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Chapter 3. Total Energy Is Not Created or Destroyed 2012/2/6 1 3.1 CONSERVATION OF ENERGY ⎛ v2 ⎞ θ = U + M ⎜⎜ + ψ ⎟⎟ ⎝2 ⎠ ACC = Input − Output ⎛ v2 ⎞ ⎪⎫ ⎛ Rate at which energy ⎞ ⎛ Rate at which energy ⎞ d ⎪⎧ U M ψ + + ⎨ ⎜ ⎟⎬ = ⎜ ⎟−⎜ ⎟ dt ⎪⎩ ⎝ 2 ⎠ ⎪⎭ ⎝ enters the system ⎠ ⎝ leaves the system ⎠ (3.1-1) Here U is the total internal energy, v2/2 is the kinetic energy per unit mass (where v is the center of mass velocity), and Ψ is the potential energy per unit mass. 2012/2/6 2 2 3 4 Q ⎛ ∧ v2 ⎞ ⎜ U + +ψ ⎟ 2 ⎝ ⎠k W = Ws + Wpv ⎛ ∧ v2 ⎞ ⎜ U + +ψ ⎟ 2 ⎝ ⎠k 1 ⎞ ⎛ v2 U + M ⎜⎜ + ψ ⎟⎟ ⎠ ⎝ 2 ⎛ ⎞ v ⎜ U + +ψ ⎟ 2 ⎝ ⎠k ∧ 2012/2/6 5 Wflow 2 ⎛ ∧ v2 ⎞ ⎜ U + +ψ ⎟ 2 ⎝ ⎠k 3 Various energy enter/leave the system (1) Energy flow accompanying mass flow K ) u2 ⎞ ⎛ & ∑ M i ⎜⎜U + 2 + ψ ⎟⎟ (3.1-2) k =1 ⎠k ⎝ ) where U is the internal energy per unit mass of the k th flow stream, and M& i is its mass flow rate. (2) Heat Q& = ∑ Q& j j 2012/2/6 where Q& j is the heat flow at the j th heat flow port. 4 Work(3, 4) W = Ws + Wpv (3) Shaft work, Ws, the mechanical energy flow that occurs without a deformation of the system boundaries. Examples, steam turbine, internal combustion engine, pump, and compressor. (4) PV work, the movement of the system boundary d ( LA ) dL dV & W = −F = − ( F / A) = −P dt dt dt 2012/2/6 (3.1-3) where P is the pressure exerted by the system at its boundaries. It represents that work done on the system in compression is positive, 5 Flow work (5) Wflow 5. Work of a flowing fluid against pressure Figure 3.1-1 Flow through a valve 2012/2/6 6 ⎛ Work done by surrounding fluid in ⎞ ) ⎜ ⎟ M PV Δ = pushing fluid element of mass ( ) 1 1ΔM 1 ⎜ 1⎟ ⎜ into the valve ⎟ ⎝ ⎠ ⎛ Work done on surrounding fluid ⎞ ⎜ ⎟ ⎜ by movement of fluid element of ⎟ ) ⎜ mass ( ΔM ) out of the valve (since ⎟ = − P V ΔM 2 2 2 2 ⎜ ⎟ ⎜ this fluid element is pushing the ⎟ ⎜ ⎟ ⎝ surrounding fluid ⎠ ⎛ ) ) on the system due to ⎞ ⎜⎜ Net work done ⎟⎟ = PV 1 1ΔM 1 − P2V2 ΔM 2 movement of fluid ⎝ ⎠ For a more general system, with several mass flow ports, ⎛ Net work done on the system due ⎜ ⎜ to the pressure forces acting on ⎜ the fluids moving into and out of ⎜ ⎝ the system ⎞ ⎟ K ) ⎟ = ΔM PV k ⎟ ∑ k =1 ⎟ ⎠ ⎛ Net rate at which work in done on ⎜ ⎜ the system due to the pressure forces ⎜ acting on the fluids moving into and out ⎜ ⎝ of the system 2012/2/6 ( )k ⎞ ⎟ K ) ⎟ = M& PV k ⎟ ∑ k =1 ⎟ ⎠ ( )k 7 Energy balance (mass basis) ⎛ v2 ⎞ ⎫⎪ K & ⎛ ˆ v 2 ⎞ d ⎧⎪ ⎨U + M ⎜ + ψ ⎟ ⎬ = ∑ M k ⎜ U + + ψ ⎟ + Q& 2 dt ⎪⎩ ⎝ 2 ⎠ ⎪⎭ k =1 ⎝ ⎠k ( ) K dV & −P + Ws + ∑ M& k PVˆ dt k =1 (3.1-4) k ⎛ v2 ⎞⎫ K & ⎛ ˆ v 2 ⎞ d⎧ ⎜ ⎟ ⎜ ⎨U + M ⎜ + ψ ⎟⎬ = ∑ M k ⎜ H + + ψ ⎟⎟ + Q& + W& 2 dt ⎩ ⎝2 ⎠⎭ k =1 ⎝ ⎠k (3.1-4a) where & − P dV W& = Ws dt 2012/2/6 8 (3.1-4a) No Acc. 2012/2/6 9 Energy Balance (molar basis) ⎛ v2 ⎞⎫ K & ⎛ ⎛ v2 ⎞⎞ d⎧ ⎨U + Nm⎜⎜ + ψ ⎟⎟⎬ = ∑ Nk ⎜⎜ H + m⎜⎜ + ψ ⎟⎟ ⎟⎟ + Q& + W& dt ⎩ ⎝2 ⎠⎭ k =1 ⎝ ⎝2 ⎠ ⎠k (3.1-4b) Since: M = Nm; M& k = N& k m H: total enthalpy ∧ H : specific enthalpy H : molar enthalpy ∧ mH = H ) v2 ) v2 ⎛ ⎞ ⎛ ⎞ & & M k ⎜⎜ H + + ψ ⎟⎟ = Nk m⎜⎜ H + + ψ ⎟⎟ 2 2 ⎝ ⎠ ⎝ ⎠ 2 2 ) ⎛ ⎛ ⎞ ⎞ ⎞⎞ ⎛ ⎛ v v & & = N k ⎜⎜ mH + m⎜⎜ + ψ ⎟⎟ ⎟⎟ = Nk ⎜⎜ H + m⎜⎜ + ψ ⎟⎟ ⎟⎟ ⎠⎠ ⎠⎠ ⎝2 ⎝2 ⎝ ⎝ 2012/2/6 10 Commonly Used Forms of Energy balance Mass basis ( ) K d {U } = ∑ M& k Hˆ k + Q& + W& dt k =1 ⎛ v2 ⎞ M ⎜⎜ + ψ ⎟⎟ << U ; ⎝2 ⎠ (3.1-5a) v2 + ψ << Hˆ 2 Molar basis K d {U } = ∑ N& k (H )k + Q& + W& dt k =1 (3.1-5b) ⎛ v2 ⎛ v2 ⎞ ⎞ Nm⎜⎜ + ψ ⎟⎟ << U ; m⎜⎜ + ψ ⎟⎟ << H ⎝2 ⎠ ⎝2 ⎠ 2012/2/6 11 To obtain its difference form from above: Integrate from t1 to t2 ⎛ v2 ⎞ ⎫⎪ K & ⎛ ˆ v 2 ⎞ d ⎧⎪ ⎨U + M ⎜ +ψ ⎟ ⎬ = ∑ M k ⎜ H + +ψ ⎟ + Q& + W& 2 dt ⎪⎩ ⎝ 2 ⎠ ⎪⎭ k =1 ⎝ ⎠k ⎧⎪ ⎛ v2 ⎞ ⎫⎪ ⎧⎪ ⎛ v2 ⎞ ⎫⎪ ⇒ ⎨U + M ⎜ +ψ ⎟ ⎬ − ⎨U + M ⎜ +ψ ⎟ ⎬ ⎪⎩ ⎝ 2 ⎠ ⎪⎭t2 ⎪⎩ ⎝ 2 ⎠ ⎪⎭t1 K t2 k =1 t1 = ∑∫ Q=∫ t2 t1 2 ⎛ ⎞ v ˆ & M k ⎜ H + +ψ ⎟ dt + Q + W 2 ⎝ ⎠k & ; Qdt W = Ws − ∫ V ( t2 ) V ( t1 ) 2012/2/6 Ws = ∫ t2 t1 W&s dt ; V ( t2 ) ∫V (t ) 1 (3.1-6) t2 PdV = ∫ P t1 dV dt dt PdV 12 The first term on the right side of Eq. 3.1-6 is usually the most troublesome to evaluate because the mass flow rate and/or the thermodynamic properties of the flowing fluid may change with time. However, if the thermodynamic properties of the fluids entering and leaving the system are independent of time, we have K ∑ ∫t t2 k =1 1 2 2 K ⎛ ⎛ ⎞ ⎞ t2 & v v & ˆ ˆ M k ⎜⎜ H + + ψ ⎟⎟ dt = ∑ ⎜⎜ H + + ψ ⎟⎟ ∫ M k dt t 2 2 k =1 ⎝ ⎝ ⎠k ⎠k 1 ⎞ ⎛ ˆ v2 = ∑ ΔM k ⎜⎜ H + + ψ ⎟⎟ 2 k =1 ⎠k ⎝ K 2012/2/6 (3.1-7) 13 When PE, KE<<0, Ws=0, for a single flow system, (3.1-5a) becomes: dM = M& dt dU dV & & ˆ = MH + Q − P dt dt dU dM ˆ & dV = H +Q− P dt dt dt For a time interval dt dU = Hˆ dM + Q − PdV 2012/2/6 (3.1-8) (3.1-9a) 14 For a closed system dU = Q − PdV (3.1-9b) Closed system No K.E. and no P.E. No Ws A single flow NOTE: Positive Q: Added on system Negative Q: removed from system -PdV: PV work done on system in compression is positive PdV: PV work done by system in expansion is negative 2012/2/6 15 (3.1-6) (3.1-7) 2012/2/6 16 3.2 SEVERAL EXAMPLES OF USING THE ENERGY BALANCE The energy balance equations can be used for the description of any process. First, define the system, describe the system boundary. The system may be selected using different boundary definitions, such as closed or open system. This may result in different formulation in the energy balance equation. However, processes occurring in nature are in no way influenced by our mathematical description of them. Therefore, if our descriptions are correct, they must lead to the same final result for the system and its surroundings regardless of which system choice is made. Result (open system) = Result (closed system) This is demonstrated in the following example, where the same result is obtained by choosing for the system first a given mass of material (closed) and then a specified region in space (open). 2012/2/6 17 ILLUSTRATION 3.2-1 Showing That the Final Result Should Not Depend on the Choice of System A compressor is operating in a continuous, steady-state manner to produce a gas at temperature T2 and pressure P2 from one at T1 and P1. Show that for the time interval Δt ( ) ) ) Q + Ws = H 2 − H1 ΔM where ΔM is the mass of gas that has flowed into or out of the system in the time Δt. Establish this result by (1) first writing the balance equations for a closed system consisting of some convenient element of mass, and then (2) by writing the balance equations for the compressor and its contents, which is an open system. 2012/2/6 18 3.2-1 (1) The closed-system analysis t t +Δt (V2)t = (V1) t+ Δt (1 + 2012/2/6 C) portion (C + 2) portion 19 (1) The mass balance for the closed system M 2 ( t + Δt ) + M c ( t + Δt ) = M 1 ( t ) + M c ( t ) The compressor is in stady-state operation, M c ( t + Δt ) = M c ( t ) . Thus, M 2 ( t + Δt ) = M 1 ( t ) = ΔM The energy balance for this system, neglecting the potential and kinetic energy terms, is ( M Uˆ ) 2 2 t +Δt ( + M cUˆ c ) t +Δt ( ) ( − M 1Uˆ1 − M cUˆ c t Q ) t ˆ ˆ = Ws + Q + PV 1 1 M 1 − PV 2 2 M 2 (a) { } { − ∫ PdV = − P1 ∫ dVinlet − P2 ∫ dVoutlet = − P1 (V1 )t +Δt − (V1 )t − P2 (V2 )t +Δt − (V2 )t ˆ ˆ − ∫ pdV = PV 1 1 − PV 2 2 = PV 1 1 M 1 − PV 2 2M 2 L Net PV -work The compressor is in steady-state operation, ∴ (a ) ⇒ ACC= 2012/2/6 ( ΔM (Uˆ ) ( M Uˆ ) c c t +Δt = ( M Uˆ ) c c } & − P dV W& = Ws dt In (3.1-4a) t ˆ ˆ ΔM Uˆ 2 − Uˆ1 = Ws + Q + PV 1 1ΔM − PV 2 2 ΔM 2 ˆ ˆ ˆ + PV 2 2 − U1 − PV 1 1 ) ( ) = ΔM Hˆ 2 − Hˆ 1 = Ws + Q Proved 20 (2) The mass balance for the open system System: the gas within the compressor (3.1-6) Open steady-state process, constant energy, volume ⎧⎪ ⎧⎪ ⎛ v2 ⎞ ⎫⎪ ⎛ v2 ⎞ ⎫⎪ − ⎨U + M ⎜ + ψ ⎟ ⎬ ⎨U + M ⎜ + ψ ⎟ ⎬ ⎪⎩ ⎝ 2 ⎠ ⎪⎭t +Δt ⎪⎩ ⎝ 2 ⎠ ⎪⎭t 2 2 ⎛ ⎞ ⎛ ⎞ t +Δt v v =∫ M& 1 ⎜ Hˆ 1 + 1 + ψ 1 ⎟ dt + ∫ M& 2 ⎜ Hˆ 2 + 2 + ψ 2 ⎟ dt + Q + W t t 2 2 ⎝ ⎠ ⎝ ⎠ ⎛ ˆ v12 ⎞ ⎛ ˆ ⎞ v22 ACC= 0 = ΔM1 ⎜ H1 + + ψ 1 ⎟ + ΔM 2 ⎜ H 2 + + ψ 2 ⎟ + Q + Ws 2 2 ⎝ ⎠ ⎝ ⎠ Neglecting EKE , EP , t +Δt ( ) ( ) = ( Hˆ − Hˆ ) ΔM 0 = ΔM1 Hˆ 1 + ΔM 2 Hˆ 2 + Q + Ws Q + Ws 2012/2/6 2 1 Proved 21 COMMENT The final results are exactly the same for both closed and open system calculations. Result (open case) = Result (closed case) For closed case, the volume of the system changes, the PV work is involved; for open case, the enthalpy which contain the flow work is involved. Somehow, PV work is related to flow work. PV work (closed) = Flow work (open) For this particular example, the choice of open case is easier than the choice of closed case since the process is open steadystate operation. However, in other case, the closed system may be easy to apply such as for a piston cylinder system. Easier choice based on the nature of the operation: 2012/2/6 Open steady state process Î using open system Gas within piston and cylinder Î using closed system 22 ILLUSTRATION 3.2-2 – – 2012/2/6 Showing That Processes in Closed and Open Systems Are Different A mass M of gas is to be compressed from temperature T1 and pressure P1 to T2 and P2 in a one-step process in a frictionless piston and cylinder and a continuous process in which the mass M of gas is part of the feed stream to the compressor of the previous illustration(3.2-1). Compute the sum Q + W for each process. 23 a. The piston-and-cylinder process Here we take the gas within the piston and cylinder as the system. The energy balance for this closed system is ( ) M Uˆ 2 − Uˆ1 = Q + W = Q − ∫ PdV 2012/2/6 24 b. The flow compressor process System: the contents of the compressor Recalled illustration(3.2-1) Open steady-state process 3.1-4a becomes: ( ) 0 = M Hˆ 1 − Hˆ 2 + Q + W ( ) M Hˆ 2 − Hˆ 1 = Q + W = Q + Ws H2 H1 dV/dt = 0 2012/2/6 25 COMMENT Flow and nonflow systems For closed (nonflow) system, the changes of Q and PV-work results in the internal energy change of the system ( Q − ∫ PdV = M Uˆ 2 − Uˆ1 ) For flow system, the changes of Q and Ws result in the enthalpy changes between the inlet/outlet streams ( ) M Hˆ 2 − Hˆ 1 = Q + W = Q + Ws 2012/2/6 26 ILLUSTRATION 3.2-3 A Joule-Thomson or Isenthalpic Expansion A gas at pressure P1 and temperature T1 is steadily exhausted to the atmosphere at pressure P2 through a pressure-reducing valve. Find an expression relating the downstream gas temperature T2 to P1, P2 and T1. Since the gas flows through the valve rapidly one can assume that there is no heat transfer to the gas. Also, the potential and kinetic energy terms can be neglected. 2012/2/6 27 SOLUTION System: gas within (involving) the pressure-reducing valve The flow is steady. No heat (Q) or work (Ws) flows No PV-work, constant V Valve is not a shaft equipment (neither a compressor nor a pump) Neglecting the kinetic and potential energy changes Mass balance (Molar) 0 = N&1 + N& 2 Energy balance 0 = N&1 H1 + N& 2 H 2 Thus, (2.2-1b) H1 = H 2 H 1 (T1, P1 ) = H 2 (T2 , P2 ) The relation between T1, P1, T2 , P2 can be found. 2012/2/6 28 COMMENTS No heat and energy flows, the thermodynamic properties of the system remain the same values; thus, enthalpies in the upstream and downstream states must be the same. Knowing H2 = H1 and either one of T2 (or P2), then, the outlet condition P2 (or T2) can be found out. 2012/2/6 29 3.3 THE THERMODYNAMIC PROPERTIES OF MATTER The internal energy (U) and enthalpy (H) in energy balance equation are thermodynamic properties which depend on the state of the system. The thermodynamic properties can be obtained from PVT data, graphical figures, or Tables (e.g. steam tables) Equations of state calculations (f (P,V,T) = 0) 2012/2/6 Ideal gas law van der Waals equation 30 An ideal gas An ideal gas is