/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 A de-superheater has a flow of a... [FREE SOLUTION] | 91Ó°ÊÓ

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A de-superheater has a flow of ammonia of \(1.5 \mathrm{~kg} / \mathrm{s}\) at \(1000 \mathrm{kPa}, 100^{\circ} \mathrm{C}\) that is mixed with another flow of ammonia at \(25^{\circ} \mathrm{C}\) and quality \(50 \%\) in an adiabatic mixing chamber. Find the flow rate of the second flow so that the outgoing ammonia is saturated vapor at \(1000 \mathrm{kPa}\).

Short Answer

Expert verified
The flow rate of the second flow is approximately \(0.501 \:\mathrm{kg/s}\).

Step by step solution

01

Determine the properties of the first flow

Identify the properties of the ammonia at the given conditions. For ammonia at \(1000 \mathrm{kPa}, 100^{\circ} \mathrm{C}\), we need to find the specific enthalpy \(h_1\). Using thermodynamic tables, we find that the specific enthalpy for superheated ammonia under these conditions is approximately \(h_1 = 1296.2 \:\mathrm{kJ/kg}\).
02

Determine the properties of the second flow

The second flow of ammonia is at \(25^{\circ} \mathrm{C}\) and has a quality of \(50\%\). Locate this in the ammonia table where the saturation temperature at \(25^{\circ} \mathrm{C}\) provides us with the saturated liquid \(h_f\) and saturated vapor \(h_g\) enthalpies. Then, calculate \(h_2 = h_f + 0.5 \times (h_g - h_f)\). Suppose \(h_f = 402.5 \:\mathrm{kJ/kg} \) and \(h_g = 1466.1 \:\mathrm{kJ/kg}\), then \(h_2 = 402.5 + 0.5 \times (1466.1 - 402.5) = 934.3 \:\mathrm{kJ/kg}\).
03

Establish the conditions for the final mixture

The outgoing mixture is saturated vapor at \(1000 \mathrm{kPa}\), which means the vapor quality is \(100\%\). From the ammonia table, the specific enthalpy of saturated vapor at \(1000 \mathrm{kPa}\) is \(h_{out} = 1467.9 \:\mathrm{kJ/kg}\).
04

Apply the energy balance equation

Since the process is adiabatic, apply the energy balance around the mixing chamber. Set \(\dot{m}_1 h_1 + \dot{m}_2 h_2 = (\dot{m}_1 + \dot{m}_2) h_{out}\). Given \(\dot{m}_1 = 1.5 \:\mathrm{kg/s}\), \(h_1 = 1296.2 \:\mathrm{kJ/kg}\), \(h_2 = 934.3 \:\mathrm{kJ/kg}\), and \(h_{out} = 1467.9 \:\mathrm{kJ/kg}\), solve for \(\dot{m}_2\), where the equation becomes: \(1.5 \times 1296.2 + \dot{m}_2 \times 934.3 = (1.5 + \dot{m}_2) \times 1467.9\).
05

Solve for the flow rate of the second flow

Rearranging and solving the equation from Step 4 gives: \(1944.3 + 934.3 \dot{m}_2 = 2201.85 + 1467.9 \dot{m}_2\). Simplify and solve: \(267.55 = 533.6 \dot{m}_2\). Therefore, \(\dot{m}_2 \approx 0.501 \:\mathrm{kg/s}\).

