Thermodynamics: An Engineering Approach Chapter 5 PDF

Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by...

Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi  Objectives Develop the conservation of mass principle. Apply the conservation of mass principle to various systems including steady-flow control volumes. Apply the first law of thermodynamics as the statement of the conservation of energy principle to control volumes. Identify the energy carried by a fluid stream crossing a control surface as the sum of internal energy, flow work, kinetic energy, and potential energy of the fluid and to relate the combination of the internal energy and the flow work to the property enthalpy. Solve energy balance problems for common steady-flow devices such as nozzles, compressors, turbines, throttling valves, Apply the energy balance to general unsteady-flow processes with particular emphasis on the uniform-flow process as the model for commonly encountered charging and discharging processes. 2 CONSERVATION OF MASS Conservation of mass: Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume. Mass m and energy E can be converted to each other according to where c is the speed of light in a vacuum, which is c = 2.9979  108 m/s. The mass change due to energy change is negligible. 3 Conservation of Energy for Control volumes The conservation of mass and the conservation of energy principles for open systems or control volumes apply to systems having mass crossing the system boundary or control surface. In addition to the heat transfer and work crossing the system boundaries, mass carries energy with it as it crosses the system boundaries. Thus, the mass and energy content of the open system may change when mass enters or leaves the control volume. Thermodynamic processes involving control volumes can be considered in two groups: steady-flow processes and unsteady-flow processes. During a steady-flow process, the fluid flows through the control volume steadily, experiencing no change with time at a fixed position. 4 A pressurized water nuclear reactor steam power plant has many examples of open system operation. Some of these are the pressure vessel, steam generator, turbine, condenser, and pumps. Photo courtesy of Progress Energy Carolinas, Inc. 5 A heat exchanger, the heater core from a 1966 289 V8 Mustang, is another example of an open system. Cooling water flows into and out of the tubes and air is forced through the fin sturucture. 6 A hair drier is an example of a one entrance, one exit open system that receives electrical work input to drive the fan and power the resistance heater. 7 Let’s review the concepts of mass flow rate and energy transport by mass. One should study the development of the general conservation of mass presented in the text. Here we present an overview of the concepts important to successful problem solving techniques. Mass Flow Rate Mass flow through a cross-sectional area per unit time is called the mass flow rate m. Note the dot over the mass symbol indicates a time rate of change. It is expressed as  where Vn is the velocity normal to the cross-sectional flow area. 8 If the fluid density and velocity are constant over the flow cross-sectional area, the mass flow rate is V A m  Vave A  ave v where  is the density, kg/m3 ( = 1/v), A is the cross-sectional area, m2; and Vave is the average fluid velocity normal to the area, m/s. Example 5-1 Refrigerant-134a at 200 kPa, 40% quality, flows through a 1.1-cm inside diameter, d, tube with a velocity of 50 m/s. Find the mass flow rate of the refrigerant-134a. At P = 200 kPa, x = 0.4 we determine the specific volume from table A12, Page 918 v  v f  xv fg  0.0007533  0.4(0.0999  0.0007533) m3  0.0404 kg Vave A Vave  d 2 m  v v 4 50 m / s  (0.011 m) 2  0.0404 m3 / kg 4 kg 9  0.117 s The fluid volume flowing through a cross-section per unit time is called the volume flow rate V. The volume flow rate is given by integrating the product of the velocity normal to the flow area and the differential flow area over the flow area. If the velocity over the flow area is a constant, the volume flow rate is given by (note we are dropping the “ave” subscript on the velocity)   V  VA (m3 / s) The mass and volume flow rate are related by V   V  m ( kg / s) Example 5-2 v Air at 100 kPa, 50oC, flows through a pipe