Thermodynamics: Heat and Work Transfer, Specific Heats, and Conservation of Energy, Study notes of Thermodynamics

Download Thermodynamics: Heat and Work Transfer, Specific Heats, and Conservation of Energy and more Study notes Thermodynamics in PDF only on Docsity! Chapter 5, ECE 309, Spring 2016. 1 . Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy gained by the system must be exactly equal to the amount of energy lost by the surroundings. A closed system can exchange energy with its surroundings through heat and work transfer. In other words, work and heat are the forms that energy can be transferred across the system boundary. Based on kinetic theory, heat is defined as the energy associated with the random motions of atoms and molecules. Heat Transfer Heat is defined as the form of energy that is transferred between two systems by virtue of a temperature difference. Note: there cannot be any heat transfer between two systems that are at the same temperature. Note: It is the thermal (internal) energy that can be stored in a system. Heat is a form of energy in transition and as a result can only be identified at the system boundary. Heat has energy units kJ (or BTU). Rate of heat transfer is the amount of heat transferred per unit time. Heat is a directional (or vector) quantity; thus, it has magnitude, direction and point of action. Notation: – Q (kJ) amount of heat transfer – Q° (kW) rate of heat transfer (power) – q (kJ/kg) - heat transfer per unit mass – q° (kW/kg) - power per unit mass Sign convention: Heat Transfer to a system is positive, and heat transfer from a system is negative. It means any heat transfer that increases the energy of a system is positive, and heat transfer that decreases the energy of a system is negative. Chapter 5, ECE 309, Spring 2016. 2 . Fig. 5-1: Sign convention: positive if to the system, negative if from the system. Modes of Heat Transfer Heat can be transferred in three different modes conduction, convection, and radiation. All modes of heat transfer require the existence of a temperature difference. Conduction: is the transfer of energy from the more energetic particles to the adjacent less energetic particles as a result of interactions between particles. In solids, conduction is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons. Conduction will be discussed in more details in Ch 10 & 11. Convection: is the mode of energy transfer between a solid surface and the adjacent liquid or gas which is in motion, and it involves the combined effects of conduction and fluid motion (advection). Convection is called forced if the fluid is forced to flow by external means such as a fan or a pump. It is called free or natural if the fluid motion is caused by buoyancy forces that are induced by density differences due to the temperature variation in a fluid. Convection will be discussed in more details in Ch 12, 13 & 14. Radiation: is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. Radiation will be discussed in more details in Ch 15. Work Work is the energy interaction between a system and its surroundings. More specifically, work is the energy transfer associated with force acting through a distance. Q = 5 kJ Q = - 5 kJ Heat in Heat out System Chapter 5, ECE 309, Spring 2016. 5 . Mechanical work There are several ways of doing work, each in some way related to a force acting through a distance. )(. kJsFW = If the force is not constant, we need to integrate: ∫= 2 1 )(. kJdsFW There are two requirements for a work interaction: there must be a force acting on the boundary the boundary must move Therefore, the displacement of the boundary without any force to oppose or drive this motion (such as expansion of a gas into evacuated space) is not a work interaction, W=0. Also, if there are no displacements of the boundary, even if an acting force exists, there will be no work transfer W = 0 (such as increasing gas pressure in a rigid tank). Moving Boundary Work The expansion and compression work is often called moving boundary work, or simply boundary work. We analyze the moving boundary work for a quasi-equilibrium process, see Chapter 3. Consider the gas enclosed in a piston-cylinder at initial P and V. If the piston is allowed to move a distance ds in a quasi-equilibrium manner, the differential work is: PdVPAdsdsFWb === .