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EGR 334 Thermodynamics Chapter 8: Sections 3-4 Lecture 32: Superheat, Reheat, and Cogeneration Quiz Today? Today’s main concepts: • Explain how a superheater changes the Rankine cycle model • Explain how a reheat line changes the Rankine cycle model. • Explain how regenerative heating changes the Rankine cycle model. • Explain how cogeneration and process steam lines change the Rankine cycle model. • Be able to use mass balance, energy balance, and entropy balance to solve systems with modified Rankine cycle models. Reading Assignment: Read Chapter 9, Sections 1-2 Homework Assignment: Problems from Chap 8: 21,29, 49, 60 Sec 8.1 : Modeling Vapor Power Systems Vapor Power System Model 3 Sec 8.2 : Rankine Cycle 4 Tools: Conservation of mass Conservation of energy (First Law) Entropy balance (Second Law) Thermodynamic properties Energy balances: For the turbine: W h1 h 2 m For the condenser: Q m Q h 2 h3 m c .w . h 6 h5 For the pump: W m h 4 h3 For the boiler: Q m h1 h 4 4 3 WP vd P v3 p 4 p 3 m in t . rev Sec 8.2 : Rankine Cycle 5 Process 1-2: Isentropic expansion of the fluid through the turbine from saturated vapor to the condenser pressure. Process 3-4: Isentropic compression of fluid to compressed liquid. Process 2-3: Heat transfer at constant pressure through the condenser to saturated liquid. Process 4-1: Heat transfer at constant pressure through the boiler to saturated vapor W cycle Q in W T W P Q in Sec 8.2.2 : Modified Rankine Cycle In addition to being able to work with the Ideal Rankine Cycle, you should also be prepared to evaluate performance that includes the following modifications -- Superheating: Heating of the fluid past a saturated vapor and well into the superheated region by the steam generator. -- Reheating: Reheating fluid after it has passed through one turbine and then passing through a second turbine at lower pressure. -- Regenerative Feedwater Heating: A process where some of the steam is bled off and used to partially reheat the condensate from the condenser. 6 Sec 8.2.2 : Modified Rankine Cycle In addition to being able to work with the Ideal Rankine Cycle, you should also be prepared to evaluate performance that includes the following modifications -- Non-Ideal Rankine (with irreversibilities): Similar to the Ideal Rankine cycle except the processes through the turbine and pump are not isentropic. Turbine and pump efficiencies need to be used to determine work in each of the components. -- Cogeneration: A process where some of the steam is bled off for use by other parts of the plant, such as for heating, cleaning, etc. 7 Sec 8.3 : Superheat and Reheat 8 How can the efficiency of the Rankine cycle be increased? Raise the Average High Temperature. As the temperature is raised, note that the sat. vapor location also shifts to the left, so that when the fluid expands through the turbine, a lower quality of the fluid is obtained. Why is this bad? Net work is decreased as area under the curve get narrower and there are physical effects on the turbine that involve more blade wear. It is more desirable to have the exit quality of the steam, x2, be above 90% How can the cycle maintain high boiler Pressure and Temperature and still maintain a higher exit quality? Superheat the steam. 