Isentropic Efficiencies

Outline Unit Ten – Isentropic Efficiencies • Review maximum adiabatic work • Goals for unit ten – Isentropic Efficiencies – Second law for open systems • Problem solving with isentropic efficiencies for work input and output devices • Second law for open systems 2 Review Maximum Adiabatic Work • Maximum adiabatic work is for s = 0 • As a result of studying this unit you should be able to use the concept of isentropic efficiency, s – This is also minimum adiabatic work input • Solve using tables – From given inlet (initial) conditions, find the initial state properties including sinitial – From sfinal = sinitial and one other property of final state get all final state properties • Solve for ideal gases – Isentropic equations for constant heat capacity or ideal gas tables give final state • Ideal gas tables or cT for energy or enthalpy dScv Q   m o so   m i si  cv 1 d LW    Sirr dt outlet T inlet T dt  dScv Q 1 d LW    m o so   m i si  cv  dt outlet T T dt inlet  m s   m s o o i i 4 What is s? • As a result of studying this unit you should be able to use the second law for open systems outlet – Empirical correction factor to get actual work from maximum work – One condition on final state (P or v) is the same for both actual and maximum work – The s equations for work input and work output are different – Actual and ideal processes differ in final temperature giving difference in work 3 Second Unit Ten Goal For steady, adiabatic, reversible systems First Unit Ten Goal 0 inlet 5 • Empirical correction factor, based on past design experience, to obtain actual work output from isentropic work • It is not a loss of energy • Different definitions for work input and work output devices • Compares ideal work, |ws|, and actual work, |wa| or |w|, with same inlet state and same final pressure (or volume) • Plot this on h-s coordinates 6 1 h-s Coordinates? Water h-s Isobars 6000 • We have displayed properties on P-v and T-v charts • We can use any coordinates to display properties • T-s coordinates used because area under curve is reversible heat transfer • h-s coordinates illustrate actual and ideal open-system work amounts • Slope of isobar (constant P line) on h-s diagram is the temperature, T P = 2 Mpa Isobar slope is temperature on an h-s diagram, which always slopes upward. Slope is constant in the mixed region and increases in the gas region. 5000 4000 h 3000 2000 Both curves start at saturated liquid and use SI units, h in kJ/kg and s in kJ/kg·K 1000 0 0 2 4 6 7 1 ws h2 h2s h 2 h2s Actual work: |w| = |h1 – h2| w 2s s = |w|/|ws| s h2 2s |ws| Isentropic efficiency for work output device: 2 10 |w| h1 Ideal work: |ws| = |h1 – h2s| Actual work: |w| = |h1 – h2| Isentropic efficiency for work input device: 1 s s = |ws|/|w| 9 Compare Isentropic Efficiencies h h 2 h2s 1 h1 ws w h2 h2s h2 2s |ws| |w| 2 2s Output s = |w|/|ws| h1 s 1 Input s = |ws|/|w| 12 8 Input Isentropic Efficiency Ideal work: |ws| = |h1 – h2s| h1 8 s Output Isentropic Efficiency h P = 20 kPa s 10 Typical Problem • Closed system or steady open system with one inlet and one outlet, given – Inlet (or initial) state – One property (such as pressure or specific volume) for outlet (or final) state – Isentropic efficiency • Process is adiabatic • Find work and actual outlet (final) state 12 2 Solution to Typical Problem Example Problem One • Find maximum work, ws, for isentropic process • An adiabatic steam turbine has an inlet pressure and temperature of 5 MPa and 500oC. Its outlet pressure is 50 kPa and its isentropic efficiency is 84% • Find the actual work and the actual exit state of the turbine • Assumptions: steady flow with negligible changes in kinetic and potential energies • Equations: w = hin – hout ws = hin – hout,s – property tables or ideal gas – ws = hin – hout,s or ws = uinitial – ufinal,s • Actual work from isentropic efficiency – Output: w = s ws Input: w = ws / s • Actual final state from first law – Steady open system: hout = hin – w – Closed system: ufinal = uinitial – w 13 14 Solving Example One Example One Solution • Find isentropic work for H2O with Pin = 5 • Same first law for both maximum and actual work: ws = hin – hout,s w = hin – hout • Use property tables for water: MPa, Tin = 500oC and Pout = 50 kPa • Actual work is s (84%) times isentropic work for work output device • Get actual outlet state as hout = hin – wactual • Actual outlet state important when it is input to subsequent device – At initial state of 5 MPa and 500oC, hin = 3434.7 kJ/kg and sin = 6.9781 kJ/kg∙K – Outlet Pout = 50 kPa, sout,s = sin is mixed – xout,s = (s – sf)/sfg = (6.9781 – 1.0912)/6.5019 – hout,s = 340.54 + (0.905)2304.7 = 2427.7 kJ/kg 15 16 Example One Concluded • Isentropic work, ws = hin – hout,s = 3434.7 kJ/kg – 2427.7 kJ/kg = 1007.0 kJ/kg • Actual work, w = s ws = (0.84) (1007.0 kJ/kg) = 845.9 kJ/kg • Actual outlet state – hout = hin – w = 3434.7 kJ/kg – 845.9 kJ/kg = 2588.8 kJ/kg – xout = (hout – hf)/hfg = (2588.8 kJ/kg– 340.54 kJ/kg)/(2304.7 kJ/kg) = 0.974 is final state (with Pout = 50 kPa) 17 Two Important Equations • • • • • • • • Second law: ds = (du + Pdv)/T Multiply by T: Tds = du + Pdv Enthalpy: h = u + Pv or u = h – Pv Derivative: du = dh – Pdv – vdP Substitute for du in Tds equation to get Tds = dh – vdP or dh = Tds + vdP For ds = 0, dh = vdP  h    T Slope of isobar on h-s  s  P 18 3 Integrating vdP Example Two • dh = vdP in an isentropic process • See last two charts (not usually covered in lecture) that integrate this for an ideal gas with isentropic constant heat capacity relation: v = CP1/k for ds = 0 • Main application is for liquids in which v is essentially constant • For liquids in an isentropic flow process, h = vp or w = hin – hout = vin(Pin – Pout) • Pump inlet saturated liquid water at 20 kPa and outlet pressure of 6 MPa has isentropic efficiency of 80%. Find work. • ws = vin (Pin – Pout) = (0.001017 m3/kg) (20 kPa – 6000 kPa) = –6.08 kJ/kg – vin used here is vf(20 kPa) • w = ws / s = (–6.08 kJ/kg) / 0.80 = –7.60 kJ/kg • hout = hin – w = 251.42 – (–7.60) = 259.02 kJ/kg 19 20 A Note on dh = vdP Example Three • Cannot do isentropic calculation for liquid from equations • Even computer programs have problems because of convergence • Doing problem on previous chart as isentropic calculation on computer gives h = 5.77 kJ/kg instead of 6.08 kJ/kg • Computer calculation at final state gives v = 0.001015 m3/kg a difference of 0.2% • Given: adiabatic air compressor has an inlet pressure and temperature of 100 kPa and 300 K. Outlet pressure is 1.5 MPa. Isentropic efficiency of 82%. • Find: work and the outlet temperature • Assumptions: Steady system, one inlet and one outlet, negligible changes in kinetic and potential energies • Properties: Ideal gas with variable heat capacity using tables 21 22 Solving Example Three Example Three Solution • Use ideal gas tables for air • Find minimum work input required to compress air from Pin = 100 kPa, Tin = 300 K, to Pout = 1.5 MPa • Find actual work if isentropic efficiency is 82% • Find actual outlet state for actual work and determine the temperature at that state – At Tin = 300 K, Pr = 1.3860, h = 300.19 kJ/kg – Maximum work in isentropic process so that Pr(Tout,s) = Pr(Tin) Pout/ Pin = 20.790 => Tout,s = 641.2 K – Interpolate to get hout,s = 654.94 kJ/kg • • • • – Important if used as inlet to subsequent device 23 First law: ws = hin – hout,s = –354.75 kJ/kg w = ws/s = –354.75/0.82 = –432.6 kJ/kg hout = hin – w = 732.82 kJ/kg Interpolate to find Tout = 718.6 K 24 4 Example Three with Constant cp • Use properties at T = 450 K: cp = 1.020 kJ/kg∙K, k = 1.391 • For constant s in ideal gas with constant cp, T2,s = T1(P2/P1)(k – 1)/k = (300 K)(1500 kPa / 100 kPa)(1.391 – 1)/1.391 = 642.26 K • ws = cp(T1 – T2,s) = (1.020 kJ/kg∙K)(300 K – 642.26) = -349.1 kJ/kg (input) • w = ws/s = (-349.1 kJ/kg)/82% = -425.74 kJ/kg = cp(T1 – T2) Constant cp Example II • T2 = T1 – w/cp = T1 – ws/scp = T1 – cp(T1 – T2,s)/scp = T1 – (T1 – T2,s)/s • T2 = 300 K – (-425.74 kJ/kg)/(1.020 kJ/kg∙K) = 717.4 K • Actual outlet state calculations important if outlet from one device is input to another device 25 Control Volume Entropy 26 Reversible Processes • Reversible process is = part of  • General equations dScv Q   m o so   m i si  cv dt outlet T inlet dS cv Q   m o so   m i si  cv dt outlet T inlet • Reversible adiabatic process dS cv Q L Wcv   m o so   m i si  cv  dt outlet T T inlet dScv   m o so   m i si  0 dt outlet inlet • Steady reversible adiabatic process • Lost work (LW) is lost opportunity to do work due to irreversible processes  m s   m s o o outlet i i 0 inlet 27 Transient Processes • Integrate inequality over time  dS Q   m o so  inlet  m i si  Tcv dt   dtcv  outlet  dScv  dt dt   dScv  Scv  m2 s2  m1s1 cv      m k sk dt    m k sk dt   mk sk m2 s2  m1s1 cv   mo so   mi si   outlet inlet notation notation, used in derivation only, as reminder that we assume inlet and outlet entropies are constant or at Q cv dt some average T value 28 Example Problem  • A steam turbine with given Tin, Pin, and m has two outlets at Pout,1 and pout,2, with given outlet mass flow rates • What is the maximum work? • Find by saying that sout,1 = sout,2 = sin • Satisfies result that maximum adiabatic work is in isentropic process  m s   m s o o outlet i i 0 inlet 30 5 Derivation Details Isentropic ideal gas dh = vdP • Next two slides show application to isentropic work as vdP • Limited to ideal gas with constant heat capacities • Isentropic path for v(P) generally not known • Note general trend of vdP • To integrate this we must know v(P) for an isentropic process • For an ideal gas with constant heat capacities, Pvk = constant so v = CP-1/k • This gives following integration   – Want large v for work output (dP < 0) – Want small v for work input (dP > 0) 31  P2 2  CP11/ k    CP 1/ k dP    const s  1  1 / k  P1 P1 1/ k 11/ k 1/ k 11 / k  P1 v1 P1 P v P k P2v2  P1v1   2 2 2  k 1 k 1 32 k h   dh   vdP P  Integrating Ideal Gas dh = vdP II • The ratio k/(k – 1) is cp/R h  cp cp cp k c  v   k 1 cp c p  cv R 1 cv c k P2v2  P1v1   p RT2  RT1   c p T2  T1  k 1 R • The result is correct as h = cpT, but does not predict T2,which has to be found from another calculation 33 6