Question

Steam enters an adiabatic nozzle at \(3.5 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) with a low velocity and leaves at \(1.6 \mathrm{MPa}\) and \(250^{\circ} \mathrm{C}\) at a rate of \(0.4 \mathrm{kg} / \mathrm{s}\). If the ambient state is \(100 \mathrm{kPa}\) and \(18^{\circ} \mathrm{C}\) determine \((a)\) the exit velocity, \((b)\) the rate of exergy destruction, and \((c)\) the second-law efficiency.

Short answer

Question: Determine the exit velocity (a), the rate of exergy destruction (b), and the second-law efficiency (c) for an adiabatic nozzle. Step 1: Find the initial and final specific enthalpy values. Initial specific enthalpy, h₁ Final specific enthalpy, h₂ Step 2: Find the exit velocity using the conservation of energy principle. Exit velocity, v₂ Step 3: Find the initial and final specific entropy values. Initial specific entropy, s₁ Final specific entropy, s₂ Step 4: Determine the rate of exergy destruction. Rate of exergy destruction, Ė_D Step 5: Calculate the second-law efficiency. Second-law efficiency, e₂

Step-by-step solution

Find the initial and final specific enthalpy

To find the exit velocity, we need to find the initial and final specific enthalpy values. Use steam tables to find the specific enthalpy at the given pressure and temperature conditions: Initial specific enthalpy, \(h_1\) at \(3.5\ \mathrm{MPa}\) and \(300^\circ\mathrm{C}\): Final specific enthalpy, \(h_2\) at \(1.6\ \mathrm{MPa}\) and \(250^\circ\mathrm{C}\):

Find the exit velocity using the conservation of energy principle

Since the nozzle is adiabatic and there's no work interaction, we can ignore the heat and work terms in the energy equation. The conservation of energy equation for the nozzle is: \(h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}\) Given the low initial velocity, we can neglect \(v_1\): \(v_2^2 = 2(h_1 - h_2)\) Solve for the exit velocity, \(v_2\): \(v_2 = \sqrt{2(h_1 - h_2)}\)

Find the initial and final specific entropy values

Use steam tables to find the specific entropy at the given pressure and temperature conditions: Initial specific entropy, \(s_1\) at \(3.5\ \mathrm{MPa}\) and \(300^\circ\mathrm{C}\): Final specific entropy, \(s_2\) at \(1.6\ \mathrm{MPa}\) and \(250^\circ\mathrm{C}\):

Determine the rate of exergy destruction

We can find the rate of exergy destruction using the following formula: \(\dot{E}_{D} = T_0 \cdot (\dot{m} \cdot (s_2 - s_1)) - \dot{m} \cdot (h_2 - h_1) + \frac{1}{2} \cdot \dot{m} \cdot v_2^2\) Where \(T_0 = 18^\circ\mathrm{C} + 273.15 = 291.15\mathrm{K}\) is the ambient temperature. Calculate the rate of exergy destruction, \(\dot{E}_{D}\).

Calculate the second-law efficiency

The second-law efficiency is the ratio of the useful work output to the reversible work input or the exergy destroyed in the process. In our case, the useful work output is the kinetic energy of the steam, and we can calculate it using the following formula: \(e_2 = \frac{(\dot{m} \cdot h_1) - (\dot{m} \cdot h_2) + (\dot{m} \cdot v_1^2) / 2 - (\dot{m} \cdot v_2^2) / 2 - \dot{E}_D}{\dot{m} \cdot (h_1 - h_2) + (\dot{m} \cdot v_1^2) / 2 - (\dot{m} \cdot v_2^2) / 2}\) Calculate the second-law efficiency, \(e_2\). Now we have determined the exit velocity \((a)\), the rate of exergy destruction \((b)\), and the second-law efficiency \((c)\) for the adiabatic nozzle.

Key concepts

Specific Enthalpy

In thermodynamics, specific enthalpy is a property that reflects the energy content per unit mass of a substance. This energy comprises both the internal energy due to molecular motion and the energy required to push the surrounding environment to make room for the substance.

When dealing with an adiabatic nozzle like in our exercise, specific enthalpy is vital for analyzing the energy changes as steam travels through the nozzle. By consulting steam tables, students can locate the specific enthalpy at particular pressures and temperatures, important for subsequent calculations such as exit velocity.

For example, at 3.5 MPa and 300°C, the specific enthalpy of steam, denoted as \( h_1 \) can be found in the steam tables. Similarly, the final specific enthalpy \( h_2 \) at 1.6 MPa and 250°C is also determined this way. The difference in specific enthalpy values (\( h_1 - h_2 \) ) represents the energy available to be converted into kinetic energy, which dictates the exit velocity of the steam from the nozzle.

Exit Velocity Calculation

The exit velocity calculation for a fluid streaming through an adiabatic nozzle is a straightforward application of the first law of thermodynamics, particularly when the nozzle performs no work and exchanges no heat with its surroundings.

Using the equation \( v_2^2 = 2(h_1 - h_2) \) and assuming negligible initial velocity, we can derive the exit velocity (\( v_2 \) ) as the square root of twice the difference in specific enthalpy. This calculation demonstrates the conversion of thermal energy into kinetic energy—which is, quite literally, the speed at which the steam exits the nozzle.

Improving comprehension of this concept requires clarity on energy conservation principles and familiarity with manipulating algebraic equations to isolate the desired variable—in this case, the exit velocity.

Exergy Destruction Rate

The exergy destruction rate is a measure of the irreversible loss of work potential during a process due to inefficiencies. In our example, we can calculate this rate by considering both the increase in entropy and the change in specific enthalpy, as these are indicators of energy dispersion that can no longer be harnessed to perform work.

The formula \( \dot{E}_{D} = T_0 \cdot (\dot{m} \cdot (s_2 - s_1)) - \dot{m} \cdot (h_2 - h_1) + \frac{1}{2} \cdot \dot{m} \cdot v_2^2 \) incorporates ambient temperature (\( T_0 \) ), mass flow rate (\( \dot{m} \) ), and differences in specific entropy (\( s_1 \) and \( s_2 \) ) and specific enthalpy (\( h_1 \) and \( h_2 \) ). This calculation is imperative in real-world applications as it highlights the potential for improvement in energy conversion processes.

Second-Law Efficiency

Lastly, the second-law efficiency is a critical parameter quantifying how closely a thermal system approaches the ideal, reversible operation. It's a ratio that compares the useful work (or in our case, the change in kinetic energy of steam) against the maximum possible work output under reversible conditions.

The calculation of second-law efficiency, represented by the formula \( e_2 \) provided in the exercise, can be nuanced because it requires accurate figures for specific enthalpy, velocities, and also the rate of exergy destruction.

It serves as a benchmark for performance evaluation, providing insights on whether and where improvements can be made in the thermodynamic cycle to approach theoretical limits. Teaching this concept is enhanced by drawing parallels to everyday inefficient processes, emphasizing the impact of irreversibilities on practical energy utilization.

In this context, the second-law efficiency can direct engineers and students alike towards more sustainable practices and designs that align closer with the ideals of energy optimization and conservation.