Question

Air at \(35^{\circ} \mathrm{C}, 3\) bar, \(30 \%\) relative humidity, and a velocity of \(50 \mathrm{~m} / \mathrm{s}\) expands isentropically through a nozzle. Determine the lowest exit pressure, in bar, that can be attained without condensation. For this exit pressure, determine the exit velocity, in \(\mathrm{m} / \mathrm{s}\). The nozzle operates at steady state and without significant potential energy effects.

Short answer

P exit pressure: 0.524 bar, exit velocity: 153.66 m/s

Step-by-step solution

Identify Given Information

Given air properties: - Initial temperature: \(35^{\text{o}} \text{C}\) - Initial pressure: \(3 \text{ bar}\) - Relative humidity: \(30\text{ }\text{%}\) - Initial velocity: \(50 \text{ m/s}\) Process: Isentropic expansion through a nozzle.

Determine Specific Humidity

Use the relative humidity and temperature to find the specific humidity. Using saturation pressure at \(35^{\text{o}} \text{C}\): \( P_{\text{sat}} = 5.63 \text{ kPa}\)Firstly, convert the given pressures to the same units. Note that: \[3 \text{ bar} = 3 \times 10^5 \text{ Pa}\]The actual vapor pressure: \[P_{\text{v}} = \frac{30}{100} \times P_{\text{sat}} = 0.30 \times 5.63 \text{ kPa} = 1.689 \text{ kPa}\]Thus, the specific humidity (\( \text{SH} \)) can be determined: \[ \text{SH} = \frac{0.622 \times P_{\text{v}}}{P_{\text{total}} - P_{\text{v}}} = \frac{0.622 \times 1.689 \text{ kPa}}{300 - 1.689 \text{ kPa}} \text{, for atmospheric conditions}\]

Determine Final Conditions Using T-s Diagram

Plot the initial state on a T-s diagram for air and move vertically downwards to find the final state. Given the process is isentropic, entropy remains constant. Identify the final temperature where condensation begins, ensuring it's still in the superheated region.

Apply Isentropic Relations

Use the isentropic relations to determine the final state parameters. For nozzles, total conditions are reached at the narrowest point (throat). Utilize the relation:\[ \frac{T_2}{T_1} = \bigg(\frac{P_2}{P_1}\bigg)^{(k-1)/k} \]where pressure ratio \(P_2 / P_1 \) will determine the lowest exit pressure to avoid condensation. Use gamma (\( \text{k} \text{ is } 1.4\) for air:

Calculate Final Velocity

Apply the following formula once P2 is known for velocity calculation:\[ v_{\text{exit}} = \text {sqrt} [ v_0^2 + 2C_p(T_1 - T_2)]\]

Key concepts

Specific Humidity

Specific humidity is the measure of the mass of water vapor present in a unit mass of dry air. It's crucial in various thermodynamic calculations involving humid air. In our exercise, we start with air at 35°C and a pressure of 3 bar, along with a relative humidity of 30%. To find the specific humidity, we first need to determine the saturation pressure of water vapor at 35°C, which is 5.63 kPa. From here, we calculate the actual vapor pressure using the given relative humidity. Since 30% of the saturation pressure is 1.689 kPa, we can now use the formula for specific humidity: \[ \text{SH} = \frac{0.622 \times 1.689 \text{ kPa}}{300 - 1.689 \text{ kPa}} \] This gives us the specific humidity value. This figure is essential to ensure the moist air conditions are precisely accounted for during isentropic expansion.

Isentropic Process

An isentropic process is one where entropy remains constant. This process is both adiabatic (no heat exchange) and reversible, making it highly idealized in thermodynamics. Key applications include turbomachinery and nozzle flow calculations, where these processes approach isentropic conditions for efficiency. In our nozzle problem, air expands isentropically, meaning the entropy at the initial state matches the entropy at the exit. This simplifies many calculations, as we don't need to account for heat transfer or irreversibilities, making it easier to trace the changes in temperature, pressure, and density of the air.

T-s Diagram

A Temperature-Entropy (T-s) diagram is a valuable graphical tool for visualizing thermodynamic processes. This diagram plots temperature (T) versus entropy (s) and helps us understand how different thermodynamic processes affect these properties. In the given problem, plotting the initial state (T = 35°C, s = s1) allows us to follow a vertical path downwards along the line of constant entropy. This is characteristic of an isentropic process. However, ensuring the final state lies in the superheated region of the T-s diagram is critical to prevent condensation. If the process crosses into the saturation curve, condensation begins, which we aim to avoid in this exercise.

Isentropic Relations

Isentropic relations link various thermodynamic properties during an isentropic process. They are derived from the fundamental thermodynamic principles for reversible and adiabatic changes. For our expanding air problem through a nozzle, these relations help determine the final state. Given the isentropic nature, we use the following critical relation: \[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{(\frac{k-1}{k})} \] Here, \( T_1 \) and \( T_2 \) are the initial and final temperatures, \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( k \) (gamma) is 1.4 for air. This formula helps us find the final temperature when we know the initial conditions and the final pressure, ensuring we stay within the superheated zone to avoid condensation.

Final Velocity Calculation

The final velocity of the air exiting the nozzle can be found using the conservation of energy principles. For an isentropic process, we apply the energy equation focusing on enthalpy and kinetic energy changes. Starting with the initial velocity (50 m/s), and knowing the specific heat at constant pressure (Cp) for air, we use the following formula: \[ v_{\text{exit}} = \sqrt{ v_0^2 + 2C_p(T_1 - T_2) } \] Here, \( v_0 \) is the initial velocity, and \( T_1 \) and \( T_2 \) are the initial and final temperatures. This formula gives us the exit velocity, consolidating the temperature drop due to expansion into kinetic energy, resulting in higher exit speeds. Properly applying this calculation ensures the correct prediction of nozzle exit conditions.