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APPLIED HERMODYNAMICS UORIAL REVISION OF ISENROPIC EFFICIENCY ADVANCED SEAM CYCLES INRODUCION his tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more advanced

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APPLIED HERMODYNAMICS UORIAL REVISION OF ISENROPIC EFFICIENCY ADVANCED SEAM CYCLES INRODUCION his tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more advanced level with ractical alications. Before you start this tutorial you should be familiar with the following. he basic rinciles of thermodynamics equivalent to level. Basic steam cycles, mainly the Rankine and Carnot cycles. Fluid roerty tables and charts mainly a set of standard thermodynamic tables and a h - s chart for steam which you must have in your ossession. he use of entroy. On comletion of the tutorial you should be able to understand isentroic efficiency for turbines and comressors. describe the use of rocess steam. describe the use of back ressure turbines. describe the use of ass out turbines. solve steam cycles involving ass out and back ressure turbines. describe the use of feed heating and suerheating in steam cycles. solve roblems involving feed heating and re-heating. You may be very familiar with all these studies in which case you should roceed directly to section. For those who wish to revise the basics, section should be comleted. his covers entroy. isentroic rocesses. roerty diagrams. isentroic efficiency. D.J.Dunn . REVISION OF ENROPY. DEFINIION Entroy is a roerty which measures the usefulness of energy. It is defined most simly as ds dq/d where S is entroy is temerature Q is heat transfer he units of entroy is hence J/k. he units of secific entroy are J/kg K... ISENROPIC PROCESS ISENROPIC means constant entroy. Usually (but not always) this means a rocess with no heat transfer. his follows since if dq is zero so must be ds. A rocess with no heat transfer is called ADIABAIC. An adiabatic rocess with no friction is hence also ISENROPIC... PROPERY DIAGRAMS he two most commonly used roerty diagrams are i. Enthaly - Entroy (h - s) diagrams and ii. emerature - Entroy ( - s) diagrams. h-s diagrams are commonly used for steam work. he diagram will hence show the saturation curve. You should familiarise yourself with the h-s diagram for steam and ensure that you can use it to find values of h and s for any ressure, temerature or dryness fraction. -s diagrams are commonly used for gas..4. ISENROPIC EFFICIENCY Real exansion and comression rocesses have a degree of friction and this will generate heat which is in effect a heat transfer. increase the entroy. make the final enthaly bigger than it would otherwise be. make the final temerature bigger than it would otherwise be if it is a gas or suerheated vaour. D.J.Dunn An adiabatic rocess with friction has no external heat transfer (ΦWatts or Q Joules) but the internal heat generated causes an increase in entroy. Consider the exansion and comression rocesses on fig. and. fig. fig. he ideal change in enthaly is h ' - h he actual change is h - h he isentroic efficiency is defined as h (actual) h h η is for an exansion. h (ideal) h h ' h (ideal) h ' h η is for a comression. h (actual) h h In the case of a erfect gas h c hence η is for an exansion ' ' η is for a comression Note that for an exansion this roduces a negative number on the to and bottom lines that cancels out. D.J.Dunn WORKED EXAMPLE No. A steam turbine takes steam at 70 bar and 500oC and exands it to 0. bar with an isentroic efficiency 0.9. he rocess is adiabatic. he ower outut of the turbine is 5 MW. Determine the enthaly at exit and calculate the flow rate of steam in kg/s. Note you need the tables and h-s chart for steam. SOLUION h 40 kj/kg (tables) s kj/kg K for an ideal exansion ss' Assuming that the steam becomes wet during the exansion, then s' sf +x'sfg. at 0. bar x' (tables) x' Note if x' is larger than then the steam is still suerheated and the solution does not involve x. Now find h'. h ' h f + x' hfg. at 0. bar h ' 9 + (0.896)(9) 5. kj/kg. Ideal change in enthaly actual change in enthaly h' kj/kg h 0.9(-57.5) -.7 kj/kg actual change in enthaly h (h - h ) -.7 h h 78. kj/kg From the steady flow energy equation (with which you should already be familiar) we have Φ + P H/s Since there is no heat transfer then this becomes P H/s m (h - h) P m(-.7) kw hence m 0.96 kg/s (Note the sign convention used here is negative for energy leaving the system) D.J.Dunn 4 WORKED EXAMPLE No. A gas turbine exands gas from MPa ressure and 600 o C to 00 kpa ressure. he isentroic efficiency 0.9. he mass flow rate is kg/s. Calculate the exit temerature and the ower outut. ake cv 78 J/kg K and c 005 J/kg K SOLUION he rocess is adiabatic so the ideal temerature ' is given by ' (r ) -/ r is the ressure ratio r / 0. c /c v.005/ '87(0.) -/ K Now we use the isentroic efficiency to find the actual final temerature. η is ( - )/( ' - ) 0.9 ( - 87)/( ) K Now we use the SFEE to find the ower outut. Φ + P m c ( - ) he rocess is adiabatic Φ 0. P (.005)( ) kw (out of system) D.J.Dunn 5 SELF ASSESSMEN EXERCISE No.. Steam is exanded adiabatically in a turbine from 00 bar and 600oC to 0.09 bar with an isentroic efficiency of he mass flow rate is 40 kg/s. Calculate the enthaly at exit and the ower outut. (Ans. 5 MW). A gas comressor comresses gas adiabatically from bar and 5oC to 0 bar with an isentroic efficiency of he gas flow rate is 5 kg/s. Calculate the temerature after comression and the ower inut. (Ans. -.5 MW) ake cv 78 J/kg K and c 005 J/kg K D.J.Dunn 6 . BACK-PRESSURE AND PASS-OU URBINES It is assumed that the student is already familiar with steam cycles as this is necessary for this tutorial. If an industry needs sufficient quantities of rocess steam (e.g. for sugar refining), then it becomes economical to use the steam generated to roduce ower as well. his is done with a steam turbine and generator and the rocess steam is obtained in two ways as follows. By exhausting the steam at the required ressure (tyically bar) to the rocess instead of to the condenser. A turbine designed to do this is called a BACK-PRESSURE URBINE. By bleeding steam from an intermediate stage in the exansion rocess. A turbine designed to do this is called a PASS-OU URBINE. he steam cycle is standard excet for these modifications... BACK-PRESSURE URBINES he diagram shows the basic circuit. he cycle could use reheat as well but this is not normal. Figure D.J.Dunn 7 WORKED EXAMPLE No. For a steam circuit as shown reviously, the boiler roduces suerheated steam at 50 bar and 400oC. his is exanded to bar with an isentroic efficiency of 0.9. he exhaust steam is used for a rocess. he returning feed water is at bar and 40oC. his is umed to the boiler. he water leaving the um is at 40oC and 50 bar. he net ower outut of the cycle is 60 MW. Calculate the mass flow rate of steam. SOLUION Referring to the cycle sketch revious for location oints in the cycle we can find: h 96 kj/kg s kj/kg K For an ideal exansion s s s f +x' sfg at bar x'(5.) x' 0.95 h4 hf + x'hfg at bar h (64) h kj/kg ideal change in enthaly actual change in enthaly kj/kg 0.9(-6) kj/kg he ower outut of the turbine is found from the steady flow energy equation so : P m(-550.9) kw P m kw (outut) Next we examine the enthaly change at the um. h 68 kj/kg at bar and 40 o C h 7 kj/kg at 50 bar and 40oC. Actual change in enthaly 7-69 kj/kg he ower inut to the um is found from the steady flow energy equation so : P -m() kw P - m kw(inut) Net Power outut of the cycle 60 MW hence m - m m 09.5 kg/s D.J.Dunn 8 SELF ASSESSMEN EXERCISE No. A back ressure steam cycle works as follows. he boiler roduces 8 kg/s of steam at 40 bar and 500oC. his is exanded to bar with an isentroic efficiency of he um is sulied with feed water at 0.5 bar and 0oC and delivers it to the boiler at oc and 40 bar. Calculate the net ower outut of the cycle. (Answer 5.4 MW) D.J.Dunn 9 .. PASS-OU URBINES he circuit of a simle ass-out turbine lant is shown below. Steam is extracted between stages of the turbine for rocess use. he steam removed