Question

Water vapor enters an insulated nozzle operating at steady state at \(0.7 \mathrm{MPa}, 320^{\circ} \mathrm{C}, 35 \mathrm{~m} / \mathrm{s}\) and expands to \(0.15 \mathrm{MPa}\). If the isentropic nozzle efficiency is \(94 \%\), determine the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\).

Short answer

First calculate the specific enthalpies then use efficiency and the energy balance equation to find the final velocity.

Step-by-step solution

Determine Initial and Final States

Identify the initial and final states of the water vapor. Initially, the vapor is at 0.7 MPa and 320°C with a velocity of 35 m/s. The final pressure is 0.15 MPa.

Find Properties of Steam

Use steam tables or Mollier diagram to find the specific enthalpies at the initial and final states. For the initial state (0.7 MPa, 320°C), find the specific enthalpy, \(h_1\). For the final state (0.15 MPa), determine the isentropic enthalpy, \(h_2s\).

Determine Isentropic Final Enthalpy

Using the isentropic assumption, find the specific enthalpy \(h_2s\) at the final state (0.15 MPa) where the entropy remains constant. This requires us to use the specific entropy at the initial state \(s_1\) and find the corresponding enthalpy at 0.15 MPa.

Calculate Actual Final Enthalpy

Apply the isentropic efficiency \(\eta_s\) to find the actual specific enthalpy at the final state, \(h_2\). The formula is \[h_2 = h_1 - \eta_s \cdot (h_1 - h_2s)\]

Apply Energy Balance in the Nozzle

Using the energy balance equation for a nozzle, relate the enthalpies and velocities. The equation is: \[h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}\] Find the velocities \(v_2\) at the exit by substituting the known values of enthalpies and initial velocity.

Solve for Exit Velocity \(v_2\)

Rearrange the energy balance equation to solve for the exit velocity \(v_2\): \[v_2 = \sqrt{2(h_1 - h_2) + v_1^2}\] Substitute the values of initial enthalpy \(h_1\), final enthalpy \(h_2\), and initial velocity \(v_1\) to find the final velocity \(v_2\).

Key concepts

Isentropic Process

An isentropic process is an idealized thermodynamic process that occurs at a constant entropy. This means there are no changes in entropy from the initial to the final state of the system. In practical terms, it helps in simplifying the analysis of thermodynamic processes, particularly in devices like nozzles, turbines, and compressors. During an isentropic process, the system is perfectly insulated, and there are no irreversibilities such as friction or turbulence.
In our given exercise, determining the isentropic enthalpy \( h_{2s} \) at the final state is crucial. Since the process is isentropic, the entropy at the initial state \( s_1 \) is equal to the entropy at the final state \( s_2 \). This allows us to use steam tables to find the specific enthalpy at given pressure and entropy conditions.

Specific Enthalpy

Specific enthalpy \( h \) is a property of a substance, which is the measure of the total energy content per unit mass. It’s commonly used in thermodynamic calculations to determine the energy interactions in heat exchangers, turbines, and nozzles.
In the context of the nozzle exercise, specific enthalpy values at different states are required to evaluate energy transitions and to apply the energy balance equation. These values can be found using steam tables given the pressure and temperature of the steam. For example, at an initial condition of 0.7 MPa and 320°C, we look up the specific enthalpy in the steam tables. Similarly, for the final state enthalpy at a different pressure, steam tables help again in finding the appropriate values.
The formula used in our exercise to relate isentropic and actual enthalpies is:
\[ h_{2} = h_{1} - \eta_{s} (h_{1} - h_{2s}) \]
This relation incorporates the isentropic efficiency to determine the actual enthalpy after accounting for real-world inefficiencies.

Steam Tables

Steam tables are essential tools in thermodynamics that list the properties of water and steam, such as temperature, pressure, specific enthalpy, and specific entropy. They allow engineers and students to quickly find the thermodynamic properties of steam at various conditions, crucial for performing energy calculations.
In our nozzle efficiency exercise, we use steam tables to determine the specific enthalpy at both the initial and final states. Here’s how we use them:

• First, obtain the specific enthalpy \( h_{1} \) at the initial state with given conditions (0.7 MPa, 320°C).
• Next, assuming an isentropic process, find the specific entropy \( s_{1} \) and use it to determine the specific enthalpy \( h_{2s} \) at the final pressure (0.15 MPa).

Steam tables are invaluable as they provide these thermodynamic properties directly, without requiring complex calculations, thus simplifying the analysis of thermodynamic cycles and processes.