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ENERGY ANALYSIS OF
 CLOSED SYSTEMS
                                 Lecture 7

        Keith Vaugh BEng (AERO) MEng



Reference text: Chapter 5 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition
                   Yunus A. Cengel, Robert H. Turner, John M. Cimbala
                                   McGraw-Hill, 2008
                                                                                  KEITH VAUGH
OBJECTIVES
             }

             KEITH VAUGH
OBJECTIVES
Examine the moving boundary work or P dV work
commonly encountered in reciprocating devices
such as automotive engines and compressors.
                                                }

                                                KEITH VAUGH
OBJECTIVES
Examine the moving boundary work or P dV work
commonly encountered in reciprocating devices
such as automotive engines and compressors.
                                                        }
Identify the first law of thermodynamics as simply a
statement of the conservation of energy principle for
closed (fixed mass) systems.




                                                        KEITH VAUGH
OBJECTIVES
Examine the moving boundary work or P dV work
commonly encountered in reciprocating devices
such as automotive engines and compressors.
                                                        }
Identify the first law of thermodynamics as simply a
statement of the conservation of energy principle for
closed (fixed mass) systems.

Develop the general energy balance applied to
closed systems.




                                                        KEITH VAUGH
OBJECTIVES
Examine the moving boundary work or P dV work
commonly encountered in reciprocating devices
such as automotive engines and compressors.
                                                        }
Identify the first law of thermodynamics as simply a
statement of the conservation of energy principle for
closed (fixed mass) systems.

Develop the general energy balance applied to
closed systems.

Define the specific heat at constant volume and the
specific heat at constant pressure.


                                                        KEITH VAUGH
}

KEITH VAUGH
Relate the specific heats to the calculation of the
changes in internal energy and enthalpy of ideal
gases.
                                                     }

                                                     KEITH VAUGH
Relate the specific heats to the calculation of the
changes in internal energy and enthalpy of ideal
gases.
                                                     }
Describe incompressible substances and determine
the changes in their internal energy and enthalpy.




                                                     KEITH VAUGH
Relate the specific heats to the calculation of the
changes in internal energy and enthalpy of ideal
gases.
                                                         }
Describe incompressible substances and determine
the changes in their internal energy and enthalpy.

Solve energy balance problems for closed (fixed
mass) systems that involve heat and work
interactions for general pure substances, ideal gases,
and incompressible substances.




                                                         KEITH VAUGH
MOVING BOUNDARY WORK
                                      Moving boundary work (PdV work)
                                      The work associated with a moving
                                      boundary is called boundary work
                                                                            }
                                      The expansion and compression work in a
                                      piston cylinder device is given by




A gas does a differential amount
of work δWb as it forces the piston
to move by a differential amount
ds.
                                                                                KEITH VAUGH
MOVING BOUNDARY WORK
                                      Moving boundary work (PdV work)
                                      The work associated with a moving
                                      boundary is called boundary work
                                                                            }
                                      The expansion and compression work in a
                                      piston cylinder device is given by

                                        δ Wb = Fds = PAds = PdV




A gas does a differential amount
of work δWb as it forces the piston
to move by a differential amount
ds.
                                                                                KEITH VAUGH
MOVING BOUNDARY WORK
                                      Moving boundary work (PdV work)
                                      The work associated with a moving
                                      boundary is called boundary work
                                                                            }
                                      The expansion and compression work in a
                                      piston cylinder device is given by

                                        δ Wb = Fds = PAds = PdV
                                               2
                                         Wb = ∫ P dV
                                               1
                                                           ( kJ )

A gas does a differential amount
of work δWb as it forces the piston
to move by a differential amount
ds.
                                                                                KEITH VAUGH
MOVING BOUNDARY WORK
                                      Moving boundary work (PdV work)
                                      The work associated with a moving
                                      boundary is called boundary work
                                                                             }
                                      The expansion and compression work in a
                                      piston cylinder device is given by

                                        δ Wb = Fds = PAds = PdV
                                               2
                                         Wb = ∫ P dV
                                               1
                                                           ( kJ )

A gas does a differential amount      Quasi-equilibrium process
of work δWb as it forces the piston
to move by a differential amount
                                      A process during which the system remains
ds.                                   nearly in equilibrium at all times.
                                                                                KEITH VAUGH
KEITH VAUGH
The area under the process curve on
a P-V diagram represents the
boundary work.
               2         2
Area = A = ∫ dA = ∫ P dV
               1        1



                                      KEITH VAUGH
The boundary work
                                      done during a process
                                        depends on the path
                                         followed as well as
                                              the end states.




The area under the process curve on
a P-V diagram represents the
boundary work.
               2         2
Area = A = ∫ dA = ∫ P dV
               1        1



                                                                KEITH VAUGH
The boundary work
                                      done during a process
                                        depends on the path
                                         followed as well as
                                              the end states.




The area under the process curve on
                                          The net work done
a P-V diagram represents the
                                        during a cycle is the
boundary work.
                                      difference between the
               2         2                  work done by the
Area = A = ∫ dA = ∫ P dV                system and the work
               1        1
                                         done on the system.

                                                                KEITH VAUGH
POLYTROPIC, ISOTHERMAL
   & ISOBARIC PROCESSES

P = CV − n   Polytropic process: C, n (polytropic exponent) constants   }

                                                                        KEITH VAUGH
POLYTROPIC, ISOTHERMAL
   & ISOBARIC PROCESSES

P = CV − n   Polytropic process: C, n (polytropic exponent) constants


                                    V2− n+1 − V1− n+1 P2V2 − P1V1
                                                                             }
       2          2                                                 Polytropic
Wb = ∫ P dV = ∫       CV − n dV = C                  =              process
      1         1                        −n + 1          1− n




                                                                                 KEITH VAUGH
POLYTROPIC, ISOTHERMAL
   & ISOBARIC PROCESSES

P = CV − n   Polytropic process: C, n (polytropic exponent) constants


                                    V2− n+1 − V1− n+1 P2V2 − P1V1
                                                                             }
       2          2                                                 Polytropic
Wb = ∫ P dV = ∫       CV − n dV = C                  =              process
      1         1                        −n + 1          1− n

     mR (T2 − T1 )         Polytropic and
Wb =                       for ideal gas
        1− n




                                                                                 KEITH VAUGH
POLYTROPIC, ISOTHERMAL
   & ISOBARIC PROCESSES

P = CV − n    Polytropic process: C, n (polytropic exponent) constants


                                    V2− n+1 − V1− n+1 P2V2 − P1V1
                                                                             }
       2          2                                                 Polytropic
Wb = ∫ P dV = ∫       CV − n dV = C                  =              process
      1          1                       −n + 1          1− n

     mR (T2 − T1 )            Polytropic and
Wb =                          for ideal gas
        1− n

          2           2
                             −1         ⎛ V2 ⎞    When n=1
Wb = ∫ P dV = ∫           CV dV = PV ln ⎜ ⎟
       1             1                  ⎝ V1 ⎠    (Isothermal process)




