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Vation Is Left As an Cise 408 Chapter 9 Gas Power Systems 70 60 ~ I=" 50 ,:; u 40 'u"'" .....!.::: 30 0;" E 20 ~ ~ 10 0 5 10 15 20 .... Figure 9.6 Thermal efficiency of the cold air· Compression ralio, r standard Diesel cycle, k = 1.4. where r is the compression ratio and rc the cutoff ratio. The derivation is left as an cise. This relationship is shown in Fig. 9,6 for k = 1.4. Equation 9.13 for the Diesel differs from Eq. 9.8 for the Otto cycle only by the term in brackets, which for " > greater than unity. Thus, when the compression ratio is the same, the thermal the cold air-standard Diesel cycle would be less than that of the cold air-standard cycle. In the next example, we illustrate the analysis of the air-standard Diesel cycle. 1 II. I Q., ~ h 'I ....., ~',"1""": " .} ....I.-J,. l_ L ' ...... h,. "L b~ f .... l. Qt\VJ(",. / 'I".... tl.. ? r -J L...,.., U I'~J l.'''''/',; 1;1' J ... ,," j.,,>-I. '" At the beginning of the compression process of an air-standard Diesel cycle operating with a compression ratio of 18 , the perature is 300 K and the pressure is 0.1 MPa. The cutoC:( ratio for the cycle is 2. Determine (a) the temperature and at the end of each process of the cycle, (b) the thermal ' fficiency, (c) the mean effective pressure, in MPa. SOLUTION f'{UW ~ ' .... m c ..'( J ;(....clv4.. 'j ....;Y'\ j~..ll ~ ; ......'c.. Known: An air-standard Diesel cycle is executed with specified conditions at the beginning of the compression stroke. compression and cutoff ratios are given. Find: Determine the temperature and pressure at the end of each process, the thermal efficiency, and mean pressure. ..I" :J f£! I,..,,"",, ~..... I ., j ell\(.\ Schematic and Given Data: \1) ". 1.... "v'" rr J1. ' PI T I 4 v s .... Figure E9.2 9.J Air-Standard Diesel Cycle 409 the piston- cylinder assembly IS the closed system. 71pre:SSlcm and expimsibn processes are adiabatic. are internally reversible. '1 .la r"'O ,'~/1I 1. V:,,~""'\. modeled as an ideal gas. ~ potential energy effects are negligible. J) T~!J.., L1l"-, / . ' LJrv::.I1-T .4T.J<:, /ot1 -lillt;/v vr - JJ'" 1::: begins by detemtining proRMJies-at each_prin.cipal state of the cycle. With Tl = 300 K, Table ~giVeS ( " t.,h and vrI = 621 .2. For tl e isentropic compression process 1-2 _ I. ~ I ') L -r DP 0/ f ~ ?I~ if; r 6"'I • IJ' (: .l ,.,.. Y V2 Vrl 621.2 - :: ~ - I, 2 J - I It "1. Va = -v I = ­ = -­= 34.51 Y. v.. it" 1 '<J " "'-' T ' V r r 18 ~ - 1, v '-' 1 I ~ ~ • , V'­ , Vr - ~,I l in Tab~ we get T2 = 898.3 K and h2 = 930.98 kJ/kg. With ~ ideal g a on of state I • I • A r -; ~ f) tJ -~- - -----" _ . _ T2 VI _ (898.3~ _ 1­ '-, j~ r: '! :. P, i'!.L.,- Ii P2 - PI V - (0.1) 300 (18) - 5.39 MPa , , ~, tJ • ..A.- ~ 'j..:.-r , " . Tl 2 , r ~ .:. -;' I 1. ~ 'hI (. .' " . ' r"~ '" " e,I , . • at state 2 can be evaluated alternatively lising the isentrHpic relationship, P2 ' =' P'I'(Pr2/PrI)' .i ~ (1 ) e. ?rocess 2-3 occurs at constant pressure, the ideal gas equation of state gives 11-, ilJ'} .J; ~ )( f8 _ __1_ - ~ J..JJ"I "<- WI.' 11 ~I__ V3 t! 1,Ji J ('J. - :F,t~l T3 = -T2 ~7o . V2 17 I.: I,,~ . .k;4j • -,J •• '" t L("I-n~.L , ." I ~ .. r'1 , ... l. the compression ratio r, and ~e cutoff ratio re, we have r~~ t::V~ ~8 _ J.t(?/"'): L V) V,4 - Vr3 - . ) - 35.73 l. p-1, (... -:; VI. ('1" fd~l. J w,'!..., in Table .~ with Vr4' we get li4 =.§.64d1OTkg and T4 =~ K. The pressure at state 4 can be found using N ~)- , relationship P4 = P3(Pr4/Pr3 ) or the ideal gas equation of state applied at states 1 and 4. With V~= V, the ideal bl.... v"dJ of state gives 8- ~/fL JCJ.. MP"I 'I, .. .. ~ LIt.... (J I, ,-, ' V P.. PI T. (0.1 MPa) = = (~ K) = n ~J a 1~.'; ~~ , TI \ 300 K 7 "" therrnai efficiency is found using , ~ CJ"... f-­ 'Y/ = 1- Q4Jm =['1 __ ~li. - lil ) J> Q23/m h3 - h2 ~ 664.3 - 214.07 = 1 - 1999.1 _ 930.98 = 0.578 (57.8%) mean effective pressure written in terms of specific volumes is 41 0 Chapter 9 Gas Power Systems The net work of the cycle equals the net heat added [w . ~.= ;;J ~ycle Q23 Q41 . ... -- = - - - = (h3 - h2) - (U4 - Ul) m m Tn = (1999.1 - 930.98) - ~- 214.07) = ~ kJ/kg ' 1k1 ,J The specific volume at state 1 is 'lO,i,) j r.\~1.