Principle of Energy Conservation

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Principle of Energy Conservation
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• 4.1 Energy, Work and Heat
• 5.1 1st Law for Control Mass
• 5.2 Specific Heats for Ideal Gas
• 5.3 1st Law for Control Volume
• 5.4 Steady-State, Steady-Flow Process
• 5.5 Uniform-State, Uniform-Flow Process

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4.1 Energy, Work and Heat

• Energy, Work and Heat
• Two major concepts of thermodynamics are energy
• and entropy. They are both abstract concepts in
• classical thermodynamics, which are derived from
• the 1st and 2nd law of thermodynamics.

Energy, Work and Heat
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Energy, Work and Heat
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• Energy
• Macroscopic view
• 1st Law
• Energy can be transferred in two different forms,
• heat and work. Energy is an abstract concept
  given
• as a thermodynamic property, defined in terms of
• the 1st law of thermodynamics.
• Microscopic view
• Energy can be stored in many different forms
• intermolecular potential energy, molecular
  kinetic
• energy and intermolecular energy including
  rotational
• and vibrational energy.

Energy, Work and Heat
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• Work and Heat
• There are only two forms of energy transfer, work
  and
• heat.
• Work a form of energy transfer due to an
  externally
• applied force and resulting displacement of the
  system
• F can be any generalized force and x is the
• corresponding generalized displacement.
• Conventionally work is defined as positive if
  energy is
• transferred from the system to the surrounding.

Energy, Work and Heat
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• Heat a form of energy transfer from a higher
  temperature
• to a lower temperature. Conventionally heat is
  defined as
• positive if energy is transferred to the system
  from the
• surrounding.
• The system contains energy, not heat. Heat is
  identified
• at the boundary, as energy transfer occurs across
  the
• boundary due to difference in T.
• ,

Energy, Work and Heat
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• Units for work are same as those for energy.
• 1 J 1 Nm
• Power
• 1 W 1 J/s
• 1 hp 550 lbf ft/s
• Specific work work per unit mass of the system

Energy, Work and Heat
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• Units of heat are same as those for work.
• 1 btu 1.055056 kJ
• 1 cal 4.1868 J
• Specific heat transfer heat transfer per unit
  mass of
• the system
• Adiabatic process
• Q0

Energy, Work and Heat
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• Example of work crossing the boundary of a system

Energy, Work and Heat
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• Example of work crossing the boundary of a system
• because of a flow of an electric current across
  the
• system boundary

Energy, Work and Heat
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Energy, Work and Heat
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• Schematic arrangement showing
• work done on a surface film

Energy, Work and Heat
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• Schematic arrangement fir magnetic cooling

Energy, Work and Heat
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• Work done at the moving boundary in a quasi-
• equilibrium process
• Quasi-equilibrium process internally reversible
  process.
• Infinitesimal change to maintain equilibrium
  throughout
• the process.
• Non-equilibrium process the intermediate states
  can
• not be defined. The force exerted by the pressure
  of
• gas does not equal the external force.

Energy, Work and Heat
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• Use of pressure-volume diagram to show work done
• at the moving boundary of a system in a quasi -
• equilibrium process

Energy, Work and Heat
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• Work done depends not only on the two end
• states but also on the path followed.
• W
  path function
• 
  inexact differential
• Various quasi-equilibrium processes between two
• given states, indicating that work is a path
  function.

Energy, Work and Heat
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Energy, Work and Heat
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point function
Energy, Work and Heat
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• Point Functions
• thermodynamic properties
• e.g. V, U, H, ?
• The change in volume depends only on the two end
• states.
• dV exact differential

Energy, Work and Heat
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• The relationship between P and V is required for
• integration.
• Polytropic Process
• where n1(isothermal process),

Energy, Work and Heat
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• Other Modes of Work
• ? ? ??
• There is only one mode of heat, while there can
  be
• many different modes of work.

Energy, Work and Heat
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• Non-equilibrium Process No Work is involved.
• Example of a process involving a change
• of volume for which work is zero

Energy, Work and Heat
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• Heat and Work
• (1) Both are transient phenomena.
• They cross the boundary as the system undergoes a
  
• change in its state.
• (2) Both are boundary phenomena.
• They represent energy crossing the boundary of
  the
• system.
• (3) Both are path functions and inexact
  differentials.
• Thermodynamic properties are state or point
  functions,
• which do not depend on the previous path or
  history.

