/*! 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 47 Superheated steam at \(T_{1}\lef... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Superheated steam at \(T_{1}\left(^{\circ} \mathrm{C}\right)\) and 20.0 bar is blended with saturated steam at \(T_{2}\left(^{\circ} \mathrm{C}\right)\) and 10.0 bar in a ratio (1.96 kg of steam at 20 bar)/(1.0 kg of steam at 10 bar). The product stream is at 250^'C and 10.0 bar. The process operates at steady state. (a) Calculate \(T_{1}\) and \(T_{2},\) assuming that the blender operates adiabatically. (b) If in fact heat is being lost from the blender to the surroundings, is your estimate of \(T_{1}\) too high or too low? Briefly explain.

Short Answer

Expert verified
The values of \(T_{1}\) and \(T_{2}\) will depend on the specific enthalpies for steam at the given pressures in the steam tables. However, if heat is being lost to the surroundings, the estimate of \(T_{1}\) will be too high because heat loss will reduce the final temperature and require a lower initial temperature for \(T_{1}\).

Step by step solution

01

Understand the given information

The problem provides us with certain information - pressure and respective ratios of two steams: One is superheated at \(T_{1}\) degrees Celsius and 20.0 bar and another one is saturated at \(T_{2}\) degrees Celsius and 10.0 bar. The mass flow rates of the two streams are in the ratio 1.96:1.0.
02

Use the energy balance equation to solve for \(T_{1}\) and \(T_{2}\)

In an adiabatic, steady-state process such as this, the energy balance equation can be written as: \(m_1 \cdot h_1 + m_2 \cdot h_2 = m_3 \cdot h_3\) We know the mass flow rates \(m_1\) and \(m_2\) are in the ratio 1.96:1.0. Thus, we can rewrite the equation as: \(1.96 \cdot h_{T_1, 20 \text{ bar}} + h_{T_2, 10 \text{ bar}} = h_{250^{\circ}C, 10\text{ bar}}\) Here, \(h_1\), \(h_2\), and \(h_3\) are enthalpies at given temperatures and pressures which can be obtained from steam tables.
03

Solve for \(T_{1}\) and \(T_{2}\)

To find \(T_{1}\) and \(T_{2}\), we need to refer to steam tables for given pressures and find the enthalpy at these temperatures. From the steam tables, we obtain \(h_{250C, 10 bar} = 2974 \text{ kJ/kg}\). Then, we substitute the values into our balance equation and solve for \(T_1\) and \(T_2\).

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.

Steam Enthalpy
Steam enthalpy is a crucial concept in the understanding of steam properties and energy transfer systems, particularly within the context of chemical engineering thermodynamics. Enthalpy, denoted by 'H', represents the total heat content of a system and is a measure of energy in a thermodynamic system. It is the sum of the internal energy plus the product of pressure and volume.
For steam, a form of water in its gaseous state, the enthalpy is significant for calculations involving the energy balance of processes where steam is used for heating, power generation, or as a reactant. In the textbook exercise, steam enthalpy is used to solve for temperatures of two different streams of steam when they are mixed. The enthalpy of steam varies with its temperature and pressure, and it can be classified based on the steam condition; for instance, saturated steam has a specific enthalpy that differs from superheated steam.
To find the specific enthalpy values (usually in kJ/kg) for different conditions, engineers make use of steam tables or thermodynamic software. These tables provide data on different steam conditions such as temperature and pressure and allow us to determine the enthalpy without the need for complex calculations. By understanding steam enthalpy, one can efficiently solve for unknown temperatures or energies in a system by using the energy balance equation, a method we see employed in the given exercise.
Adiabatic Process
An adiabatic process is a thermodynamic process in which no heat transfer takes place between the system and its surroundings. This means that the system is perfectly insulated from the environment, so the total heat Q within the system remains constant during the process. In such a scenario, any changes in the internal energy of the system are due solely to work done by or on the system.

Application in Chemical Engineering

In chemical engineering thermodynamics, adiabatic processes are often assumed for simplifying calculations or designing processes where minimal heat loss is expected. The assumption of adiabatic conditions allows one to focus on the work aspects of a thermodynamic cycle without worrying about the complexities introduced by heat exchange. This is especially helpful in teaching basic principles of energy balance and thermodynamics.
When solving the textbook exercise, the assumption of adiabatic mixing simplifies the energy balance to consider only the enthalpy of the steam and the mass flow rates. The adiabatic condition implies that the sum of enthalpies of the incoming streams must be equal to the enthalpy of the outgoing stream, which is foundational to arriving at the correct temperatures for the steam streams being mixed.
Chemical Engineering Thermodynamics
Chemical Engineering Thermodynamics is the study of energy, entropy, and the equilibrium states in chemical processes. It is a foundational subject for chemical engineers, as it governs the principles required to design and analyze energy-related processes prevalent in the industry.

Equations and Laws

This field predominantly revolves around the laws of thermodynamics, which detail how energy is transferred within systems and between systems and their environment. The energy balance equation is a direct application of the first law of thermodynamics, often called the law of energy conservation, stating that energy cannot be created or destroyed but only transformed.
In the context of the exercise provided, chemical engineering thermodynamics principles guide how one should approach solving for unknown temperatures when mixing two different streams of steam. The energy balance equation is a powerful tool that relates the mass, enthalpy, and energy interaction of the system. By using the enthalpy data from steam tables, and applying the principles of adiabatic processes and energy conservation, students of chemical engineering can predict the outcomes of complex systems.
Understanding these principles is critical not only academically but also practically, as real-world chemical engineering often requires adjustments for non-ideal conditions, heat loss, or gains and understanding of system dynamics beyond idealized assumptions as suggested by an adiabatic assumption in textbook problems.

