MASS & ENERGYANALYSIS OF CONTROL     SYSTEMS                                  Lecture 8         Keith Vaugh BEng (AERO) ME...
OBJECTIVES             }             KEITH VAUGH
OBJECTIVESDevelop the conservation of mass principle.                                              }                      ...
OBJECTIVESDevelop the conservation of mass principle.Apply the conservation of mass principle to varioussystems including ...
OBJECTIVESDevelop the conservation of mass principle.Apply the conservation of mass principle to varioussystems including ...
OBJECTIVESDevelop the conservation of mass principle.Apply the conservation of mass principle to varioussystems including ...
}KEITH VAUGH
Solve energy balance problems for common steady-flowdevices such as nozzles, compressors, turbines, throttlingvalves, mixer...
Solve energy balance problems for common steady-flowdevices such as nozzles, compressors, turbines, throttlingvalves, mixer...
CONSERVATION OF MASS   Conservation of mass                                                           }   Mass, like energ...
Mass is conservedeven during chemicalreactions                KEITH VAUGH
Mass is conserved                                                        even during chemical                             ...
MASS & VOLUME FLOW RATES  &δ m = ρVn dAc  &       &  m = ∫ δ m = ∫ ρVn dAc         Ac                Ac                   ...
MASS & VOLUME FLOW RATES  &δ m = ρVn dAc  &       &  m = ∫ δ m = ∫ ρVn dAc         Ac                Ac                   ...
MASS & VOLUME FLOW RATES  &δ m = ρVn dAc  &       &  m = ∫ δ m = ∫ ρVn dAc         Ac                Ac                   ...
MASS & VOLUME FLOW RATES  &δ m = ρVn dAc  &       &  m = ∫ δ m = ∫ ρVn dAc         Ac                Ac                   ...
MASS & VOLUME FLOW RATES  &δ m = ρVn dAc  &       &  m = ∫ δ m = ∫ ρVn dAc         Ac                Ac                   ...
CONSERVATION OF MASS                    PRINCIPLE    The conservation of mass principle for a control volume    The net ma...
min − mout = Δmcv   ( kg )                                                              dmcv                              ...
min − mout = Δmcv               ( kg )                                                              dmcv                  ...
min − mout = Δmcv                    ( kg )                                                              dmcv             ...
MASS BALANCE FOR STEADY        FLOW PROCESSES During a steady-flow process, the total amount of mass contained within a con...
For steady-flow processes, we areinterested in the amount of massflowing per unit time, that is, the massflow rate &in      &...
Special case: Incompressible FlowThe conservation of mass relations can be simplified even further when thefluid is incompre...
FLOW WORK & THE ENERGY     OF A FLOWING FLUIDFlow work, or flow energyThe work (or energy) required to push the mass into o...
FLOW WORK & THE ENERGY     OF A FLOWING FLUIDFlow work, or flow energyThe work (or energy) required to push the mass into o...
F = PA                         W flow = FL = PAL = PV   ( kJ )                         W flow = Pv    ( kg)               ...
TOTAL ENERGY OF A                    FLOWING FLUIDe = u + ke + pe = u +                      V2                      2    ...
TOTAL ENERGY OF A                    FLOWING FLUIDe = u + ke + pe = u +                      V2                      2    ...
TOTAL ENERGY OF A                    FLOWING FLUIDe = u + ke + pe = u +                      V2                      2    ...
The total energy consists of three parts for a non-flowing fluid and four parts for aflowing fluid.                           ...
ENERGY TRANSPORT BY                    MASSAmount of energy transport        Emass                         ⎛             ...
When the kinetic and potential                                     energies of a fluid stream are                          ...
ENERGY TRANSPORT BY              MASS         Many engineering systems         such as power plants operate         under ...
Under steady-flow conditions, the mass     Under steady-flow conditions, the fluidand energy contents of a control volume    ...
Mass and energy balances for a steady-flow process       &    &      ∑m = ∑m       in     out         & &        m1 = m2   ...
Mass and energy balances for a steady-flow process         &    &        ∑m = ∑m          in          out          & &     ...
Mass and energy balances for a steady-flow process            &    &           ∑m = ∑m            in           out         ...