Energy, Work and Heat
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• Identification of works involved is an important
• part of many thermodynamic problems.
• Example showing how
• selection of the system
• determines whether work
• is involved in a process

Energy, Work and Heat
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• An example showing the difference
• between heat and work

Energy, Work and Heat
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• Example 4.1
• Consider as a system the gas in the cylinder. The
  cylinder
• is fitted with a piston on which a number of
  small weights
• are placed. The initial pressure is 200kPa, and
  the initial
• volume of the gas is 0.4m3.
• Let a Bunsen burner be placed under the cylinder,
• and let the volume of the gas increase to
  0.1m3
• while the pressure remains constant.
  Calculate
• the work done by the system during this
  process.

Energy, Work and Heat
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• 2. Consider the same system and initial
  conditions,
• but at same time the Bunsen burner is under
  the
• cylinder and the piston is rising, let
  weights be
• removed from the piston at such a rate that,
  during
• the process, the temperature of the gas
  remains
• constant.
• n1

Energy, Work and Heat
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• 3. Consider the same system, but during the heat
• transfer let the weights be removed at such a
  rate
• that the expression
  describes the
• relation between pressure and volume during
  the
• process. Again the final volume is 0.1m3.
• Calculate the work.

Energy, Work and Heat
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• 4. Consider the system and initial state in the
  first
• three examples, but let the piston be held by
  a pin
• so that the volume remains constant. In
  addition,
• let heat be transferred from the system until
  the
• pressure drops to 100kPa. Calculate the work.
• The work is zero, because there is no change
  in
• volume.

Energy, Work and Heat
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• Pressure-volume diagram

Energy, Work and Heat
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• Example 4.2
• Consider a slightly different piston cylinder
  arrangement
• as shown in Fig.4.8. In this example the piston
  is loaded
• with a mass, mp, the outside atmosphere P0, a
  linear
• spring and a single point force F1. The piston is
  restricted
• in its motion by lower and upper stops trapping
  the gas
• with a pressure P. A force balance on the piston
  in the
• direction of motion yields
• with a zero acceleration in a quasi-equilibrium
  process.
• The forces, when the piston between the stops,
  are

Energy, Work and Heat
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• with the linear spring constant, ks. The piston
  position
• for a relaxed spring is x0, which depends on how
  the
• spring is installed. The force balance then gives
  the
• gas pressure by division with the area, A as
• To illustrate the process in a P-V diagram, the
  distance
• x is converted to volume by multiplication with
  A

Energy, Work and Heat
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• This relation gives the pressure as a linear
  function of
• the volume, with the line having a slope of
  C2ks/A2.
• With the stops installed, the minimum and maximum
  
• volumes limit the possible states of the system
  to the
• Combination of P and V as shown in Fig.4.9.
  Regardless
• of what substance is inside, any process must
  proceed
• along the lines in the P-V diagram. The work term
  in a
• quasi-equilibrium process then follows as

Energy, Work and Heat
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• with and , subject to the
  constraint
• that
• These limits show that only the part of the
  process
• that follows the sloped line, when the piston
  moves,
• contributes to the work. Any part of the process
  with
• a pressure smaller than Pmin or larger than Pmax
  does
• not involve work as the piston is held in fixed
  piston
• by one of the stops. The maximum work then
  arises,
• when the piston travels the total distance
  between
• the two stops.

Energy, Work and Heat
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• The process curve showing possible P-V combination

Energy, Work and Heat
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5.1 1st Law for Control Mass

• The 1st Law of Thermodynamics
• (Conservation of Energy)
• Example of a control mass undergoing a cycle

1st Law for Control Mass
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• Control Mass Undergoing a Process
• 
  E all energy of
• 
  the system
• Energy Conservation
• The net change of the energy is always equal to
  the
• net transfer of energy across the boundary as
  heat
• and work.

dE
1st Law for Control Mass
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• Internal Energy
• E U KE PE
• U internal energy
• E represents all the energy of the system in the
  given
• state.
• Kinetic and potential energy(KE and PE) depend on
  
• the coordinate frame that we choose to select.
  They
• can be specified by the macroscopic properties of
• mass, velocity and elevation.
• Internal energy(U) is a thermodynamic property,
  which
• may be uniquely given from the thermodynamic
  state
• of the system.

