PTT 201/4 THERMODYNAMICS SEM 1 ( 2012/2013). CHAPTER 2: First Law of Thermodynamics, Energy Transfer and General Energy Analysis. Forms of energy. Energy can exist in numerous forms:. light. Macroscopic forms of energy. light. Microscopic forms of energy
PTT 201/4THERMODYNAMICS SEM 1 (2012/2013) CHAPTER 2: First Law of Thermodynamics, Energy Transfer and General Energy Analysis
Forms of energy Energy can exist in numerous forms: light Macroscopic forms of energy light Microscopic forms of energy Those related to the molecular structure of a system and the degree of the molecular activity. light Thermal Mechanical Their sum constitutes the total energy, E of a system Kinetic Potential Electric Magnetic Chemical Nuclear Related to the motion & the influence of some external effects such as : gravity, magnetion, electricity & surface tension light Internal energy, U The sum of all the microscopic forms of energy. light The total energy of a system on a unit mass: Usually ignored. light light Kinetic energy, KE The energy that a system possesses as a result of its motion relative to some reference frame. Potential energy, PE The energy that a system possesses as a result of its elevation in a gravitational field.
Forms of energy Kinetic energy, KE Kinetic energy per unit mass: light Potential energy, PE Kinetic energy per unit mass: light Velocity Gravitational acceleration Elevation of the center of gravity of a system relative to some arbitrarily selected reference level Total energy of a system Total energy of a system per unit mass: light
Forms of energy Flow of steam in a pipe light Energy flow rate light Mass flow rate
Some physical insight to internal energy The internal energy of a system is the sum of all forms of the microscopic energies The various forms of microscopic energies that make up sensible energy. light light Sensible energy The portion of the internal energy of a system associated with the kinetic energies of the molecules. light Sensible & latent energy Thermal Energy Latent energy The internal energy associated with the phase of a system. light Chemical energy The internal energy associated with the atomic bonds in a molecule. light Chemical energy Nuclear energy Nuclear energy The tremendous amount of energy associated with the strong bonds within the nucleus of the atom itself. light INTERNAL ENERGY
Example : Wind Energy A site evaluated for a wind farm is observed to have steady winds at a speed of 8.5 m/s. Determine the wind energy (a) per unit mass, (b) for a mass of 10 kg, and (c) for a flow rate of 1154 kg/s for air. SOLUTION :
Some physical insight to internal energy The total energy of a system, can be contained or stored in a system, and thus can be viewed as the static forms of energy. The forms of energy not stored in a system can be viewed as: the dynamic forms of energy @ energy interactions. light are recognized at the system boundary as they cross it The macroscopic kinetic energy is an organized form of energy and is much more useful than the disorganized microscopic kinetic energies of the molecules. light Represent the energy gained or lost by a system during a process. • The only two forms of energy interactions associated with a closed system are : • heat transfer • work. An energy interaction is heat transfer if its driving force is a temperature difference. Otherwise it is work. The difference is
Energy transfer by heat light light Heat The form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. light Temperature difference is the driving force for heat transfer. The larger the temperature difference, the higher is the rate of heat transfer. Energy is recognized as heat transfer only as it crosses the system boundary. Energy can cross the boundaries of a closed system in the form of heat and work.
Energy transfer by heat light Heat transfer per unit mass light Adiabatic process light A process during which there is no heat transfer Amount of heat transfer when heat transfer rate changes with time light light light Both the system and surroundings are at the same temperature and therefore there is no driving force (temp diff) for heat transfer The system is well insulated so that only a negligible amount of heat can pass through the boundary Amount of heat transfer when heat transfer rate is constant light During an adiabatic process, a system exchanges no heat with its surroundings.
Historical Background on Heat • Kinetic theory: Treats molecules as tiny balls that are in motion and thus possess kinetic energy. • Heat: The energy associated with the random motion of atoms and molecules. light Heat Transfer Mechanisms light light Radiation: The transfer of energy due to the emission of electromagnetic waves (or photons). light Convection: The transfer of energy between a solid surface and the adjacent fluid that is in motion, and it involves the combined effects of conduction and fluid motion. Conduction: The transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles.
ENERGY TRANSFER BY WORK Work:The energy transfer associated with a force acting through a distance. • A rising piston, a rotating shaft,andan electric wire crossing the system boundaries are all associated with work interactions Work done per unit mass Power is the work done per unit time (kW) Specifying the directions of heat and work.
THE FIRST LAW OF THERMODYNAMICS • The first law of thermodynamics (the conservation of energy principle) provides a sound basis for studying the relationships among the various forms of energy and energy interactions. • The first law states that energy can be neither created nor destroyed during a process; it can only change forms. Energy cannot be created or destroyed; it can only change forms.
Energy balance The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process. (Qin – Qout) + Wsh = ∆Esystem
Energy Change of a System, Esystem (Qin – Qout )+ Wsh= ∆U + ∆KE + ∆PE (Energy Balance for Closed Systems) light Internal, kinetic, and potential energy changes
Example : Energy Balance for Closed Systems Water flows over a waterfall 100 m in height. Take 1 kg of the water as the system, and assume that it does not change energy with its surroundings. What is the potential energy of the water at the top of the falls with respect to the base of the falls? What is the kinetic energy of the water just before it strikes bottom? 16
SOLUTION : The 1 kg of water exchanges no energy with the surroundings. Thus, for each part of the process ∆Esystem = ∆U + ∆KE + ∆PE= 0 Determine the potential energy, ∆PE ∆PE= mg (z2 – z1) = = Determine the kinetic energy, ∆KE ∆U + ∆KE + ∆PE = 0 During the free fall of the water no mechanism exists for conversion of potential or kinetic energy into internal energy. Thus, ∆U must be zero. = 0 + ∆KE + ∆PE = 0 = 0 + (KE2 – KE1) + (PE2 – PE1) = 0 Let KE1 and PE2= 0 Thus, we get KE2= 981 J 981 kg m2/s2 = 981 N.m = 981 J 17
Example : Cooling of a Hot Fluid in a Tank A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel. Ans: U2=400 kJ
Mechanisms of Energy Transfer, Ein and Eout • Heat transfer • Work transfer • Mass flow Energy Balanced for Open Systems: SI unit: (Qin – Qout) + Wsh= ∆H + ∆u2+ g ∆z (kJ/kg) 2 English system unit: (Qin – Qout) + Wsh= ∆H + ∆u2+ g∆z (Btu/Ibm) 2gc gc The energy content of a control volume can be changed by mass flow as well as heat and work interactions. Energy balance for cycle When ∆E =0, Qnet=Wnet Wnet,out = Qnet,in
Example : Energy Balance for Open Systems Air at 1 bar and 25˚C enters a compressor at low velocity, discharges at 3 bar, and enters a nozzle in which it expands to a final velocity of 600 m/s at the initial conditions of pressure and temperature. If the work of compression is 240 kJ per kilogram of air, how much heat must be removed during compression? Write energy balance for open system: Q + Wsh = ∆H + ∆u2+ g ∆z (kJ/kg) 2 Assumptions: 1. No enthalpy change, ∆H, of air due to the air returns to its initial temperature and pressure. 2. The potential energy, ∆PE, is presumed negligible due to no change in elevation involved 3. Neglecting the initial kinetic energy; u12 2 Ans: Q= -60 kJ/kg 20 20
Class discussion EXAMPLE 2-11 EXAMPLE 2-12 21