Mass and energy balances for a steady-flow process            &    &           ∑m = ∑m            in           out         ...
Energy balance relations with sign conventions (i.e., heat input and workoutput are positive)                    ⎛      &...
STEADY FLOW    ENGINEERING DEVICESMany engineering devices operate essentially under the sameconditions for long periods o...
Nozzles and DiffusersCommonly utilised in jet engines, rockets,spacecraft, and even garden hoses.A nozzle is a device that...
Turbines and compressorsTurbine drives the electric generator in powerplants. As the fluid passes through the turbine,work ...
Throttling valvesThrottling valves are any kind of flow-restricting devices that cause a significantpressure drop in the flui...
Mixing chambersIn engineering applications, the section            60°Cwhere the mixing process takes place iscommonly ref...
Heat exchangers  Heat exchangers are devices where two  moving fluid streams exchange heat without  mixing. Heat exchangers...
Mass and energy balances for theadiabatic heat exchanger in thefigure are:      & &      Ein = Eout       & & &       m1 = ...
Pipe and duct flowThe transport of liquids or gases in pipesand ducts is of great importance in manyengineering application...
Energy balance for the pipe flow   shown in the figure is        & &        Ein = Eout         &       &      &      &      ...
ENERGY ANALYSIS OFUNSTEADY FLOW PROCESSES   Many processes of interest, however, involve   changes within the control volu...
Charging of a rigid tank from a supplyline is an unsteady-flow process sinceit involves changes within the controlvolume.  ...
Charging of a rigid tank from a supplyline is an unsteady-flow process sinceit involves changes within the controlvolume.  ...
Charging of a rigid tank from a supply   The shape and size of a control volumeline is an unsteady-flow process since      ...
Mass balance   & &           &   min − mout = Δmsystem    &   Δmsystem = m final − minitial                               ...
Mass balance    & &           &    min − mout = Δmsystem     &    Δmsystem = m final − minitial                           ...
Mass balance    & &           &    min − mout = Δmsystem     &    Δmsystem = m final − minitial                           ...
Mass balance    & &           &    min − mout = Δmsystem     &    Δmsystem = m final − minitial                           ...
The energy equation of a uniform-flow systemreduces to that of a closed system when allthe inlets and exits are closed.    ...
The energy equation of a uniform-flow systemreduces to that of a closed system when allthe inlets and exits are closed.    ...
The energy equation of a uniform-flow system       A uniform flow system may involvereduces to that of a closed system when ...
Conservation of mass    Mass and volume flow rates    Mass balance for a steady flow process    Mass balance for incompre...
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SSL8 Mass & Energy Analysis of Control Systems

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Conservation of mass
Mass and volume flow rates
Mass balance for a steady flow process
Mass balance for incompressible flow
Flow work and the energy of a flowing fluid
Energy transport by mass
Energy analysis of steady flow systems
Steady flow engineering devices
Nozzles and diffusers
Turbines and compressors
Throttling valves
Mixing chambers and heat exchangers
Pipe and duct flow
Energy analysis of unsteady flow processes

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  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
  • note the use of symbols here. DON’T confuse Volume (V) and velocity (V)\n
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  • What is the difference between a turbine and a throttling valve?\n\n