1st Law for Control Mass
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• No information about the absolute magnitude of
  the
• internal energy.
• U extensive property
• u intensive property

1st Law for Control Mass
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• Example 5.2
• Consider a stone having a mass of 10kg and a
  bucket
• containing 100kg of liquid water. Initially the
  stone is
• 10.2m above the water, and the stone and the
  water
• are at the same temperature, state 1. The stone
  then
• falls into the water.
• Determine ?U, ?KE, ?PE, Q, and W for the
  following
• changes of state, assuming standard gravitational
• acceleration of 9.80665m/s2
• a. The stone is about to enter the water,
  state 2.
• b. The stone has just come to rest in the
  bucket,
• state 3.

1st Law for Control Mass
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• c. Heat has been transferred to the
  surroundings
• in such an amount that the stone and
  water
• are at the temperature, , state 4.
• 1. The stone has fallen from to , and we
  assume no heat transfer as it falls.
• For the process from state 1 to state 2,
• and

1st Law for Control Mass
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• 2. For the process from state 2 to state 3 with
  zero
• kinetic energy
• 3. In final state,
• 
  

1st Law for Control Mass
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• Example 5.3
• A vessel having a volume of 5m3, contains 0.05m3
  of
• saturated liquid water and 4.95m3 of saturated
  water
• vapor at 0.1MPa. Heat is transferred until the
  vessel is
• filled with saturated vapor. Determine the heat
  transfer
• for this process.
• Control mass All the water inside the
  vessel.
• Initial state Pressure, volume of liquid,
  volume of
• vapor State 1 is fixed.
• Final state Somewhere along the saturated-
• vapor curve the water
  was heated,
• so P2gtP1.

1st Law for Control Mass
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• process constant volume and mass, therefore,
  
• constant specific volume.

1st Law for Control Mass
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• Then

1st Law for Control Mass
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• Problem Analysis
• Define the system. Control mass or control
  volume?
• (2) Find out the initial and final states.
• (3) Description of the process.
• (4) Diagram of the process helps.
• (5) How to obtain thermodynamic properties.
• (6) Analysis of the problem.
• 1st or 2nd laws? Relevant work modes?
• (7) Solution technique.
• Numerical solution? Trial-and-error?

1st Law for Control Mass
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• Enthalpy
• H U PV
• h u Pv (hH/m)
• Enthalpy is also a thermodynamic property, which
  may
• be defined as either an extensive or an intensive
• property. (H or h)
• In a constant pressure process

1st Law for Control Mass
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• therefore
• The combination, UPV, appears often in control
• volume analysis of the 1st law of thermodynamics.

1st Law for Control Mass
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• Example 5.4
• A cylinder fitted with a piston has a volume of
  0.1m3
• and contains 0.5kg of steam at 0.4MPa. Heat is
• transferred to the steam until the temperature is
  300?C,
• while the pressure remains constant.
• Determine the heat transfer and the work for this
  process.
• Control mass Water inside cylinder.
• Initial state P1, V1, m therefore v1 is known,
  state 1 is
• fixed(at P1, v1, check steam table - two-phase
  region).
• Final state P2, T2 therefore state 2 is
  fixed(superheated).
• Process Constant pressure.
• Model Steam tables.

