/*! 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 62 A \(0.75 \mathrm{~m}^{3}\) capac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(0.75 \mathrm{~m}^{3}\) capacity tank contains a two-phase liquid-vapor mixture at \(240^{\circ} \mathrm{C}\) and a quality of \(0.6 .\) Heat is transferred to the mixture and saturated vapor from the tank are withdrawn at \(240^{\circ} \mathrm{C}\) to maintain the constant pressure inside the tank by means of a pressure regulating value. This process continues until the tank has only saturated vapor left in it. Neglect the effect of kinetic and potential energies. Determine the amount of heat transfer in \(\mathrm{kJ}\).

Short Answer

Expert verified
Approximately 728 kJ.

Step by step solution

01

Understand the Given Data

The tank has a capacity of \(0.75 \,\text{m}^3\), contains a two-phase mixture at \(240^{\circ}\text{C}\) with a quality \(x = 0.6\). We need to transfer heat until only saturated vapor remains at \(240^{\circ}\text{C}\).
02

State Definitions and Find Necessary Properties

From the steam tables, determine the properties of water at \(240^{\circ}\text{C}\): specific volumes \(v_f\) (saturated liquid) and \(v_g\) (saturated vapor), and the enthalpies \(h_f\) , \(h_g\).
03

Calculate Initial Mass of the Mixture

Use the specific volumes and quality to find the initial specific volume of the mixture \(v = v_f + x(v_g - v_f)\). Then, use the tank volume to find the total mass of the mixture: \[m = \frac{V}{v}\].
04

Determine Initial Enthalpy

Calculate the initial enthalpy of the mixture: \[h_1 = h_f + x(h_g - h_f)\]. The total initial energy \(U_1 = mh_1\).
05

Calculate Final Enthalpy

When only saturated vapor remains, \(h_2 = h_g\). The total final energy \(U_2 = m_{final}h_2\). Since the tank is at constant pressure and temperature, \(m_{final} = \frac{V}{v_g}\).
06

Calculate Heat Transfer

Using conservation of energy: \(\Delta U = U_2 - U_1\). The heat transfer \(Q = \Delta U\) since there is no work being done. Substitute the values from previous steps to find \(Q\).
07

Perform Final Computation

Plug in the values into the above formulas and calculate the final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

two-phase liquid-vapor mixture
When we talk about a two-phase liquid-vapor mixture, we are referring to a substance that exists in both liquid and vapor states simultaneously. In thermodynamics, this typically occurs when dealing with water and steam. The liquid-vapor mixture is characterized by a quality factor, denoted by \( x \). This quality value represents the fraction of the mass that is in the vapor phase. For instance, if the quality is 0.6, it means 60% of the mass is vapor, and 40% is liquid.

In our exercise, the tank holds such a mixture at a quality of 0.6 at a constant temperature of \( 240^{\text{C}} \). The process of heating and removing vapor ensures that the pressure inside the tank remains constant.

Understanding two-phase mixtures is crucial for many applications in engineering, such as power plants, refrigeration, and heating systems. These systems often operate at conditions where liquid and vapor coexist, making it important to know how to calculate the behavior and properties of the mixtures accurately. The interplay between the liquid and vapor phases affects how energy is transferred and how the system's overall performance can be optimized.
specific volumes
Specific volume is a property of substances that describes the volume occupied by a unit mass. It is denoted typically by \( v \) and is the reciprocal of density. In our context, specific volumes are critical to understanding the states and transformations of the liquid-vapor mixture.

The specific volume of the mixture can be calculated using the specific volume of the saturated liquid \( v_f \) and the specific volume of the saturated vapor \( v_g \), combined with the quality \( x \). The formula to calculate the specific volume of the mixture is:

\[ v = v_f + x(v_g - v_f) \]

In simpler terms, this equation accounts for the contributions of both the liquid and vapor phases based on their respective specific volumes and the quality.

For example, in our exercise, the initial specific volume must be calculated to determine the mixture's total mass. This is done by dividing the tank's volume by the calculated specific volume. Understanding specific volumes helps engineers and scientists determine how a two-phase mixture will behave under different conditions, which is essential for designing and analyzing thermal systems.
enthalpies
Enthalpy is a measure of the total energy of a thermodynamic system. It includes internal energy, which is the energy required to create the system, and the energy required to make room for it by displacing its environment. Enthalpy is denoted by \( h \) and is particularly useful when dealing with energy transfer processes like heat and work in thermodynamic cycles.

For a two-phase liquid-vapor mixture, the enthalpy is a mix of the enthalpies of the liquid and vapor states. The initial enthalpy of the mixture can be calculated as:

\[ h_1 = h_f + x(h_g - h_f) \]

Where \( h_f \) is the enthalpy of the saturated liquid, \( h_g \) is the enthalpy of the saturated vapor, and \( x \) is the quality.

In the given exercise, we start with a mixture and heat it until only saturated vapor remains. The enthalpy when there is only saturated vapor left is simply \( h_g \). By using these enthalpies, we can determine the change in energy and thus the amount of heat transferred into the tank. This change is fundamentally important in designing and analyzing thermodynamic processes, such as those found in heating systems and steam engines. Enthalpy allows for a straightforward calculation of changes in energy under constant pressure, simplifying the complex processes involved in heat transfer.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(\mathrm{kW}\).

Refrigerant 134 a enters an air-conditioner compressor at \(3.2\) bar, \(10^{\circ} \mathrm{C}\), and is compressed at steady state to 10 bar, \(70^{\circ} \mathrm{C}\). The volumetric flow rate of refrigerant entering is \(3.0 \mathrm{~m}^{3} / \mathrm{min}\). The power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

Shows a cooling tower operating at steady state. Warm water from an air- conditioning unit enters at \(49^{\circ} \mathrm{C}\) with a mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\). Dry air enters the tower at \(21^{\circ} \mathrm{C}, 1\) bar with a volumetric flow rate of \(1.41 \mathrm{~m}^{3} / \mathrm{s}\). Because of evaporation within the tower, humid air exits at the top of the tower with a mass flow rate of \(1.764 \mathrm{~kg} / \mathrm{s}\). Cooled liquid water is collected at the bottom of the tower for return to the air-conditioning unit together with makeup water. Determine the mass flow rate of the makeup water, in \(\mathrm{kg} / \mathrm{s}\).

Vegetable oil for cooking is dispensed from a cylindrical can fitted with a spray nozzle. According to the label, the can is able to deliver 560 sprays, each of duration \(0.25 \mathrm{~s}\) and each having a mass of \(0.25 \mathrm{~g}\). Determine (a) the mass flow rate of each spray, in \(\mathrm{g} / \mathrm{s}\). (b) the mass remaining in the can after 560 sprays, in \(\mathrm{g}\), if the initial mass in the can is \(170 \mathrm{~g}\).

Helium gas flows through a well-insulated nozzle at steady state. The temperature and velocity at the inlet are \(333 \mathrm{~K}\) and \(53 \mathrm{~m} / \mathrm{s}\), respectively. At the exit, the temperature is \(256 \mathrm{~K}\) and the pressure is \(345 \mathrm{kPa}\). The mass flow rate is \(0.5 \mathrm{~kg} / \mathrm{s}\). Using the ideal gas model, and neglecting potential energy effects, determine the exit area, in \(\mathrm{m}^{2}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.