Question

The pressure in a pipeline that transports helium gas at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) is maintained at \(14.5\) psia by venting helium to the atmosphere through a \(0.25\)-in-internal-diameter tube that extends \(30 \mathrm{ft}\) into the air. Assuming both the helium and the atmospheric air to be at \(80^{\circ} \mathrm{F}\), determine \((a)\) the mass flow rate of helium lost to the atmosphere through the tube, (b) the mass flow rate of air that infiltrates into the pipeline, and \((c)\) the flow velocity at the bottom of the tube where it is attached to the pipeline that will be measured by an anemometer in steady operation.

Short answer

Question: Determine the mass flow rate of helium lost to the atmosphere, the mass flow rate of air infiltrating the pipeline, and the flow velocity at the bottom of the tube. Answer: The mass flow rate of helium lost to the atmosphere is 1 kg/s. The mass flow rate of air infiltrating the pipeline is 1 kg/s. The flow velocity at the bottom of the tube is 100.6 m/s.

Step-by-step solution

Convert necessary values to SI units

Before solving, we need to convert all values to SI units. Pressure will be in Pascals (Pa), temperature in Kelvin (K), diameter in meters (m), and length in meters (m). 1. Convert \(14.5\) psia to Pa: \(1 \text{ psia} = 6895 \text{ Pa}\), so \(14.5 \text{ psia} = 14.5 \times 6895 = 100,017.5 \text{ Pa}\) 2. Convert \(80^\circ \text{F}\) to Kelvin: \(K = \frac{5}{9}(^\circ \text{F} - 32) + 273.15\), so \(80^\circ \text{F} = \frac{5}{9}(80 - 32) + 273.15 = 299.72 \text{ K}\) 3. Convert \(0.25\) in to meters: \(1 \text{ in} = 0.0254 \text{ m}\), so \(0.25 \text{ in} = 0.25 \times 0.0254 = 0.00635 \text{ m}\) 4. Convert \(30 \text{ ft}\) to meters: \(1 \text{ ft} = 0.3048 \text{ m}\), so \(30 \text{ ft} = 30 \times 0.3048 = 9.144 \text{ m}\)

Find density of helium and air using the Ideal Gas Law

1. The specific gas constant for helium is \(R_{g1} = 2077 \frac{\text{J}}{\text{kg K}}\) and for air is \(R_{g2} = 287 \frac{\text{J}}{\text{kg K}}\). 2. Calculate the density of helium: \(\rho_1 = \frac{P}{R_{g1} \times T} = \frac{100,017.5 \text{ Pa}}{2077 \frac{\text{J}}{\text{kg K}} \times 299.72 \text{ K}} = 0.163 \frac{\text{kg}}{\text{m}^3}\) 3. Calculate the density of air: \(\rho_2 = \frac{P}{R_{g2} \times T} = \frac{100,017.5 \text{ Pa}}{287 \frac{\text{J}}{\text{kg K}} \times 299.72 \text{ K}} = 1.176 \frac{\text{kg}}{\text{m}^3}\)

Calculate cross-sectional area of the tube

The cross-sectional area of the tube is: \(A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.00635}{2}\right)^2 = 3.176 \times 10^{-5} \text{m}^2\)

Calculate mass flow rate of helium lost to the atmosphere (a)

The mass flow rate of helium lost to the atmosphere can be found by calculating the mass flow rate at the top of the tube: 1. Using the given mass flow rate in the pipeline: \(Q_\text{pipeline} = 5 \frac{\text{lbm}}{\text{s}} = 2.268 \frac{\text{kg}}{\text{s}}\) (conversion: \(1 \text{ lbm} = 0.4536 \text{ kg}\)) 2. \(\Delta Q = Q_\text{pipeline} - Q_\text{lost}\) 3. Mass conservation implies the mass flow rate at the top of the tube and pipeline are equal, so: \(Q_\text{lost} = Q_\text{pipeline} - \Delta Q = 2.268 - (5 - \Delta Q) \implies \Delta Q = 2.268 - 1.268 = 1 \frac{\text{kg}}{\text{s}}\)

Calculate mass flow rate of air infiltrating the pipeline (b)

The mass flow rate of air infiltrating the pipeline is equal to the mass flow rate of helium lost to the atmosphere: \(Q_\text{air} = Q_\text{lost} = 1 \frac{\text{kg}}{\text{s}}\)

Calculate flow velocity at the bottom of the tube (c)

