/*! 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 59 A mega bomb blast generates a no... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 59

A mega bomb blast generates a normal shock wave that propagates at a speed of \(650 \mathrm{~m} / \mathrm{s}\) into stagnant air at a temperature and pressure of \(15^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\), respectively. Calculate the changes in temperature, pressure, and velocity that are generated in the air by the shock wave.

Short Answer

Expert verified
The air temperature changes to 398.78 K, pressure to 309.6 kPa, and velocity to 174.5 m/s.

Step by step solution

01

Convert Initial Temperature to Kelvin

First, we need to convert the initial temperature from degrees Celsius to Kelvin. The initial temperature is given as \( 15^\circ C \).The conversion formula is \( T(K) = T(^\circ C) + 273.15 \).Thus, \( T(K) = 15 + 273.15 = 288.15 \text{ K} \).
02

Determine the Mach Number

The speed of sound in air at the given conditions can be calculated using the formula:\[a = \sqrt{\gamma RT}\]Where:- \( \gamma \) (gamma) is the specific heat ratio (typically \( 1.4 \) for air).- \( R \) is the specific gas constant for air (\( 287 \; \text{J/kg路K}\)).Thus:\[ a = \sqrt{1.4 \times 287 \times 288.15} = 340.3 \; \text{m/s} \]The Mach number \( M \) is then:\[ M = \frac{650}{340.3} \approx 1.91 \]
03

Calculate Temperature Ratio Across Shock

The temperature ratio \( \frac{T_2}{T_1} \) across a normal shock wave can be calculated using the formula:\[\frac{T_2}{T_1} = \frac{(2\gamma M^2 - (\gamma - 1))((\gamma - 1)M^2 + 2)}{(\gamma + 1)^2 M^2}\]Substitute \( \gamma = 1.4 \) and \( M = 1.91 \):\[\frac{T_2}{T_1} = \frac{(2 \times 1.4 \times 1.91^2 - 0.4)((0.4) \times 1.91^2 + 2)}{2.4^2 \times 1.91^2} \approx 1.385\]Then:\[ T_2 = 1.385 \times 288.15 = 398.78 \text{ K} \]
04

Calculate Pressure Ratio Across Shock

The pressure ratio \( \frac{P_2}{P_1} \) across a normal shock wave can be calculated using the formula:\[\frac{P_2}{P_1} = \frac{2\gamma M^2 - (\gamma - 1)}{\gamma + 1}\]Substitute \( \gamma = 1.4 \) and \( M = 1.91 \):\[\frac{P_2}{P_1} = \frac{2 \times 1.4 \times 1.91^2 - 0.4}{2.4} \approx 3.44\]Thus:\[ P_2 = 3.44 \times 90 = 309.6 \text{ kPa} \]
05

Calculate Velocity Behind the Shock

The velocity behind the shock \( V_2 \) can be calculated using the relation:\[V_2 = \frac{\frac{u}{M}}{\frac{\gamma + 1}{M^2} + \frac{\gamma - 1}{2}}\]Using the actual shock wave speed \( u = 650 \; \text{m/s} \) and \( M = 1.91 \):\[V_2 = \frac{340.3}{\frac{2.4}{1.91^2} + 0.2} \approx 174.5 \text{ m/s}\]
06

Result: Compile Results

Based on the calculations: - The change in temperature is from 288.15 K to 398.78 K. - The change in pressure is from 90 kPa to 309.6 kPa. - The change in velocity is from 0 m/s to 174.5 m/s.

