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The air conditioner in a car uses \(R-134 a\), and the compressor power input is \(1.5 \mathrm{kW}\), bringing the R-134a from 201.7 kPa to 1200 kPa by compression. The cold space is a heat exchanger that cools \(30^{\circ} \mathrm{C}\) atmospheric air from the outside down to \(10^{\circ} \mathrm{C}\) and blows it into the car. What is the mass flow rate of the \(\mathrm{R}-134 \mathrm{a}\), and what is the low-temperature heat-transfer rate? What is the mass flow rate of air at \(10^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The steps described above can be used to calculate the mass flow rate of R-134a and air and the low-temperature heat-transfer rate. Actual values would depend on the specific internal energy and enthalpy values obtained from the R-134a tables, and the specific heat capacity of air.

Step by step solution

01

Calculate the work done on R-134a

The work \(W\) done on the R-134a by the compressor is given by its power input \(P\) and the time in seconds \(t\), as \( W = P \times t \). Given that \( P = 1.5 \, \mathrm{kW} = 1500 \, \mathrm{W} \) and we want the work done per second, we find \( W = 1500 \, \mathrm{W} \times 1 \, \mathrm{second} = 1500 \, \mathrm{J}\). This is the work done on the R-134a per second.
02

Calculate the mass flow rate of R-134a

The work done on R-134a per unit mass is given by the change in specific internal energy, which is the difference between the specific internal energy at states 1 and 2. We can find these values from the R134a table for the given pressures. Let \( u_1 \) and \( u_2 \) be the specific internal energies at 201.7 kPa and 1200 kPa respectively. The mass flow rate \( \dot{m} \) of the R-134a can then be found using \( \dot{m} = \frac{W}{u_2 - u_1} \).
03

Determine the heat-transfer rate at low temperature

The low-temperature heat-transfer rate \( \dot{Q}_L \) can be obtained from the first law of thermodynamics knowing the work input and the mass flow rate of the R-134a. The law states that \( \dot{Q}_L = \dot{m} (h_1 - h_2) + W \), where \( h_1 \) and \( h_2 \) are the specific enthalpies at states 1 and 2. These values can be obtained from the R-134a table at the corresponding pressures.
04

Calculate the mass flow rate of air

Assuming the air behaves as an ideal gas and knowing the cooling capacity \( \dot{Q}_L \), the mass flow rate of air \( \dot{m}_{air} \) is given by \( \dot{m}_{air} = \dot{Q}_L / (c_p (T_{in} - T_{out})) \), where \(T_{in}\) and \(T_{out}\) are the air temperatures before and after cooling (given as 30°C and 10°C respectively), and \(c_p\) is the specific heat capacity of air at constant pressure.