                                                                                 KEITH VAUGH
POLYTROPIC, ISOTHERMAL
   & ISOBARIC PROCESSES

P = CV − n    Polytropic process: C, n (polytropic exponent) constants


                                    V2− n+1 − V1− n+1 P2V2 − P1V1
                                                                               }
       2          2                                                   Polytropic
Wb = ∫ P dV = ∫       CV − n dV = C                  =                process
      1          1                       −n + 1          1− n

     mR (T2 − T1 )              Polytropic and
Wb =                            for ideal gas
        1− n

          2           2
                               −1       ⎛ V2 ⎞      When n=1
Wb = ∫ P dV = ∫           CV dV = PV ln ⎜ ⎟
       1             1                  ⎝ V1 ⎠      (Isothermal process)

          2                2                       Constant pressure process
Wb = ∫ P dV = P0 ∫ dV = P0 (V2 − V1 )
      1                   1                        (Isobaric process)
                                                                                   KEITH VAUGH
ENERGY BALANCE FOR
    CLOSED SYSTEMS
                 }

                     KEITH VAUGH
ENERGY BALANCE FOR
                         CLOSED SYSTEMS
   Ein − Eout
   14 2 4 3
  Net energy transfer
by heat, work and mass
                         =          ΔEsystem
                                    123
                             Change in internal, kinetic,
                              potential, etc... energies
                                                            ( kJ )   Energy balance for any system
                                                                     undergoing any process          }

                                                                                                     KEITH VAUGH
ENERGY BALANCE FOR
                            CLOSED SYSTEMS
     Ein − Eout
     14 2 4 3
   Net energy transfer
 by heat, work and mass
                          =             ΔEsystem
                                        123
                              Change in internal, kinetic,
                               potential, etc... energies


                                            dEsystem
                                                                ( kJ )      Energy balance for any system
                                                                            undergoing any process          }
     & &
     Ein − Eout               =                                          ( kW )      Energy balance
     14 2 4 3                                 dt                                     in the rate form
Rate of net energy transfer                 123
 by heat, work and mass           Rate of change in internal,
                                  kinetic, potential, etc... energies




                                                                                                            KEITH VAUGH
ENERGY BALANCE FOR
                            CLOSED SYSTEMS
     Ein − Eout
     14 2 4 3
   Net energy transfer
 by heat, work and mass
                          =             ΔEsystem
                                        123
                              Change in internal, kinetic,
                               potential, etc... energies


                                            dEsystem
                                                                ( kJ )      Energy balance for any system
                                                                            undergoing any process            }
     & &
     Ein − Eout               =                                          ( kW )      Energy balance
     14 2 4 3                                 dt                                     in the rate form
Rate of net energy transfer                 123
 by heat, work and mass           Rate of change in internal,
                                  kinetic, potential, etc... energies


                                                                                                  The total quantities are
         &                     &                              ⎛ dE ⎞
     Q = QΔt,             W = W Δt,                  and ΔE = ⎜ ⎟ Δt
                                                              ⎝ dt ⎠
                                                                                         ( kJ )   related to the quantities
                                                                                                  per unit time




                                                                                                                 KEITH VAUGH
ENERGY BALANCE FOR
                            CLOSED SYSTEMS
     Ein − Eout
     14 2 4 3
   Net energy transfer
 by heat, work and mass
                          =             ΔEsystem
                                        123
                              Change in internal, kinetic,
                               potential, etc... energies


                                            dEsystem
                                                                ( kJ )      Energy balance for any system
                                                                            undergoing any process            }
     & &
     Ein − Eout               =                                          ( kW )      Energy balance
     14 2 4 3                                 dt                                     in the rate form
Rate of net energy transfer                 123
 by heat, work and mass           Rate of change in internal,
                                  kinetic, potential, etc... energies


                                                                                                  The total quantities are
         &                     &                              ⎛ dE ⎞
     Q = QΔt,             W = W Δt,                  and ΔE = ⎜ ⎟ Δt
                                                              ⎝ dt ⎠
                                                                                         ( kJ )   related to the quantities
                                                                                                  per unit time

     ein − eout = Δesystem                         ( kg)
                                                     kJ                               Energy balance per
                                                                                      unit mass basis




                                                                                                                 KEITH VAUGH
ENERGY BALANCE FOR
                            CLOSED SYSTEMS
     Ein − Eout
     14 2 4 3
   Net energy transfer
 by heat, work and mass
                          =             ΔEsystem
                                        123
                              Change in internal, kinetic,
                               potential, etc... energies


                                            dEsystem
                                                                ( kJ )      Energy balance for any system
                                                                            undergoing any process            }
     & &
     Ein − Eout               =                                          ( kW )      Energy balance
     14 2 4 3                                 dt                                     in the rate form
Rate of net energy transfer                 123
 by heat, work and mass           Rate of change in internal,
                                  kinetic, potential, etc... energies


                                                                                                  The total quantities are
         &                     &                              ⎛ dE ⎞
     Q = QΔt,             W = W Δt,                  and ΔE = ⎜ ⎟ Δt
                                                              ⎝ dt ⎠
                                                                                         ( kJ )   related to the quantities
                                                                                                  per unit time

     ein − eout = Δesystem                         ( kg)
                                                     kJ                               Energy balance per
                                                                                      unit mass basis

    δ Ein − δ Eout = dEsystem                      or δ ein − δ eout = desystem         Energy balance in
                                                                                        differential form

                                                                                                                 KEITH VAUGH
&           &
Wnet , out = Qnet , in or Wnet , out = Qnet , in   Energy balance
                                                   for a cycle




                                                                    KEITH VAUGH
&           &
  Wnet , out = Qnet , in or Wnet , out = Qnet , in   Energy balance
                                                     for a cycle

Energy balance when sign convention is used (i.e. heat input and
work output are positive; heat output and work input are negative

  Qnet ,in − Wnet ,out = ΔEsystem                             For a cycle ΔE = 0
  or Q − W = ΔE                                               thus Q = W




                                                                                   KEITH VAUGH
&           &
  Wnet , out = Qnet , in or Wnet , out = Qnet , in   Energy balance
                                                     for a cycle

Energy balance when sign convention is used (i.e. heat input and
work output are positive; heat output and work input are negative

  Qnet ,in − Wnet ,out = ΔEsystem                             For a cycle ΔE = 0
  or Q − W = ΔE                                               thus Q = W


  Q = Qnet ,in = Qin − Qout
  W = Wnet ,out = Wout − Win




                                                                                   KEITH VAUGH
&           &
  Wnet , out = Qnet , in or Wnet , out = Qnet , in    Energy balance
                                                      for a cycle

Energy balance when sign convention is used (i.e. heat input and
work output are positive; heat output and work input are negative

  Qnet ,in − Wnet ,out = ΔEsystem                                 For a cycle ΔE = 0
  or Q − W = ΔE                                                   thus Q = W