a gas at such a low pressure that there are no interactions among its molecules. Also, an ideal gas occupies negligible molecule volume. The ideal gas law can be expressed as PV = RT (3.3-1) Since there is no intermolecular forces, the internal energy and enthalpy of an ideal gas are function of temperature only. U = f (T) and H = f (T) 2012/2/6 31 Constant-volume heat capacity definition If a sufficiently small quantity of heat is added to the system, it is observed that the temperature rise produced, ΔT, is linearly related to the heat (Q) added and inversely proportional to the number of moles (N) in the system: Q = C ΔT N For constant V process (N =1 mol), dU = dQV ⎛ ∂U (T ,V ) ⎞ dQV ⎛ ∂U ⎞ CV = =⎜ ⎟ ⎟ =⎜ dT ⎝ ∂T ⎠V ⎝ ∂T ⎠V 2012/2/6 (3.3-2) 32 Constant-pressure heat capacity definition Q = C ΔT N At constant pressure (N = 1 mol), dH = dQP ⎛ ∂ H (T , P ) ⎞ dQP ⎛ ∂ H ⎞ =⎜ CP = ⎟ ⎟ =⎜ ∂T dT ⎝ ∂T ⎠ P ⎝ ⎠P 2012/2/6 (3.3-3) 33 Ideal gas heat capacity CP* = a + bT + cT 2 + dT 3 + L CP* = H IG U IG d H d (U + RT ) = = CV* (T ) + R dT dT (T2 ) = H IG (T1 ) + ∫ CP* (T ) dT T2 T1 (T2 ) = U (T1 ) + ∫T IG T2 CV* (T ) dT 1 Set T1 = TR = 0 K, 2012/2/6 H IG U IG H IG (0 K) = 0 constant C → CP* (T ) T (T ) = ∫0 CP* (T ) dT ⎯⎯⎯⎯ T P constant C → CV* (T ) T (T ) = ∫0 CV* (T ) dT ⎯⎯⎯⎯ T V 34 PVT obtained from graphical figures How to sharp see H-S, T-S, P-H diagrams? H-S: P-solid; T-dash T-S: P-solid; H-solid; V-dash P-H: T-solid; V-big dash; S-dash 2012/2/6 35 P, k Pa Figure 3.3-1 (a) Enthalpy Entropy of Mollier diagram For steam (H-S diagram) T, oC H 2012/2/6 S 36 Figure 3.3-1 (b) Temperature Entropy diagram for steam (T-S diagram) T 2012/2/6 S 37 Figure 3.3-2 Pressure-enthalpy diagram for methane (P-H) P 2012/2/6 H 38 Figure 3.3-3 Pressure-enthalpy diagram for nitrogen P 2012/2/6 H 39 Figure 3.3-4 Pressure-enthalpy diagram for HFC-134a P 2012/2/6 H 40 Properties of a two-phase mixture: the lever rule θˆ = ω Iθˆ I + ω IIθˆ II = ω Iθˆ I + (1 − ω I )θˆ II θˆ, an intensive property (3.3-9) ω I , the mass fraction of the system that is in phase I θˆ I , the value of the variable in phase I For mixtures of steam and water, the mass fraction of steam is termed the quality and is frequently expressed as a percent %. 2012/2/6 41 Energy difference between two phases Δ vapHˆ = Hˆ V − Hˆ L = enthalpyof vaporization per unit mass or on a molar basis Δ vap H = H − H = molar enthalpyof vaporization V L Δfus H = H − H = molar enthalpyof melting or fusion L S Δsub H = H − H = molar enthalpyof sublimation V 2012/2/6 S 42 Simplifications Solid or liquids at low pressure H ≈U (3.3-10) Idealized incompressible fluid or solid ⎛ ∂V ⎞ ⎜ ⎟ =0 ⎝ ∂P ⎠T 2012/2/6 (3.3-11) 43 3.4 APPLICATIONS OF THE MASS AND ENERGY BALANCES 2012/2/6 Some examples as follows 44 ILLUSTRATION 3.4-1 Joule-Thomson Calculation Using a Mollier Diagram and Steam Tables Steam at 400 bar and 500 oC undergoes a Joule-Thomson expansion to 1 bar. Determine the temperature of the steam after the expansion using a. Fig. 3.3-1a b. Fig. 3.3-1b c. The steam tables in Appendix A.III 2012/2/6 45 Solution (3.4-1a, b) We start from Illustration 3.2-3, where it was shown that Hˆ 1 = Hˆ (T1 , P1 ) = Hˆ (T2 , P2 ) = Hˆ 2 for a Joule-Thomson expansion. Since T1 and P1 are known, H1 can be found from either Fig. 3.3-1 or the steam tables. Then, since H2 = H1, and P2 are known, T2 can be found. a. Using Fig. 3.3-1a, we first locate the point P = 400 bar = 40 000 kPa and T = 500oC, which corresponds to H1 = 2900 kJ/kg. Following a line of constant enthalpy to P = 1 bar = 100 kPa, we find that the final temperature is about 214oC. b. Using Fig. 3.3-1b, we locate the point P = 400 bar and T = 500oC and follow the curved line of constant enthalpy to a pressure of 1 bar to see that T2 = 214oC. 2012/2/6 46 Solid line 40,000 kPa Dash line Figure 3.3-1 (a) Enthalpy