δ The quasi-equilibrium expansion process is shown in Fig. 5-4. On this diagram, the differential area dA under the process curve in P-V diagram is equal to PdV, which is the differential work. Note: a gas can follow several different paths from state 1 to 2, and each path will have a different area underneath it (work is path dependent). The net work or cycle work is shown in Fig. 5-5. In a cycle, the net change for any properties (point functions or exact differentials) is zero. However, the net work and heat transfer depend on the cycle path. ∆U = ∆P = ∆T = ∆(any property) = 0 Chapter 5, ECE 309, Spring 2016. 6 . Fig. 5-4: the area under P-V diagram represents the boundary work. Fig. 5-5: network done during a cycle. Polytropic Process During expansion and compression processes of real gases, pressure and volume are often related by PVn=C, where n and C are constants. the moving work for a polytropic process can be found: n VPVPdVCVPdVW n polytopic − − === ∫ ∫ − 1 1122 2 1 2 1 Since CVPVP nn == 2211 . For an ideal gas (PV= mRT) it becomes: V P Wnet 1 2 V1 V2 Chapter 5, ECE 309, Spring 2016. 7 . ( ) )(1, 1 12 kJn n TTmR Wpolytropic ≠ − − = The special case n =1 is the isothermal expansion P1V1 = P2V2 = mRT0= C, which can be found from: )(1,ln 1 2 11 2 1 2 1 , kJn V V VPdV V CPdVW isothermalb =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ === ∫∫ Since for an ideal gas, PV=mRT0 at constant temperature T0, or P=C/V. Example 5-2: Polytropic work A gas in piston-cylinder assembly undergoes a polytropic expansion. The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3. Determine the work for the process, in kJ, if a) n=1.5, b) n=1.0, and c) n=0. Solution: Assume that i) the gas is a closed system, ii) the moving boundary is only work mode, and iii) the expansion is polytropic. a) n =1.5 ∫ − − == 2 1 1122 1 V V n VPVP PdVW We need P2 that can be found from nn VPVP 2211 = : ( ) barbar V VPP n 06.1 2.0 1.03 5.1 2 1 12 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ( )( ) ( )( ) kJ mN kJ bar mNmbarW 6.17 .10 1 1 /10 5.11 1.032.006.1 3 253 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = b) n =1, the pressure volume relationship is PV = constant. The work is: ( )( ) kJ mN kJ bar mNmbarW V VVPPdVW 79.20 1.0 2.0ln .10 1 1 /101.03 ln 3 25 3 1 2 11 2 1 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == ∫ c) For n = 0, the pressure-volume relation reduces to P=constant (isobaric process) and the integral become W= P (V2-V1). Substituting values and converting units as above, W=30 kJ. Chapter 5, ECE 309, Spring 2016. 10 . Q° – W° = dE / dt (kW) For a cyclic process, the initial and final states are identical, thus ∆E=0. The first law becomes: Q – W = 0 (kJ) Note: from the first law point of view, there is no difference between heat transfer and work, they are both energy interactions. But from the second law point of view, heat and work are very different. Example 5-4: Fist law Air is contained in a vertical piston-cylinder assembly fitted with an electrical resistor. The atmospheric pressure is 100 kPa and piston has a mass of 50 kg and a face area of 0.1 m2. Electric current passes through the resistor, and the volume of air slowly increases by 0.045 m3. The mass of the air is 0.3 kg and its specific energy increases by 42.2 kJ/kg. Assume the assembly (including the piston) is insulated and neglect the friction between the cylinder and piston, g = 9.8 m/s2. Determine the heat transfer from the resistor to air for a system consisting a) the air alone, b) the air and the piston. Fig. 5-7: Schematic for problem 5-4. Assumptions: two closed systems are under consideration, as shown in schematic. the only heat transfer is from the resistor to the air. ∆KE = ∆PE= 0 (for air) the internal energy is of the piston is not affected by the heat transfer. Air Piston Air Piston System boundary part b System boundary part a Chapter 5, ECE 309, Spring 2016. 