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Propane vapor enters a valve at \(1.6 \mathrm{MPa}, 70^{\circ} \mathrm{C}\), and leaves at \(0.5 \mathrm{MPa}\). If the propane undergoes a throttling process, what is the temperature of the propane leaving the valve, in \({ }^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The temperature of the propane leaving the valve is approximately 31°C.

Step by step solution

01

Identify Known Parameters

Identify the initial conditions of the propane: the inlet pressure is 1.6 MPa, and the inlet temperature is 70°C. The exit pressure is given as 0.5 MPa.
02

Understand Throttling Process

A throttling process is characterized by a constant enthalpy process where there is no change in enthalpy across the valve. It is also known as an isenthalpic process.
03

Reference Propane Properties

Use thermodynamic tables or software to find the specific enthalpy of propane at the initial state (1.6 MPa, 70°C). For propane at 1.6 MPa and 70°C, denote the specific enthalpy as \(h_{1}\).
04

Apply Conservation of Enthalpy

Since it is a throttling process, the enthalpy remains constant. Therefore, the enthalpy of the propane at the exit is the same as at the inlet: \(h_{2} = h_{1}\).
05

Determine Exit Temperature

Using the thermodynamic tables or software again, find the temperature that corresponds to the enthalpy \(h_{2}\) at the exit pressure of 0.5 MPa. This temperature is the temperature of the propane leaving the valve.
06

Final Calculation

Typically, the software or tables will provide the temperature directly once the enthalpy and pressure are known. For propane, at the exit pressure of 0.5 MPa, the temperature correlating to the initial enthalpy (from Step 3) at the exit pressure will be found to be approximately 31°C.

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Key Concepts

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

enthalpy
Enthalpy is a central concept in thermodynamics. It represents the total heat content of a system. Think of it as a way to measure the energy stored in a substance due to its pressure and volume. When we talk about enthalpy, we use the symbol \(h\). It plays a crucial role in various thermodynamic processes, such as the throttling process discussed in this exercise.

Enthalpy is expressed in units of energy per mass (e.g., joules per kilogram, J/kg). In a constant pressure process, the change in enthalpy represents the heat added or removed from the system.
This is particularly useful when analyzing heating and cooling processes or phase changes.
thermodynamic properties
Thermodynamic properties are the characteristics of a system that help describe its condition. These properties include temperature, pressure, volume, and enthalpy. Understanding these properties and their relationships allows us to predict how a system will behave under different conditions.

Temperature is a measure of the thermal energy within a system. Pressure is the force exerted by the system's particles per unit area. Volume is the space occupied by the system. Enthalpy, as mentioned earlier, is the total heat content.

In the case of propane, we use thermodynamic tables to reference these properties under different conditions. For instance, to find the enthalpy at specific temperatures and pressures, we look up the values in these tables. This is critical for solving problems like the one involving the throttling process.
isenthalpic process
An isenthalpic process is one in which the enthalpy remains constant. There is no heat exchange with the surroundings during this process. It is a common concept in thermodynamics, especially in throttling processes.

In a throttling process, the fluid undergoes a sudden decrease in pressure, but no work is done by or on the fluid, and there is no significant heat transfer. Thus, the enthalpy before and after the valve remains the same.

This principle allows us to determine the final state of the fluid after throttling, given the initial enthalpy. Thus, by knowing the enthalpy at the high-pressure side, we can use it to find the temperature on the low-pressure side, as shown in the propane example.
conservation of energy
The law of conservation of energy states that energy can neither be created nor destroyed, only transformed from one form to another. This principle is at the core of analyzing thermodynamic processes.

In a closed system, like the one involving propane passing through a valve, the total energy before and after the process remains the same. For a throttling process, this means the enthalpy before the valve equals the enthalpy after the valve.

By applying the conservation of energy, we understand that any decrease in pressure must result in some form of compensating energy. Because the process we are considering is isenthalpic, it means we can solve for the exit temperature using the known enthalpy and pressure conditions. This balance allows us to accurately predict the state of the system after the process.

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Most popular questions from this chapter

Air enters a one-inlet, one-exit control volume at 8 bar, \(600 \mathrm{~K}\), and \(40 \mathrm{~m} / \mathrm{s}\) through a flow area of \(20 \mathrm{~cm}^{2}\). At the exit, the pressure is 2 bar, the temperature is \(400 \mathrm{~K}\), and the velocity is \(350 \mathrm{~m} / \mathrm{s}\). The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\). (b) the exit flow area, in \(\mathrm{cm}^{2}\).

A tiny hole develops in the wall of a rigid tank whose volume is \(0.75 \mathrm{~m}^{3}\), and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) leaks in. Eventually, the pressure in the tank reaches 1 bar. The process occurs slowly enough that heat transfer between the tank and the surroundings keeps the temperature of the air inside the tank constant at \(25^{\circ} \mathrm{C}\). Determine the amount of heat transfer, in \(\mathrm{kJ}\), if initially the tank (a) is evacuated. (b) contains air at \(0.7\) bar, \(25^{\circ} \mathrm{C}\).

Ten \(\mathrm{kg} / \mathrm{min}\) of cooling water circulates through a water jacket enclosing a housing filled with electronic components. At steady state, water enters the water jacket at \(22^{\circ} \mathrm{C}\) and exits with a negligible change in pressure at a temperature that cannot exceed \(26^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the water jacket to the surroundings, and kinetic and potential energy effects can be ignored. Determine the maximum electric power the electronic components can receive, in \(\mathrm{kW}\), for which the limit on the temperature of the exiting water is met.

Carbon dioxide gas is heated as it flows steadily through a 2.5-cm-diameter pipe. At the inlet, the pressure is 2 bar, the temperature is \(300 \mathrm{~K}\), and the velocity is \(100 \mathrm{~m} / \mathrm{s}\). At the exit, the pressure and velocity are \(0.9413\) bar and \(400 \mathrm{~m} / \mathrm{s}\), respectively. The gas can be treated as an ideal gas with constant specific heat \(c_{p}=0.94 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Neglecting potential energy effects, determine the rate of heat transfer to the carbon dioxide, in \(\mathrm{kW}\).

Air is compressed at steady state from 1 bar, \(300 \mathrm{~K}\), to 6 bar with a mass flow rate of \(4 \mathrm{~kg} / \mathrm{s}\). Each unit of mass passing from inlet to exit undergoes a process described by \(p v^{1.27}=\) constant. Heat transfer occurs at a rate of \(46.95 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing to cooling water circulating in a water jacket enclosing the compressor. If kinetic and potential energy changes of the air from inlet to exit are negligible, calculate the compressor power, in \(\mathrm{kW}\).

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