9 How is steam superheated? Sec 8.3 : Superheat and Reheat 10 Superheating: It is not necessary to have the exit from the boiler be a saturated vapor (xboiler exit = 1). Operate the boiler, such that the exit of the boiler is a superheated vapor. The only difference in calculating cycle performance from the Ideal Rankine Cycle, is that state 3 is no longer expected to fall along the saturated vapor curve, but will be found on the superheated steam table. Sec 8.3 : Superheat and Reheat When Reheating is implemented, there is a multistage stage turbine (often identified as a High Pressure and Low Pressure turbine.) Before vapor starts to condense (reaches saturation), the vapor is sent back to the boiler and reheated. 11 Sec 8.3 : Superheat and Reheat 12 This affects the Ideal Rankine Cycle model because there are additional states to determine properties for, a second pass through the turbine, and a second heat exchange with the boiler. Therefore, identify additional properties states (1, 2, 3, and 4) and then rework the 1st law equations as applied to the turbine and boiler. 0 W H P _ turbine m ( h1 h 2 ) and 0 W L P _ turbine m ( h3 h 4 ) Boiler Reheat Equation 0 Q reheat m ( h 2 h3 ) Sec 8.4 : Regenerative Vapor Power Cycle 13 Regenerative Feedwater Heating. (regeneration) Steam is diverted to use in a mixing chamber to preheat the condensate prior to entering boiler. This is useful since water is typically cooled below saturated level out of the condensor. Note: This rerouting will diminish the net work output. The reduction in Qboiler should be less than the reduction of Wcycle. Sec 8.4 : Regenerative Vapor Power Cycle 14 Open feedwater analysis: #2 Mass Balance: m 1 m 2 m 3 #1 Often will use the term “blend fraction” y m 2 consequently, 1 y m 1 Energy Balance: 0 #3 m 3 m 1 m h i i 0 m 2 h 2 m 3 h3 m 1 h1 m 2 h2 m 3 h5 m 1 h6 0 m 2 m 1 h2 m 3 m 1 h5 m 1 m 1 h 6 yh 2 1 y h5 h 6 Thus: y h6 h5 h 2 h5 Sec 8.4 : Regenerative Vapor Power Cycle 15 Energy balances: For the turbines: W T 1 h1 h 2 m and For the condenser: Q out m 1 y h3 h 4 For the pump: W P m h 7 h 6 1 y h5 h 4 For the boiler: Q in m h1 h 7 W T 2 m 1 y h 2 h3 16 Example (8.44): An Ideal Rankine cycle with the state properties given below includes one open feedwater heater operating at 100 psi. Saturated liquid exits the open feedwater heater at 100 psi. The mass flow rate of steam into the first turbine state is 1.4 x 106 lbm/hr. Determine (a) The net power developed, in Btu/hr. (b) The thermal efficiency. (c) The mass flow rate of cooling water, if T = 20°F. State 1 T (°F) 1100 p (psi) 1600 x h (Btu/lbm) s (Btu/lbm K) 2 3 4 5 6 7 100 1 1 100 100 1600 17 Example (8.44) continued: Start by finding the state properties State 1: Using Table A4E with T = 1100 F and p = 1600 psi, find h1 = 1547.7 and s1 = 1.6315 State 2: Using Table A4E with s2=s1 and p = 100 psi, find h2 = 1210.7 State 3: Using Table A3E with s3=s1, and p = 1 psi, find x3 = 81.2% and then, find h3 = 911.2 State 4: Using Table A3E with sat. liquid and p = 1 psi, find v4 = vf = 0.01614, h4 = hf = 69.74, and s4 = sf = 0.1327 State 1 T (°F) 1100 p (psi) 1600 x h (Btu/lbm) s (Btu/lbm K) 2 3 4 5 6 7 100 1 1 100 100 1600 18 Example (8.44) continued: Start by finding the state properties State 5: Using h5 - h4 =v4 (p5 - p4) find h5 = 70.04 State 6: Using Table A4E with sat. liq. and p = 100 psi, find v6 = 0.01774 h6 = 298.6, and s6 = 0.4744 State 7: Using h7 - h6 =v6 (p7 - p6) find h7 = 303.5 State 1 2 3 4 5 6 7 T (°F) 1100 1100 p (psi) 1600 1600 100 100 1 1 100 100 100 100 1600 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 19 Example (8.44) continued: Find heat in provided in boiler: 0 Q in m 1 ( h 7 h1 ) then Q in m 1 ( h1 h 7 ) (1.4 10 lb m / hr )(1547.7 303.5) B tu / lb m 6 1.74 10 B tu / hr 9 State 1 2 3 4 5 6 7 T (°F) 1100 p (psi) 1600 100 1 1 100 100 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 20 Example (8.44) continued: Now find how the mass flow