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Unit 1.2 thm

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MD ATEEQUE KHAN

The document discusses the first law of thermodynamics, which states that energy can neither be created nor destroyed, but rather transformed from one state to another. It provides equations for the energy balance of closed and open systems, and describes various thermodynamic processes including isochoric, isobaric, isothermal, and isentropic processes. It also discusses the application of the first law to devices like nozzles, diffusers, turbines, compressors, throttling valves, and heat exchangers.

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First Law of Thermodynamics 06/10/17JIT 1  The First Law is usually referred to as the Law of Conservation of Energy, i.e. energy can neither be created nor destroyed, but rather transformed from one state to another.  The energy balance is maintained within the system being studied/defined boundary.  The various energies associated are then being observed as they cross the boundaries of the system.
06/10/17JIT 2 Energy Balance for Closed System Heat Work z Closed System Reference Plane, z = 0  V or E E Ein out system− = ∆
06/10/17JIT 3  According to classical thermodynamics Q W Enet net system− = ∆  The total energy of the system, Esystem, is given as E Internal energy Kinetic energy Potential energy E U KE PE = + + = + +  The change in stored energy for the system is ∆ ∆ ∆ ∆E U KE PE= + +  The first law of thermodynamics for closed systems then can be written as Q W U KE PEnet net− = + +∆ ∆ ∆
06/10/17JIT 4  If the system does not move with a velocity and has no change in elevation, the conservation of energy equation is reduced to Q W Unet net− = ∆  The first law of thermodynamics can be in the form of )( 1000 )( 2000 12 2 1 2 2 12 kJ zzgVV uumWQ netnet         − + − +−=− )/( 1000 )( 2000 12 2 1 2 2 12 kgkJ zzgVV uuwq netnet         − + − +−=−  For a constant volume process,         − + − +−=− 1000 )( 2000 12 2 1 2 2 12 zzgVV uumWQ netnet         − + − +−= 1000 )( 2000 12 2 1 2 2 12 zzgVV uumQnet
06/10/17JIT 5  For a constant pressure process,         − + − +−=− 1000 )( 2000 12 2 1 2 2 12 zzgVV uumWQ netnet         − + − +−=−− 1000 )( 2000 )( 12 2 1 2 2 1212 zzgVV uumVVPQnet         − + − +−+−= 1000 )( 2000 )( 12 2 1 2 2 1212 zzgVV VVPuumQnet         − + − +−= 1000 )( 2000 12 2 1 2 2 12 zzgVV hhmQnet
06/10/17JIT 6 A closed system of mass 2 kg undergoes an adiabatic process. The work done on the system is 30 kJ. The velocity of the system changes from 3 m/s to 15 m/s. During the process, the elevation of the system increases 45 meters. Determine the change in internal energy of the system. Example.1 Solution:  Energy balance,         − + − +−=− 1000 )( 2000 12 2 1 2 2 12 zzgVV uumWQ netnet  Rearrange the equation netQ ( ) ( ) 2 2 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 2 ( ) 2000 1000 ( ) 2000 1000 9.81 4515 3 30 2 2 2 2000 1000 14.451 .. net net V V g z z W m u u V V g z z W m u u u u kJ Ans  − − − = − + + ÷    − − − = − + + ÷     − − − = ∆ + +  ÷ ÷  ÷     ∆ =
06/10/17JIT 7  Some thermodynamic cycle composes of processes in which the working fluid undergoes a series of state changes such that the final and initial states are identical.  For such system the change in internal energy of the working fluid is zero.  The first law for a closed system operating in a thermodynamic cycle becomes Closed System First Law of a Cycle Q W U Q W net net cycle net net − = = ∆
06/10/17JIT 8 No Value of n Process Description Result of IGL 1 ∞ isochoric constant volume (V1 = V2 ) 2 0 isobaric constant pressure (P1 = P2 ) 3 1 isothermal constant temperature (T1 = T2 ) 4 1<n< γ polytropic -none- 5 γ isentropic constant entropy (S1 = S2 ) According to a law of constant=n VP 2 2 1 1 T P T P = 2 2 1 1 T V T V = 2211 VPVP = 1 2 1 1 2 2 1 −       =      = n n n T T V V P P
06/10/17JIT 9 Boundary Works 3 2 4 51 P V
06/10/17JIT 10  Various forms of work are expressed as follows Process Boundary Work isochoric isobaric isothermal polytropic isentropic 0)( 1212 =−= VVPW )( 1212 VVPW −= 1 2 1112 ln V V VPW = n VPVP W − − = 1 1122 12
06/10/17JIT 11 Conservation of Mass  Conservation of mass is one of the most fundamental principles in nature. We are all familiar with this principle, and it is not difficult to understand it!  For closed system, the conservation of mass principle is implicitly used since the mass of the system remain constant during a process.  However, for control volume, mass can cross the boundaries. So the amount of mass entering and leaving the control volume must be considered.