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Key Concepts

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

R-134a refrigerant
R-134a is a type of refrigerant commonly used in automotive air conditioners as well as in domestic appliances like refrigerators. It is a hydrofluorocarbon (HFC), recognized for its low toxicity and non-flammability. R-134a does not contribute to ozone depletion, making it a safer alternative compared to previous refrigerants such as CFCs. In thermodynamic analyses involving R-134a, two key properties often need to be considered: pressure and specific internal energy. These properties, along with specific enthalpy, are crucial for calculations related to refrigeration cycles, such as determining the work input and heat exchange parameters. With R-134a, understanding its phase behavior and thermodynamic tables is fundamental to solving engineering problems related to air conditioning systems.
Compressor work
In refrigeration systems, the compressor plays a critical role by compressing the refrigerant, which increases its pressure and temperature. The work done by the compressor is the energy needed to perform this compression and can be calculated from the power input to the compressor. This power input is often given in kilowatts (kW) and can be converted to joules (J) when multiplied by time in seconds. It is important in these calculations to note that the work done on the refrigerant depends on its change in specific internal energy. By compressing the refrigerant from a low-pressure state to a high-pressure state, the compressor ensures the refrigerant can effectively absorb heat from the vehicle's cabin, thereby providing cooling.
Mass flow rate
The mass flow rate, noted as \(\dot{m}\), is a fundamental concept in thermodynamics, representing how much mass passes through a certain point or surface per unit of time. For R-134a refrigerant in an air conditioning system, determining the mass flow rate is essential to understand how efficiently the refrigerant circulates through the system. Knowing the mass flow rate allows engineers to tailor the system's performance to specific cooling requirements. It can be calculated using the formula \(\dot{m} = \frac{W}{u_2 - u_1}\), which relates the compressor work and the change in specific internal energy of the refrigerant between two states. This calculation helps in assessing the capacity of the air conditioning system and ensuring that it operates within optimal parameters.
Heat transfer
Heat transfer in the context of refrigeration involves moving heat away from a space to achieve cooling. In this exercise, heat removal from the atmospheric air is crucial for reducing its temperature from 30°C to 10°C before it is circulated within the car. The heat transfer rate \(\dot{Q}_L\) is central to evaluating the effectiveness of the refrigeration cycle. According to the first law of thermodynamics, \(\dot{Q}_L\) encompasses both the work done by the compressor and the change in enthalpies of the refrigerant. This process underscores the core principle of how air conditioners cool down the interior climate of vehicles, ensuring passenger comfort. Calculating the heat transfer rate can also help engineers and students estimate the energy efficiency of such systems.
Ideal gas behavior
When analyzing air in heat exchange processes, assuming ideal gas behavior simplifies calculations and provides a good approximation under many conditions. The assumption relies on the relations described by the ideal gas law, which ties together pressure, volume, and temperature of a gas. With ideal gas behavior, properties like the specific heat capacity at constant pressure \(c_p\) are used to calculate energy changes, such as those needed for cooling and heating. In the case of the air being cooled from 30°C to 10°C, assuming ideal gas behavior allows the calculation of the mass flow rate of air using \(\dot{m}_{air} = \dot{Q}_L / (c_p (T_{in} - T_{out}))\). This calculation is integral to understanding how much air must be cycled through the system to achieve the desired temperature reduction.

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Most popular questions from this chapter

The refrigerant \(R-22\) is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at \(50 \mathrm{F} ;\) its exit temperature from the compressor is measured and found to be 185 F. If the compressor exit is 300 psia, what is the isentropic efficiency of the compressor and the coefficient of performance of the heat pump?

A \(10-\mathrm{kg} / \mathrm{s}\) steady supply of saturated-vapor steam at \(500 \mathrm{kPa}\) is required for drying a wood pulp slurry in a paper mill (see Fig. \(\mathrm{P} 11.64\) ). It is decided to supply this steam by cogeneration; that is, the steam supply will be the exhaust from a steam turbine. Water at \(20^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) is pumped to a pressure of \(5 \mathrm{MPa}\) and then fed to a steam generator with an exit at \(400^{\circ} \mathrm{C}\). What is the additional heat-transfer rate to the steam generator beyond what would have been required to produce only the desired steam supply? What is the difference in net power?

A Brayton cycle produces \(14 \mathrm{MW}\) with an inlet state of \(17^{\circ} \mathrm{C}, 100 \mathrm{kPa}\), and a compression ratio of \(16: 1 .\) The heat added in the combustion is \(960 \mathrm{kJ} / \mathrm{kg} .\) What is the highest temperature and the mass flow rate of air, assuming cold air properties?

A refrigerator in a laboratory uses \(R-22\) as the working substance. The high pressure is 1200 \(\mathrm{kPa},\) the low pressure is \(201 \mathrm{kPa},\) and the compressor is reversible. It should remove \(500 \mathrm{W}\) from a specimen currently at \(-20^{\circ} \mathrm{C}\) (not equal to \(T_{L}\) in the cycle) that is inside the refrigerated space. Find the cycle COP and the electrical power required.

In one type of nuclear power plant, heat is transferred in the nuclear reactor to liquid sodium. The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water. Saturated vapor steam at \(5 \mathrm{MPa}\) exits this heat exchanger and is then superheated to \(600^{\circ} \mathrm{C}\) in an external gas-fired superheater. The steam enters the turbine, which has one (opentype) feedwater extraction at 0.4 MPa. The isentropic turbine efficiency is \(87 \%,\) and the condenser pressure is \(7.5 \mathrm{kPa}\). Determine the heat transfer in the reactor and in the superheater to produce a net power output of \(1 \mathrm{MW}\).

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