with a volume flow rate of 40 m3/min. Find the mass flow rate through the pipe, in kg/s. Assume air to be an ideal gas, so From table A1, page 898 , we find the gas constant (R) RT kJ (50  273) K m3 kPa v  0.287 P kg  K 100kPa kJ m3  0.9270 kg 10  V 40m3 / min 1 min m   v 0.9270m3 / kg 60s kg  0.719 s Conservation of Mass for General Control Volume The conservation of mass principle for the open system or control volume is expressed as or  m   m in out  m  system ( kg / s) Steady-State, Steady-Flow Processes Most energy conversion devices operate steadily over long periods of time. The rates of heat transfer and work crossing the control surface are constant with time. The states of the mass streams crossing the control surface or boundary are constant with time. Under these conditions the mass and energy content of the control volume are constant with time. 11  dmCV  mCV  0 dt Steady-state, Steady-Flow Conservation of Mass: Since the mass of the control volume is constant with time during the steady-state, steady-flow process, the conservation of mass principle becomes or  m in  m  out ( kg / s) 12 Flow work and the energy of a flowing fluid Energy flows into and from the control volume with the mass. The energy required to push the mass into or out of the control volume is known as the flow work or flow energy. The fluid up steam of the control surface acts as a piston to push a unit of mass into or out of the control volume. Consider the unit of mass entering the control volume shown below. As the fluid upstream pushes mass across the control surface, work done on that unit of mass is 13 FLOW WORK AND THE ENERGY OF A FLOWING FLUID Flow work, or flow energy: The work (or energy) required to push the mass into or out of the control volume. This work is necessary for maintaining a continuous flow through a control volume. 14 Total Energy of a Flowing Fluid The flow energy is automatically taken care of by enthalpy. h = u + Pv In fact, this is the main reason for defining the property enthalpy. The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid. 15 Energy Transport by Mass When the kinetic and potential energies of a fluid stream are negligible When the properties of the mass at each inlet or exit change with time as well as over the cross section 16 ENERGY ANALYSIS OF STEADY-FLOW SYSTEMS Steady-flow process: A process during which a fluid flows through a control volume steadily. 17 The term Pv is called the flow work done on the unit of mass as it crosses the control surface. The total energy of flowing fluid The total energy carried by a unit of mass as it crosses the control surface is the sum of the internal energy, flow work, potential energy, and kinetic energy. 2 V   u  Pv   gz 2 2 V  h  gz 2 Here we have used the definition of enthalpy, h = u + Pv. Energy transport by mass Amount of energy transport across a control surface:  V2  Emass  m  m  h   gz  (kJ)  2  18 Rate of energy transport across a control surface:  V2  Emass  m  m  h   gz  (kW )  2  Conservation of Energy for General Control Volume The conservation of energy principle for the control volume or open system has the same word definition as the first law for the closed system. Expressing the energy transfers on a rate basis, the control volume first law is E in  E out  E system ( kW ) or      Rate of net energy transfer Rate change in internal, kinetic, by heat, work, and mass potential, etc., energies Considering that energy flows into and from the control volume with the mass, energy enters because net heat is transferred to the control volume, and energy leaves because the control volume does net work on its surroundings, the open system, or control volume, the first law becomes 19 where  is the energy per unit mass flowing into or from the control volume. The energy per unit mass, , flowing across the control surface that defines the control volume is composed of four terms: the internal energy, the kinetic energy, the potential energy, and the flow work. The total energy carried by a unit of mass as it crosses the control surface is 2 V   u  Pv   gz 2 2 V  h  gz 2 Where the time rate change of the energy of the control volume has been written as E CV 20 Steady-State, Steady-Flow Processes Most energy conversion devices operate steadily over long periods of time. The rates of heat