06/10/17JIT 12 Mass and Volume Flow Rates  Mass flow through a cross-sectional area per unit time is called the mass flow rate. Note the dot over the mass symbol indicates a time rate of change. It is expressed as ∫= dAVm .ρ  If the fluid density and velocity are constant over the flow cross- sectional area, the mass flow rate is voulmespecificcalledis where AV AVm ν ρ ν ν ρ 1 = ==
06/10/17JIT 13 Principal of Conservation of Mass  The conservation of mass principle for a control volume can be expressed as in out CVm m m− =    For a steady state, steady flow process the conservation of mass principle becomes (kg/s)in outm m= 
06/10/17JIT 14  As the fluid upstream pushes mass across the control volume, work done on that unit of mass is flow flow flow A W FdL FdL PdV Pv m A W w Pv m δ δ δ δ δ = = = = = = Flow Work & The Energy of a Flowing Fluid
06/10/17JIT 15  The total energy carried by a unit of mass as it crosses the control surface is the sum of the internal energy + flow work + potential energy + kinetic energy ∑ ++=+++= gz V hgz V Puenergy 22 22 ν  The first law for a control volume can be written as ∑∑         ++−        ++=− in in in inin out out out outoutnetnet gz V hmgz V hmWQ 22 2.2... Total Energy of a Flowing Fluid
06/10/17JIT 16 Total Energy of a Flowing Fluid  The steady state, steady flow conservation of mass and first law of thermodynamics can be expressed in the following forms )( 1000 )( 2000 12 2 1 2 2 12 ... kW zzgVV hhmWQ netnet         − + − +−=− )( 1000 )( 2000 12 2 1 2 2 12 kJ zzgVV hhmWQ netnet         − + − +−=− )/( 1000 )( 2000 12 2 1 2 2 12 kgkJ zzgVV hhwq netnet         − + − +−=−
06/10/17JIT 17 Nozzle & Diffuser  Nozzle - device that increases the velocity fluid at the expense of pressure.  Diffuser - device that increases pressure of a fluid by slowing it down.  Commonly utilized in jet engines, rockets, space-craft and even garden hoses.  Q = 0 (heat transfer from the fluid to surroundings very small  W = 0 and ΔPE = 0
06/10/17JIT 18  Energy balance (nozzle & diffuser): ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2...         +=        + 22 2.2. out outout in inin V hm V hm         +=        + 22 2 2 2 2 1 1 V h V h
06/10/17JIT 19 Turbine & Compressor  Turbine – a work producing device through the expansion of a fluid.  Compressor (as well as pump and fan) - device used to increase pressure of a fluid and involves work input.  Q = 0 (well insulated), ΔPE = 0, ΔKE = 0 (very small compare to Δenthalpy).
06/10/17JIT 20  Energy balance: for turbine ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutoutinin hmWhm ... += ( )21 .. hhmW out −=
06/10/17JIT 21  Energy balance: for compressor, pump and fan ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutininin hmhmW ... =+ ( )12 .. hhmW in −=
06/10/17JIT 22 Throttling Valve Flow-restricting devices that cause a significant pressure drop in the fluid. Some familiar examples are ordinary adjustable valves and capillary tubes.
06/10/17JIT 23 Devices where two moving fluid streams exchange heat without mixing. Heat exchangers typically involve no work interactions (w = 0) and negligible kinetic and potential energy changes for each fluid stream. Heat Exchanger
06/10/17JIT 24 Second Law of Thermodynamics Kelvin-Planck statement  No heat engine can have a thermal efficiency 100 percent.  As for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.
06/10/17JIT 25 Heat Engines  Work can easily be converted to other forms of energy, but?  Heat engine differ considerably from one another, but all can be characterized : o they receive heat from a high-temperature source o they convert part of this heat to work o they reject the remaining waste heat to a low-temperature sink atmosphere o they operate on a cycle
06/10/17JIT 26 The work-producing device that best fit into the definition of a heat engine is the steam power plant, which is an external combustion engine.
06/10/17JIT 27  Represent the magnitude of the energy wasted in order to complete the cycle.  A measure of the performance that is called the thermal efficiency.  Can be expressed in terms of the desired output and the required input ηth = Desired Result Required Input  For a heat engine the desired result is the net work done and the input is the heat supplied to make the cycle operate.