Entropy of Mollier diagram For steam H1 = H2 = 2900kJ/kg 2012/2/6 100 kP 500 oC 214.0oC 47 Solution (3.4-1c) c. Using the steam tables of Appendix A.III, we have that at P = 400 bar = 40 MPa and T = 500oC, H = 2903.3 kJ/kg. At P = 1 bar = 0.1 MPa, H = 2875.3 kJ/kg at T = 200oC and H = 2974.3 kJ/kg at T = 250oC. Assuming that the enthalpy varies linearly with temperature between 200 and 250 oC at P = 1 bar, we have by interpolation 2903.3 − 2875.3 T = 200 + (250 − 200 ) × = 214.1o C 2974.3 − 2875.3 2012/2/6 T, oC P, MPa H, kJ/kg 200 0.1 2875.3 (214.1) 0.1 2903.3 250 0.1 2974.3 48 COMMENT For many problems a graphical representation of thermodynamic data, such as Figure 3.3-1, is easiest to use, although the answers obtained are approximate and certain parts of the graphs may be difficult to read accurately. The use of tables of thermodynamic data, such as the steam tables, generally leads to the most accurate answers; however, one or more interpolations may be required. For example, if the initial conditions of the steam had been 475 bar and 530oC instead of 400 bar and 500oC, the method of solution using Fig. 3.3-1 would be unchanged; however, using the steam tables, we would have to interpolate with respect to both temperature and pressure to get the initial enthalpy of the steam, shown in next page. 2012/2/6 49 Interpolation between temperatures and *: first cal’d; **: second cal’d pressures Enthalpy kJ/kg 500oC 530oC 550oC 40 MPa 2903.3 (3050.8)* 3149.1 47.5 MPa 50 MPa 2012/2/6 [2924.0]** 2720.1 (2899.7)* 3019.5 50 ILLUSTRATION 3.4-2 Application of the Complete Energy Balance Using the Steam Tables An adiabatic steady-state turbine is being designed to serve as an energy source for a small electrical generator. The inlet to the turbine is steam at 600 oC and 10 bar, with a mass flow rate of 2.5 kg/s through an inlet pipe that is 10 cm in diameter. The conditions at the turbine exit are T = 400 oC and P = 1 bar. Since the steam expands through the turbine, the outlet pipe is 25 cm in diameter. Estimate the rate at which work can be obtained from this turbine. 2012/2/6 51 Solution System: the turbine and its contents Open, steady-state flow process Constant mass, volume (no PV work), and energy in the system Mass Balance dM = 0 = M& 1 + M& 2 dt (2.2-1b) Energy balance 2 2 ⎛ ⎞ ⎛ ⎞ & ⎛ v2 ⎞⎫ v v d ⎧ 1 2 & & ˆ ˆ ⎜ ⎟ ⎜ ⎟⎟ + Ws ⎜ ⎟ U M gh = = M H + + M H + + + 0 ⎬ ⎨ 1⎜ 1 2⎜ 2 ⎟ ⎜ ⎟ dt ⎩ 2⎠ 2⎠ ⎝ ⎝ ⎝2 ⎠⎭ (3.1-4a) 2012/2/6 52 Energy balance 2 2 ⎛ ⎞ ⎛ ⎞ & v v 1 2 & & ˆ ˆ 0 = M 1 ⎜ H1 + ⎟ + M 2 ⎜ H 2 + ⎟ + Ws 2⎠ 2⎠ ⎝ ⎝ 2 2 ⎛ ⎞ ⎛ ⎞ v v 1 2 & & & ˆ ˆ Ws = − M 1 ⎜ H1 + ⎟ − M 2 ⎜ H 2 + ⎟ 2⎠ 2⎠ ⎝ ⎝ 2 3 3 d π kg m m 2 m & ˆ v [ =] Volumetric flow rate = MV = [ =] [ =] m 4 s kg s s kg m3 4 ⋅ 2.5 ⋅ 0.4011 s kg 4M& 1Vˆ1 v1 = = 2 2 π din 3.14159 ⋅ ( 0.1m ) 2012/2/6 m m = 127.7 ; v2 = 158.0 s s 53 2 2 ⎛ ⎞ ⎛ ⎞ v v 1 2 & & & ˆ ˆ Ws = − M 1 ⎜ H1 + ⎟ − M 2 ⎜ H 2 + ⎟ 2⎠ 2⎠ ⎝ ⎝ kg ⎧ ˆ 1 2 2 ⎫ kJ ˆ = −2.5 ⎨ H1 − H 2 + v1 − v2 ⎬ ; s ⎩ 2 ⎭ kg ( ) ( ) m2 1 J =1 kg 2 s J ⎛ ⎞ 2 1 ⎜ kg kJ 1 kg 1kJ ⎟ 2 2 m ⎟ = −2.5 ⎜ 419.7 + 127.7 − 158.0 ⋅ 2 ⋅ 2 m 1000 J ⎟ s ⎜ kg 2 s ⎜ ⎟ 2 s ⎝ ⎠ kg kJ kJ = −2.5 ( 419.7 − 4.3) = −1038.5 ( = -1329 hp) s kg s ( 2012/2/6 ) 54 COMMENT If we had completely neglected the kinetic energy terms in this calculation, the error in the work term would be 4.3 kJ/kg, or about 1%. Generally, the contribution of kinetic and potential energy terms can be neglected when there is a significant change in the fluid temperature. 