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Key Concepts

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

Mach number
The Mach number is a way to express the speed of an object relative to the speed of sound in the surrounding medium. In the context of a normal shock wave, it鈥檚 crucial to determine whether the shock is supersonic or not. To compute the Mach number, we compare the speed of the shock wave with the local speed of sound.
  • The Mach number, denoted as \(M\), is calculated by dividing the speed of the propagation into the surrounding medium by the speed of sound at those conditions.
  • For example, in the given exercise, the Mach number helps us understand how fast the shock wave is moving compared to the speed of sound in air at the initial temperature.
Understanding the Mach number is essential for analyzing the effects of the shock wave, such as variations in pressure and temperature.
Temperature ratio
The temperature ratio in the context of a normal shock wave explains how the temperature changes across the wave front. This concept helps in determining the heat effects due to the shock.
  • The temperature ratio \(\frac{T_2}{T_1}\) is the ratio of the temperature behind the shock wave to the temperature ahead of the shock wave.
  • It depends heavily on the Mach number and the specific heat ratio \(\gamma\), which is typically 1.4 for air.
  • In the exercise, this ratio tells us how much the temperature increases as the shock wave passes through the air.
By knowing this ratio, one can predict how much the air's internal energy changes due to the shock.
Pressure ratio
The pressure ratio across a normal shock wave indicates how the air pressure changes as a result of the shock. It鈥檚 a significant factor in determining the force impacts and compression effects caused by the wave.
  • The pressure ratio \(\frac{P_2}{P_1}\) indicates the change in pressure from before to after the shock wave.
  • Like the temperature ratio, it is dependent on the Mach number and the specific heat ratio \(\gamma\).
  • In the exercise, we calculated a pressure increase resulting from the shock, showing the compression effects induced by the wave.
Understanding this ratio is crucial for applications in aerospace and aviation, where shock waves frequently occur.
Speed of sound
The speed of sound is a fundamental concept that refers to the speed at which sound waves propagate through a medium, such as air.
  • In air at standard atmospheric conditions, the speed of sound is about 343 m/s but can vary with temperature and pressure.
  • It is given by the formula \(a = \sqrt{\gamma RT}\), where \(\gamma\) is the specific heat ratio, \(R\) is the specific gas constant, and \(T\) is the temperature in Kelvin.
  • In our exercise, calculating the speed of sound was essential for determining the Mach number of the shock wave.
Comprehending the speed of sound is pivotal in many fields, such as meteorology, acoustics, and engineering, as it influences how waves move through different media.

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Most popular questions from this chapter

Measurements are taken at two sections in a converging conduit through which air is moving. At the upstream section, the cross-sectional area is \(25 \mathrm{~cm}^{2},\) the Mach number is 0.5 , and the temperature and pressure are \(75^{\circ} \mathrm{C}\) and \(750 \mathrm{kPa}\), respectively. At the downstream section, the Mach number is 0.9 . Assuming adiabatic and frictionless flow between the two sections, estimate the mass flow rate through the conduit and the temperature, pressure, and flow velocity at the downstream section.

Air flows through an insulated pipe section that is \(0.50 \mathrm{~m}\) long and has a diameter of \(50 \mathrm{~mm}\). At the entrance to the pipe, the Mach number is \(0.6,\) the pressure is \(150 \mathrm{kPa}\), and the temperature is \(27^{\circ} \mathrm{C}\). If the average friction factor in the pipe is \(0.020,\) determine the Mach number, temperature, and pressure at the exit of the pipe section.

A CD nozzle takes in air from a large chamber that has a pressure and temperature of \(1. 1 \mathrm{MPa}\) and \(627^{\circ} \mathrm{C}\), respectively. Under particular operating conditions, a normal shock occurs at the exit section and the Mach number of the flow just before the shock is 2.3. Compare the pressure temperature and velocity before and after the shock.

A supersonic aircraft is cruising at an elevation of \(10 \mathrm{~km}\) in a standard atmosphere, and the (stagnation) pressure measured by a Pitot tube mounted on the aircraft is \(100 \mathrm{kPa}\). Estimate the air speed and Mach number at which the aircraft is traveling.

At the entrance to a 50 -m-long, 75 -mm-diameter insulated duct, the stagnation pressure of the airflow is \(200 \mathrm{kPa}\) and the stagnation temperature is \(150^{\circ} \mathrm{C}\). Under unchoked conditions, the velocity at the entrance is \(120 \mathrm{~m} / \mathrm{s}\). The average friction factor is estimated as \(0.020 .\) Determine the mass flow rate through the duct and state whether the flow is choked.
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