transfer and work crossing the control surface are constant with time. The states of the mass streams crossing the control surface or boundary are constant with time. Under these conditions the mass and energy content of the control volume are constant with time. dmCV  m CV  0 dt dECV  E CV  0 dt Steady-state, Steady-Flow Conservation of Mass:  m in  m  out ( kg / s) Steady-state, steady-flow conservation of energy Since the energy of the control volume is constant with time during the steady-state, steady-flow process, the conservation of energy principle becomes 21 or E in  E out  0 E system ( kW )      Rate of net energy transfer Rate change in internal, kinetic, by heat, work, and mass potential, etc., energies or Considering that energy flows into and from the control volume with the mass, energy enters because heat is transferred to the control volume, and energy leaves because the control volume does work on its surroundings, the steady-state, steady-flow first law becomes 22 Often this result is written as Q net   Q in   Q out where Wnet   Wout   Win Steady-state, steady-flow for one entrance and one exit A number of thermodynamic devices such as pumps, fans, compressors, turbines, nozzles, diffusers, and heaters operate with one entrance and one exit. The steady- state, steady-flow conservation of mass and first law of thermodynamics for these systems reduce to 23   is the where the entrance to the control volume is state 1 and the exit is state 2 and m mass flow rate through the device. When can we neglect the kinetic and potential energy terms in the first law? Consider the kinetic and potential energies per unit mass. 2 V ke  2  m (45m / s) 2 1kJ / kg kJ For V = 45 ke   1 s 2 1000m2 / s2 kg  m (140m / s) 2 1kJ / kg kJ V = 140 ke   10 s 2 1000m2 / s2 kg pe  gz m 1kJ / kg kJ For z  100m pe  9.8 100m  0.98 s2 1000m2 / s2 kg m 1kJ / kg kJ z  1000m pe  9.8 2 1000m  9.8 s 1000m2 / s2 kg 24 When compared to the enthalpy of steam (h  2000 to 3000 kJ/kg) and the enthalpy of air (h  200 to 6000 kJ/kg), the kinetic and potential energies are often neglected. When the kinetic and potential energies can be neglected, the conservation of energy equation becomes Q  W  m  (h2  h1 ) ( kW ) We often write this last result per unit mass flow as q  w  (h2  h1 ) ( kJ / kg) Q W where q  and w . m m Some Steady-Flow Engineering Devices Below are some engineering devices that operate essentially as steady-state, steady- flow control volumes. 25 We may be familiar with nozzles, diffusers, turbines, compressors, compressors, heat exchangers, and mixing devices. However, the throttle valve may be a new device to many of us. The Throttle may be a simple as the expansion tube used in automobile air conditioning systems to cause the refrigerant pressure drop between the exit of the condenser and the inlet to the evaporator. 26 SOME STEADY-FLOW ENGINEERING DEVICES Many engineering devices operate essentially under the same conditions for long periods of time. The components of a steam power plant (turbines, compressors, heat exchangers, and pumps), for example, operate nonstop for months before the system is shut down for maintenance. Therefore, these devices can be conveniently analyzed as steady-flow devices. A modern land-based gas turbine used for electric power production. This is a General Electric LM5000 turbine. It has a length of 6.2 m, it weighs 12.5 tons, and produces 55.2 MW at 3600 rpm with steam injection. 27 Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even garden hoses. A nozzle is a device that increases the velocity of a fluid at the expense of pressure. A diffuser is a device that increases the pressure of a fluid by slowing it down. The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows. The reverse is true for diffusers. Energy balance for a nozzle or diffuser: 28 Deceleration of Air in a Diffuser 29 Acceleration = 2.8 kJ/kg of Steam in a Nozzle Steam 5 kg/s 2 1 1.4 MPa 1.8 MPa, 400C 275 m/s 0.02 m2 30 Turbines and Compressors Turbine drives the electric generator In steam, gas, or hydroelectric power plants. As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work. Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas. A compressor is capable of compressing the gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases. 31 Compressing Air by a Compressor 32 Power Generation by a Steam Turbine 33 Throttling Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid. valves What is the difference between a turbine and a throttling valve? The pressure drop in the fluid is often accompanied by a large drop in temperature, and for that reason throttling devices are commonly used in refrigeration and air-conditioning applications. Energy balance 34 Expansion of Refrigerant-134a in a Refrigerator 35 Mixing chambers In engineering applications, the section where the mixing process takes place is commonly referred to as a mixing chamber. The mixing chamber does not have to be a distinct “chamber.” An ordinary T- elbow or a Y-elbow in a shower, for example, serves as the mixing chamber for the cold- and hot-water streams. The conservation of mass principle for a mixing chamber requires that the sum of the incoming mass flow rates equal the mass flow rate of the outgoing mixture. The conservation of energy equation is analogous to the conservation of mass equation. 36 60C Mixing of Hot and Cold Waters in a Shower 150 kPa 10C 45C 37  Heat exchangers Heat exchangers are devices where two moving fluid streams exchange heat without mixing. Heat exchangers are widely used in various industries, and they come in various designs. The heat transfer associated with a heat exchanger may be zero or nonzero depending on how the control volume is selected. 38 Cooling of Refrigerant-134a by Water 39 40 Pipe and duct flow The transport of liquids or gases in pipes and ducts is of great importance in many engineering applications. Flow through a pipe or a duct usually satisfies the steady-flow conditions. 41 Electric Heating of Air in a House 42 Nozzles and Diffusers    V1 V2  V1    V1 V2  V1 For flow through nozzles, the heat transfer, work, and potential energy are normally neglected, and nozzles have one entrance and one exit. The conservation of energy becomes 43   Solving for V2  2 V2  2(h1  h2 )  V1 Example 5-4 Steam at 0.4 MPa, 300oC, enters an adiabatic nozzle with a low velocity and leaves at 0.2 MPa with a quality of 90%. Find the exit velocity, in m/s. Control Volume: The nozzle Property Relation: Steam tables Process: Assume adiabatic, steady-flow Conservation Principles: Conservation of mass: For one entrance, one exit, the conservation of mass becomes  m in   m out m 1  m 2  m 44  Conservation of energy: According to the sketched control volume, mass crosses the control surface, but no work or heat transfer crosses the control surface. Neglecting the potential energies, we have Neglecting the inlet kinetic energy, the exit velocity is  V2  2(h1  h2 ) Now, we need to find the enthalpies from the steam tables Table A6, Page 908. Superheated  Saturated Mix.  kJ  T1  300 C  h1  3067.1 o P2  0.2 MPa  h2  kg  P1  0.4 MPa  x2  0.90  45 At 0.2 MPa hf = 504.7 kJ/kg and hfg = 2201.6 kJ/kg from table A-5, Page 906. h2 = h f + x2 h fg kJ = 504.7 + (0.90)(2201.6) = 2486.1 kg kJ 1000 m 2 / s 2 V2  2(3067.1  2486.1) kg kJ / kg m  1078.0 s 46 Turbines If we neglect the changes in kinetic and potential energies as fluid flows through an adiabatic turbine having one entrance and one exit, the conservation of mass and the steady-state, steady-flow first law becomes 47  Example 5-5 High pressure air at 1300 K flows into an aircraft gas turbine and undergoes a steady-state, steady-flow, adiabatic process to the turbine exit at 660 K. Calculate the work done per unit mass of air flowing through the turbine when (a) Temperature-dependent data are used. (b) Cp,ave at the average temperature is used. (c) Cp at 300 K is used. Control Volume: The turbine. Property Relation: Assume air is an ideal gas and use ideal gas relations. Process: Steady-state, steady-flow, adiabatic process 48 Conservation Principles:  m   m out Conservation of mass: in m 1  m 2  m Conservation of energy: According to the sketched control volume, mass and work cross the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic, we have 0  m 1h1  Wout  m 2 h2 W  m (h  h ) out 1 2 49 The work done by the air per unit mass flow is Wout wout   h1  h2 m Notice that the work done by a fluid flowing through a turbine is equal to the enthalpy decrease of the fluid. (a) Using the air tables, Table A-17 (ideal gas properties of air), page 925 at T1 = 1300 K, h1 = 1395.97 kJ/kg at T2 = 660 