Ch-4 Energy Analysis of Closed Systems PDF

Ch-4: Energy Analysis of Closed Systems Book: ▪ Thermodynamics: An Engineering Approach by Yunus A. Cengel, Michael A. Boles, 5th / 8th Ed. (Chap 4 and 5) ▪ Applied Thermodynamics by TD Eastop and A McConkey, 5th Ed. Mechanical Engineering Dept....

Ch-4: Energy Analysis of Closed Systems Book: ▪ Thermodynamics: An Engineering Approach by Yunus A. Cengel, Michael A. Boles, 5th / 8th Ed. (Chap 4 and 5) ▪ Applied Thermodynamics by TD Eastop and A McConkey, 5th Ed. Mechanical Engineering Dept. HITEC Univ. 1  Energy Analysis of Closed Systems Moving Boundary Work ▪ Moving Boundary Work (P dV Work): The expansion and compression work in a piston-cylinder device → Moving Boundary Work associated with Real Engines or Compressors cannot be determined exactly from a Thermodynamic Analysis alone → Because piston usually moves at very high speeds, making it difficult for the gas inside to maintain equilibrium o States through which the system passes during the process cannot be specified, and no process path can be drawn o Boundary Work in real engines or compressors is determined by direct measurements ▪ In this section, we analyze the moving boundary work for a quasi-equilibrium process ▪ Quasi-Equilibrium Process: A process during which the system remains nearly in equilibrium at all times ▪ Ifthe piston is allowed to move a distance ds in a quasi- equilibrium manner, Differential Work done during this process is: Mechanical Engineering Dept. HITEC Univ. 2  Energy Analysis of Closed Systems Moving Boundary Work ▪ Total Boundary Work done during the entire process as the piston moves is obtained by adding all the differential works from the initial state to the final state Wb is Positive → for Expansion Wb is Negative → for Compression integral can be evaluated only if we know the functional relationship between P and V during the process i.e., → P = f ( V ) should be available ▪ Total Area A under the process curve 1–2 is obtained by adding differential areas dA: ▪ Relationshipbetween P and V during an Expansion or a Compression Process based on experimental data can always be plotted on P-V Diagram of the process → to calculate the Area underneath graphically to determine the Work Done Mechanical Engineering Dept. HITEC Univ. 3  Energy Analysis of Closed Systems Moving Boundary Work ▪ A gas can follow several different paths as it expands from state 1 to state 2 o each path will have a different area underneath it, and this area represents the magnitude of the work, the work done will be different for each process ▪ Ifwork were not a path function, no Cyclic Devices (Car Engines, Power Plants) could operate as work-producing devices o The work produced by these devices during one part of the cycle would have to be consumed during another part, and there would be no net work output boundary work done The net work done during a Cycle is the during a process depends difference between the work done by the on the path followed as system and the work done on the well as the end states system. Mechanical Engineering Dept. HITEC Univ. 4  Energy Analysis of Closed Systems Moving Boundary Work ▪ Pressure P in Eq. of Wb is actually the pressure at the Inner Surface of the piston o It becomes equal to the pressure of the gas in the cylinder only if the process is Quasi-Equilibrium ⇒ Entire Gas in the cylinder is at the same pressure at any given time ▪ Above Eq. can also be used for Non-Quasi-Equilibrium processes provided that the pressure at the inner face of the piston Pi is used for P Generalizing the Boundary Work Relation as: o Wb represents the amount of energy transferred from the system during an Expansion Process (or to the system during a Compression Process) o Therefore, it has to appear somewhere else and we must be able to account for it since Energy is conserved Mechanical Engineering Dept. HITEC Univ. 5  Energy Analysis of Closed Systems Moving Boundary Work ▪ E.g., Boundary Work done by the Expanding Hot Gases in a Car Engine is used to overcome Friction between the piston and the cylinder, to push Atmospheric Air out of the way, and to rotate the Crankshaft => Energy transferred by the system as work must equal the energy received by the crankshaft, the atmosphere, and the energy used to overcome friction ▪ Use of the Boundary Work Relation is not limited to the Quasi-equilibrium Processes of gases only → can also be used for Solids and Liquids Mechanical Engineering Dept. HITEC Univ. 6  Energy Analysis of Closed Systems