onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd http://teacher.buet.ac.bd/zahurul/ ME 20: Basic hermodynamics If a system is taken through a cycle and produces work, it must be exchanging heat with at least two reservoirs at 2 different temperatures. If a system is taken through a cycle while exchanging heat with a single reservoir, work must be zero or negative. Heat can never be converted continuously and completely into work, but work can always be converted continuously and completely into heat. Work is a more valuable form of energy than heat. For a cycle and single reservoir, W net 0. lausius Inequity Example: lausius Inequality: Steam Power Plant: c Dr. Md. Zahurul Haq BUE ME 20 202 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 2 / 26 lausius Inequality δq 0 4 δw net δw rev +δw sys = δq R du δq R δq = R δq δw net = R du W net = δq R = δq 0 as W net 0 { δq 0 reversible process < irreversible process 037 t 0.7 MPa sat = H = 65 o t 0.5 kpa sat = L = 56 o Q H = Q 2 = h 2 h = 2066.3 kj/kg Q L = Q 34 = h 3 h 4 = 898.4 kj/kg δq = Q H H + Q L L =.087 kj/kg c Dr. Md. Zahurul Haq BUE ME 20 202 3 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 4 / 26
S: hermodynamic Property 038 For reversible process: δq = 0 = + = + : = 2 = 0 B = 0 B 2 = Since / is same for all reversible processes/paths between state & 2, this quantity is independent of path and is a function of end states only. his property is called, S. ds rev = s = s 2 s = rev = ds rev is a property, hence change in entropy between two end states is same for all processes, both reversible and irreversible. If no irreversibilities occur within the system boundaries of the system during the process, the system is internally reversible. 87 For an internally reversible process, the change in the entropy is due solely for heat transfer. So, heat transfer across a boundary associated with it the transfer of entropy as well. transfer or flux rev c Dr. Md. Zahurul Haq BUE ME 20 202 5 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 6 / 26 Example: Steam generation in Boiler. Example: Internally reversible heating in a piston-cylinder assembly. s 2 s = = rev sat s 2 s = q2 sat = h2 h sat = h fg sat h fg = s fg sat q 23 = 3 = 3 2 ds 52 q 2 = sat s 2 s = area 2 b a q 23 = 3 2 ds = area 2 3 c b q net = q 2 + q 23 = area 2 3 c a 69 w = g f Pdv = Pv g v f = 0.325.673 0.00044 = 70 kj/kg q = g f ds = s g s f = s fg = 373.5 6.0486 = 2256.8 kj/kg lso note that, h fg = 2257 kj/kg c Dr. Md. Zahurul Haq BUE ME 20 202 7 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 8 / 26
arnot ycle H 2 3 = ds 56 L 54 s 4 s 53-2 : Isentropic compression. 2-3 : Isothermal heat addition and expansion. 3-4 : Isentropic expansion. 4- : Isothermal heat rejection and compression. q 23 = 3 2 ds = 3 H 2 ds = Hs 3 s 2 = H s q 4 = 4 ds = L 4 ds = Ls s 4 = L s w net = q net = q 23 + q 4 = H L s q in = q 23 η arnot = w net q in = H L s H s = L H c Dr. Md. Zahurul Haq BUE ME 20 202 9 / 26 hange for Irreversible M Process c Dr. Md. Zahurul Haq BUE ME 20 202 0 / 26 04 For reversible process: = 0 = + For reversible process: < 0 = + : > 2 Since path is reversible, and since entropy is a property = ds = ds = ds > Since path is arbitrary, for irreversible process ds > = s 2 s > rev B = 0 B < 0 2 70 he work for an irreversible process is not equal to Pdv and the heat transfer is not equal to ds. herefore, the area underneath the path does not represent work and heat on the P v and s diagrams, respectively. In many situations we are not certain of the exact state through which a system passes when it undergoes an irreversible process. For this reason it is advantageous to show irreversible processes as dashed lines and reversible processes as solid lines. hus, the area underneath the dashed line will never represent work or heat. c Dr. Md. Zahurul Haq BUE ME 20 202 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 2 / 26