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  • SSL8 Mass & Energy Analysis of Control Systems

    1. 1. MASS & ENERGYANALYSIS OF CONTROL SYSTEMS Lecture 8 Keith Vaugh BEng (AERO) MEng Reference text: Chapter 6 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008 KEITH VAUGH
    2. 2. OBJECTIVES } KEITH VAUGH
    3. 3. OBJECTIVESDevelop the conservation of mass principle. } KEITH VAUGH
    4. 4. OBJECTIVESDevelop the conservation of mass principle.Apply the conservation of mass principle to varioussystems including steady- and unsteady-flow control }volumes. KEITH VAUGH
    5. 5. OBJECTIVESDevelop the conservation of mass principle.Apply the conservation of mass principle to varioussystems including steady- and unsteady-flow control }volumes.Apply the first law of thermodynamics as the statement ofthe conservation of energy principle to control volumes. KEITH VAUGH
    6. 6. OBJECTIVESDevelop the conservation of mass principle.Apply the conservation of mass principle to varioussystems including steady- and unsteady-flow control }volumes.Apply the first law of thermodynamics as the statement ofthe conservation of energy principle to control volumes.Identify the energy carried by a fluid stream crossing acontrol surface as the sum of internal energy, flow work,kinetic energy, and potential energy of the fluid and torelate the combination of the internal energy and the flowwork to the property enthalpy. KEITH VAUGH
    7. 7. }KEITH VAUGH
    8. 8. Solve energy balance problems for common steady-flowdevices such as nozzles, compressors, turbines, throttlingvalves, mixers, heaters, and heat exchangers. } KEITH VAUGH
    9. 9. Solve energy balance problems for common steady-flowdevices such as nozzles, compressors, turbines, throttlingvalves, mixers, heaters, and heat exchangers. }Apply the energy balance to general unsteady-flowprocesses with particular emphasis on the uniform-flowprocess as the model for commonly encountered chargingand discharging processes. KEITH VAUGH
    10. 10. 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. KEITH VAUGH
    11. 11. Mass is conservedeven during chemicalreactions KEITH VAUGH
    12. 12. Mass is conserved even during chemical reactionsMass m and energy E can be converted to each other according to E = mc 2where 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 absolutely negligible. KEITH VAUGH
    13. 13. MASS & VOLUME FLOW RATES &δ m = ρVn dAc & & m = ∫ δ m = ∫ ρVn dAc Ac Ac } 1 Definition ofVavg = ∫ Vn dAc Ac Ac average velocity The average velocity Vavg is defined as the average speed through a cross section. KEITH VAUGH
    14. 14. MASS & VOLUME FLOW RATES &δ m = ρVn dAc & & m = ∫ δ m = ∫ ρVn dAc Ac Ac } 1 Definition ofVavg = ∫ Vn dAc Ac Ac average velocity The average velocity Vavg is defined as the average speed through a cross section. & m = ρVavg Ac ( s) kg KEITH VAUGH
    15. 15. MASS & VOLUME FLOW RATES &δ m = ρVn dAc & & m = ∫ δ m = ∫ ρVn dAc Ac Ac } 1 Definition ofVavg = ∫ Vn dAc Ac Ac average velocity The average velocity Vavg is defined as the average speed through a cross section. & m = ρVavg Ac ( s) kg & V & & m = ρV = Mass flow rate v KEITH VAUGH
    16. 16. MASS & VOLUME FLOW RATES &δ m = ρVn dAc & & m = ∫ δ m = ∫ ρVn dAc Ac Ac } 1 Definition ofVavg = ∫ Vn dAc Ac Ac average velocity The average velocity Vavg is defined as the average speed through a cross section. & m = ρVavg Ac ( s) kg & V & & m = ρV = Mass flow rate v Volume flow rate & V = ∫ Vn dAc = Vavg Ac = VAc Ac ( s) m3 KEITH VAUGH
    17. 17. MASS & VOLUME FLOW RATES &δ m = ρVn dAc & & m = ∫ δ m = ∫ ρVn dAc Ac Ac } 1 Definition ofVavg = ∫ Vn dAc Ac Ac average velocity The average velocity Vavg is defined as the average speed through a cross section. & m = ρVavg Ac ( s) kg & V & & m = ρV = Mass flow rate v Volume flow rate & V = ∫ Vn dAc = Vavg Ac = VAc Ac ( s) m3 The volume flow rate is the volume of fluid flowing through a cross section per unit time. KEITH VAUGH
