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Chapter 2: Problem 10

A supersonic tunnel has a test section (T.S.) Mach number of \(M_{\mathrm{T} . s}=2.0\). The reservoir for this tunnel is the room with \(T_{\text {room }}=15^{\circ} \mathrm{C}\) and \(p_{\text {room }}=100 \mathrm{kPa}\). The test section has two windows (each \(10 \times 20 \mathrm{~cm}\) ). Calculate (a) the speed of sound in the test section (b) the force on each glass window Assume \(R=287 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \gamma=1.4\).

Short Answer

Expert verified
The speed of sound in the test section and the force on each window can be calculated following the ideal gas law and the isentropic flow relations considering the flow is supersonic. The principles of fluid dynamics are applied here, specifically the Mach number, which is a ratio of the speed of an object moving through a medium to the speed of sound in that medium.

Step by step solution

01

Calculate the speed of sound in the test section

The Mach number in the test section is given, which is defined as the ratio between the object speed and the speed of sound (M = v/a). Hence, we first need to calculate the speed of sound in the test section (a), using the formula: \(a = \sqrt{\gamma R T}\), where T is the temperature (in Kelvin), R is the specific gas constant (in J/(kg*K)), and 纬 is the adiabatic constant. The temperature T needs to be converted to Kelvin from Celsius: \(T = T_{room} + 273 = 15 + 273 = 288 K\). Hence, then we plug the value into the formula: \(a = \sqrt{1.4 * 287 * 288}\).
02

Calculation of the velocity in the test section

Knowing the Mach number and the speed of sound, we can calculate the flow velocity in the test section: \(v = M * a = 2.0 * a\). Here, a was calculated in the previous step.
03

Determining the pressure in the test section

To calculate the force on the windows, we need the pressure in the test section. This can be found from the isentropic flow relations. Using the relation \(p = p_{room} * (1 + 0.5*(纬 - 1) * M^2)^(-纬 / (纬 - 1))\), where p_room is the pressure in the room and M is the Mach number. We can calculate: \(p = 100 * (1 + 0.5 * (1.4 - 1)*2.0^2)^(-1.4 / (1.4 - 1))\).
04

Measuring the force on each window

To calculate the force on the window, the pressure difference across the window and window area need to be considered. Here, it's assumed that the outside pressure is the room pressure. The force (\(F = p*A\), where A is the area of the window) is calculated as: \(F = p * 0.10 * 0.20\), where p is the pressure, 0.20 m and 0.10 m are the window dimensions.

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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, denoted as M, is a dimensionless quantity that represents the ratio of the speed of an object in a fluid (usually air) to the speed of sound in that fluid. It's a critical parameter in aerodynamics and plays a vital role in understanding supersonic and subsonic flows.

When the Mach number is above 1.0, the flow is supersonic, which means the object is moving faster than the speed of sound. In the practice problem, the Mach number for the tunnel's test section is given as 2.0, indicating that the flow within the tunnel is supersonic and hence, the shock waves and compressibility effects become significant.

In the context of the tunnel, engineers and scientists use the Mach number to determine important aspects of the airflow, such as pressure and temperature changes, as well as the forces acting on objects within the flow, like the windows. Understanding the Mach number allows them to design tunnels that can simulate the conditions craft or models would experience in actual flight.
Speed of Sound Calculation
The speed of sound is the distance traveled per unit time by a sound wave propagating through an elastic medium. In air, this speed is dependent upon the temperature of the air; specifically, it increases as the temperature increases because the air molecules move faster at higher temperatures.

To calculate the speed of sound, the formula used is:
\[a = \sqrt{\gamma R T}\]
Where:
  • \(\gamma\) is the adiabatic constant, or heat capacity ratio,
  • R is the specific gas constant for air, and
  • T is the absolute temperature in Kelvin (K).
In the given exercise, setting \(\gamma = 1.4\) for air and the specific gas constant R as 287 J/(kg*K), the speed of sound in the test section can be found after converting the given room temperature from Celsius to Kelvin. The calculated speed of sound is essential for determining the velocity of the flow in the test section of the supersonic tunnel, as well as for further calculations regarding the forces on the tunnel elements, such as the windows.
Isentropic Flow Relations
Isentropic flow relations are equations that describe the properties of a fluid moving adiabatically鈥攚ith no heat transfer into or out of the fluid鈥攖hrough an area where the flow is assumed to be reversible and entropy is constant.

In the context of supersonic flight and wind tunnels, these relations provide a way to calculate changes in flow properties like pressure, temperature, and density as a fluid moves through varying cross-sectional areas at different Mach numbers.

