Energy Systems Analy&Dgn
Energy Systems Analy&Dgn ME 4315
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Metals mks THE CONSERVATION NONCONSERVATION LAWS USED BY THE MECHANICAL ENGINEER Dennis Ballou INTRODUCTION The purpose of this paper is to examine the various laws underlying mechanical engineering from two viewpoints that of a system and that of a control volume A further purpose is to restate a generalized method the Reynolds Transport Theorem for transforming the laws from their formulation in terms of a system to that equally applicable to a control volume First let us be clear about the de nitions of system and control volume A system1 sometimes called closed system is a collection of identi able masses whose composition remains constant over time Mass cannot cross the boundary of a system although energy might The system boundary however is subject to change eg movement of a piston in a cylinder If we wish to rrther restrict the system to prevent energy from crossing the boundary an extremely use rl concept for modeling mechanical engineering systems then we refer to the system as an isolated system We can model an entire steam power plant as a system because the mass of the water circulating through the system does not change with time If we increase the system boundary to include a fuel tank a portion of the atmosphere to provide combustion air a portion to receive the ue gas and a portion of say a lake to provide cooling water then we have an isolated system that is enough isolated for useful engineering calculations But many engineering components such as turbines and compressors are open with mass constantly crossing the boundary These components are analyzed as control volumes or open systems A control volume or open system is an identi able space whose boundaries remain xed We should get used to both de nitions as they are used throughout our literature A problem is immediately presented however because our fundamental laws are formulated with respect to systems We need therefore a method for transforming the laws from applicability to a system to that of a control volume The Reynolds Transport Theorem is the tool for accomplishing this task Let us rst tabulate our laws in a form useful for analysis After Munson let the variable B stand for the parameters further de ned in the table to follow We note that B is an extensive property pertaining to the entire system control volume The parameter b is the corresponding intensive property such that B mb 1 where m is the mass of the system and B Ipr rys CV I See eg Munson p 184 2 See eg Black and Hartley p 3 A dm dt drh pI7 dA pI7dA7 Figure 1 Control Surface an dA exiting mass ow or We see from the gure and the de nition of 393 that outward mass ow occurs when 90 lt 0 lt 90 Furthermore because the cosine of angles within this range is positive outward ow rate is a positive quantity On the other hand inward ow rate occurs for angles greater than 90 and less than 90 90 lt G lt 90 Thus inward ow rate is a negative quantity Now consider transport across the boundary of any extensiVe property We express this transport by multiplying the intensive property by the mass ow rate For the di erential flow rate the transport of a property may be generalized by the expression dBbdmprdZ For transport into and out of the entire control surface we have B I b p17 d2 2 cs or if we assume that properties across the ports are constant so that they can be placed outside the integral and if we postulate a nite number of ports then we can write B an39z an z 3 at in Recall that ow into the control volume is negative and ow out positive Finally let us consider the transformation 39om the system viewpoint to that of the control volume at the point in time when their boundaries coincide We note that the system passes into through and out of the control volume over time We perceive then that at coincidence the time rate of change of B for the system is equal to the time rate of change of B for the control volume plus in the limit the instantaneous transport of B across the coincident boundaries In equation form we have DB y 53C Ipr dA 3 4 Dt 6t CS By convention the capital D is used to denote the rate of change from the system viewpoint It is the total derivative as