06/10/17JIT 28 The thermal efficiency is always less than 1 or less than 100 percent. ηth net out in W Q = , W W W Q Q net out out in in net , = − ≠ where
06/10/17JIT 29  Applying the first law to the cyclic heat engine Q W U W Q W Q Q net in net out net out net in net out in out , , , , , − = = = − ∆  The cycle thermal efficiency may be written as ηth net out in in out in out in W Q Q Q Q Q Q = = − = − , 1
06/10/17JIT 30 A thermodynamic temperature scale related to the heat transfers between a reversible device and the high and low-temperature reservoirs by Q Q T T L H L H = The heat engine that operates on the reversible Carnot cycle is called the Carnot Heat Engine in which its efficiency is ηth rev L H T T , = −1
06/10/17JIT 31 Heat Pumps and Refrigerators  A device that transfers heat from a low temperature medium to a high temperature one is the heat pump.  Refrigerator operates exactly like heat pump except that the desired output is the amount of heat removed out of the system  The index of performance of a heat pumps or refrigerators are expressed in terms of the coefficient of performance.
06/10/17JIT 32
06/10/17JIT 33 COP Q W Q Q Q HP H net in H H L = = −, COP Q W R L net in = ,
06/10/17JIT 34 Process Description 1-2 Reversible isothermal heat addition at high temperature 2-3 Reversible adiabatic expansion from high temperature to low temperature 3-4 Reversible isothermal heat rejection at low temperature 4-1 Reversible adiabatic compression from low temperature to high temperature Carnot Cycle
06/10/17JIT 35 Execution of Carnot cycle in a piston cylinder device
06/10/17JIT 36
06/10/17JIT 37  The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows  The coefficients of performance of actual and reversible refrigerators operating between the same temperature limits compare as follows
06/10/17JIT 38 Example1 A steam power plant produces 50 MW of net work while burning fuel to produce 150 MW of heat energy at the high temperature. Determine the cycle thermal efficiency and the heat rejected by the cycle to the surroundings. Solution: ηth net out H W Q MW MW = = = , . 50 150 0 333 or 33.3% W Q Q Q Q W MW MW MW net out H L L H net out , , = − = − = − = 150 50 100
06/10/17JIT 39 A Carnot heat engine receives 500 kJ of heat per cycle from a high- temperature heat reservoir at 652ºC and rejects heat to a low- temperature heat reservoir at 30ºC. Determine : (a) The thermal efficiency of this Carnot engine (b) The amount of heat rejected to the low-temperature heat reservoir Example.2 QL WOUT QH TH = 652o C TL = 30o C HE ηth rev L H T T K K or , ( ) ( ) . . = − = − + + = 1 1 30 273 652 273 0 672 67 2% Q Q T T K K Q kJ kJ L H L H L = = + + = = = ( ) ( ) . ( . ) 30 273 652 273 0 328 500 0 328 164 Solution:
06/10/17JIT 40 An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2ºC while operating in a room where the temperature is 25ºC and has a COP of 13.5. Is there any truth to his claim? Example.3 Solution: QL Win QH TH = 25o C TL = 2o C R COP Q Q Q T T T K K R L H L L H L = − = − = + − = ( ) ( ) . 2 273 25 2 1196 this claim is also false!
06/10/17JIT 41 • Identifies the direction of a process. (e.g.: Heat can only spontaneously transfer from a hot object to a cold object, not vice versa) • Used to determine the “Quality” of energy. (e.g.: A high-temperature energy source has a higher quality since it is easier to extract energy from it to deliver useable work.) • Used to exclude the possibility of constructing 100% efficient heat engine and perpetual-motion machines. (violates the Kevin-Planck and the Clausius statements of the second law) • Used to introduce concepts of reversible processes and irreversibilities. • Determines the theoretical performance limits of engineering systems. (e.g.: A Carnot engine is theoretically the most efficient heat engine; its performance can be used as a standard for other practical engines)
06/10/17JIT 42 • A process can not happen unless it satisfies both the first and second laws of thermodynamics. The first law characterizes the balance of energy which defines the “quantity” of energy. The second law defines the direction which the process can take place and its “quality”. • Define a “Heat Engine”: A device that converts heat into work while operating in a cycle. Heat engine QH QL TH TL Wnet DQ-Wnet=DU (since DU=0 for a cycle) ⇒Wnet= QH-QL Thermal efficiency, hth is defined as hth=Wnet/QH=(QH-QL)/QH =1-(QL/QH) Question: Can we produce an 100% heat engine, i.e. a heat engine where Q =0?
