Question

An ideal gas mixture with \(k=1.31\) and a molecular weight of 23 is supplied to a converging nozzle at \(p_{\mathrm{o}}=5\) bar, \(T_{\mathrm{o}}=\) \(700 \mathrm{~K}\), which discharges into a region where the pressure is 1 bar. The exit area is \(30 \mathrm{~cm}^{2}\). For steady isentropic flow through the nozzle, determine (a) the exit temperature of the gas, in \(\mathrm{K}\). (b) the exit velocity of the gas, in \(\mathrm{m} / \mathrm{s}\). (c) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\).

Short answer

Exit Temperature: Solve the temperature equation to get \(T_e\approx 470 \ \mathrm{K}\). Exit Velocity: Using the previous steps, calculate \(V_e\approx 632 \ \mathrm{m/s}\). Mass Flow Rate: Combine steps to find \(\dot{m}\approx 0.349 \ \mathrm{kg/s}\).

Step-by-step solution

Calculate the Exit Temperature

To find the exit temperature of the gas, use the isentropic relation for temperatures in terms of the stagnation temperature and pressure ratio: \(\frac{T}{T_0} = \left(\frac{p}{p_0}\right)^\frac{k-1}{k}\).Given values: - Stagnation temperature \(T_0 = 700 \ \mathrm{K}\) - Stagnation pressure \(p_0 = 5 \ \mathrm{bar}\) - Exit pressure \(p = 1 \ \mathrm{bar}\) - Specific heat ratio \(k = 1.31\).Substitute the given values into the formula to find the exit temperature: \[\frac{T}{700 \ \mathrm{K}} = \left( \frac{1 \ \mathrm{bar}}{5 \ \mathrm{bar}} \right)^\frac{1.31-1}{1.31}\]Solve for \(T\): \[T = 700 \ \mathrm{K} \times \left( \frac{1}{5} \right)^\frac{0.31}{1.31}\].Evaluate the expression to find \(T\).

Calculate the Exit Velocity

Utilize the isentropic relation for velocity under steady isentropic flow: \[V_{e} = \sqrt{\frac{2k}{k-1} R T_0 \left( 1 - \left( \frac{p}{p_0} \right)^{\frac{k-1}{k}} \right)} \].Given values: - Specific heat ratio \(k = 1.31\) - Gas constant for the gas \(R = \frac{R_u}{\text{molecular weight}} = \frac{8.314}{23} \ \mathrm{J/(mol \ \cdot \ K)} \) - Temperature at the inlet \(T_0 = 700 \ \mathrm{K}\) - Stagnation and exit pressures as provided.Substitute the given values to find the exit velocity: \[ V_{e} = \sqrt{\frac{2 \times 1.31}{1.31-1} \times \frac{8.314}{23} \times 700 \ \left( 1 - \left( \frac{1}{5} \right)^\frac{0.31}{1.31} \right) } \].Evaluate the expression to find \(V_e\).

Calculate the Mass Flow Rate

To determine the mass flow rate, use the following continuity equation in terms of density, velocity, and area: \[\dot{m} = \rho \cdot V \cdot A \].The density (\(\rho\)) can be calculated from the ideal gas law: \(\rho = \frac{p}{RT}\).Given values: - Exit pressure \( p = 1 \ \mathrm{bar} = 100 \ \mathrm{kPa} \) - Gas constant \( R = \frac{8.314}{23} \ \mathrm{J/(mol \ \cdot \ K)} \) - Exit temperature calculated previously.Calculate \(\rho\):\[ \rho_e = \frac{100 \times 10^3}{ \frac{8.314}{23} \times T_e} \].Then, use the exit velocity, the density, and the exit area (converted to square meters):\[ A_e = 30 \ \mathrm{cm}^2 = 0.003 \ \mathrm{m}^2 \].Combine them to find \(\dot{m}\):\[\dot{m} = \rho_e \times V_e \times 0.003 \].Evaluate the expression to find the mass flow rate.

Key concepts

isentropic relation for temperature

The isentropic relation for temperature is a key concept when dealing with flows through nozzles. This concept connects the temperature of a gas at one point to its pressure at another point, assuming the process is isentropic (no heat transfer and reversible). The relation is given by the equation:
\(\frac{T}{T_0} = \bigg(\frac{p}{p_0}\bigg)^{\frac{k-1}{k}}\).

Here:
- \T\ is the exit temperature.
- \T_0\ is the stagnation temperature (700 K in our problem).
- \p\ is the exit pressure (1 bar).
- \p_0\ is the stagnation pressure (5 bar).
- \k\ is the specific heat ratio (1.31 for our ideal gas mixture).
You simply need to substitute the known values into the formula and solve for \T\ using basic algebra. It's a straightforward, yet fundamental step in evaluating the conditions at the nozzle's exit. This relationship will give us insight into how temperature drops as the gas expands through the nozzle.

exit velocity calculation

In isentropic flow, the exit velocity can be calculated using the energy equation. This equation accounts for the conversion of thermal energy into kinetic energy as the gas flows through the nozzle. The formula used is:
\[V_{e} = \sqrt{ \frac{2k}{k-1}RT_0 \bigg(1 - (\frac{p}{p_0})^{\frac{k-1}{k}}\bigg)} \]

Here:
- \V_e\ is the exit velocity.
- \k\ is the specific heat ratio (1.31).
- \R\ is the specific gas constant, which is derived by dividing the universal gas constant by the molecular weight: \R = \frac{8.314}{23} J/(mol \cdot K)\.
- \T_0\ is the stagnation temperature (700 K).
- \p\ is the exit pressure (1 bar).
- \p_0\ is the stagnation pressure (5 bar).
By substituting given values and solving the above equation, you can determine the exit velocity of the gas. The rapid decrease in pressure results in an increase in velocity, illustrating the conversion of thermal energy into kinetic energy.

mass flow rate

The mass flow rate is an essential concept in fluid dynamics. It denotes the amount of mass passing through a given section of the nozzle per unit time. The formula used for calculating the mass flow rate is:
\[\backslashdot{m} = \rho \V \A \]

Here:
- \rho\ is the density of the gas.
- \V\ is the velocity of the gas.
- \A\ is the cross-sectional area of the nozzle exit.

To find the density, you can use the ideal gas law:
\[\rho = \frac{p}{RT} \]

Substitute the known values:
- Exit pressure \p = 100\ kPa (1 bar).
- Gas constant \R = \frac{8.314}{23} \ J/(mol \cdot K).
- Exit temperature \T \ found from the isentropic relation.
After calculating the density, you can integrate it with the exit velocity and the area (which should be converted to square meters: 30 cm^2 = 0.003 m^2) to calculate the mass flow rate. This quantity helps in understanding the total mass ejected by the nozzle per second.

ideal gas law

The ideal gas law is a fundamental relation connecting pressure, volume, and temperature of an ideal gas. This law can also be used to find the density of the gas once the pressure and temperature are known. The formula is:
\[\rho = \frac{p}{RT} \]

Here:
- \rho\ is the density of the gas.
- \p\ is the pressure.
- \R\ is the specific gas constant.
- \T\ is the temperature.
By substituting known values (like pressure at the exit and exiting temperature), you can compute the density of the gas.

The ideal gas law is essential in linking macroscopic properties of gases and aiding other calculations in the context of fluid flows through nozzles, ensuring that all thermodynamic properties are consistently aligned. This assists in comprehensive understanding and accurate computations.