Lecture 7 ME300 Thermodynamics II Page 1 Lecture 7: Exergy Analysis Closed System Reading Assignment: Sections 7.1-7.3 Problems: P7, 7.15, 7.29 Any questions on Lecture 6? 7.1 Introducing Exergy Exergy: Property that quantifies potential for use. 7.2 Conceptualizing Exergy How to make best use of the situation when a system at a state different from that of the environment is allowed to interact with the environment and eventually comes to equilibrium state (called dead state) during its transient time? Examples: (a) Spontaneous cooling of hot body or warming of a cold body in a room (b) Braking of vehicles and locomotives in motion N a t ur a l ga s i s us e d f or he a t i ng bui l di ng. B ut hot a i r ( ~ 6 5 - 75 o F ) ca nnot r e pr oduc e t he N G i t us e d. D i d w e f ul l y ut i l i z e t he pot e nt i a l of N G f or he a t i ng t he bui l di ng ? H ow ca n w e m i ni m i z e i r r e v e r s i bi l i t i e s ? Lecture 7 ME300 Thermodynamics II Page 2 (c) Heat being dissipated from the heat exchangers behind the refrigerators (d) Simple gas turbine engine?s exhaust energy leading to co-generation power plant (e) Use of low-grade heat coming out of co-gen power plants (f) On-the-road vehicles with internal combustion engines making use of exhaust gas energy for turbochargers -----. Overall system (includes system + environment), system and environment used for evaluating exergy Exergy of a system (1) E x e r gy r e f e r e nce PE M e cha ni ca l W or k E l e ct r i ci t y E x e r gy i s t he m a x i m um t he or e t i ca l w or k obt a i na bl e f r om a n ov e r a l l s y s t e m co ns i s t i ng of a s y s t e m a nd t he e nv i r on m e nt a s t he s y s t e m c om e s i nt o e qui l i br i um w i t h t he e nv i r onm e nt ( pa s s e s t o de a d s t a t e ) . 25 o C , 1 a t m . Lecture 7 ME300 Thermodynamics II Page 3 The closed overall system boundary exergy reference environment is drawn so that only work crosses its boundary, Q=0 (thermally isolate system ? e.g. Joule?s experiment); V0 is constant and the environment is at 25oC and 1 atm. Neglecting ?PE and ?KE, change in the overall system energy, . E, Uo: Initial and final (dead state) energy of the system S, So: Initial and final (dead state) entropy of the system V, Vo: Initial and final (dead state) volume of the system Total volume of the overall system is constant. ?Ue: Change in internal energy of the environment Apply 1st T dS equation ( to replace ?Ue Rearranging gives: , rearrange for , substitute in Equation 2 to get: Replace E by E=U+KE+PE, giving: , here E, exergy is given by Exergy Aspects: 1. Property of a system and environment combined (overall system). Once environment is specified, then it represents property of the system. 2. The maximum work (exergy) cannot be negative. 3. Exergy is not conserved. It is destroyed by irreversibilities. In the limiting case, the system is allowed to change spontaneously to the dead state without producing any work ? potential to produce work destroyed. 4. Exergy ? maximum theoretical work obtainable. Conversely, it can be regarded as minimum theoretical work required to restore the system from dead state to its initial state. 5. A system in dead state means that it is in thermal and mechanical equilibrium with its environment and its value is 0 and no more useful work. ?Thermo-static implies no more work out, thermodynamics implies converting energy to work until it reaches equilibrium? Mongia?s spin in order to Lecture 7 ME300 Thermodynamics II Page 4 minimize struggle with the use of the word thermodynamics (quasi-static assumption) to describe relationship between W, Q and E. Specific Exergy (2) Please, go through example on page 336 and Example 7.1 Calculate specific exergy of saturated water vapor at 120oC, having a velocity 30 m/s and an elevation of 6 m relative to an exergy reference environment where To = 298 K (25oC), po = 1 atm, and g = 9.8 m/s. Recall for saturated liquids: Use Table A-2 to get the properties. Use Eq. (2) to calculate e: Example 7.1: Exergy of an internal combustion engine: 7 bar, 867oC; reference environment: 300 K, 1.013 bar Ignore PE and KE. Ideal gas properties from Table A-22: Let us look at the individual components of e: (a) (b) (c) Focus on u and s to make the most impact on e. Exergy Exchange: Initially the system state is: Interaction with the surrounding occurs via work and heat transfer leading to different state: Subtracting, we get exergy change in the system: Lecture 7 ME300 Thermodynamics II Page 5 Exergy Transfer Exergy destruction Problem 7.31 Known: 2 lb of ammonia at 20 psi with 80% quality is heated by an electrical heating element until its volume increases by 25% with negligible increase in pressure. Find: Calculate: (a) The energy transferred by electrical work and the accompanying exergy transfer. (b) The amount of energy transfer by work to the piston and the accompanying exergy transfer. (c) The change in exergy of the ammonia. (d) The amount of exergy destruction. Engineering Model: 1. NH3 in a closed system with the environment conditions as shown. 2. Adiabatic system with negligible KE and PE. 3. Negligible mass of the electrical resistor. Analysis: Properties of NH3 from Table A-13E and A-14E Lecture 7 ME300 Thermodynamics II Page 6 Schematic & Given Data: Which are nearly saturated conditions at 20 lbf/in.2 20 lb f / i n 2 T S 1 a tm 1 2 0 60 o F Lecture 7 ME300 Thermodynamics II Page 7 Exergy change is given by: Exergy Transfer Exergy destruction Negligible KE, PE and for piston only the equation reduces to: = 5.25 Btu There is no change in volume. Exergy change: 10.8451778?5201.3687?1.106] Energy destruction Lecture 7 ME300 Thermodynamics II Page 8 Lecture 8: Exergy Analysis Closed System Reading Assignment: Section 7.4 Problems: P8, 7.37, 7.44 Mechanical Engineering
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