  Q = Qnet ,in = Qin − Qout
  W = Wnet ,out = Wout − Win




     General                 Q − W = ΔE
     Stationary systems      Q − W = ΔU
     Per unit mass           q − w = Δe          Various forms of the first-law
                                                 relation for closed systems
     Differential form       δ q = δ w = de
                                                 when sign convention is used.
                                                                                       KEITH VAUGH
KEITH VAUGH
The first law cannot be proven mathematically, but no
process in nature is known to have violated the first
law, therefore this can be taken as sufficient proof




                                               KEITH VAUGH
Energy balance for a constant pressure
expansion or compression process
General analysis for a closed system
undergoing a quasi-equilibrium constant-
pressure process Q is to the system and
W is from the system

Example 5-5 page 168
    Ein − Eout            =          ΔEsystem
    14 2 4 3                         123
   Net energy transfer        Change in internal, kinetic,
 by heat, work and mass        potential, etc... energies




                                                             KEITH VAUGH
Energy balance for a constant pressure
expansion or compression process
                                                                                      Z 0            Z 0
General analysis for a closed system                         Q − W = ΔU + Δ KE              + Δ PE
undergoing a quasi-equilibrium constant-                     Q − Wother − Wb = U 2 − U1
pressure process Q is to the system and
                                                             Q − Wother − P0 (V2 − V1 ) = U 2 − U1
W is from the system
                                                             Q − Wother = (U 2 + P2V2 ) − (U1 − P1V1 )
Example 5-5 page 168
                                                             H = U + PV
    Ein − Eout            =          ΔEsystem
    14 2 4 3                         123                     Q − Wother = H 2 − H 1
   Net energy transfer        Change in internal, kinetic,
 by heat, work and mass        potential, etc... energies




                                                                                             KEITH VAUGH
Energy balance for a constant pressure
expansion or compression process
                                                                                       Z 0            Z 0
General analysis for a closed system                          Q − W = ΔU + Δ KE              + Δ PE
undergoing a quasi-equilibrium constant-                      Q − Wother − Wb = U 2 − U1
pressure process Q is to the system and
                                                              Q − Wother − P0 (V2 − V1 ) = U 2 − U1
W is from the system
                                                              Q − Wother = (U 2 + P2V2 ) − (U1 − P1V1 )
Example 5-5 page 168
                                                              H = U + PV
    Ein − Eout            =          ΔEsystem
    14 2 4 3                         123                      Q − Wother = H 2 − H 1
   Net energy transfer        Change in internal, kinetic,
 by heat, work and mass        potential, etc... energies

                                                             For a constant pressure expansion
                                                             or compression process:

                                                                     ΔU + Wb = ΔH




                                                                                              KEITH VAUGH
Energy balance for a constant pressure
expansion or compression process
                                                                                       Z 0            Z 0
General analysis for a closed system                          Q − W = ΔU + Δ KE              + Δ PE
undergoing a quasi-equilibrium constant-                      Q − Wother − Wb = U 2 − U1
pressure process Q is to the system and
                                                              Q − Wother − P0 (V2 − V1 ) = U 2 − U1
W is from the system
                                                              Q − Wother = (U 2 + P2V2 ) − (U1 − P1V1 )
Example 5-5 page 168
                                                              H = U + PV
    Ein − Eout            =          ΔEsystem
    14 2 4 3                         123                      Q − Wother = H 2 − H 1
   Net energy transfer        Change in internal, kinetic,
 by heat, work and mass        potential, etc... energies

                                                             For a constant pressure expansion
                                                             or compression process:

                                                                     ΔU + Wb = ΔH

                                                               We,in − Qout − Wb = ΔU
                                                               We,in − Qout = ΔH = m ( h2 − h1 )
                                                                                              KEITH VAUGH
SPECIFIC HEAT
The energy required to raise the temperature of a unit mass of a
substance by one degree                                            }
          Specific heat is the energy
          required to raise the
          temperature of a unit mass
          of a substance by one
          degree in a specified way.




                                                                   KEITH VAUGH
SPECIFIC HEAT
The energy required to raise the temperature of a unit mass of a
substance by one degree                                            }
          Specific heat is the energy
          required to raise the
          temperature of a unit mass
          of a substance by one
          degree in a specified way.


In thermodynamics we are interested in two kinds of specific heats
 ✓ Specific heat at constant volume, cv
 ✓ Specific heat at constant pressure, cp

                                                                   KEITH VAUGH
Specific heat at constant volume, cv
The energy required to raise the temperature of the unit mass of a substance by
one degree as the volume is maintained constant.

Specific heat at constant pressure, cp
The energy required to raise the temperature of
the unit mass of a substance by one degree as the
pressure is maintained constant.




                                    Constant-volume and
                                constant-pressure specific
                                           heats cv and cp
                              (values are for helium gas).

                                                                          KEITH VAUGH
Formal definitions of cv and cp

           ⎛ ∂u ⎞
      cv = ⎜
           ⎝ ∂T ⎟ v
                 ⎠


            ⎛ ∂h ⎞
      c p = ⎜
            ⎝ ∂T ⎟ p
                  ⎠




                                 KEITH VAUGH
Formal definitions of cv and cp


                         }
           ⎛ ∂u ⎞          the change in internal energy
      cv = ⎜                with temperature at constant
           ⎝ ∂T ⎟ v
                 ⎠                    volume



                         }
            ⎛ ∂h ⎞         the change in internal enthalpy
      c p = ⎜
            ⎝ ∂T ⎟ p
                  ⎠          with temperature at constant
                                        pressure




                                                               KEITH VAUGH
Formal definitions of cv and cp


                          }
            ⎛ ∂u ⎞          the change in internal energy
       cv = ⎜                with temperature at constant
            ⎝ ∂T ⎟ v
                  ⎠                    volume



                          }
             ⎛ ∂h ⎞         the change in internal enthalpy
       c p = ⎜
             ⎝ ∂T ⎟ p
                   ⎠          with temperature at constant
                                         pressure


The equations in the figure are valid for any
substance undergoing any process.

cv and cp are properties.

cv is related to the changes in internal energy
and cp to the changes in enthalpy.

A common unit for specific heats is kJ/kg · °C or                The specific heat of a substance
kJ/kg · K
                                                                     changes with temperature
                                                                                    KEITH VAUGH
INTERNAL ENERGY, ENTHALPY &
SPECIFIC HEATS OD IDEAL GASES

 An Ideal gas can be defined as a gas whose temperature, pressure
 and specific volume are related by
                                                                   }
      PV = RT

 Joule demonstrated mathematically and
 experimentally in 1843, that for an Ideal
 gas the internal energy is a function of
 the temperature only;

        u = u (T )




                                                                   KEITH VAUGH
Using the definition of enthalpy and the equation of
                         state of an ideal gas

                              h = u + PV ⎫
                                         ⎬ h = u + RT
                              PV = RT ⎭

For ideal gases, u, h,        u = u (T ) , h = h (T ) , du = cv (T ) dT ,
cv, and cp vary with
temperature only.
                              dh = c p (T ) dT




                                                                            KEITH VAUGH
Using the definition of enthalpy and the equation of
                         state of an ideal gas

                              h = u + PV ⎫
                                         ⎬ h = u + RT
                              PV = RT ⎭

