/*! 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 42 Jets of high-speed steam are use... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 42

Jets of high-speed steam are used in spray cleaning. Steam at 15.0 bar with \(150^{\circ} \mathrm{C}\) of superheat is fed to a well-insulated valve at a rate of \(1.00 \mathrm{kg} / \mathrm{s}\). As the steam passes through the valve, its pressure drops to 1.0 bar. The outlet stream may be totally vapor or a mixture of vapor and liquid. Kinetic and potential energy changes may be neglected. (a) Draw and label a flowchart, assuming that both liquid and vapor emerge from the valve. (b) Write an energy balance and use it to determine the total rate of flow of enthalpy in the outlet stream \(\left(\dot{H}_{\text {out }}=\dot{m}_{1} \hat{H}_{1}+\dot{m}_{v} \hat{H}_{v}\right) .\) Then determine whether the outlet stream is in fact a mixture of liquid and vapor or whether it is pure vapor. Explain your reasoning. (c) What is the temperature of the outlet stream? (d) Assuming that your answers to Parts (b) and (c) are correct and that the pipes at the inlet and outlet of the valve have the same inner diameter, would \(\Delta E_{\mathrm{k}}\) across the valve be positive, negative, or explain.

Short Answer

Expert verified
The process is adiabatic and steady-state, so the enthalpy per unit mass at the inlet and the outlet is the same. We need to compute this and check steam tables to identify the nature (liquid or vapor or both) of the outlet stream, which then allows to find outlet temperature. Given the inlet and outlet pipes have the same diameter, \(\Delta E_{k}\) would be zero.

Step by step solution

01

Creating a flowchart

Consider a system with control volume around the valve, excluding the inlet and exit pipes. The inflow steam is labelled as 'in' and goes to 'out', the outlet stream, which may be two-phase (Water + Vapor) or just vapor.
02

Energy balance equation

The energy balance equation for a steady-state open system with one inlet and one outlet can be written as \(\dot{H}_{in} = \dot{H}_{out}\). Where \(\dot{H}_{in}\) is the total rate of inflow of enthalpy, which is the flow rate \(\dot{m}_{in}\) times the specific enthalpy \(H_{in}\) of steam at 15 bar and \(150^{\circ}C\), and \(\dot{H}_{out}\) is total rate of outflow of enthalpy, which is the sum of flow rates times the specific enthalpies of the components of the outlet stream, assuming the outlet streams consist of both vapor and liquid.
03

Solve for Enthalpy

Enthalpy data for inlet and outlet conditions can be obtained from steam tables. For the inlet condition, we lookup the specific enthalpy of superheated steam at 15 bar and \(150^{\circ}C\). For the outlet condition, we setup an equation where the total enthalpy of the outlet stream equals the enthalpy of the inlet stream. Here, more detailed steps might be needed depending on the knowledge of the student in thermodynamics and assumptions made for the outlet condition.
04

Determine the status of the outlet stream

Based on calculated enthalpies, the state of the outlet stream can be decided. If the specific enthalpy of the outlet stream falls between the enthalpy of saturated liquid and saturated vapor at 1 bar, it indicates a mixture of vapor and liquid. Otherwise, it is purely vapor.
05

Outlet Stream Temperature

The temperature of the outlet stream is determined by the saturation temperature at the outlet pressure, as well as the enthalpy state of the outlet stream. If the outlet stream is a mixture of vapor and liquid, the temperature is the saturation temperature, otherwise, look up the temperature corresponding to the specific enthalpy of the outlet stream from superheated steam tables.
06

Determine Kinetic Energy Change

As per the problem's conditions, we assume that pipes at the inlet and outlet of the valve have the same inner diameter, and the steady-state flow, the volume flow rates at the inlet and outlet would be the same which indicates equal velocities and thus, \(\Delta E_{k}\) across the valve is zero.

