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Refrigerant-134a enters a 28-cm-diameter pipe steadily at \(200 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) with a velocity of \(5 \mathrm{m} / \mathrm{s}\). The refrigerant gains heat as it flows and leaves the pipe at \(180 \mathrm{kPa}\) and \(40^{\circ} \mathrm{C}\). Determine (a) the volume flow rate of the refrigerant at the inlet, \((b)\) the mass flow rate of the refrigerant, and \((c)\) the velocity and volume flow rate at the exit.

Short Answer

Expert verified
Question: Determine the volume flow rate at the inlet, the mass flow rate, and the velocity and volume flow rate at the exit for Refrigerant-134a in a given pipe. Answer: The volume flow rate at the inlet is 0.3079 m³/s, the mass flow rate is 1.673 kg/s, the exit velocity is 5.636 m/s, and the volume flow rate at the exit is 0.347 m³/s.

Step by step solution

01

Calculate the volume flow rate at the inlet

To calculate the volume flow rate at the inlet, we can use the equation: \(Q_{in}=A_{in} \times v_{in}\). Given the diameter, we can calculate the area with the following equation: \(A_{in}=\pi \frac{D^2}{4}\). Using the information from the exercise, the area is: \(A_{in}=\pi \frac{(0.28\,\text{m})^2}{4}=0.06157\,\text{m}^2\) Now, we can calculate the volume flow rate by multiplying the area with the velocity. \(Q_{in}=0.06157\,\text{m}^2 \times 5\,\text{m/s} = 0.3079\,\text{m}^3/\text{s}\)
02

Calculate the mass flow rate

To calculate the mass flow rate, we can use the equation: \(\dot{m} = \rho_{in} \times Q_{in}\). First, we need to find the density of the refrigerant at the entrance conditions, \(200\, \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\). Using a refrigerant-134a table or an online calculator, the density is found to be \(\rho_{in}= 5.436\,\text{kg/m}^3\). Now, we can find the mass flow rate: \(\dot{m}= 5.436\,\text{kg/m}^3 \times 0.3079\,\text{m}^3/\text{s}= 1.673\,\text{kg/s}\)
03

Calculate the exit velocity

The conservation of mass principle states that the mass flow rate at the entrance is equal to the mass flow rate at the exit. Thus, we can write the equation: \(\rho_{in} A_{in} v_{in}=\rho_{out} A_{out} v_{out}\) We are given the properties at the exit as \(180 \mathrm{kPa}\) and \(40^{\circ} \mathrm{C}\). Using a refrigerant-134a table or an online calculator, the density at the exit is found to be \(\rho_{out}= 4.826\,\text{kg/m}^3\). The area of the pipe remains unchanged. We can now solve for the exit velocity: \(v_{out}=\frac{\rho_{in} A_{in} v_{in}}{\rho_{out} A_{out}}=\frac{5.436\,\text{kg/m}^3 \times 0.06157\,\text{m}^2 \times 5\,\text{m/s}}{4.826\,\text{kg/m}^3 \times 0.06157\,\text{m}^2} = 5.636\,\text{m/s}\)
04

Calculate the volume flow rate at the exit

To calculate the volume flow rate at the exit, we can use the equation: \(Q_{out}=A_{out} \times v_{out}\). \(Q_{out}=0.06157\,\text{m}^2 \times 5.636\,\text{m/s} = 0.347\,\text{m}^3/\text{s}\) In conclusion, the volume flow rate at the inlet is \(0.3079\,\text{m}^3/\text{s}\), the mass flow rate is \(1.673\,\text{kg/s}\), the exit velocity is \(5.636\,\text{m/s}\) and the volume flow rate at the exit is \(0.347\,\text{m}^3/\text{s}\).

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Key Concepts

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

Volume Flow Rate
Volume flow rate is a crucial concept in thermodynamics and fluid mechanics, representing the quantity of fluid that passes a given point within a pipe or a channel per unit time. It's denoted by the symbol \(Q\) and commonly measured in cubic meters per second (\(m^3/s\)). To calculate the volume flow rate, one needs to know the cross-sectional area of the pipe (\(A\)) and the velocity of the fluid (\(v\)) at that point, with the relationship defined as \(Q = A \times v\).

In our exercise involving Refrigerant-134a, we utilized this principle to find the flow rate at the inlet by measuring the pipe's diameter to determine its cross-sectional area and multiplying it by the fluid's velocity. Ensuring accurate measurements and understanding the relationship between these variables is essential for engineers and students alike in the field.
Mass Flow Rate
Mass flow rate is a measure that provides insight into the amount of mass moving through a cross-section of a pipe or channel per unit time. Represented by \(\dot{m}\), it is calculated by multiplying the density of the fluid (\(\rho\)) by the volume flow rate (\(Q\)), given by the equation \(\dot{m} = \rho \times Q\).

