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Chapter 20: Problem 30

(II) A central heat pump operating as an air conditioner draws 33,000 Btu per hour from a building and operates between the temperatures of \(24^{\circ} \mathrm{C}\) and \(38^{\circ} \mathrm{C} .(a)\) If its coefficient of performance is 0.20 that of a Carnot air conditioner, what is the effective coefficient of performance? (b) What is the power \((\mathrm{kW})\) required of the compressor motor? (c) What is the power in terms of hp?

Short Answer

Expert verified
(a) COP = 4.244; (b) Power = 2.279 kW; (c) Power = 3.06 hp.

Step by step solution

01

Understand the concept of Coefficient of Performance (COP)

The Coefficient of Performance (COP) of an air conditioner is the ratio of the cooling effect (in energy units like BTU/hr) to the work input (in energy units like kW or BTU/hr). A Carnot air conditioner operates at maximum theoretical efficiency given by the reverse Carnot cycle, calculated using the formula \( COP_{Carnot} = \frac{T_{cold}}{T_{hot} - T_{cold}} \), where \( T_{cold} \) and \( T_{hot} \) are the absolute temperatures in Kelvin.
02

Convert temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin. Use the formula \( T_{K} = T_{C} + 273.15 \). Thus,- \( T_{cold} = 24^{\circ} \, C + 273.15 = 297.15 \, K \)- \( T_{hot} = 38^{\circ} \, C + 273.15 = 311.15 \, K \)
03

Calculate the COP of a Carnot Air Conditioner

Using the formula for the Carnot COP: \[ COP_{Carnot} = \frac{T_{cold}}{T_{hot} - T_{cold}} = \frac{297.15}{311.15 - 297.15} = \frac{297.15}{14} \approx 21.22 \]
04

Determine the effective COP

Given that the actual COP is 0.20 of COP of a Carnot, \[ COP_{actual} = 0.20 \times 21.22 = 4.244 \]
05

Calculate the power required in kW

The power required \( W \) can be found using the formula \( COP = \frac{Q_{c}}{W} \). Rearranging for \( W \), we have \( W = \frac{Q_{c}}{COP} \). Given \( Q_{c} = 33,000 \, BTU/hr \) and \( COP = 4.244 \), first convert \( Q_{c} \) to kW:1 BTU/hr = 0.00029307107 kW. Therefore, \( Q_{c} = 33,000 \times 0.00029307107 = 9.6713 \, kW \).Now, \( W = \frac{9.6713}{4.244} \approx 2.279 \, kW \).
06

Convert the power to horsepower

1 horsepower (hp) is approximately 0.7457 kW. Therefore, the compressor power in hp is \( \frac{2.279}{0.7457} \approx 3.06 \, hp \).

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Key Concepts

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

Carnot Cycle
The Carnot Cycle is a theoretical thermodynamic cycle that provides the maximum possible efficiency for a heat engine or a refrigerator. A heat pump acting as an air conditioner operates by removing heat from a cooler area and releasing it into a warmer one.
The efficiency of this cycle, and thus the air conditioning process, depends on the temperatures it operates between. The Carnot Cycle sets a standard for these systems using the formula for the Coefficient of Performance (COP).
  • The process involves two isothermal processes (constant temperature) and two adiabatic processes (no heat exchange).
  • The formula for COP in a Carnot air conditioner is: \( COP_{Carnot} = \frac{T_{cold}}{T_{hot} - T_{cold}} \).
Understanding the Carnot Cycle helps in determining how far real systems operate from ideal conditions and how much improvement is needed for better performance.
Btu to kW conversion
In thermodynamics, conversions between different energy units are crucial, especially in applications like refrigeration. Here, we discuss how to convert British Thermal Units (BTU/hr) to kilowatts (kW).
BTU is commonly used in the United States to describe the power of heating and cooling appliances. In many scientific and international contexts, the SI unit kW is preferred.
  • 1 BTU/hr is equal to approximately 0.000293 kW.
  • To convert 33,000 BTU/hr to kW, you multiply by 0.000293 to get about 9.671 kW.
This conversion is essential for calculating power requirements and efficiencies in terms familiar to engineers and scientists globally.
Carnot efficiency
Carnot efficiency represents the ideal efficiency of a reversible engine cycle. It is a measure of how effectively a machine can convert heat into work under ideal conditions. This concept is not just applicable to engines, but also to refrigerators and air conditioners.
The Carnot efficiency is valuable because it defines the upper limit of efficiency for any system, showcasing what is theoretically possible. The equation is:
  • Carnot efficiency: \( \eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}} \), where temperatures are in Kelvin.
The efficiency depends solely on the temperatures at which the system operates. This theoretical limit helps engineers design more efficient real-world systems by illustrating how much improvement is possible.
Refrigeration cycles
Refrigeration cycles are processes used to move heat from a cooler to a warmer area. They are fundamental in air conditioning, refrigeration, and heat pump applications. Understanding these cycles is key to optimizing energy consumption and achieving maximum efficiency.
In practice, several types of refrigeration cycles exist:
  • Vapor-compression cycle, which is the most common in household fridges and air conditioners.
  • Absorption refrigeration cycle, often used where waste heat is available.
  • The reverse Carnot cycle, which provides the basis for comparing other refrigeration cycles' efficiency.
These cycles involve compressing, condensing, expanding, and evaporating refrigerants. Analyzing these steps allows professionals to innovate and improve modern refrigeration systems.

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Most popular questions from this chapter

(II) The working substance of a certain Carnot engine is 1.0 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of \(5.7 .\) The work output of the engine is 920 \(\mathrm{J}\) in each cycle. Compute the temperatures of the two reservoirs between which this engine operates.

(II) When \(2.0 \mathrm{~kg}\) of water at \(12.0^{\circ} \mathrm{C}\) is mixed with \(3.0 \mathrm{~kg}\) of water at \(38.0^{\circ} \mathrm{C}\) in a well- insulated container, what is the change in entropy of the system? \((a)\) Make an estimate; (b) use the integral \(\Delta S=\int d Q / T\)

(II) A restaurant refrigerator has a coefficient of performance of \(5.0 .\) If the temperature in the kitchen outside the refrigerator \(32^{\circ} \mathrm{C}\), what is the lowest temperature that could be obtained inside the refrigerator if it were ideal?

(II) An ideal (Carnot) engine has an efficiency of 38\(\% .\) If it were possible to run it backward as a heat pump, what would be its coefficient of performance?

(II) A particular car does work at the rate of about \(7.0 \mathrm{~kJ} / \mathrm{s}\) when traveling at a steady \(20.0 \mathrm{~m} / \mathrm{s}\) along a level road. This is the work done against friction. The car can travel \(17 \mathrm{~km}\) on \(1 \mathrm{~L}\) of gasoline at this speed (about \(40 \mathrm{mi} / \mathrm{gal}\) ). What is the minimum value for \(T_{\mathrm{H}}\) if \(T_{\mathrm{L}}\) is \(25^{\circ} \mathrm{C} ?\) The energy available from \(1 \mathrm{~L}\) of gas is \(3.2 \times 10^{7} \mathrm{~J}\)
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