/*! 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 49 Ethene \(\left(\mathrm{C}_{2} \m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 49

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) at \(77^{\circ} \mathrm{F}\) undergoes complete combustion with stoichiometric air at \(77^{\circ} \mathrm{F}\) and \(70 \%\) humidity in an insulated steady-flow combustion chamber. Estimate the exit temperature, assuming a pressure of \(14.5\) psia.

Short Answer

Expert verified
Solve the energy balance equation to find exit temperature.

Step by step solution

01

- Write the combustion reaction

Write the balanced chemical equation for the combustion of ethene (C2H4):\[\text{C}_2\text{H}_4 + 3\text{O}_2 + 3.76 \times 3\text{N}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 3.76 \times 3\text{N}_2 \]
02

- Calculate air-fuel ratio

The stoichiometric air-fuel ratio can be found using the molar masses of the reactants. For ethene, \(\text{Molar mass of C}_2\text{H}_4 = 28 g/mol\). The required oxygen is 3 moles per mole of ethene, making the molar mass of oxygen required (32 g/mol for O2):\(3 \text{ moles of O}_2 \times 32 \text{ g/mol} = 96 \text{ g/mol}\)
03

- Find the enthalpies of reactants

Determine the enthalpy of ethene and the enthalpy of air (considering 70% humidity) at 77 掳F. Use standard enthalpy tables for the values.
04

- Write the energy balance equation

The steady-flow energy equation for an insulated system is: \(\text{Energy in} = \text{Energy out}\). Assuming no work, kinetic, or potential energy changes, this becomes: \(\text{Enthalpy of reactants} = \text{Enthalpy of products}\).
05

- Calculate the enthalpy of the products

Using the standard enthalpy values at the desired state, calculate the enthalpy of the products: CO2, H2O, and remaining nitrogen. Adjust the enthalpy values from the reference temperature to the exit temperature using the specific heat capacities.
06

- Solve for the exit temperature

Balance the enthalpies from steps 3 and 5. Use the energy balance equation to solve for the exit temperature by equating the enthalpy of reactants to the enthalpy of products.

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 91影视!

Key Concepts

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

Stoichiometric combustion
Stoichiometric combustion refers to the exact mixture of fuel and oxidizer that ensures complete combustion without any leftover fuel or oxygen. It's the theoretical ideal condition for combustion. In the case of ethene \(\text{C}_2\text{H}_4\), you'll need 3 moles of oxygen \(3\text{O}_2\) based on the balanced chemical equation: \(\text{C}_2\text{H}_4 + 3\text{O}_2 + 3.76 \times 3\text{N}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 3.76 \times 3\text{N}_2\). For every mole of ethene, this exact amount of oxygen ensures that every bit of \(\text{C}_2\text{H}_4\) is completely combusted.
Enthalpy calculation
Enthalpy calculation is about determining the energy required or released during a process, such as combustion. In our exercise, calculating enthalpy involves using standard enthalpy tables to find the enthalpy of the reactants (ethene and air) at the given temperature (77 掳F). The standard enthalpy values help us understand the energy content of each component before and after the reaction. In practice, this means looking up values from tables and applying corrective factors for things like temperature differences.
Steady-flow energy equation
The steady-flow energy equation is essential for analyzing systems where fluid (or gas) flows through a device steadily, like in a combustion chamber. The energy balance for an insulated system, as given in our problem, shows that the energy entering the system equals the energy leaving it. Mathematically, \(\text{Enthalpy of reactants} = \text{Enthalpy of products}\). This equation assumes no work is performed, and there's no change in kinetic or potential energy.
Specific heat capacity
Specific heat capacity is the amount of heat required to change a substance's temperature by one degree. It plays a crucial role in determining how much energy is needed to raise the temperature of combustion products from the reference state to the exit temperature. Each substance has its specific heat capacity, which you can find in standard tables. For example, carbon dioxide (CO2), water vapor (H2O), and nitrogen (N2) all have different specific heat capacities. These values are indispensable when performing enthalpy calculations in step 5.
Combustion reaction
A combustion reaction involves a fuel reacting with an oxidizer, resulting in the production of heat and typically light. In our ethene combustion exercise, the balanced chemical equation: \(\text{C}_2\text{H}_4 + 3\text{O}_2 + 3.76 \times 3\text{N}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 3.76 \times 3\text{N}_2\) represents the complete combustion of ethene. This equation shows the reactants transforming into carbon dioxide, water, and remaining nitrogen, illustrating the conversion of chemical energy into thermal energy. Ensuring the reaction is balanced is key to proper enthalpy calculations and achieving true stoichiometric combustion.