1st Law for Control Mass
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1st Law for Control Mass
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• No change in kinetic energy, no change in
  potential
• energy. Work is done by movement at the boundary.
• Assume the process to be quasi-equilibrium. Since
  the
• pressure is constant,
• Therefore , the first law is

1st Law for Control Mass
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1st Law for Control Mass
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5.2 Specific Heats for Ideal Gas

• Specific Heats
• Specific heat at constant volume
• Specific heat at constant pressure
• The specific heats are thermodynamic properties.
• (A function in terms of thermodynamic properties
  is
• itself a thermodynamic property.)
• For solid and liquid,

Specific Heats for Ideal Gas
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• Internal Energy, Enthalpy and Specific Heats of
• Ideal Gas
• For ideal gas the internal energy and enthalpy
  are the
• functions of temperature only.
• Ideal gas specific heats are also called as zero-
• pressure specific heats, , .
• from

Specific Heats for Ideal Gas
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• The energy in a molecule may be stored in several
• forms. Translational and rotational energies
  increase
• linearly with temperature. Vibrational energy is
• temperature dependent.
• monoatomic Ar, Ne, He (f3)
• diatomic Air, O2, H2 (f6)
• polyatomic CO2, H2O (f9)

Specific Heats for Ideal Gas
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• Specific Heats
• 
  Constant-pressure
• 
  specific heats for a
• 
  number of gases at
• zero
  pressure

Specific Heats for Ideal Gas
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• Enthalpy from Specific Heats of Ideal Gas
• - constant
• less accurate except for monatomic gases
• - as an analytical function of T
• close empirical approximation
• - Gas Table
• most accurate

Specific Heats for Ideal Gas
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• Example 5.6
• Calculate the change of enthalpy as 1kg of oxygen
  is
• heated from 300 to 1500K. Assume ideal-gas
  behavior.
• Our most accurate answer for the ideal-gas
  enthalpy
• change for oxygen between 300 and 1500K would be
• from the ideal-gas tables, Table A.13.

Specific Heats for Ideal Gas
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• The empirical equation from Table A.11 should
  give
• a good approximation to this result.

Specific Heats for Ideal Gas
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• If we assume constant specific heat, we must be
• concerned about what value we are going to use.
  If
• we use the value at 300K from Table A.10,
• On the other hand, suppose we assume that the
  specific
• heat is constant at its value at 900K, the
  average
• temperature. Substituting 900K into the equation
  for
• specific heat from Table A.11,

Specific Heats for Ideal Gas
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• Example 5.7
• A cylinder fitted with a piston has an initial
  volume of
• 0.1m3 and contains nitrogen at 150kPa, 25?C. The
• piston is moved, compressing the nitrogen until
  the
• pressure is 1MPa and the temperature is 150?C.
  During
• this compression process heat is transferred from
  the
• nitrogen, and the work done on the nitrogen is
  20kJ.
• Determine the amount of this heat transfer.

Specific Heats for Ideal Gas
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Control mass Nitrogen. Initial state P1, T1,
V1 state 1 fixed. Final state P2, T2 state 2
fixed. Process work input known. Model Ideal
gas, constant specific heat at 300K, Table A.10

• First law

Specific Heats for Ideal Gas
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• Example 5.8
• During the charging of a storage battery, the
  current
• is 20A and the voltage is 12.8V. The rate of heat
  
• transfer from the battery is 10W. At what rate is
  the
• internal energy increasing?

Specific Heats for Ideal Gas
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5.3 1st Law for Control Volume

• Control Volume Formulation Mass

1st Law for Control Volume
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• Control Volume Formulation Energy

1st Law for Control Volume
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• Control Volume Formulation
• (??)

1st Law for Control Volume
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5.4 Steady-State, Steady-Flow Process

• Steady State, Steady Flow(SSSF) Process
• CV at rest.
• SS everywhere in the CV.
• SS for flows across the CS.
• SS for heat and work.
• where ,

Steady-State, Steady-Flow Process
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• Example 5.10
• The mass rate of flow into a steam turbine is
  1.5kg/s,
• and the heat transfer from the turbine is 8.5kW.
  The
• following data are known for the steam entering
  and
• leaving the turbine.

Steady-State, Steady-Flow Process
70

• Determine the power output of the turbine.
• Control volume Turbine.
• Inlet state Fixed.
• Exit state Fixed.
• Process SSSF.
• Model Steam tables.
• First law
• with

Steady-State, Steady-Flow Process
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• hi3137.0kJ/kg
• he2675.5kJ/kg

Steady-State, Steady-Flow Process
72

• Example 5.11
• Steam at 0.6MPa, 200?C enters an insulated nozzle
  
• with a velocity of 50m/s. It leaves at a pressure
  of
• 0.15MPa and a velocity of 600m/s. Determine the
• final temperature if the steam is superheated in
  the
• final state, and the quality if it is saturated.
• Control volume Nozzle.
• Inlet state Fixed.
• Exit state Pe known.
• Process SSSF.
• Model Steam tables.

