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Chapter 3: Problem 24

A gas with a specific weight of \(7.75 \mathrm{~N} / \mathrm{m}^{3}\) flows at a rate of \(0.500 \mathrm{~kg} / \mathrm{s}\) through a \(350 \mathrm{~mm} \times 510 \mathrm{~mm}\) duct. Estimate the average velocity in the duct.

Short Answer

Expert verified
The average velocity in the duct is approximately 3.54 m/s.

Step by step solution

01

Understand the Problem

We need to find the average velocity of a gas flowing through a duct given its specific weight, mass flow rate, and duct dimensions. The specific weight of the gas is given as \(7.75 \text{ N/m}^3\), and the mass flow rate is \(0.5 \text{ kg/s}\). The duct dimensions are \(350 \text{ mm} \times 510 \text{ mm}\).
02

Convert Duct Dimensions from mm to m

Convert the dimensions of the duct from millimeters to meters for consistency in units.\[ 350 \text{ mm} = 0.35 \text{ m} \]\[ 510 \text{ mm} = 0.51 \text{ m} \]
03

Calculate Cross-Sectional Area of the Duct

Calculate the cross-sectional area (\(A\)) using the formula:\[ A = ext{width} \times ext{height} \]Substitute the values:\[ A = 0.35 \text{ m} \times 0.51 \text{ m} = 0.1785 \text{ m}^2 \]
04

Calculate Gas Density

Use the specific weight \(\gamma\) to calculate the density (\(\rho\)) of the gas:\[ \gamma = \rho \cdot g \]Given \(\gamma = 7.75 \text{ N/m}^3\) and \(g = 9.81 \text{ m/s}^2\):\[ \rho = \frac{7.75 \text{ N/m}^3}{9.81 \text{ m/s}^2} \approx 0.79 \text{ kg/m}^3 \]
05

Calculate Average Velocity

The average velocity (\(V\)) can be calculated using the formula:\[ \dot{m} = \rho \cdot A \cdot V \]Solve for \(V\):\[ V = \frac{\dot{m}}{\rho \cdot A} \]Substitute the given and calculated values:\[ V = \frac{0.5 \text{ kg/s}}{0.79 \text{ kg/m}^3 \times 0.1785 \text{ m}^2} \approx 3.54 \text{ m/s} \]

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Key Concepts

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

Specific Weight
The specific weight of a substance is essentially how much weight is concentrated in a certain volume, typically expressed in \( ext{N/m}^3\). It gives us an idea of how heavy the gas is in relation to the space it occupies. To understand this better, think of specific weight as being closely related to both density and gravity. If the specific weight is higher, it means the substance feels heavier for every cubic meter. The formula connecting these is \( ext{Specific weight} = ext{density} \times ext{gravity}\). In our case, the gas had a specific weight of \( ext{7.75 N/m}^3\). By knowing the specific weight, you can reverse-engineer to find density, as shown in the exercise. This is invaluable in fluid mechanics when dealing with different types of gases and liquids. Understanding their specific weight helps in understanding how they behave under different forces.
Mass Flow Rate
Mass flow rate describes how much mass of a gas or liquid passes through a given point or section in space per unit time. It is typically measured in \( ext{kg/s}\). In simpler terms, it tells you how much of the fluid is "flowing" per second.In the exercise, the mass flow rate was given as \( ext{0.500 kg/s}\). With this value, fluid mechanics concepts tell us that we can determine various other quantities, such as velocity or density, provided other parameters are also available.To calculate the average velocity of the gas in the duct, mass flow rate becomes pivotal when combined with the density and cross-sectional area of the duct. \[ \dot{m} = \rho \times A \times V \] is the key formula connecting them all. This formula can be rearranged to find any one of those terms if the others are known, illustrating the interconnected nature of fluid properties.
Duct Dimensions
When considering the flow of gases through a duct, the dimensions of the duct are crucial. They give us the cross-sectional area through which the gas passes. Initially provided in millimeters, these dimensions need to be converted to meters to maintain unit consistency.In the exercise, we converted the duct dimensions of \( ext{350 mm}\) and \( ext{510 mm}\) to \( ext{0.35 m}\) and \( ext{0.51 m}\), respectively. The cross-sectional area of the duct was then calculated using: \[ A = ext{width} \times ext{height} = 0.35 ext{ m} \times 0.51 ext{ m} = 0.1785 ext{ m}^2 \] With cross-sectional area, we can predict how a given mass flow will behave in terms of speed and pressure across the duct. This practically helps engineers in designing HVAC systems, pipelines, and other fluid-conveying structures.
Gas Density
Gas density is an essential property when studying fluid mechanics. It tells us how much mass is contained within a unit volume, expressed in \( ext{kg/m}^3\). The density of a gas gives us insight into its compression, expansion, and overall behavior under different conditions.In the exercise, density was calculated using the specific weight and the gravitational constant: \[ \rho = \frac{\text{Specific weight}}{\text{gravity}} \] Plugging in, \( ho = \frac{7.75 \, \text{N/m}^3}{9.81 \, \text{m/s}^2} \approx 0.79 \, \text{kg/m}^3\). Knowing this density allows us to further calculate the average velocity of the gas in the duct using our relationships from fluid mechanics. Density is not just a static property but dynamically interacts with pressure and temperature, embodying the ideal gas law. Mastery of these concepts empowers students with the knowledge to solve a range of real-world engineering problems.

