Lecture # 4

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Lecture # 4 PROPERTIES OF PURE SUBSTANCES PURE SUBSTANCE

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Lecture # 4. PURE SUBSTANCE. PROPERTIES OF PURE SUBSTANCES. PHASES OF A PURE SUBSTANCE Solid Liquid Gas. PHASE-CHANGE PROCESSES OF PURE SUBSTANCES. PROPERTY DIAGRAMS FOR PHASE-CHANGE PROCESSES :. 1 -The T - v Diagram:. - PowerPoint PPT Presentation

Transcript of Lecture # 4

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Lecture # 4PROPERTIES OF PURE SUBSTANCES

PURE SUBSTANCE

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PHASES OF A PURE SUBSTANCE

SolidLiquidGas.

PHASE-CHANGE PROCESSES OF PURE SUBSTANCES

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PROPERTY DIAGRAMS FOR PHASE-CHANGE PROCESSES:

1- The T-v Diagram:

FIGURE 1-7. T-v diagram of constant-pressure phase-change processes of a pure substance at various pressures (numerical values are for water).

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2- The P-v Diagram

FIGURE 1-7P-v diagram of a pure substance

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PROPERTY TABLES( Thermodynamic tables)•For most substances, the relationships among thermodynamic properties are too complex to be expressed by simple equations.

•Therefore, properties are frequently presented in the form of tables.

• The steam tables are used to demonstrate the use of thermodynamic property tables.

•Property tables of other substances are used in the same manner.

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•For each substance, the thermodynamic properties are listed in more than one table.

•In fact, a separate table is prepared for each region of interest such as the superheated vapor, compressed liquid, and saturated (mixture) regions.

•Property tables are given in the appendix in both SI and English units. The tables in English units carry the same number as the corresponding tables in SI followed by an identifier E.•Tables A–6 and A–6E, for example, list properties of superheated water vapor, the former in SI and the latter in English units. Before we get into the discussion of property tables

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Remember the following two properties:

• The Entropy s .• Entropy is a property associated with the

second law of thermodynamics.• The enthalpy h.• In the analysis of certain types of processes,

particularly in power generation and refrigeration, we frequently encounter the combination of properties u + Pv. which is called h or the specific enthalpy.

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•The subscript f is used to denote properties of a saturated liquid, and the subscript g to denote the properties of saturated vapor.• The subscript s fg, which denotes the difference between the saturated vapor and saturated liquid values of the same property. For example,vf = specific volume of saturated liquidvg = specific volume of saturated vaporvfg = difference between vg and vf (that is, vfg = vg - vf ) The quantity hfg is called the enthalpy of vaporization (or latent heat of vaporization). It represents the amount of energy needed to vaporize a unit mass of saturated liquid at a given temperature or pressure. It decreases as the temperature or pressure increases and becomes zero at the critical point.

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EXAMPLE 1–1

A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.

Solution: A rigid tank contains saturated liquid water. The pressure and volume of the tank are to be determined.

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Analysis: The state of the saturated liquid water is shown on a T-v diagram. Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90°C:

P = Psat @ 90°C = 70.14 kP, From Table (A–4)The specific volume of the saturated liquid at 90°C is: V = v f @ 90°C = 0.001036 m3/kg from Table (A–4)Then the total volume of the tank becomesV = mv = (50 kg ) (0.001036 m3/kg) = 0.0518 m3

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EXAMPLE 1–2A piston–cylinder device contains 2 ft3 of saturated water vapor at 50-psia pressure. Determine the temperature and the mass of the vapor inside the cylinder.Solution A cylinder contains saturated water vapor. The temperature and the mass of vapor are to be determined.Analysis The state of the saturated water vapor is shown on a P-v diagram. Since the cylinder contains saturated vapor at 50 psia, the temperature inside must be the saturation temperature at this pressure:T = T sat @ 50 psia = 281.03°F (Table A–5E)

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The specific volume of the saturated vapor at 50 psia isV = v g @ 50 psia = 8.5175 ft3/lbm (Table A–5E)Then the mass of water vapor inside the cylinder becomes:

mm

lblbft

ftVm 235.0/5175.8

23

3

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Example 1–3A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine (a) the volume change and (b) the amount of energy transferred to the water.

Solution Saturated liquid water is vaporized at constant pressure. The volume change and the energy transferred are to be determined.

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