Unt Eght Calculatons wth Entropy Mechancal Engneerng 370 Thermodynamcs Larry Caretto October 6, 010 Outlne Quz Seven Solutons Second law revew Goals for unt eght Usng entropy to calculate the maxmum work n an adabatc process Example calculaton Other calculatons wth entropy Usng heat capactes to compute entropy Checkng to see f a process satsfes the second law The Second Law There exsts an extensve thermodynamc property called the entropy, S, defned as follows (T must be absolute temperature): ds = (du + PdV)/T For any process ds dq/t For an solated system ds 0 Why??? Because t s the law! Second Law Notes Started wth mathematcal statement on prevous chart Proved that heat flows from hgher to lower temperature Proved that reversble (Carnot) cycle s most effcent for two and only two temperature reservors Hghest engne effcency Hghest coeffcent of performance for refrgeraton cycles 3 4 Uses of Entropy We wll use entropy to compute reversble processes Reversble processes are the most effcent Calculatons wth entropy wll tell us the best possble outcome we can expect Effcency may be less than 100% A reversble adabatc process has constant entropy Entropy s a Property 5 6 ME 370 Thermodynamcs 1
Entropy s a Property If we know the state of the system, we can fnd the entropy We can use the entropy as one of the propertes to defne the state Use the followng f we are gven a value of s and a value of T or P f s < s f (T or P) => compressed lqud f s > s g (T or P) => gas (superheat) regon otherwse n mxed regon wth saturaton T and P and other propertes found from Cycles wth Q H = Q L + W Engne cycle converts heat to work Refrgeraton cycle transfers heat from low to hgh Hgh Temperature Heat Source T Low Temperature Low Temperature temperature Heat Snk THeat Source qualty, x = (s s f )/s fg S k S k 7 8 Engne Cycle Refrgeraton Cycle QH W W Q COP L W W Hgh Temperature THeat Snk t S Q H QL Q Q L H As the effcency of processes n an solated system decreases, the entropy of that system ncreases What s Entropy? Hgh Temperature Heat Source T Q H Low Temperature Heat Snk S k Engne Cycle W S sol syst = Q H /T H + Q L /T L + 0 Hgh (for Temperature cycle) THeat Snk t For Q S H = 100 kj, T H = 1000 K and T L = 500 Q H K. Carnot cycle effcency s = 1 W T H /T H = 1 500/1000 = 50% Q L = Q H W = Q H Q H = Q H (1 ) S sol syst = Low QTemperature H /T H + THeat Source Q L /T L = Q S H [ 1/T k H + (1 ) / T L ] Q L Q L Refrgeraton Cycle 9 What s Entropy? II S sol syst = Q H [ 1/T H + (1 ) / T L ] For Q H = 100 kj, T H = 1000 K, T L = 500 K S sol syst = (100 kj) [ 1/(1000 K) + (1 ) / (500 K) ] = (0.1 kj/k) (1 ) Ths equaton gves S sol syst = 0 for = 50% (the maxmum effcency) and S sol syst = 0.1 for = 0 where there s no work. 10 Entropy (kj/k) 0.01 0.008 0.006 0.004 0.00 0 Entropy versus Effcency What s Entropy? II Carnot cycle effcency s = 1 T H /T H = 1 500/1000 = 50% 0 0.1 0. 0.3 0.4 0.5 Effcency Engne cycle wth two temperature reservors; Q H = 100 kj, T H = 1000 K, T L = 500 K. 11 Reversble Process In a reversble process t s possble to return an solated collecton of systems to ther ntal states wth no changes n the surroundngs Ths s an dealzaton; we cannot do better than a reversble process Ths s the = part of the sgn n ds dq/t and ds solated system 0 For a reversble process ds = dq/t and ds solated system = 0 1 ME 370 Thermodynamcs