Steady-State, Steady-Flow Process
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Steady-State, Steady-Flow Process
74

• Example 5.12
• In a refrigeration system, in which R-134a is the
  
• refrigerant, the R-134a enters the compressor at
  200kPa
• , -10?C and leaves at 1.0MPa, 70?C. The mass rate
  of
• flow is 0.015kg/s, and the power input to the
  compressor
• is 1kW.
• On leaving the compressor the refrigerant enter a
  water-
• cooled condenser at 1.0MPa, 60?C and leaves as a
  liquid
• at 0.95MPa, 35?C. Water enters the condenser at
  10?C
• and leaves at 20?C. Determine
• The heat transfer rate from the compressor.
• The rate at which cooling water flows through the
  
• condenser.

Steady-State, Steady-Flow Process
75

• First control volume Compressor.
• Inlet state Pi, Ti state fixed.
• Exit state Pe, Te state fixed.
• Process SSSF.
• Model R-134a tables.

Steady-State, Steady-Flow Process
76

• Second control volume Condenser.
• Inlet states R-134a-fixed water-fixed.
• Exit states R-134a-fixed water-fixed.
• Process SSSF.
• Model R-134a tables steam tables.

Steady-State, Steady-Flow Process
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Steady-State, Steady-Flow Process
78

• Example 5.13
• Consider the simple steam power plant. The
  following
• data are for such a power plant.

Steady-State, Steady-Flow Process
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• Simple steam power plant

Steady-State, Steady-Flow Process
80

• Determine the following quantities per kilogram
  flowing
• through the unit.
• Heat transfer in line between boiler and turbine.
• Turbine work.
• Heat transfer in condenser.
• Heat transfer in boiler.
• All processes SSSF.
• Model Steam tables.

Steady-State, Steady-Flow Process
81

• From the steam tables
• 1. Control volume Pipe line between the boiler
  and the turbine.

Steady-State, Steady-Flow Process
82

• 2. Control volume Turbine.
• 3. Control volume Condenser.
• 4. Control volume Boiler.

Steady-State, Steady-Flow Process
83

• Example 5.14
• The centrifugal air compressor of a gas turbine
  receives
• air from the ambient atmosphere where the
  pressure is
• 1bar and the temperature 300K. At the discharge
  of the
• compressor the pressure is 4bar, the temperature
  is 480
• K, and the velocity is 100m/s. The mass rate of
  flow
• into the compressor is 15 kg/s. Determine the
  power
• required to drive the compressor.
• We consider a control volume around the
  compressor,
• and locate the control volume at some distance
  from
• the compressor so that the air crossing the
  control
• surface has a very low velocity and is
  essentially at
• ambient conditions.

Steady-State, Steady-Flow Process
84

• If we located our control volume directly across
  the
• inlet section, it would be necessary to know the
• temperature and velocity at the compressor inlet.
• Inlet and exit states Both states fixed.
• Process SSSF.
• Model Ideal gas with constant specific heat,
  value
• from Table A.10(300K)

Steady-State, Steady-Flow Process
85
Steady-State, Steady-Flow Process
86

• A more accurate model for the behavior of the air
  in
• this process would be the ideal gas and air
  tables,
• A.12.

Steady-State, Steady-Flow Process
87
5.4 Uniform-State, Uniform-Flow Process

• Uniform State, Uniform Flow(USUF) Process
• CV at rest.
• Uniform state in CV. May change with time.
• SS for the state of mass across the CS.
• Mass flow rates may change with time.

Uniform-State, Uniform-Flow Process
88

• Example 5.15
• Steam at 800kPa, 300?C is throttled to 200kPa.
  Changes
• in kinetic energy are negligible for this
  process.
• Determine the final temperature of the steam,
  and the
• average Joule-Thomson coefficient.
• Control volume Throttle valve.
• Inlet state Pi, Ti known state fixed.
• Exit state Pe known.
• Process SSSF.
• Model Steam tables.

