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Unformatted text preview: 7!. Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an
adiabatic heat exchanger. The rate at which the steam is obtained, the rate of exergy destruction, and the secondlaw efﬁciency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3
Air properties are used for exhaust gases. 4 Pressure drops in the heat exchanger are negligible. Properties The gas constant of air is R = 0.287 kag.K. The speciﬁc heat of air at the average temperature
of exhaust gases (650 K) is cp = 1.063 kJ/kg.K (Table A2). Analysis (a) We denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3)
and (4). The properties of water are (Table A4) T3 = 20°C} h3 = 83.91 kJ/kg Eggécgas
x = 0 s = 0.29649 kJ/k .K 350°C
3 3 g 150 kPa Hmt
T4 = 200°C 11,. = 2792.0 1(ng Exchanger
X4 =1 S4 = 6.4302 kJ/kg.K Sat. yap. Water
200°C 20°C An energy balance on the heat exchanger gives
mah, + thwh3 = mah, + rhwh4
macpm — T2) = ”1.0014 _ h3)
(0.8 kg/s)(1.063 kJ/kg°C)(400 — 350)°C = mw(2792.o — 83.91)kJ/kg
rhw = 0.01570 kg/s (b) The speciﬁc exergy changes of each stream as it ﬂows in the heat exchanger is T
Asa = c, 1n—2 = (0.8 kg/s)(1.063 kJ/kg.K)lnw
T1 (400 + 273) K = —0.08206 kJ/kg.K Al/la : Cp(T2 — 7.l) _ TOASa
= (1.063 kJ/kg.°C)(350  400)°C  (20 + 273 K)(0.08206 kJ/kg.K)
= —29. 106 kJ/kg All/w = h4 '“ ha ‘T0(S4 ‘53)
= (2792.0 — 83.91)kJ/kg — (20 + 273 K)(6.4302 — 0.29649)kJ/kg.K = 910.913 kJ/kg The exergy destruction is determined from an exergy balance on the heat exchanger to be — xdest = WA 1/4, + n'1wA w. = (0.8 kg/s)(29.106 kJ/kg) + (0.01570 kg/s)(910.913) kJ/kg = —8.98 kW or
x,est = 8.98 kW (c) The second—law efﬁciency for a heat exchanger may be deﬁned as the exergy increase of the cold ﬂuid
divided by the exergy decrease of the hot ﬂuid. That is, _ mwww _ (0.01570kg/s)(910.913kJ/kg) —,— = 0.614
— maAl/la — (0.8 kg/s)(29.106 kJ/kg) 2 , A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature
in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output,
the secondlaw efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1 The airstandard assumptions are applicable. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg‘K, c., = 0.823 kJ/kgK, R = 0.287 kJ/kgK,
and k = 1.349 (Table A2b). Analysis ((1) The clearance volume and the total volume of the engine at the beginning of compression
process (state 1) are 3
14=Vc+0.0028m —>vc = 0.0002154 m3 = v2 : vx
V5 ‘4 1/1 = vc + vd = 0.0002154 +0.0028 = 0.003015 m3 = v4 Process 12: Isentropic compression kl
T2 = T1[%1—] = (328 K)(14)1‘349'1 = 823.9 K
2
”—1 k
P2 =P1[ ] = (95 kPaX14)1'349 =3341kPa “2 Process 2x and x3: Constantvolume and
constant pressure heat addition processes: P
Tx =T2 sz (823.9 K):—232—%= 2220K
2 q“ = cu (TX — T2) = (0.823 kJ/kg.K)(2220 — 823.9)K = 1149 kJ/kg
q2_x : qx_3 = cp (T3 —Tx ) —>1149 kJ/kg = (0.823 kJ/kg.K)(T3 — 2220)K ——>T3 = 3254 K (b) qin = q2_x + qx_3 = 1149 + 1149 = 2298 kJ/kg 3254 K
2220 K T
v3 = vx T—3 = (0.0002154 m3) x = 0.0003158 m3 Process 34: isentropic expansion. V H 00003158 3 13494
T4 =T3[—3] =(3254 Ki—l] =1481K V4 0.003015 m3
v k 0 0003158 3 1349 P4 = P3 A = (9000 kPa ———E3— = 428.9 kPa
V4 0.003015 In Process 41: constant voume heat rejection.
