20... Figure 9.6 Thermal efficiency of the cold air standard Diesel cycle, k = 1.4.

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408 Chapter 9 Gas Power Systems ~ I=",:; u '" 'u!.:::... 0; " E ~ ~ 70 60 50 40 30 20 10 0 5 10 15 Compression ralio, r 20... Figure 9.

1 408 Chapter 9 Gas Power Systems ~ I=",:; u '" 'u!.:::... 0; " E ~ ~ Compression ralio, r Figure 9.6 Thermal efficiency of the cold air 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 'j...;y'\ f'{uw ~ '... m c..'( J ;(...clv4.. 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: PI I \1) " "v'" rr J1. ' T 4 v s... Figure E9.2

$9.J Air-Standard Diesel Cycle 409 e. the piston- cylinder assembly IS the closed system. 71pre:SSlcm and expimsibn processes are adiabatic. are internally reversible.. '1.la r"'o,'~/1i 1. V:,,~""'\.$

2 9.J Air-Standard Diesel Cycle 409 e. 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 = For tl e isentropic compression process 1-2 _ I.. ~ I ') -r DP 0/ f ~?I~ if; r 6"'I L IJ' (:.l,.,.. Y V2 Vrl :: ~ - I, 2 J - I It "1. Va = -v I = = - = Y. v.. it" 1 '<J " "'-' T 1 ' V r r 18 ~ - 1, v '-' I ~ ~, V', Vr - ~,I l in Tab~ we get T2 = K and h2 = 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) 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 )?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. V 2 I.: I,,~. 17.k;4j -,J the compression ratio r, and ~e cutoff ratio r e, we have r~~ t::v~ ~8 _ J.t(?/"'): L V) V,4 - Vr3 -. ) l. '" t L("I-n~.L,." I ~.. r'1,... 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~a 1~.'; ~~, TI \ 300 K 7 J "" therrnai efficiency is found using, ~ 'Y/ = 1 - Q4Jm =['1 ~li. - lil ) J> Q23/m h3 - h2 ~ = _ = (57.8%) mean effective pressure written in terms of specific volumes is CJ"... f-

$... -- = - - - = (h3 - h 2 ) - (U4 - Ul) m m Tn = (1999.1-930.98) - ~- 214.07) = ~ kj/kg ' 1k1,J 'lo,i,) r.\~1.> ~ _., 8314 N m) (300 K) ( 28.97 kg. K _ 3 lon/m.... ~ 0' ~ \)"1 -::. t~.:h 5 2-0.$

3 41 0 Chapter 9 Gas Power Systems The net work of the cycle equals the net heat added The specific volume at state 1 is Inserting values [w. ~.= ;;J ~ycle Q23 Q = = (h3 - h 2 ) - (U4 - Ul) m m Tn = ( ) - ~ ) = ~ kj/kg ' 1k1,J 'lo,i,) r.\~1.> ~ _., 8314 N m) (300 K) ( kg. K _ 3 lon/m.... ~ 0' ~ \)"1 -::. t~.:h m /kg t.,jf'jjn me _ ~/kg 1103 N. mill MPa I '~ ~ "'In 'to p - 0,861(1, - 1/18) m 3 /kg 1 kj 10 6 N/m 2 - ~ 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 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. I j dual cycle 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 the pressure variations more closely is the air-standard dual cycle. The dual cycle is in Fig 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 2 p=c 3 v = c 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

4 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 --;;; = 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 --;;; = 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 WCYclJm Q51/m 1/ = = =.;:.-- (Q2i m + Q 3~m) (Q2im + Q34/m ) (u 5 - Ul) = ' (9.14) (113 - U 2 ) + (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. I., 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 volratio 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 1 =300K 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.

5 4! 2 Chapter 9 Gas Power Systems Schematic and Giv.en DaJa: p --- } T 2 5 2, =300K 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 = kj/kg, Tz = K, U? = kj/kg. Since Process 2-3 occurs at constant volume, ideal gas equation of state reduces to give Y,...,--""J;I,'1 Ii. 'If.:?'*!,1.. P3 T.)!.:,4 J~ T3 = P2 T2 = (1.5)(898.3) = K p) 'P "" ~ TJ Interpolating in Table.t>4we get h3 = kj/kg and u) = kj/kg. Since Process 3-4 occurs at constant pressure, the ideal gas equation of state reduces to give V { !1?1 11\ _,,}- -r,,} T4 = -=- T3 = (1.2)(1347.5)= 1617 K ~:;- 1"-. ~ 1.. V3 ;, ' ::: -r., From Tab~ h4 = kj/kg and V,4 = Process 4-5 is an isentropic expansion, so Vs v's = V,4 V 4 The volume ratio VS/V4 required by this equation can be expressed as Vs = Vs V3 V 4 V] V. - y "Y.. JI Inserting this in the above expression for V,5 Interpolating in Ta~2. we get U5 = kj/kg. r..l.l... ~~ll

9.5 Modeling Gas Turbine Power Plants 413!hennal efficiency is Q5Jm (U5 - Ul) ~=1- =1-, (Q23/m + Q34/m) (U3 - U2) + (h4 -h3) (475.96-214.07) --- - (1, '? = 1 - -----'--------'---- (1065.8-673.

6 9.5 Modeling Gas Turbine Power Plants 413!hennal efficiency is Q5Jm (U5 - Ul) ~=1- =1-, (Q23/m + Q34/m) (U3 - U2) + (h4 -h3) ( ) (1, '? = ' '---- ( )+ ( ) = (63.5%) work of the cycle equals the net heat added, so, '1 J -+ mep = ~ " (U3 - U2) + (h4 - h3) - (us - Ul) VI(l - -specific volume at state 1 is evaluated in Example 9.2 as VI = m 3 /kg. Inserting values into the above expresmep [( ) + ( ) - ( ) ) (ḵj)ii03 N'mlll 6 MPa 2 I. '. kg 1 kj 10 N/m.. =0.56 MPa 0.861(1-1/18) m 3 /kg l/r) of the chapter deals with gas turbine power plants. Gas turbines tend to be lighter more compact than the vapor power plants studied in Chap. 8. The favorable power ratio of gas turbines makes them well suited for transportation applications propulsion, marine power plants, and so on). Gas turbines are also commonly used,1.1',nn,", power generation. Modeling Gas Turbine POlNer Plants power plants may operate on either an open or closed basis. The open mode pic in Fig. 9.8a is more common. This is an engine in which atmospheric air is continuously into the compressor, where it is compressed to a high pressure. The air then enters a chamber, or combustor, where it is mixed with fuel and combustion occurs, re in combustion products at an elevated temperature. The combustion products expand the turbine and are subsequently discharged to the surroundings. Part of the turbine developed is used to drive the compressor; the remainder is available to generate elec to propel a vehicle, or for other purposes. In the system pictured in Fig. 9.8b, the fluid receives an energy input by heat transfer from an external source, for example :~as COOllea nuclear reactor. The gas exiting the turbine is passed through a heat exchanger, it is cooled prior to reentering the compressor. idealization often used in the study of open gas turbine power plants is that of an ~1lT Sla./lQlzra analysis. In an air-standard analysis two assumptions are always made: air-standard analysis: working fluid is air, which behaves as an ideal gas.g=-a_s_tu_r_h_in_,e_s

Teaching schedule *15 18

Teaching schedule *15 18 Teaching schedule Session *15 18 19 21 22 24 Topics 5. Gas power cycles Basic considerations in the analysis of power cycle; Carnot cycle; Air standard cycle; Reciprocating engines; Otto cycle; Diesel