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A \(1 \mathrm{~kg}\) steel piston/cylinder contains ammonia at \(50^{\circ} \mathrm{C}, 20 \%\) quality with a volume of \(1 \mathrm{~L}\). The ammonia expands slowly, and heat is transferred to keep constant temperature. The process stops when \(P=1600 \mathrm{kPa}\). Determine the work and heat transfer for this process.

Short Answer

Expert verified
The work done is the integral of pressure over the volume change. Heat transfer is calculated through energy balance, considering constant temperature.

Step by step solution

01

Understanding Initial Conditions

We start by analyzing the initial conditions given in the problem: temperature \(T_1 = 50^{\circ}\, \mathrm{C}\), quality \(x = 20\% = 0.2\), and an initial volume \(V_1 = 1\, \mathrm{L} = 0.001\, \mathrm{m^3}\). The mass of the ammonia is \(1\, \mathrm{kg}\). The goal is to use these conditions to find the initial pressure \(P_1\), specific volume \(v_1\), internal energy \(u_1\), and enthalpy \(h_1\).
02

Calculate Initial Pressure and Specific Volume

Using the ammonia tables for \(T_1\) and the given quality \(x\), we find: \(v_f\) and \(v_g\) corresponding to \(50^{\circ}\, \mathrm{C}\). The specific volume is calculated as \(v_1 = v_f + x (v_g - v_f)\). The pressure can be obtained by matching the initial temperature and quality in the tables.
03

Final Conditions and Phase Determination

At the end of the process, we know \(P_2 = 1600\, \mathrm{kPa}\) and temperature remains constant at \(T_2 = 50^{\circ}\, \mathrm{C}\). Using the refrigerant tables, check the phase of ammonia at these conditions to decide the appropriate properties for specific enthalpy \(h_2\) and specific volume \(v_2\).
04

Determine Volume Change

Using the initial and final specific volumes, the change in volume \(\Delta V\) is computed: \(\Delta V = m (v_2 - v_1)\). The mass \(m = 1\, \mathrm{kg}\) is used directly since it is provided.
05

Calculate Work Done

For an isothermal process in a closed system, the boundary work \(W\) can be simplified to \(W = \int_{V_1}^{V_2} P\, dV\). Since pressure is not constant, express pressure as a function of volume and integrate using known initial and final states. Use tables to find necessary entropy, quality, or interpolation to support accurate integration.
06

Energy Balance for Heat Transfer

Apply the first law of thermodynamics to determine the heat transfer \(Q\): \( \Delta U = Q - W \), rearranged to \( Q = \Delta U + W \). Here, change in internal energy \(\Delta U = m (u_2 - u_1)\). Compute \(u_1\) and \(u_2\) using properties at the initial and final states respectively.

