/*! 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 30 Acetylene gas \(\left(\mathrm{C}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Acetylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) at \(25^{\circ} \mathrm{C}, 1\) atm enters a reactor operating at steady state and burns with \(40 \%\) excess air entering at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}, 80 \%\) relative humidity. An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{O}_{2}, \mathrm{NO}\), and \(\mathrm{N}_{2}\) exits at \(2200 \mathrm{~K}, 0.9 \mathrm{~atm}\). Determine, per kmol of \(\mathrm{C}_{2} \mathrm{H}_{2}\) entering, the composition of exiting mixture.

Short Answer

Expert verified
For 1 kmol of entering C2H2, the exiting mixture composition includes CO2, H2O, O2, NO, and N2 based on equilibrium calculations.

Step by step solution

01

- Write the balanced chemical equation

The combustion of acetylene can be represented by the chemical equation:
02

- Determine the air-fuel ratio

( i.c. ( for the combustion: Given with
03

- Adjust for 40% excess air

Since we have we can add 40% more oxygen to the air-fuel ratio calculated in step 2: ( with 21% in composition: flows with 40%
04

- Account for humidity in the entering air

We know that the relative humidity (RH) is undethe air's composition%, we: determined atmospheric atures: content
05

- Calculate the volume of dry air required

Now rectify the Volume,dry air over pnorwith excess air:
06

- Determine the equilibrium constants and perform equilibrium calculations

Given the exit mixture contains we need to use thermodynamic tables for specific properties at the elusive first player the mixture, compositions temperature comes.

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.

acetylene combustion
The combustion of acetylene (C\textsubscript{2}H\textsubscript{2}) is the chemical reaction between acetylene gas and oxygen to produce carbon dioxide and water as the main products. Here, acetylene reacts with oxygen in a controlled manner. The balanced chemical equation for complete combustion of acetylene is: \[ \text{2} \text{C}_2\text{H}_2 + \text{5} \text{O}_2 \rightarrow \text{4} \text{CO}_2 + \text{2} \text{H}_2\text{O} \]This equation means that for every 2 moles of C\textsubscript{2}H\textsubscript{2}, 5 moles of O\textsubscript{2} are required, and it results in 4 moles of CO\textsubscript{2} and 2 moles of H\textsubscript{2}O. Combustion reactions are exothermic, releasing energy in the form of heat and light.
Understanding this balanced equation helps predict the products and calculate the required reactant amounts, including adjusting for real-world conditions like excess air and humidity.
excess air
Excess air in combustion processes is used to ensure complete burning of the fuel. More air than theoretically needed is supplied to guarantee there is enough oxygen available. If the theoretical air required (stoichiometric air) for the given fuel is known, excess air can be calculated as follows:\[ \text{Excess Air} = \text {Theoretical Air } \times \frac{\text{Excess Amount}}{100} \]In our exercise, 40% excess air means we add 40% more than the stoichiometric air, which helps in practical scenarios where complete mixing and combustion should take place.
Providing excess air prevents harmful pollutants and soot formation. While it ensures thorough combustion, it can also lead to heat losses because extra nitrogen is heated but not used actively in the reaction. Thus, balancing becomes crucial for efficient combustion.
relative humidity
Relative humidity (RH) measures the amount of moisture in the air relative to the maximum amount of moisture the air can hold at a specific temperature. It is expressed as a percentage. In combustion calculations, RH affects the air composition because the air may carry water vapor that participates in the reaction.For instance, in the provided exercise:- Given 80% RH, the moisture in the entering air impacts the reactant quantities since the air is not dry.- Moisture content in air can be calculated or referred to using psychrometric charts or equations based on temperature and pressure.Humidity affects the combustion process by altering the initial gas composition and possibly the combustion efficiency and temperature.
equilibrium constants
Equilibrium constants define the ratio of concentrations of products and reactants at equilibrium for reversible chemical reactions. For combustion in a steady-state reactor, this helps in predicting the composition of the mixture at the exit.For a generic reaction:\[ \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \]The equilibrium constant (\text{K}\textsubscript{eq}) is formulated as:\[ \text{K}_{eq} = \frac{\text{[C]}^c \text{[D]}^d}{\text{[A]}^a \text{[B]}^b} \]It involves the concentrations (or partial pressures for gases) of products and reactants. This is essential in combustion systems operating at high temperatures and varying pressures to determine the exact exit compositions.
steady state reactor
A steady-state reactor maintains constant conditions over time, implying that the reactants entering the reactor and products leaving it don't change with time. For the given exercise, the reactor operates under steady-state conditions, ensuring that:- The input and output flow rates and compositions remain constant.- Energy balance remains steady with continuous heat generation or absorption.- Achieving equilibrium states, relevant parameters like temperature, pressure, and concentration are stable.The steady-state assumption simplifies analysis and design calculations since it removes the time dependency from the equations, making it easier to predict the output composition based on known input conditions and reactions happening inside.

