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Summary of L1 – L4

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Summary of L1 – L4

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  1. ME 200 L5: Energy AccountingConservation of Energy for a Closed System First Law of ThermodynamicsSpring 2014 MWF 1030-1120 AMJ. P. Gore, Reilly University Chair Professorgore@purdue.eduGatewood Wing 3166, 765 494 0061Office Hours: MWF 1130-1230TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edu

  2. Summary of L1 – L4 • Kinetic Energy and Potential Energy are macro- scale mechanical energies of a mass. • Internal Energy is an extensive property of a working substance and is defined by composition, T, & P. • Internal Energy + Kinetic Energy + Potential Energy may transform amongst themselves and also change by heat transfer and work interactions, all subject to Laws of Thermodynamics! • Heat Transfer is a result of temperature difference by conduction, convection and radiation. • Work Transfer is all modes of energy transfer that are not a result of temperature difference.

  3. An object of mass 80 lb (or 36.29 kg), initially at rest, experiences a constant horizontal acceleration of 12 ft/s2 (or 3.6576 m/s2) due to the action of a resultant force that is applied for 6.5 s. Determine the work of the resultant force, in ft-lbf, in Btu, in J and kJ. Given m = 80 lb or 36.29 kg V1 = 0 ft/s or 0 m/s a = 12 ft/s2 or 3.6576 m/s2 t = 6.5 s Find W in ft-lbf and Btu? And W in J and kJ Sketch Assumptions The 80 lb (36.29 kg) mass is the system. Motion is horizontal, so the system experiences no change in potential energy. The horizontal acceleration is constant. Basic Equations R z m x Mechanical Work Example 3

  4. Solution R z m x Mechanical Work Example 0 SI System British System (1) Work does not depend on the units we use to measure it! Different numerical values are assigned to identical work in different measuring systems! (2) See Conversion factors on inside cover of book. 10.255 kJ = 9.71 Btu because 1Btu=1.0551 kJ or 1 kJ = 0.9478 Btu 4

  5. Sign Convention • Our sign convention for work is easy to remember: • Work done by a system is considered to be useful to mankind, so is defined to be positive. • Therefore work done by the accelerator on the particle is positive. • Of course, if the work for “the system” defined as “the particle” is to be calculated, then it is negative but equal in magnitude to the work done by the accelerator.

  6. Piston Cylinder Systems • Piston cylinder systems are widely used • There function is to transfer expansion and compression energy change into linear motion and eventually rotary motion. Applications: I. C. engines, hydraulic jacks, bicycle air-pump, balloon inflator etc. Stroke=Crank circle diameter. The cylinder length must clear the end to end motion of the connecting rod. The clearance volume defines the compression ratio. The BDC volume defines the cylinder capacity. The pressure, temperature and volume within the cylinder are related and determine power output.

  7. Expansion and Compression Work The above derivation is applicable to experimental pressure volume traces (see Fig. 2.5 in text) as well as theoretical approximations to processes for defining the system behavior. Work is a path function and can not be evaluated by just knowing the end states 1 and 2. Also, in writing the above equations, the assumption that the pressure in the cylinder is uniform through out the volume has been made. This makes the work a quasi-steady approximation to reality. None the less, this approximation has been found to be very useful in industry.

  8. Expansion and Compression Work If gas volume decreases, work is negative and is done on the gas. If gas volume increases, work is done by the gas on the piston and hence on the connecting rod and the crank shaft etc. Practice these derivations

  9. Examples of work functions involving different processes: Shaft Work τ – torque ω – angular velocity (rad/s)

  10. Examples of work functions involving different processes: Spring Work k – spring constant xi – displacement from equilibrium

  11. Examples of work functions involving different processes: Work done by “flowing” electrons Electric Power i – electric current (amp) – potential difference (V) R – resistance (ohms) Ohm’s Law 11

  12. Additional examples of work • Torsion of a solid bar • See eq. 2.18 • 2. Stretching of a liquid film • See eq. 2.19 • 3. Charging of Electrolytic cell, Electric Field, Magnetic Field • Work done by electromotive force • Work done by dielectric in a uniform electric field • Work done by magnetic material in a field

  13. Learning Outcomes for Lecture 5 • Applyclosed system energy balances, observing sign conventions for work and heat transfer. • Conductenergy analyses of components undergoing thermodynamic processes.

  14. Closed System Energy Balance • Energy is an extensive property that includes the internal energy, the kinetic energy and the gravitational potential energy. • For closed systems, energy is transferred in and out across the system boundary by two means only: bywork and byheat. • Energy is conserved. This is the first law of thermodynamics.