For ideal gases, u, h,        u = u (T ) , h = h (T ) , du = cv (T ) dT ,
cv, and cp vary with
temperature only.
                              dh = c p (T ) dT

                         Internal energy change of an ideal gas

                                                 2
                              Δu = u2 − u1 = ∫ cv (T ) dT
                                                 1




                                                                            KEITH VAUGH
Using the definition of enthalpy and the equation of
                         state of an ideal gas

                              h = u + PV ⎫
                                         ⎬ h = u + RT
                              PV = RT ⎭

For ideal gases, u, h,        u = u (T ) , h = h (T ) , du = cv (T ) dT ,
cv, and cp vary with
temperature only.
                              dh = c p (T ) dT

                         Internal energy change of an ideal gas

                                                 2
                              Δu = u2 − u1 = ∫ cv (T ) dT
                                                 1



                         Internal energy change of an ideal gas

                                                 2
                              Δh = h2 − h1 = ∫ c p (T ) dT
                                                 1



                                                                            KEITH VAUGH
At low pressures, all real gases
approach ideal-gas behaviour, and
therefore their specific heats depend
on temperature only.

The specific heats of real gases at low
pressures are called ideal-gas specific
heats, or zero-pressure specific heats,
and are often denoted cp0 and cv0.




                         Ideal-gas
                         constant-pressure
                         specific heats for
                         some gases (see
                         Table A–2c for cp
                         equations).
                                             KEITH VAUGH
At low pressures, all real gases             u and h data for a number of
approach ideal-gas behaviour, and            gases have been tabulated.
therefore their specific heats depend
on temperature only.                         These tables are obtained by
                                             choosing an arbitrary reference
The specific heats of real gases at low       point and performing the
pressures are called ideal-gas specific       integrations by treating state 1 as
heats, or zero-pressure specific heats,       the reference state.
and are often denoted cp0 and cv0.




                         Ideal-gas
                         constant-pressure
                         specific heats for
                         some gases (see      In the preparation of ideal-gas
                         Table A–2c for cp    tables, 0 K is chosen as the
                         equations).          reference temperature
                                                                                KEITH VAUGH
Internal energy and enthalpy change
when specific heat is taken constant
at an average value

 u2 − u1 = cv,avg (T2 − T1 )           kJ
                                            kg
  h2 − h1 = c p,avg (T2 − T1 )         kJ
                                            kg




The relation Δ u = cv ΔT is valid for any
kind of process, constant-volume or not.
                                                 KEITH VAUGH
Internal energy and enthalpy change
when specific heat is taken constant
at an average value

 u2 − u1 = cv,avg (T2 − T1 )           kJ
                                            kg
  h2 − h1 = c p,avg (T2 − T1 )         kJ
                                            kg




                                                 For small temperature intervals, the specific
                                                  heats may be assumed to vary linearly with
                                                                                temperature.




The relation Δ u = cv ΔT is valid for any
kind of process, constant-volume or not.
                                                                                   KEITH VAUGH
Three ways of calculating Δu and Δh




                                      KEITH VAUGH
Three ways of calculating Δu and Δh

By using the tabulated u and h data. This is the
easiest and most accurate way when tables are
readily available.




                                                   KEITH VAUGH
Three ways of calculating Δu and Δh

By using the tabulated u and h data. This is the
easiest and most accurate way when tables are
readily available.




                                                   KEITH VAUGH
Three ways of calculating Δu and Δh

By using the tabulated u and h data. This is the
easiest and most accurate way when tables are
readily available.

By using the cv or cp relations (Table A-2c) as a
function of temperature and performing the
integrations. This is very inconvenient for hand
calculations but quite desirable for computerised
calculations. The results obtained are very
accurate.




                                                    KEITH VAUGH
Three ways of calculating Δu and Δh

By using the tabulated u and h data. This is the
easiest and most accurate way when tables are
readily available.

By using the cv or cp relations (Table A-2c) as a
function of temperature and performing the
integrations. This is very inconvenient for hand
calculations but quite desirable for computerised
calculations. The results obtained are very
accurate.




                                                    KEITH VAUGH
Three ways of calculating Δu and Δh

By using the tabulated u and h data. This is the
easiest and most accurate way when tables are
readily available.

By using the cv or cp relations (Table A-2c) as a
function of temperature and performing the
integrations. This is very inconvenient for hand
calculations but quite desirable for computerised
calculations. The results obtained are very
accurate.

By using average specific heats. This is very
simple and certainly very convenient when
property tables are not available. The results
obtained are reasonably accurate if the
temperature interval is not very large.



                                                    KEITH VAUGH
Three ways of calculating Δu and Δh

By using the tabulated u and h data. This is the
easiest and most accurate way when tables are
readily available.

By using the cv or cp relations (Table A-2c) as a
function of temperature and performing the
integrations. This is very inconvenient for hand      Δu = u2 − u1         ( table )
calculations but quite desirable for computerised             2

calculations. The results obtained are very
                                                      Δu =   ∫ c (T ) dT
                                                              1
                                                                  v

accurate.                                             Δu ≅ cv,avg ΔT

By using average specific heats. This is very        Three ways of calculating Δu
simple and certainly very convenient when
property tables are not available. The results
obtained are reasonably accurate if the
temperature interval is not very large.



                                                                         KEITH VAUGH
Specific Heat Relations of Ideal Gases

    h = u + RT
    dh = du + RdT
    dh = c p dT and du = cv dT




                                        KEITH VAUGH
Specific Heat Relations of Ideal Gases
                                        The relationship between cp, cv and R


                                 }
    h = u + RT
    dh = du + RdT                          c p = cv + R    (   kJ
                                                                    kg ⋅ K   )
    dh = c p dT and du = cv dT




                                                                             KEITH VAUGH
Specific Heat Relations of Ideal Gases
                                        The relationship between cp, cv and R


                                 }
    h = u + RT
    dh = du + RdT                          c p = cv + R    (   kJ
                                                                    kg ⋅ K   )
    dh = c p dT and du = cv dT
                                        On a molar basis

                                           c p = cv + Ru       ( kJ kmol ⋅ K )




                                                                             KEITH VAUGH
Specific Heat Relations of Ideal Gases
                                        The relationship between cp, cv and R


                                 }
    h = u + RT
    dh = du + RdT                          c p = cv + R         (   kJ
                                                                         kg ⋅ K   )
    dh = c p dT and du = cv dT
                                        On a molar basis

                                           c p = cv + Ru            ( kJ kmol ⋅ K )
                                              cp          Specific
                                           k=
                                              cv          heat ratio




                                                                                  KEITH VAUGH
Specific Heat Relations of Ideal Gases
                                             The relationship between cp, cv and R


                                   }
     h = u + RT
     dh = du + RdT                              c p = cv + R         (   kJ
                                                                              kg ⋅ K   )
     dh = c p dT and du = cv dT
                                             On a molar basis

                                                c p = cv + Ru            ( kJ kmol ⋅ K )
                                                   cp          Specific
                                                k=
                                                   cv          heat ratio


The specific ratio varies with temperature, but this variation is very mild.