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

The Federal Clean Air Act of 1970 and succeeding Clean Air Act Amendments target the oxides of nitrogen \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), collectively known as \(\mathrm{NO}_{x}\), as significant air pollutants. \(\mathrm{NO}_{x}\) is formed in combustion via three primary mechanisms: thermal \(\mathrm{NO}_{x}\) formation, prompt \(\mathrm{NO}_{x}\) formation, and fuel \(\mathrm{NO}_{x}\) formation. Discuss these formation mechanisms, including a discussion of thermal \(\mathrm{NO}_{x}\) formation by the Zeldovich mechanism. What is the role of \(\mathrm{NO}_{x}\) in the formation of ozone? What are some \(\mathrm{NO}_{x}\) reduction strategies?

Why might oxygen contained in an iron tank be treated as inert in a thermodynamic analysis even though iron oxidizes in the presence of oxygen?

An isolated system has two phases, denoted by \(\mathrm{A}\) and B, each of which consists of the same two substances, denoted by 1 and \(2 .\) Show that necessary conditions for equilibrium are 1\. the temperature of each phase is the same, \(T_{\mathrm{A}}=T_{\mathrm{B}}\). 2\. the pressure of each phase is the same, \(p_{\mathrm{A}}=p_{\mathrm{B}}\). 3\. the chemical potential of each component has the same value in each phase, \(\mu_{1}^{\mathrm{A}}=\mu_{1}^{\mathrm{B}}, \mu_{2}^{\mathrm{A}}=\mu_{2}^{\mathrm{B}}\).

Spark-ignition engine exhaust gases contain several air pollutants including the oxides of nitrogen, \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), collectively known as \(\mathrm{NO}_{x}\). Additionally, the exhaust gases may contain carbon monoxide (CO) and unburned or partially burned hydrocarbons (HC). (a) The pollutant amounts actually present depend on engine design and operating conditions, and typically differ significantly from values calculated on the basis of chemical equilibrium. Discuss both the reasons for these discrepancies and possible mechanisms by which such pollutants are formed in an actual engine. (b) For spark-ignition engines, the average production of pollutants upstream of the catalyst, in g per mile of vehicle travel, are nitric oxides, \(1.5\); hydrocarbons, 2 ; and carbon monoxide, 20 . For a city in your locale having a population of 100,000 or more, estimate the annual amount, in \(\mathrm{kg}\), of each pollutant that would be discharged if automobiles had no emission control devices. Repeat if the vehicles adhere to current U.S. government emissions standards.

A mixture of \(1 \mathrm{kmol}\) of \(\mathrm{CO}\) and \(\frac{1}{2} \mathrm{kmol}\) of \(\mathrm{O}_{2}\) is held at ambient temperature and pressure. After 100 hours only an insignificant amount of \(\mathrm{CO}_{2}\) has formed. Why?

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.