distinguished from the rst term on the right which is a partial derivative The rst term on the right pertains to changes in the control volume with time If nothing is changing such as pressure or temperature then that term is zero and the only active term is that which accounts for the mass transport A control volume for which the rst term is zero is said to be in steady state Consider a steam power plant operating in steady state at a particular power level Temperatures pressures and ow rates as well as other properties at various locations throughout the plant are invariant Now suppose that the load demand increases To accommodate the demand the throttle valve is opened to increase the mass ow rate For a short interval of time property values throughout the system will experience uctuations until the plant settles down again in steady state It is during this and other transitory periods such as initially lling the plant with water that the rst term is operative CONSERVATION 0F MASS We start with equation 1 From the table b l and 1 0 Substituting in the transport equation 4 gives Dmsrs 077 Dr pfd2 0 CS For a control volume in steady state with a nite number of entrances and exits each of which experience constant densities and velocities across their areas we can write see equation 3 Zm Zrh0 5 where n39t pVA CONSERVATION 0F LINEAR MOMENTUM Dml7m 0117117 TTC VpVaAZFCV ZFSYS or for a nite number of ports steady state and steady ow constant properties across the ports referred to hereinafter as SEV Standard Engineering Version ZFCV Zn zI7 Zml7 6 3 See Manson p 193 4 See Equation 316 Black amp Hartley CONSERVATION OF AN GULAR MOMENTUM DFxm I7m 57xm 17 Dr 6 I017 41A Zermw gm cs or for the SEV with one entrance and one exit Tm rhrxr7w m7m 7 For rotary machines pumps turbines compressors only the tangential component of I7 contributes to torque therefore we may write Jsmn merez eral 8 where the sign of the product rV 9 is determined by the right hand rule r x V If the thumb points in the direction of the angular velocity the product is positive The numerals 1 and 2 refer respectively to the entrance and exit ports For shaft power 73mm Also rm U the tangential velocity at radius r thus Wsm mU2 V02 quot U V01 9 where the product UV 9 is positive when both U and V9 are in the same direction determined by the right hand rule On a unit mass basis WSHAFI39 UzVoz quot UlVoi 5 with signs of the products determined as above CONSERVATION OF ENERGY DE DE Ti C IepVdAQ WCV QWsys CS or for the SEV Q WZn39ze Zrhe 11 5 See Equation 554 Munson Metals holy HEAT TRANSFER AND HEAT EXCHANGERS By Dennis Ballou INTRODUCTION The engineer is frequently called upon to transfer energy from one uid to another The transfer is most often effected by a heat exchanger A steam power plant for example includes several major heat exchangers boiler superheater economizer and condenser It also includes minor heat exchangers such as the gland exhaust condenser and the lube oil cooler The principal heat exchangers in an air conditioning system are the evaporator and the condenser Other applications abound throughout the mechanical industry THEORETICAL CONSIDERATIONS AND THE HEAT TRANSFER PROBLEM The typical heat transfer problem involves energy transfer from a uid stream at an assumed average temperature through a series of resistances to a destination uid stream also at an assumed average temperature For transfer through a pipe wall the resistances consist of in order a thermal boundary layer TBL a scale or soot layer the pipe wall another scale layer and the destination TBL Each TBL is characterized by a convection or lm coef cient designated h Newton s Law of Cooling models heat transfer through the TBL qquot 11Ta T where q quot heat ux or rate of heat transfer per unit area Btu h z h convection lm coef cient Btu h fz 2 R To average temperature of the uid stream T surface temperature of the pipe Fourier s Law governs transfer by conduction through the pipe wall nk dr where k thermal conductivity Btu ft h R The negative sign results from the decreasing temperature gradient Integration yields MILO T2 q 1nr2r1 where q heat transfer rate Btu h L length of pipe through which heat transfer occurs ft An experimentally determined fouling factor R ft2 h 0 R Btu