/*! 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 27 Measurements are taken at two se... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
Mass flow rate is 3.502 kg/s, downstream temperature is 289.44 K, pressure is 373993 Pa, and velocity is 319.38 m/s.

Step by step solution

01

Convert Temperature to Kelvin

First, convert the given upstream temperature from degrees Celsius to Kelvin. The conversion formula is:\[T(K) = T(°C) + 273.15\]For this problem, the upstream temperature is:\[T_1 = 75 + 273.15 = 348.15 \, K\]
02

Convert Cross-sectional Area to Square Meters

Convert the given upstream cross-sectional area from \(\text{cm}^2\) to \(\text{m}^2\):\[A_1 = 25 \, \text{cm}^2 = 25 \, \times 10^{-4} \, \text{m}^2 = 0.0025 \, \text{m}^2\]
03

Calculate Upstream Velocity

Use the Mach number to find the velocity at the upstream section. The formula is:\[v_1 = M_1 \cdot a_1\]where \(a_1\) is the speed of sound given by:\[a_1 = \sqrt{\gamma \cdot R \cdot T_1}\]Assuming the air is an ideal gas, use \(\gamma = 1.4\) and \(R = 287 \, \text{J/(kgâ‹…K)}\):\[a_1 = \sqrt{1.4 \times 287 \times 348.15} = \sqrt{139925.61} = 374.11 \, \text{m/s}\]\[v_1 = 0.5 \times 374.11 = 187.06 \, \text{m/s}\]
04

Calculate Upstream Density

Use the ideal gas law to find the density at the upstream section:\[\rho_1 = \frac{P_1}{R \cdot T_1} = \frac{750000}{287 \times 348.15} \approx 7.492 \, \text{kg/m}^3\]
05

Calculate Mass Flow Rate

The mass flow rate \( \dot{m} \) is calculated using the continuity equation:\[\dot{m} = \rho_1 \times A_1 \times v_1\]\[\dot{m} = 7.492 \times 0.0025 \times 187.06 = 3.502 \, \text{kg/s}\]
06

Calculate Downstream Velocity

Use the Mach number at the downstream section to find the velocity:\[v_2 = M_2 \cdot a_2\]We need \(a_2\), the speed of sound at the downstream section. First, calculate the temperature using isentropic relations:\[T_2 = T_1 \cdot \left(\frac{1 + \frac{\gamma - 1}{2}M_1^2}{1 + \frac{\gamma - 1}{2}M_2^2}\right)\]\[T_2 = 348.15 \cdot \left(\frac{1 + 0.4 \cdot 0.5^2}{1 + 0.4 \cdot 0.9^2}\right) = 348.15 \cdot \frac{1.1}{1.324} = 289.44 \, \text{K}\]Now calculate \(a_2\):\[a_2 = \sqrt{1.4 \cdot 287 \cdot 289.44} = 354.87 \, \text{m/s}\]\[v_2 = 0.9 \times 354.87 = 319.38 \, \text{m/s}\]
07

Calculate Downstream Pressure

Use the isentropic flow assumption to find the downstream pressure:\[P_2 = P_1 \cdot \left(\frac{T_2}{T_1}\right)^\frac{\gamma}{\gamma-1}\]\[P_2 = 750000 \cdot \left(\frac{289.44}{348.15}\right)^{3.5} = 373993 \, \text{Pa}\]

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Key Concepts

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

Adiabatic Flow
In an adiabatic flow, no heat is transferred between the fluid and its surroundings. This means there is no thermal energy entering or leaving the system as the fluid moves. Because of this, the energy of the fluid is conserved, and any changes in temperature or pressure are due solely to work done by or on the fluid as it flows.
  • Adiabatic processes are common in many engineering applications, such as nozzles or certain segments of pipelines, where heat exchange with the surroundings is negligible.
  • In this type of flow, even though the temperature changes, the heat content remains unchanged, leading to sole reliance on internal energy and work aspects.
When analyzing such flows, it's crucial to apply the first law of thermodynamics to consider the work-energy principles without the need for heat transfer equations.
Mach Number
The Mach number tells us how fast a flow is moving relative to the speed of sound in the same medium. It's a unitless number defined as the ratio of the object's speed to the speed of sound.
  • If the Mach number is less than 1, the flow is subsonic.
  • If it is exactly 1, the flow is sonic.
  • If it is greater than 1, we call the flow supersonic.
In the problem, the upstream Mach number is 0.5, showing subsonic flow, while it increases to 0.9 at the downstream section, which is still subsonic but closer to reaching sonic conditions.
This concept is vital for understanding how changes in velocity affect different parameters like pressure and temperature in a fluid flow.
Isentropic Relations
Isentropic relations are used to describe idealized processes that are both adiabatic and reversible, meaning no entropy is generated and the process is fully efficient. For such flows, isentropic relationships help predict outcomes like pressure, temperature, and density.
  • These relationships are derived from the compression and expansion of fluids under those ideal conditions.
  • They enable the calculation of downstream conditions from upstream conditions by connecting these variables through specific mathematical equations.
For example, in our problem, the temperature and pressure at the downstream section were determined using isentropic relations. This allows engineers to predict how a gas will behave when it is compressed or expanded while minimizing losses due to friction or heat.
Ideal Gas Law
The ideal gas law is a simple relationship connecting pressure, volume, and temperature of a gas. It is given by the equation: \[ PV = nRT \]where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the ideal gas constant,
  • \( T \) is the temperature in Kelvin.
In practice, when dealing with flowing gases like air in our exercise, it is more convenient to use the version involving density \( \(\rho\)\): \[ P = \rho R T \]This equation allows us to calculate the density of the gas when the pressure and temperature are known, assuming ideal gas behavior. In the given problem, the ideal gas law was used to find the upstream density, which is fundamental for determining the mass flow rate. The ideal gas assumption also simplifies the analysis by considering air as a perfect gas with constant properties over the range of conditions considered here.

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Most popular questions from this chapter

A particular supersonic aircraft is designed to fly at a maximum Mach speed of 2.2 at an altitude where the temperature is \(-45^{\circ} \mathrm{C}\). It is expected that the temperature on the nose cone of the aircraft will be approximately equal to the stagnation temperature. Determine the stagnation temperature and compare the temperature on the nose cone to the static temperature of the surrounding air.

Air enters a conduit with a Mach number of 0.3 , a pressure of \(100 \mathrm{kPa}\), and a temperature of \(27^{\circ} \mathrm{C}\). If heat is added to the air at a rate of \(80 \mathrm{~kJ} / \mathrm{kg}\) and the flow is frictionless, estimate the Mach number, pressure, and temperature of the air at the exit of the conduit.

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.

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.

A CD nozzle with a throat diameter of \(40 \mathrm{~mm}\) connects air in a source reservoir to a discharge reservoir, where the diameter of the nozzle exit is \(90 \mathrm{~mm}\). When the pressure and temperature in the source reservoir are \(101 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\), respectively, a normal shock occurs at the nozzle exit. Determine the back pressure on the nozzle and the mass flow rate through the nozzle under this condition.

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