qout = cu (T4 — T1): (0.823 kJ/kg  K)(1481— 328)K = 948.7 kJ/kg The net work output and the thermal efficiency are net,out W
”m = net,out _ 1349 kJ/kg =°.587 qin _ 2298 kJ/kg (c) The mean effective pressure is determined to be _P1(/1_ (95kPa)(0.003015m3) _0003043kg RT1 (0.287 kPa m3/kgKi328 K) m MEP = mwnew _ (0.003043 kg)(1349 kJ/kg) kPam3
VlVz (0.003015—0.0002154)m3 kJ )2 1466 kPa (d) The power for engine speed of 3500 rpm is Wnet = mwnet % = (0.003043 kg)(1349 kJ/kg)————3500 (”V/mm) [1 mm )2120kW
605 (2 rev/cycle) Note that there are two revolutions in one cycle in fourstroke engines. (6) The secondlaw efficieny of the cycle is defined as the ratio of actual thermal efficiency to the
maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure
to be 25°C and 100 kPa. T0 _ (25 + 273) K — = 0.908
77““ T3 3254 K and — ’7‘“ =——0'587 =0.646 0.908 max The rate of exergy of the exhaust gases is determined as follows T P
x4 = 144 —u0 —T0(s4 —sO) = c‘,(T4 —T0)—T0[cp ln—4—Rln—i]
T0 P0
= (0.823)(1481— 298)— (298) (1.110 kJ/kg.Kln 1481 —0.287 In 428'9 = 567.6 kag
298 100
X4 = mx4 1 = (0.003043 kg)(567.6 kJ/kgﬁsﬂOﬂ/m—m)£1 mm) = 50.4 kW
2 (2 rev/cycle) 60 s 5 ,1 The total exergy destruction associated with the Brayton cycle described in Prob. 9119 and the
exergy at the exhaust gases at the turbine exit are to be determined. Properties The gas constant of air is R = 0.287 kJ/kgK (Table Al).
Analysis From Prob. 9119, qin = 480.82, go“, = 372.73 kJ/kg, and T1 = 310 K —> s; =1.73498 kJ/kgK h2 = 618.26 kJ/kg —> s; = 2.42763 kJ/kg K 1150 K
T3 =1150 K —> s; = 3.12900 kJ/kgK h4 = 803.14 kJ/kg —» s; = 2.69407 kJ/kgK h5 = 738.43 kJ/kg ——> s; = 2.60815 kJ/kgK 310 K and, from an energy balance on the heat exchanger, h5 —h2 = h4 —h6 ——>h6 = 803.14—(738.43—618.26) = 682.97 kJ/kg
——> s; = 2.52861 kJ/kg ~ K Thus, P _ _ _ o o 2
xdestroyed,12 — TOSgen,12 —T0(52 “S1)— To[52 ‘51 —R1n T,
1 = (310 K)(2.42763 —1.73498 — (0.287 kJ/kg . K)1n(7)) = 41.59 kI/kg P _ _ _ o o 4 xdestroyed,34 _ TOSgen,34 — T0 (S4 _S3)_ T0 [S4 _S3 —Rll’l P
3 = (310 K)(2.69407—3.12900 —(0.287kJ/kg  K)1n(1/7)) = 38.30 kJ/kg xdestroyed,regen = TOSgen,regen = T0 [(55 _ $2 )+ (S6 _ S4 )] : T0 [(SS‘3 _ 52 )+ (S; _ S2 )1
= (310 K)(2.60815 — 2.42763 + 2.52861— 2.69407) = 4.67 kJ/kg 4R 53 P3 &0 q
xdestroyed,53 = TOSgen,53 = To 53 ‘55 ” T, = To 53 ‘55 ‘Rh‘? ‘ Tm
R 5 H 480.82 kJ/kg = 310 K 3.12900— 2.60815 — = 78.66
