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Chapter 4: Problem 10

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

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

Step by step solution

01

- 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掳C, and at exit, at 9 bar and 30掳C.
02

- Convert Mass Flow Rate to kg/s

Given the mass flow rate is 6 kg/min, convert this to kg/s: \[ \dot{m} = \frac{6 \, \text{kg/min}}{60 \, \text{s/min}} = 0.1 \, \text{kg/s} \]
03

- 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 \( V_1 \).
04

- 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: \[ A_1 = \frac{\pi (0.025)^2}{4} \]
05

- Solve for Inlet Velocity

Substitute values into the continuity equation to find the inlet velocity \( V_1 \): \[ V_1 = \frac{\dot{m}}{\rho_1 A_1} = \frac{0.1}{\rho_1 \cdot \frac{\pi (0.025)^2}{4}} \]
06

- 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 \( A_2 \).
07

- Calculate Exit Pipe Diameter

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 m鲁/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.

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Most popular questions from this chapter

A rigid tank of volume \(0.75 \mathrm{~m}^{3}\) is initially evacuated. \(\mathrm{A}\) hole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) flows in until the pressure in the tank reaches \(1 \mathrm{bar}\). Heat transfer between the contents of the tank and the surroundings is negligible. Determine the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\).

Pumped-hydraulic storage power plants use relatively inexpensive off-peak baseload electricity to pump water from a lower reservoir to a higher reservoir. During periods of peak demand, electricity is produced by discharging water from the upper to the lower reservoir through a hydraulic turbinegenerator. A single device normally plays the role of the pump during upper-reservoir charging and the turbine-generator during discharging. The ratio of the power developed during discharging to the power required for charging is typically much less than \(100 \%\). Write a report describing the features of the pump-turbines used for such applications and their size and cost. Include in your report a discussion of the economic feasibility of pumped-hydraulic storage power plants,

An insulated rigid tank is initially evacuated. A valve is opened and atmospheric air at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) enters until the pressure in the tank becomes 1 bar, at which time the valve is closed. Is the final temperature of the air in the tank equal to, greater than, or less than \(20^{\circ} \mathrm{C}\) ?

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Would it be desirable for a coolant circulating inside the engine of an automobile to have a large or a small specific heat \(c_{p} ?\) Discuss.
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