Problem 17 Evaluate the potential energy \(... [FREE SOLUTION] (2024)

Chapter 8: Problem 17

Evaluate the potential energy \(u\) per unit mass as a function of \(p\) for aperfect gas of molecular weight \(M\) at temperature \(T\). For the steadyisothermal flow of this gas through a pipe of varying cross section andvarying height above the earth, find expressions for the pressure, density,and velocity of the gas as functions of the cross section \(S\) of the pipe, theheight \(h\), and the pressure \(p_{0}\) and velocity \(v_{0}\) at a point in thepipe at height \(h=0\) where the cross section is \(S_{0}\). Assume \(p, v\), and\(\rho\) uniform over the cross section.

Short Answer

Expert verified

Pressure, density, and velocity can be derived using Bernoulli's equation, the continuity equation, and the ideal gas law. Specifically, they are functions of the cross-section area, height, reference pressure, and velocity.

Step by step solution

- Define Potential Energy per Unit Mass

For a perfect gas, the potential energy per unit mass, \(u\), due to gravity is given by \(u = gh\), where \(g\) is the gravitational acceleration and \(h\) is the height above the reference point.

- Bernoulli's Equation Application

For steady isothermal flow of a gas, Bernoulli’s equation can be applied: \[ \frac{v^2}{2} + gh + \frac{p}{\rho} = \text{constant} \] Here, \(v\) is the velocity, \(g\) is the acceleration due to gravity, and \(p\) is the pressure. We use this equation to relate the variables at different points in the pipe.

- Apply Continuity Equation

The continuity equation for incompressible flow states that \[ \rho v S = \text{constant} \] where \(S\) is the cross-sectional area. This implies that \[ vS = \frac{v_0 S_0}{\rho / \rho_0} \]

- Ideal Gas Law

Recall the ideal gas law: \[ p = \rho R T / M \] where \(R\) is the universal gas constant, \(T\) is the temperature, and \(M\) is the molecular weight of the gas.

- Combine Equations to Find Pressure

Combine Bernoulli's equation and the continuity equation with the ideal gas law. After algebraic manipulations, solve to find the expression for pressure, \(p\), as a function of \(S\), \(h\), \(p_0\), \(v_0\), and \(S_0\).

- Compute Density

Using the ideal gas law, express density, \(\rho\), in terms of the newly derived pressure, temperature \(T\), and molecular weight \(M\) of the gas.

- Calculate Velocity

Using the continuity equation, solve for velocity \(v\) as a function of the given variables. With pressure and density known, substitute these expressions back into the equation.

Key Concepts

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

potential energy

Potential energy is the energy possessed by an object due to its position in a gravitational field. For a perfect gas at height \(h\) above the earth, the potential energy per unit mass is given by \(u = gh\). Here, \(g\) represents the acceleration due to gravity. This energy increases as the height \(h\) increases. It's important to consider potential energy when analyzing gas flow in a pipe with varying heights.

In our exercise, understanding potential energy helps in applying Bernoulli's equation to determine how pressure and velocity change with height.

Bernoulli's equation

Bernoulli's equation is a principle of fluid dynamics that describes the behavior of a moving fluid. For a steady isothermal flow, the equation is given by:
\[ \frac{v^2}{2} + gh + \frac{p}{\rho} = \text{constant} \]
Where:

\(v\) is the velocity of the gas
\(gh\) is the potential energy per unit mass
\(p\) is the pressure
\(\rho\) is the density

This equation implies that in a steady flow, the sum of kinetic energy (\( \frac{v^2}{2} \)), potential energy (\(gh\)), and pressure energy (\( \frac{p}{\rho} \)) remains constant.
In our context, Bernoulli's equation helps relate the pressure, velocity, and height of the gas at different points in the pipe.

continuity equation

The continuity equation is a fundamental principle that states the mass flow rate of a fluid remains constant from one cross-section to another in a pipe. For incompressible flow, it is expressed as:
\[ \rho v S = \text{constant} \]
Where:

\( \rho \) is the density of the gas
\( v \) is the velocity of the gas
\( S \) is the cross-sectional area of the pipe

Since our flow is steady and the gas is effectively incompressible over short distances, the volume flow rate \( vS \) at any two points in the pipe must be equal:
\[ vS = \frac{v_0 S_0}{\rho / \rho_0} \]
Where \(v_0\), \(S_0\), and \(\rho_0\) are the initial velocity, cross-sectional area, and density, respectively. This helps us understand how velocity changes with cross-sectional area and density.

ideal gas law

The ideal gas law is a fundamental principle used to relate the pressure, volume, temperature, and amount of gas. It is given by:
\[ p = \rho \frac{RT}{M} \]
Where:

\(p\) is the pressure
\( \rho \) is the density of the gas
\( R \) is the universal gas constant
\( T \) is the temperature
\( M \) is the molecular weight of the gas

This law tells us that for a given temperature and molecular weight, the pressure of the gas is directly proportional to its density.
In our problem, we use the ideal gas law to express the density \( \rho \) in terms of pressure, temperature, and molecular weight, allowing us to connect it with Bernoulli's equation and the continuity equation to find the pressure, density, and velocity at different points in the pipe.

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