3. In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68 °F to 3200 °F. The incoming air-gas mixture and the products of combustion may both be considered ideal gases with a molecular weight = 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other?
3. In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68 °F to 3200 °F. The incoming air-gas mixture and the products of combustion may both be considered ideal gases with a molecular weight = 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other?
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
Problem 1LR
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![### Problem Statement
In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68°F to 3200°F. The incoming air-gas mixture and the products of combustion are considered ideal gases with a molecular weight of 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other?
### Equations and Analysis
1. **Continuity Equation:**
\[
\dot{m}(V_{in} - V_{out}) + F = 0
\]
2. **Mass Flow Rate:**
\[
\dot{m} = \rho_{in}AV_{in}
\]
3. **Force Balance:**
\[
F = A(P_{in} - P_{out})
\]
4. **Velocity Relation:**
\[
\frac{V_{out}}{V_{in}} = \frac{P_{out}}{P_{in}}
\]
5. **Pressure and Velocity Relation:**
\[
m[V_{in} - V_{out}] + F = 0
\]
6. **Ideal Gas Law Relation:**
\[
\frac{P_{out}}{P_{in}} \cong \frac{T_2}{T_1}
\]
### Temperature Conversion
- Convert temperatures from Fahrenheit to Rankine:
\[
T(R^\circ) = 68F + 460 = 528R^\circ
\]
\[
3200F + 460 = 3660R^\circ
\]
This set of equations and conversions helps in determining the pressure difference across the flame using properties of ideal gases and principles of fluid dynamics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63f5865a-7713-44c4-be73-1c5678c4385b%2F17cd0fc6-bd3e-4da7-a0ce-c8a2ff584923%2Fvilr10c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68°F to 3200°F. The incoming air-gas mixture and the products of combustion are considered ideal gases with a molecular weight of 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other?
### Equations and Analysis
1. **Continuity Equation:**
\[
\dot{m}(V_{in} - V_{out}) + F = 0
\]
2. **Mass Flow Rate:**
\[
\dot{m} = \rho_{in}AV_{in}
\]
3. **Force Balance:**
\[
F = A(P_{in} - P_{out})
\]
4. **Velocity Relation:**
\[
\frac{V_{out}}{V_{in}} = \frac{P_{out}}{P_{in}}
\]
5. **Pressure and Velocity Relation:**
\[
m[V_{in} - V_{out}] + F = 0
\]
6. **Ideal Gas Law Relation:**
\[
\frac{P_{out}}{P_{in}} \cong \frac{T_2}{T_1}
\]
### Temperature Conversion
- Convert temperatures from Fahrenheit to Rankine:
\[
T(R^\circ) = 68F + 460 = 528R^\circ
\]
\[
3200F + 460 = 3660R^\circ
\]
This set of equations and conversions helps in determining the pressure difference across the flame using properties of ideal gases and principles of fluid dynamics.
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