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Chap. 3 (Sec. 3-5 to End of Chapter) Mass Flow Rate = r A V (kg/s) Volume Flow Rate = A V (m3 /s) V = velocity Conservation of Mass (Continuity Principle) min mout min mout msystem m in m out dmsystem / dt Conservation of Mass (Continuity Principle) Steady state ? Change of Any quantity with time = 0 m in m out dmsystem / dt 0 m in m out (r A V)in = (r A V)out Incompressible fluid ? Flow Work Wflow = PV wflow = Pv Total energy of moving fluid = e + Pv = u + p.e + k.e + Pv = h + p.e + k.e First Law of Thermodynamics (ENERGY BALANCE) also known as Conservation of Energy Principle * CLOSED SYSTEMS * OPEN SYSTEMS (CONTROL VOLUME) • STEADY FLOW • UNSTEADY FLOW * CLOSED SYSTEMS (e.g: piston-cylinder etc.) Ein – Eout = Esystem Net Energy transfer by heat, work, and mass Change in internal, KE, PE etc Q – W = E Qnet, in – Wnet,out = Esystem Moving boundary work, Shaft work, Paddle Work etc Q – W = E Qnet, in – Wnet,out = Esystem Moving boundary work, Shaft work, Paddle Work etc How the above equation simplifies for different situations ? for for for for stationary systems constant volume process constant pressure process many other situations given in your text book Q – W = E Qnet, in – Wnet,out = Esystem Moving boundary work, Shaft work, Paddle Work etc For example: Based on problem statement if you having the following information Closed System Stationary Adiabatic Constant Volume Process Paddle Work (wnet, in) Final Temperature ? Wnet,in = Usystem wnet,in = usystem What will be my approach R134a if (a) Substances like Steam, if (b) ideal gases if (a) Substances like Steam, R134a U = m (u2-u1) As long as you have any two properties, you can find rest How? State (Saturated liquid , mixture, superheated) if (b) ideal gases U = m (u2-u1) = m Cv (T2 – T1) PV = mRT Pv = RT (PV/T)1 = (PV/T)2 STEADY – FLOW SYSTEMS E in E out Rate of Net Energy transfer in by heat, work, and mass = Rate of Net Energy transfer out by heat, work, and mass 2 2 V V -W m e (he e gze ) m i (hi i gzi ) Q 2 2 for each exit Mass balance ? for each inlet One inlet (1) and one exit (2) 2 2 V V 2 1 Q - W mh2 h1 g(z2 z1 ) 2 On a unit mass basis V22 V12 q - w h2 h1 g(z2 z1 ) 2 2 2 V V 2 1 Q - W mh2 h1 g(z2 z1 ) 2 In general, how above equation simplifies for (how to judge - based on function) Nozzles Diffusers Compressors, Pumps Turbines Mixtures, Heat Exchangers Throttling devices UNSTEADY – FLOW SYSTEMS Mass Balance min – mout = msystem min – mout = (m2-m1)system Energy Balance Ein – Eout = Esystem (Q in Win miθi ) (Q out Wout me θ e ) (m2 e 2 m1e1 ) sy stem θ h ke pe e u ke pe (Q in Win mihi ) (Q out Wout mehe ) (m2u2 m1u1 ) sy stem Chap. 5 The Second Law of Thermodynamics (What is second law and how it helps) Statements (in words or schematic) Some concepts Heat Engine, Refrigerator, Heat Pump Reversible and Irreversible processes * The Carnot Cycle and it’s importance 1ST LAW Qnet – Wnet = E for cyclic devices E = 0 Therefore Wnet = Qnet Wnet = Qin - Qout HIGH TEMPERATURE RESERVOIR AT TH QH W HE net, out QL LOW TEMPERATURE SINK AT TL PERFORMANCE OR EFFICIENCY THERMALEFFICIENCY (ηth ) ηth Wnet,out QH DESIRED OUTPUT REQUIRED INPUT NET WORK OUTPUT TOTALHEAT INPUT Q H QL Q 1 L QH QH WARM ENVIRONMENT AT TH>TL WARM ENVIRONMENT AT TH>TL QH W REFRI. COPR QL DESIRED OUTPUT QL REQUIRED INPUT Wnet,in QL 1 QH QL QH / QL 1 QH W HEAT PUMP net, IN COLD REFRIGERATED SPACE AT TEMPERATURE TL COPR (HOME) net, IN QL COLD ENVIRONMENT AT TEMPERATURE TL COPHP DESIRED OUTPUT QH REQUIRED INPUT Wnet,in COPHP QH 1 QH Q L 1 QL / QH CARNOT PRINCIPLES 1. The efficiency of an irreversible heat engine is always less than of a reversible one operating between the same two reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same (Independent of working fluid and it’s properties, the way cycle is executed, or the type of reversible engine) FOR REVERSIBLE PROCESS ALONE QH QL T H rev TL TL ηth, r ev 1 TH 1 COPR, rev TH / TL 1 1 COPHP,rev 1 TL / TH