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Key Concepts

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

Flowchart Creation
Creating a flowchart is essential for visualizing thermodynamic processes. It provides a clear representation of the flow and transformations happening in the system. In this context, we focus on a valve with an inlet and outlet stream. Imagine the 'in' as the steam entering the valve and the 'out' as the steam leaving it. This simple schematic helps in understanding whether the outlet consists of a mixture (liquid + vapor) or just vapor alone.

Add labels:
  • Indicate direction of flow.
  • Show states of the steam coming in and going out.
Such visual aids simplify the problem by clearly identifying which areas you need to focus on for calculations and analysis.
Steam Enthalpy Calculation
Steam enthalpy calculation involves determining the energy within the inlet and outlet streams. You'll use the concept of specific enthalpy, which is energy per unit mass. For steam at 15 bar with 150掳C superheat, we refer to steam tables to obtain the specific enthalpy value. This is crucial, as it serves as the foundation for the energy balance equation.

Here's how to proceed:
  • Find specific enthalpy for the inlet condition.
  • Apply it to calculate \(\dot{H}_{in} = \dot{m}_{in} H_{in}\).
Once we have the inlet enthalpy, we compare it against possible outlet conditions. This comparison aids in understanding whether or not the steam exits as a mixture or is purely vapor.
Phase Determination
Determining the phase of the outlet stream hinges on comparing enthalpy values. The specific enthalpy at the outlet defines whether the steam is purely vapor or a mix of liquid and vapor. For instance, we check if the outlet's specific enthalpy is between the saturated liquid and vapor enthalpies at 1 bar.

This involves:
  • Using steam tables for 1 bar to get enthalpies for saturated liquid and vapor.
  • Identifying where the outlet enthalpy falls.
If it's in between, you have a two-phase mixture; otherwise, it's all vapor. This method ensures a clear understanding of the phase present.
Temperature of Outlet Stream
The temperature of the outlet stream can be deduced once we know the state of the stream. If it is a mix of liquid and vapor at 1 bar, the temperature will be the saturation temperature at that pressure.

Here's a simple guide:
  • Refer to steam tables for the saturation temperature at 1 bar.
  • If purely vapor, check superheated steam tables for that specific enthalpy.
This approach helps in pinpointing the exact temperature efficiently, essential for complete thermodynamic analysis.
Kinetic Energy Change Analysis
In analyzing kinetic energy changes, we consider the steady-state condition and same diameter pipes. This implies that the mass flow rate and hence the velocity are maintained from inlet to outlet.

Key takeaways:
  • Under these conditions, \(\Delta E_{k}\) remains zero.
  • No net change in kinetic energy due to consistent velocity.
This assumption simplifies the problem and focuses the analysis on the thermal aspects of the steam transformation, easing the calculation process.

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Most popular questions from this chapter

A steam trap is a device to purge steam condensate from a system without venting uncondensed steam. In one of the crudest trap types, the condensate collects and raises a float attached to a drain plug. When the float reaches a certain level, it "pulls the plug," opening the drain valve and allowing the liquid to discharge. The float then drops down to its original position and the valve closes, preventing uncondensed steam from escaping. (a) Suppose saturated steam at 25 bar is used to heat \(100 \mathrm{kg} / \mathrm{min}\) of an oil from \(135^{\circ} \mathrm{C}\) to \(185^{\circ} \mathrm{C}\). Heat must be transferred to the oil at a rate of \(1.00 \times 10^{4} \mathrm{kJ} / \mathrm{min}\) to accomplish this task. The steam condenses on the exterior of a bundle of tubes through which the oil is flowing. Condensate collects in the bottom of the exchanger and exits through a steam trap set to discharge when 1200 g of liquid is collected. How often does the trap discharge? (b) Especially when periodic maintenance checks are not performed, steam traps often fail to close completely and so leak steam continuously. Suppose a process plant contains 1000 leaking traps (not an unrealistic supposition for some plants) operating at the condition of Part (a), and that on the average 10\% additional steam must be fed to the condensers to compensate for the uncondensed steam venting through the leaks. Further suppose that the cost of generating the additional steam is \$7.50 per million Btu, where the denominator refers to the enthalpy of the leaking steam relative to liquid water at \(20^{\circ} \mathrm{C}\). Estimate the yearly cost of the leaks based on \(24 \mathrm{h} /\) day, 360 day/yr operation.