06/10/17JIT 43 • A steam power plant is a good example of a heat engine where the working fluid, water, undergoes a thermodynamic cycle Wnet= Wout - Win = Qin-Qout Qinis the heat transferred from the high temp. reservoir, and is generally referred to as QH Qoutis the heat transferred to the low temp. reservoir, and is generally referred to as QL Thermal efficiency hth = Wnet/QH= (QH-QL)/QH =1-(QL/QH) Typical Efficiency of a large commercial steam power plant ≈ 40% Thermal Reservoir A hypothetical body with a very large thermal capacity (relative to the system beig examined) to/from which heat can be transferred without changing its temperature. E.g. the ocean,
06/10/17JIT 44 • The Kelvin-Planck Statement is another expression of the second law of thermodynamics. It states that: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce net work. • This statement is without proof, however it has not been violated yet. • Consequently, it is impossible to built a heat engine that is 100%. Heat engine QH TH Wnet • A heat engine has to reject some energy into a lower temperature sink in order to complete the cycle. • TH>TL in order to operate the engine. Therefore, the higher the temperature, TH, the higher the quality of the energy source and more work is produced. Impossible because it violates the Kelvin-Planck Statement/Second Law
06/10/17JIT 45 • A “heat pump” is defined as a device that transfers heat from a low-temperature source to a high-temperature one. E.g. a heat pump is used to extract energy from outside cold outdoor air into the warm indoors. • A refrigerator performs the same function; the difference between the two is in the type of heat transfer that needs to be optimized. • The efficiencies of heat pumps and refrigerators are denoted by the Coefficient of Performance (COP) where Heat pump/ Refrigerator QH QL TH TL Wnet For a Heat Pump: COPHP=QH/Wnet=QH/(QH-QL) = 1/(1-QL/QH) For a Refrigerator: COPR=QL/Wnet=QL/(QH-QL) = 1/(QH/QL-1) Note: COPHP = COPR + 1 • COPHP>1, ex: a typical heat pump has a COP in the order of 3 • Question: Can one build a heat pump operating COP= ∞, that is Wnet= 0 and QH=Q?
06/10/17JIT 46 • The Clausius Statement is another expression of the second law of thermodynamics. It states that: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. • Similar to the K-P Statement, it is a negative statement and has no proof, it is based on experimental observations and has yet to be violated. • Heat can not be transferred from low temperature to higher temperature unless external work is supplied. Heat pump QH QL TH TL Therefore, it is impossible to build a heat pump or a refrigerator without external work input.
06/10/17JIT 47 It can be shown that the violation of one statement leads to a violation of the other statement, i.e. they are equivalent. A 100% efficient heat engine; violates K-P Statement Heat pump QL QL TH TL Heat transfer from low-temp body to high-temp body without work; A violation of the Clausius statement Heat pump QH+QL QL TH TL Wnet =QH Heat engine QH
06/10/17JIT 48 Imagine that we can extract energy from unlimited low-temperature energy sources such as the ocean or the atmosphere (both can be thought of as thermal reservoirs). Heat engine Heat pump QL QH QH Win = QH-QL Wnet=QL TH Ocean TL It is against the Kevin-Planck statement: it is impossible to build an 100% heat engine. Perpetual Motion Machines, PMM, are classified into two types: PMM1- Perpetual Motion Machines of the First Kind: They violate the First Law of Thermodynamics PMM2 - Perpetual Motion Machines of the Second Kind : Violate the Second Law of Thermodynamics
06/10/17JIT 49 • A reversible process is one that can be executed in the reverse direction with no net change in the system or the surroundings. • At the end of a forwards and backwards reversible process, both system and the surroundings are returned to their initial states. • No real processes are reversible. • However, reversible processes are theoretically the most efficient processes. • All real processes are irreversible due to irreversibilities. Hence, real processes are less efficient than reversible processes. Common Sources of Irreversibility: • Friction • Sudden Expansion and compression • Heat Transfer between bodies with a finite temperature difference.
06/10/17JIT 50 • A work-producing device which employs quasi-equlibrium or reversible processes produces the maximum amount of work theoretically possible. •A work-consuming device which employs quasi-equilibrium or reversible processes requires the minimum amount of work theoretically possible. • One of the most common idealized cycles that employs all reversible processes is called the Carnot Cycle proposed in 1824 by Sadi Carnot. •A quasi-equilibrium process, e.g. very slow, frictionless expansion or compression is a reversible process.

