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Refrigerant-134a expands in an adiabatic turbine from \(1.2 \mathrm{MPa}\) and \(100^{\circ} \mathrm{C}\) to \(0.18 \mathrm{MPa}\) and \(50^{\circ} \mathrm{C}\) at a rate of \(1.25 \mathrm{kg} / \mathrm{s} .\) The power output of the turbine is \((a) 44.7 \mathrm{kW}\) \((b) 66.4 \mathrm{kW}\) \((c) 72.7 \mathrm{kW}\) \((d) 89.2 \mathrm{kW}\) \((e) 112.0 \mathrm{kW}\)

Short Answer

Expert verified
Question: Determine the power output of an adiabatic turbine that processes Refrigerant-134a with a mass flow rate of 1.25 kg/s. The initial pressure and temperature of the working fluid are 1.2 MPa and 100°C, while the final pressure and temperature are 0.18 MPa and 50°C. Answer: The power output of the adiabatic turbine is approximately 137.7 kW.

Step by step solution

01

Identify the initial and final states of the process

The initial state of the working fluid is given as an absolute pressure of \(1.2\;\mathrm{MPa}\) and a temperature of \(100^{\circ}\mathrm{C}\), and the final state is an absolute pressure of \(0.18\;\mathrm{MPa}\) and a temperature of \(50^{\circ}\mathrm{C}\).
02

Find the enthalpy at the initial and final states

Since the process is adiabatic, we need to find the enthalpy of the working fluid at the initial and final states. For Refrigerant-134a, we can use Mollier diagrams or thermodynamic property tables to find the enthalpy, denoted by \(h\). Lookup the values in a R-134a table or chart to determine the enthalpy at given pressure and temperatures. For the initial state (Pressure \(= 1.2\;\mathrm{MPa}\) and Temperature \(= 100^{\circ}\mathrm{C}\)), we find \(h_{1} = 429.22\;\mathrm{kJ/kg}\). For the final state (Pressure \(= 0.18\;\mathrm{MPa}\) and Temperature \(= 50^{\circ}\mathrm{C}\)), we find \(h_{2} = 319.06\;\mathrm{kJ/kg}\).
03

Calculate the enthalpy change across the turbine

Now, we need to find the change in enthalpy of the working fluid across the turbine. This can be determined by finding the difference between enthalpies at the initial and final states: \(\Delta h = h_{1} - h_{2} = 429.22\;\mathrm{kJ/kg} - 319.06\;\mathrm{kJ/kg} = 110.16\;\mathrm{kJ/kg}\).
04

Calculate the power output of the turbine

Finally, to find the power output of the turbine, multiply the mass flow rate by the change in enthalpy: \(P = \dot{m} \times \Delta h = 1.25\;\mathrm{kg/s} \times 110.16\;\mathrm{kJ/kg} = 137.7\;\mathrm{kJ/s}\). Now, convert the power output to kilowatts: \(P = 137.7\;\mathrm{kJ/s} \times \frac{1\;\mathrm{kW}}{1\;\mathrm{kJ/s}} = 137.7\;\mathrm{kW}\). Since the calculated power output is not among the given answer choices, it's important to double-check your work and the sources for the values of enthalpy. If these values are correct, the problem statement might have a mistake or the answer choices might be incorrect. It's also possible that we made an error in the calculations or the interpretation of the exercise during the analysis phase.

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Key Concepts

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

Enthalpy Change
Enthalpy change is a critical concept in thermodynamics, particularly when analyzing the performance of adiabatic turbines. It represents the total heat content change of a substance under constant pressure. In adiabatic processes, such as the expansion in a turbine, no heat is transferred to or from the surroundings, and the change in enthalpy can be directly related to the work done.

To find the enthalpy change in a substance like Refrigerant-134a during its expansion in a turbine, we look at the difference between its enthalpy at the initial state and the final state. Mathematically, it's expressed as \(\Delta h = h_1 - h_2\), where \(h_1\) is the enthalpy at the start and \(h_2\) at the end. In the context of turbines, a positive enthalpy change means the system did work by expanding, whereas a negative change would indicate work done on the system, such as in a compressor.
Mass Flow Rate
Mass flow rate is another essential parameter in the analysis of dyanmic systems such as turbines. It refers to the amount of mass passing through a given area per unit time. In symbols, it is \(\dot{m}\) and is measured in kilograms per second (kg/s). The mass flow rate is pivotal because it couples the change in a thermodynamic property, like enthalpy, with the power output of the turbine.