For monatomic gases (helium, argon, etc.), its value is essentially constant at 1.667.

Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room
temperature.
                                                                                       KEITH VAUGH
INTERNAL ENERGY, ENTHALPY &
SPECIFIC HEATS OF SOLIDS & LIQUIDS
           Incompressible substance
           A substance whose specific volume (or density) is constant.
           Solids and liquids are incompressible substances.
                                                                                  }

    The specific volumes of               The cv and cp values of incompressible
    incompressible substances            substances are identical and are
    remain constant during a             denoted by c.
    process.
                                                                            KEITH VAUGH
Internal energy changes
   du = cv dT = c (T ) dT
                     2
   Δu = u2 − u1 = ∫ c (T ) dT
                     1              ( kg)
                                    kJ

   Δu ≅ cv,avg (T2 − T1 )       ( kg)
                                 kJ




                                            KEITH VAUGH
Enthalpy changes
   h = u + PV




                   KEITH VAUGH
Enthalpy changes
   h = u + PV
                          Z   0

   dh = du + VdP + P dV           = du + VdP




                                               KEITH VAUGH
Enthalpy changes
   h = u + PV
                          Z   0

   dh = du + VdP + P dV           = du + VdP

  Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP              ( kg)
                                               kJ




                                                       KEITH VAUGH
Enthalpy changes
   h = u + PV
                          Z   0

   dh = du + VdP + P dV           = du + VdP

   Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP             ( kg)
                                               kJ

For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT
For liquids, two specical cases are commonly encountered
1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT
2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP




                                                                            KEITH VAUGH
Enthalpy changes
   h = u + PV
                              Z   0

   dh = du + VdP + P dV               = du + VdP

   Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP                 ( kg)
                                                    kJ

For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT
For liquids, two specical cases are commonly encountered
1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT
2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP

                                                   The enthalpy of a
   h@P,T ≅ h f @T + V f @T ( P − Psat @T )
                                                   compressed liquid




                                                                            KEITH VAUGH
Enthalpy changes
   h = u + PV
                              Z   0

   dh = du + VdP + P dV               = du + VdP

   Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP                 ( kg)
                                                    kJ

For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT
For liquids, two specical cases are commonly encountered
1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT
2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP

                                                   The enthalpy of a
   h@P,T ≅ h f @T + V f @T ( P − Psat @T )
                                                   compressed liquid


   A more accurate relation than



                                                                            KEITH VAUGH
Enthalpy changes
   h = u + PV
                              Z   0

   dh = du + VdP + P dV               = du + VdP

   Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP                 ( kg)
                                                    kJ

For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT
For liquids, two specical cases are commonly encountered
1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT
2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP

                                                   The enthalpy of a
   h@P,T ≅ h f @T + V f @T ( P − Psat @T )
                                                   compressed liquid


   A more accurate relation than h@P,T ≅ h f @T



                                                                            KEITH VAUGH
Moving boundary work
    Wb for an isothermal process
    Wb for a constant-pressure process
    Wb for a polytropic process
Energy balance for closed systems
    Energy balance for a constant-pressure expansion or compression process
Specific heats
    Constant-pressure specific heat, cp
    Constant-volume specific heat, cv
Internal energy, enthalpy and specific heats of ideal gases
    Energy balance for a constant-pressure expansion or compression process
Internal energy, enthalpy and specific heats of incompressible substances
(solids and liquids)

                                                                               KEITH VAUGH

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SSL7 Energy Analysis of Closed Systems