11 . a) taking the air as the system, (∆KE + ∆PE + ∆U)air = Q – W Q = W + ∆Uair For this system work is done at the bottom of the piston. The work done by the system is (at constant pressure): ( )∫ −== 2 1 12 V V VVPPdVW The pressure acting on the air can be found from: PApiston = mpiston g + Patm Apiston ( )( ) ( ) kPakPa Pa kPa mN Pa m smkgP P A gm P atm 91.104100 1000 1 /1 1 1.0 /81.950 22 2 piston piston =+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= += Thus, the work is W = (104.91 kPa)(0.045m3) = 4.721 kJ With ∆Uair = mair ∆uair, the heat transfer is Q = W + mair ∆uair = 4.721 kJ + (0.3 kg)(42.2 kJ/kg) = 17.38 kJ b) system consisting the air and the piston. The first law becomes: (∆KE + ∆PE + ∆U)air + (∆KE + ∆PE + ∆U)piston = Q – W where (∆KE = ∆PE)air = 0 and (∆KE = ∆U)piston= 0. Thus, it simplifies to: (∆U)air + (∆PE)piston = Q – W For this system, work is done at the top of the piston and pressure is the atmospheric pressure. The work becomes W = Patm ∆V = (100 kPa)(0.045m3) = 4.5 kJ The elevation change required to evaluate the potential energy change of the piston can be found from the volume change: ∆z = ∆V / Apiston = 0.045 m3/ 0.1 m2 = 0.45 m (∆PE)piston = m piston g ∆z = (50 kg)(9.81 m/s2)(0.45 m) = 220.73 J = 0.221 kJ Q = W + (∆PE)piston + mair ∆uair Q = 4.5 kJ + 0.221 kJ + (0.3 kg)(42.2 kJ/kg) = 17.38 kJ Note that the heat transfer is identical in both systems. Chapter 5, ECE 309, Spring 2016. 12 . Specific Heats The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. There are two kinds of specific heats: specific heat at constant volume, Cv (the energy required when the volume is maintained constant) specific heat at constant pressure, Cp (the energy required when the pressure is maintained constant) The specific heat at constant pressure Cp is always higher than Cv because at constant pressure the system is allowed to expand and energy for this expansion must also be supplied to the system. Let’s consider a stationary closed system undergoing a constant-volume process (wb = 0). Applying the first law in the differential form: δq – δw = du at constant volume (no work) and by using the definition of Cv, one can write: v v v T uC or dudTC ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ = = Similarly, an expression for the specific heat at constant pressure Cp can be found. From the first law, for a constant pressure process (wb + ∆u = ∆h). It yields: p p T hC ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ = specific heats (both Cv and Cp) are properties and therefore independent of the type of processes. Cv is related to the changes in internal energy u, and Cp to the changes in enthalpy, h. It would be more appropriate to define: Cv is the change in specific internal energy per unit change in temperature at constant volume. Cp is the change in specific enthalpy per unit change in temperature at constant pressure. Specific heats for ideal gases It has been shown mathematically and experimentally that the internal energy is a function of temperature only. u = u(T) Using the definition of enthalpy (h = u + Pv) and the ideal gas equation of state (Pv = RT), we have: Chapter 5, ECE 309, Spring 2016. 15 . The final pressure can be found from ideal gas equation of state: ( ) 21 21 21 VV RTmm VV RTm P fft f + + = + = For tank 1 and 2, we can write: V1 = m1RT1/P1 and V2 = m2RT2/P2. Thus, the final pressure, Pf becomes: ( ) ( ) ( )( ) ( )( ) ( )( ) bar bar Kkg bar Kkg KkgP P Tm P Tm Tmm P RTm P RTm RTmm P f ff f 05.1 2.1 3008 7.0 3502 31510 2 22 1 11 21 2 22 1 11 21 = + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = b) the heat transfer can be found from an energy balance: ∆U = Q – W With W = 0, Q = Uf –Ui where initial internal energy is: Ui= m1 u(T1) + m2 u(T2) The final internal energy is: Uf = (m1 + m2) u(Tf) The energy balance becomes: Q = m1 [u(Tf) – u(T1)] + m2 [u(Tf) – u(T2)] Since the specific heat Cv is constant Q = m1 Cv [Tf – T1] + m2 Cv [Tf – T2] ( ) ( ) ( ) ( ) kJKK Kkg kJkgKK Kkg kJkgQ 25.37300315 . 