In the case of the exercise, the isentropic flow relation applicable to pressure is:
\[p = p_{\text{room}} \times \left(1 + \frac{1}{2}(\gamma - 1) M^2\right)^{\frac{-\gamma}{\gamma - 1}}\]
By utilizing the provided reservoir conditions (room temperature and pressure), the Mach number in the tunnel, and the adiabatic constant, one can calculate the pressure in the test section. This pressure is then used to determine the force exerted on the windows by the supersonic flow, an essential factor for structural design and safety considerations in the operation of supersonic test tunnels.

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Most popular questions from this chapter

Consider the flow of air (perfect gas) in a constantarea duct with heat transfer. Air enters the duct at a supersonic Mach number, \(M_{1}=2.6\). The gas properties are: \(\gamma=1.4=\) constant and \(c_{p}=1004 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Assuming the fluid is inviscid, and the rate of heat transfer per unit mass flow rate is \(q=\) \(572 \mathrm{~kJ} / \mathrm{kg}\), calculate (a) exit total temperature, \(T_{\mathrm{t} 2}(\mathrm{~K})\) (b) exit Mach number, \(M_{2}\) (c) exit static pressure, \(p_{2}\), in \(\mathrm{kPa}\) (d) total pressure loss, \(\Delta p_{1} / p_{\mathrm{tl}}(\%)\) (e) critical heat transfer rate to choke the inlet flow,

A supersonic combustion is modeled as Rayleigh flow. The inlet Mach number is \(M_{1}=3.0\), the total temperature and pressure at the inlet are \(T_{\mathrm{tl}}=1,500 \mathrm{~K}\) and \(p_{\mathrm{t} 1}=\) \(100 \mathrm{kPa}\), respectively. Calculate (a) minimum \(q\) to choke the flow at the exit (b) exit temperature, \(T_{2}\) (c) if the fuel is hydrogen with a heating value of \(120,000 \mathrm{~kJ} / \mathrm{kg}\), calculate the fuel-to-air ratio, \(f\). Assume \(\gamma=1.4\) and \(c_{p}=1,004 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)

Air enters a frictionless, constant-area pipe at \(p_{1}=\) 60 psia., \(T_{1}=500^{\circ} \mathrm{R} /\) and \(M_{1}=0.6 .\) If heat is transferred to the air in the pipe at \(q=300 \mathrm{BTU} / \mathrm{lbm}\) of air, calculate (a) the exit Mach number (b) static and total pressure and temperature at the exit, \(p_{2}, T_{2}, P_{12}, T_{2}\) (c) the critical heat flux \(q^{*}\) that will thermally choke the pipe. $$ c_{p}=0.24 \mathrm{BTU} / \mathrm{lbm} .{ }^{\circ} R, \gamma=1.4 $$

A scramjet combustor has a supersonic inlet condition and a choked exit. The combustor flow area increases linearly in the flow direction, as shown. The inlet and exit flow conditions are $$ \begin{aligned} M_{1} &=3.0 \\ p_{1} &=1 \mathrm{bar} \\ T_{1} &=1000 \mathrm{~K} \\ A_{1} &=1 \mathrm{~m}^{2} \\ M_{2} &=1.0 \\ A_{2} &=1.4 \mathrm{~m}^{2} \\ \gamma &=1.4, R=287 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \end{aligned} $$ The total heat release due to combustion, per unit flow rate in the duct, is initially assumed to be \(15 \mathrm{MJ} / \mathrm{kg}\). If we divide the combustor into three constant-area sections, with stepwise jumps in the duct area, we may apply Rayleigh flow principles to each segment, as shown. The heat release per segment is then \(1 / 3\) of the total heat release in the duct, i.e., \(5,000 \mathrm{~kJ} / \mathrm{kg}\). As the exit condition of a segment needs to be matched to the inlet condition of the following segment, we propose to satisfy continuity equation at the boundary through an isentropic step area expansion, i.e., \(p_{1}, T_{\mathrm{t}}\) remain the same and only the Mach number jumps isentropically through area expansion. If we march from the inlet condition toward the exit with the assumed heat release rates, we calculate the exit Mach number \(M_{2}\). Since the exit flow is specified to be choked, then we need to adjust the total heat release in order to get a choked exit. Calculate the critical heat release in the above duct that leads to thermal choking of the flow.

A duct with square cross-section is shown. The inlet flow conditions are known to be: \(M_{1}=0.3, p_{1}=150 \mathrm{kPa}\), \(T_{1}=300 \mathrm{~K}, \gamma=1.4, R=287 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The duct is insulated but the flow is frictional, with the wall friction coefficient \(C_{i}=0.005\). Assuming the flow is steady, calculate (a) the mass flow rate, \(m\) in \(\mathrm{kg} / \mathrm{s}\) and \(\mathrm{lbm} / \mathrm{s}\) (b) the hydraulic diameter of the duct in centimeter and feet (c) the exit Mach number, \(M_{2}\) (d) the percentage total pressure loss, \(\left(\Delta p_{1} / p_{\mathrm{t}}\right) \times 100\), across the duct
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