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Most popular questions from this chapter

An orifice meter consists of an arrangement in which a plate containing a circular orifice at its center is placed normal to the flow in a conduit, and the flow rate in the conduit is related to the measured difference in pressure across the orifice. Consider the orifice meter shown in Figure \(3.52,\) where the orifice has a diameter of \(50 \mathrm{~mm}\) and a discharge coefficient of 0.62 and the conduit has a diameter of \(100 \mathrm{~mm} .\) The fluid is water at \(20^{\circ} \mathrm{C},\) and the differential pressure gauge shows a pressure difference of \(30 \mathrm{kPa}\). What is the volume flow rate in the conduit?

The temperature, \(T,\) and the vertical component, \(w,\) of the wind velocity on the side of a very steep cliff are approximated by the relations $$ T(z, t)=20\left(1-0.3 z^{2}\right) \sin \left(\frac{\pi t}{6}\right){ }^{\circ} \mathrm{C}, \quad w=2.1\left(1+0.5 z^{2}\right) \mathrm{m} / \mathrm{s} $$ where \(z\) is the elevation above sea level in \(\mathrm{km}\) and \(t\) is the time in seconds. The horizontal components of the wind velocity are negligible along the cliff. Estimate the rate of change of temperature in the wind at \(z=1.2 \mathrm{~km}\) and \(t=5400 \mathrm{~s}(1.5 \mathrm{~h})\).

Measurements in the outer regions of a hurricane indicate a pressure of \(101 \mathrm{kPa}\) and velocities of \(6 \mathrm{~m} / \mathrm{s}\) and \(3 \mathrm{~m} / \mathrm{s}\) in the radial and tangential directions, respectively. If the pressure at the center of the hurricane is estimated as \(98 \mathrm{kPa}\), what wind speed do you expect at the center?

The velocity distribution in a pipe with a circular cross section under turbulent flow conditions can be estimated by the relation $$ v(r)=V_{0}\left(1-\frac{r}{R}\right)^{\frac{1}{7}} $$ where \(v(r)\) is the velocity at a distance \(r\) from the centerline of the pipe, \(V_{0}\) is the centerline velocity, and \(R\) is the radius of the pipe. (a) Calculate the average velocity and the volume flow rate in the pipe in terms of \(V_{0} .\) Express your answers in rational form. (b) Based on the result in part (a), assess the extent to which the velocity can be assumed to be constant across the cross section.

Experiments indicate that the shear stress, \(\tau_{0}\), on the wall of a 200 -mm pipe can be related to the flow in the pipe using the relation $$ \tau_{0}=0.04 \rho V^{2} $$ where \(\rho\) and \(V\) are the density and velocity, respectively, of the fluid in the pipe. If water at \(20^{\circ} \mathrm{C}\) flows in the pipe at a flow rate of \(60 \mathrm{~L} / \mathrm{s}\) and the pipe is horizontal, estimate the pressure drop per unit length along the pipe.
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