  • 1. ENERGY ANALYSIS OF CLOSED SYSTEMS Lecture 7 Keith Vaugh BEng (AERO) MEng Reference text: Chapter 5 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008 KEITH VAUGH
  • 2. OBJECTIVES } KEITH VAUGH
  • 3. OBJECTIVES Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. } KEITH VAUGH
  • 4. OBJECTIVES Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. } Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems. KEITH VAUGH
  • 5. OBJECTIVES Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. } Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems. Develop the general energy balance applied to closed systems. KEITH VAUGH
  • 6. OBJECTIVES Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. } Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems. Develop the general energy balance applied to closed systems. Define the specific heat at constant volume and the specific heat at constant pressure. KEITH VAUGH
  • 8. Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases. } KEITH VAUGH
  • 9. Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases. } Describe incompressible substances and determine the changes in their internal energy and enthalpy. KEITH VAUGH
  • 10. Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases. } Describe incompressible substances and determine the changes in their internal energy and enthalpy. Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances. KEITH VAUGH
  • 11. MOVING BOUNDARY WORK Moving boundary work (PdV work) The work associated with a moving boundary is called boundary work } The expansion and compression work in a piston cylinder device is given by A gas does a differential amount of work δWb as it forces the piston to move by a differential amount ds. KEITH VAUGH
  • 12. MOVING BOUNDARY WORK Moving boundary work (PdV work) The work associated with a moving boundary is called boundary work } The expansion and compression work in a piston cylinder device is given by δ Wb = Fds = PAds = PdV A gas does a differential amount of work δWb as it forces the piston to move by a differential amount ds. KEITH VAUGH
  • 13. MOVING BOUNDARY WORK Moving boundary work (PdV work) The work associated with a moving boundary is called boundary work } The expansion and compression work in a piston cylinder device is given by δ Wb = Fds = PAds = PdV 2 Wb = ∫ P dV 1 ( kJ ) A gas does a differential amount of work δWb as it forces the piston to move by a differential amount ds. KEITH VAUGH
  • 14. MOVING BOUNDARY WORK Moving boundary work (PdV work) The work associated with a moving boundary is called boundary work } The expansion and compression work in a piston cylinder device is given by δ Wb = Fds = PAds = PdV 2 Wb = ∫ P dV 1 ( kJ ) A gas does a differential amount Quasi-equilibrium process of work δWb as it forces the piston to move by a differential amount A process during which the system remains ds. nearly in equilibrium at all times. KEITH VAUGH
  • 16. The area under the process curve on a P-V diagram represents the boundary work. 2 2 Area = A = ∫ dA = ∫ P dV 1 1 KEITH VAUGH
  • 17. The boundary work done during a process depends on the path followed as well as the end states. The area under the process curve on a P-V diagram represents the boundary work. 2 2 Area = A = ∫ dA = ∫ P dV 1 1 KEITH VAUGH
  • 18. The boundary work done during a process depends on the path followed as well as the end states. The area under the process curve on The net work done a P-V diagram represents the during a cycle is the boundary work. difference between the 2 2 work done by the Area = A = ∫ dA = ∫ P dV system and the work 1 1 done on the system. KEITH VAUGH
  • 19. POLYTROPIC, ISOTHERMAL & ISOBARIC PROCESSES P = CV − n Polytropic process: C, n (polytropic exponent) constants } KEITH VAUGH
  • 20. POLYTROPIC, ISOTHERMAL & ISOBARIC PROCESSES P = CV − n Polytropic process: C, n (polytropic exponent) constants V2− n+1 − V1− n+1 P2V2 − P1V1 } 2 2 Polytropic Wb = ∫ P dV = ∫ CV − n dV = C = process 1 1 −n + 1 1− n KEITH VAUGH
  • 21. POLYTROPIC, ISOTHERMAL & ISOBARIC PROCESSES P = CV − n Polytropic process: C, n (polytropic exponent) constants V2− n+1 − V1− n+1 P2V2 − P1V1 } 2 2 Polytropic Wb = ∫ P dV = ∫ CV − n dV = C = process 1 1 −n + 1 1− n mR (T2 − T1 ) Polytropic and Wb = for ideal gas 1− n KEITH VAUGH
  • 22. POLYTROPIC, ISOTHERMAL & ISOBARIC PROCESSES P = CV − n Polytropic process: C, n (polytropic exponent) constants V2− n+1 − V1− n+1 P2V2 − P1V1 } 2 2 Polytropic Wb = ∫ P dV = ∫ CV − n dV = C = process 1 1 −n + 1 1− n mR (T2 − T1 ) Polytropic and Wb = for ideal gas 1− n 2 2 −1 ⎛ V2 ⎞ When n=1 Wb = ∫ P dV = ∫ CV dV = PV ln ⎜ ⎟ 1 1 ⎝ V1 ⎠ (Isothermal process) KEITH VAUGH
  • 23. POLYTROPIC, ISOTHERMAL & ISOBARIC PROCESSES P = CV − n Polytropic process: C, n (polytropic exponent) constants V2− n+1 − V1− n+1 P2V2 − P1V1 } 2 2 Polytropic Wb = ∫ P dV = ∫ CV − n dV = C = process 1 1 −n + 1 1− n mR (T2 − T1 ) Polytropic and Wb = for ideal gas 1− n 2 2 −1 ⎛ V2 ⎞ When n=1 Wb = ∫ P dV = ∫ CV dV = PV ln ⎜ ⎟ 1 1 ⎝ V1 ⎠ (Isothermal process) 2 2 Constant pressure process Wb = ∫ P dV = P0 ∫ dV = P0 (V2 − V1 ) 1 1 (Isobaric process) KEITH VAUGH
  • 24. ENERGY BALANCE FOR CLOSED SYSTEMS } KEITH VAUGH
  • 25. ENERGY BALANCE FOR CLOSED SYSTEMS Ein − Eout 14 2 4 3 Net energy transfer by heat, work and mass = ΔEsystem 123 Change in internal, kinetic, potential, etc... energies ( kJ ) Energy balance for any system undergoing any process } KEITH VAUGH
  • 26. ENERGY BALANCE FOR CLOSED SYSTEMS Ein − Eout 14 2 4 3 Net energy transfer by heat, work and mass = ΔEsystem 123 Change in internal, kinetic, potential, etc... energies dEsystem ( kJ ) Energy balance for any system undergoing any process } & & Ein − Eout = ( kW ) Energy balance 14 2 4 3 dt in the rate form Rate of net energy transfer 123 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies KEITH VAUGH
  • 27. ENERGY BALANCE FOR CLOSED SYSTEMS Ein − Eout 14 2 4 3 Net energy transfer by heat, work and mass = ΔEsystem 123 Change in internal, kinetic, potential, etc... energies dEsystem ( kJ ) Energy balance for any system undergoing any process } & & Ein − Eout = ( kW ) Energy balance 14 2 4 3 dt in the rate form Rate of net energy transfer 123 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies The total quantities are & & ⎛ dE ⎞ Q = QΔt, W = W Δt, and ΔE = ⎜ ⎟ Δt ⎝ dt ⎠ ( kJ ) related to the quantities per unit time KEITH VAUGH
  • 28. ENERGY BALANCE FOR CLOSED SYSTEMS Ein − Eout 14 2 4 3 Net energy transfer by heat, work and mass = ΔEsystem 123 Change in internal, kinetic, potential, etc... energies dEsystem ( kJ ) Energy balance for any system undergoing any process } & & Ein − Eout = ( kW ) Energy balance 14 2 4 3 dt in the rate form Rate of net energy transfer 123 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies The total quantities are & & ⎛ dE ⎞ Q = QΔt, W = W Δt, and ΔE = ⎜ ⎟ Δt ⎝ dt ⎠ ( kJ ) related to the quantities per unit time ein − eout = Δesystem ( kg) kJ Energy balance per unit mass basis KEITH VAUGH
  • 29. ENERGY BALANCE FOR CLOSED SYSTEMS Ein − Eout 14 2 4 3 Net energy transfer by heat, work and mass = ΔEsystem 123 Change in internal, kinetic, potential, etc... energies dEsystem ( kJ ) Energy balance for any system undergoing any process } & & Ein − Eout = ( kW ) Energy balance 14 2 4 3 dt in the rate form Rate of net energy transfer 123 by heat, work and mass Rate of change in internal, kinetic, potential, etc... energies The total quantities are & & ⎛ dE ⎞ Q = QΔt, W = W Δt, and ΔE = ⎜ ⎟ Δt ⎝ dt ⎠ ( kJ ) related to the quantities per unit time ein − eout = Δesystem ( kg) kJ Energy balance per unit mass basis δ Ein − δ Eout = dEsystem or δ ein − δ eout = desystem Energy balance in differential form KEITH VAUGH