    18. 18. CONSERVATION OF MASS PRINCIPLE The conservation of mass principle for a control volume The net mass transfer to or from a control volume during a time interval Δt is equal to the net change (increase or decrease) in } the total mass within the control volume during Δt.⎛ Total mass entering ⎞ ⎛ Total mass leaving ⎞ ⎛ Net change in mass ⎞⎜ the CV during Δt ⎟ − ⎜ the CV during Δt ⎟ = ⎜ within the CV during Δt ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ KEITH VAUGH
    19. 19. min − mout = Δmcv ( kg ) dmcv & & min − mout = dt ( s) kgConservation of mass principle for an ordinarybathtub KEITH VAUGH
    20. 20. min − mout = Δmcv ( kg ) dmcv & & min − mout = dt ( s) kg General conservation of mass d r r dt ∫ cv ρ dV + ∫ cs ( ) ρ V ⋅ n dA = 0Conservation of mass principle for an ordinarybathtub KEITH VAUGH
    21. 21. min − mout = Δmcv ( kg ) dmcv & & min − mout = dt ( s) kg General conservation of mass d r r dt ∫ cv ρ dV + ∫ cs ( ) ρ V ⋅ n dA = 0 General conservation of mass in rate form d ∫ & & ρ dV = ∑ m − ∑ m dt cv in outConservation of mass principle for an ordinarybathtub dmcv or & = ∑m − ∑m & dt in out KEITH VAUGH
    22. 22. MASS BALANCE FOR STEADY FLOW PROCESSES During a steady-flow process, the total amount of mass contained within a control volume does not change with time (mCV = constant). } Then the conservation of mass principle requires that the total amount of mass entering a control volume equal the total amount of mass leaving it. KEITH VAUGH
    23. 23. For steady-flow processes, we areinterested in the amount of massflowing per unit time, that is, the massflow rate &in &∑m = ∑m out ( s) kg Multiple inlets and exits& &m1 = m2 → ρV1 A1 = ρV2 A2 Single streamMany engineering devices such asnozzles, diffusers, turbines,compressors, and pumps involve asingle stream (only one inlet and one Conservation of mass principle for a twooutlet). inlet - one outlet steady flow system KEITH VAUGH
    24. 24. Special case: Incompressible FlowThe conservation of mass relations can be simplified even further when thefluid is incompressible, which is usually the case for liquids. & & ∑V = ∑V in out ( s) m3 & & V1 = V2 → V1 A1 = V2 A2 There is no such thing as a “conservation of volume” principle. However, for steady flow of liquids, the volume flow rates, as well as the mass flow rates, remain constant since liquids are essentiallyDuring a steady-flow process, incompressible substances.volume flow rates are notnecessarily conservedalthough mass flow rates are. KEITH VAUGH
    25. 25. FLOW WORK & THE ENERGY OF A FLOWING FLUIDFlow work, or flow energyThe work (or energy) required to push the mass into or out of thecontrol volume. This work is necessary for maintaining acontinuous flow through a control volume. }In the absence of acceleration, the force applied on a fluid by a piston is equal to the force appliedon the piston by the fluid. KEITH VAUGH
    26. 26. FLOW WORK & THE ENERGY OF A FLOWING FLUIDFlow work, or flow energyThe work (or energy) required to push the mass into or out of thecontrol volume. This work is necessary for maintaining acontinuous flow through a control volume. }In the absence of acceleration, the force applied on a fluid by a piston is equal to the force appliedon the piston by the fluid. KEITH VAUGH
    27. 27. F = PA W flow = FL = PAL = PV ( kJ ) W flow = Pv ( kg) kJSchematic for flow work KEITH VAUGH
    28. 28. TOTAL ENERGY OF A FLOWING FLUIDe = u + ke + pe = u + V2 2 + gz ( kg) kJ }θ = Pv + e = Pv + ( u + ke + pe) KEITH VAUGH
    29. 29. TOTAL ENERGY OF A FLOWING FLUIDe = u + ke + pe = u + V2 2 + gz ( kg) kJ }θ = Pv + e = Pv + ( u + ke + pe)But the combination Pv + u has been previously defined as theenthalpy h KEITH VAUGH
