Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 9: Problem 7
The mass flow through a rocket engine is \(25 \mathrm{~kg} / \mathrm{s}\). If the exit area, velocity, and pressure are \(2 \mathrm{~m}^{2}, 4000 \mathrm{~m} / \mathrm{s}\), and \(2 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\), respectively, calculate the thrust at a standard altitude of \(50 \mathrm{~km}\).
Short Answer
Expert verified
The thrust at a standard altitude of 50 km is 140,000 N.
Step by step solution
01
Understand the Thrust Equation
The thrust generated by a rocket engine is given by the equation \( F = \dot{m} v_e + (p_e - p_a) A_e \), where \dot{m} is the mass flow rate, \( v_e \) is the exit velocity, \( p_e \) is the exit pressure, \( p_a \) is the ambient pressure, and \( A_e \) is the exit area. Identify these variables from the given problem.
02
Identify Known Values
From the problem statement, we have the mass flow rate \( \dot{m} = 25 \, \text{kg/s} \), exit velocity \( v_e = 4000 \, \text{m/s} \), exit pressure \( p_e = 2 \, \times 10^4 \, \text{N/m}^2 \), and the exit area \( A_e = 2 \, \text{m}^2 \).
03
Determine Ambient Pressure
At 50 km altitude, the standard atmospheric pressure \( p_a \) is negligible and can be approximated as 0 N/m虏 for calculation simplicity, although it can vary slightly based on specific atmospheric conditions.
04
Apply the Thrust Equation
Substitute the known values into the thrust equation: \[ F = (25 \, \text{kg/s})(4000 \, \text{m/s}) + (2 \, \times \, 10^4 \, \text{N/m}^2 - 0) (2 \, \text{m}^2) \].
05
Calculate Each Component of Thrust
Calculate the momentum term: \[ \dot{m} v_e = 25 \, \text{kg/s} \times 4000 \, \text{m/s} = 100,000 \, \text{N} \].Calculate the pressure term:\[ (p_e - p_a) A_e = (2 \, \times \, 10^4 \, \text{N/m}^2) \times 2 \, \text{m}^2 = 40,000 \, \text{N} \].
06
Sum the Components to Find Total Thrust
Add the momentum and pressure components to find the total thrust:\[ F = 100,000 \, \text{N} + 40,000 \, \text{N} = 140,000 \, \text{N} \].
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.
Thrust Equation
Rocket propulsion is propelled fundamentally by the thrust equation. This equation encapsulates how force is generated to move the rocket. It is expressed as \( F = \dot{m} v_e + (p_e - p_a) A_e \). Let's break this down:
The thrust equation essentially calculates two core components: momentum thrust and pressure thrust. The first part of the equation (\( \dot{m} v_e \)) accounts for momentum. The second part (\( (p_e - p_a) A_e \)) reflects the additional thrust resulting from pressure differences. Understanding this equation gives insight into how rockets achieve flight.
- \( \dot{m} \): Represents the mass flow rate, i.e., the amount of mass being expelled by the rocket's engine each second.
- \( v_e \): The exit velocity, indicating how fast the mass leaves the nozzle.
- \( p_e \) and \( p_a \): The exit pressure and the ambient pressure, respectively. These contribute to the thrust when combined with the area \( A_e \), representing a "pressure thrust".
- \( A_e \): The exit area, essentially the cross-sectional area of the rocket's nozzle where gases are expelled.
The thrust equation essentially calculates two core components: momentum thrust and pressure thrust. The first part of the equation (\( \dot{m} v_e \)) accounts for momentum. The second part (\( (p_e - p_a) A_e \)) reflects the additional thrust resulting from pressure differences. Understanding this equation gives insight into how rockets achieve flight.
Mass Flow Rate
Mass flow rate is a crucial concept in understanding how rockets work. It is symbolized as \( \dot{m} \) and describes how much mass is exiting the rocket per second.
The mass flow rate is critical for calculating the momentum term in the thrust equation (\( \dot{m} v_e \)), making it a vital parameter in rocket propulsion engineering. Without knowing \( \dot{m} \), engineers cannot accurately compute the thrust or predict a rocket鈥檚 performance.
- In our case, the mass flow rate is \( 25 \text{ kg/s} \). This tells us that every second, 25 kilograms of propellant are expelled from the rocket.
- This measure is essential because it directly impacts the amount of thrust generated. More mass flow rate usually means more thrust, assuming other factors remain constant.
The mass flow rate is critical for calculating the momentum term in the thrust equation (\( \dot{m} v_e \)), making it a vital parameter in rocket propulsion engineering. Without knowing \( \dot{m} \), engineers cannot accurately compute the thrust or predict a rocket鈥檚 performance.
Exit Velocity
Exit velocity, denoted as \( v_e \), is the speed at which exhaust gases are expelled from a rocket's nozzle. It is a critical factor in determining the rocket's thrust.
Exit velocity is pivotal in determining the effectiveness of a rocket's propulsion system, and optimizing \( v_e \) is central to many aerospace engineering tasks.
- For our problem, \( v_e \) is \( 4000 \text{ m/s} \), meaning gases are released at an extremely high speed. This velocity is achieved by the design and function of the rocket nozzle, converting thermal energy into kinetic energy.
- High exit velocities lead to more significant momentum thrust, enhancing a rocket's ability to rise against Earth's gravity.
- The higher the exit velocity, the more efficient the propulsion system, resulting in better performance for a given amount of fuel.
Exit velocity is pivotal in determining the effectiveness of a rocket's propulsion system, and optimizing \( v_e \) is central to many aerospace engineering tasks.
Atmospheric Pressure at Altitude
Atmospheric pressure at altitude significantly impacts rocket thrust calculations. It is represented as \( p_a \) in the thrust equation.
Therefore, understanding atmospheric pressure changes with altitude is vital for accurate thrust calculations, as it influences how effectively a rocket can perform in varying environmental conditions.
- At sea level, atmospheric pressure is about 101,325 N/m虏, but it decreases with altitude.
- At an altitude of 50 km, as in this problem, the atmospheric pressure can be considered negligible. For simplification, it is often assumed to be nearly 0 N/m虏 in calculations.
- This reduction affects the pressure thrust term (\( (p_e - p_a) A_e \)), as the difference in pressures becomes more pronounced, potentially increasing total thrust output.
Therefore, understanding atmospheric pressure changes with altitude is vital for accurate thrust calculations, as it influences how effectively a rocket can perform in varying environmental conditions.