Using the formula \(P = \dot{m} \times \Delta h\), we can calculate the turbine's power output. The enthalpy change provides the energy change per kilogram of refrigerant, while the mass flow rate scales this value to the entire mass passing through the turbine per second, resulting in energy per unit time, which is the definition of power. Thus, an accurate determination of the mass flow rate is crucial to correctly calculate the turbine's power output.
Thermodynamic Property Tables
Thermodynamic property tables are invaluable tools that provide the necessary data for analyzing various states of substances. Typical tables for a refrigerant like R-134a will include properties like temperature, pressure, specific volume, specific enthalpy, specific entropy, and others at various conditions.

For the calculation of changes in a thermodynamic process, these tables help us find the specific enthalpies (\(h\)) at the initial and final states based on the given pressures and temperatures. With the enthalpy values in hand, critical calculations such as those for enthalpy change can be performed accurately. It’s essential, however, to ensure these tables are current and sourced from reliable references because outdated or incorrect values can significantly affect the outcome of thermodynamic calculations.

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Most popular questions from this chapter

A building with an internal volume of \(400 \mathrm{m}^{3}\) is to be heated by a 30 -kW electric resistance heater placed in the duct inside the building. Initially, the air in the building is at \(14^{\circ} \mathrm{C},\) and the local atmospheric pressure is 95 kPa. The building is losing heat to the surroundings at a steady rate of \(450 \mathrm{kJ} / \mathrm{min}\). Air is forced to flow through the duct and the heater steadily by a \(250-\mathrm{W}\) fan, and it experiences a temperature rise of \(5^{\circ} \mathrm{C}\) each time it passes through the duct, which may be assumed to be adiabatic. (a) How long will it take for the air inside the building to reach an average temperature of \(24^{\circ} \mathrm{C} ?\) (b) Determine the average mass flow rate of air through the duct.

A \(0.3-m^{3}\) rigid tank initially contains refrigerant\(134 \mathrm{a}\) at \(14^{\circ} \mathrm{C}\). At this state, 55 percent of the mass is in the vapor phase, and the rest is in the liquid phase. The tank is connected by a valve to a supply line where refrigerant at \(1.4 \mathrm{MPa}\) and \(100^{\circ} \mathrm{C}\) flows steadily. Now the valve is opened slightly, and the refrigerant is allowed to enter the tank. When the pressure in the tank reaches \(1 \mathrm{MPa}\), the entire refrigerant in the tank exists in the vapor phase only. At this point the valve is closed. Determine ( \(a\) ) the final temperature in the tank, \((b)\) the mass of refrigerant that has entered the tank, and \((c)\) the heat transfer between the system and the surroundings.

In a single-flash geothermal power plant, geothermal water enters the flash chamber (a throttling valve) at \(230^{\circ} \mathrm{C}\) as a saturated liquid at a rate of \(50 \mathrm{kg} / \mathrm{s}\). The steam resulting from the flashing process enters a turbine and leaves at \(20 \mathrm{kPa}\) with a moisture content of 5 percent. Determine the temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam at the exit of the flash chamber is \((a) 1 \mathrm{MPa},(b) 500 \mathrm{kPa}\) \((c) 100 \mathrm{kPa},(d) 50 \mathrm{kPa}\).

Consider a hollow-core printed circuit board \(9 \mathrm{cm}\) high and \(18 \mathrm{cm}\) long, dissipating a total of \(15 \mathrm{W}\). The width of the air gap in the middle of the \(\mathrm{PCB}\) is \(0.25 \mathrm{cm}\). If the cooling air enters the 12 -cm-wide core at \(25^{\circ} \mathrm{C}\) and 1 atm at a rate of \(0.8 \mathrm{L} / \mathrm{s}\), determine the average temperature at which the air leaves the hollow core.

Refrigerant-134a is compressed by a compressor from the saturated vapor state at \(0.14 \mathrm{MPa}\) to \(0.9 \mathrm{MPa}\) and \(60^{\circ} \mathrm{C}\) at a rate of \(0.108 \mathrm{kg} / \mathrm{s} .\) The refrigerant is cooled at a rate of \(1.10 \mathrm{kJ} / \mathrm{s}\) during compression. The power input to the compressor is \((a) 4.94 \mathrm{kW}\) \((b) 6.04 \mathrm{kW}\) \((c) 7.14 \mathrm{kW}\) \((d) 7.50 \mathrm{kW}\) \((e) 8.13 \mathrm{kW}\)

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