Question

An ideal gas with constant specific heat ratio \(k\) enters a nozzle operating at steady state at pressure \(p_{1}\), temperature \(T_{1}\), and velocity \(\mathrm{V}_{1} .\) The air expands isentropically to a pressure of \(p_{2}\) (a) Develop an expression for the velocity at the exit, \(\mathrm{V}_{2}\), in terms of \(k, R, \mathrm{~V}_{1}, T_{1}, p_{1}\), and \(p_{2}\), only. (b) For \(\mathrm{V}_{1}=0, T_{1}=1000 \mathrm{~K}\), plot \(\mathrm{V}_{2}\) versus \(p_{2} / p_{1}\) for selected values of \(k\) ranging from \(1.2\) to \(1.4\).

Short answer

The exit velocity \( \mathrm{V}_2 \) is \( \mathrm{V}_2 = \sqrt{ 2 \left( \frac{kR}{k-1} \right) T_1 \left(1 - \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \right) + \mathrm{V}_1^2 } \).Plotting \( \mathrm{V}_2 \) vs \( \frac{p_2}{p_1} \) for \( V_1 = 0 \) and \( T_1 = 1000 \ensuremath{K} \), use \( k = 1.2 \) to \( k = 1.4 \).

Step-by-step solution

Apply the Energy Equation

For a steady flow process, the energy equation for the nozzle is given by \[ h_1 + \frac{1}{2} \mathrm{V}_1^2 = h_2 + \frac{1}{2} \mathrm{V}_2^2 \]where \(h\) is the specific enthalpy. Since the process is isentropic, we can use \( h = c_p T\).

Substitute the enthalpy terms

Using \( h = c_p T \), the energy equation becomes \[ c_p T_1 + \frac{1}{2} \mathrm{V}_1^2 = c_p T_2 + \frac{1}{2} \mathrm{V}_2^2 \]. Since the process is isentropic, relate the temperatures as part of the isentropic relations.

Use the Isentropic Relations

For an isentropic process, the following relation holds: \[ \frac{T_2}{T_1} = \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \] Substituting for \( T_2 \) gives: \[ c_p T_1 + \frac{1}{2} \mathrm{V}_1^2 = c_p T_1 \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} + \frac{1}{2} \mathrm{V}_2^2 \].

Express \( T_1 \)

Rearrange the equation to solve for \( \mathrm{V}_2 \): \[ \mathrm{V}_2 = \sqrt{ 2c_p T_1 \left(1 - \left(\frac{p_2}{p_1}\right)^{\frac{k-1}{k}} \right) + \mathrm{V}_1^2 }. \]To simplify further, use \( c_p = \frac{kR}{k-1} \).

Substitute \( c_p \) with \( \frac{kR}{k-1} \)

Substitute \( c_p \) in the equation: \[ \mathrm{V}_2 = \sqrt{ 2 \left( \frac{kR}{k-1} \right) T_1 \left(1 - \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \right) + \mathrm{V}_1^2 }. \]

Plot \( \mathrm{V}_2 - \frac{p_2}{p_1} \) for different \( k \)

Given \( V_1 = 0 \) and \( T_1 = 1000 \) K, use different values of \( k \) (from 1.2 to 1.4) to plot \( V_2 \) versus \( \frac{p_2}{p_1} \) using the derived formula: \[ \mathrm{V}_2 = \sqrt{ 2 \left( \frac{kR}{k-1} \right) 1000 \left(1 - \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \right) }. \]

Key concepts

nozzle flow

When air or any ideal gas flows through a nozzle, its pressure, temperature, and velocity change. A nozzle is a device designed to control the speed and direction of this gas flow. In many practical applications, understanding how these parameters transform is crucial for performance and efficiency.
Throughout nozzle flow, we apply thermodynamic principles to predict changes in state properties. For steady-state operations, where conditions no longer change with time, specific equations help solve for these state properties. This forms the backbone of working with nozzles in aerospace, mechanical engineering, and other fields.

steady state

Steady state refers to a condition where variables (like velocity, pressure, and temperature in a nozzle) do not change with time. This simplification allows us to focus on spatial changes. For a nozzle operating at steady state, we leverage the energy equation to understand how kinetic and internal energies transform.
In other words, while the gas flows through the nozzle, its properties might vary from entry to exit, but at any point in time, these properties remain constant. This assumption is crucial as it allows us to use mathematical relations, such as the steady-state energy equation for our calculations.

isentropic process

An isentropic process is a reversible adiabatic process, meaning no heat is transferred in or out of the system, and entropy remains constant. In our nozzle flow scenario, assuming an isentropic expansion simplifies the analysis by using specific straightforward relationships.
For instance, for an ideal gas undergoing an isentropic process, the relation between temperatures and pressures is given by:
\[ \frac{T_2}{T_1} = \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \],
where \(T_1\) and \(T_2\) are the temperatures at entry and exit, and \(p_1\) and \(p_2\) are the corresponding pressures. This relationship helps us to calculate the temperature at the nozzle's exit, given the entry conditions and pressure changes.

specific enthalpy

Specific enthalpy (h) is a measure of energy content per unit mass in a thermodynamic system. For an ideal gas, it's related to temperature and specific heat at constant pressure (\(c_p\)), given by \( h = c_p T\).
In nozzle flow analysis, we use specific enthalpy in the energy equation to determine how much kinetic energy change occurs due to the gas expansion: \[ h_1 + \frac{1}{2} \mathrm{V}_1^2 = h_2 + \frac{1}{2} \mathrm{V}_2^2 \]
By substituting appropriate values and isentropic relations, this equation enables us to solve for the exit velocity of the gas.

ideal gas

An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. The ideal gas law, given by\( PV = nRT \), relates pressure \( P \), volume \( V \), amount of gas \( n \), gas constant \( R \), and temperature \( T \).
In our nozzle flow problem, assuming air behaves as an ideal gas allows us to utilize simple mathematical relations and specific calculations. For instance, using constants and specific heat ratios \( k \) helps derive and simplify expressions for temperature, pressure, and velocity changes in an isentropic process.
Remember, while real gases may exhibit different behaviors under extreme conditions, the ideal gas assumption holds robust and simplifies many practical problems efficiently.