AME 436. Energy and Propulsion. Lecture 7 Unsteady-flow (reciprocating) engines 2: Using P-V and T-s diagrams

Transcription

1 AME 46 Energy and ropulsion Lecture 7 Unsteady-flow (reciprocating) engines : Using - and -s diagrams Outline! Air cycles! What are they?! Why use - and -s diagrams?! Using - and -s diagrams for air cycles!!!!!! Seeing heat, work and KE Constant and processes Inferring efficiencies Compression & expansion component efficiencies Correspondence between processes on - and -s Hints & tricks 1

2 Air-cycles - what are they?! Mostly we ll analyze air cycles - working fluid is air (or other ideal gas) with constant properties! Changes in C, M i, γ, etc. due to changes in composition, temperature, etc. are neglected; simplifies analysis, leads to simple analytic expressions for efficiency, power, etc.! Later we ll analyze fuel-air cycles (using GASEQ) where the changes in C, M i, γ, etc. due to changes in composition, temperature, etc. are considered! Fuel-air cycles still can t account for slow burn, heat loss, etc. since it s still a thermodynamic analysis that tells us nothing about reaction rates, burning velocities, etc.! - and -s diagrams will be used provide a clear visual representation of cycles Why use - diagrams?! ressure () vs. time and cylinder volume () vs. time are easily measured in reciprocating-piston engine experiments! d = work! d over whole cycle = net work transfer + net change in KE + net change in E = net heat transfer! Heat addition is usually modeled as constant or constant, so show as straight lines on - diagram! Note that on - diagram is cylinder volume (units m )! NO a property of the gas (it s a property of the cylinder)! Specific volume (v = /m) (units m /kg) IS a property of the gas! We need to use not v since cycle work = d! Can t use v since mass in the cylinder (m) changes during intake / blowdown / exhaust 4

3 Why use -s diagrams?! Idealized compression & expansion modeled as constant S! ds δq/ (an important consequence of the nd Law)! For adiabatic process δq = 0, for reversible = (not >) sign applies, thus ds = 0! Note that ds = 0 still allows for any amount of work transfer (W) to / from the gas! Note nd law does NO lead to ds δq/ - δw/ because there is no entropy change due to work transfer! For reversible process, ds = Q, thus area under -S curves show amount of heat transferred! ds over whole cycle = net heat transfer = net work transfer + net change in KE + net change in E! -S diagrams show the consequences of non-ideal compression or expansion (ds > 0)! For ideal gases, Δ ~ heat transfer or work transfer or ΔKE! Efficiency can be determined by breaking any cycle into Carnot-cycle strips, each strip (i) having η th,i = 1 - L,i / H,i 5 -S & - for control mass: work, heat & KE! For an ideal gas with constant C & C v : Δh = C p Δ, Δu = C v Δ 1st Law for a control mass with ΔE = 0 (OK assumption for IC engines) de = d(u + KE) = d[m(u + KE)] = δq - δw (using KE rather than u / avoid confusion with u {internal energy}) q = Q/m; w = W/m (heat or work transfer per unit mass)! If no work transfer (δw = 0) or KE change (ΔKE = 0), du = C v d = δq q 1 = C v ( - 1 )! If no heat transfer (δq = 0) or ΔKE du = C v d = δw w 1 = -C v ( - 1 )! If no work or heat transfer (δq = δw = 0) ΔKE = KE KE 1 = -C v ( - 1 ) For a control mass containing an ideal gas with constant C v, Δ ~ heat transfer - work transfer - ΔKE! Note that the nd law was not invoked, thus the above statements are true for any process, reversible or irreversible 6

4 -s & -v for control volume: work, heat & KE! 1st Law for a control volume, steady flow, with ΔE = 0 0 = Q W + m "# (h in h out )+ ( KE in KE out ) $ %! For ideal gas with constant C, dh = C d h - h 1 = C ( - 1 )! If = outlet, 1 = inlet, and noting Q / m = q 1 ; W / m = w 1 h h 1 = q 1 w 1 KE KE 1 ( )! If no work transfer KE change (w 1 = ΔKE = 0) dh = C d = dq q 1 = C ( - 1 )! If no heat transfer (dq = 0) or ΔKE dh = C d = δw w 1 = -C ( - 1 )! If no work or heat transfer ΔKE = KE KE 1 = -C ( - 1 ) For a control volume containing an ideal gas with constant C, Δ ~ heat transfer - work transfer - ΔKE (Same statement as control mass with C replacing C v )! Again true for any process, reversible or irreversible 7 -s & -v diagrams: work, heat & KE w 4 + (KE 4 KE ) = -C v ( 4 - ) (control mass) w 4 + (KE 4 KE ) = -C ( 4 - ) (control vol.) q = C v ( - ) (const. ) q = C ( - ) (const. ) 4 1 w 1 + (KE KE 1 ) = -C v ( - 1 ) (control mass) w 1 + (KE KE 1 ) = -C ( - 1 ) (control vol.) Case shown: Const. ol. heat input, r c = r e =, γ = 1.4, Δ comb = fq R /C v = 68K, 1 = atm 8 4