Generation Example: Irreversible process of water. ds δq = δσ ds δq 0 σ produced generated by internal irreversibilities. 73 > 0 irreversible process σ : = 0 internally reversible process < 0 impossible process he entropy of a system can increase in only two ways, either by heat addition or by the presence of irreversibility. he entropy of a system can decrease only in only by heat removal. Reversible process: ds = / & adiabatic process: = 0 s = constant: for reversible adiabatic process. ll isentropic processes are not necessarily reversible and adiabatic. he entropy can remain constant during a process if the heat removal balances the irreversibility. c Dr. Md. Zahurul Haq BUE ME 20 202 3 / 26 72 du + dke + dpe = δw du = δw W m = g f du = u g u f = 2087.56 kj/kg Note that, the work input by stirring is greater in magnitude than the work done by the water as it expands 70 kj/kg. δσ/m = ds = ds 0 = ds σ m = s g s f = 6.048 kj/kg.k c Dr. Md. Zahurul Haq BUE ME 20 202 4 / 26 Principle of Increase of 7 diabatic or isolated system ontrol Mass, temperature = Surroundings, temperature = o δw ds For control mass: ds M. For the surroundings at o, < 0, and reversible heat transfer: ds surr = o. [ ] ds net = ds sys + ds surr o If > o < 0 ds net 0 If < o > 0 ds net 0 ds net 0 he entropy change for an isolated system cannot be negative. 88 he entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero. 89 system and its surroundings form an isolated system. ds isolated 0 c Dr. Md. Zahurul Haq BUE ME 20 202 5 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 6 / 26
of a Pure Substance of a Pure Substance Example: Suppose that kg of saturated water vapor at 00 o is to a saturated liquid at 00 o in a constant-pressure process by heat transfer to the surrounding air, which is at 25 o. What is the net increase in entropy of the water plus surroundings? s net = s sys + s surr s sys = s fg = 6.048 kj/kg.k s surr = q o = h fg o = 2257 298 = 7.574 kj/kg.k s net =.533 kj/kg.k So, increase in net entropy. 50 emperature-entropy diagram for steam. of a Pure Substance of a Pure Substance of Ideal Gas c Dr. Md. Zahurul Haq BUE ME 20 202 7 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 8 / 26 5 Enthalpy-entropy diagram Mollier diagram for steam. First Law: δw = du Reversible process: δw = Pdv : = ds Ideal gas: Pv = R : du = c v d : dh = c P d du = Pdv + ds ds = du + Pdv : st ds Equation. ds = c V d + P dv = c V d + R dv v : For ideal gas. = s 2, s,v = 2 d c V + R ln v2 v = s 2 s = c V ln 2 + R ln v2 v : Ideal gas with c V = constant. h = u + Pv dh = du + Pdv + vdp = rev + vdp = ds + vdp dh = ds + vdp ds = dh vdp : 2 nd ds Equation. = s 2,P 2 s,p = 2 d c P R ln P2 P = s 2 s = c P ln 2 R ln P2 P : Ideal gas with c P = constant. c Dr. Md. Zahurul Haq BUE ME 20 202 9 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 20 / 26
of a Pure Substance Example: Determine the change in specific entropy, in KJ/kg-K, of air as an ideal gas undergoing a process from 300 K, bar to 400 K, 5 bar. Because of the relatively small temperature range, we assume a constant value of c P =.008 KJ/kg-K. of a Pure Substance Example: ir is contained in one half of an insulated tank. he other side is completely evacuated. he membrane is punctured and air quickly fills the entire volume. alculate the specific entropy change of the isolated system. s = c P ln 2 R ln P 2 P =.008 kj ln kg.k = 0.79 kj/kg.k 400K 300K 8.34 28.97 kj kg.k ln 5bar bar Note that, for isentropic compression, 2s = P 2 /P k /k = 475 K. Hence, entropy change is - ve because of cooling of air from 475 K to 400 K. c Dr. Md. Zahurul Haq BUE ME 20 202 2 / 26 022 du = δw s 2 s = c v ln 2 + R ln v2 v w = 0,q = 0 du = 0 c V d = 0 2 =. v2 s 2 s = c v ln 2 + R ln v = 0 + 287 ln2 = 98.93 kj/kg.k Note that: s 2 s = 98.93 kj/kg.k > = s > 0 }{{} 0 c Dr. Md. Zahurul Haq BUE ME 20 202 22 / 26 of a Pure Substance Isentropic Process: s = constant s = 0 v2 v s 2 s = c P ln 2 + R ln k ln 2 = R c V ln v2 v = k ln v2 v = ln v of a Pure Substance Example: ir is compressed in a car engine from 22 o and 95 kpa in a reversible and adiabatic manner. If the compression ratio, r c = V /V 2 of this engine is 8, determine the final temperature of the air. = k 2 = v s = s 2,constant k s 2 s = c P ln 2 R ln ln 2 = R c P ln P2 P = P2 P = k k ln k 2 = P2 k P k P2 P = ln P2 k P s = s 2,constant k = Pv k = constant s = s 2,constant k 90 2 = v k k v 2 = = 2958.4 = 677.7 K c Dr. Md. Zahurul Haq BUE ME 20 202 23 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 24 / 26
Second-Law of hermodynamics for V Systems Rate Balance for ontrol Volumes Second-Law of hermodynamics for V Systems ds cv dt = j Q j j + ms i ms e + σ cv 042 For M systems: m i = 0, m e = 0 = dscm dt = j Q j j + σ For steady-state steady-flow SSSF process: ds dv /dt = 0. For -inlet & -outlet SSSF process: m i = m e. s e s i = j q j j + σcv m For adiabatic -inlet & -outlet SSSF process: s e s i = σcv m = s e s i 75 c Dr. Md. Zahurul Haq BUE ME 20 202 25 / 26 c Dr. Md. Zahurul Haq BUE ME 20 202 26 / 26