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

Prove that for an ideal gas, \(\hat{U}\) and \(\hat{H}\) are related as \(\hat{H}=\hat{U}+R T\), where \(R\) is the gas constant. Then: (a) Taking as given that the specific internal energy of an ideal gas is independent of the gas pressure, justify the claim that \(\Delta \hat{H}\) for a process in which an ideal gas goes from \(\left(T_{1}, P_{1}\right)\) to \(\left(T_{2}, P_{2}\right)\) equals \(\Delta \hat{H}\) for the same gas going from \(T_{1}\) to \(T_{2}\) at a constant pressure of \(P_{1}\) (b) Calculate \(\Delta H(\text { cal })\) for a process in which the temperature of 2.5 mol of an ideal gas is raised by \(50^{\circ} \mathrm{C},\) resulting in a specific internal energy change \(\Delta \hat{U}=3500 \mathrm{cal} / \mathrm{mol}\)

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.

A rigid 6.00-liter vessel contains 4.00 L of liquid water in equilibrium with 2.00 L of water vapor at \(25^{\circ} \mathrm{C} .\) Heat is transferred to the water by means of an immersed electrical coil. The volume of the coil is negligible. Use the steam tables to calculate the final temperature and pressure (bar) of the system and the mass of water vaporized (g) if 3915 kJ is added to the water and no heat is transferred from the water to its surroundings. (Note: A trial-and-error calculation is required.)

A liquid mixture of benzene and toluene is to be separated in a continuous single-stage equilibrium flash tank. The pressure in the unit may be adjusted to any desired value, and the heat input may similarly be adjusted to vary the temperature at which the separation is conducted. The vapor and liquid product streams both emerge at the temperature \(T\left(^{\circ} \mathrm{C}\right)\) and pressure \(P(\mathrm{mm} \mathrm{Hg})\) maintained in the vessel. Assume that the vapor pressures of benzene and toluene are given by the Antoine equation, Table B.4 or APEx; that Raoult's law- -Equation 6.4-1 - applies; and that the enthalpies of benzene and toluene liquid and vapor are linear functions of temperature. Specific enthalpies at two temperagiven here for each substance in each phase. \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l}) \quad\left(T=0^{\circ} \mathrm{C}, \quad \hat{H}=0 \mathrm{kJ} / \mathrm{mol}\right) \quad\left(T=80^{\circ} \mathrm{C}, \quad \hat{H}=10.85 \mathrm{kJ} / \mathrm{mol}\right)\) \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{v}) \quad\left(T=80^{\circ} \mathrm{C}, \hat{H}=41.61 \mathrm{kJ} / \mathrm{mol}\right) \quad\left(T=120^{\circ} \mathrm{C}, \hat{H}=45.79 \mathrm{kJ} / \mathrm{mol}\right)\) \(\mathrm{C}_{7} \mathrm{H}_{8}(\mathrm{l}) \quad\left(T=0^{\circ} \mathrm{C}, \quad \hat{H}=0 \mathrm{kJ} / \mathrm{mol}\right) \quad\left(T=111^{\circ} \mathrm{C}, \hat{H}=18.58 \mathrm{kJ} / \mathrm{mol}\right)\) \(\mathrm{C}_{7} \mathrm{H}_{8}(\mathrm{v}) \quad\left(T=89^{\circ} \mathrm{C}, \hat{H}=49.18 \mathrm{kJ} / \mathrm{mol}\right) \quad\left(T=111^{\circ} \mathrm{C}, \hat{H}=52.05 \mathrm{kJ} / \mathrm{mol}\right)\) (a) Suppose the feed is equimolar in benzene and toluene \(\left(z_{\mathrm{B}}=0.500\right) .\) Take a basis of 1 mol of feed and do the degree-of-freedom analysis on the unit to show that if \(T\) and \(P\) are specified, you can calculate the molar compositions of each phase \(\left(x_{\mathrm{B}} \text { and } y_{\mathrm{B}}\right),\) the moles of the liquid and vapor products \(\left(n_{\mathrm{L}} \text { and } n_{\mathrm{V}}\right),\) and the required heat input \((Q) .\) Don't do any numerical calculations in this part. (b) Do the calculations of Part (a) for \(T=90^{\circ} \mathrm{C}\) and \(P=652 \mathrm{mm} \mathrm{Hg}\). (c) For \(z_{B}=0.5\) and \(T=90^{\circ} \mathrm{C},\) there is a range of feasible operating pressures for the evaporator, \(P_{\min }P_{\max } ?[\) Hint: Look at your solution to Part (b) and think about how it would change if you lowered \(P .]\) (d) Set up a spreadsheet to perform the calculation of Part (b) and then use it to determine \(P_{\max }\) and \(P_{\text {min. }}\) The spreadsheet should appear as follows (some solutions are shown): Additional columns may be used to store other calculated variables (e.g., specific enthalpies). Briefly explain why \(Q\) is positive when \(P=652 \mathrm{mm}\) Hg and negative when \(P=714 \mathrm{mm} \mathrm{Hg}\). (e) In successive rows, repeat the calculation for the same \(z_{\mathrm{B}}\) and \(T\) at several pressures between \(P_{\min }\) and \(P_{\text {max. Generate a plot }}\) (using the spreadsheet program itself, if possible) of \(n_{\mathrm{V}}\) versus \(P . \mathrm{At}\) approximately what pressure is half of the feed stream vaporized?

See all solutions

Recommended explanations on Chemistry 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.