accounts for scale resistancel THE OVERALL HEAT TRANSFER COEFFICIENT Heat transfer through the various layers can be represented by an overall heat transfer coe z39cz39ent U such that q UAD where A heat transfer surface area based on either the inside or outside pipe surface U overall heat transfer coefficient BtuhftZR D log mean temperature difference derived below In analogy with electrical resistance we can express the Newton and Fourier laws as T1 T 2 Ii quot T 2 qconv Room qcond Room where 1 111r r Room R 21 hA 0quot anL We can then write T002 T001 Tooz R11nquot2 1 R2 1 1 See eg Incropera p 585 for representative fouling factors Selecting either the inside or outside area as a basis factoring out IA for example and letting A1 ZarlL we have qAUD 1 U 1121 quot1 moir1 1122 i hl k r2 5 LOG MEAN TEMPERATURE DIFFERENCE What temperature difference is to be used for D Because the temperature varies on each side of the heat exchanger as heat transfer occurs we must derive a sort of average temperature difference Consider a counter ow heat exchanger which we represent as shown in the gure below Let T temperature of the hot uid t temperature of the cold uid h hot uid c cold uid C heat capacity rate the i entrance condition 0 outlet condition D log mean temperature difference Ch qqc qh qh Chg T qc Ccto t dqCthCcdtUATdA dq dq 1 l A dT dt UATdA d cm a c h c h AT ln 2 T a r A11 A For multipass and cross ow heat exchangers the expression for D is modi ed by a multiple F which depends upon heat exchange geometry number of tube passes and whether the uid is mixed ow transverse to the uid direction is not prevented or unmixed transverse ow ix prevented by ns or separate tubes The F factor can be found in the technical 2 literature 40 v 400 9quotquot Wu K Ma39s EXTENDED SURFACES FIN S Extended surfaces or ns are frequently used to increase the heat transfer rate Fins can be justi ed economically when n effectiveness de ned by the below approximate equation is greater than about 2 where k thermal conductivity P perimeter of n h convection coef cient Ac cross sectional area of n 2 See eg Incropera and DeWitt 4 ed pp 59294 mu 4 I 211 MH a v M M H adv wyo 0quot v 9vI 42 mn iv WI Ah 9 l mm M wrmod z amp 91x p mbmw M r 3 o sw to W I Q fgly 9 39 Ofquot Because the temperature varies from the base of the n to its tip an adjustment is made to the total heat transfer surface area which consists of the uncovered base area plus the n area Multiplying the total area A by an overall surface efficiency 1103 which is less than unity gives NA noqt q t qmax hAtgb A where N number of ns Af area of a single n A total area including base 71 n efficiency q total heat transfer rate from A h convection coef cient 0 T Tao T b base temperature Tao bulk temperature Fin ef ciency is a mction of n geometry convection coef cient and thermal conductivity Values of n efficiency for various shapes can be found in the technical literature4 Having found both n and overall surface ef ciency the overall heat transfer coefficient is modi ed by multiplying each area n plus uncovered base by the overall surface ef ciency This step results in the following relation for U l U 1 Zirllnr2r r R r 7701quot 7701 k 7392 7702 770272 12 From the de nition of n efficiency it is deduced that ns with a high thermal conductivity low ratio of perimeter to cross sectional area and small convection coef cient are desirable Because a small convection coef cient conduces to n performance ns are frequently placed on the gas side of a gastoliquid heat exchanger Automobile radiators and air conditioning evaporators and condensers are common examples Fins are often placed on economizers in boiler designs 3 See Incropera Section 365 4 See eg Incropera p 123 ME H313 Me Application of Psychrometrics to Heating and Cooling There are several cases of interest which we will consider1 First we can heat or cool the air without changing the moisture content of the air We could heat with humidi cation or cool with dehumidi cation We can use evaporative cooling adiabatic mixing of airstreams and will look at cooling tower processes Humidi cation means that we are adding moisture content to the air and dehumidi cation means that we are removing moisture content from the air 1 Heating and Cooling