Moving Boundary Work ▪ Boundary Work for a Constant-Volume Process -- Isochoric Mechanical Engineering Dept. HITEC Univ. 7  Energy Analysis of Closed Systems Moving Boundary Work ▪ Boundary Work for a Constant-Pressure Process -- Isobaric ⇒ ▪ Boundary Work for Isothermal Process ⇒ ⇒ Mechanical Engineering Dept. HITEC Univ. 8  Energy Analysis of Closed Systems Moving Boundary Work Polytropic Process ▪ During Actual Expansion and Compression processes of gases, pressure, and volume are often related by: ⇒ Polytropic process: C, n ( Polytropic Exponent ) constants n may take on any value from + ∞ to - ∞ depending on the particular process Above Eq. is valid for any exponent n except n = 1 ∵ For an ideal gas (PV = mRT) Polytropic and for Ideal Gas Schematic and P-V diagram ⇒ for a Polytropic Process Mechanical Engineering Dept. HITEC Univ. 9  Energy Analysis of Closed Systems Moving Boundary Work Polytropic Process -- contd -- ▪ state 1 to state A is Constant Pressure Cooling (n = 0 ) ▪ state 1 to state B is Isothermal Compression (n = 1 ) ▪ state 1 to state C is Reversible Adiabatic Compression (n = γ) ▪ state 1 to state D is Constant Volume Heating (n = ∞) Similarly; ▪ 1 to A' is Constant Pressure Heating; ▪ 1 to B' is Isothermal Expansion; ▪ 1 to C' is Reversible Adiabatic Expansion; ▪ 1 to D' is Constant Volume Cooling Mechanical Engineering Dept. HITEC Univ. 10  Energy Analysis of Closed Systems Moving Boundary Work Example A piston–cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, Determine: (a) the final pressure inside the cylinder (b) the total work done by the gas, and (c) the fraction of this work done against the spring to compress it. ⇒ Mechanical Engineering Dept. HITEC Univ. 11  Energy Analysis of Closed Systems Moving Boundary Work A piston–cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, Determine: (a) the final pressure inside the cylinder (b) the total work done by the gas, and (c) the fraction of this work done against the spring to compress it. ⇒ Mechanical Engineering Dept. HITEC Univ. 12  Energy Analysis of Closed Systems ENERGY BALANCE FOR CLOSED SYSTEMS Example A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25 °C, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25°C. Determine: (a) the Volume of the tank (b) the Final Pressure, and (c) the Heat Transfer for this process. ⇒ ⇒ Mechanical Engineering Dept. HITEC Univ. 13  Energy Analysis of Closed Systems ENERGY BALANCE FOR CLOSED SYSTEMS Example A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25 °C, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25°C. Determine: (a) the Volume of the tank (b) the Final Pressure, and (c) the Heat Transfer for this process. ⇒ Mechanical Engineering Dept. HITEC Univ. 14 Mechanical Engineering Dept. HITEC Univ. 15  Energy Analysis of Closed Systems Moving Boundary Work Example a. Unit mass of a fluid at a pressure of 3 bar, and with a specific volume of 0.18 m3/kg, contained in a cylinder behind a piston expands to a pressure of 0.6 bar according to a law p = c/v2, where c is a constant. Calculate the work done during the process. b. If the fluid is then cooled at constant pressure until the piston regains its original position; heat is then supplied with the piston firmly locked in position until the pressure rises to the original value of 3 bar. Calculate the network done by the fluid. Mechanical Engineering Dept. HITEC Univ. 16  Energy Analysis of Closed Systems Moving Boundary Work Example a. Unit mass of a fluid at a pressure of 3 bar, and with a specific volume of 0.18 m3/kg, contained in a cylinder behind a piston expands to a pressure of 0.6 bar according to a law p = c/v2, where c is a constant. Calculate the work done during the process. b. If the fluid is then cooled at constant pressure until the piston regains its original position; heat is then supplied with the piston firmly locked in position until the pressure rises to the original value of 3 bar. Calculate the network done by the fluid. Mechanical Engineering Dept. HITEC Univ. 17  Energy Analysis of Closed Systems Specific Heats ▪ Specific Heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way ▪ Specific Heat at Constant Volume, Cv: 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: Energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant cp is always greater than cv → at constant pressure the system is Constant-volume and allowed to expand and the energy for constant-pressure this expansion work must also be specific heats cv and cp supplied to