2012/2/6 55 ILLUSTRATION 3.4-3 Use of Mass and Energy Balances with an Ideal Gas A compressed-air tank is to be pressurized to 40 bar by being connected to a high-pressure line containing air at 50 bar and 20 oC. The pressurization occurs so quickly that the process can be assumed to be adiabatic; also, there is no heat transfer from the air to the tank. Assuming air to be an ideal gas with CV = 21 J/(mol K). a. If the tank initially contains air at 1 bar and 20 oC, what will be the temperature of the air in the tank at the end of the filling process? b. After a sufficiently long period of time, the gas in the tank is found to be at room temperature (20 oC) because of heat exchange with the tank and the atmosphere. What is the new pressure of air in the tank? 2012/2/6 56 Solution System: the contents of the tank 1. The kinetic and potential energies are small and neglected. 2. Since the tank is connected to a source of gas at constant temperature and pressure, Hin is constant. 3. The initial process is adiabatic, so Q = 0, and the system (the contents of the tank) is of constant volume, so ΔV=0. mass balance: (3.1-5a) energy balance: 2012/2/6 N2 − N1 = ΔN N2U 2 − N1U 1 = (ΔN )H in 57 Ideal gas, U = CVT , H = CPT , N = PV RT Energy balance N 2CVT2 − N1CVT1 = ( ΔN ) CPTin = ( N 2 − N1 ) CPTin ⎛ P2V2 PV ⎞ P2V2 PV 1 1 1 1 CVT2 − CVT1 = ⎜ − ⎟ CPTin RT2 RT1 R RT T 2 1⎠ ⎝ Eliminate V2 = V1 and R ⎛P P⎞ P2CV − P1CV = ⎜ 2 − 1 ⎟ CPTin ⎝ T2 T1 ⎠ CV ( P2 − P1 ) P1 P2 + = CPTin T1 T2 T2 = P2 CV ( P2 − P1 ) CPTin 2012/2/6 + P1 T1 P2= 40bar Tin= 293.15K P1= 1bar T1= 293.15K CV= 21 J/(molK) CP= CV+8.314 58 2012/2/6 (a) T2 = 404.2 K = 132.05 oC (b) The tank is the system: N2 =N1 T2 = 20oC, P2/P1= (N2RT2/V2) / (N1RT1/V1) = P2 =P1 (T2 /T1)= 40(293.15/404.2)= 28.94 bar Discussion: (1) Note that T2 = 132.05 oC >> 20 oC, because pressurization increases the temperature. In other words, due to H(20 oC) = U (20 oC) + PV(20 oC) Î U (20 oC) increase to final value of U (132.05 oC). 59 ILLUSTRATION 3.4-4 Example of a Thermodynamics Problem That Cannot Be Solved with Only the Mass and Energy Balances A compressor is a gas pumping device that takes in gas at low pressure and discharges it at a higher pressure. Since this process occurs quickly compared with heat transfer, it is usually assumed to be adiabatic: that is, there is no heat transfer to or from the gas during its compression. Assuming that the inlet to the compressor is air [which we will take to be an ideal gas with CP* = 29.3 J/(mol K)] at 1 bar and 290 K. and that the discharge is at a pressure of 10 bar, estimate the temperature of the exit gas and the rate at which work is done on the gas for a gas flow of 2.5 mol/s. 2012/2/6 60 Solution (Since there are two unknown thermodynamic quantities, the final temperature and the rate at which work is being done, we can anticipate that the mass and energy balances will not be sufficient to solve this problem.) The system will be taken to be the gas contained in the compressor. We have used the subscript 1 to indicate the flow stream into the compressor and 2 to indicate the flow stream out of the compressor. Since the compressor operates continuously, the process may be assumed to be in a steady state, 2012/2/6 61 (3.1-5a) dN = N& 1 + N& 2 = 0; N& 1 = − N& 2 dt dU = N& 1 H 1 + N& 2 H 2 + Q& + W& = 0; dt * & & & & Ws = N1 H 1 − N1 H 2 = N1CP (T2 - T1 ) W s = C (T2 - T1 ) * P Two unknows : T2 and W s 2012/2/6 62 We need additional information about the system before we can solve the problem. This additional information will be obtained using an additional (Entropy) balance equation developed in the next chapter (Chapter 4). 2012/2/6 63 ILLUSTRATION 3.4-5 Use of Mass and Energy Balances to Solve an Ideal Gas Problem A gas cylinder of 1 m3 volume containing nitrogen initially at a pressure of 40 bar and a temperature of 200 K is connected to another cylinder of 1 m3 volume that is evacuated. A valve between the two cylinders is opened until the pressures in the cylinders equalize. Find the final temperature and pressure in each cylinder if there is no heat flow into or out of the cylinders or between the gas and the cylinder. You may assume that the gas is ideal with a constant-pressure heat capacity of 29.3 J/(mol K). 