/*! 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 40 Liquid water at 60 bar and \(250... [FREE SOLUTION] | 91Ó°ÊÓ

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Liquid water at 60 bar and \(250^{\circ} \mathrm{C}\) passes through an adiabatic expansion valve, emerging at a pressure \(P_{\mathrm{f}}\) and temperature \(T_{\mathrm{f}} .\) If \(P_{\mathrm{f}}\) is low enough, some of the liquid evaporates. (a) If \(P_{\mathrm{f}}=1.0\) bar, determine the temperature of the final mixture \(\left(T_{\mathrm{f}}\right)\) and the fraction of the liquid feed that evaporates \(\left(y_{\mathrm{v}}\right)\) by writing an energy balance about the valve and neglecting \(\Delta \dot{E}_{\mathrm{k}}\) (b) If you took \(\Delta \dot{E}_{\mathrm{k}}\) into account in Part (a), how would the calculated outlet temperature compare with the value you determined? What about the calculated value of \(y_{\mathrm{v}} ?\) Explain. (c) What is the value of \(P_{\mathrm{f}}\) above which no evaporation would occur? (d) Sketch the shapes of plots of \(T_{\mathrm{f}}\) versus \(P_{\mathrm{f}}\) and \(y_{\mathrm{v}}\) versus \(P_{\mathrm{f}}\) for 1 bar \(\leq P_{\mathrm{f}} \leq 60\) bar. Briefly explain your reasoning.

Short Answer

Expert verified
The final temperature and evaporated fraction depend on the initial conditions, parameters of the process, and properties of water at given conditions. The neglect of kinetic energy changes will result in insignificant error for low-speed flows while for high-speed flows, it might change the values considerably. No evaporation will occur when the pressure is high enough so that the system remains in a liquid phase. The plots will show a downward slope for \(T_{\mathrm{f}}\) versus \(P_{\mathrm{f}}\) and an upward slope for \(y_{\mathrm{v}}\) versus \(P_{\mathrm{f}}\) until no evaporation pressure is achieved and then, both remain constant.

Step by step solution

01

Apply The Energy Balance Principle

Since it is given that the part (a) of the problem neglects the kinetic energy change \(\Delta \dot{E}_{\mathrm{k}}\), start with applying the first law of thermodynamics for an open system considering the initial and final states: \( \Delta \dot{E_{\mathrm{u}}} + \Delta \dot{E_{\mathrm{p}}} + \Delta \dot{E_{\mathrm{k}}} = 0 \). This simplifies to \( \Delta h = 0 \) for the isenthalpic process since \( \Delta \dot{E_{\mathrm{k}}} \) is neglected and the system is adiabatic.
02

Find The Final Temperature

Now, knowing the initial properties of water and using the steam tables, we can find the total enthalpy at the inlet. We can equate this to the enthalpy of the mixture at the outlet to find the final temperature \(T_{\mathrm{f}}\). For this, it's also important to note that enthalpy is a function of both temperature and pressure.
03

Calculate The Evaporated Fraction

For calculating the evaporated fraction \(y_{\mathrm{v}}\), we can use the specific enthalpies of the liquid and vapor phases obtained from the steam tables and set up a mass balance around the valve. The fraction can be obtained by dividing the mass of the vapor by the total mass.
04

Analyse The Effect Of Kinetic Energy

In part (b), we need to discuss the possible effect of taking into account \(\Delta \dot{E_{\mathrm{k}}\). As the valve is a restriction to the flow, we can expect an increase in velocity, hence, an increase in kinetic energy after expansion. This would mean a decrease in internal energy and a lower \(T_{\mathrm{f}}\) and \(y_{\mathrm{v}}\).
05

Find The Critical Pressure

To find the value of \(P_{\mathrm{f}}\) at which no evaporation would occur, we need to find the pressure at which inlet enthalpy corresponds to a saturated liquid. Above this pressure, water remains in a liquid phase due to the higher boiling point.
06

Interpret The Plots

To sketch the plots of \(T_{\mathrm{f}}\) versus \(P_{\mathrm{f}}\) and \(y_{\mathrm{v}}\) versus \(P_{\mathrm{f}}\), it's important to understand that \(T_{\mathrm{f}}\) decreases with an increase in \(P_{\mathrm{f}}\), while \(y_{\mathrm{v}}\) increases with the decrease in \(P_{\mathrm{f}}\). The values of \(P_{\mathrm{f}}\) above which no evaporation occurs will be the points where \(T_{\mathrm{f}}\) remains constant and \(y_{\mathrm{v}}\) becomes zero.

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Key Concepts

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

Adiabatic expansion
Adiabatic expansion occurs when a fluid such as liquid water, passes through a valve or some other restriction without any heat exchange with its surroundings. This means that during the expansion process, the system neither gains nor loses heat.

In thermodynamics, this is classified under the isenthalpic process, where the enthalpy before and after expansion remains constant (\( \Delta h = 0 \)). Here, as water expands from 60 bar to 1 bar pressure, the absence of external heat exchange results in some of the initial internal energy being converted into work done in expansion instead.

This can lead to a temperature drop in the fluid, since as pressure decreases, the capacity for kinetic motion and thus thermal energy also may change. This particular exercise focuses on how expansion might lead to partial evaporation due to pressure decrease, potentially changing some of the water from liquid to vapor phase.
Energy balance
Energy balance in thermodynamics is a crucial principle that applies to any system undergoing change. In our exercise, a simplified energy balance is applied, which serves to assess what's happening inside the adiabatic expansion valve.

Given that the exercise deals with an adiabatic process without kinetic energy consideration (\( \Delta \dot{E}_{k} \) is neglected), the energy equation boils down to maintaining constant enthalpy (\( \Delta h = 0 \)). This means that whatever energy is residing in the liquid water at the initial state must equal the sum of energies in the resulting mix of phases after passing through the valve.