    30. 30. TOTAL ENERGY OF A FLOWING FLUIDe = u + ke + pe = u + V2 2 + gz ( kg) kJ }θ = Pv + e = Pv + ( u + ke + pe)But the combination Pv + u has been previously defined as theenthalpy h The flow energy is automatically V2θ = h + ke + pe = h + 2 + gz ( kg) kJ taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy. KEITH VAUGH
    31. 31. The total energy consists of three parts for a non-flowing fluid and four parts for aflowing fluid. KEITH VAUGH
    32. 32. ENERGY TRANSPORT BY MASSAmount of energy transport Emass ⎛ = mθ = m ⎜ h + ⎝ V2 2 ⎞ + gz ⎟ ⎠ ( kJ ) }Rate of energy transport & ⎛ V2 ⎞ Emass & & = mθ = m ⎜ h + + gz ⎟ ( kW ) ⎝ 2 ⎠ The product ṁiθi is the energy transported into control volume by mass per unit time KEITH VAUGH
    33. 33. When the kinetic and potential energies of a fluid stream are negligible Emass = mh & & Emass = mhThe product ṁiθi is the energy When the properties of the mass attransported into control volume by each inlet or exit change with timemass per unit time. as well as over the cross section ⎛ Vi 2 ⎞ Ein,mass = ∫ θ i δ mi = ∫ ⎜ hi + + gzi ⎟ δ mi mi mi ⎝ 2 ⎠ KEITH VAUGH
    34. 34. ENERGY TRANSPORT BY MASS Many engineering systems such as power plants operate under steady conditions. } KEITH VAUGH
    35. 35. Under steady-flow conditions, the mass Under steady-flow conditions, the fluidand energy contents of a control volume properties at an inlet or exit remainremain constant. constant (do not change with time). KEITH VAUGH
    36. 36. Mass and energy balances for a steady-flow process & & ∑m = ∑m in out & & m1 = m2 Mass ρ1V1 A1 = ρ2V2 A2 balance A water heater in steady operation. KEITH VAUGH
    37. 37. Mass and energy balances for a steady-flow process & & ∑m = ∑m in out & & m1 = m2 Mass ρ1V1 A1 = ρ2V2 A2 balanceEnergy balances Z 0 dEsystem & & Ein − Eout = =0 14 2 4 3 dt 3 Rate of net energy transfer 14 2 4 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies A water heater in steady operation. KEITH VAUGH
    38. 38. Mass and energy balances for a steady-flow process & & ∑m = ∑m in out & & m1 = m2 Mass ρ1V1 A1 = ρ2V2 A2 balanceEnergy balances Z 0 dEsystem & & Ein − Eout = =0 14 2 4 3 dt 3 Rate of net energy transfer 14 2 4 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies A water heater in steady operation. & Ein = & Eout ( kW ) { { Rate of net energy transfer Rate of net energy out, by heat, work and mass by heat, work and mass KEITH VAUGH
    39. 39. Mass and energy balances for a steady-flow process & & ∑m = ∑m in out & & m1 = m2 Mass ρ1V1 A1 = ρ2V2 A2 balanceEnergy balances Z 0 dEsystem & & Ein − Eout = =0 14 2 4 3 dt 3 Rate of net energy transfer 14 2 4 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies A water heater in steady operation. & Ein = & Eout ( kW ) { { Rate of net energy transfer Rate of net energy out, by heat, work and mass by heat, work and mass ⎛ & + Win + ∑ m h + & V2 ⎞ & ⎛ & + ∑m h + V2 ⎞ Qin &⎜ + gz ⎟ = Qout + Wout &⎜ + gz ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ 1 4 44 2 4 4 4 in 3 1 4 44 2 4 4 4 out 3 for each inlet for each exit KEITH VAUGH
    40. 40. Energy balance relations with sign conventions (i.e., heat input and workoutput are positive) ⎛ &− W = + ∑ m h + & V2 ⎞ ⎛ V2 ⎞ Q &⎜ & + gz ⎟ − ∑ m ⎜ h + + gz ⎟ ⎝ 2 ⎠ in ⎝ 2 ⎠ 1 4 44 2 4 4 4 out 3 1 4 44 2 4 4 4 3 for each exit for each inlet ⎡ &− W = m h2 − h1 + & & V22 − V12 ⎤ Q ⎢ + g ( z2 − z1 ) ⎥ ⎣ 2 ⎦ V22 − V12 q − w = h2 − h1 + + g ( z2 − z1 ) 2 q − w = h2 − h1 Q& q= W& W& w= & m Refer to pages, 212 - 214 KEITH VAUGH