40 bar/200K 2012/2/6 0 bar 1 m3 1 m3 Cylinder 1 Cylinder 2 20 bar / ???K 20 bar / ???K 64 Solution This problem is more complicated than the previous ones because we are interested in changes that occur in two separate cylinders. We can try to obtain a solution to this problem in two different ways. First, we could consider each tank to be a separate system, and so obtain two mass balance equations and two energy balance equations, which are coupled by the fact that the mass flow rate and enthalpy of the gas leaving the first cylinder are equal to the like quantities entering the second cylinder. Alternatively, we could obtain an equivalent set of equations by choosing a composite system of the two interconnected gas cylinders to be the first system and the second system to be either one of the cylinders. In this way the first (composite) system is closed and the second system is open. We will use the composite system here: you are encouraged to explore the first system choice independently and to verify that the same solution is obtained. 2012/2/6 65 Mass balance: N1i = N1f + N 2f (a) Energy balance: N1i U 1 = N1f U 1 + N 2f U 2 i f f (b) The ideal gas equation, N = PV RT P1iV1i P1 f V1 f P2 f V2 f i f f = + N1 = N1 + N 2 ⇒ RT1i RT1 f RT2 f P1i P1 f P2 f Same values for R and V , = f + f i T1 T1 T2 (a') QU = CV T ,EQ( b ) ⇒ ⎛ P1iV1i ⎞ ⎛ P1 f V1 f i ⎜ RT i ⎟ CV T1 = ⎜ RT f ⎝ 1 ⎠ ⎝ 1 Q ⎞ ⎛ P2 f V2 f f ⎟ CV T1 + ⎜ T f T f ⎠ ⎝ 2 2 P1i = P1 f + P2 f ⎞ f C T ⎟ V 2 ⎠ (c) 1 = P1i = 20 bar 2 ∴ EQ (a') ⇒ P1 f = P2 f 2 1 1 = + T1i T1 f T2 f 2012/2/6 (c') 66 the open but not S.S system: the gas initially filled LHS cylinder dN1 & =N dt d ( N1U 1 ) & ACC = = N H 1 which is not ZERO dt d ( N1U 1 ) d (U 1 ) d ( N1 ) dN = N1 +U1 = H1 1 dt dt dt dt d (U 1 ) dN N1 = ( H1 −U1 ) 1 dt dt For ideal gas, ⎛ PV ⎞ d⎜ 1 1⎟ d (T1 ) RT PV 1 1 CV = ⎡⎣( CP − CV )T1 ) ⎤⎦ ⎝ 1 ⎠ dt RT1 dt ⎛P⎞ d⎜ 1⎟ d (T1 ) T P1 CV = (T1 ) ⎝ 1 ⎠ RT1 dt dt ⎛P⎞ d ln ⎜ 1 ⎟ CV d lnT1 ⎝ T1 ⎠ = R dt dt 2012/2/6 67 ⎛ P1 ⎞ d ln ⎜ ⎟ CV d lnT1 ⎝ T1 ⎠ = R dt dt Integrate from i to f for the LHS cyclinder: ⎛ T1 f ⎞ ⎜ Ti ⎟ ⎝ 1 ⎠ CV R ⎛ P1 f ⎞⎛ T1i ⎞ = ⎜ i ⎟⎜ f ⎟ ⎝ P1 ⎠⎝ T1 ⎠ CP R ⎛ T1 f ⎞ ⎛ P1 f ⎞ ⎜ T i ⎟ = ⎜ Pi ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ T1 f = ( 200 )e ⎛ ln 0.5 ⎞ ⎜ ⎟ ⎝ 3.488 ⎠ EQ( f ) = 164.3 K; Substituted into EQ(a'): T2 f = 255.6 K Using ideal gas EOS: N1f = 1.464 mol; N 2f = 0.941 mol 2012/2/6 68 ILLUSTRATION 3.4-6 Showing That the Change in State Variables between Fixed Initial and Final States Is Independent of the Path Followed ΔU=f (state); ΔH=f (state) Q, W = f (path) One mole of a gas at a temperature of 25 oC and a pressure of 1 bar (the initial state) is to be heated and compressed in a frictionless piston (reversible process) and cylinder to 300 oC and 10 bar (the final state). 2012/2/6 69 ILLUSTRATION 3.4-6 (continued) Compute the heat and work required along each of the following paths. 2012/2/6 Path A. Isothermal (constant temperature) compression to 10 bar, and then isobaric (constant pressure) heating to 300 oC. Path B. Isobaric heating to 300 oC followed by isothermal compression to 10 bar. Path C. Adiabatic compression in which PVγ = constant, where γ = Cp/Cv followed by an isobaric cooling or heating, if necessary, to 300 oC. For simplicity, the gas is assumed to be ideal with CP = 38 J/(mol K). 