Using the mass flow rate equation \(Q = \rho A v\), we can solve for the velocity at the bottom of the tube: 1. The density of the mixture at the bottom of the tube can be found as the average of the densities of helium and air: \(\rho_\text{mixture} = \frac{\rho_1 + \rho_2}{2} = \frac{0.163 + 1.176}{2} = 0.669 \frac{\text{kg}}{\text{m}^3}\) 2. Calculate the flow velocity: \(v = \frac{Q_\text{pipeline}}{\rho_\text{mixture} \times A} = \frac{2.268 \frac{\text{kg}}{\text{s}}}{0.669 \frac{\text{kg}}{\text{m}^3} \times 3.176 \times 10^{-5} \text{m}^2} = 100.6 \frac{\text{m}}{\text{s}}\) The answers to the problem are as follows: (a) The mass flow rate of helium lost to the atmosphere is \(1 \frac{\text{kg}}{\text{s}}\). (b) The mass flow rate of air infiltrating the pipeline is \(1 \frac{\text{kg}}{\text{s}}\). (c) The flow velocity at the bottom of the tube is \(100.6 \frac{\text{m}}{\text{s}}\).

Key concepts

Helium Mass Flow Rate

Understanding how to calculate the helium mass flow rate in a pipeline is crucial when trying to determine how much of the gas is being used or lost. In the context of our exercise, the helium mass flow rate lost to the atmosphere is essential for maintaining the pressure within a pipeline.

To calculate the helium mass flow rate lost, we employ the principle of mass conservation. This law states that mass cannot be created or destroyed in a closed system. Considering the pipeline as a closed system, the mass flow rate at the entry must equal the exit, assuming no other losses. Therefore, the mass flow rate lost can be found by taking the initial mass flow rate within the pipeline and subtracting the flow rate of helium that remains after some of it has been vented to the atmosphere.

In the exercise solution, this was illustrated as \( Q_\text{lost} = Q_\text{pipeline} - \Delta Q \), which simplifies to \( Q_\text{lost} = 1 \frac{\text{kg}}{\text{s}} \) after applying known mass flow rates and performing the appropriate subtraction.

Air Infiltration Mass Flow

When considering a system where one gas is lost to the atmosphere, like helium in our exercise, we must also account for air that potentially infiltrates the system. This air infiltration mass flow can impact the overall system's pressure and composition.

In our problem, we've calculated the mass flow rate of helium that was lost and, by applying the same mass conservation principle, we can deduce that an equivalent amount of air has infiltrated the pipeline to balance the loss. Therefore, the mass flow rate of air infiltrating (\textbf{b}) is considered equal to the mass flow rate of the lost helium, which is \( Q_\text{air} = 1 \frac{\text{kg}}{\text{s}} \).

Flow Velocity Measurement

The measurement of flow velocity in a pipeline or tube is a common task in fluid dynamics and is directly related to the mass flow rate and density of the fluid. The flow velocity is a measure of how fast the fluid or gas is moving within a conduit. To determine the flow velocity at a particular point in a system, one can use the relationship \( Q = \rho A v \), where \( Q \) is the mass flow rate, \( \rho \) is the fluid's density, \( A \) is the cross-sectional area of the flow, and \( v \) is the flow velocity.

In the case of our exercise (\textbf{c}), the flow velocity at the bottom of the tube is calculated using the average density of the helium-air mixture and the known cross-sectional area of the tube, resulting in a flow velocity of \( 100.6 \frac{\text{m}}{\text{s}} \). This value would be the reading obtained by an anemometer placed at the bottom of the tube.

Ideal Gas Law Application

The Ideal Gas Law is a cornerstone of thermodynamics and fluid mechanics, crucial for calculating various properties of gases under certain conditions. The law relates pressure (\textbf{P}), volume (\textbf{V}), and temperature (\textbf{T}) to the amount of substance (in moles) and can be represented as \( PV = nRT \) in its simplest form. For our purposes, we use a form that helps to find the density (\textbf{\( \rho \)}): \( \rho = \frac{P}{RT} \), where \textbf{R} is the specific gas constant for the gas in question.

In the given exercise, we applied the Ideal Gas Law to calculate densities of both helium and air, which were then used to determine the mass flow rate and velocity. Calculating the density of the gases at known pressures and temperatures is vital for further calculations and the Ideal Gas Law provides an accessible way to do so.

SI Unit Conversion

In scientific calculations, particularly in fluid dynamics, it is a common practice to convert all measurements to the International System of Units (SI) to ensure consistency and accuracy. This system includes units like meters for length, kilograms for mass, and seconds for time.

In our exercise, we performed conversions for pressure from pounds per square inch absolute (psia) to pascals (Pa), Fahrenheit to Kelvin for temperature, inches to meters for diameter, and feet to meters for length. Being comfortable with SI unit conversions is essential since it allows for a standardized approach to problem-solving and prevents possible errors that might arise from unit inconsistencies.