Uniform-State, Uniform-Flow Process
89
Uniform-State, Uniform-Flow Process
90

• Example 5.16
• Consider the throttling process across the
  expansion valve
• or through the capillary tube in a
  vapor-compression
• refrigeration cycle. In this process the pressure
  of the
• refrigerant drops from the high pressure in the
  condenser
• to the low pressure in the evaporator, and during
  this
• process some of the liquid flashes into vapor. If
  we
• consider this process to be adiabatic, the
  quality of the
• refrigerant entering the evaporator can be
  calculated.
• Consider the following process, in which ammonia
  is the
• refrigerant. The ammonia enters the expansion
  valve at a
• pressure of 1.50MPa and a temperature of 32?C.
  Its
• pressure on leaving the expansion valve is
  268kPa.

Uniform-State, Uniform-Flow Process
91

• Calculate the quality of the ammonia leaving the
• expansion valve.
• Control volume Expansion valve or capillary
  tube.
• Inlet state Pi, Ti known state fixed.
• Exit state Pe known.
• Process SSSF.
• Model Ammonia tables.

Uniform-State, Uniform-Flow Process
92

• Example 5.17
• Steam at a pressure of 1.4MPa, 300?C is flowing
  in a pipe
• . Connected to this pipe through a valve is an
  evacuated
• tank. The valve is opened and the tank fills with
  steam
• until the pressure is 1.4MPa, and then the valve
  is closed.
• The process takes place adiabatically and kinetic
  energies
• and potential energies are negligible. Determine
  the final
• temperature of the steam.
• Control volume Tank.
• Initial state(in tank) Evacuated, mass m10.
• Final state P2 known.
• Inlet state Pi, Ti (in tank) known.
• Process USUF.

Uniform-State, Uniform-Flow Process
93

• Flow into an evacuated vessel-control volume
  analysis

Uniform-State, Uniform-Flow Process
94

• The temperature corresponding to a pressure of
• 1.4MPa and an internal energy of 3040.4kJ/kg is
  found
• to be 452?C.

Uniform-State, Uniform-Flow Process
95

• Example 5.18
• Let the tank of the previous example have a
  volume 0.4
• m3 and initially contain saturated vapor at
  350kPa. The
• valve is then opened and steam from the line at
  1.4MPa,
• 300?C flows into the tank until the pressure is
  1.4MPa.
• Calculate the mass of steam that flows into the
  tank.
• Control volume Tank.
• Initial state P1, saturated vapor state fixed.
• Final state P2.
• Inlet state Pi, Ti state fixed.
• Process USUF.
• Model Steam tables.

Uniform-State, Uniform-Flow Process
96
Uniform-State, Uniform-Flow Process
97
Uniform-State, Uniform-Flow Process
98

• Example 5.19
• A tank of 2m3 volume contains saturated ammonia
  at a
• temperature of 40?C. Initially the tank contains
  50 liquid
• and 50 vapor by volume. Vapor is withdrawn from
  the
• top of the tank until the temperature is 10?C.
  Assuming
• that only vapor(i.e., no liquid) leaves and that
  the process
• is adiabatic, calculate the mass of ammonia that
  is
• withdrawn.

Uniform-State, Uniform-Flow Process
99

• Control volume Tank.
• Initial state T1, Vliq, Vvap state fixed.
• Final state T2.
• Exit state Saturated vapor(temperature
  changing).
• Process USUF.
• Model Ammonia tables.

Uniform-State, Uniform-Flow Process
100

Uniform-State, Uniform-Flow Process
101
Uniform-State, Uniform-Flow Process
102

• For ideal gas the internal energy is a function
  of the temperature only.

Uniform-State, Uniform-Flow Process
103

• from specific heats.

Uniform-State, Uniform-Flow Process
104

• Throttling Process
• SS, no work, no heat transfer, no change in PE
  and KE.
• Joule-Thomson coefficient
• T drops during throttling.
• T drops during throttling.

Uniform-State, Uniform-Flow Process