Simple heating and cooling simply involve changing the dry bulb temperature of the moist air without changing the moisture content The humidity ratio remains unchanged and so we use a horizontal line on the psychrometric chart to represent this process This provides no control over the relative humidity and may produce conditions that are uncomfortable Heating will result in lower relative humidity and cooling will result in a higher one 2 Cooling with Dehumidification We can use dehumidi cation to get rid of some of the moisture in the air which will lower the relative humidity This can be accomplished by lowering the air temperature below the dew point and letting some of the moisture condense out Typically the moisture condenses on the outside of the refrigerant tubing This process is assumed to occur as simple cooling rst and then condensation While the moisture is condensing the air is assumed to remain saturated The air is assumed to leave the cooling section at the nal dewpoint temperature The process as seen on the psychrometric chart is shown below 100 quot39 100 I l 4 T3 T2 T1 b Process Representation for Cooling and Dehum di cation Figure 117 Cooling with dehumidi cation 1 Black W2 and Hartley JG Themwdynamics Harper Collins 1995 To analyze this rst we realize that the mass of the dry air does not change throughout this process The water mass does change however The conservation of mass for water gives that 03 quot1413 me 01 mill or because ma1 mag mwz 01 12371391a An energy balance gives no work Q Ihaa ha rith hwz ma h1 75111013 hl quot39 ma 01 w3hw2 3 Heating with humidi cation Adding humidi cation is often necessary to make a heated space comfortable One example of humidi cation is the adiabaticsaturation process when air is passed over a moist surface to gain moisture by evaporation In this case the dry bulb temperature could be increased or decreased depending on the temperature of the evaporating water This analysis is very similar to that used for the dehumidi cation and is left to the student or their textbooks l 4 Evaporative Cooling Here the energy removed from the warm air is used to evaporate water This method adds moisture to the air stream while cooling it and so would be useful in regions where the warm air has a low humidity This process is considered to be adiabatic because little heat is lost to the environment Again we can apply conservation of mass of air and water and conservation of energy We nd that the mass of water added is just the mass of air times the difference in humidity ratio identical to that for a humidi er The difference is that here the water is liquid and has low enthalpy therefore not changing that of the air stream much Thus this process is considered constant enthalpy or constant wet bulb temperature 5 Adiabatic Mixing Airconditioning processes often require the mixing of two air streams especially in the case where outside fresh air is mixed with the return air in the system The two streams are assumed to have different dry bulb temperatures and different humidity ratios Usually this mixing is adiabatic with regard to heat transfer to the surroundings We can find the conditions of the mixed air stream by again looking at a mass balance of the air and water For the air quot1111 maz ma3 and for the water wlrha1 02 maz mamas We can ellmlnate rim to get 03 w 771111 1112 We can also obtain E M maS hl hz Note that condensation can occur sometimes when the two streams are mixed 6 Wet Cooling Tower A wetcooling tower is a way of cooling down water that has been used to remove heat such as in a power plant This might be water that must be returned to a river or lake or perhaps the water must be reused Below is a schematic of a wetcooling tower Warmair exit From condenser System bonndary Atmospheric air Makeup water To condenser Figure 1112 Schematic of a wetcooling tower The water is cooled by evaporation into the air in the tower The air continuously ows upward through the tower Some water is carried off by the air stream from evaporation and must be replaced Here the mass of the air is unchanged as is passes through the tower The mass of the condenser water leaving is usually the same as at the entrance That is m3m4mw N 343IS holes NOZZLE EQUATIONS FLOW OF STEAM IN