  15. Closed System Transient Energy Balance Time rate of change of energy contains kinetic, potential and internal energy time rate form of the energy balance net rate at which energy is being transferred by work at time t time rate of change of the energy contained within the system at time t net rate at which energy enters via heat transfer at time t 15

  16. The time rate form of the closed system energy balance is (Eq. 2.37) • The rate form expressed in words is time rate of change of the energy contained within the system at time t net rate at which energy is being transferred in by heat transfer at timet net rate at which energy is being transferred out by work at time t Closed System Transient Energy Balance

  17. E2 – E1 = (U2 – U1) + (KE2 – KE1) + (PE2 – PE1) DE = DU + DKE + DPE (Eq. 2.27b) Change in Energy of a System • The changes in energy of a system from state 1 to state 2 consist of internal, kinetic and potential energy changes. (Eq. 2.27a) • Energy at state 1 or state 2 or any other state is defined in reference to a standard state. • Definition of energy at all states must have identical standard base state. • Changes in the energy of a system between states, defined with identical standard state have significance.

  18. net amount of energy transferred in across the system boundary by heattransfer during the time interval net amount of energy transferred out across the system boundary by work during the time interval change in the amount of energy contained within a system during some time interval • Using previously defined symbols, this can be expressed as: E2 – E1 = Q – W (Eq. 2.35a) • Alternatively, DKE + DPE + DU = Q – W (Eq. 2.35b) Closed System Energy Balance • The energy concepts introduced thus far are summarized in words as follows: In Eqs. 2.35, a minus sign appears before W because energy transfer by work from the system to the surrounding is taken as positive.

  19. Example Problem Imagine a party at a college location as sketched below. Bob goes to the refrigerator door to get a soda… Music speakers Well insulated party room A/C Vent TV Electrical supply cable Refrigerator (fridge) door open Door locked

  20. An electric generator coupled to a windmill produces an average power of 15 kW. The power is used to charge a storage battery. Heat transfer from the battery to the surroundings occurs at a constant rate of 1.8 kW. For 8 h of operation, determine the total amount of energy stored in the battery, in kJ. Find: ΔE in kJ? System Given W = -15 kW Q = -1.8 kW Δt = 8 h Assumptions The battery is a closed system. The work and heat transfer rates are constant. • W = ?15 kW • Q = ?1.8 kW • Δt = 8 h storage battery Example 1 20

  21. Example 1 Basic Equations Solution 21 21

  22. Example 2 An electric motor draws a current of 10 amp with a voltage of 110 V. The output shaft develops a torque of 10.2 N-m and a rotational speed of 1000 RPM. For operation at steady state, determine for the motor, each in kW. the electric power required. the power developed by the output shaft. the rate of heat transfer. Find Welectric in kW? Wshaft in kW? Q in kW? Sketch Given I = 10 amp V = 110 V τ = 10.2 N-m ω = 1000 RPM Assumptions The motor is a closed system. The system is at steady state. Basic Equations + - • I = 10 amp • V = 110 V • τ = 10.2 N-m • ω = 1000 RPM motor 22 22

  23. Example 2 Given I = 10 amp V = 110 V τ = 10.2 N-m ω = 1000 RPM Find Welectric in kW? Wshaft in kW? Q in kW? Sketch Basic Equations Solution + - • I = 10 amp • V = 110 V • τ = 10.2 N-m • ω = 1000 RPM motor 0 23 23 23

  24. Example 3 A gas within a piston-cylinder assembly (undergoes a thermodynamic cycle consisting of) three processes: Process 1-2: Constant volume, V = 0.028 m3, U2 – U1 = 26.4 kJ. Process 2-3: Expansion with pV = constant, U3 = U2. Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ. There are no significant changes in kinetic or potential energy. Sketch the cycle on a p-V diagram. Calculate the net work for the cycle, in kJ. Calculate the heat transfer for process 2-3, in kJ. Calculate the heat transfer for process 3-1, in kJ. 24

  25. Find p-V diagram Wnet = ? in kJ Q23 = ? in kJ Q31 = ? in kJ System Given 1-2: V = 0.028 m3, U2 – U1 = 26.4 kJ 2-3: pV = constant, U3 = U2 3-1: p = 1.4 bar, W31 = -10.5 kJ Assumptions The gas is the closed system. For the system, ΔKE = ΔPE = 0. Volume change is the only work mode. Basic Equations gas Example 3 25

  26. Solution Example 3 0 26

  27. Solution Example 3 27

  28. Solution Example 3 0 0 0 0 0 0 28

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