Unt Eght Goals As a result of studyng ths unt you should be able to feel more comfortable wth the noton of entropy as property fnd entropy values n property tables use entropy and one other property to defne a state that you fnd n a table be able to pronounce entropy and enthalpy clearly to dstnghsh between the two More Unt Eght Goals understand the meanng of equatons for a reversble process recognze that the maxmum work s done n a reversble process compute S = dq/t for a reversble process snce ds = dq/t for reversble processes compute s surr = Q surr /T surr = -Q syst /T surr prove that a process n an solated system satsfes the second law because S IS 0 13 14 Stll More Unt Eght Goals recognze that fndng the maxmum work for an adabatc process s the same as fndng the work n a constant entropy (sentropc) process be able to solve the followng class of problems Gven: Intal Condtons, and one fnal property Fnd: Maxmum work for adabatc process Soluton: Compute sentropc process Maxmum Work between two states occurs n a reversble process Look at dfferental analyss of reversble and rreversble process Intal and fnal state the same so any property change s the same du rev = du rrev ds rev = ds rrev 15 16 Maxmum Work II Maxmum Work III between two states occurs n a reversble process Apply second law: ds dq/t ds rev = dq rev /T ds rrev > dq rrev /T But we just sad that ds rev = ds rrev ds rev = dq rev /T = ds rrev > dq rrev /T between two states occurs n a reversble process Apply frst law du rev = dq rev dw rev du rrev = dq rrev dw rrev Have shown du rev = du rrev ; equate the frst law expressons for these du values dq rev dw rev = dq rrev dw rrev Concluson: dq rev > dq rrev 17 18 ME 370 Thermodynamcs 3
Maxmum Work IV between two states occurs n a reversble process Have followng results from frst and second law dq rev dw rev = dq rrev dw rrev dq rev > dq rrev Combnng these gves dw rev dw rrev = dq rev dq rrev > 0 19 Maxmum Work V between two states occurs n a reversble process Have just shown that dw rev dw rrev > 0 so dw rev > dw rrev Remember that work output s postve and work nput s negatve If work output s 00, whch s larger a work output of 100 or a work output of 300? 0 Maxmum Work for Input The maxmum work (nput or output) occurs n a reversble process Sgn conventon: work output s postve and work nput s negatve For a work nput, the maxmum work s negatve; the actual work s a negatve number wth a larger magntude than the maxmum work For a work nput, the maxmum work s the mnmum work nput 1 Maxmum Adabatc Work (S = 0) From gven nlet condtons, fnd the ntal state propertes ncludng s ntal The maxmum work n an adabatc process occurs when s fnal = s ntal From s fnal = s ntal and one other property of fnal state get all fnal state propertes Fnd work from frst law We have been workng problems lke ths wthout realzng that we were fndng the maxmum work Example Problem A steady-flow steam turbne has an nlet P and T of 10 MPa and 450 o C What s the maxmum work f Q = 0 and P out = 50 kpa? Soluton: apply frst law for adabatc steady flow wth one nlet and one outlet and KE = PE neglgble For maxmum work n an adabatc Example Problem II Frst law reduces to w u = h n h out At 10 MPa and 450 o C nlet, s n = 6.419 kj/kg K and h n = 34.4 kj/kg At P out = 50 kpa and s out = 6.419 kj/kg K, x out = (6.419 kj/kg K 1.091 kj/kg K) / 6.5019 kj/kg K = 0.81987 h out = 340.54 kj/kg + 0.81987(645. kj/kg) = 509.5 kj/kg w u = h n h out = 34.4 kj/kg 509.5 process, s out = s n kj/kg = 733.15 kj/kg 3 4 ME 370 Thermodynamcs 4