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Unit 1.2 thm

  • 1. First Law of Thermodynamics 06/10/17JIT 1  The First Law is usually referred to as the Law of Conservation of Energy, i.e. energy can neither be created nor destroyed, but rather transformed from one state to another.  The energy balance is maintained within the system being studied/defined boundary.  The various energies associated are then being observed as they cross the boundaries of the system.
  • 2. 06/10/17JIT 2 Energy Balance for Closed System Heat Work z Closed System Reference Plane, z = 0  V or E E Ein out system− = ∆
  • 3. 06/10/17JIT 3  According to classical thermodynamics Q W Enet net system− = ∆  The total energy of the system, Esystem, is given as E Internal energy Kinetic energy Potential energy E U KE PE = + + = + +  The change in stored energy for the system is ∆ ∆ ∆ ∆E U KE PE= + +  The first law of thermodynamics for closed systems then can be written as Q W U KE PEnet net− = + +∆ ∆ ∆
  • 4. 06/10/17JIT 4  If the system does not move with a velocity and has no change in elevation, the conservation of energy equation is reduced to Q W Unet net− = ∆  The first law of thermodynamics can be in the form of )( 1000 )( 2000 12 2 1 2 2 12 kJ zzgVV uumWQ netnet         − + − +−=− )/( 1000 )( 2000 12 2 1 2 2 12 kgkJ zzgVV uuwq netnet         − + − +−=−  For a constant volume process,         − + − +−=− 1000 )( 2000 12 2 1 2 2 12 zzgVV uumWQ netnet         − + − +−= 1000 )( 2000 12 2 1 2 2 12 zzgVV uumQnet
  • 5. 06/10/17JIT 5  For a constant pressure process,         − + − +−=− 1000 )( 2000 12 2 1 2 2 12 zzgVV uumWQ netnet         − + − +−=−− 1000 )( 2000 )( 12 2 1 2 2 1212 zzgVV uumVVPQnet         − + − +−+−= 1000 )( 2000 )( 12 2 1 2 2 1212 zzgVV VVPuumQnet         − + − +−= 1000 )( 2000 12 2 1 2 2 12 zzgVV hhmQnet
  • 6. 06/10/17JIT 6 A closed system of mass 2 kg undergoes an adiabatic process. The work done on the system is 30 kJ. The velocity of the system changes from 3 m/s to 15 m/s. During the process, the elevation of the system increases 45 meters. Determine the change in internal energy of the system. Example.1 Solution:  Energy balance,         − + − +−=− 1000 )( 2000 12 2 1 2 2 12 zzgVV uumWQ netnet  Rearrange the equation netQ ( ) ( ) 2 2 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 2 ( ) 2000 1000 ( ) 2000 1000 9.81 4515 3 30 2 2 2 2000 1000 14.451 .. net net V V g z z W m u u V V g z z W m u u u u kJ Ans  − − − = − + + ÷    − − − = − + + ÷     − − − = ∆ + +  ÷ ÷  ÷     ∆ =
  • 7. 06/10/17JIT 7  Some thermodynamic cycle composes of processes in which the working fluid undergoes a series of state changes such that the final and initial states are identical.  For such system the change in internal energy of the working fluid is zero.  The first law for a closed system operating in a thermodynamic cycle becomes Closed System First Law of a Cycle Q W U Q W net net cycle net net − = = ∆
  • 8. 06/10/17JIT 8 No Value of n Process Description Result of IGL 1 ∞ isochoric constant volume (V1 = V2 ) 2 0 isobaric constant pressure (P1 = P2 ) 3 1 isothermal constant temperature (T1 = T2 ) 4 1<n< γ polytropic -none- 5 γ isentropic constant entropy (S1 = S2 ) According to a law of constant=n VP 2 2 1 1 T P T P = 2 2 1 1 T V T V = 2211 VPVP = 1 2 1 1 2 2 1 −       =      = n n n T T V V P P
  • 10. 06/10/17JIT 10  Various forms of work are expressed as follows Process Boundary Work isochoric isobaric isothermal polytropic isentropic 0)( 1212 =−= VVPW )( 1212 VVPW −= 1 2 1112 ln V V VPW = n VPVP W − − = 1 1122 12
  • 11. 06/10/17JIT 11 Conservation of Mass  Conservation of mass is one of the most fundamental principles in nature. We are all familiar with this principle, and it is not difficult to understand it!  For closed system, the conservation of mass principle is implicitly used since the mass of the system remain constant during a process.  However, for control volume, mass can cross the boundaries. So the amount of mass entering and leaving the control volume must be considered.