K, h2 = 670.47 kJ/kg wout  h1  h2 kJ  (1395.97  670.47) kg kJ  7255. kg 50 (b) Using Table A-2(b), page 900 by interpolation; at Tave = 980 K, Cp, ave = 1.138 kJ/kgK wout  h1  h2  C p , ave (T1  T2 ) kJ  1138. (1300  660) K kg  K kJ  728.3 kg (c) Using Table A-2(b) page 900, at T = 300 K, Cp = 1.005 kJ/kg K wout  h1  h2  C p (T1  T2 ) kJ  1005. (1300  660) K kg  K kJ  643.2 kg 51 Compressors and fans Compressor blades in a ground station gas turbine Photo courtesy of Progress Energy Carolinas, Inc. 52 Compressors and fans are essentially the same devices. However, compressors operate over larger pressure ratios (P2/P1) than fans. Axial flow compressors are made of several “fan blade” like stages as shown on the pervious slide. If we neglect the changes in kinetic and potential energies as fluid flows through an adiabatic compressor having one entrance and one exit, the steady-state, steady-flow first law or the conservation of energy equation becomes 53 Example 5-6 Nitrogen gas is compressed in a steady-state, steady-flow, adiabatic process from 0.1 MPa, 25oC. During the compression process the temperature becomes 125oC. If the mass flow rate is 0.2 kg/s, determine the work done on the nitrogen, in kW. Control Volume: The compressor (see the compressor sketched above) Property Relation: Assume nitrogen is an ideal gas and use ideal gas relations Process: Adiabatic, steady-flow 54 Conservation Principles:  m   m out Conservation of mass: in m 1  m 2  m Conservation of energy: According to the sketched control volume, mass and work cross the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic, we have for one entrance and one exit 0  m 1 (h1  0  0)  ( Win )  m 2 (h2  0  0) W  m (h  h ) in 2 1 55  The work done on the nitrogen is related to the enthalpy rise of the nitrogen as it flows through the compressor. The work done on the nitrogen per unit mass flow is Win win   h2  h1  m Assuming constant specific heats at 300 K from Table A-2(b) page 900, we write the work as win  C p (T2  T1 ) kJ  1039. (125  25) K kg  K kJ  103.9 kg 56 Throttling devices Consider fluid flowing through a one-entrance, one-exit porous plug. The fluid experiences a pressure drop as it flows through the plug. No net work is done by the fluid. Assume the process is adiabatic and that the kinetic and potential energies are neglected; then the conservation of mass and energy equations become 57 This process is called a throttling process. What happens when an ideal gas is throttled? When throttling an ideal gas, the temperature does not change. We will see later in Chapter 11 that the throttling process is an important process in the refrigeration cycle. A throttling device may be used to determine the enthalpy of saturated steam. The steam is throttled from the pressure in the pipe to ambient pressure in the calorimeter. The pressure drop is sufficient to superheat the steam in the calorimeter. Thus, the temperature and pressure in the calorimeter will specify the enthalpy of the steam in the pipe. 58 Example 5-7 One way to determine the quality of saturated steam is to throttle the steam to a low enough pressure that it exists as a superheated vapor. Saturated steam at 0.4 MPa is throttled to 0.1 MPa, 100oC. Determine the quality of the steam at 0.4 MPa. Throttling orifice 1 2 Control Surface Control Volume: The throttle Property Relation: The steam tables Process: Steady-state, steady-flow, no work, no heat transfer, neglect kinetic and potential energies, one entrance, one exit Conservation Principles: Conservation of mass:  m in   m out m 1  m 2  m 59  Conservation of energy: According to the sketched control volume, mass crosses the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic with no work, we have for one entrance and one exit, From table A-6, page 908 we found h2 0  m 1 (h1  0  0)  0  m 2 (h2  0  0) m 1h1  m 2 h2 h1  h2 T2  100o C  kJ  h2  2675.8 P2  0.1 MPa  kg 60 Therefore, as long as we have to determine the quality , means I have two phases (Vap+Liq). Then I have to use Table A-5, Page 906 for 100 kPa kJ h1  h2  2675.8 kg   h f  x1h fg @ P 0.4 MPa 1 h1  h f x1  h fg 2675.8  604.66  2133.4  0.971 61 Conservation Principles: Conservation of mass: 0(steady) m in  m out  m system (kg / s) For one entrance, one exit, the conservation of mass becomes m in  m out m 1  m 2  m 62 63