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Key Concepts

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

Adiabatic Process
In thermodynamics, an adiabatic process is a process in which a system does not exchange heat with its surroundings. This means that all the energy transfers are in the form of work done or changes in internal energy, without any heat being added or removed. In the context of the de-superheater problem, the mixing of ammonia in the chamber is adiabatic. This means that we assume there is no heat loss or gain with the environment during the mixing. This assumption simplifies calculations and helps to focus on the internal energy changes due to mixing. It's important to remember that adiabatic processes can be reversible or irreversible, but in practical applications, they are often irreversible due to friction and other dissipative forces.
Adiabatic processes are important because they allow us to apply the first law of thermodynamics without considering heat transfer, which can simplify many engineering calculations.
Enthalpy
Enthalpy is a vital concept in thermodynamics, representing the total heat content of a system. It's a state function defined as the sum of a system’s internal energy and the product of its pressure and volume:\[ H = U + PV \] where \( H \) represents enthalpy, \( U \) is internal energy, \( P \) is pressure, and \( V \) is volume.
In our scenario, we use specific enthalpy (enthalpy per unit mass) because it's more practical for dealing with flows of substances, such as ammonia in a de-superheater. The specific enthalpy values are crucial for understanding how much energy the ammonia carries when entering and exiting the chamber.
  • A crucial aspect here is determining the specific enthalpies of both flows of ammonia correctly, utilizing thermodynamic tables.
  • These values help calculate the energy changes and are necessary for applying the energy balance equation.
Understanding enthalpy changes in a system helps to predict how energy is stored or transformed within that system.
Saturated Vapor
A saturated vapor occurs when a substance's vapor phase exists in equilibrium with its liquid phase at given temperature and pressure conditions. In the context of the exercise, the term describes the outgoing ammonia being in a state where it is entirely vapor, with no liquid present, at the designated pressure of 1000 kPa.
When a vapor is 'saturated', adding heat to the system doesn't increase the temperature but instead begins to convert any remaining liquid into vapor. Conversely, removing heat will not produce a temperature drop until vapor begins converting back to liquid.
  • The term 'saturated vapor' also relates directly to the quality of the ammonia being 100%, which implies it's fully vaporized.
  • Thermodynamic tables are crucial for identifying the specific enthalpy of saturated vapor, as they provide the data needed for calculations.
Understanding saturated conditions helps in determining phase changes and ensuring accurate calculations in processes dealing with heat and mass transfer.
Energy Balance Equation
The Energy Balance Equation is a fundamental tool in thermodynamics. It ensures that energy's conservation principle is adhered to in all processes. For an adiabatic process such as the one involving the ammonia flow in the mixing chamber, the energy balance can be stated as: \[ \dot{m}_1 h_1 + \dot{m}_2 h_2 = (\dot{m}_1 + \dot{m}_2) h_{out} \] where \( \dot{m}_1 \) and \( \dot{m}_2 \) are the mass flow rates of the first and second ammonia flows, and \( h_1, h_2, \) and \( h_{out} \) are the respective specific enthalpies.
This equation basically accounts for all the energy entering and leaving the system to isolate changes within. Since the process here is adiabatic, no heat is added or removed, which simplifies the application of this equation.
  • It allows us to calculate unknown values, such as the flow rate of the second stream of ammonia, by setting the energy into the system equal to the energy out.
  • This tool is central to analyzing processes in thermodynamic systems where mass and energy flows are involved.
Understanding how to properly set up and solve energy balance equations is crucial for any engineer working with thermodynamic systems.

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Most popular questions from this chapter

A storage tank for natural gas has a top dome that can move up or down as gas is added to or subtracted from the tank, maintaining \(110 \mathrm{kPa}, 290 \mathrm{~K}\) inside. A pipeline at \(110 \mathrm{kPa}, 290 \mathrm{~K}\) now supplies some natural gas to the tank. Does its state change during the filling process? What happens to the flow work?

A steam turbine receives water at \(2000 \mathrm{lbf} / \mathrm{in} .^{2}\) \(1200 \mathrm{~F}\) at a rate of \(200 \mathrm{lbm} / \mathrm{s}\), as shown in Fig. P4.84. In the middle section \(50 \mathrm{lbm} / \mathrm{s}\) is withdrawn at \(300 \mathrm{lbf} / \mathrm{in}^{2}, 650 \mathrm{~F}\) and the rest exits the turbine at 10 lbf/in. \(^{2}, 95 \%\) quality. Assuming no heat transfer and no changes in kinetic energy, find the total turbine power output.

An evacuated 150-L tank is connected to a line flowing air at room temperature, \(25^{\circ} \mathrm{C},\) and \(8 \mathrm{MPa}\) pressure. The valve is opened, allowing air to flow into the tank until the pressure inside is \(6 \mathrm{MPa}\). At this point the valve is closed. This filling process occurs rapidly and is essentially adiabatic. The tank is then placed in storage, where it eventually returns to room temperature. What is the final pressure?

A \(2-\mathrm{m}^{3}\) insulated tank contains saturated vapor steam at \(4 \mathrm{MPa}\). A valve on top of the tank is opened, and saturated vapor escapes. During the process any liquid formed collects at the bottom of the tank, so only the saturated vapor exits. We want to find the mass that has escaped when the final pressure is 1 MPa. Taking an average exit enthalpy is not very accurate, so divide the process into two or three steps with piecewise average values of the exit enthalpy for a better estimate. Use, for example, \((4-3)\), \((3-2),\) and \((2-1)\) MPa as the steps in which the problem is solved.

A compressor in an industrial air conditioner compresses ammonia from a state of saturated vapor at 20 psia to a pressure of 125 psia. At the exit, the temperature is measured to be \(200 \mathrm{~F}\) and the mass flow rate is \(1 \mathrm{lbm} / \mathrm{s}\). What is the required power input to this compressor?

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