70 2012/2/6 71 Q, W, ΔU, and ΔH for an ideal gas system Reversible process Isothermal (CONSTANT T) Isobaric (CONSTANT P) Isochoric (CONSTANT V) Q W ΔU (=CV ΔT) - RTln(P2/P1) RTln(P2/P1) 0 CP ΔT = ΔH - R ΔT CV ΔT = ΔU 0 0 CV ΔT Adiabatic (Q = 0) T2 ⎛ P2 ⎞ =⎜ ⎟ T1 ⎝ P1 ⎠ (γ −1) γ 2012/2/6 C where γ = P CV ΔH (=CP ΔT) 0 72 Path A i. Isothermal compression at 298.15K Wi = − ∫V PdV = − ∫V V2 V2 1 1 RT V P dV = − RT ln 2 = RT ln 2 V V1 P1 = 8.314 J/ ( mol K ) × 298.15 K × ln 10 = 5707.7 J/mol 1 ΔU i = CV* ΔT = 0 = Δ H i Qi = ΔU i − Wi = 0 − 5707.7 = −5707.7 J/mol ii. Isobaric heating Wii = − ∫V P2 dV = − P2 ∫V dV = − P2 (V 3 − V 2 ) V3 V3 2 2 = − R (T3 − T2 ) = −8.314 J/ ( mol K ) × 275 = −2286.3 J/mol Qii = Δ H i = CP* ΔT = 38 J/ ( mol K ) × 275 = 10450 J/mol For the overall process Q = Qi + Qii = 4742.3 J/mol W = Wi + Wii = 3421.4 J/mol ΔU = 8163.7 J/mol 2012/2/6 73 Path B i. Isobaric heating at 1 bar Qi = Δ H i = CP* ΔT =10450 J/mol Wi =-R (T2 -T1 ) = −2286.3 J/mol; (work done by system) ii. Isothermal compression at 300K P2 10 Wii = RT ln = 8.314(300+273.15) ln = 10972.2 J/mol 1 P1 Qii = −10972.2 J/mol Q = Qi + Qii = -552.2 J/mol W = Wi + Wii = 8685.9 J/mol ΔU = 8163.7 J/mol 2012/2/6 74 Path C γ i. Compression with P V = constant γ γ ⎛ RT1 ⎞ ⎛ RT2 ⎞ γ γ PV = P = P V = P ⎟ ⎟ 1 1 1⎜ 2 2 2⎜ P P ⎝ 1 ⎠ ⎝ 2 ⎠ T2 ⎛ P2 ⎞ =⎜ ⎟ T1 ⎝ P1 ⎠ ( γ −1) / γ T2 = 298.15 K (10 ) 0.28/1.28 =493.38 K Wi = CV (T2 − T1 ) = 5795.6 J/mol Qi = 0 ΔU i = Wi = 5795.6 J/mol ii. Isobaric heating Qii = C p (T3 − T2 ) = 3031.3 J/mol Wii = − R (T3 − T2 ) = - 663.2 J/mol; T3 =573.15K (work done by system) Q =3031.3 J/mol W =5132.4 J/mol 2012/2/6 75 Overall process 2012/2/6 Path Q(J/mol) W(J/mol) Q+W=ΔU (J/mol) A 4742.3 3421.4 8163.7 B -522.2 8685.9 8163.7 C 3031.3 5132.4 8163.7 76 Homework assignments Problems 3.3, 3.5, 3.13, 3.15, 3.19, 3.29 2012/2/6 77 Illistration 3-4.1 Regression of Steam Table Data at 47.5MPa + 500C Using MATHCAD Method − ( 1 ) ⎛⎜ 40 40 Mxy := ⎜ ⎜ 50 ⎜ 50 ⎝ 500 ⎞ ⎟ 550 ⎟ 500 ⎟ ⎟ 550 ⎠ ⎛⎜ 2903.3 ⎞⎟ 3149.1 ⎟ vz := ⎜ ⎜ 2720.1 ⎟ ⎜ 3019.5 ⎟ ⎝ ⎠ v := ⎛ 47.5 ⎞ ⎜ ⎟ ⎝ 530 ⎠ cc := regress( Mxy , vz , 1 ) z := interp( cc , Mxy , vz , v ) z = 2.9362 × 10 3 ANSWER −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Method − ( 2 ) Since doing the first two interpolations, doing the last interpolations as follows: d := 3050.8 + 7.5⋅ ⎛⎜ ⎝ 2899.7 − 3050.8⎞ 10 3 d = 2.9375× 10 ⎟ ⎠ 3 2903.3 + 0.6⋅ ( 3149.1 − 2903.3) = 3.0508× 10 3 2720.1 + 0.6⋅ ( 3019.5 − 2720.1) = 2.8997× 10 vvx := ⎛ 500 ⎞ ⎜ ⎟ ⎝ 550 ⎠ vvy := ⎛ 2903.3⎞ ⎜ ⎟ ⎝ 3149.1⎠ vvs := regress( vvx , vvy , 1 ) dd1 := interp( vvs , vvx , vvy , 530 ) vvx := ⎛ 500 ⎞ ⎜ ⎟ ⎝ 550 ⎠ vvy := dd1 = 3.0508 × 10 3 ⎛ 2720.1 ⎞ ⎜ ⎟ ⎝ 3019.5 ⎠ vvs := regress( vvx , vvy , 1 ) dd2 := interp( vvs , vvx , vvy , 530 ) dd2 = 2.8997 × 10 3 Using dd1 and dd2, put into the following regression (REGRESS) and interpolation (INTERP): vvx := ⎛ 40 ⎞ ⎜ ⎟ ⎝ 50 ⎠ vvy := ⎛ dd1 ⎞ ⎜ ⎟ ⎝ dd2 ⎠ vvs := regress( vvx , vvy , 1 ) dd := interp( vvs , vvx , vvy , 47.5) 3 dd = 2.9375 × 10 ANSWER Summary for Mass and Energy Balance Equations Referred to 4th Ed., Ch 3. pp50 (or 2nd Ed.; 3rd Ed., pp25-30) of Sandler textbook General mass and energy balance equations in general form: Accumulation = In − Out + Generation 1. Mass balance equation in mass rate of change form, defined Table 2.2-1 (3rd Ed.): dM K & = ∑ Mk dt k K: total # of entry ports; k: an arbitrary port 2. Conservation of energy balance equation: Defined in 2.3 (3rd); 3.1(4th) K v2 (1) ∑ M& k ( U + of kth flow stream {energy come into system} +ψ ) 2 k =1 k v2 d (2) ( U + M ( + ψ )) dt 2 & (3) Q of the system {energy accumulation} {Heat added to the system} K (4) ∑ M& k ( PVˆ ) k {Net work done on the system due to pressure forces} k =1 (5) − P (6) Ws dV dt {Work done by the system through an expansion} {Net shaft work done by the system through shaft work} Differential form of energy balances (i.e. mass rate of change form) become: (2) = (1) + (4) + (3) + (6) + (5): 2 K ⎞ K ⎛ ⎞ d ⎛⎜ v2 ⎟ = ∑ M& k ⎜Û + v + ψ ⎟ + ∑ M& k ( PV̂ )k + Q& + W& s + ( − P dV ) U M ( ψ ) + + ⎟ k =1 ⎜ ⎟ k =1 2 2 dt ⎜⎝ dt ⎠ ⎝ ⎠k ⇒ 2 ⎞ K ⎛ ⎞ d ⎛⎜ v2 ⎟ = ∑ M& k ⎜ Ĥ + v + ψ ⎟ + Q& + W& U M ( ψ ) + + ⎟ k =1 ⎜ ⎟ dt ⎜⎝ 2 2 ⎠ ⎝ ⎠k where W& = W& s + ( − P dV ) dt Table 2.3-1 (3rd); 3.1-4a (4th) 78