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Chapter 2: Problem 147

Give the phase and the specific volume for the following: a. \(\mathrm{R}-410 \mathrm{a}, T=-25 \mathrm{~F}, P=30 \mathrm{lbf} / \mathrm{in} .^{2}\) b. \(\mathrm{R}-410 \mathrm{a}, T=-25 \mathrm{~F}, P=40 \mathrm{lbf} / \mathrm{in} .^{2}\) c. \(\mathrm{H}_{2} \mathrm{O}, T=280 \mathrm{~F}, P=35 \mathrm{lbf} / \mathrm{in} .^{2}\) d. \(\mathrm{NH}_{3}, T=60 \mathrm{~F}, P=15 \mathrm{lbf} / \mathrm{in} .^{2}\)

Short Answer

Expert verified
Refer to property tables: a) R-410a possibly superheated, b) R-410a likely subcooled, c) Steam likely superheated, d) NH鈧� mixture or slightly superheated.

Step by step solution

01

Identify Property Tables

To determine the phase and specific volume, use property tables for each substance. For applications involving the refrigerant R-410a or water (H鈧侽) and ammonia (NH鈧�), consult the specific refrigerant or steam tables that include temperature and pressure data.
02

Case a: R-410a at T = -25掳F, P = 30 lbf/in虏

Use the R-410a refrigerant table to check the saturation pressure at -25掳F. If P is less than saturation pressure, it's superheated; if P matches, it's saturated; if greater, it's subcooled. Check specific volumes and interpolate if necessary.
03

Case b: R-410a at T = -25掳F, P = 40 lbf/in虏

Check with the R-410a refrigerant table again. Compare given pressure with the saturation pressure at -25掳F. Classify the phase, and find the specific volume using interpolations if needed from the tables.
04

Case c: H鈧侽 at T = 280掳F, P = 35 lbf/in虏

Using the steam tables, check the saturation properties of water at 280掳F. Determine the phase by comparing given pressure with saturation pressure. Determine the correct state and interpolate to find the specific volume if the conditions lie between two entries.
05

Case d: NH鈧� at T = 60掳F, P = 15 lbf/in虏

Refer to NH鈧� refrigerant tables for saturation properties at 60掳F. Check the phase by comparing the given pressure with the saturation pressure. Determine if the state is vapour, liquid, or a mixture, and interpolate specific volume from relevant data.