745.08350315 . 745.02 =−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = The plus sign indicates that the heat transfer is into the system. CO2 8 kg, 27C, 1.2 bar CO2 2 kg, 77C, 0.7 bar Valve Chapter 6, ECE 309, Spring 2016. 1 . Chapter 6: The First Law: Control Volumes The first law is discussed for closed systems in Chapter 5. In this Chapter, we extend the conservation of energy to systems that involve mass flow across their boundaries, control volumes. Any arbitrary region in space can be selected as control volume. There are no concrete rules for the selection of control volumes. The boundary of control volume is called a control surface. Conservation of Mass Like energy, mass is a conserved property, and it cannot be created or destroyed. Mass and energy can be converted to each other according to Einstein’s formula: E = mc2, where c is the speed of light. However, except for nuclear reactions, the conservation of mass principle holds for all processes. For a control volume undergoing a process, the conservation of mass can be stated as: total mass entering CV – total mass leaving CV = net change in mass within CV CVei mmm ∆=−∑∑ Fig. 6-1: Conservation of mass principle for a CV. The conservation of mass can also be expressed in the rate form: dtdmmm CVei /=−∑∑ •• The amount of mass flowing through a cross section per unit time is called the mass flow rate and is denoted by m°. The mass flow rate through a differential area dA is: dm°= ρVn dA m°i m°o Control volume Chapter 6, ECE 309, Spring 2016. 2 . where V is the velocity component normal to dA. Thus, the mass flow rate for the entire cross-section is obtained by: )kg/s(dAVm A n∫=• ρ Assuming one-dimensional flow, a uniform (averaged or bulk) velocity can be defined: m°= ρ V A (kg/s) where V (m/s) is the fluid velocity normal to the cross sectional area. The volume of the fluid flowing through a cross-section per unit time is called the volumetric flow, V°: /s)(m3VAdAVV A n == ∫• The mass and volume flow rate are related by: m°=ρV°= V°/ v. Conservation of Energy For control volumes, an additional mechanism can change the energy of a system: mass flow in and out of the control volume. Therefore, the conservation of energy for a control volume undergoing a process can be expressed as total energy crossing boundary as heat and work + total energy of mass entering CV – total energy of mass leaving CV = net change in energy of CV CVmassoutmassin EEEWQ ∆=++− ∑∑ ,, This equation is applicable to any control volume undergoing any process. This equation can also be expressed in rate form: dtdEdtdEdtdEWQ CVmassoutmassin /// ,, =++− ∑∑•• Fig. 6-2: Energy content of CV can be changed by mass flow in/out and heat and work interactions. Mass in Mass out Q W Control volume Chapter 6, ECE 309, Spring 2016. 5 . ∆PE = 0, no change in fluid elevation. Mass equation for nozzles becomes: m°1 = m°2 or ρ1A1V1 = ρ2A2V2 Energy balance: Q° - W° = m°1 θ1 - m°2 θ2 (h + V2 / 2)at inlet = (h + V2 / 2)at exit Diffusers are exactly the same device as nozzles; the only difference is the direction of the flow. Thus, all equations derived for nozzles hold for diffusers. Fig. 6-5: Schematic for diffuser. Example 6-1: Nozzle Steam enters a converging-diverging nozzle operating at steady state with P1 = 0.05 MPa, T1 = 400 °C and a velocity of 10 m/s. The steam flows through the nozzle with negligible heat transfer and no significant change in potential energy. At the exit, P2 = 0.01 MPa, and the velocity is 665 m/s. The mass flow rate is 2 kg/s. Determine the exit area of the nozzle, in m2. Assumptions: 1. Steady state operation of the nozzle 2. Work and heat transfer are negligible, Q° = W°= 0. 