This equilibrium allows us to predict the temperature and degree of vaporization in the output state. Such an analysis shows how one can deduce both the temperature change and the evaporation fraction using principles of energy conservation effectively.
Phase change
Phase change is a fascinating thermodynamic process, especially when substances like water change from liquid to vapor. In this exercise, the phase change occurs during adiabatic expansion when water at high pressure passes through a valve, dropping significantly in pressure.

At a lower pressure of 1 bar, as seen in the exercise, some of the liquid water in this condition evaporates. This transition happens because the water reaches its boiling point and changes into vapor, a process that requires a certain amount of energy known as the latent heat of vaporization. This energy requirement can cause the water temperature to drop, leading to partial evaporation.

The fraction of liquid that evaporates (\( y_{v} \)) can be calculated through mass balance, where the amount of initial mass lost as vapor is taken from the initial mass, indicative of the phase change extent.
Steam tables
Steam tables are an essential tool in thermodynamics used widely to find properties of water and steam under different conditions of pressure and temperature. They become particularly useful when dealing with the energy calculations in phase changes.

In this problem, the steam tables help identify the initial enthalpy of water at 60 bar and 250°C and help find the enthalpies at the lower pressure and mixed state.

By referring to steam tables, you can find values such as specific enthalpy and entropy, which permit accurate determination of temperature (\( T_{f} \)) and evaporation fraction (\( y_{v} \)) at the final state after expansion. These tables thus play a crucial role in the energy balance calculations and help accurately gauge phase transitions with changing pressure.

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Most popular questions from this chapter

Your friend has asked you to help move a 60 inch \(\times 78\) inch mattress with a mass of 75 Ib \(_{\mathrm{m}}\). The two of you position it horizontally in an open flat-bed trailer that you hitch to your car. There is nothing available to tie the mattress to the trailer, but you know there is a risk of the mattress being lifted from the trailer by the air flowing over it and perform the following calculations: (a) Although the conditions do not exactly match those for which the Bemoulli equation is applicable, use the equation to get a rough estimate of how fast you can drive (miles/h) before the mattress is lifted. Assume the velocity of air above the mattress equals the velocity of the car, the pressure difference between the top and bottom of the mattress equals the weight of the mattress divided by the mattress cross-sectional area, and air has a constant density of \(0.075 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\). What is your result? (b) You see that your friend also has several boxes of books. Since you would like to drive at 60 miles per hour, what weight of books ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) do you need to put on the mattress to hold it in place?

Steam at \(260^{\circ} \mathrm{C}\) and 7.00 bar absolute is expanded through a nozzle to \(200^{\circ} \mathrm{C}\) and 4.00 bar. Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. The specific enthalpy of steam is \(2974 \mathrm{kJ} / \mathrm{kg}\) at \(260^{\circ} \mathrm{C}\) and 7 bar and \(2860 \mathrm{kJ} / \mathrm{kg}\) at \(200^{\circ} \mathrm{C}\) and 4 bar. Use the open-system energy balance to calculate the exit steam velocity.

Methane enters a 3 -cm ID pipe at \(30^{\circ} \mathrm{C}\) and 10 bar with an average velocity of \(5.00 \mathrm{m} / \mathrm{s}\) and emerges at a point 200 m lower than the inlet at \(30^{\circ} \mathrm{C}\) and 9 bar. (a) Without doing any calculations, predict the signs ( \(+\) or \(-\) ) of \(\Delta \dot{E}_{\mathrm{k}}\) and \(\Delta \dot{E}_{\mathrm{p}},\) where \(\Delta\) signifies (outlet - inlet). Briefly explain your reasoning. (b) Calculate \(\Delta \dot{E}_{\mathrm{k}}\) and \(\Delta \dot{E}_{\mathrm{p}}(\mathrm{W}),\) assuming that the methane behaves as an ideal gas. (c) If you determine that \(\Delta \dot{E}_{\mathrm{k}} \neq-\Delta \dot{E}_{\mathrm{p}},\) explain how that result is possible.

Air is heated from \(25^{\circ} \mathrm{C}\) to \(140^{\circ} \mathrm{C}\) prior to entering a combustion furnace. The change in specific enthalpy associated with this transition is \(3349 \mathrm{J} / \mathrm{mol}\). The flow rate of air at the heater outlet is \(1.65 \mathrm{m}^{3} / \mathrm{min}\) and the air pressure at this point is \(122 \mathrm{kPa}\) absolute. (a) Calculate the heat requirement in \(\mathrm{kW}\), assuming ideal-gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible. (b) Would the value of \(\Delta \dot{E}_{k}\) [which was neglected in Part (a)] be positive or negative, or would you need more information to be able to tell? If the latter, what additional information would be needed?

A Thomas flowmeter is a device in which heat is transferred at a measured rate from an electric coil to a flowing fluid, and the flow rate of the stream is calculated from the measured increase of the fluid temperature. Suppose a device of this sort is inserted in a stream of nitrogen, the current through the heating coil is adjusted until the wattmeter reads \(1.25 \mathrm{kW},\) and the stream temperature goes from \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) before the heater to \(34^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) after the heater. (a) If the specific enthalpy of nitrogen is given by the formula \(\hat{H}(\mathrm{kJ} / \mathrm{kg})=1.04\left[T\left(^{\circ} \mathrm{C}\right)-25\right]\) what is the volumetric flow rate of the gas (L/s) upstream of the heater (i.e., at \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) )? (b) List several assumptions made in the calculation of Part (a) that could lead to errors in the calculated flow rate.

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