Question

Refrigerant 134 a enters the condenser of a refrigeration system operating at steady state at 9 bar, \(50^{\circ} \mathrm{C}\), through a 2.5-cm-diameter pipe. At the exit, the pressure is 9 bar, the temperature is \(30^{\circ} \mathrm{C}\), and the velocity is \(2.5 \mathrm{~m} / \mathrm{s}\). The mass flow rate of the entering refrigerant is \(6 \mathrm{~kg} / \mathrm{min}\). Determine (a) the velocity at the inlet, in \(\mathrm{m} / \mathrm{s}\). (b) the diameter of the exit pipe, in \(\mathrm{cm}\).

Short answer

(a) Compute velocity using continuity. (b) Use continuity to find diameter.

Step-by-step solution

- Determine Specific Volumes at Inlet and Exit

Using standard refrigerant tables or software, determine the specific volumes (u) of Refrigerant 134a at the given entry and exit conditions. For entry, at 9 bar and 50°C, and at exit, at 9 bar and 30°C.

- Convert Mass Flow Rate to kg/s

Given the mass flow rate is 6 kg/min, convert this to kg/s: \[ \dot{m} = \frac{6 \, \text{kg/min}}{60 \, \text{s/min}} = 0.1 \, \text{kg/s} \]

- Apply Continuity Equation at Inlet

Using the continuity equation for the inlet: \[ \dot{m} = \rho_1 A_1 V_1 \] Where \( \rho_1 \) (density at inlet) is the inverse of specific volume at inlet. Area \( A_1 \) is the cross-sectional area of the inlet pipe (\( \frac{\pi d^2}{4} \)) and \( d \) is the diameter. Rearrange to solve for the velocity \( V_1 \).

- Calculate Cross-Sectional Area of Inlet Pipe

Calculate the area of the cross-section of the inlet pipe using the formula for the area of a circle: \[ A_1 = \frac{\pi d_1^2}{4} \] where the diameter \( d_1 = 2.5 \, \text{cm} = 0.025 \, \text{m} \), thus: \[ A_1 = \frac{\pi (0.025)^2}{4} \]

- Solve for Inlet Velocity

Substitute values into the continuity equation to find the inlet velocity \( V_1 \): \[ V_1 = \frac{\dot{m}}{\rho_1 A_1} = \frac{0.1}{\rho_1 \cdot \frac{\pi (0.025)^2}{4}} \]

- Apply Continuity Equation at Exit

Using the continuity equation at the exit: \[ \dot{m} = \rho_2 A_2 V_2 \] Where \( \rho_2 \) is the inverse of specific volume at exit and \( V_2 = 2.5 \, \text{m/s} \). Rearrange to solve for area \( A_2 \).

- Calculate Exit Pipe Diameter

Using area calculated: \[ A_2 = \frac{\dot{m}}{\rho_2 V_2} \]. The diameter can be found as: \[ d_2 = \sqrt{\frac{4A_2}{\pi}} \]

Key concepts

mass flow rate conversion

Converting the mass flow rate from one unit to another is crucial in thermodynamics and fluid mechanics. In this example, the mass flow rate of the refrigerant is given as 6 kg/min. To better suit our calculations, we need to convert this to kg/s. This can be done by dividing by 60, since there are 60 seconds in a minute. The formula used is: \[ \dot{m} = \frac{6 \, \text{kg/min}}{60 \, \text{s/min}} = 0.1 \, \text{kg/s} \] This ensures that all our subsequent calculations are in the correct unit system, simplifying further steps.

continuity equation

The continuity equation is a fundamental principle in fluid mechanics, stating that mass must be conserved in a steady flow system. This is expressed as: \[ \dot{m} = \rho AV \] For a given pipe, \(\rho\) is the density of the fluid, \(A\) is the cross-sectional area, and \(V\) is the velocity. At the inlet, the formula becomes: \[ \dot{m} = \rho_1 A_1 V_1 \] At the exit, the formula is: \[ \dot{m} = \rho_2 A_2 V_2 \] By rearranging these equations, we can solve for unknowns such as the inlet velocity and the exit pipe diameter. This principle is fundamental to ensuring that our system adheres to physical laws.

specific volume determination

Specific volume (\(u\)) is the volume occupied by a unit mass of a fluid, typically given in m³/kg. It is the inverse of density \(\rho\), and can be obtained from standard refrigerant tables or specialized software. For R-134a at 9 bar and \(50^{\circ}\mathrm{C}\), and at 9 bar and \(30^{\circ}\mathrm{C}\), look up or calculate the specific volumes. These values are essential for converting between mass flow rate and volumetric flow rate, as well as for applying the continuity equation correctly.

velocity calculation

Velocity at both the inlet and exit can be derived using the continuity equation. For the inlet, we use: \[ \dot{m} = \rho_1 A_1 V_1 \] Rearrange it to solve for \(V_1\): \[ \ V_1 = \frac{\dot{m}}{\rho_1 A_1} \] Substitute the known values to get \(V_1\). For the exit, if \(V_2\) is given (such as 2.5 m/s), apply it directly. Understanding how velocity changes through different sections of the refrigeration system helps identify potential bottlenecks or design inefficiencies.

pipe diameter calculation

Calculating the diameter of pipes at different points in the system involves the cross-sectional area derived from the continuity equation. For the inlet area calculation: \[ \ A_1 = \frac{\pi d_1^2}{4} \] where \(d_1\) is the inlet diameter. For the exit, after finding \(A_2\) using the continuity equation, compute the diameter: \[ \ d_2 = \sqrt{\frac{4A_2}{\pi}} \] This ensures accurate piping, preventing flow restrictions or inefficiencies.