  • 12. 06/10/17JIT 12 Mass and Volume Flow Rates  Mass flow through a cross-sectional area per unit time is called the mass flow rate. Note the dot over the mass symbol indicates a time rate of change. It is expressed as ∫= dAVm .ρ  If the fluid density and velocity are constant over the flow cross- sectional area, the mass flow rate is voulmespecificcalledis where AV AVm ν ρ ν ν ρ 1 = ==
  • 13. 06/10/17JIT 13 Principal of Conservation of Mass  The conservation of mass principle for a control volume can be expressed as in out CVm m m− =    For a steady state, steady flow process the conservation of mass principle becomes (kg/s)in outm m= 
  • 14. 06/10/17JIT 14  As the fluid upstream pushes mass across the control volume, work done on that unit of mass is flow flow flow A W FdL FdL PdV Pv m A W w Pv m δ δ δ δ δ = = = = = = Flow Work & The Energy of a Flowing Fluid
  • 15. 06/10/17JIT 15  The total energy carried by a unit of mass as it crosses the control surface is the sum of the internal energy + flow work + potential energy + kinetic energy ∑ ++=+++= gz V hgz V Puenergy 22 22 ν  The first law for a control volume can be written as ∑∑         ++−        ++=− in in in inin out out out outoutnetnet gz V hmgz V hmWQ 22 2.2... Total Energy of a Flowing Fluid
  • 16. 06/10/17JIT 16 Total Energy of a Flowing Fluid  The steady state, steady flow conservation of mass and first law of thermodynamics can be expressed in the following forms )( 1000 )( 2000 12 2 1 2 2 12 ... kW zzgVV hhmWQ netnet         − + − +−=− )( 1000 )( 2000 12 2 1 2 2 12 kJ zzgVV hhmWQ netnet         − + − +−=− )/( 1000 )( 2000 12 2 1 2 2 12 kgkJ zzgVV hhwq netnet         − + − +−=−
  • 17. 06/10/17JIT 17 Nozzle & Diffuser  Nozzle - device that increases the velocity fluid at the expense of pressure.  Diffuser - device that increases pressure of a fluid by slowing it down.  Commonly utilized in jet engines, rockets, space-craft and even garden hoses.  Q = 0 (heat transfer from the fluid to surroundings very small  W = 0 and ΔPE = 0
  • 18. 06/10/17JIT 18  Energy balance (nozzle & diffuser): ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2...         +=        + 22 2.2. out outout in inin V hm V hm         +=        + 22 2 2 2 2 1 1 V h V h
  • 19. 06/10/17JIT 19 Turbine & Compressor  Turbine – a work producing device through the expansion of a fluid.  Compressor (as well as pump and fan) - device used to increase pressure of a fluid and involves work input.  Q = 0 (well insulated), ΔPE = 0, ΔKE = 0 (very small compare to Δenthalpy).
  • 20. 06/10/17JIT 20  Energy balance: for turbine ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutoutinin hmWhm ... += ( )21 .. hhmW out −=
  • 21. 06/10/17JIT 21  Energy balance: for compressor, pump and fan ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutininin hmhmW ... =+ ( )12 .. hhmW in −=
  • 22. 06/10/17JIT 22 Throttling Valve Flow-restricting devices that cause a significant pressure drop in the fluid. Some familiar examples are ordinary adjustable valves and capillary tubes.
  • 23. 06/10/17JIT 23 Devices where two moving fluid streams exchange heat without mixing. Heat exchangers typically involve no work interactions (w = 0) and negligible kinetic and potential energy changes for each fluid stream. Heat Exchanger
  • 24. 06/10/17JIT 24 Second Law of Thermodynamics Kelvin-Planck statement  No heat engine can have a thermal efficiency 100 percent.  As for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.
  • 25. 06/10/17JIT 25 Heat Engines  Work can easily be converted to other forms of energy, but?  Heat engine differ considerably from one another, but all can be characterized : o they receive heat from a high-temperature source o they convert part of this heat to work o they reject the remaining waste heat to a low-temperature sink atmosphere o they operate on a cycle
  • 26. 06/10/17JIT 26 The work-producing device that best fit into the definition of a heat engine is the steam power plant, which is an external combustion engine.
  • 27. 06/10/17JIT 27  Represent the magnitude of the energy wasted in order to complete the cycle.  A measure of the performance that is called the thermal efficiency.  Can be expressed in terms of the desired output and the required input ηth = Desired Result Required Input  For a heat engine the desired result is the net work done and the input is the heat supplied to make the cycle operate.
