Presentation on theme: "CHAPTER 4: Energy Analysis of Closed Systems"— Presentation transcript:
1CHAPTER 4: Energy Analysis of Closed Systems PTT 201/4 THERMODYNAMICSEM 1 (2013/2014)CHAPTER 4:Energy Analysis of Closed Systems
2MOVING BOUNDARY WORKMoving boundary work (P dV work): The expansion and compression work in a piston-cylinder device.Quasi-equilibrium process: A process during which the system remains nearly in equilibrium at all times.Wb is positive for expansionWb is negative for compressionThe work associated with a movingboundary is called boundary work.A gas does a differential amount of work Wb as it forces the piston to move by a differential amount ds.
3The boundary work done during a process depends on the path followed as well as the end states. The area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system.
4Polytropic, Isothermal, and Isobaric processes Polytropic process: C, n (polytropic exponent) constantsPolytropic processPolytropic and for ideal gas, n ≠ 1When n = 1Constant temp. process (isothermal )Constant pressure process (isobaric)What is the boundary work for a constant-volume process?Schematic and P-V diagram for a polytropic process.
5Boundary Work for a Constant-Volume Process Example 4-1:Boundary Work for a Constant-Volume ProcessA rigid tank contains air at 500 kPa and 150˚C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65˚C and 400 kPa, respectively. Determine the boundary work done during this process.Wb= 0 kJ
6Boundary Work for a Constant-Pressure Process Example 4-2:Boundary Work for a Constant-Pressure ProcessA frictionless piston-cylinder device contains 5 kg of steam at 400 kpa and 200˚C. Heat is now transferred to the steam until the temperature reaches 250˚C. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process.Wb= 122 kJ
7Home TaskPlease try the following examples:EXAMPLE 4-3EXAMPLE 4-4
8Energy balance for a constant-pressure expansion or compression process General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process. Q is to the system and W is from the system.An example of constant-pressure process
9SPECIFIC HEATSSpecific heat at constant volume, cv: The energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant.Specific heat at constant pressure, cp: The energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant.True or False?cp is always greater than cvConstant-volume and constant-pressure specific heats cv and cp(values are for helium gas).
10Formal definitions of cv and cp. The equations in the figure are valid for any substance undergoing any process.cv and cp are properties.cv is related to the changes in internal energy and cp to the changes in enthalpy.A common unit for specific heats is kJ/kg·°C or kJ/kg·K. Are these units identical?Formal definitions of cv and cp.
11INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES Joule showed using this experimental apparatus that u=u(T)Internal energy and enthalpy change of an ideal gasFor ideal gases, u, h, cv, and cp vary with temperature only.
12At low pressures, all real gases approach ideal-gas behavior, and therefore their specific heats depend on temperature only.The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted cp0 and cv0.u and h data for a number of gases have been tabulated.These tables are obtained by choosing an arbitrary reference point and performing the integrations by treating state 1 as the reference state.Ideal-gas constant-pressure specific heats for some gases (see Table A–2c for cp equations).In the preparation of ideal-gas tables, 0 K is chosen as the reference temperature.
13Internal energy and enthalpy change when specific heat is taken constant at an average value (kJ/kg)
14Three ways of calculating u and h By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available.By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate.By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large.Three ways of calculating u.
15Specific Heat Relations of Ideal Gases The relationship between cp, cv and Rdh = cpdT and du = cvdTOn a molar basisSpecific heat ratioThe specific ratio varies with temperature, but this variation is very mild.For monatomic gases (helium, argon, etc.), its value is essentially constant atMany diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature.The cp of an ideal gas can be determined from a knowledge of cv and R.
16Evaluation of the ∆u of an Ideal Gas Example 4-7:Evaluation of the ∆u of an Ideal GasAir at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using (a) data from the air table (Table A-17), (b) the functional form of the specific heat (Table A-2c), and (c) the average specific heat value (Table A-2b).∆u = kJ/kg(b) ∆u = kJ/kg(c) ∆u = 220 kJ/kg
17Heating of a Gas in a Tank by Stirring Example 4-8:Heating of a Gas in a Tank by StirringAn insulated rigid tank initially contains 0.7 kg of helium at 27˚C and 350 kPa. A paddle wheel with a power rating of kW is operated within the tank for 30 min. Determine (a) the final temperature and (b) the final pressure of the helium gas.T2 = 39.4˚C(b) P2 = 364 kPa
18Heating of a Gas at Constant Pressure Example 4-10:Heating of a Gas at Constant PressureA piston-cylinder device initially contains air at 150 kPa and 27˚C. At this state, the piston is resting on a pair of stops, as shown in Fig. 4-32, and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the air.T3= 1400 K(b) W13 = 140 kJ(c) Qin = 767 kJ
19Home TaskPlease try the following examples:EXAMPLE 4-9
20INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Incompressible substance: A substance whose specific volume (or density) is constant. Solids and liquids are incompressible substances.
21Internal Energy Changes Enthalpy ChangesThe enthalpy of a compressed liquidUsually a more accurate relation than
22Cooling of an Iron Block by Water Example 4-12:Cooling of an Iron Block by WaterA 50-kg iron block at 80˚C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25˚C. Determine the temperature when thermal equilibrium is reached.T2 = 25.6˚C
23Heating of Aluminum Rods in a Furnace Example 4-13:Heating of Aluminum Rods in a FurnaceLong cylindrical aluminum rods (ρ = 2700 kg/m3 and cp = kj/kg.K) of 5-cm diameter are heat treated from 20˚C to an average temperature of 400˚C by drawing them at a velocity of 8 m/min through a long oven. Determine the rate of heat transfer to the rods in the oven.
24Home TaskPlease try the following examples:EXAMPLE 4-11
25Summary Moving boundary work Specific heats Wb for an isothermal processWb for a constant-pressure processWb for a polytropic processSpecific heatsConstant-pressure specific heat, cpConstant-volume specific heat, cvInternal energy, enthalpy, and specific heats of ideal gasesSpecific heat relations of ideal gasesInternal energy, enthalpy, and specific heats of incompressible substances (solids and liquids)