> ~ _., 8314 N· (300 K) ( m) 28.97 kg . K _ 3 5 2 - 0.861 m /kg t. lON/m ...... ,Jf'Jjn Inserting values ~ ~ \)"1 -::. ­ t~.:h. 0'I me _ ~/kg 1103 N . mill MPa I '~ ~ "'In 'to 3 2 p - 0,861(1, - 1/18) m /kg 1 kJ 106 N/m - ~ ftI ''y ( d ~ I():I .... = 0.76 MPa 0, 'J) b f kt' ,. It) lit '''' 1. o This solution uses the air tables, which account explicitly for the variation of the specific heats with that Eg. 9.13 based onthe assumption of constant specific heats has not been used to determine the thennal The cold air-standard solution of this example is left as an exercise. Air-Standard Dual Cycle The pressure-volume diagrams of actual internal combustion engines are not well by the Otto and Diesel cycles. An air-standard cycle that can be made to dual cycle the pressure variations more closely is the air-standard dual cycle. The dual cycle is in Fig. 9.7. As in the Otto and Diesel cycles, Process 1-2 is an isentropic The heat addition occurs in two steps, however: Process 2~ 3 is a constant-volume dition; Process 3-4 is a constant-pressure heat addition. Process 3- 4 also fIrst part of the power stroke. The isentropic expansion from state 4 to state 5 is .. mainder of the power stroke. As in the Otto and Diesel cycles, the cycle is a constant-volume heat rejection process, Process 5-1. Areas on the T-s and p-u can be interpreted as heat and work, respectivel)', as in the cases of the Otto c'lc\e';',. P 3 T 4 2 p=c 3 v = c 5 2 v =c v A. Figure 9.7 p-v and T-s diagrams of the air-standard dual cycle. 0.4 Air-Standard Dual Cycle 4 I I Since the dual cycle is composed of the same types of processes as the Diesel cycles, we can simply write down the appropriate work and heat transfer ex­ by reference to the corresponding earlier developments. Thus, during the isentropic process 1-2 there is no heat transfer, and the work is W l2 --;;; = 11 2 - III the corresponding process of the Otto cycle, in the constant-volume portion of the process, Process 2-3, there is no work, and the heat transfer is Q23 -;;; = u3 ­ U2 constant-pressure portion of the heat addition process, Process 3-4, there is both work transfer, as for the corresponding process of the Diesel cycle and the isentropic expansion process 4-5 there is no heat transfer, and the work is W45 --;;; = 114 - U5 the constant-volume heat rejection process 5-1 that completes the cycle involves but no work thennal efficiency is the ratio of the net work of the cycle to the total heat added WCYcl Jm Q51/m 1/ = = 1 - ----=.;:.--­ (Q2i m + Q 3~m) (Q2im + Q34/m ) (u 5 - Ul) = 1 - ---'------ (9 .14) I (113 - U2 ) + (h 4 - h3) ., example to follow provides an illustration of the analysis of an air-standard dual cycle. exhibits many of the features found in the Otto and Diesel cycle examples previously. ihe beginning of the compression process of an air-standard dual cycle with a compression ratio of 18, the temperature is K and the pressure is 0.1 MPa. The pressure ratio for the constant volume part of the heating process is 1.5:1. The vol­ ratio for the constant pressure part of the heating process is 1.2:1. Determine (a) the thermal efficiency and (b) the mean . An air-standard dual cycle is executed in a piston-cylinder assembly. Conditions are known at the beginning of the : om~)res~;ion' process, and necessary volume and pressure ratios are specified. Determine the thermal efficiency and the mep, in MPa. 4 ! 2 Chapter 9 Gas Power Systems Schematic and Giv.en DaJa: p --- } T 2 5 2 , - -7 =300K 1 1 v 5 .... Figure E9.3 AssumptWns: 1. The air in the piston-cylinder assembly is the closed system. 2. The compression and expansion processes are adiabatic. I 3. All processes are internally reversible. J o :> ~ I ",' ., PI _11 (. , I t 4. The air is modeled as an ideal gas. (;, :; 2.flrl (J1 Vr. = 62.1, 2.. S. Kinetic and potential energy effects are negligible. I 1 V'r 2 bJ('I'L ~ 1'1,s Analysis: The analysis begins by determining properties at each principal state of the cycle. States 1 and 2 are the same in Example 9.2. so UI = 214.07 kJ/kg, Tz = 898.3 K, U? = 673.2 kJ/kg. Since Process 2-3 occurs at constant volume, ideal gas equation of state reduces to give Y,...,-- ""J;I,'1 Ii.
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