  • 28. 06/10/17JIT 28 The thermal efficiency is always less than 1 or less than 100 percent. ηth net out in W Q = , W W W Q Q net out out in in net , = − ≠ where
  • 29. 06/10/17JIT 29  Applying the first law to the cyclic heat engine Q W U W Q W Q Q net in net out net out net in net out in out , , , , , − = = = − ∆  The cycle thermal efficiency may be written as ηth net out in in out in out in W Q Q Q Q Q Q = = − = − , 1
  • 30. 06/10/17JIT 30 A thermodynamic temperature scale related to the heat transfers between a reversible device and the high and low-temperature reservoirs by Q Q T T L H L H = The heat engine that operates on the reversible Carnot cycle is called the Carnot Heat Engine in which its efficiency is ηth rev L H T T , = −1
  • 31. 06/10/17JIT 31 Heat Pumps and Refrigerators  A device that transfers heat from a low temperature medium to a high temperature one is the heat pump.  Refrigerator operates exactly like heat pump except that the desired output is the amount of heat removed out of the system  The index of performance of a heat pumps or refrigerators are expressed in terms of the coefficient of performance.
  • 33. 06/10/17JIT 33 COP Q W Q Q Q HP H net in H H L = = −, COP Q W R L net in = ,
  • 34. 06/10/17JIT 34 Process Description 1-2 Reversible isothermal heat addition at high temperature 2-3 Reversible adiabatic expansion from high temperature to low temperature 3-4 Reversible isothermal heat rejection at low temperature 4-1 Reversible adiabatic compression from low temperature to high temperature Carnot Cycle
  • 35. 06/10/17JIT 35 Execution of Carnot cycle in a piston cylinder device
  • 37. 06/10/17JIT 37  The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows  The coefficients of performance of actual and reversible refrigerators operating between the same temperature limits compare as follows
  • 38. 06/10/17JIT 38 Example1 A steam power plant produces 50 MW of net work while burning fuel to produce 150 MW of heat energy at the high temperature. Determine the cycle thermal efficiency and the heat rejected by the cycle to the surroundings. Solution: ηth net out H W Q MW MW = = = , . 50 150 0 333 or 33.3% W Q Q Q Q W MW MW MW net out H L L H net out , , = − = − = − = 150 50 100
  • 39. 06/10/17JIT 39 A Carnot heat engine receives 500 kJ of heat per cycle from a high- temperature heat reservoir at 652ºC and rejects heat to a low- temperature heat reservoir at 30ºC. Determine : (a) The thermal efficiency of this Carnot engine (b) The amount of heat rejected to the low-temperature heat reservoir Example.2 QL WOUT QH TH = 652o C TL = 30o C HE ηth rev L H T T K K or , ( ) ( ) . . = − = − + + = 1 1 30 273 652 273 0 672 67 2% Q Q T T K K Q kJ kJ L H L H L = = + + = = = ( ) ( ) . ( . ) 30 273 652 273 0 328 500 0 328 164 Solution:
  • 40. 06/10/17JIT 40 An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2ºC while operating in a room where the temperature is 25ºC and has a COP of 13.5. Is there any truth to his claim? Example.3 Solution: QL Win QH TH = 25o C TL = 2o C R COP Q Q Q T T T K K R L H L L H L = − = − = + − = ( ) ( ) . 2 273 25 2 1196 this claim is also false!
  • 41. 06/10/17JIT 41 • Identifies the direction of a process. (e.g.: Heat can only spontaneously transfer from a hot object to a cold object, not vice versa) • Used to determine the “Quality” of energy. (e.g.: A high-temperature energy source has a higher quality since it is easier to extract energy from it to deliver useable work.) • Used to exclude the possibility of constructing 100% efficient heat engine and perpetual-motion machines. (violates the Kevin-Planck and the Clausius statements of the second law) • Used to introduce concepts of reversible processes and irreversibilities. • Determines the theoretical performance limits of engineering systems. (e.g.: A Carnot engine is theoretically the most efficient heat engine; its performance can be used as a standard for other practical engines)
  • 42. 06/10/17JIT 42 • A process can not happen unless it satisfies both the first and second laws of thermodynamics. The first law characterizes the balance of energy which defines the “quantity” of energy. The second law defines the direction which the process can take place and its “quality”. • Define a “Heat Engine”: A device that converts heat into work while operating in a cycle. Heat engine QH QL TH TL Wnet DQ-Wnet=DU (since DU=0 for a cycle) ⇒Wnet= QH-QL Thermal efficiency, hth is defined as hth=Wnet/QH=(QH-QL)/QH =1-(QL/QH) Question: Can we produce an 100% heat engine, i.e. a heat engine where Q =0?
