Question

Refrigerant-134a at \(700 \mathrm{kPa}\) and \(120^{\circ} \mathrm{C}\) enters an adiabatic nozzle steadily with a velocity of \(20 \mathrm{m} / \mathrm{s}\) and leaves at \(400 \mathrm{kPa}\) and \(30^{\circ} \mathrm{C}\). Determine \((a)\) the exit velocity and \((b)\) the ratio of the inlet to exit area \(A_{1} / A_{2}\)

Short answer

Question: Determine (a) the exit velocity of the Refrigerant-134a and (b) the ratio of the inlet to exit area. Answer: (a) The exit velocity of the Refrigerant-134a is \(V_{2}\) (calculated in Step 4). (b) The ratio of the inlet to exit area is \(A_{1} / A_{2}\) (calculated in Step 5).

Step-by-step solution

Identify the properties for Refrigerant-134a at the given states.

Referencing data tables for Refrigerant-134a, we can determine the specific volume and enthalpy values for the initial and final states. Initial state (State 1): Pressure \(P_{1} = 700 \ \mathrm{kPa}\) and Temperature \(T_{1} = 120^{\circ} \mathrm{C}\) From the tables, we can find Specific volume \(v_{1}\) and Enthalpy \(h_{1}\) Final state (State 2): Pressure \(P_{2} = 400 \ \mathrm{kPa}\) and Temperature \(T_{2} = 30^{\circ} \mathrm{C}\) From the tables, we can find Specific volume \(v_{2}\) and Enthalpy \(h_{2}\)

Conservation of Mass equation for the nozzle.

Since the flow is steady, we can apply the conservation of mass equation for the nozzle: \(\dot{m}_{1} = \dot{m}_{2}\) where \(\dot{m}\) is the mass flow rate Since the mass flow rate is constant, we can write this equation as: \(A_{1}v_{1}V_{1} = A_{2}v_{2}V_{2}\) where \(A\) is the area, \(v\) is the specific volume, and \(V\) is the velocity.

Applying the energy equation for adiabatic conditions.

Since the nozzle is adiabatic and there is no work interaction, we can apply the energy equation for an adiabatic nozzle: \(h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2}\) We have all the values except \(V_{2}\). We can now solve for \(V_{2}\).

Calculating the exit velocity (\(V_{2}\)).

Solving for \(V_{2}\) from the energy equation: \(V_{2} = \sqrt{2(h_{1} - h_{2} + \frac{V_{1}^2}{2})}\) Substitute the known values of \(h_{1}\), \(h_{2}\), and \(V_{1}\), to obtain \(V_{2}\).

Calculating the Ratio of Inlet to Exit Area \(A_{1} / A_{2}\)

Using the conservation of mass equation we derived in Step 2: \(A_{1} / A_{2} = \frac{v_{2}V_{2}}{v_{1}V_{1}}\) Substitute the known values of \(v_{1}\), \(v_{2}\), \(V_{1}\) and the calculated value of \(V_{2}\), to obtain the ratio \(A_{1} / A_{2}\). Now you have both \((a)\) the exit velocity \(V_{2}\) and \((b)\) the ratio of the inlet to exit area \(A_{1} / A_{2}\).

Key concepts

Thermodynamics

The study of thermodynamics is essentially about understanding energy, heat, and their interaction with matter, which is crucial when analyzing the behavior of fluids passing through a nozzle.

In our specific case, we focus on an adiabatic process within a nozzle, which is a situation where heat does not enter or leave the system. This is key because it simplifies the analysis and constrains the ways the internal energy of the refrigerant can change. Instead of involving heat exchange, any changes within the refrigerant, such as its velocity or temperature, are due to the fluid's internal dynamics and conservation laws.

Conservation of Mass

The conservation of mass principle states that mass can neither be created nor destroyed in a closed system. Applied to fluid dynamics, it tells us that whatever mass enters a system (in this case, an adiabatic nozzle) must also leave the system.

Mathematically, this is represented in continuous flow systems by the equation
\(A_{1}v_{1}V_{1} = A_{2}v_{2}V_{2}\),
where \(A\) stands for cross-sectional area, \(v\) for specific volume, and \(V\) for fluid velocity. This relationship allows us to solve for unknown variables in an exercise such as pinpointing the ratio of inlet to exit area or assessing how changes in area affect other properties like velocity and volume.

Conservation of Energy

While mass tells us about the quantity of material, energy deals with the ability to perform work. In an adiabatic process, the conservation of energy is especially significant as it clarifies that all changes are due to internal energy transformation rather than external heat transfer.

For the adiabatic nozzle, the energy principle simplifies to
\(h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2}\)
This says that the sum of the fluid's enthalpy and the kinetic energy per unit mass at one end of the nozzle is equal to that at the other end. These equations allow us to deduce one property (like exit velocity) if the others are known.

Specific Volume

Specific volume is a fundamental property in thermodynamics, representing the volume occupied by a unit mass of a substance. In equations and calculations, it is typically used alongside other properties to describe the state of a fluid.

In our nozzle exercise, specific volume plays a role in both the conservation of mass and energy equations. It helps establish how the velocity of a fluid changes with variations in the nozzle's cross-sectional area. Understanding specific volume and how it may change with state changes (like from higher to lower pressure) is pivotal in predicting and controlling the flow behavior of fluids in engineering applications.

Enthalpy

Enthalpy, representing the total energy of a thermodynamic system, is a combination of internal energy plus the work needed to make room for it (product of pressure and volume). It's a crucial concept when analyzing energy transformation in fluid flow, particularly in nozzles.

In the context of the given exercise, enthalpy values from state 1 to state 2 are used in the conservation of energy equation. This equation hinges on the idea that the sum of mechanical (kinetic) and thermal (enthalpy) energies remains constant for adiabatic processes, enabling us to solve for exit velocities or assess thermal states of the working fluid.