must be relaced by make u water at oint 6. Figure 4 In order to solve roblems you need to study the energy balance at the feed ums more closely so that the enthaly at inlet to the boiler can be determined. Consider the ums on their own, as below. he balance of ower is as follows. Figure 5 P + P increase in enthaly er second. mchc - maha -mbhb From this the value of h C or the mass m C may be determined. his is best shown with a worked examle. D.J.Dunn 0 WORKED EXAMPLE No.4 he circuit below shows the information normally available for a feed um circuit. Determine the enthaly at entry to the boiler. SOLUION h A h f 9 kj/kg at 0. bar Figure 6 P(ideal) (5)(0.00)(80-)(00) 9.5 kw P (actual) 9.5/ kw P(ideal) (40)(0.00)(80-0.)(00) 9.6 kw P (actual) 9.6/ kw otal ower inut kw hb 84 kj/kg (from water tables or aroximately hf at 0 o C) hence h C - 40 h A - 5h B hc - 40(9) - 5(84) hc 90 kj/kg D.J.Dunn WORKED EXAMPLE No.5 he following worked examle will show you to solve these roblems. A assout turbine lant works as shown in fig. 4. he boiler roduces steam at 60 bar and 500 o C which is exanded through two stages of turbines. he first stage exands to bar where 4 kg/s of steam is removed. he second stage exands to 0.09 bar. he isentroic efficiency is 0.9 for the overall exansion. Assume that the exansion is a straight line on the h - s chart. he condenser roduces saturated water. he make u water is sulied at bar and 0oC. he isentroic efficiency of the ums is 0.8. he net ower outut of the cycle is 40 MW. Calculate:. the flow rate of steam from the boiler.. the heat inut to the boiler.. the thermal efficiency of the cycle. SOLUION URBINE EXPANSION h 4 kj/kg from tables. h 5 ' 65 kj/kg using isentroic exansion and entroy. 0.9 (4 - h5) / (4-65) hence h5 9 kj/kg Sketching the rocess on the h - s chart as a straight line enables h4 to be icked off at bar. h4 770 kj/kg. POWER OUPU Pout m(h - h4 ) + (m - 4)(h4 - h5) P out m(4-770) + (m - 4)(770-9) P out 65 m m - 96 Figure 7 D.J.Dunn POWER INPU he ower inut is to the two feed ums. Figure 8 h 6 84 kj/kg (water at bar and 0oC) h h f at 0.09 bar 8 kj/kg. P (ideal) change in flow energy 4 x 0.00 x (60 - ) x 00 kw.6 kw P (actual).6 / kw P(actual) (m-4) x 0.00 x ( ) x 00/ m kw NE POWER ENERGY BALANCE ON PUMPS kw P out - P - P m m m m - 96 hence m 7.4 kg/s hence HEA INPU P 9.5 kw P 49.4 kw (using the value of m just found) m h (m-4) h + P + P 7. h.4 x h 7 kj/kg Heat inut m(h - h) 55 kw EFFICIENCY Efficiency η 40/. % D.J.Dunn SELF ASSESSMEN EXERCISE No.. A steam turbine lant is used to suly rocess steam and ower. he lant comrises an economiser, boiler, suerheater, turbine, condenser and feed um. he rocess steam is extracted between intermediate stages in the turbine at bar ressure. he steam temerature and ressure at outlet from the suerheater are 500oC and 70 bar, and at outlet from the turbine the ressure is 0. bar. he overall isentroic efficiency of the turbine is 0.87 and that of the feed um is 0.8. Assume that the exansion is reresented by a straight line on the h-s chart. he makeu water is at 5oC and bar and it is umed into the feed line with an isentroic efficiency 0.8 to relace the lost rocess steam. If due allowance is made for the feed um-work, the net mechanical ower delivered by the lant is 0 MW when the rocess steam load is 5 kg/s. Calculate the rate of steam flow leaving the suerheater and the rate of heat transfer to the boiler including the economiser and suerheater. Sketch clear - s and h-s and flow diagrams for the lant. (9.46 kg/s 95. MW). he demand for energy from an industrial lant is a steady load of 60 MW of rocess heat at 7 o C and a variable demand of u to 0 MW of ower to drive electrical generators. he steam is raised in boilers at 70 bar ressure and suerheated to 500oC. he steam is exanded in a turbine and then condensed at 0.05 bar. he rocess heat is rovided by the steam bled from the turbine at an aroriate ressure, and the steam condensed in the rocess heat exchanger is returned to the feed water line. Calculate the amount of steam