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Chapter 4: Problem 2

Refrigerant \(134 \mathrm{a}\) exits a heat exchanger through \(0.75\)-in.diameter tubing with a mass flow rate of \(0.9 \mathrm{lb} / \mathrm{s}\). The temperature and quality of the refrigerant are \(-15^{\circ} \mathrm{F}\) and \(0.05\), respectively. Determine the velocity of the refrigerant, in \(\mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
Use the mass flow equation to find velocity: V = 峁� / (A蟻). Insert values for mass flow, area, and density to find velocity.

Step by step solution

01

Convert Units

Initially, convert the diameter from inches to meters. Since 1 inch equals 0.0254 meters: Diameter in meters = 0.75 inches 脳 0.0254 meters/inch = 0.01905 meters.
02

Calculate Cross-Sectional Area

Calculate the cross-sectional area ( A ) of the tubing using the formula for the area of a circle A = 蟺d虏 / 4 : A = (3.1416 脳 (0.01905 meters)虏 ) / 4 A 鈮� 2.85 脳 10^-4 square meters.
03

Obtain Specific Volume

Using the given temperature and quality, obtain the specific volume ( 谓 ) of the refrigerant from refrigerant 134a properties tables at -15掳F and quality 0.05. Assume 谓 鈮� 0.05 脳 谓_g + 0.95 脳 谓_f. We calculate 谓, ensuring values are consistent with this mixture.
04

Calculate Density

Calculate the density ( 蟻 ) of the refrigerant. Density 蟻 = 1 / 谓.
05

Apply Continuity Equation

Use the mass flow rate ( 峁� 鈭� = A 蟻 V ) to solve for the velocity ( V ) of the refrigerant: V = 峁� / (A 蟻). Substitute the values of mass flow rate, area, and density to determine the velocity in meters per second ( m/s ).

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Key Concepts

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

unit conversion
Unit conversion is the first step in solving many engineering problems because it ensures that all measurements are in compatible units. In this exercise, we start by converting the diameter of the tubing from inches to meters. Since 1 inch is equivalent to 0.0254 meters, we can calculate the diameter in meters by multiplying the given diameter in inches by 0.0254. Converting units systematically helps avoid errors and ensures consistency in further calculations.
cross-sectional area calculation
Calculating the cross-sectional area of the tubing is crucial because it helps in determining the flow characteristics of the refrigerant. The tubing's cross-sectional area is derived from the formula for the area of a circle, which is \(\frac{\text{蟺d}^2}{4}\). Plugging in the converted diameter ensures that the area is calculated correctly in square meters. This area is vital for further steps like applying the continuity equation.
specific volume determination
Specific volume is an essential property in thermodynamics, especially for fluid flow calculations. Here, we determine the specific volume of refrigerant 134a by using its temperature (-15掳F) and quality (0.05). The quality signifies the proportion of the refrigerant that is vapor. The specific volume is calculated using a weighted average of the specific volumes of the saturated liquid (谓_f) and vapor (谓_g). This step is crucial for density calculation.
density calculation
Density is the mass per unit volume of a substance. Once we have the specific volume (谓), calculating the density (蟻) is straightforward using the formula \(\rho = \frac{1}{谓}\). This relationship is straightforward because specific volume is the inverse of density. Knowing the density of the refrigerant is essential for applying the continuity equation to determine its velocity.
continuity equation
The continuity equation is a fundamental principle in fluid dynamics, stating that the mass flow rate must remain constant from one cross-section to another. The equation is \(\text{峁� = A * 蟻 * V}\), where 峁� is the mass flow rate, A is the cross-sectional area, 蟻 is the density, and V is the velocity. By rearranging this equation to solve for velocity (V), we substitute the known values of mass flow rate, area, and density to find the refrigerant's velocity in meters per second. This final step ties all the previous steps together to provide the desired solution.

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Most popular questions from this chapter

Liquid water enters a valve at \(300 \mathrm{kPa}\) and exits at \(275 \mathrm{kPa}\). As water flows through the valve, the change in its temperature, stray heat transfer with the surroundings, and potential energy effects are negligible. Operation is at steady state. Modeling the water as incompressible with constant \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), determine the change in kinetic energy per unit mass of water flowing through the valve, in \(\mathrm{kJ} / \mathrm{kg}\).

Air at \(600 \mathrm{kPa}, 330 \mathrm{~K}\) enters a well-insulated, horizontal pipe having a diameter of \(1.2 \mathrm{~cm}\) and exits at \(120 \mathrm{kPa}\), \(300 \mathrm{~K}\). Applying the ideal gas model for air, determine at steady state (a) the inlet and exit velocities, each in \(\mathrm{m} / \mathrm{s}\), and (b) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\).

Air enters a compressor operating at steady state with a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) and a volumetric flow rate of \(8 \mathrm{ft}^{3} / \mathrm{s}\). The air velocity in the exit pipe is \(225 \mathrm{ft} / \mathrm{s}\) and the exit pressure is \(150 \mathrm{lbf} /\) in. \(^{2}\) If each unit mass of air passing from inlet to exit undergoes a process described by \(p v^{1.3}=\) constant, determine the diameter of the exit pipe, in inches.

Refrigerant \(134 \mathrm{a}\) enters a well-insulated nozzle at \(200 \mathrm{lbf} / \mathrm{in} .{ }^{2}, 220^{\circ} \mathrm{F}\), with a velocity of \(120 \mathrm{ft} / \mathrm{s}\) and exits at 20 lbf/in. \({ }^{2}\) with a velocity of \(1500 \mathrm{ft} / \mathrm{s}\). For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in \({ }^{\circ} \mathrm{F}\).

Steam enters a turbine operating at steady state at \(700^{\circ} \mathrm{F}\) and \(450 \mathrm{lbf} / \mathrm{in}^{2}\) and leaves as a saturated vapor at \(1.2 \mathrm{lbf} / \mathrm{in} .^{2}\) The turbine develops \(12,000 \mathrm{hp}\), and heat transfer from the turbine to the surroundings occurs at a rate of \(2 \times 10^{6} \mathrm{Btu} / \mathrm{h}\). Neglecting kinetic and potential energy changes from inlet to exit, determine the volumetric flow rate of the steam at the inlet, in \(\mathrm{ft}^{3} / \mathrm{s}\).
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