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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 \({ }^{\circ} \mathrm{F}\).

Short Answer

Expert verified
The exit temperature is approximately 94°F.

Step by step solution

01

Identify Given Information

The initial conditions are: Pressure: \(P_1 = 200 \text{ lbf/in}^2\) Temperature: \(T_1 = 220^{\circ} \text{F}\) Velocity: \(V_1 = 120 \text{ ft/s}\) The exit conditions are: Pressure: \(P_2 = 20 \text{ lbf/in}^2\) Velocity: \(V_2 = 1500 \text{ ft/s}\)
02

Apply the Steady-Flow Energy Equation

The Steady-Flow Energy Equation for a control volume with negligible potential energy effects is: \[ h_1 + \frac{1}{2} V_1^2 = h_2 + \frac{1}{2} V_2^2 \] where \( h \) is the specific enthalpy and \( V \) is the velocity.
03

Calculate Change in Kinetic Energy

Calculate the change in kinetic energy: \[ \Delta KE = \frac{1}{2} (V_2^2 - V_1^2) \] Substitute the given values: \[ \Delta KE = \frac{1}{2} (1500^2 - 120^2) \text{ ft}^2/\text{s}^2 = 1,122,240 \text{ ft}^2/\text{s}^2 \]
04

Convert to Consistent Units

Convert the kinetic energy term to BTU/lbm: \[ \frac{1,122,240 \text{ ft}^2/\text{s}^2}{32.174 \text{ ft}/\text{s}^2 \cdot 778.169 \text{ ft} \cdot \text{lbf}/\text{BTU}} \approx 4.51 \text{ BTU}/\text{lbm} \]
05

Determine Initial Enthalpy

Use the given initial conditions (\(P_1\) and \(T_1\)) to find the initial enthalpy \(h_1\) from the refrigerant tables. For \(P_1 = 200 \text{ lbf/in}^2\) and \(T_1 = 220^{\circ} \text{F}\), assume \(h_1 = 116.4 \text{ BTU}/\text{lbm}\) (as obtained from refrigerant tables).
06

Solve for Final Enthalpy

Use the equation from Step 2, solving for the final enthalpy \(h_2\): \[ h_2 = h_1 + \frac{1}{2} V_1^2 - \frac{1}{2} V_2^2 \approx 116.4 \text{ BTU}/\text{lbm} - 4.51 \text{ BTU}/\text{lbm} = 111.89 \text{ BTU}/\text{lbm} \]
07

Determine the Exit Temperature

Using the exit pressure \(P_2\) and the enthalpy \(h_2\) found in Step 6, refer to the refrigerant tables again to find the corresponding exit temperature \(T_2\). For \(P_2 = 20 \text{ lbf/in}^2\) and \(h_2 = 111.89 \text{ BTU}/\text{lbm}\), \(T_2\) can be interpolated as approximately \(T_2 = 94^{\circ} F\).

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Key Concepts

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

steady-flow energy equation
Understanding the steady-flow energy equation is crucial for many thermodynamic problems, such as determining the exit temperature of refrigerants. This equation is based on the first law of thermodynamics applied to a control volume, where energy can enter and leave through mass flow. The equation essentially balances the energy entering and leaving the system, considering kinetic, potential, and specific enthalpy changes. Here's the equation in its simplest form: \[ h_1 + \frac{1}{2} V_1^2 = h_2 + \frac{1}{2} V_2^2 \]In this case, potential energy changes are neglected. This makes calculations a bit simpler since we only focus on specific enthalpy (h) and kinetic energy (\(\frac{1}{2} V^2\)). By using this equation, you can solve for the unknowns when initial and final states are given.
specific enthalpy
The concept of specific enthalpy is essential when dealing with thermodynamic systems. Specific enthalpy (h) is a measure of the energy content of a substance per unit mass, often expressed in BTU/lbm or kJ/kg. In nozzle problems, this property helps us understand how the energy stored in a fluid changes as it moves through different states. When refrigerant 134a moves from state 1 (entry) to state 2 (exit), its enthalpy changes due to pressure and temperature variations. To find specific enthalpies at different states, we refer to refrigerant tables, which provide these values based on known pressures and temperatures. In our problem, we use these tables to get \(h_1\) and \(h_2\) to eventually determine the exit temperature.
kinetic energy
Kinetic energy plays a significant role in nozzle flow problems, where changes in fluid velocity are prominent. The formula to calculate kinetic energy (KE) is:\[ \Delta KE = \frac{1}{2} (V_2^2 - V_1^2) \]Here, \(V_1\) and \(V_2\) are the velocities at the entry and exit of the nozzle, respectively. In the given problem, substituting values: \[ \Delta KE = \frac{1}{2} (1500^2 - 120^2) \text{ ft}^2/\text{s}^2 = 1,122,240 \text{ ft}^2/\text{s}^2 \]To use this result in our steady-flow energy equation, we convert it to consistent units like BTU/lbm. Properly accounting for kinetic energy changes helps us understand how much the speed of the fluid changes the specific enthalpy, which is necessary for solving the exit temperature.
refrigerant tables
Refrigerant tables are invaluable for finding properties like pressure, temperature, enthalpy, and entropy of refrigerants like R-134a. These tables catalog thermodynamic properties at different states, enabling you to solve complex problems. When working on exercises like our nozzle temperature determination problem, you'll use refrigerant tables to find the specific enthalpy (h) at given pressures and temperatures. For example, with initial conditions of 200 lbf/in² and 220°F, we find the initial enthalpy \( h_1 \) to be 116.4 BTU/lbm. Using these tables again at the final state (20 lbf/in² and the calculated final enthalpy), we determine the exit temperature, completing our problem-solving cycle.
unit conversion
Unit conversion is an essential skill in thermodynamics. Different units like feet, seconds, BTU, and pounds might be mixed into one problem, so converting them correctly is crucial. In the given nozzle problem, the kinetic energy was initially calculated in \( \text{ft}^2/\text{s}^2 \). To effectively apply this to our steady-flow energy equation, we converted it to BTU/lbm. Here's how unit conversion was done: \[ \frac{1,122,240 \text{ ft}^2/\text{s}^2}{32.174 \text{ ft}/\text{s}^2 \cdot 778.169 \text{ ft} \cdot \text{lbf}/\text{ BTU}} = 4.51 \text{ BTU}/\text{ lbm} \]In problems involving multiple physical quantities, always ensure units are consistent. This avoids calculation errors and ensures the accuracy of results.

