 60°F with a possible outside temperature of -10°F. The heat pump's performance should counter the energy loss from the shelter, which is 0.3 Btu/s per degree difference between the inside and the outside. The pump operates at 50% of the Carnot efficiency.
02

Calculate Temperature Difference

The temperature difference () between inside and outside is given by:\[\Delta T = 60 - (-10) = 70 \text{ °F}\]
03

Calculate Heat Loss Rate

The energy loss rate in Btu/s is:\[\text{Energy loss rate} = 0.3 \times \Delta T = 0.3 \times 70 = 21 \text{ Btu/s}\]
04

Convert Heat Pump Power to Btu/s

Convert the heat pump power from kW to Btu/s:\[\text{1 kW} = 3412.14 \text{ Btu/h}\] Therefore:\[5 \text{ kW} = 5 \times 3412.14 = 17060.7 \text{ Btu/h}\]Convert to Btu/s:\[\frac{17060.7}{3600} \approx 4.739 \text{ Btu/s}\]
05

Calculate Maximum COP of Carnot Cycle

The COP of a Carnot cycle () is given by:\[\text{COP}_{\text{Carnot}} = \frac{T_{\text{hot}}}{T_{\text{hot}} - T_{\text{cold}}}\]Convert temperatures to Rankine (Fahrenheit to Rankine:  =  + 459.67):\[T_{\text{hot}} = 60 + 459.67 = 519.67 \text{ R}\] and\[T_{\text{cold}} = -10 + 459.67 = 449.67 \text{ R}\]Calculate:\[\text{COP}_{\text{Carnot}} = \frac{519.67}{519.67 - 449.67} = \frac{519.67}{70} \approx 7.42\]
06

Calculate COP of the Heat Pump

The heat pump works at 50% of the COP of a Carnot heat pump:\[\text{COP of Heat Pump} = 0.5 \times \text{COP}_{\text{Carnot}} = 0.5 \times 7.42 \approx 3.71\]
07

Calculate the Heat Pump's Heating Capacity

The heating capacity of the heat pump is the product of its power input and its COP:\[\text{Heating capacity} = 4.739 \times 3.71 = 17.58 \text{ Btu/s}\]
08

Compare Heating Capacity and Heat Loss

Compare the heat pump's heating capacity with the heat loss: - Heat loss = 21 Btu/s - Heating capacity = 17.58 Btu/s The heating capacity is less than the heat loss.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carnot Heat Pump
The Carnot heat pump is an idealized heat pump operating on the Carnot cycle, which is the most efficient thermodynamic cycle known for converting heat energy. The efficiency of a heat pump is often compared to this model to show how well it operates under certain conditions. In the case of our exercise, the Carnot heat pump provides the benchmark for the maximum possible coefficient of performance (COP). This allows us to compare real-world systems to an ideal scenario.

A Carnot heat pump operates between two temperatures, commonly referred to as the hot reservoir (temperature inside the shelter) and the cold reservoir (outside temperature). The temperature difference in this exercise is calculated from 60°F to -10°F. By converting these temperatures to an absolute scale (Rankine), we can derive the theoretical COP of a Carnot heat pump. Since the real-world heat pump in question only achieves 50% of the efficiency of a Carnot heat pump, understanding this idealized model's performance helps us assess the given pump's effectiveness.
Coefficient of Performance (COP)
The coefficient of performance, or COP, in the context of a heat pump, is a measure of its efficiency. It is defined as the amount of heat delivered to the hot space divided by the work input. The formula for calculating the COP of a heat pump using the Carnot model is:
  • Theoretical COP (Carnot): \(\text{COP}_\text{Carnot} = \frac{T_\text{hot}}{T_\text{hot} - T_\text{cold}}\)
where \(T_\text{hot}\) and \(T_\text{cold}\) are the absolute temperatures of the hot and cold reservoirs, respectively.

For the heat pump query in the exercise, it has a COP that is only 50% of the Carnot COP, which impacts its overall energy efficiency. By understanding and calculating the COP, we can intuitively ascertain if a heat pump is viable for specific thermal conditions. We find that our heat pump COP is around 3.71, indicating that for every unit of work, it delivers 3.71 units of heat — an essential metric when determining if the heat pump can offset the heat loss in the cold environment.
Energy Conversion
Energy conversion in thermodynamics refers to the process of converting one form of energy into another. In our context, it's about how the mechanical energy from a motor (in kW) is transformed to achieve heating (in Btu/s) inside a shelter using a heat pump. This step involves both conversion and understanding the rate at which energy must be supplied to meet specific conditions, such as keeping a living space warm.

In the exercise, the heat pump's electrical input of 5 kW is converted to heating power in Btu/s. This involves using the conversion factor where 1 kW equals approximately 3412.14 Btu/h. However, due to the necessity of providing real-time conversions for faster calculations, this power is then transformed into Btu/s. Ultimately, the heating capacity delivered by the heat pump is influenced by both this conversion and the efficiency at which the pump operates (its COP). Thus, energy conversion is crucial to figuring out if the heating capacity is sufficient against energy loss experienced by the explorers' shelter.

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Most popular questions from this chapter

Two heat engines operate between the same two energy reservoirs, and both receive the same \(Q_{H}\). One engine is reversible and the other is not. What can you say about the two \(Q_{L}\) 's?

A heat pump cools a house at \(70 \mathrm{~F}\) with a maximum of 4000 Btu/h power input. The house gains 2000 Btu/h per degree temperature difference from the ambient, and the heat pump's \(\mathrm{COP}\) is \(60 \%\) of the theoretical maximum. Find the maximum outside temperature for which the heat pump provides sufficient cooling.

A water cooler for drinking water should cool 10 gal/h water from \(65 \mathrm{~F}\) to \(50 \mathrm{~F}\) using a small refrigeration unit with a COP of 2.5 . Find the rate of cooling required and the power input to the unit.

A car engine burns 5 kg fuel (equivalent to addition of \(Q_{H}\) ) at \(1500 \mathrm{~K}\) and rejects energy to the radiator and the exhaust at an average temperature of \(750 \mathrm{~K}\). If the fuel provides \(40000 \mathrm{~kJ} / \mathrm{kg}\), what is the maximum amount of work the engine can provide?

A \(10-\mathrm{m}^{3}\) tank of air at \(500 \mathrm{kPa}, 600 \mathrm{~K}\) acts as the high-temperature reservoir for a Carnot heat engine that rejects heat at \(300 \mathrm{~K}\). A temperature difference of \(25^{\circ} \mathrm{C}\) between the air tank and the Carnot cycle high temperature is needed to transfer the heat. The heat engine runs until the air temperature has dropped to \(400 \mathrm{~K}\) and then stops. Assume constant specific heat for air and find how much work is given out by the heat engine.

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