In practical scenarios like our textbook example, where the mass flow rate of Refrigerant-134a was needed, you must first determine the fluid's density at the given pressure and temperature. With this information, it's possible to calculate how much mass of refrigerant flows per second, which is fundamental for applications in heating, cooling, and refrigeration systems where precise control of mass is required for system efficiency and effectiveness.
Conservation of Mass
The conservation of mass, also known as the principle of mass conservation, asserts that mass cannot be created or destroyed within a closed system; it's conserved over time. This fundamental concept is leveraged in fluid dynamics to equate the mass entering a system to the mass exiting it, when there are no leaks or accumulation, described mathematically as \(\rho_{in} A_{in} v_{in} = \rho_{out} A_{out} v_{out}\).

In the context of our exercise, we applied the principle to determine the velocity and volume flow rate of the refrigerant at the exit of the pipe. By using the densities and area of the pipe, we ensured the mass remained constant from the inlet to the outlet. Understanding how this principle operates is invaluable for engineers in designing systems where changes in fluid properties and flow characteristics are expected.
Refrigerant Properties
Refrigerants are substances used in heat pumps and refrigeration cycles, with properties that can change significantly under varying temperatures and pressures. The thermodynamic properties of refrigerants, including density, specific heat, and boiling point, are essential for calculating the flow rates and for the design process of refrigeration systems.

For our exercise solution, we referenced refrigerant-134a property tables to find the exact density at specified conditions. It's critical to have accurate data about refrigerant properties, as these directly influence calculations related to mass and volume flow rates, and consequently impact the performance of refrigeration and HVAC systems. Professionals in the field often consult standardized tables or use software tools to obtain such detailed information about refrigerant properties.

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Most popular questions from this chapter

A tank with an internal volume of \(1 \mathrm{m}^{3}\) contains air at \(800 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A valve on the tank is opened allowing air to escape and the pressure inside quickly drops to \(150 \mathrm{kPa}\), at which point the valve is closed. Assume there is negligible heat transfer from the tank to the air left in the tank. (a) Using the approximation \(h_{e} \approx\) constant \(=h_{e, \mathrm{avg}}=\) \(0.5\left(h_{1}+h_{2}\right),\) calculate the mass withdrawn during the process. (b) Consider the same process but broken into two parts. That is, consider an intermediate state at \(P_{2}=400 \mathrm{kPa}\), calculate the mass removed during the process from \(P_{1}=800 \mathrm{kPa}\) to \(P_{2}\) and then the mass removed during the process from \(P_{2}\) to \(P_{3}=150 \mathrm{kPa},\) using the type of approximation used in part \((a),\) and add the two to get the total mass removed. (c) Calculate the mass removed if the variation of \(h_{e}\) is accounted for.

A \(0.06-m^{3}\) rigid tank initially contains refrigerant- 134 a at \(0.8 \mathrm{MPa}\) and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant-134a at \(1.2 \mathrm{MPa}\) and \(36^{\circ} \mathrm{C}\). Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has entered the tank and (b) the amount of heat transfer. Answers: (a) \(64.8 \mathrm{kg}\), (b) \(627 \mathrm{kJ}\)

During the inflation and deflation of a safety airbag in an automobile, the gas enters the airbag with a specific volume of \(15 \mathrm{ft}^{3} / \mathrm{lbm}\) and at a mass flow rate that varies with time as illustrated in Fig. P5-165E. The gas leaves this airbag with a specific volume of \(13 \mathrm{ft}^{3} / \mathrm{lbm},\) with a mass flow rate that varies with time, as shown in Fig. P5-165E. Plot the volume of this bag (i.e., airbag size) as a function of time, in \(\mathrm{ft}^{3}\).

A \(2-\mathrm{ft}^{3}\) rigid tank contains saturated refrigerant \(134 \mathrm{a}\) at 160 psia. Initially, 5 percent of the volume is occupied by liquid and the rest by vapor. A valve at the top of the \(\operatorname{tank}\) is now opened, and vapor is allowed to escape slowly from the tank. Heat is transferred to the refrigerant such that the pressure inside the tank remains constant. The valve is closed when the last drop of liquid in the tank is vaporized. Determine the total heat transfer for this process.

A constant-pressure \(R-134\) a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams. Determine the flow power needed to pass \(6 \mathrm{L} / \mathrm{s}\) of \(\mathrm{R}-134 \mathrm{a}\) at \(320 \mathrm{kPa}\) and 55 percent quality through this unit. What is the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), of the two outlet streams?

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