Question

Refrigerant- 134 a enters a diffuser steadily as saturated vapor at \(600 \mathrm{kPa}\) with a velocity of \(160 \mathrm{m} / \mathrm{s}\), and it leaves at \(700 \mathrm{kPa}\) and \(40^{\circ} \mathrm{C}\). The refrigerant is gaining heat at a rate of \(2 \mathrm{kJ} / \mathrm{s}\) as it passes through the diffuser. If the exit area is 80 percent greater than the inlet area, determine ( \(a\) ) the exit velocity and \((b)\) the mass flow rate of the refrigerant.

Short answer

Question: Determine the exit velocity and mass flow rate of a refrigerant (R-134a) in a diffuser given the initial state (saturated vapor at 600 kPa, velocity 160 m/s), final state (700 kPa, 40°C), and the rate at which it gains heat (2 kJ/s). The exit area of the diffuser is 80% greater than the inlet area.

Step-by-step solution

Determine refrigerant properties at the initial and final states

Since R-134a is entering the diffuser as saturated vapor at 600 kPa, we can find its specific volume (\(v_{1}\)) and specific enthalpy (\(h_{1}\)) using a R-134a property table. Similarly, we can find the specific volume (\(v_{2}\)) and specific enthalpy (\(h_{2}\)) for the refrigerant at the exit state (700 kPa, 40°C).

Apply conservation of mass

For steady flow, the mass flow rate at the inlet and outlet will be equal. So, we can set up the equation: \(\dot{m}_{1} = \dot{m}_{2}\) Since the mass flow rate is the product of density, velocity, and cross-sectional area: \(\rho_{1} A_{1} V_{1} = \rho_{2} A_{2} V_{2}\) We also know that the cross-sectional area at the exit (\(A_{2}\)) is 80% greater than the area at the inlet (\(A_{1}\)): \(A_{2} = 1.8A_{1}\) Applying the ideal gas law, we can determine the densities of the refrigerant at the inlet and outlet: \(\rho_{1} = \frac{P_{1}}{R_{134a} T_{1}}\) \(\rho_{2} = \frac{P_{2}}{R_{134a} T_{2}}\) Now, we substitute the above equations into the mass flow rate equation and solve for \(V_{2}\) (the exit velocity).

Apply conservation of energy

The steady-flow energy equation for the diffuser can be written as: \(\dot{m}(h_{1}+\frac{V_{1}^{2}}{2}) + Q_{in} = \dot{m}(h_{2}+\frac{V_{2}^{2}}{2})\) We know the heat input rate: \(Q_{in} = 2\,\text{kJ/s}\) Substituting the known values and the expression of \(V_{2}\) obtained from the mass conservation equation, we can solve for the mass flow rate (\(\dot{m}\)).

Calculate the exit velocity and mass flow rate

Plug in the values obtained in Steps 1, 2, and 3, and solve for \(V_{2}\) and \(\dot{m}\): Exit velocity (\(V_{2}\)): (calculate from equations in Step 2) Mass flow rate (\(\dot{m}\)): (calculate from equations in Step 3) With these calculations, we have determined the exit velocity and mass flow rate of the refrigerant in the diffuser.

Key concepts

Conservation of Mass

Understanding the conservation of mass is fundamental for analyzing systems like diffusers. The basic principle of conservation of mass states that mass cannot be created or destroyed within a closed system. For steady-flow devices, such as diffusers, this principle simplifies to the concept that the mass flow rate of a fluid entering the device must equal the mass flow rate of the fluid exiting the device.
This concept is epitomized in the equation
\[\begin{equation}\dot{m}_1 = \dot{m}_2\end{equation}\]
, indicating that the mass flow rate \(\dot{m}\) remains constant. When analyzing a diffuser in thermodynamics, it's essential to remember that the conservation of mass helps us determine the relationship between the inlet and outlet velocities and areas. Given that a diffuser typically increases the cross-sectional area to reduce velocity and increase pressure, the equation
\[\begin{equation}\rho_1 A_1 V_1 = \rho_2 A_2 V_2\end{equation}\]
can be used to associate these quantities, where \(\rho\) represents density, \(A\) is the cross-sectional area, and \(V\) is the velocity of the refrigerant.

Conservation of Energy

The conservation of energy, also known as the first law of thermodynamics, asserts that energy cannot be created or destroyed in an isolated system. Instead, it can only change forms. In the context of a steady-flow process like that in a diffuser, this concept helps us track the energy transformation from kinetic energy (due to velocity) to internal energy (temperature and enthalpy).
For a diffuser, the conservation of energy is expressed by the steady-flow energy equation:
\[\begin{equation}\dot{m}(h_1+\frac{V_1^2}{2}) + Q_{in} = \dot{m}(h_2+\frac{V_2^2}{2})\end{equation}\]
where \(h\) denotes specific enthalpy, \(V\) represents velocity, and \(Q_{in}\) is the rate of heat transfer to the refrigerant. The equation balances the energy entering and leaving the system, accounting for the work done by or on the fluid and the heat transfer. It is essential to note that specific enthalpy incorporates refrigerant properties, and for precise calculations, these should be obtained from accurate refrigerant tables or equations of state.

Refrigerant Properties

Refrigerant properties are essential in thermodynamic analysis. These properties include specific volume, density, specific enthalpy, specific heat capacities, and thermal conductivity, among others. For the refrigerant R-134a mentioned in the diffuser analysis exercise, properties like specific volume (\(v\)) and specific enthalpy (\(h\)) will vary with temperature and pressure, and they are crucial to our calculations involving mass and energy balances.

Understanding Specific Volume and Enthalpy

Specific volume is related to the density of the refrigerant and is inverse to density, while specific enthalpy represents the energy content per unit mass of the refrigerant. These properties affect the behavior of the refrigerant as it passes through different components in a system, such as compressors, condensers, and diffusers.
Accurate refrigerant property data allow for precise mass flow rate and energy transfer calculations. Thus, consulting the R-134a property tables or software-based calculations for the correct values at given temperatures and pressures is vital for accurate thermodynamic analysis.

Steady-Flow Process

A steady-flow process is inherent to devices like diffusers, where fluids flow continuously without fluctuation over time. In such processes, the fluid properties at any given section of the system remain constant during the operation period. This implies that within the system's bounds, variables like pressure, temperature, velocity, and others remain unchanging at any specific point.

Applying Steady-Flow to Diffusers

In the analysis of a diffuser, which is essentially a diverging tube allowing the gas to expand and slow down, the steady-flow assumption simplifies the equations needed for mass and energy balances. For example, it means that the mass flow rate and energy at the entrance and exit are consistent, with differences accounted for by fluid properties and heat transfer.
In practice, we take advantage of the steady-flow assumption to apply the continuity equation for mass conservation and the first law of thermodynamics for energy conservation. This enables us to establish relationships between inlet and outlet velocities, determine changes in enthalpy, and conclude the overall heat transfer and work interactions across the diffuser.