First Law of Thermodynamics. Control Mass (Closed System) Conservation of Mass. Conservation of Energy

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Kerry Allison
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1 First w of hermodynmics Reding Problems , 3-5, , 5-5, 5-9, 5-37, 5-0, 5-, 5-63, 5-7, 5-8, , 6-5, 6-60, 6-80, 6-9, 6-, 6-68, 6-73 Control Mss (Closed System) In this section we will exmine the cse of control surfce tht is closed to mss flow, so tht no mss cn escpe or enter the defined control region. Conservtion of Mss Conservtion of Mss, which sttes tht mss cnnot be creted or destroyed, is implicitly stisfied by the definition of control mss. Conservtion of Energy he first lw of thermodynmics sttes Energy cnnot be creted or destroyed it cn only chnge forms. energy entering - energy leving = chnge of energy within the system Sign Convention Cengel Approch et rnsfer: ork rnsfer: het trnsfer to system is positive nd het trnsfer from system is negtive. work done by system is positive nd work done on system is negtive. For instnce: moving boundry work is defined s: b = P dv During compression process, work is done on the system nd the chnge in volume goes negtive, i.e. dv < 0. In this cse the boundry work will lso be negtive.

2 Culhm Approch Using my sign convention, the boundry work is defined s: b = P dv During compression process, the chnge in volume is still negtive but becuse of the negtive sign on the right side of the boundry work eqution, the boundry work directed into the system is considered positive. Any form of energy tht dds to the system is considered positive. Exmple: A Gs Compressor Performing st lw energy blnce: Initil Energy E + { Energy gin Energy loss } = F inl Energy E E + = E E = A first lw blnce for control mss cn lso be written in differentil form s follows: de = δ δ Note: d or for chnge in property nd δ for pth function

3 he differentil form of the energy blnce cn be written s rte eqution by dividing through by dt, differentil time, nd then letting dt 0 in the limit to give de dt = δ dt δ dt de dt = Ẇ where de dt = rte of energy increse within the system, du dt + d KE dt + d P E dt = rte of het trnsfer Ẇ = rte of work done, power most closed systems encountered in prctice re sttionry i.e. the velocity nd the elevtion of the center of grvity of the system remin constnt during the process for sttionry systems we cn ssume tht d KE dt = 0 nd d P E dt = 0 Exmple 3-: During stedy-stte opertion, gerbox receives Ẇ in = 60 k through the input shft nd delivers power through the output shft. For the gerbox s the system, the rte of energy trnsfer by het is given by Newton s w of Cooling s Ẇ = ha( b f ). where h, the het trnsfer coefficient, is constnt (h = 0.7 k/m K) nd the outer surfce re of the gerbox is A =.0 m. he temperture of the outer surfce of the gerbox is b = 300 K nd the mbient temperture surrounding the gerbox is = 93 K. Evlute the rte of het trnsfer, nd the power delivered through the output shft, Ẇ out.. in = 60 k = 300 K b = 93 K f. = ha(b-) f. out A =.0 m 3

4 Forms of Energy rnsfer ork Versus et ork is mcroscopiclly orgnized energy trnsfer. et is microscopiclly disorgnized energy trnsfer. more on this when we discuss entropy et Energy het is defined s form of energy tht is trnsferred solely due to temperture difference (without mss trnsfer) het trnsfer is directionl (or vector) quntity with mgnitude, direction nd point of ction modes of het trnsfer: ork Energy conduction: diffusion of het in sttionry medium (Chpters 0, & ) convection: it is common to include convective het trnsfer in trditionl het trnsfer nlysis. owever, it is considered mss trnsfer in thermodynmics. (Chpters 3 & ) rdition: het trnsfer by photons or electromgnetic wves (Chpter 5) work is form of energy in trnsit. One should not ttribute work to system. work (like het) is pth function (mgnitude depends on the process pth) work trnsfer mechnisms in generl, re force cting over distnce Mechnicl ork = F ds if there is no driving or resisting force in the process (e.g. expnsion in vcuum) or the boundries of the system do not move or deform, = 0. Moving Boundry ork = F ds = P A ds = P dv decrese in the volume, dv ve results in work ddition (+ve) on the system

5 consider compression in piston/cylinder, where A is the piston cross sectionl re (frictionless) the re under the process curve on P V digrm is proportionl to the work is: +ve for compression ve for expnsion P dv sometimes clled P dv work or compression /expnsion work polytropic processes: where P V n = C exmples of polytropic processes include: Isobric process: if n = 0 then P = C nd we hve constnt pressure process Isotherml process: if n = then from the idel gs eqution P V = R nd P V is only function of temperture Isometric process: if n then P /n V = C /n nd we hve constnt volume process Isentropic process: if n = k = C p /C v then we hve n isentropic process. (tbulted vlues for k re given in ble A-) If we combine P v k = C with P v = R we get the isentropic equtions, given s: = ( ) k ( ) (k )/k v P = v P Cse : for n idel gs with n = = C ln V V Cse : for n = P V P V n (in generl) = mr( ) n (idel gs) 5