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Key Concepts

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

Refrigerant Tables
Refrigerant tables are essential tools in thermodynamics, especially for engineers and technicians working with refrigeration systems and HVAC. These tables contain vital data about refrigerants, which are substances used in cooling mechanisms. Different refrigerants have specific properties depending on the thermodynamic conditions they're placed under, such as pressure and temperature.
  • A typical refrigerant table will list values for temperature, pressure, specific volume, enthalpy, and entropy for both saturated liquid and vapor states.
  • By understanding these tables, one can determine the state (phase) of the refrigerant, whether it is in liquid, vapor, or a mixture state.
  • For example, common refrigerants like R-410a have specific tables which provide crucial information for conditions outside the typical range of household systems.
Using these tables involves matching given pressure and temperature conditions to find corresponding saturation pressures and specific volumes. Knowing how to utilize these tables is fundamental for anyone operating or designing refrigeration systems.
Saturation Pressure
Saturation pressure is a critical concept in understanding the phase behavior of substances. It is the pressure at which a liquid boils and turns into a vapor at a given temperature. This property is prominently featured in refrigerant and steam tables.
  • If the pressure in a system is lower than the saturation pressure at a specific temperature, the substance is in a superheated state, meaning it is entirely in a vapor form.
  • If it matches saturation pressure, the substance is in a saturated state, where both liquid and vapor coexist.
  • When pressure exceeds the saturation pressure, the substance is subcooled, existing purely as a liquid.
Understanding these distinctions is essential for precisely controlling industrial and refrigeration processes. For instance, in a scenario where R-410a is at -25掳F, determining whether its state is saturated, superheated, or subcooled involves checking its pressure against its saturation pressure at that temperature.
Specific Volume
Specific volume is a thermodynamic property that refers to the volume occupied by a unit mass of a substance. It is an important parameter when analyzing phases and properties of refrigerants and working fluids in general. Specific volume is often recorded in property tables, giving insights into the state of the substance under different pressures and temperatures.
  • For example, in refrigerant tables, specific volume data helps identify whether a substance exists as a liquid or vapor under the given conditions.
  • It is typically expressed in units of cubic feet per pound (ft鲁/lb) or cubic meters per kilogram (m鲁/kg).
  • Specific volume is inversely proportional to density; so, the larger the specific volume, the lower the density, indicating a vapor phase, and vice versa for a liquid phase.
Examining specific volume values alongside temperature and pressure allows clear understanding and prediction of the phase behavior, which is vital in designing efficient systems.
Phase Classification
Phase classification is determining whether a substance is in the liquid, vapor, or mixture (saturated) state based on its thermodynamic properties like temperature and pressure. It directly affects the way a substance behaves and how it can be manipulated for work in processes such as heating, cooling, and refrigeration.
  • Classifying the phase often starts by identifying the known temperature or pressure and using that to find the corresponding saturation point on a property table.
  • If you're dealing with R-410a or NH鈧�, for instance, and you know both its temperature and pressure, you can quickly determine whether the phase is predominantly liquid, vapor, or a combination by consulting the appropriate tables.
  • The phase information helps in optimizing and controlling equipment like compressors and turbines, enabling better energy efficiency and performance.
Once you've determined the phase, using interpolation between given table values can allow the calculation of properties like specific volume, making accurate predictions and adjustments possible.

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Most popular questions from this chapter

Refrigerant- \(410 \mathrm{a}\) in a piston/cylinder arrangement is initially at \(60 \mathrm{~F}, x=1\). It is then expanded in a process so that \(P=C v^{-1}\) to a pressure of \(30 \mathrm{lbf} / \mathrm{in} .^{2} .\) Find the final temperature and specific volume.

A \(35-\mathrm{ft}^{3}\) rigid tank has air at 225 psia and ambient \(600 R\) connected by a valve to a piston/cylinder. The piston of area \(1 \mathrm{ft}^{2}\) requires 40 psia below it to float (see Fig. \(\mathrm{P} 2.166 \mathrm{E}\) ). The valve is opened, the piston moves slowly \(7 \mathrm{ft}\) up, and the valve is closed. During the process, air temperature remains at \(600 R\). What is the final pressure in the tank?

Is it reasonable to assume that at the given states the substance behaves as an ideal gas? a. Oxygen at \(30^{\circ} \mathrm{C}, 3 \mathrm{MPa}\) b. Methane at \(30^{\circ} \mathrm{C}, 3 \mathrm{MPa}\) c. Water at \(30^{\circ} \mathrm{C}, 3 \mathrm{MPa}\) d. \(\mathrm{R}-134 \mathrm{a}\) at \(30^{\circ} \mathrm{C}, 3 \mathrm{MPa}\) e. \(\mathrm{R}-134 \mathrm{a}\) at \(30^{\circ} \mathrm{C}, 100 \mathrm{kPa}\)

Helium in a steel tank is at 36 psia, \(540 R\) with a volume of \(4 \mathrm{ft}^{3}\). It is used to fill a balloon. When the pressure drops to 18 psia, the flow of helium stops by itself. If all the helium is still at \(540 R\), how big a balloon is produced?

Use a linear interpolation to estimate \(T_{\text {sat }}\) at 900 kPa for nitrogen. Sketch by hand the curve \(P_{\text {sat }}(T)\) by using a few table entries around \(900 \mathrm{kPa}\) from Table B.6.1. Is your linear interpolation above or below the actual curve?
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