  • 30. & & Wnet , out = Qnet , in or Wnet , out = Qnet , in Energy balance for a cycle KEITH VAUGH
  • 31. & & Wnet , out = Qnet , in or Wnet , out = Qnet , in Energy balance for a cycle Energy balance when sign convention is used (i.e. heat input and work output are positive; heat output and work input are negative Qnet ,in − Wnet ,out = ΔEsystem For a cycle ΔE = 0 or Q − W = ΔE thus Q = W KEITH VAUGH
  • 32. & & Wnet , out = Qnet , in or Wnet , out = Qnet , in Energy balance for a cycle Energy balance when sign convention is used (i.e. heat input and work output are positive; heat output and work input are negative Qnet ,in − Wnet ,out = ΔEsystem For a cycle ΔE = 0 or Q − W = ΔE thus Q = W Q = Qnet ,in = Qin − Qout W = Wnet ,out = Wout − Win KEITH VAUGH
  • 33. & & Wnet , out = Qnet , in or Wnet , out = Qnet , in Energy balance for a cycle Energy balance when sign convention is used (i.e. heat input and work output are positive; heat output and work input are negative Qnet ,in − Wnet ,out = ΔEsystem For a cycle ΔE = 0 or Q − W = ΔE thus Q = W Q = Qnet ,in = Qin − Qout W = Wnet ,out = Wout − Win General Q − W = ΔE Stationary systems Q − W = ΔU Per unit mass q − w = Δe Various forms of the first-law relation for closed systems Differential form δ q = δ w = de when sign convention is used. KEITH VAUGH
  • 35. The first law cannot be proven mathematically, but no process in nature is known to have violated the first law, therefore this can be taken as sufficient proof KEITH VAUGH
  • 36. Energy balance for a constant pressure expansion or compression process General analysis for a closed system undergoing a quasi-equilibrium constant- pressure process Q is to the system and W is from the system Example 5-5 page 168 Ein − Eout = ΔEsystem 14 2 4 3 123 Net energy transfer Change in internal, kinetic, by heat, work and mass potential, etc... energies KEITH VAUGH
  • 37. Energy balance for a constant pressure expansion or compression process Z 0 Z 0 General analysis for a closed system Q − W = ΔU + Δ KE + Δ PE undergoing a quasi-equilibrium constant- Q − Wother − Wb = U 2 − U1 pressure process Q is to the system and Q − Wother − P0 (V2 − V1 ) = U 2 − U1 W is from the system Q − Wother = (U 2 + P2V2 ) − (U1 − P1V1 ) Example 5-5 page 168 H = U + PV Ein − Eout = ΔEsystem 14 2 4 3 123 Q − Wother = H 2 − H 1 Net energy transfer Change in internal, kinetic, by heat, work and mass potential, etc... energies KEITH VAUGH
  • 38. Energy balance for a constant pressure expansion or compression process Z 0 Z 0 General analysis for a closed system Q − W = ΔU + Δ KE + Δ PE undergoing a quasi-equilibrium constant- Q − Wother − Wb = U 2 − U1 pressure process Q is to the system and Q − Wother − P0 (V2 − V1 ) = U 2 − U1 W is from the system Q − Wother = (U 2 + P2V2 ) − (U1 − P1V1 ) Example 5-5 page 168 H = U + PV Ein − Eout = ΔEsystem 14 2 4 3 123 Q − Wother = H 2 − H 1 Net energy transfer Change in internal, kinetic, by heat, work and mass potential, etc... energies For a constant pressure expansion or compression process: ΔU + Wb = ΔH KEITH VAUGH
  • 39. Energy balance for a constant pressure expansion or compression process Z 0 Z 0 General analysis for a closed system Q − W = ΔU + Δ KE + Δ PE undergoing a quasi-equilibrium constant- Q − Wother − Wb = U 2 − U1 pressure process Q is to the system and Q − Wother − P0 (V2 − V1 ) = U 2 − U1 W is from the system Q − Wother = (U 2 + P2V2 ) − (U1 − P1V1 ) Example 5-5 page 168 H = U + PV Ein − Eout = ΔEsystem 14 2 4 3 123 Q − Wother = H 2 − H 1 Net energy transfer Change in internal, kinetic, by heat, work and mass potential, etc... energies For a constant pressure expansion or compression process: ΔU + Wb = ΔH We,in − Qout − Wb = ΔU We,in − Qout = ΔH = m ( h2 − h1 ) KEITH VAUGH
  • 40. SPECIFIC HEAT The energy required to raise the temperature of a unit mass of a substance by one degree } Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way. KEITH VAUGH
  • 41. SPECIFIC HEAT The energy required to raise the temperature of a unit mass of a substance by one degree } Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way. In thermodynamics we are interested in two kinds of specific heats ✓ Specific heat at constant volume, cv ✓ Specific heat at constant pressure, cp KEITH VAUGH
  • 42. Specific heat at constant volume, cv The energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. Specific heat at constant pressure, cp The energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant. Constant-volume and constant-pressure specific heats cv and cp (values are for helium gas). KEITH VAUGH
  • 43. Formal definitions of cv and cp ⎛ ∂u ⎞ cv = ⎜ ⎝ ∂T ⎟ v ⎠ ⎛ ∂h ⎞ c p = ⎜ ⎝ ∂T ⎟ p ⎠ KEITH VAUGH
  • 44. Formal definitions of cv and cp } ⎛ ∂u ⎞ the change in internal energy cv = ⎜ with temperature at constant ⎝ ∂T ⎟ v ⎠ volume } ⎛ ∂h ⎞ the change in internal enthalpy c p = ⎜ ⎝ ∂T ⎟ p ⎠ with temperature at constant pressure KEITH VAUGH
  • 45. Formal definitions of cv and cp } ⎛ ∂u ⎞ the change in internal energy cv = ⎜ with temperature at constant ⎝ ∂T ⎟ v ⎠ volume } ⎛ ∂h ⎞ the change in internal enthalpy c p = ⎜ ⎝ ∂T ⎟ p ⎠ with temperature at constant pressure The equations in the figure are valid for any substance undergoing any process. cv and cp are properties. cv is related to the changes in internal energy and cp to the changes in enthalpy. A common unit for specific heats is kJ/kg · °C or The specific heat of a substance kJ/kg · K changes with temperature KEITH VAUGH
  • 46. INTERNAL ENERGY, ENTHALPY & SPECIFIC HEATS OD IDEAL GASES An Ideal gas can be defined as a gas whose temperature, pressure and specific volume are related by } PV = RT Joule demonstrated mathematically and experimentally in 1843, that for an Ideal gas the internal energy is a function of the temperature only; u = u (T ) KEITH VAUGH
  • 47. Using the definition of enthalpy and the equation of state of an ideal gas h = u + PV ⎫ ⎬ h = u + RT PV = RT ⎭ For ideal gases, u, h, u = u (T ) , h = h (T ) , du = cv (T ) dT , cv, and cp vary with temperature only. dh = c p (T ) dT KEITH VAUGH
  • 48. Using the definition of enthalpy and the equation of state of an ideal gas h = u + PV ⎫ ⎬ h = u + RT PV = RT ⎭ For ideal gases, u, h, u = u (T ) , h = h (T ) , du = cv (T ) dT , cv, and cp vary with temperature only. dh = c p (T ) dT Internal energy change of an ideal gas 2 Δu = u2 − u1 = ∫ cv (T ) dT 1 KEITH VAUGH
  • 49. Using the definition of enthalpy and the equation of state of an ideal gas h = u + PV ⎫ ⎬ h = u + RT PV = RT ⎭ For ideal gases, u, h, u = u (T ) , h = h (T ) , du = cv (T ) dT , cv, and cp vary with temperature only. dh = c p (T ) dT Internal energy change of an ideal gas 2 Δu = u2 − u1 = ∫ cv (T ) dT 1 Internal energy change of an ideal gas 2 Δh = h2 − h1 = ∫ c p (T ) dT 1 KEITH VAUGH
  • 50. At low pressures, all real gases approach ideal-gas behaviour, and therefore their specific heats depend on temperature only. The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted cp0 and cv0. Ideal-gas constant-pressure specific heats for some gases (see Table A–2c for cp equations). KEITH VAUGH
  • 51. At low pressures, all real gases u and h data for a number of approach ideal-gas behaviour, and gases have been tabulated. therefore their specific heats depend on temperature only. These tables are obtained by choosing an arbitrary reference The specific heats of real gases at low point and performing the pressures are called ideal-gas specific integrations by treating state 1 as heats, or zero-pressure specific heats, the reference state. and are often denoted cp0 and cv0. Ideal-gas constant-pressure specific heats for some gases (see In the preparation of ideal-gas Table A–2c for cp tables, 0 K is chosen as the equations). reference temperature KEITH VAUGH