5 -s & -v diagrams: work, heat, KE & E! How do I know that work shown on the - (via d) diagram is the same as that shown (via C v Δ) on the -s diagram? As an example, for isentropic compression γ W = d = d = γ 1 γ γ 1 γ 1 1 γ 1 = 1 γ 1 γ 1 γ 1 γ 1 γ 1 = mr γ 1 γ 1 1 γ 1 1 γ 1 = mc v 1 = mc v 1 = mc v ( ) 9 -s & -v diagrams: work, heat, KE & E! Going back to the 1st Law again de = δq δw de = δq δw! Around a closed path, since E = U + KE + E; since U is a property of the system, du = 0, thus around a closed path, i.e. a complete thermodynamic cycle (neglecting E again) δq = δw + ΔKE But wait - does this mean that the thermal efficiency η th = δw + ΔKE =1? δq No, the definition of thermal efficiency is η th = What you get What you pay for =! For a reversible process, δq = ds, thus δq = ds and ds = δq = δw + ΔKE /m δw + ΔKE = Heat in Heat in heat out Heat in! hus, for a reversible process, the area inside a cycle on a -s diagram is equal to both (net work transfer + net KE) and net heat transfer 10 <1 5

6 -s & -v diagrams: work, heat & KE! Animation: using -s diagram to determine heat & work Net heat transfer = net work + ΔKE Heat transfer in out 11 Constant and curves! For ideal gas with constant specific heats, review of thermodynamics (1st lecture) showed that S S 1 = C ln Rln ; S S = C ln 1 + Rln 1 1! If = constant, ln( / 1 ) = 0 = 1 exp[(s -S 1 )/C ]! If = constant, ln( / 1 ) = 0 = 1 exp[(s -S 1 )/C v ] constant or curves are exponentials on a -S diagram! Since constant or curves are exponentials, as S increases, the Δ between two constant- or constant- curves increases; this ensures that compression work is less than expansion work for ideal Otto (const. ) or Brayton (const. ) cycles! Since C = C v + R, C > C v or 1/C < 1/C v, constant curves rise faster than constant curves on a -S diagram! Constant or constant lines cannot cross (unless they correspond to cycles with different C or C )

7 Constant and curves (s) = exp[(s-s )/C v ] (constant ) (s) = exp[(s-s )/C ] (constant ) 4 1 Constant v or curves spread out as s increases - 4 > - 1 (s) = 1 exp[(s-s 1 )/C v ] (constant ) (s) = 1 exp[(s-s 1 )/C ] (constant ) 1 Constant and curves ayback for compression work and KE decrease ( - 4 ) - ( - 1 ) ~ net work + KE increase ~ Work input + KE decrease during compression Net work + KE decrease ~ ( - 4 ) - ( - 1 ) = ( - 1 )[exp(δs/c v ) - 1] > 0 or ( - 1 )[exp(δs/c ) - 1] >

8 Constant and curves Constant-v heat addition & rejection (Otto cycle) emperature (K) Double-click chart to open Excel spreadsheet Constant- heat addition & rejection (Brayton cycle) Entropy (J/kg-K) Constant-v curves are steeper than constant- curves on the -s Both cases: / 1 = 5, γ = 1.4, fq R = 4.5 x 10 5 J/kg, 1 = atm he two cycles shown also have the same thermal efficiency (η th ) 15 Inferring efficiencies Constant-v heat addition & rejection (Otto cycle) emperature (K) H,i Double-click chart to open Excel spreadsheet L,i Entropy (J/kg-K) Carnot cycle strip η th,i = 1 - L,i / H,i Carnot cycles appear as rectangles on the -s diagram; any cycle can be broken into a large number of tall skinny Carnot cycle strips, each strip (i) having η th,i = 1 - L,i / H,i 16 8

9 Compression & expansion efficiency! If irreversible compression or expansion, ds > δq/; if still adiabatic (δq = 0) then ds > 0! Causes more work input (more Δ) during compression, less work output (less Δ) during expansion! Define compression efficiency η comp & expansion efficiency η exp Reversible adiabatic work input for given or ratio η comp Actual work input required for same or ratio η exp = C [ 1 ( 1 / ) γ 1 1 ] C ( 1 ) = ( 1 / ) γ 1 1 / 1 1 (control mass, specified volume ratio) Actual work output for given or ratio Reversible adiabatic work output for same or ratio = C ( 1 ) C [ 1 ( 1 / ) γ 1 1 ] = / 1 1 ( 1 / ) γ 1 1 (control mass, specified volume ratio) 17 Compression & expansion efficiency η comp = C [ 1 ( / 1 ) (γ 1)/γ 1 ] C ( 1 ) η exp = = ( / 1 ) (γ 1)/γ 1 / 1 1 (control mass, specified pressure ratio) C ( 1 ) C [ 1 ( / 1 ) (γ 1)/γ 1 ] = / 1 1 ( / 1 ) (γ 1)/γ 1 (control mass, specified pressure ratio)! Control volume: replace C v with C, but it cancels out so definitions are same! hese relations give us a means to quantify the efficiency of an engine component (e.g. compressor, turbine, ) or process (compression, expansion) as opposed to the whole cycle 18 9