splits from the turbine y h 6 h5 h 2 h5 298.6 70.04 1210.7 70.04 0.200 c therefore: m 2 y m1 (0.200)(1.4 10 lb m / hr ) 0.28 10 lb m / hr 6 6 m 3 (1 y ) m 1 (1 0.200)(1.4 10 lb m / hr ) 1.12 10 lb m / hr 6 State 1 T (°F) 1100 p (psi) 6 2 3 4 5 6 7 1600 100 1 1 100 100 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 21 Example (8.44) continued: Find work from turbine 0 W turbine m 2 ( h1 h 2 ) m 3 ( h1 h3 ) W turbine m 2 ( h1 h 2 ) m 3 ( h1 h3 ) (0.28 10 lb m / hr )(1547.7 1210.7 ) B tu / lb m 6 (1.12 10 lb m / hr )(1547.7 911.2) B tu / lb m 6 94.36 10 712.88 10 807.24 10 B tu / hr 6 6 State 1 T (°F) 1100 p (psi) 6 2 3 4 5 6 7 1600 100 1 1 100 100 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 22 Example (8.44) continued: Find heat provided to pumps: Pump 1: 0 W pum p 1 m 3 ( h 4 h5 ) W pum p 1 m 3 ( h 4 h5 ) (1.12 10 lb m / hr )(69.74 70.04) B tu / lb m 0.336 10 B tu / hr Pump 2: 6 6 W pum p 2 m 1 ( h 6 h 7 ) (1.4 10 lb m / hr )(298.6 303.5) B tu / lb m 6.860 10 B tu / hr 6 6 State 1 2 3 4 5 6 7 T (°F) 1100 p (psi) 1600 100 1 1 100 100 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 23 Example (8.44) continued: therefore the Net Work W net W turbine W pum p 1 W pum p 2 807.24 10 6.860 10 0.336 10 800.0 10 B tu / hr 6 6 6 6 Plant thermal efficiency: W net 800 10 B tu / hr 6 Q in 1740 10 B tu / hr 6 State 1 T (°F) 1100 p (psi) 45.98% 2 3 4 5 6 7 1600 100 1 1 100 100 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 24 Example (8.44) continued: for the condenser: 0 m 3 ( h3 h 4 ) m cw ( h cw In h cw O ut ) m cw m 3 h3 h 4 h cw In hcw O ut m3 (1.4 10 lb m / hr ) 6 State 1 T (°F) 1100 p (psi) h3 h 4 c H 2 0 (T cw In Tcw O ut ) (911.2 69.74) B tu / lb m (1 B tu / lbm R )( 20 R ) 58.9 10 lb m / hr 6 2 3 4 5 6 7 1600 100 1 1 100 100 1600 s.h. s.h. 0.812 0.00 liq. 0.00 liq. h (Btu/lbm) 1547.7 1210.7 911.2 69.74 70.04 298.6 303.5 s (Btu/lbm K) 1.6315 1.6315 1.6315 0.1327 x 0.4744 Sec 8.2.4 : Principal Irreversibilities and Losses 25 NonIdeal Rankine Cycle. For real processes, there is some heat transfer with the surroundings, and thus an increase in entropy. In the turbine and pump, this can be quantified using the isentropic efficiencies. For Turbine: W T m real h1 h 2 T W T m S h1 h 2 S For Pump: P W P m S W P m real h 4 S h3 h 4 h3 But, for a real process, the inefficiencies associated with combustion are more significant. These, however, are external to the power cycle, so are not analyzed as part of the Rankine Cycle analysis. Cogeneration Systems Cogeneration systems provide both electrical work out and process steam or hot water for use by an end user which is often a commerical, industrial, or governmental user. Electricity provided to the community Steam exported to the community Cogeneration Systems --Exporting useful steam to the community limits the electricity that also can be provided from a given fuel input, however. --For instance, to produce saturated vapor at 100oC (1 atm) for export to the community water circulating through the power plant will condense at a higher temperature and thus at a higher pressure. --In such an operating mode thermal efficiency is less than when condensation occurs at a pressure below 1 atm, as in a plant fully dedicated to power production. T > 100oC p > 1 atm Sec 8.4.3 : Multiple Feedwater Heaters As closing comment: You now have most of the tools you would need to analyze some relatively complicated vapor power generation units and be able to discuss their respective efficiencies or how to suggest improvements in their performance... (not that you would want to). 28 29 End of Lecture 32 slides So ends Chap 8!