that has to be raised in the boiler. Assume an overall isentroic efficiency of 0.88 in the turbine. he exansion is reresented by a straight line on the h-s diagram. Neglect the feed um work. (Answer 6 kg/s). D.J.Dunn 4 . ADVANCED SEAM CYCLES In this section you will extend your knowledge of steam cycles in order to show that the overall efficiency of the cycle may be otimised by the use of regenerative feed heating and steam re-heating. Regenerative feed heating is a way of raising the temerature of the feed water before it reaches the boiler. It does this by using internal heat transfer within the ower cycle. Steam is bled from the turbines at several oints and used to heat the feed water in secial heaters. In this way the temerature of the feed water is raised along with the ressure in stages so that the feed water is nearly always saturated. he heat transfers in the heaters and in the boiler are conducted aroximately isothermally. Studies of the Carnot cycle should have taught you that an isothermal heat transfer is reversible and achieves maximum efficiency. he ultimate way of conducting feed heating is to ass the feed water through a heat exchanger inside the turbine casing. In this way the temerature of the steam on one side of heat exchanger tubes is equal to the temerature of the water on the other side of the tubes. Although the temerature is changing as water and steam flow through heat exchanger, at any one oint, the heat transfer is isothermal. If no suerheating nor undercooling is used then the heat transfers in the boiler and condenser are also isothermal and efficiencies equal to those of the Carnot cycle are theoretically ossible. Figure 9 here are several reasons why this arrangement is imractical. Most of them are the same reasons why a Carnot cycle is imractical. i. he steam would be excessively wet in the turbine. ii. Placing a heat exchanger inside the turbine casing is mechanically imossible. iii. he ower outut would be small even though the cycle efficiency would be high. D.J.Dunn 5 Steam reheating is another way of imroving the thermodynamic efficiency by attemting to kee the steam temerature more constant during the heat transfer rocess inside the boiler. Figure 0 Suerheated steam is first assed through a high ressure turbine. he exhaust steam is then returned to the boiler to be reheated almost back to its original temerature. he steam is then exanded in a low ressure turbine. In theory, many stages of turbines and reheating could be done thus making the heat transfer in the boiler more isothermal and hence more reversible and efficient. If a steam cycle used many stages of regenerative feed heating and many stages of reheating, the result would be an efficiency similar to that of the Carnot cycle. Although racticalities revent this haening, it is quite normal for an industrial steam ower lant to use several stages of regenerative feed heating and one or two stages of reheating. his roduces a significant imrovement in the cycle efficiency. here are other features in advanced steam cycles which further imrove the efficiency and are necessary for ractical oeration. For examle air extraction at the condenser, steam recovery from turbine glands, de-suerheaters, de-aerators and so on. hese can be found in details in textbooks devoted to ractical steam ower lant. D.J.Dunn 6 4. FEED HEAING 4.. PRACICAL DESIGNS Practical feed heaters may be heat exchangers with indirect contact. he steam is condensed through giving u its energy and the hot water resulting may be inserted into the feed system at the aroriate ressure. he tye which you should learn is the oen or direct contact mixing tye. he bled steam is mixed directly with the feed water at the aroriate ressure and condenses and mixes with the feed water. Comare a basic Rankine cycle with a similar cycle using one such feed heater. Figure Figure D.J.Dunn 7 4.. ENERGY BALANCE FOR MIXING FEED HEAER Consider a simle mixing tye feed heater. he bled steam at () is mixed directly with incoming feed water (6) resulting