NOZZLES NATURE OF PROCESS IN AN IDEAL NOZZLE FUNDAMENTAL ENERGY RELATION Assume Frictionless Adiabatic Flow In nite Reservoir vaoJBJB Ln 314 r Neglecting potential energy conservation of energy requires V32 V12 ho 2 hl 2 V2 V2 47L where V absolute velocity ho h1 Ahs Available energy Now consider the reservoir stagnation condition VR 0 2 hR0hoV 2 2 K29 hR hoAhs and hRhaV7 VELOCITY OF FLOW and QUANTITY DISCHARGED If be neglected V12 ho h1 1 Adapted om E F Church Steam Turbines McGrawHill 1950 V1 i hl 2238lho h1 s the constant includesJand gc with an approach velocity V2 V 2238 h quot l h 22382 Volume discharge from equation 1 Q AV 2238Ah0 h1 7h 2238 hO hl v1 V1 Mass Velocity G 752 2238 G v1 ho hl To allow for losses a velocity coefficient kn is introduced V 2238kmh0 hl QAVn391v AJ X V v Now for constant ow the area will depend on the ratio V As the steam expands through the nozzle both v and Vwill increase Calculations for Nozzle with Ideal Frictionless Steam Flow Start with steam at 100 psia and 440 F expanding through a nozzle to P 90 psia and let m llbm s h0 12486Btulbm so 16756Btulbm 0R y 40 Q x4 h 12382Btulbm I 418 F v1 562 3 lbm V K i Ahs 104 Btulb Available energy VI 2238x104 721 fts 11bm 562 ft3 5 144nz 2 s lbm 721 ftz 1122 A1 If the calculation is repeated using a drop from 100 to 80 psia A will found to be 08641712 Refer to the following table for additional calculations P1 100 90 80 70 60 55 50 40 30 20 10 2 1 M 12486 12382 12275 1215 12018 11942 11865 11689 11471 11175 10698 10261 973 936 del hs 0 104 211 336 468 544 621 797 1015 1311 1788 2225 2756 3126 C1 0 721 1028 1296 1529 1649 1763 1996 2253 2561 2991 3336 3714 3956 NOZZLE CALCULATIONS v1 A in2 D in 52 562 1122 1195 616 0864 1049 68 0756 0981 767 0722 0959 82 0716 0955 881 072 0958 1045 0755 098 135 0863 1049 1924 1082 1174 3567 1718 1479 656 283 1899 149 577 2711 279 1017 36 MININIUM SECTION OR THROAT OF NOZZLE Variations in Nozzle Area and Form Variations in Flow Consider a valve in the exit line As the valve is opened P will drop and 752 will increase until P1 N 55 psia At that point a maximum ow is reached which is not exceeded no matter how low the nal pressure is made The ow will always reach a maximum at P1 R E 055 These statements however do not apply completely if there is a divergent section beyond the minimum diameter or if the entrance is not well rounded For an ori ce plate the ow continues to increase as P1 is lowered The critical pressure ratio is the value of P1 P0 below which no further increase of quot391 will occur Flow of an Ideal Gas Critical Pressure Rate 10 rP1Po 05 kaC V Our problem is now to nd 739 I for which m I is maximum v1 2 uo voV ulRv1I 2 2 LetVo 0 V12 7uo ui39l39 Vo Iiv1 Noting that 11 u1 is equivalent to isentropic work W0 151 Viz Invoiv1 k Pv Pv P Pv 2 kl o o 11 kl 0V0 11 k1 Rm 1 I p 2 11 but 1 5 and r 1 therefore Vl k IZVO l r v0 1 Po 2 k l 3 a 2 a 2k Prquot rquot Aconst rquot rquot J 1 kl quot 2 k1 2er1 kZ1rT 1 0 For air k141 rm 0527 For superheated steam k E 128 r 0546 For dry saturated steam k 2 1135 r E 0577 Critical Velocity 2 KL k ve l rquot 2 k l V2 k ic Pv 2 k1 V 2quot av 2quot R71 k1 k1 Ifk14 and R5334 12 11me R VB 4475 7 Velocity of Sound k k For an adiabatlc process Pave Povo therefore 1 k PC k1 Povo P v P v rcr 1 P v then substitutinginthe above or or P cr or or or 2 0 expression for critical velocity VCI ngrvcr V kRIL r which is the expression of the velocity of sound in an ideal gas V is often called the acoustic or sonic velocity For air Vc 4902 Tc For superheated steam Vc E 596 Tc Useful Relations For superheated steam and approximately for saturated steam P mm 2 03144146 v for 1 30547130 0 For wet steam P mm 030444 for Pl s 057710 rough estimate V0 For superheated steam 04104o m ax m m All of the foregoing equations hold where r lt critical For r gt critical lit for steam with k13 is m 13914 ffiorm r177o o Nozzle Velocity Coef cient kn and Q n39wo 0242Aquot E The following empirical formula based on experimental results was adapted 39om Church2 kn 102 0164x 0165362 00671x3 000884 where x V31 1000 Vl ans DESIGN OF IDEAL STEAM NOZZLES Convergent Nozzle n39z 160bm sPo l30psiag 106psia l 420quotFs0 16309BtulbmquotR PC 130 0546 71 Opsz39a which is less than 106 psia therefore a convergent nozzle