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Key Concepts

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

Throttling Process
The throttling process is a type of thermodynamic process where a fluid flows through a restriction, like a valve or porous plug, resulting in a drop in pressure. Surprisingly, this process occurs without any change in enthalpy ( h ), meaning it is isenthalpic.
This does not mean that other properties remain unchanged - typically, the temperature and internal energy may vary substantially.
In many engineering applications, including refrigeration systems, throttling is used to cool down fluids with high pressure and temperature.
Key characteristics of the throttling process include:
  • No work is done by or on the fluid, as there are no moving parts within the throttling mechanism.
  • No heat transfer takes place with the surroundings, making it an adiabatic process.
  • Since enthalpy remains constant, the initial and final states have the same total enthalpy values.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, is a fundamental principle governing energy interaction. It states that energy cannot be created or destroyed in an isolated system. The total energy of a system undergoing any process remains constant.
This principle can be expressed mathematically as:\( Q - W = \Delta U \)
where \( Q \) represents the heat added to the system, \( W \) is the work done by the system, and \(\Delta U \) is the change in internal energy.
When applied to a process, such as the expansion of ammonia in a piston-cylinder system, it helps quantify the work done and heat transferred. By evaluating the internal energy change between start and end states, you can solve for unknown heat or work terms, assuming one is known. This knowledge is key in understanding energy transfer and efficiency in thermodynamic cycles.
Phase Determination
Phase determination in thermodynamics involves identifying the state of a substance at a given temperature and pressure.
Ammonia, like many substances, can exist in different phases: solid, liquid, or vapor.
When solving thermodynamic problems, it is crucial to know the phase because it affects the physical properties like enthalpy and specific volume.
To determine the phase, we often lookup tables or charts:
  • We use the given temperature and pressure to find the relative position of the substance in the table.
  • If the pressure and temperature lie within the saturation dome, the substance is a mixture of phases (liquid-vapor).
  • If the pressure is above or below the saturation pressure at a given temperature, the substance is either a superheated vapor or a compressed liquid, respectively.
Accurately determining phase is necessary to apply correct equations for property calculations, making it a vital skill in fluid mechanics and thermodynamics.
Refrigerant Tables
Refrigerant tables are important tools used for determining the properties of refrigerants at various temperatures and pressures. These tables list properties such as enthalpy, entropy, specific volume, and internal energy.
They help engineers and students find thermodynamic properties without lengthy calculations, which is essential for solving problems efficiently.
  • Typically, refrigerant tables have entries for both saturated and superheated states of the refrigerant.
  • The saturated table entries provide data about the properties at the boiling or condensation points.
  • Superheated tables offer additional data at higher temperatures where the substance exists purely in vapor form.
Using these tables allows for quick integration into equations of state or other thermodynamic relationships required in process calculations, as seen in tasks like determining work and heat flow in expansion processes.

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Most popular questions from this chapter

A resistor in a heating element is a total of \(0.5 \mathrm{~kg}\) with specific heat of \(0.8 \mathrm{~kJ} / \mathrm{kgK}\). It is now receiving \(500 \mathrm{~W}\) of electric power, so it heats from \(20^{\circ} \mathrm{C}\) to \(180^{\circ} \mathrm{C}\). Neglect external heat loss and find the time the process took and the entropy generation.

Helium in a piston/cylinder at \(20^{\circ} \mathrm{C}, 100 \mathrm{kPa}\) is brought to \(400 \mathrm{~K}\) in a reversible polytropic process with exponent \(n=1.25\). Find the final pressure and both the specific heat transfer and the specific work.

Consider a small air pistol (Fig. P6.52) with a cylinder volume of \(1 \mathrm{~cm}^{3}\) at \(250 \mathrm{kPa}, 27^{\circ} \mathrm{C}\). The bullet acts as a piston initially held by a trigger. The bullet is released, so the air expands in an adiabatic process. If the pressure should be \(120 \mathrm{kPa}\) as the bullet leaves the cylinder, find the final volume and the work done by the air.

A cylinder/piston contains \(4 \mathrm{ft}^{3}\) air at 16 lbf/in. \({ }^{2}\), \(77 \mathrm{~F}\). The air is compressed in a reversible polytropic process to a final state of \(120 \mathrm{lbf} / \mathrm{in} .^{2}, 400\) F. Assume the heat transfer is with the ambient at \(77 \mathrm{~F}\) and determine the polytropic exponent \(n\) and the final volume of the air. Find the work done by the air, the heat transfer, and the total entropy generation for the process.

A small halogen light bulb receives electrical power of \(50 \mathrm{~W}\). The small filament is at \(1000 \mathrm{~K}\) and gives out \(20 \%\) of the power as light and the rest as heat transfer to the gas, which is at \(500 \mathrm{~K}\); the glass is at \(400 \mathrm{~K}\). All the power is absorbed by the room walls at \(25^{\circ} \mathrm{C}\). Find the rate of generation of entropy in the filament, in the total bulb including the glass, and in the total room including the bulb.

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