in hotter feed water (7). Figure Mass of bled steam y kg Mass of feed water entering - y kg Doing an energy balance we find y h + (-y)h 6 h 7 WORKED EXAMPLE No.6 A feed heater is sulied with condensate at 0. bar. he bled steam is taken from the turbine at 0 bar and 0.95 dry. Calculate the flow rate of bled steam needed to just roduce saturated water at outlet. SOLUION Assumtions Figure 4. Energy inut from um is negligible.. No energy is lost..he heater ressure is the same as the bled ressure. In this case h6 hf at 0. bar 9 kj/kg h 7 h f at 0 bar 008 kj/kg h hf + xhfg at 0 bar h (795) 7. kj/kg ENERGY BALANCE y(7.) (-y)(9) hence y 0.44 kg Note that it is usual to calculate these roblems initially on the basis of kg coming from the boiler and returning to it. D.J.Dunn 8 4.. CYCLE WIH ONE FEED HEAER Figure 5 If only one feed heater is used, the steam is bled from the turbine at the oint in the exansion where it just becomes dry saturated and the saturation temerature is estimated as follows. t s (bleed) {t s (high ressure) + t s (low ressure)}/ For examle a cycle oerating between 40 bar and 0.05 bar. t s (40 bar ) 50. oc ts (0.05 bar ) 6.7 o C ts (bleed ) ( )/ 8.5 o C he ressure corresonding to this is.5 bar so this is the bleed ressure. D.J.Dunn 9 WORKED EXAMPLE No.7 A Rankine cycle works between 40 bar, 400oC at the boiler exit and 0.05 bar at the condenser. Calculate the efficiency with no feed heating. Assume isentroic exansion. Ignore the energy term at the feed um. SOLUION Figure 6 h 4 kj/kg s kj/kg K ss x x h h f + x hfg (48) 04.6 kj/kg h 4 h f at 0.05 bar kj/kg Φ h - h 0 kj/kg into boiler. P h - h 89.4 kj/kg (out of turbine) η P/ Φ 8. % D.J.Dunn 0 WORKED EXAMPLE No.8 Reeat the last examle but this time there is one feed heater. SOLUION Figure 7 he bleed ressure was calculated in an earlier examle and was.5 bar. s s kj/kg K x x (not quite dry). h h f + x hfg (48) h 66 kj/kg h 7 h f at.5 bar 584 kj/kg Neglecting um ower h6 h5 hf kj/kg h h kj/kg Conducting an energy balance we have yh + (-y) h 6 h 7 hence y 0.85 kg Φ h - h 60 kj/kg into boiler. Rather than work out the ower from the turbine data, we may do it by calculating the heat transfer rate from the condenser as follows. Φout (-y)(h4 - h5) 0.85( ) kj/kg P Φin - Φout 07 kj/kg (out of turbine) η P/ Φin 40.8 % Note that the use of the feed heater roduced an imrovement of.5 % in the thermodynamic efficiency. D.J.Dunn SELF ASSESSMEN EXERCISE No.4 A simle steam lant uses a Rankine cycle with one regenerative feed heater. he boiler roduces steam at 70 bar and 500oC. his is exanded to 0. bar isentroically. Making suitable assumtions, calculate the cycle efficiency. (4.8%) CYCLE WIH WO FEED HEAERS When two (or more) feed heaters are used, the efficiency is further increased. the rinciles are the same as those already exlained. he mass of bled steam for each heater must be determined in turn starting with the high ressure heater. It is usual to assume isentroic exansion that enables you to ick off the enthaly of the bled steam from the h-s chart at the ressures stated. D.J.Dunn WORKED EXAMPLE No.9 A steam ower lant works as follows. he boiler roduces steam at 00 bar and 600oC. his is exanded isentroically to 0.04 bar and condensed. Steam is bled at 40 bar for the h.. heater and 4 bar for the l.. heater. Solve the thermodynamic efficiency. SOLUION Figure 8 Figure 9 Ignoring the energy inut from the um we find: h h0 hf 40 bar 087 kj/kg h9 h8 hf 4 bar 605 kj/kg h7 h6 hf 0.04 bar kj/kg H.P. HEAER xh + (-x)h 9 h 0 0x + 605(-x) 087 hence x 0.78 kg D.J.Dunn L.P. HEAER (-x)h8 yh4 + (-x-y)h7 0.8(605) 740y + (0.8-y)() y 0.5 kg BOILER heat inut Φin h - h kj/kg CONDENSER heat outut Φout (-x-y))(h5 - h6 ) Φout 0.67(080-).5 kj/kg POWER OUPU P Φin - Φout 4.5 kj/kg η P/ Φin 48. % D.J.Dunn 4 SELF ASSESSMEN EXERCISE No. 5. Exlain how it is theoretically ossible to arrange a regenerative steam cycle which has a cycle efficiency equal to that of a Carnot cycle. In a regenerative steam cycle steam

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