is indicated ho h1 12335 12151 184Btulbm E available energy V51 2238V184 960 ft S E isentropic velocity 2 Id p 82 Fig 79 m1 1604508 96 10821312 E isentropic area A S 51 ConvergentDivergent Nozzle At the throat P0 130 psiaB P6 71 psia s0 51 16309Btulbm quotR Ahs ho hs 12335 11812 523Btulbm K 2238J523 1618 s A W0372m2 1618 At the exit Let P1 16 psiaP P0 71 0psiavs 22703 3 lbm Ahs 12335 107161619Btulbm V1 2238sl619 2848fts S W 183739 2 1 2848 m EFFECT OF NOZZLE FRICTION ON THERMODYNAMIC PROPERTIES 1 Entropy is increased 2 Available energy is decreased 3 Velocity of ow at throat is decreased 4 Volume of owing steam is decreased 5 Throat area necessary to discharge a given mass of steam is increased DESIGN OF ACTUAL NOZZLES Extending the previous calculations Convergent Nozzle K 960 fts kn 0965 VI VmE 2238knAhs 22380965J184 926 fts h1 ho 77nAhs 12335 09562 184 12164Btulbm A 1606192144 1 39 2 1 926 1 3m ConvergentDivergent Nozzle At the throat Vs 960 k o965 V 22380965 J523 1562 ft s h 12335 O9652 523 118SBtulbm 1606192144 091339 2 1562 m A At the exit Vs 2848fts k 0922 VI 2238 0922 J16 9 2626 ft s hl 12335 09222 1619 1096Btulbm A 1602333144 1 2051712 1562 STEAM TURBINE DESIGN3 CONVERSION OF KINETIC ENERGY OF THE GASSTEAM INTO BLADE WORK Consider a frictionless blade that turns the steam through 180 and exits with zero absolute velocity This condition represents the greatest possible conversion of kinetic energy of the entering jet into blade work We proceed to develop a relation between the absolutejclocity of the jet entering the blade V and the blade speed V1 For a given blade speed this relation will permit us to design a nozzle such that the exiting velocity will provide for maximum energy conversion or in different words maximum ef ciency V Wl Let W be the veloc1ty of the Jet relative to the blade The positive direction is to the right Vb Kmn VWV VL J Because the blade is frictionless W2 Wl Furthermore because energy conversion in the blade is complete V2 0 Substituting and combining equations we get V1 K W 39Ws 2V VI 2V 1 As we shall see later the centrifugal force of rotation and the strength of the blade material limit the blade speed Given the blade speed however we can determine the ideal absolute velocity entering the blade ACTUAL NOZZLE ANGLE We must now modify this result to account for the geometry restrictions of a real turbine In our derivation the acute angle between V and the tangential direction called the nozzle angle is zero In an actual turbine because of physical constraints the nozzle angle must be greater than zero but not so great as to cause an appreciable loss in ef ciency Nor should the angle be so small as to cause an excessively long nozzle that would increase friction and decrease ef ciency quotThe values used in practice range om 10 to 30 deg 12 to 20 deg being common The larger angles are used only when necessary and usually at the lowpressure end of large turbines quot4 Equation 1 corrected for a nite nozzle angle a becomes HIM xx j V1 cosa 2Vb 2 M Because of disk friction and farming losses V1 is usually increased somewhat say 10 over the theoretical value 3 Adapted from E F Church Jr Steam Turbines McGrawHill 1950 4 Id p91 BLADE WORK AND POWER First write the Reynolds transport theorem for angular momentum DFxmi7sys a 7me CV r Dt at LSrxVpVdAZerTW Assuming steady state and steady ow with one entrance 1 and one exit 2 the equation reduces to TShai h72xV271x11 For the turbine blade the mean radius is constant between entrance and exit Furthermore the tangential component of velocity is the only contributor to torque The radial and axial components affect bearing loads but have no effect on torque thus TSWt 17er rV1 The shaft work then is WSha T meraz Vin But Vb cor therefore WSW Ith VB2 V191 3 On a unit mass basis WW Vb n2 V91 4 This result is most easily visualized by constructing entering and leaving velocity triangles IMPULSE BLADING VELOCITY TRIANGLES AND BLADE WORK Having determined blade speed 39om strength considerations nozzle angle om fabrication and e iciency considerations and V from equation 2 we proceed to construct the velocity triangles From these triangles we can nd the change in absolute tangential velocity and calculate the shaft work Entrance Triangle We rst draw