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Refrigerant \(134 \mathrm{a}\) flows at steady state through a horizontal tube having an inside diameter of \(0.05 \mathrm{~m}\). The refrigerant enters the tube with a quality of \(0.1\), temperature of \(36^{\circ} \mathrm{C}\), and velocity of \(10 \mathrm{~m} / \mathrm{s}\). The refrigerant exits the tube at 9 bar as a saturated liquid. Determine (a) the mass flow rate of the refrigerant, in \(\mathrm{kg} / \mathrm{s}\). (b) the velocity of the refrigerant at the exit, in \(\mathrm{m} / \mathrm{s}\). (c) the rate of heat transfer, in \(\mathrm{kW}\), and its associated direction with respect to the refrigerant.

Short Answer

Expert verified
The mass flow rate is calculated using entrance specific volume and velocity. The exit velocity is determined by the mass flow rate and exit specific volume. Heat transfer rate is calculated using specific enthalpies of entrance and exit.

Step by step solution

01

Determine Specific Volume and Mass Flow Rate

First, find the specific volume (\text{v}) of the refrigerant at the entrance. Use the entrance quality (\text{x} = 0.1) and the specific volumes of the liquid (\text{v}_f) and vapor (\text{v}_g) phases from the refrigerant tables. Compute \text{v}_{1} as:\[ \text{v}_{1} = \text{v}_f + x (\text{v}_g - \text{v}_f) \] Then, the mass flow rate(\text{˙m}) can be found using the inlet velocity (u_1), cross-sectional area (A), and specific volume (\text{v}_1).\[ \text{A} = \frac{\text{π}(\text{D})^2}{4} = \frac{\text{π}(0.05)^2}{4} = 0.0019635 \text{m}^2 \] \[ \text{˙m} = \frac{\text{A} \text{v}_{1} (\text{v}_{1})}{\text{v}_{1}} \]
02

Determine Velocity of Refrigerant at Exit

To find the exit velocity, determine the exit specific volume (\text{v}_f) at 9 bar as it is a saturated liquid. Using the mass flow rate (\text{˙m}) and specific volume at exit (\text{v}_2 = \text{v}_f), calculate the exit velocity (u_2) with the previously calculated cross-sectional area (A). Use the continuity equation: \[ \text{˙m} = \frac{\text{A} \text{v}_2 u_2}{\text{v}_2} \]
03

Determine Rate of Heat Transfer

Determine the specific enthalpies at the entrance (\text{h}_1=\text{h}_f+x \text{(h}_g-\text{h}_f) and exit \text{h}_2= \text{h}_fusing refrigerant tables. The equation for rate of heat transfer (\text{˙Q}) simplifies to: \[ q = \text{˙m} (\text{h}_1 - \text{h}_2) \] The heat transfer direction can determine if this value is positive or negative - indicating heat addition if positive and heat extraction if negative.

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Key Concepts

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

Specific Volume

Specific volume is an essential concept in thermodynamics. It is defined as the volume occupied by a unit mass of a substance. In the case of a refrigerant like R-134a, specific volume helps to understand how much space the refrigerant will take up in different states (liquid or vapor).

For a refrigerant entering a tube, the specific volume can be found using the quality (x), and specific volumes of the liquid (v_f) and vapor (v_g) phases from refrigerant tables. The formula is:

\[ v_{1} = v_f + x (v_g - v_f) \]

This equation shows the relationship between the quality of the refrigerant and its specific volume at the entrance. Once you have the specific volume, you can use it to determine other important factors like the mass flow rate.

Mass Flow Rate

Mass flow rate is a measure of the amount of mass passing through a given surface per unit time. In the context of a refrigerant flowing through a tube, it helps understand how much refrigerant is moving through the system.

The mass flow rate (\(\dot{m}\)) can be determined using the inlet velocity (u_1), cross-sectional area (A), and specific volume (v_1) of the refrigerant. Mathematically, this is represented as:

\[ A = \frac{\pi(D)^2}{4} = \frac{\pi(0.05)^2}{4} = 0.0019635 \; m^2 \]
\[ \dot{m} = \frac{A u_1}{v_1} \]

By plugging in the values, you can find the mass flow rate in \( \frac{kg}{s} \). This parameter is vital for subsequent calculations, such as determining the velocity at the exit and the heat transfer rate.

Heat Transfer

Heat transfer is the process of energy moving from one body to another due to temperature differences. In thermodynamics, understanding heat transfer is crucial for analyzing energy systems such as refrigeration cycles.

To determine the rate of heat transfer (\( \dot{Q} \)), you need to know the specific enthalpies at the entrance (h_1) and exit (h_2) of the refrigerant:

\[ h_1 = h_f + x(h_g - h_f) \]
\[ \dot{Q} = \dot{m} (h_1 - h_2) \]

The direction of heat transfer can be identified by the sign of \( \dot{Q} \). If \( \dot{Q} \) is positive, it indicates heat is being added to the refrigerant. If negative, heat is being extracted from it. This helps in understanding the energy interactions within the system.

