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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 the same as the initial specific enthalpy at 0.5 MPa.

Step by step solution

01

- Understand the Throttling Process

In a throttling process, the enthalpy remains constant (h_in = h_out). This means that the enthalpy of propane entering the valve is equal to the enthalpy of propane leaving the valve.
02

- Locate Initial State of Propane

Find the specific enthalpy for the propane at the initial state, given: Pressure (\(P_1\)) = 1.6 MPa and Temperature (\(T_1\)) = 70°C. Use the propane property tables to find this specific enthalpy (h_in).
03

- Determine Final State Enthalpy

Since the enthalpy remains constant during the throttling process, the final state enthalpy (h_out) is equal to the initial state enthalpy (h_in).
04

- Locate Final State Temperature

Using the propane property tables again, find the temperature corresponding to the final state with Pressure (\(P_2\)) = 0.5 MPa, and the specific enthalpy found in the previous step (h_out). This final temperature will be the temperature of the propane leaving the valve.

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Key Concepts

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

enthalpy
Enthalpy is a key concept in thermodynamics, notably essential in processes like throttling. Enthalpy, denoted as \(h\), combines the internal energy of a system with its pressure-volume work: \[ h = u + Pv \].

Here, \(u\) is the internal energy, \(P\) is the pressure, and \(v\) is the specific volume. Enthalpy helps to simplify the energy balance equations in open systems, like valves where fluid flows in and out.

During a throttling process, often found in refrigeration and heating systems, the enthalpy remains constant. This is crucial, as understanding this principle can help you analyze the initial and final states of the fluid involved. Thus, in the given propane exercise, knowing that \( h_{\text{in}} = h_{\text{out}} \) allows us to track the energy changes throughout the process.
specific enthalpy
Specific enthalpy, \(h\), refines the concept of enthalpy on a per mass basis. It is an intensive property, meaning it does not depend on the size of the system. Specific enthalpy is typically expressed as \( \text{kJ/kg} \).

To understand it better, if enthalpy measures the total energy in a system, specific enthalpy measures the energy per unit mass. This characteristic makes it convenient to use in practical calculations, where dealing with per unit mass simplifies equations.
In the propane exercise, specific enthalpy is used to determine the system's energy state at both the initial and final stages. By keeping the specific enthalpy constant (since it is a throttling process), we can find the final state's temperature using the property tables.
propane property tables
Propane property tables are essential tools in thermodynamic calculations. These tables provide crucial data about propane's thermodynamic properties, such as temperature, pressure, specific volume, and specific enthalpy, at various states.

When using propane property tables, start by identifying the given conditions (e.g., pressure and temperature) to find other properties. For instance, in the exercise, to find the specific enthalpy of propane at 1.6 MPa and 70°C, you refer to the tables' corresponding entries.
  • You repeat this process at the final state (0.5 MPa) to verify what temperature aligns with the unchanged specific enthalpy. This data helps bridge gaps between theoretical knowledge and practical applications in the real world.
pressure-temperature relationship
The pressure-temperature relationship is a fundamental aspect of thermodynamics, especially when dealing with phase changes and properties of substances. For a given fluid, these two properties are often interrelated and can be charted on phase diagrams or found in property tables.
  • In a throttling process, while the pressure drops, the temperature may change depending on the fluid's properties and initial state.


In the propane exercise, initially, the fluid is at 1.6 MPa and 70°C. After throttling, the pressure drops to 0.5 MPa. Using the property tables, we need to find the temperature corresponding to the new pressure, ensuring the specific enthalpy remains constant. This pressure-temperature relationship helps us understand the behavior of fluids during thermodynamic processes and predict outcomes like temperature changes post-throttling.

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Most popular questions from this chapter

A water heater operating under steady flow conditions receives water at the rate of \(5 \mathrm{~kg} / \mathrm{s}\) at \(80^{\circ} \mathrm{C}\) temperature with specific enthalpy of \(320.5 \mathrm{~kJ} / \mathrm{kg}\). Water is heated by mixing steam at temperature \(100.5^{\circ} \mathrm{C}\) and specific enthalpy of 2650 \(\mathrm{kJ} / \mathrm{kg}\). The mixture of water and steam leaves the heater in the form of liquid water at temperature \(100^{\circ} \mathrm{C}\) with specific enthalpy of \(421 \mathrm{~kJ} / \mathrm{kg}\). Calculate the required steam flow rate to the heater per hour.

At steady state, a stream of liquid water at \(20^{\circ} \mathrm{C}, 1 \mathrm{bar}\) is mixed with a stream of ethylene glycol \((M=62.07)\) to form a refrigerant mixture that is \(50 \%\) glycol by mass. The water molar flow rate is \(4.2 \mathrm{kmol} / \mathrm{min}\). The density of ethylene glycol is \(1.115\) times that of water. Determine (a) the molar flow rate, in \(\mathrm{kmol} / \mathrm{min}\), and volumetric flow rate, in \(\mathrm{m}^{3} / \mathrm{min}\), of the entering ethylene glycol. (b) the diameters, in \(\mathrm{cm}\), of each of the supply pipes if the velocity in each is \(2.5 \mathrm{~m} / \mathrm{s}\).

A well-insulated turbine operating at steady state develops \(23 \mathrm{MW}\) of power for a steam flow rate of \(40 \mathrm{~kg} / \mathrm{s}\). The steam enters at \(360^{\circ} \mathrm{C}\) with a velocity of \(35 \mathrm{~m} / \mathrm{s}\) and exits as saturated vapor at \(0.06\) bar with a velocity of \(120 \mathrm{~m} / \mathrm{s}\). Neglecting potential energy effects, determine the inlet pressure, in bar.

Wind turbines and hydraulic turbines develop mechanical power from moving streams of air and water, respectively. In each case, what aspect of the stream is tapped for power?

Figure P4.74 shows a turbine operating at a steady state that provides power to an air compressor and an electric generator. Air enters the turbine with a mass flow rate of \(5.6 \mathrm{~kg} / \mathrm{s}\) at \(517^{\circ} \mathrm{C}\) and exits the turbine at \(117^{\circ} \mathrm{C}, 2\) bar. The turbine provides power at a rate of \(800 \mathrm{~kW}\) to the compressor and at a rate of \(1200 \mathrm{~kW}\) to the generator. Air can be modeled as an ideal gas, and kinetic and potential energy changes are negligible. Determine (a) the volumetric flow rate of the air at the turbine exit, in \(\mathrm{m}^{3} / \mathrm{s}\), and (b) the rate of heat transfer between the turbine and its surroundings, in \(\mathrm{kW}\).

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