a horizontal line representing the tangential direction Then we construct a vector representing V at angle a a er which we complete the entering triangle using the vector relation V1 11 The angle between the relative velocity and the tangential direction is designated VEll V1 Wal W01 Vb The Exit Triangle Draw W at angle y to the tangent Reducing y somewhat 39om the calculated value for B will result in increased blade ef ciency quotValues of 39y in use vary from 15 to 30 deg at high and intermediate pressures and from 30 to 40 deg at the lowpressure end of the turbine sometimes reaching 40 to 50 deg in large turbines where maximum ow area is neededquot5 W 2 is found by multiplying W by the velocity 7 coe z cient k1 which accounts for friction and turbulence The velocity coef cient is a function of the total change of direction of the steam in the W2 Vaz V2 blade 180 6 7 the blade width to radius W32 39 7 ratio and the relative velocity and density at blade W92 V92 entrance Because suf cient data are not available at the beginning of the design the following empirical formula adapted from Church for a one inch blade width is suggested6 12 k 0892 600x10 5W The triangles are easily solved for needed values as follows V Kcosa W ka Va Wu Vl sina axial component Va2 W2 Sin 7 Wi91V01Vb 2WZCOS WIMW VVW tanquotE reJn n We The Reheat Factor and the Condition Curve Only a portion of the available energy to a stage is turned into work The remainder termed reheat q shows up as an increase in the enthalpy of the steam Because the constant pressure lines on an hs chart Mollier chart diverge the summation of the individual isentropic drops for the total stages is greater than the isentropic drop between the initial and nal steam conditions We account for this variation using a reheat factor R which has been precalculated by various investigators 2W4 R Ahs total 51d p 153 6 Id p 16869 For preliminary design R can be estimated from the following chart taken from Church7 i Reheat Factor R Mildim Fm Boar Reheat facinga for various enthalpy drape and initial annexheats and on an in nite number of stages of 80 per mt utage ef ciency Walkman The value from the chart must be corrected for the actual number of stages and stage ef ciency 1en 1R 11 17773 4 A line connecting the initial and nal states plus the intervening states found by adding the reheat at constant pressure is called the condition line 71d p240 HE4313 mus EXAMPLE STEAM TURBINE DESIGN8 quotThe design of a steam turbine like that of any other important machine involves a judicious combination of theory with the results of experience governed to a great extent by the commercial element cost The progress of a particular design involves a continuous series of compromises between what is most efficient what will operate most reliably and what will cost the leastquot The client will usually specify steam conditions condenser vacuum rotational speed and capacity in kilowatts or horsepower The client may also specify a maximum cost and minimum ef ciency Calculate the principal dimensions of the nozzles and blading of a turbine given the following speci cations Power delivered at the shaft coupling 5000 kW Revolutions per minute 2400 rpm Maximum blade speed 570 s Initial steam pressure 150 psia Initial steam temperature 540 F Condenser pressure 1 in Hg Constant mean blade diameter for all stages Blade Work Select a nozzle angle equal to 20 then 3 2V V b 1213 t s I dequot cos 20 f Increase this value by about 10 the example uses 1186 to follow Church to account for disk friction and fanning VI 1357 s The Entrance Triangle V011357c0520 1257 s m1 VI91 Vb 1275 570705 s W1V11357sin20 464 s Wl 4642 7052 844 s tan l 3335 705 AtX W 8 Id Chapter VIII L The Exit Triangle 5 12 5 12 kb0892 6x10 WI 0892 6x10 x844 0917 W2 ka1 0917 844 774 s Assume that y 6 3335 A Y 7 3 8 We2 774cos3335 647 s 8 V32 V92 647570 765 s W W Y5 Wu2 K2 774sin3335 426 s y g I 7 7 I Vb V92 1393 6tan Qtan l 798 lt 3 V 765 N57 b Blade Work Per Unit Mass 39 570m 765 1275i Btu Ill 72 Wbquot UV62 Vai x x 778 Ib f lbm 322 7 wb 307SBtu lb Actual Energy Available to Blade 2 12 1357 W B 15539 39Xz 3675BtuZb 2 2x2 778QIx lbm 322 quot39 Blade Efficiency wb 3075 0836 77 AEbW 3675 Nozzle Velocity Coef cient k The following empirical