S from Heat Capacty Orgnal defnton s that dq = C x dt n a constant x process For deal gases du = c v dt and dh = c p dt for any process For deal gases, we also have dq = c v dt for constant v and dq = c p dt for constant P For reversble processes we have ds = dq/t = C x dt/t or S = C x dt/t for a constant X process Checkng the Second Law Ten klograms of a flud havng a heat capacty, c p = 3 kj/kgk s cooled from 500 to 300 K at constant pressure wth heat rejected to surroundngs at 310 K Is ths process possble? Answer s found by seeng f S of solated system consstng of flud plus surroundngs s greater than zero 5 6 Checkng the Second Law II Ideas to use here For reversble heat transfer ds = dq/t If pressure or volume s constant, dq = C p dt or C v dt For a fnte, reversble process, S = ds = dq/t For constant P, S = C p dt/t and Q = C p dt For constant V, S = C v dt/t and Q = C v dt For any process wth no nteractons except surroundngs, Q system = Q surroundngs Checkng the Second Law II Apply equatons from prevous page to see f S sol syst = S flud + S surr 0 S flud = mc p dt/t = mc p ln(t /T 1 ) = (10 kg)(3 kj/ kgk)ln(300/500) = 15.3 kj/k S surr = Q surr /T surr = Q flud /T surr Q flud = m c p dt = mc p (T T 1 ) = (10 kg)(3 kj/kgk)(300 500)K = 6000 kj We can now fnd S surr 7 8 Checkng the Second Law III S surr = Q surr /T surr = Q flud /T surr = ( 6000 kj)/(310 K) = 19.35 kj/k S sol syst = S flud + S surr = ( 15.3 + 19.35) kj/k = 4.03 kj/k Snce S sol syst 0 the process satsfes the second law How can we cool a body from 500 to 300 K f the surroundngs are at 310 K? Transfer heat through engne and refrgerator 9 Class Exercse An R-134a compressor, wth zero heat transfer, has an nlet stream of saturated vapor at 10 psa. It s outlet pressure s 160 psa. What s the mnmum work nput to the compressor? Soluton: For an adabatc process, the maxmum work output (the mnmum work nput) s n an sentropc process Fnd outlet state at P out and s out = s n 30 ME 370 Thermodynamcs 5
Exercse Soluton Gven: R-134a compressor wth Q = 0 has nlet wth saturated vapor nlet at P n = 10 psa and outlet wth P out = 160 psa Fnd: Mnmum work nput Assumptons and observatons: Steady process -- one nlet and one outlet No heat transfer, so mnmum work nput s for sentropc process Neglect knetc and potental energy terms Propertes: R-134a tables 31 Apply Frst Law Frst Law for ths case Steady-state open system One nlet and one outlet Heat transfer s gven as zero Assume knetc and potental energy changes are neglgble Steady desystem V Q W u m h gz m h dt nlet One Inlet outlet One Outlet V gz 3 W u Get w u and Propertes m h m h m h h n n out out W u wu hn hout m Inlet at 10 psa s saturated vapor: s n = s g (10 psa) = 0.948 Btu/lb m R; h n = h g (10 psa) = 98.68 Btu/lb m At sentropc outlet, P out = 160 psa, s out = s n = 0.948 Btu/lb m R s between 10 o F and 140 o F; nterpolate for h n out 33 Answer 15.3 Btu 10.06 Btu 10.06 Btu lbm lbm hout lb 0.330 Btu 0.337 Btu m lbm R lbm R 0.948 Btu 0.337 Btu 13.66 Btu lbm R lbm R lbm w u = h n h out = 98.68 Btu/lb m 13.66 Btu/lb m = 4.98 Btu/lb m s maxmum work for sentropc process For work nput, ths s mnmum w 34 Another Exercse Steam n a pston-cylnder devce has an ntal P and T of 10 MPa and 450 o C What s the maxmum work f the pston expands wth Q = 0 untl P fnal = 50 kpa? Soluton: apply frst law for adabatc closed system: Q = U + W = 0 so W = U or w = W/m = U/m = u For maxmum work n an adabatc Example Problem II Frst law reduces to w = u 1 u Intal state 10 MPa and 450 o C nlet, s 1 = 6.419 kj/kg K and u 1 = 944.5 kj/kg At P = 50 kpa and s = s 1 = 6.419 kj/kg K, x = (6.419 kj/kg K 1.091 kj/kg K) / 6.5019 kj/kg K = 0.81987 u = 340.49 kj/kg + 0.81987(14.7 kj/kg) = 097. kj/kg w = u 1 u = 944.5 kj/kg 097. process, s fnal = s ntal or s = s 1 kj/kg = 847.8 kj/kg 35 36 ME 370 Thermodynamcs 6