3. Change in potential energy from inlet to exit is negligible, ∆PE = 0. The exit area can be calculated from the mass flow rate m°: A2 = m°v2 / V2 We need the specific volume at state 2. So, state 2 must be found. The pressure at the exit is given; to fix the state 2 another property is required. From the energy equation enthalpy can be found: ( )2 2 2 112 2 1 1 2 2 2 2 1 22 VVhhVhVh −+=⇒+=+ Diffuser 1 2 A1 < A2 V1 > V2 P1 < P2 Control Volume Q° = 0 W° = 0 Chapter 6, ECE 309, Spring 2016. 6 . From Table A-6, h1 = 3278.9 kJ/kg and velocities are given. kgkJ mN kJ smkg N s mkgkJh /84.3057 .10 1 /.1 1 2 66510/9.3278 32 222 2 =⎥⎦ ⎤ ⎢⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − += From Table A-6, with P2 = 0.01 MPa, and h2 = 3057.84 kJ/kg, (using interpolation) one finds T2 = 300 °C (approximately) and v2 = 26.445 m3/kg. The exit area is: ( )( ) 2 3 2 07955.0 /665 /445.26/2 m sm kgmskgA == Fig. 6-6: Converging-diverging nozzle. Turbines and Compressors In steam, gas, or hydroelectric power plants, the device that derives the electric generator is turbine. The work of turbine is positive since it is done by the fluid. Compressors, pumps, and fans are devices used to increase the pressure of the fluid. Work is supplied to these devices, thus the work term is negative. Common assumptions for turbines and compressors: Q° = 0. ∆PE = ∆KE = 0. Example 6-2: Turbine Steam enters a turbine at steady state with a mass flow rate of 4600 kg/h. The turbine develops a power output of 1000 kW. At the inlet the pressure is 0.05 MPa, the temperature is 400 °C, and the velocity is10 m/s. At the exit, the pressure is 10 kPa, the quality is 0.9, and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and surroundings, in kW. Chapter 6, ECE 309, Spring 2016. 7 . Assumptions: Steady-state operation. The change in potential energy is negligible. The energy balance for the turbine is: ( ) ( )⎥⎦ ⎤ ⎢⎣ ⎡ −+−+= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++=− ••• •••• 2 1 2 212 1 2 1 12 2 2 2 2 1 22 VVhhmWQ gzVhmgzVhmWQ CVCV CVCV Fig. 6-7: Schematic for turbine. Using Table A-6, h1 = 3278.9 kJ/kg, using Table A-5 with x = 0.9, h2 = hf + x2 hfg = 191.83 kJ/kg + 0.9(2392.8 kJ/kg) = 2345.35 kJ/kg Therefore, h2 – h1 = 2345.35 – 3278.9 = -933.55 kJ/kg The specific kinetic energy change is: ( ) ( ) kgkJ mN kJ smkg N s mVV /2.1 .10 1 /.1 110505.0 2 1 32 2 222 1 2 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛−=− Q°CV = (1000 kW) + 4600 kg/h (-831.8 + 1.2 kJ/kg) (1h / 3600 s) (1kW / 1 kJ/s) = -61.3 kW Turbine 1 2 W° Q° = 0 (negligible) Shaft Control Volume Chapter 6, ECE 309, Spring 2016. 10 . 41213311 4321 and hmhmhmhm mmmm •••• •••• +=+ == Fig. 6-10: Schematic for heat exchanger. Example 6-3: Heat exchanger Engine oil is to be cooled by water in a condenser. The engine oil enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70°C and leaves at 35°C. The cooling water enters at 300 kPa and 15°C and leaves at 25°C. Neglecting any pressure drops; determine a) the mass flow rate of the cooling water required, and b) the heat transfer rate from the engine oil to water. We choose the entire heat exchanger as our control volume, thus work transfer and heat transfer to the surroundings will be zero. From mass balance: Heat Water 15°C 1 2 3 4 Water 25°C Oil 70°C Oil 35°C Control Volume Heat Fluid A Fluid A Fluid B Fluid B 1 2 3 4 Chapter 6, ECE 309, Spring 2016. 11 . •••••• ==== OilW mmmmmm 4321 and The conservation of energy equation is: •• •••• •• •••• − − = +=+ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++=− ∑∑ ∑∑ OilW OilWOilW eeee i i iie e ee m hh hh m hmhmhmhm hmhm gz V hmgz V hmWQ 12 43 4231 22 22 Assuming constant specific heat for both the oil and water at their average temperature, ( ) ( ) ( ) ( ) ( ) min/1.10min/6 1525 . 18.4 3570 . 016.2 12, 43, kgkg Ckg kJ Ckg kJ TTC TTC m WaterP OilP W = −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − − =• b) To determine the heat transfer from the oil to water, choose the following CV. The energy equation becomes: ( ) ( ) ( ) ( ) kgkJ Ckg kJkgQ TTCmhmWQ Oil OilPOilOilOil /36.4237035 . 016.2min/6 34, −=−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = −=∆=− • •••• Oil 70°C 3 4 Oil 35°C Control volume Q