10 Compression & expansion efficiency! Animation: comparison of ideal Otto cycle with non-ideal compression & expansion! Same parameters as before but with η comp = η exp = emperature (K) charts on top of each other; double-click each to open Excel spreadsheets Entropy (J/kg-K) 19 Correspondence between - & -s ressure (/o) olume (/o) emperature (/o) ressure increasing Entropy ((s-so)/r) ressure (/o) olume (/o) emperature (/o) olume increasing Entropy ((s-so)/r) 0 10

11 Correspondence between - & -s ressure (/o) emperature increasing olume (/o) emperature (/o) Constant temperature Entropy ((s-so)/r).5.5 ressure (/o) olume (/o) Entropy increasing emperature (/o) Constant entropy Entropy ((s-so)/r) 1 Example! Why do internal combustion engines compress before burning? Is it possible to produce work or thrust without compression? o generate positive area (thus net work) on a -s diagram (left), the preheat-addition process must be more nearly vertical than the heat addition process, otherwise there is no area thus no work (middle). he best way to do this is with isentropic compression followed by heat addition (left) - but it s not required. You could have (for example) constant- heat addition followed by isentropic expansion back to ambient (right); work would be generated (although at lower efficiency) without a compressor Net work No area, no work! S S S S 11

12 Example! Why is it necessary to add heat to generate work or thrust? Without heat transfer, dq = 0 and thus (for reversible cycles) ds = 0, thus ds = d = Net work = 0 (left figure). If the process is irreversible, ds < 0, thus ds < 0 and thus = d = Net work < 0 (right figure). No area, no work! Negative work S S Example Consider the Lenoir cycle engine shown on the - diagram a) Sketch the -s diagram corresponding to the - diagram b) Sketch modified - and -s diagrams if the initial temperature is increased by 10% (same 1, 1 and ) s s 1 const., increasing heat addition, s increases const. s, decreasing expansion, decreases 1 const., decreasing v heat rejection, s decreases - plot unchanged (but higher means less mass processed) Higher, same s 1, s increases Still need 1 const. 4 1

13 Example c) Sketch modified - and -s diagrams if the compression ratio / 1 is increased by 0% (same 1, 1 ) 1 const., larger than base cycle 1, 1 1,1 On -, curves of const. slope less steeply than const. s, thus > Still need 1 const. d) Sketch modified - and -s diagrams if a gas with higher γ is used (same 1, 1,, R) γ curves steeper if γ increases 1, 1, 1,1 s s C = R/(γ-1); if R = const. and γ increases, C decreases, thus const.-v curves on s are steeper doesn t change since 1 =, 1, and R don t change 5 Using -s and -v diagrams - summary! hermodynamic cycles as they occur in IC engines are often approximated as a series of processes occurring in an ideal gas! -S and - diagrams are very useful for inferring how changes in a cycle affect efficiency, power, peak &, etc.! he Δ (on -S diagrams) and areas (both -S & -) are very useful for inferring heat & work transfers 6 1

14 Using -s and -v diagrams - summary! Each process (curve or line) on a -s or -v diagram has parts! An initial state! A process (const.,,, S, area, etc.)! A final state» For compression and expansion processes in piston engines, a specified (i.e., a particular compression or expansion ratio)» For compressors in propulsion cycles, a specified pressure ratio» For turbines in propulsion cycles, a specified temperature that makes the work output from the turbine equal the work required to drive compressor and/or fan» For diffusers in propulsion cycles, a specified Mach number» For nozzles in propulsion cycles, the pressure after expansion (usually ambient pressure)» For heat addition processes, either a specified heat input = ds (i.e. a mixture having a specified FAR and Q R ) thus a given area on the -s diagram, or a specified temperature (i.e. for temperature limited turbines in propulsion cycles)» he constant and constant exponential curves on the -S diagram are very useful for determining end states 7 Using -s and - diagrams - summary! hree or more processes combine to make a complete cycle! When drawing - or -S diagrams, ask yourself! What is the,, and S of the initial state? Is it different from the baseline case?! For each subsequent process» What is the process? Is it the same as the baseline cycle, or does it change from (for example) reversible to irreversible compression or expansion? Does it change from (for example) constant pressure heat addition to heat addition with pressure losses?» When the process is over? Is the target a specified pressure, volume, temperature, heat input, work output, etc.?» Is a new process (afterburner, extra turbine work for fan, etc.) being added or is existing one being removed?» In gas turbine cycles, be sure to make work output of turbines = work input to compressors and fans in gas! Be sure to close the cycle by having (for reciprocating piston cycles) the final volume = initial volume or (for propulsion cycles) (usually) the final pressure = ambient pressure 8 14