( l 1800 K ] kJ/kg <90 5111,61 0 a P1 qout xdestroyed,61 = TOSgen,61 = T0[Sl _S6 + T 2 T0 51 _SG _Rln? + T
R 6 L 372.73 kJ/kg : (310 K{1.73498— 2.52861+ 10 K J: 126.7 kJ/kg Noting that kg = [email protected] 310 K = 310.24 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is
determined from V2 &0
$0
¢6 = (’16 “h0)—T0(56 ‘ 50)+—§— + 826
P (:90
where s6 —s0 = s6 —s1 = sg —sf 42m?6 = 2.52861—1.73498 = 0.79363 kJ/kgK
1 Thus, 6 = 682.97—310.24—(310 K)(0.79363 kJ/kg  K): 126.7 kJ/kg A__ ..... Amnr’s m, T. = We, P1 = 2.0m, Ta,A=..._599:'sL_W
_, P4 = 7.51%qu 151 = 2630 Sj/ﬁrL TH: 1500 K, TLfﬂoo kl #
1:, = 300K ,7 ,, , _
iinnd: (ca dun am.“ (0 Wm): (00W (*9 Xd “3 '11
A Solution _ , _ﬁ
A K; anaim ﬁgi‘wrngzg gggﬁm:w*gv 53th 3mg? _ _____ _ ______ ,
; _____ ”=1 Piﬁnﬂfujl::,,,5_99:EV=3_,5:f.....33?§;95"£3,553I S _= 6.za§____§J/eggx
A __ “a"; 3’1 = 2.0 Wu, 3: = S. ==> R2 = 2999.45 53/53}
_ “‘53 Pa = Pa, T3= 500‘c m:=> 83: 39616 R3153, 53 = “an H/RjK
“4» , P4 = 7.5 Rm, 5“ g 53 a 2‘, = 0393; 5‘, = 2317.13 133$qu
y "5" 2» sat. 33'3“”?! 95 = P4: ==> ’55 = 158.99 Big, 55 595747 RJ/fz‘gk ,, 175 = 0.0mm 1313/61
v, ,,,,, M ._.:§L?6= PI = 8 ”Pa ”Ti—“mi “5 *.,£’5$,‘,,’9:.m'§>ii]? 35 Mg; aim = '{nEiﬁr $0105  31)]: o. 7306 I(ssqs.osV—I1e.3s}+ (3mg .;k<n9.45)] mam: w) = if: = 03306 3313 ‘ 4,
f M.  W _
i v , {1 m g {a ”m Aeuiw/ W “Ninja = mMRS) = 07306133” ~ “899) ,  iikigm ﬁg‘w WM ” ,, ,, ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
‘ mm = dim ~ (ﬂout = 1115.69 R»;
#1? 39% “g” ﬁrm? ﬁwnggr 11%;.” , ““““““ W ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
bond: ”35.67
”’14:, = 61;" 39,553 = 0.4176 {44.76%}
, “23m ﬁwxesgiémmmxaéérmﬁwm
H 2d»; = 39,“. fa“, = 0 ‘ Renting: procesSes _____________________________________________________________
_, xdna = (1'43“)6‘223 ”V 7h(u;“$’3) wﬁete: Wa'ué = (31‘33)~?o{32*33)
= [g— ~%:)(m—as) + ($2.53)  Tom—33)} m
E = ~%{£,~Ba)To(s,sa)]mi
‘ = 86.93 SW
0
__ _ > = 61.03 ﬁw
. 1 u. A _ _ TH ' “ I
7777777777777777777777777777777777777 T ,, M94 ‘0’?)6164'4' m($6” 3*)
= [("":3,;'Xﬁu'ﬁe) + (ﬁeS.)— $5“;ng
= Hagm  1006st VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV
.............. T
: 813.83 aw
$21»sz ~§ 39‘ La 43%: g 12:
‘gé' at a? {3} 1:29:31? 3’ I 7&1; = (1 E3643 = (I 3” o.) 3615.53 7 52.27. I) ...
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 Spring '18
 Energy, Internal combustion engine, Diesel engine

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