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

Butane gas \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) undergoes complete combustion with \(40 \%\) excess air; both are at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\). Calculate the heat transfer from the steady-flow combustor if the products are at \(1000 \mathrm{~K}\) and the humidity of the combustion air is (a) \(90 \%\), (b) \(70 \%\), and (c) \(50 \%\).

A fuel which is \(60 \%\) ethane and \(40 \%\) octane by volume undergoes complete combustion with \(200 \%\) theoretical air. Find \((a)\) the air-fuel ratio, \((b)\) the percent by volume of \(\mathrm{N}_{2}\) in the products, and \((c)\) the dew- point temperature of the products if the pressure is \(98 \mathrm{kPa}\).

Ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is burned with dry air which contains \(5 \mathrm{~mol} \mathrm{O}_{2}\) for each mole of fuel. Calculate \((a)\) the percent of excess air, \((b)\) the air-fuel ratio, and \((c)\) the dew-point temperature. SOLUTION The stoichiometric equation is \(\mathrm{C}_{2} \mathrm{H}_{6}+3.5\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}+6.58 \mathrm{~N}_{2}\). The required combustion equation is $$ \mathrm{C}_{2} \mathrm{H}_{6}+5\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}+1.5 \mathrm{O}_{2}+18.8 \mathrm{~N}_{2} $$ (a) There is excess air since the actual reaction uses \(5 \mathrm{~mol} \mathrm{O}{ }_{2}\) rather than \(3.5\) mol. The percent of excess air is \(\%\) excess air \(=\left(\frac{5-3.5}{3.5}\right)(100 \%)=42.9 \%\) (b) The air-fuel ratio is a mass ratio. Mass is found by multiplying the number of moles by the molecular weight: $$ A F=\frac{(5)(4.76)(29)}{(1)(30)}=23.0 \mathrm{~kg}_{a} / \mathrm{kg}_{f} $$ (c) The dew-point temperature is found using the partial pressure of the water vapor in the combustion products. Assuming atmospheric pressure of \(100 \mathrm{kPa}\) (since it was not given), we find $$ P_{D}=y_{\mathrm{H}_{2} \mathrm{O}} P_{\mathrm{atm}}=\left(\frac{3}{25.3}\right)(100)=1.86 \mathrm{kPa} $$ Using Table C.2, in Appendix C, we interpolate and find \(T_{\text {d.p. }}=49^{\circ} \mathrm{C}\).

Methane \(\left(\mathrm{CH}_{4}\right)\) is burned with stoichiometric air and the products are cooled to \(20^{\circ} \mathrm{C}\) assuming complete combustion at \(100 \mathrm{kPa}\). Calculate \((a)\) the air-fuel ratio, \((b)\) the percentage of \(\mathrm{CO}_{2}\) by weight of the products, (c) the dew- point temperature of the products, and \((d)\) the percentage of water vapor condensed.

Carbon reacts with oxygen to form carbon dioxide in a steady-flow chamber. Calculate the energy involved and state the type of reaction. Assume the reactants and products are at \(25^{\circ} \mathrm{C}\) and 1 atm. The reaction equation is \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}\). The first law and Table B.6 (Appendix B) give $$ \begin{aligned} Q &=H_{P}-H_{R}=\sum_{\text {prod }} N_{i}\left(\bar{h}_{f}^{\circ}\right)_{i}-\sum_{\text {react }} N_{i}\left(\bar{h}_{f}^{\circ}\right)_{i} \\ &=(1)(-393520)-0-0=-393520 \mathrm{~kJ} / \mathrm{kmol} \end{aligned} $$ The reaction is exothermic (negative \(Q\) ).
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Astrophysics

Read Explanation

Torque and Rotational Motion

Read Explanation

Oscillations

Read Explanation

Kinematics Physics

Read Explanation

Electrostatics

Read Explanation
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.