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Most popular questions from this chapter

The procedure to inflate a hot-air balloon requires a fan to move an initial amount of air into the balloon envelope followed by heat transfer from a propane burner to complete the inflation process. After a fan operates for 10 minutes with negligible heat transfer with the surroundings, the air in an initially deflated balloon achieves a temperature of \(80^{\circ} \mathrm{F}\) and a volume of \(49,100 \mathrm{ft}^{3}\). Next the propane burner provides heat transfer as air continues to flow into the balloon without use of the fan until the air in the balloon reaches a volume of \(65,425 \mathrm{ft}^{3}\) and a temperature of \(210^{\circ} \mathrm{F}\). Air at \(77^{\circ} \mathrm{F}\) and \(14.7 \mathrm{lb} / 1 n^{2}\) surrounds the balloon. The net rate of heat transfer is \(7 \times 10^{6} \mathrm{Btu} / \mathrm{h}\). Ignoring effects due to kinetic and potential energy, modeling the air as an ideal gas, and assuming the pressure of the air inside the balloon remains the same as that of the surrounding air, determine (a) the power required by the fan, in hp. (b) the time required for full inflation of the balloon, in min.

At steady state, air at \(200 \mathrm{kPa}, 325 \mathrm{~K}\), and mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\) enters an insulated duct having differing inlet and exit cross-sectional areas The inlet cross-sectional area is \(6 \mathrm{~cm}^{2}\). At the duct exit, the pressure of the air is \(100 \mathrm{kPa}\) and the velocity is \(250 \mathrm{~m} / \mathrm{s}\). Neglecting potential energy effects and modeling air as an ideal gas with constant \(c_{p}=\) \(1.008 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the velocity of the air at the inlet, in \(\mathrm{m} / \mathrm{s}\). (b) the temperature of the air at the exit, in \(\mathrm{K}\). (c) the exit cross-sectional area, in \(\mathrm{cm}^{2}\).

A rigid, insulated tank, initially containing \(0.4 \mathrm{~m}^{3}\) of saturated water vapor at \(3.5\) bar, is connected by a valve to a large vessel holding steam at 15 bar, \(320^{\circ} \mathrm{C}\). The valve is opened only as long as required to bring the tank pressure to 15 bar. For the tank contents, determine the final temperature, in \({ }^{\circ} \mathrm{C}\), and final mass, in \(\mathrm{kg}\).

An open feedwater heater operates at steady state with liquid water entering inlet 1 at 10 bar, \(50^{\circ} \mathrm{C}\), and a mass flow rate of \(60 \mathrm{~kg} / \mathrm{s}\). A separate stream of steam enters inlet 2 at 10 bar and \(200^{\circ} \mathrm{C}\). Saturated liquid at 10 bar exits the feedwater heater at exit 3 . Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), of the steam at inlet 2 .

Air enters a horizontal, constant-diameter heating duct operating at steady state at \(290 \mathrm{~K}, 1\) bar, with a volumetric flow rate of \(0.25 \mathrm{~m}^{3} / \mathrm{s}\), and exits at \(325 \mathrm{~K}, 0.95\) bar. The flow area is \(0.04 \mathrm{~m}^{2}\). Assuming the ideal gas model with \(k=1.4\) for the air, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), (b) the velocity at the inlet and exit, each in \(\mathrm{m} / \mathrm{s}\), and (c) the rate of heat transfer, in \(\mathrm{kW}\).

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