Question

Air enters a nozzle steadily at \(200 \mathrm{kPa}\) and \(65^{\circ} \mathrm{C}\) with a velocity of \(35 \mathrm{m} / \mathrm{s}\) and exits at \(95 \mathrm{kPa}\) and \(240 \mathrm{m} / \mathrm{s}\). The heat loss from the nozzle to the surrounding medium at \(17^{\circ} \mathrm{C}\) is estimated to be \(3 \mathrm{kJ} / \mathrm{kg}\). Determine \((a)\) the exit temperature and ( \(b\) ) the exergy destroyed during this process. Answers: (a) \(34.0^{\circ} \mathrm{C}\) \((b) 36.9 \mathrm{kJ} / \mathrm{kg}\)

Short answer

Additionally, there are heat losses from the nozzle to the surrounding medium of 3 kJ/kg.

Step-by-step solution

Calculate initial total enthalpy

To find the initial total enthalpy, we need to determine the specific enthalpy and specific kinetic energy at the entrance. Given the initial temperature and pressure, we can find the specific enthalpy from air properties table. Then, we can calculate the specific kinetic energy using the formula: $$ KE = \frac{1}{2} \cdot v_1^2 $$ Where \(v_1\) is the initial velocity. Finally, the initial total enthalpy, \(h_1+\frac{1}{2}v_1^2\), can be obtained by adding the specific enthalpy and specific kinetic energy.

Apply the steady-flow energy equation

The steady-flow energy equation for a nozzle can be expressed as: $$ h_1 + \frac{1}{2}v_1^2 + Q = h_2 + \frac{1}{2}v_2^2 $$ Where subscript 2 refers to the final state, and \(Q\) is the heat loss from the nozzle to the surrounding medium, given as \(-3 \ \text{kJ/kg}\). Rearrange the equation in terms of the exit specific enthalpy, \(h_2\): $$ h_2 = h_1 + \frac{1}{2}v_1^2 - \frac{1}{2}v_2^2 - Q $$ After calculating \(h_1 + \frac{1}{2}v_1^2\), substitute the values to find \(h_2\).

Determine the exit temperature

With the obtained value of \(h_2\), we can now use the air property table to find the exit temperature (\(T_2\)), corresponding to the final specific enthalpy and pressure.

Calculate initial and final exergies

The exergy at any state can be calculated using the following formula: $$ E = (h - h_0) - T_0 (s - s_0) + \frac{1}{2}v^2 $$ Where subscript 0 refers to the dead state (ambient) conditions. Calculate the exergies at initial and final states, \(E_1\) and \(E_2\), using the given data and air property table to find \(h_0\), \(s_0\), \(h\), and \(s\).

Calculate exergy destroyed

The exergy destroyed during the process, denoted as \(E_D\), can be found using the exergy balance equation, which states that the difference between the initial and final exergies equals the exergy destroyed: $$ E_D = E_1 - E_2 $$ Substitute the previously calculated values for \(E_1\) and \(E_2\) to find the exergy destroyed during the process.

Key concepts

Steady-Flow Energy Equation

Picture a nozzle as a highway for air particles, where the rules of the road are governed by the principles of thermodynamics. One such rule is the steady-flow energy equation, a vital tool used in analyzing devices like nozzles, turbines, and compressors that have constant flow rates of fluid through them.

In the context of our nozzle problem, this equation dictates that the sum of a particle's specific enthalpy and kinetic energy at the entrance equals that at the exit, minus any heat lost to the surroundings. This is represented mathematically as:
\[h_1 + \frac{1}{2}v_1^2 + Q = h_2 + \frac{1}{2}v_2^2\]where \(h\) refers to specific enthalpy, \(v\) to velocity of the fluid, and \(Q\) to heat transfer.

Applying the Steady-Flow to Nozzles

In nozzles, air or any fluid is accelerated due to a decrease in area, turning pressure into kinetic energy. This process is adiabatic, meaning ideally, no heat is transferred. However, in reality, some heat loss occurs, and the equation above accounts for it, ensuring that the conservation of energy principle holds true throughout the process. It's this fundamental equation that aids in unraveling mysteries hidden within nozzles—including exit temperatures and velocities.

Specific Enthalpy

Delving deeper into our thermodynamic journey, we arrive at a concept crucial for understanding the energy changes in fluids: specific enthalpy. Specific enthalpy, \(h\), signifies the total energy content per unit mass of a fluid, amalgamating internal energy with the flow work done by the fluid.

This concept helps us quantify the energy changes as the fluid moves through the nozzle. In essence, it can be seen as the 'energy passport' for each fluid particle, providing a record of its energetic journey.

Energy Passport in the Context of Nozzles

Within the realms of a nozzle, the specific enthalpy accounts for both the intrinsic energy within the particles (relating to temperature) and the work done to push them through (owing to pressure). By consulting tables and using the temperature and pressure at the initial state, one can find the 'entry stamp' in this energy passport. As the fluid flows and transforms within the nozzle, its new 'departure stamp'—the exit specific enthalpy—is determined by the energy equation, which reflects the fluid's latest thermodynamic state.

Exergy Destruction

Thermodynamics isn't just about energy; it's also about quality—and that's where exergy comes into play. Exergy destruction is a measure of the potential work irreversibly lost due to inefficiencies within a system, and it's caused by factors like friction, mixing, chemical reactions, and heat transfer with the surroundings.

In our nozzle scenario, exergy destruction provides insight into the performance of the nozzle by quantifying how much useful energy is being squandered.

Understanding Exergy in a Nozzle

To calculate the exergy at a certain state, we use the formula:
\[E = (h - h_0) - T_0 (s - s_0) + \frac{1}{2}v^2\]which accounts for the specific enthalpy \(h\), ambient enthalpy \(h_0\), temperature \(T\), entropy \(s\), ambient entropy \(s_0\), and velocity \(v\). Exergy destruction therefore becomes a crucial indicator of energy efficiency. Specifically, it helps pinpoint where and how improvements can be made in the nozzle design to minimize wasted potential and maximize its thrust-producing capabilities.