formula based on experimental results was adapted from Church9 H kn 1021 0164x 01652 006713 00088x4 where x Vs 1000 9 Id p 82 Fig 79 14 Ideal Isentropic Nozzle ExitBlade Entrance Velocity V51 13i1406fws kn 0965 Nozzle Ef ciency 2 2 2 7quot VIZ2 91 2 k O9652 0931 1412 1412 Combined Nozzle and Blade Ef ciency 77quot 7W 0931O836 0778 Assume an average loss from disk friction and fanning of 4 and from leakage of 15 Stage Efficiency 773 077s1 004 0015 0735 Let 775 073 This is a provisional value It will be modi ed as the design proceeds and more precise information becomes available Number of Stages 5 4 Ideal Available Energy to Blade 5 4 f quot V2 14062 39 M5 31 395Btulbm 2 2322 778 From the given steam conditions the total isentropic drop in enthalpy is found to be Magma 12937 9029 3908Btulbm R00 10465 Reheat from above graph Trial Number of Stages n n M5 R 3908 1 0465 Ah 395 S 1035 Select 10 stages then correct for actual value of reheat R110465 11 i1 07310565 10 02 15 Trial Isentropic Drop Per Stage M 390810565 413Bt lb S u m The trial value is 18 Btulbm greater than 395 and will cause a slight increase in V which will be accounted for later Stage Reheat q q Ahs any 4131 073 11ISBtulbm Using a tabular format we proceed in turn for each stage to subtract Ahs from the entrance enthalpy and then add back the reheat to determine the stage end point The needed thermodynamic state properties including speci c volume are found as the process proceeds The Engineering Equation Solver facilitates preparation of the table Laying out the work on a Mollier chart is helpful both as a check and to determine the stages that are in the wet region thus requiring different input states Connecting all of the end points yields the condition line If the desired nal pressure 0491 psia is not reached a new trial Ahs and corresponding q are found and additional iterations run as needed To estimate the required change divide the error in enthalpy by the number of stages and by the stage ef ciency and then add or subtract this quantity from Ahs to get a new trial value A er the correct value for Ahs is found the nal velocity triangles can be constructed and corresponding values calculated The corrected value for V1 is V1 22380931413 1387 s from the velocity triangle W 1 is 873 W3 and the blade velocity coef cient is kb 0892 6x10395873 0840 This value is close enough to the previously calculated value of 0836 Proceeding Total Internal Work Per Pound of Steam w 10413 111530153tu1b Internal Ef ciency of the Turbine m 3015 M715 3908 Mechanical Losses Church states the following rough rule for total mechanical losses including bearing friction and gland pump and governor resistances 4 4 18 waIOOO J50001000 39 Mechanical loss in per cent at normal rating Assume a radiation loss of about 02 the combined mechanical and radiation losses amount to about 2 Engine Efficiency 7 m 07715098 0756 Ideal Steam Rate MA 873le kWh kWh 3908Bt1z ISR Brake Steam Rate BSR 8730756 1 1551bm kWh Turbine Mass Flow Rate quot1 1 155 1b 5000 W 1604lb W 3600s 39 s SIMILARITY LAWS AND MODELING Dimensional analysis leads to parameters which are useful for predicting performance when pumps are operated at other than the design conditions These parameters are also useful for scaling from a model to a prototype1 Head Rise Coefficient CH Power Coef cient C P Flow Coef cient C Ef ciency 77 r C39QC39HC1 Where WW SSOBHP ha actual head rise For two geometrically similar pumps we can equate the above coef cientsef ciencies for each pump to predict performance parameters Consider the following examples 0 If we change the speed of a particular pump what happens to the ow rate Head rise Shaft horsepower 2 3 0 a 0 Q2 Q1 haZ ha WYM Z Shaft o How does a change in pump diameter affect pump characteristics 3 2 5 D D D 2 1 D1 2 l D Sha Z Sh 11 For economic reasons manufacturers will frequently install impellers of different diameters in one pump casing Although the scaling laws are not speci cally applicable Munson notes that diameter changes of less than about 20 will render them useful Also if inplace testing of a pump results in a head rise higher than desired the problem might be correctable by machining the impeller diameter by an amount predicted by these laws 1 See Munson p 807