  • 52. Internal energy and enthalpy change when specific heat is taken constant at an average value u2 − u1 = cv,avg (T2 − T1 ) kJ kg h2 − h1 = c p,avg (T2 − T1 ) kJ kg The relation Δ u = cv ΔT is valid for any kind of process, constant-volume or not. KEITH VAUGH
  • 53. Internal energy and enthalpy change when specific heat is taken constant at an average value u2 − u1 = cv,avg (T2 − T1 ) kJ kg h2 − h1 = c p,avg (T2 − T1 ) kJ kg For small temperature intervals, the specific heats may be assumed to vary linearly with temperature. The relation Δ u = cv ΔT is valid for any kind of process, constant-volume or not. KEITH VAUGH
  • 54. Three ways of calculating Δu and Δh KEITH VAUGH
  • 55. Three ways of calculating Δu and Δh By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. KEITH VAUGH
  • 56. Three ways of calculating Δu and Δh By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. KEITH VAUGH
  • 57. Three ways of calculating Δu and Δh By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerised calculations. The results obtained are very accurate. KEITH VAUGH
  • 58. Three ways of calculating Δu and Δh By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerised calculations. The results obtained are very accurate. KEITH VAUGH
  • 59. Three ways of calculating Δu and Δh By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerised calculations. The results obtained are very accurate. By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large. KEITH VAUGH
  • 60. Three ways of calculating Δu and Δh By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand Δu = u2 − u1 ( table ) calculations but quite desirable for computerised 2 calculations. The results obtained are very Δu = ∫ c (T ) dT 1 v accurate. Δu ≅ cv,avg ΔT By using average specific heats. This is very Three ways of calculating Δu simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large. KEITH VAUGH
  • 61. Specific Heat Relations of Ideal Gases h = u + RT dh = du + RdT dh = c p dT and du = cv dT KEITH VAUGH
  • 62. Specific Heat Relations of Ideal Gases The relationship between cp, cv and R } h = u + RT dh = du + RdT c p = cv + R ( kJ kg ⋅ K ) dh = c p dT and du = cv dT KEITH VAUGH
  • 63. Specific Heat Relations of Ideal Gases The relationship between cp, cv and R } h = u + RT dh = du + RdT c p = cv + R ( kJ kg ⋅ K ) dh = c p dT and du = cv dT On a molar basis c p = cv + Ru ( kJ kmol ⋅ K ) KEITH VAUGH
  • 64. Specific Heat Relations of Ideal Gases The relationship between cp, cv and R } h = u + RT dh = du + RdT c p = cv + R ( kJ kg ⋅ K ) dh = c p dT and du = cv dT On a molar basis c p = cv + Ru ( kJ kmol ⋅ K ) cp Specific k= cv heat ratio KEITH VAUGH
  • 65. Specific Heat Relations of Ideal Gases The relationship between cp, cv and R } h = u + RT dh = du + RdT c p = cv + R ( kJ kg ⋅ K ) dh = c p dT and du = cv dT On a molar basis c p = cv + Ru ( kJ kmol ⋅ K ) cp Specific k= cv heat ratio The specific ratio varies with temperature, but this variation is very mild. For monatomic gases (helium, argon, etc.), its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature. KEITH VAUGH
  • 66. INTERNAL ENERGY, ENTHALPY & SPECIFIC HEATS OF SOLIDS & LIQUIDS Incompressible substance A substance whose specific volume (or density) is constant. Solids and liquids are incompressible substances. } The specific volumes of The cv and cp values of incompressible incompressible substances substances are identical and are remain constant during a denoted by c. process. KEITH VAUGH
  • 67. Internal energy changes du = cv dT = c (T ) dT 2 Δu = u2 − u1 = ∫ c (T ) dT 1 ( kg) kJ Δu ≅ cv,avg (T2 − T1 ) ( kg) kJ KEITH VAUGH
  • 68. Enthalpy changes h = u + PV KEITH VAUGH
  • 69. Enthalpy changes h = u + PV Z 0 dh = du + VdP + P dV = du + VdP KEITH VAUGH
  • 70. Enthalpy changes h = u + PV Z 0 dh = du + VdP + P dV = du + VdP Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP ( kg) kJ KEITH VAUGH
  • 71. Enthalpy changes h = u + PV Z 0 dh = du + VdP + P dV = du + VdP Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP ( kg) kJ For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT For liquids, two specical cases are commonly encountered 1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT 2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP KEITH VAUGH
  • 72. Enthalpy changes h = u + PV Z 0 dh = du + VdP + P dV = du + VdP Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP ( kg) kJ For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT For liquids, two specical cases are commonly encountered 1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT 2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP The enthalpy of a h@P,T ≅ h f @T + V f @T ( P − Psat @T ) compressed liquid KEITH VAUGH
  • 73. Enthalpy changes h = u + PV Z 0 dh = du + VdP + P dV = du + VdP Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP ( kg) kJ For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT For liquids, two specical cases are commonly encountered 1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT 2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP The enthalpy of a h@P,T ≅ h f @T + V f @T ( P − Psat @T ) compressed liquid A more accurate relation than KEITH VAUGH
  • 74. Enthalpy changes h = u + PV Z 0 dh = du + VdP + P dV = du + VdP Δh = Δu + V ΔP ≅ cavg ΔT + V ΔP ( kg) kJ For solids, the term V ΔP is insignificant thus, Δh = Δu ≅ cavg ΔT For liquids, two specical cases are commonly encountered 1. Constant pressure process, as in heaters ( Δp = 0 ) : h = Δu ≅ cavg ΔT 2. Constant temperature processes as in pumps ( ΔT ≅ 0 ); Δh = V ΔP The enthalpy of a h@P,T ≅ h f @T + V f @T ( P − Psat @T ) compressed liquid A more accurate relation than h@P,T ≅ h f @T KEITH VAUGH
  • 75. Moving boundary work  Wb for an isothermal process  Wb for a constant-pressure process  Wb for a polytropic process Energy balance for closed systems  Energy balance for a constant-pressure expansion or compression process Specific heats  Constant-pressure specific heat, cp  Constant-volume specific heat, cv Internal energy, enthalpy and specific heats of ideal gases  Energy balance for a constant-pressure expansion or compression process Internal energy, enthalpy and specific heats of incompressible substances (solids and liquids) KEITH VAUGH

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  48. Refer to Isobaric process (constant pressure) on slide 6.\n
  49. Refer to Isobaric process (constant pressure) on slide 6.\n
  50. Refer to Isobaric process (constant pressure) on slide 6.\n
  51. Refer to Isobaric process (constant pressure) on slide 6.\n
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  54. True or False, cp is always greater then cv\n
  55. True or False, cp is always greater then cv\n
  56. True or False, cp is always greater then cv\n
  57. True or False, cp is always greater then cv\n
  58. True or False, cp is always greater then cv\n
  59. True or False, cp is always greater then cv\n
  60. Refer to Joule experimental arrangement on page 174 of the text\n
  61. Refer to Joule experimental arrangement on page 174 of the text\n
  62. Refer to Joule experimental arrangement on page 174 of the text\n
  63. Refer to Joule experimental arrangement on page 174 of the text\n
  64. Refer to Joule experimental arrangement on page 174 of the text\n
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