the system (values are for helium gas). Mechanical Engineering Dept. HITEC Univ. 18  Energy Analysis of Closed Systems Specific Heats ▪ For a Fixed Mass in a Stationary Closed System undergoing a Constant-volume Process (no expansion or compression work is involved) → conservation of energy principle (ein - eout = Δesystem), for this process can be expressed in the differential form → Net Amount of Energy Transferred to the system → this energy must be equal to cv dT, where dT is the Differential Change in temperature ⇒ OR ▪ For a Constant-pressure Expansion or Compression Process → Constant pressure, for which the work term can be integrated and the resulting PV terms at the initial and final states can be associated with the internal energy terms Mechanical Engineering Dept. HITEC Univ. 19  Energy Analysis of Closed Systems Specific Heats ▪ Above equations are Property Relations and are independent of the type of processes The specific heat of a substance o are valid for any substance undergoing any process changes with temperature ▪ cv and cp are Properties ▪ cv → changes in Internal Energy ▪ cp → changes in Enthalpy ▪ Unit for specific heats is kJ/kg·°C or kJ/kg· K Mechanical Engineering Dept. HITEC Univ. 20  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES ▪ It has been demonstrate experimentally (Joule, 1843) that for an Ideal Gas the Internal Energy is a function of the Temperature only ▪ Initially, one tank contained Air at a high pressure and the other tank was evacuated o valve was opened to let Air pass from one tank to the other until the pressures equalized o Joule observed no change in the temperature of the water bath and assumed that no heat was transferred to or from the Air o Since there is no work done → Joule concluded that the Internal Energy of the Air did not change even though the volume and the pressure changed o Internal Energy is a function of temperature only and not a function of pressure or specific volume o for gases that deviate significantly from ideal-gas behavior, the Internal Energy is not a function of Temperature alone Mechanical Engineering Dept. HITEC Univ. 21  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES ▪ For a low-density gas → u depends primarily on T and much less on the second property, P or v ▪ Dependence of u on P is less at low pressure and is much less at high temperature → i.e., as the Density decreases, so does dependence of u on P (or v) Mechanical Engineering Dept. HITEC Univ. 22  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES ▪ Based on Definition of Enthalpy and the Equation of state of an Ideal Gas R is constant and u = u(T) ⇒ u and h depend only on temperature for an ideal gas Specific Heats cv and cp also depend, at most, on temperature only For ideal gases, u, h, cv, and cp vary with temperature only. ▪ At low pressures, all Real Gases approach Ideal-gas Behavior → their specific heats depend on temperature only ▪ Specific heats of real gases at low pressures are called Ideal-gas Specific Heats, or zero-pressure specific heats, and are often denoted as cp0 and cMechanical v0 Engineering Dept. HITEC Univ. 23  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES ▪ u and h data for a number of gases have been tabulated ▪ 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 (Table A–2c for cp equations). ▪ specific heats of gases with complex molecules (with two or more atoms) are higher and increase with temp. ▪ Principal Factor causing specific heat to vary with temperature is Molecular Vibration ▪ More complex molecules have multiple vibrational modes and therefore show In the preparation of ideal- greater temperature dependency gas tables, 0 K is chosen as ▪ Variation of Cp with T is smooth and may the Reference Temperature be approximated as linear over small Mechanical Engineering Dept. HITEC Univ. 24 temperature intervals  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES ▪ Internal Energy and Enthalpy Change when specific heat is taken constant at an average value over a certain range: Three ways of calculating u and h 1. By using the tabulated u and h data → easiest and most accurate way when tables are readily available 2. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations → very Inconvenient for hand calculations but quite desirable for computerized calculations → Results For small temperature obtained are very accurate intervals, the specific 3. By using average specific heats → very simple and heats may be assumed to certainly very convenient when property tables are not vary linearly with available → results obtained are reasonably accurate if temperature interval is not very large Mechanical Engineering Dept. HITEC Univ. temperature 25 Mechanical Engineering Dept. HITEC Univ. 26  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES Example A 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., 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. Mechanical Engineering Dept. HITEC Univ. 27  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES Specific Heat Relations of Ideal Gases Relationship between cp, cv and R On a molar basis dh = cpdT and du = cvdT Specific Heat Ratio o Specific Heat Ratio varies with temperature, but this variation is very mild o For Monatomic Gases (Helium, Argon, etc.), its value is essentially constant at 1.667 o Many 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 Mechanical Engineering Dept. HITEC Univ. from a knowledge of cv and R. 28  Energy Analysis of Closed Systems INTERNAL 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 o Constant-volume Assumption should be taken to imply that the energy associated with the volume change is negligible compared with other forms of energy o constant-volume and constant- pressure specific heats are identical for incompressible substances The specific volumes of The cv and cp values of incompressible substances remain incompressible substances are constant during a process. identical and are denoted by c. Mechanical Engineering Dept. HITEC Univ. 29  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Internal Energy Changes ▪ Like ideal gases, specific heats of Incompressible Substances depend on temperature only ⇒ For small temperature intervals, c value at the average temperature can be used and treated as a constant, yielding: Enthalpy Changes ⇒ = ▪ For solids, the term v ΔP is insignificant and thus: Δh = Δu ≅ cavg ΔT Mechanical Engineering Dept. HITEC Univ. 30  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Enthalpy Changes – contd -- ▪ For liquids, Two Special Cases are commonly encountered: 1. Constant-pressure processes, as in Heaters (ΔP = 0): Δh = Δu ≅ cavg ΔT 2. Constant-temperature processes, as in Pumps (ΔT = 0): Δh = v ΔP → For a process between states 1 and 2 : Mechanical Engineering Dept. HITEC Univ. 31  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Example Consider a 1000-W iron whose base plate is made of 0.5-cm-thick aluminum alloy 2024-T6 (ρ = 2770 kg/m3 and cp = 875 J/kg °C). The base plate has a surface area of 0.03 m2. Initially, the iron is in thermal equilibrium with the ambient air at 22 °C. Assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine the minimum time needed for the plate temperature to reach 140 °C. ⇒ ⇒ Mechanical Engineering Dept. HITEC Univ. 32  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Example 680 kg of fish at 5°C are to be frozen and stored at – 12°C. The specific heat of fish above freezing point is 3.182, and below freezing point is 1.717 kJ/kg K. The freezing point is – 2°C, and the latent heat of fusion is 234.5 kJ/kg. How much heat must be removed to cool the fish, and what per cent of this is latent heat? Mechanical Engineering Dept. HITEC Univ. 33  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Example A mass of 0.05 kg of a fluid is heated at a constant pressure of 2 bar until the volume occupied is 0.0658 m3. calculate the heat supplied and the work done: i. When the fluid is steam , initially dry saturated, ii. When the fluid is air, initially at 130 oC. ⇒ ⇒ Mechanical Engineering Dept. HITEC Univ. 34  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Example A mass of 0.05 kg of a fluid is heated at a constant pressure of 2 bar until the volume occupied is 0.0658 m3. calculate the heat supplied and the work done: i. When the fluid is steam , initially dry saturated, ii. When the fluid is air, initially at 130 oC. Mechanical Engineering Dept. HITEC Univ. 35 Mechanical Engineering Dept. HITEC Univ. 36  Energy Analysis of Closed Systems INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Example A 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. ⇒ ⇒ ⇒ ⇒ ⇒ Mechanical Engineering Dept. HITEC Univ. 37 Mechanical Engineering Dept. HITEC Univ. 38  MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Practice Problems: Book: Yunus Cengel 8th Ed. ▪ Examples: 4.1 – 4.4, 4.7 – 4.13, ▪ Problems: 4.1 – 4.9, 4.14, 4.18 – 4.22, 4.45 – 4.50, 4.54, 4.55, 4.58, 4.63 – 4.65, 4.67 – 4.72, 4.76. Mechanical Engineering Dept. HITEC Univ. 39