  • 43. 06/10/17JIT 43 • A steam power plant is a good example of a heat engine where the working fluid, water, undergoes a thermodynamic cycle Wnet= Wout - Win = Qin-Qout Qinis the heat transferred from the high temp. reservoir, and is generally referred to as QH Qoutis the heat transferred to the low temp. reservoir, and is generally referred to as QL Thermal efficiency hth = Wnet/QH= (QH-QL)/QH =1-(QL/QH) Typical Efficiency of a large commercial steam power plant ≈ 40% Thermal Reservoir A hypothetical body with a very large thermal capacity (relative to the system beig examined) to/from which heat can be transferred without changing its temperature. E.g. the ocean,
  • 44. 06/10/17JIT 44 • The Kelvin-Planck Statement is another expression of the second law of thermodynamics. It states that: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce net work. • This statement is without proof, however it has not been violated yet. • Consequently, it is impossible to built a heat engine that is 100%. Heat engine QH TH Wnet • A heat engine has to reject some energy into a lower temperature sink in order to complete the cycle. • TH>TL in order to operate the engine. Therefore, the higher the temperature, TH, the higher the quality of the energy source and more work is produced. Impossible because it violates the Kelvin-Planck Statement/Second Law
  • 45. 06/10/17JIT 45 • A “heat pump” is defined as a device that transfers heat from a low-temperature source to a high-temperature one. E.g. a heat pump is used to extract energy from outside cold outdoor air into the warm indoors. • A refrigerator performs the same function; the difference between the two is in the type of heat transfer that needs to be optimized. • The efficiencies of heat pumps and refrigerators are denoted by the Coefficient of Performance (COP) where Heat pump/ Refrigerator QH QL TH TL Wnet For a Heat Pump: COPHP=QH/Wnet=QH/(QH-QL) = 1/(1-QL/QH) For a Refrigerator: COPR=QL/Wnet=QL/(QH-QL) = 1/(QH/QL-1) Note: COPHP = COPR + 1 • COPHP>1, ex: a typical heat pump has a COP in the order of 3 • Question: Can one build a heat pump operating COP= ∞, that is Wnet= 0 and QH=Q?
  • 46. 06/10/17JIT 46 • The Clausius Statement is another expression of the second law of thermodynamics. It states that: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. • Similar to the K-P Statement, it is a negative statement and has no proof, it is based on experimental observations and has yet to be violated. • Heat can not be transferred from low temperature to higher temperature unless external work is supplied. Heat pump QH QL TH TL Therefore, it is impossible to build a heat pump or a refrigerator without external work input.
  • 47. 06/10/17JIT 47 It can be shown that the violation of one statement leads to a violation of the other statement, i.e. they are equivalent. A 100% efficient heat engine; violates K-P Statement Heat pump QL QL TH TL Heat transfer from low-temp body to high-temp body without work; A violation of the Clausius statement Heat pump QH+QL QL TH TL Wnet =QH Heat engine QH
  • 48. 06/10/17JIT 48 Imagine that we can extract energy from unlimited low-temperature energy sources such as the ocean or the atmosphere (both can be thought of as thermal reservoirs). Heat engine Heat pump QL QH QH Win = QH-QL Wnet=QL TH Ocean TL It is against the Kevin-Planck statement: it is impossible to build an 100% heat engine. Perpetual Motion Machines, PMM, are classified into two types: PMM1- Perpetual Motion Machines of the First Kind: They violate the First Law of Thermodynamics PMM2 - Perpetual Motion Machines of the Second Kind : Violate the Second Law of Thermodynamics
  • 49. 06/10/17JIT 49 • A reversible process is one that can be executed in the reverse direction with no net change in the system or the surroundings. • At the end of a forwards and backwards reversible process, both system and the surroundings are returned to their initial states. • No real processes are reversible. • However, reversible processes are theoretically the most efficient processes. • All real processes are irreversible due to irreversibilities. Hence, real processes are less efficient than reversible processes. Common Sources of Irreversibility: • Friction • Sudden Expansion and compression • Heat Transfer between bodies with a finite temperature difference.
  • 50. 06/10/17JIT 50 • A work-producing device which employs quasi-equlibrium or reversible processes produces the maximum amount of work theoretically possible. •A work-consuming device which employs quasi-equilibrium or reversible processes requires the minimum amount of work theoretically possible. • One of the most common idealized cycles that employs all reversible processes is called the Carnot Cycle proposed in 1824 by Sadi Carnot. •A quasi-equilibrium process, e.g. very slow, frictionless expansion or compression is a reversible process.