    41. 41. STEADY FLOW ENGINEERING DEVICESMany engineering devices operate essentially under the sameconditions for long periods of time. The components of a steampower plant (turbines, compressors, heat exchangers, and }pumps), for example, operate nonstop for months before thesystem is shut down for maintenance. Therefore, these devicescan be conveniently analysed 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. KEITH VAUGH
    42. 42. Nozzles and DiffusersCommonly utilised in jet engines, rockets,spacecraft, and even garden hoses.A nozzle is a device that increases thevelocity of a fluid at the expense of pressure.A diffuser is a device that increases thepressure of a fluid by slowing it down.The cross-sectional area of a nozzle decreasesin the flow direction for subsonic flows andincreases for supersonic flows. The reverse is Nozzles and diffusers are shaped sotrue for diffusers. that they cause large changes in fluid velocities and thus kinetic energies. & & Ein = Eout ⎛ V12 ⎞ ⎛ V22 ⎞ & & & & m ⎜ h1 + ⎟ = m ⎜ h2 + ⎟ Since Q ≅ 0, W ≅ 0, and Δpe ≅ 0 ⎝ 2 ⎠ ⎝ 2 ⎠ Refer to Example 6-4 & 6-5, page 216/7 KEITH VAUGH
    43. 43. Turbines and compressorsTurbine drives the electric generator in powerplants. As the fluid passes through the turbine,work is done against the blades, which areattached to the shaft. The shaft rotates, and theturbine produces work.Compressors, as well as pumps and fans,increase the pressure of a fluid. Work issupplied to these devices from an externalsource through a rotating shaft. Energy balance for the compressor in this figure:A fan increases the pressure of a gas slightlyand is mainly used to mobilise a gas. & & Ein = Eout & & & & Win + mh1 = Qout + mh2A compressor is capable of compressing the gasto very high pressures. Since Δke = Δpe ≅ 0Pumps work very much like compressors Refer to Example 6-6 & 6-7, page 218/9except that they handle liquids instead of gases. KEITH VAUGH
    44. 44. Throttling valvesThrottling valves are any kind of flow-restricting devices that cause a significantpressure drop in the fluid.The pressure drop in the fluid is oftenaccompanied by a large drop intemperature and for that reason throttling The temperature of an ideal gas does not change during a throttling (h = constant)devices are commonly used in refrigeration process since h = h(T).and air-conditioning applications. h2 ≅ h1 Energy balance u1 + P1v1 = u2 + P2 v2Internal energy + Flow Energy = Constant Refer to Example 6-8, page 221 During a throttling process, the enthalpy of a fluid remains constant. But internal and flow energies may be converted to each other. KEITH VAUGH
    45. 45. Mixing chambersIn engineering applications, the section 60°Cwhere the mixing process takes place iscommonly referred to as a mixingchamber. 140 kPa 10°C 43°C Energy balance for the adiabatic mixing chamber in the figure & & Ein = Eout & & & m1h1 + m2 h2 = m3h3 The T-elbow of an ordinary Since shower serves as the mixing & & Q ≅ 0, W ≅ 0, Δke ≅ 0 and Δpe ≅ 0 chamber for the hot and the cold-water streams. Refer to Example 6-9, page 223 KEITH VAUGH
    46. 46. Heat exchangers Heat exchangers are devices where two moving fluid streams exchange heat without mixing. Heat exchangers are widely used in various industries and come in a variety of designs. A heat exchanger can be as simple as two concentric pipes The heat transfer associated with a heatexchanger may be zero or nonzero depending on how the control volume is selected. KEITH VAUGH
    47. 47. Mass and energy balances for theadiabatic heat exchanger in thefigure are: & & Ein = Eout & & & m1 = m2 = mW & & & m3 = m4 = m R & & & & m1h1 + m3h3 = m2 h2 + m4 h4 Refer to Example 6-10, page 224 KEITH VAUGH
    48. 48. Pipe and duct flowThe transport of liquids or gases in pipesand ducts is of great importance in manyengineering applications. Flow through apipe or a duct usually satisfies the steady-flow conditions.Heat losses from a hot fluid flowing through Pipe or duct flow may involve morean uninsulated pipe or duct to the cooler than one form of work at the same time.environment may be very significant. KEITH VAUGH