6 Exmple 3-: A pneumtic lift s shown in the figure below undergoes qussi-equilibrium process when the vlve is opened nd ir trvels from tnk A to tnk B. A ir stte tm. stte B Ap A mp B vlve closed vlve open control mss system he initil conditions re given s follows nd the finl tempertures cn be ssumed to be the sme s the initil tempertures. P tm = 00 kp P = N/m m p = 500 kg A p = 0.05 m V A, = 0. m 3 V B, = 0. m 3 P A, = 500 kp P B, = 00 kp A, = 98 K B, = 98 K = 5 C Find the finl pressures P A, nd P B, nd the work,, in going from stte to stte. Control Volume (Open System) he mjor difference between Control Mss nd nd Control Volume is tht mss crosses the system boundry of control volume. CONSERVAION OF MASS: Unlike control mss pproch, the control volume pproch does not implicitly stisfy conservtion of mss, therefore we must mke sure tht mss is neither creted nor destroyed in our process. { rte of increse of mss within the CV } = { net rte of mss flow IN } { net rte of mss flow OU } 6

7 CONSERVAION OF ENERGY: E CV (t) + δ + δ shft + ( E IN E OU )+ (δ IN δ OU ) = E CV (t + t) () ht is flow work? his is the work required to pss the flow cross the system boundries. m IN = ρ IN volume {}}{ A IN V IN t δ IN = F distnce = P IN A IN }{{} F V IN t }{{} s = P IN m IN ρ IN since v = /ρ 7

8 δ IN = (P v m) IN flow work () Similrly δ OU = (P v m) OU (3) Substituting Eqs. nd 3 into Eq. gives the st lw for control volume E CV (t + t) E CV (t) = δ + δ shft + m IN (e + P v) IN m OU (e + P v) OU () Eqution cn lso be written s rte eqution divide through by t nd tke the limit s t 0 d dt E CV = + Ẇ shft + [ṁ(e + P v)] IN [ṁ(e + P v)] OU where: e + P v = u + P v }{{} + (V) + gz = h(enthlpy) + KE + P E Exmple 3-3: Determine the het flow rte,, necessry to sustin stedy flow process where liquid wter enters boiler t 0 C nd 0 MP nd exits the boiler t 0 MP nd qulity of for mss flow rte is kg/s. he effects of potentil nd kinetic energy re ssumed to be negligible. stem out in O ( l) G 8

9 Exmple 3-: Stem with mss flow rte of.5 kg/s enters stedy-flow turbine with flow velocity of 50 m/s t MP nd 350 C nd leves t 0. MP, qulity of, nd velocity of 00 m/s. he rte of het loss from the uninsulted turbine is 8.5 k. he inlet nd exit to the turbine re positioned 6 m nd 3 m bove the reference position, respectively. Determine the power output from the turbine. Note: include the effects of kinetic nd potentil energy in the clcultions. in out he Crnot Cycle If the het engine is reversible system where no entropy is generte internlly, we refer to the cycle s the Crnot cycle. s s s 3 in out where the efficiency is given s: η = Crnot efficiency 9

10 Prcticl Problems t stte point the stem is wet t nd it is difficult to pump wter/stem (two phse) to stte point cn we devise Crnot cycle to operte outside the wet vpor region P P 3 s s s between stte points nd 3 the vpor must be isotherml nd t different pressures - this is difficult to chieve the high temperture nd pressure t nd 3 present metllurgicl limittions he Idel Rnkine Cycle urbine 3 Boiler s Pump P Condenser wter is typiclly used s the working fluid becuse of its low cost nd reltively lrge vlue of enthlpy of vporiztion 0

11 Device st w Blnce Boiler h + q = h 3 q = h 3 h (in) urbine h 3 = h + w w = h 3 h (out) Condenser h = h + q q = h h (out) Pump h + w P = h w P = h h (in) he net work output is given s w w p = (h 3 h ) (h h ) = (h 3 h ) + (h h ) he Rnkine efficiency is η R = net work output het supplied to the boiler = (h 3 h ) + (h h ) (h 3 h ) Exmple 3-5: For the stem power plnt shown below, urbine 3 Boiler s Pump P Condenser find:,, Ẇ net = Ẇ Ẇ P, nd the overll cycle efficiency, η R given the following conditions: P = 0 kp P = 0 MP P 3 = 0 MP P = 0 kp = 0 C 3 = 530 C = 0 C x = 0.9 ṁ = 5 kg/s

Psychrometric Applications

Psychrometric Applications Psychrometric Applictions The reminder of this presenttion centers on systems involving moist ir. A condensed wter phse my lso be present in such systems. The term moist irrefers to mixture of dry ir nd