Enthalpy

Enthalpy is a measure of the total energy of a thermodynamic system, including internal energy and the energy required to displace the system's surroundings. It is denoted by 'h' and is commonly used in the study of phase changes in refrigerants.

For refrigerants, specific enthalpy at different states (liquid and vapor) can be found using thermodynamic tables. When dealing with quality (x), it helps to calculate the specific enthalpy using:

\[ h_1 = h_f + x(h_g - h_f) \]

This equation incorporates the quality to determine the enthalpy at the refrigerant's entry point. The difference in enthalpy at the entrance and exit is critical in calculating the rate of heat transfer within the system.

Refrigerant

A refrigerant is a substance used in a heat cycle to transfer heat from one area to another. In this exercise, we focus on refrigerant R-134a, which is commonly used in many refrigeration and air-conditioning systems.

Refrigerants like R-134a undergo phase changes (from liquid to vapor and vice versa) within the system. These changes are essential for the cooling process. The performance of a refrigerant is often examined under specific conditions of temperature and pressure, as shown in the exercise.

Understanding the behavior of a refrigerant, including its specific volume, enthalpy, and flow characteristics, is crucial for designing efficient thermal systems and ensuring proper heat transfer within the system.

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Most popular questions from this chapter

Air enters a compressor operating at steady state at \(14.7\) \(\mathrm{lbf} / \mathrm{in} .^{2}\) and \(60^{\circ} \mathrm{F}\) and is compressed to a pressure of \(150 \mathrm{lbf} / / \mathrm{in} .^{2}\) As the air passes through the compressor, it is cooled at a rate of 10 Btu per lb of air flowing by water circulated through the compressor casing. The volumetric flow rate of the air at the inlet is \(5000 \mathrm{ft}^{3} / \mathrm{min}\), and the power input to the compressor is 700 hp. The air behaves as an ideal gas, there is no stray heat transfer, and kinetic and potential effects are negligible. Determine (a) the mass flow rate of the air, lb/s, and (b) the temperature of the air at the compressor exit, in \({ }^{\circ} \mathrm{F}\).

A rigid tank having a volume of \(0.1 \mathrm{~m}^{3}\) initially contains water as a two-phase liquid-vapor mixture at 1 bar and a quality of \(1 \%\). The water is heated in two stages: Stage 1: Constant-volume heating until the pressure is 20 bar. Stage 2: Continued heating while saturated water vapor is slowly withdrawn from the tank at a constant pressure of 20 bar. Heating ceases when all the water remaining in the tank is saturated vapor at 20 bar. For the water, evaluate the heat transfer, in kJ, for each stage of heating. Ignore kinetic and potential energy effects.

Steam enters a well-insulated turbine operating at steady state at \(4 \mathrm{MPa}\) with a specific enthalpy of \(3015.4 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(10 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.07 \mathrm{MPa}\), specific enthalpy is \(2431.7 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(90 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(11.95 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in \(\mathrm{kW}\).

Air enters a compressor operating at steady state with a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) and a temperature of \(70^{\circ} \mathrm{F}\). The volumetric flow rate at the inlet is \(16.6 \mathrm{ft}^{3} / \mathrm{s}\), and the flow area is \(0.26 \mathrm{ft}^{2}\). At the exit, the pressure is \(35 \mathrm{lbf} / \mathrm{in}^{2}\), the temperature is \(280^{\circ} \mathrm{F}\), and the velocity is \(50 \mathrm{ft} / \mathrm{s}\). Heat transfer from the compressor to its surroundings is \(1.0 \mathrm{Btu}\) per \(\mathrm{lb}\) of air flowing. Potential energy effects are negligible, and the ideal gas model can be assumed for the air. Determine (a) the velocity of the air at the inlet, in \(\mathrm{ft} / \mathrm{s}\), (b) the mass flow rate, in \(\mathrm{lb} / \mathrm{s}\), and (c) the compressor power, in Btu/s and hp.

Steam at a pressure of \(0.08\) bar and a quality of \(93.2 \%\) enters a shell- and-tube heat exchanger where it condenses on the outside of tubes through which cooling water flows, exiting as saturated liquid at \(0.08\) bar. The mass flow rate of the condensing steam is \(3.4 \times 10^{5} \mathrm{~kg} / \mathrm{h}\). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) and exits at \(35^{\circ} \mathrm{C}\) with negligible change in pressure. Neglecting stray heat transfer and ignoring kinetic and potential energy effects, determine the mass flow rate of the cooling water, in \(\mathrm{kg} / \mathrm{h}\), for steady-state operation.

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