    49. 49. Energy balance for the pipe flow shown in the figure is & & Ein = Eout & & & & We, in + m1h1 = Qout + m2 h2 & & & We, in − Qout = m1c p (T2 − T1 )Refer to Example 6-11, page 227 KEITH VAUGH
    50. 50. ENERGY ANALYSIS OFUNSTEADY FLOW PROCESSES Many processes of interest, however, involve changes within the control volume with time. Such processes are called unsteady-flow, or transient-flow, processes. } Most unsteady-flow processes can be represented reasonably well by the uniform-flow process. Uniform-flow process The fluid flow at any inlet or exit is uniform and steady, and thus the fluid properties do not change with time or position over the cross section of an inlet or exit. If they do, they are averaged and treated as constants for the entire process. KEITH VAUGH
    51. 51. Charging of a rigid tank from a supplyline is an unsteady-flow process sinceit involves changes within the controlvolume. KEITH VAUGH
    52. 52. Charging of a rigid tank from a supplyline is an unsteady-flow process sinceit involves changes within the controlvolume. KEITH VAUGH
    53. 53. Charging of a rigid tank from a supply The shape and size of a control volumeline is an unsteady-flow process since may change during an unsteady-flowit involves changes within the control process.volume. KEITH VAUGH
    54. 54. Mass balance & & & min − mout = Δmsystem & Δmsystem = m final − minitial { i = inlet e = exit mi − me = ( m2 − m1 )CV 1 = initial state 2 = final state KEITH VAUGH
    55. 55. Mass balance & & & min − mout = Δmsystem & Δmsystem = m final − minitial { i = inlet e = exit mi − me = ( m2 − m1 )CV 1 = initial state 2 = final stateEnergy balance Ein − Eout = Esystem 14 2 4 3 { Net energy transfer Change in internal, by heat, work and mass kinetic, potential, etc... energies KEITH VAUGH
    56. 56. Mass balance & & & min − mout = Δmsystem & Δmsystem = m final − minitial { i = inlet e = exit mi − me = ( m2 − m1 )CV 1 = initial state 2 = final stateEnergy balance Ein − Eout = Esystem 14 2 4 3 { Net energy transfer Change in internal, by heat, work and mass kinetic, potential, etc... energies ⎛ ⎞ ⎛ ⎞ ⎜ Qin + Win + ∑ mθ ⎟ − ⎜ Qout + Wout + ∑ mθ ⎟ = ( m2 e2 − m1e1 )system ⎝ ⎠ ⎝ ⎠ in out KEITH VAUGH
    57. 57. Mass balance & & & min − mout = Δmsystem & Δmsystem = m final − minitial { i = inlet e = exit mi − me = ( m2 − m1 )CV 1 = initial state 2 = final stateEnergy balance Ein − Eout = Esystem 14 2 4 3 { Net energy transfer Change in internal, by heat, work and mass kinetic, potential, etc... energies ⎛ ⎞ ⎛ ⎞ ⎜ Qin + Win + ∑ mθ ⎟ − ⎜ Qout + Wout + ∑ mθ ⎟ = ( m2 e2 − m1e1 )system ⎝ ⎠ ⎝ ⎠ in out θ = h + ke + pe e = u + ke + pe KEITH VAUGH
    58. 58. The energy equation of a uniform-flow systemreduces to that of a closed system when allthe inlets and exits are closed. KEITH VAUGH
    59. 59. The energy equation of a uniform-flow systemreduces to that of a closed system when allthe inlets and exits are closed. KEITH VAUGH
    60. 60. The energy equation of a uniform-flow system A uniform flow system may involvereduces to that of a closed system when all electrical, shaft and boundary work allthe inlets and exits are closed. at once KEITH VAUGH
    61. 61. Conservation of mass  Mass and volume flow rates  Mass balance for a steady flow process  Mass balance for incompressible flowFlow work and the energy of a flowing fluid  Energy transport by massEnergy analysis of steady flow systemsSteady flow engineering devices  Nozzles and diffusers  Turbines and compressors  Throttling valves  Mixing chambers and heat exchangers  Pipe and duct flowEnergy analysis of unsteady flow processes KEITH VAUGH

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