Horatio Meshuggeneh has his own ideas of how to do things. For instance, when given the task of determining an oven temperature, most people would use a thermometer. Being allergic to doing anything most people would do, however, Meshuggeneh instead performs the following experiment. He puts acopper bar with a mass of 5.0 kg in the oven and puts an identical bar in a well- insulated 20.0-liter vessel containing 5.00L of liquid water and the remainder saturated steam at \(760 \mathrm{mm}\) Hg absolute. He waits long enough for both bars to reachthermal equilibrium with their surroundings, then quickly takes the first bar out of the oven, removes the second bar from the vessel, drops the first bar in its place, covers the vessel tightly, waits for the contents to come to equilibrium, and notes the reading on a pressure gauge built into the vessel. The value he reads is 50.1 mm Hg. He then uses the facts that copper has a specific gravity of 8.92 and a specific internal energy given by the expression \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.36 T\left(^{\circ} \mathrm{C}\right)\) to calculate the oven temperature. (a) The Meshuggeneh assumption is that the bar can be transferred from the oven to the vessel without any heat being lost. If he makes this assumption, what oven temperature does Meshuggeneh calculate? How many grams of water evaporate in the process? (Neglect the heat transferred to the vessel wall- -i.e., assume that the heat lost by the bar is transferred entirely to the water in the vessel. Also, remember that you are dealing with a closed system once the hot bar goes into the vessel.) (b) In fact, the bar lost 8.3 kJ of heat between the oven and the vessel. What is the true oven temperature? (c) The experiment just described was actually Meshuggeneh's second attempt. The first time he tried it, the final gauge pressure in the vessel was negative. What had he forgotten to do?

One thousand liters of a 95 wt\% glycerol- \(5 \%\) water solution is to be diluted to \(60 \%\) glycerol by adding a \(35 \%\) solution pumped from a large storage tank through a \(5-\mathrm{cm}\) ID pipe at a steady rate. The pipe discharges at a point 23 m higher than the liquid surface in the storage tank. The operation is carried out isothermally and takes 13 min to complete. The friction loss ( \(\hat{F}\) of Equation \(7.7-2\) ) is \(50 \mathrm{J} / \mathrm{kg}\). Calculate the final solution volume and the shaft work in \(\mathrm{kW}\) that the pump must deliver, assuming that the surface of the stored solution and the pipe outlet are both at 1 atm. Data: \(\quad \rho_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{kg} / \mathrm{L}, \rho_{\mathrm{gly}}=1.26 \mathrm{kg} / \mathrm{L} .\) (Use to estimate solution densities.)

If a system expands in volume by an amount \(\Delta V\left(\mathrm{m}^{3}\right)\) against a constant restraining pressure \(P\left(\mathrm{N} / \mathrm{m}^{2}\right),\) a quantity \(P \Delta V(\mathrm{J})\) of energy is transferred as expansion work from the system to its surroundings. Suppose that the following four conditions are satisfied for a closed system: (a) the system expands against a constant pressure (so that \(\Delta P=0\) ); (b) \(\Delta E_{\mathrm{k}}=0 ;\) (c) \(\Delta E_{\mathrm{p}}=0 ;\) and (d) the only work done by or on the system is expansion work. Prove that under these conditions, the energy balance simplifies to \(Q=\Delta H\)

Liquid water at \(30.0^{\circ} \mathrm{C}\) and liquid water at \(90.0^{\circ} \mathrm{C}\) are combined in a ratio \((1 \mathrm{kg} \text { cold water/2 } \mathrm{kg}\) hot water). (a) Use a simple calculation to estimate the final water temperature. For this part, pretend you never heard of energy balances. (b) Now assume a basis of calculation and write a closed-system energy balance for the process, assuming that the mixing is adiabatic. Use the balance to calculate the specific internal energy and hence (from the steam tables) the final temperature of the mixture. What is the percentage difference between your answer and that of Part (a)?
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