ESCUELA SUPERIOR DE
.
INGENIERIA
MECANICA Y ELECTRICA UNIDAD TICOMAN
Sistemas Propulsivos Propulsivos
“Análisis Térmico” turbo-fan
PROFESOR: Enri"ue
Arellanos
aca
E!uar!o
A#$% A#$%&O &O : '$A& '$A& ($)# ($)##E #ER% R%O O ($T) ($T)ER ERRE RE* * + + RO,R)($E* (R$PO: A. A.
IAE V2500
El V2500, pronunciar Victor -Dos Mil Quinientos ("V" en número romano simboliza las cinco compañías ue se unieron ori!inalmente para la abricaci#n $el motor%, es un turbo&n ue proporciona potencia $e empu'e a los aiones $e la amilia )irbus )*20 ()*20, )*2+, )*+ el )irbus .orporate /et%, el McDonnell Dou!las MD-0 , m&s recientemente, elEmbraer .-*01
Fan tip diameter: 63.5 in Length, flange to flange: 126 in Takeoff thrust: 23,000–32,000 lb Flat rated temperature: 86–131 F Bypass ratio: 4.5–5.4 Overall pressure ratio: 26.9–33.4°
See more at: http://www.pw.utc.com/V2500_Engine#sthash.vdS0XX!.dpu"
Facts
T/o-s0aft turbofan en1ine Fan /it0out snubbers Po/!er-metal turbine !iscs Sin1le-cr2stal turbine bla!es Active clearance control %o!ular !esi1n ,i1ital en1ine control unit On-con!ition maintenance
Development
#o/-pressure turbine Accessories an! E3ternals
En1ine testin1
Production
#o/-pressure turbine 4various parts5
MTU share
.6 7
Partners
Pratt 8 90itne2 'apanese Aero En1ines orporation
Applications
Airbus A;.< Airbus A;=> Airbus A;=. ?oein1 %,-<> @-;<>
Technical data Max. thrust
33,000 lbf
Bypass ratio
4.5:1
Pressure ratio
33.4:1
Length
12 in
!an "ia#eter
3.5 in
$eight
5,0%3 lbs
1. Type Design Definition: V2500- 5 V2522-5& V252'-5& V252(-5& V252(E-5& V252(!-5& V25)0-5& V25))-5 V2500-.5 V2525-5& V252-5
2. Description: ua rotor& aia "ow& high 1pass ratio turo"an engine:
$arts %ist '* 5+,
$arts %ist '* 5+,,
singe stage "an& '-stage ow pressure compressor %$34& +0-stage high pressure compressor $34 annuar comustion chamer 2-stage high pressure turine $64& 5-stage ow pressure turine %$64 dua channe "u authorit1 digita engine contro 7E34
. !"uipment: 6he engine starter is part o" the engine t1pe design. 8e"er to the engine $arts %ist "or detais.
#. Dimensions: 9vera %ength
*idth
!aimum 8adia $roection
)20+ mm
+;2 mm
+0,) mm V2500-54 ++0, mm V2500-54
$. %eight: 2'0'
>e& and power source "or the ignition s1stem "or the -5 series& and ehaust no>>e "or the -5 series
En el análisis termodinámico del motor turbofan se analizaran las siguientes fases:
&omenclatura
'onstantes 3onstantes de "uido de traao
Vaor de as constantes
?a@ Eponente de proceso adiaAtico de aire
?a@ +.'
?g@ Eponente de proceso adiaAtico de gases de saida
?g@+.))
3pa@ 3aor especB"ico a presiCn constante de aire
3pa@+.005 ?D/?g?
3pg@ 3aor especB"ico a presiCn constante de aire
3pg@ +.+'/ ?g?
8a@3onstante Fniversa de aire
8 [email protected](
8g@ 3onstante universa de gases
8 [email protected]'
&''0()*''& B+5.4 -+ 44.4444#s
&''''''''&''''''''''&'''''''''''''&'''''''''''&''''&''''''&'''''''''''''''&'''''''''''''& 0.)
0.)5
0.)
0.)%
&'''''''''''/P'''''''''''''& 2).* -+ 10 #hr 6+ 25,000lb #ax + 1300-7+ 15%3.1587i+ 14.% ⋀ Pe+ 0.) η #e1+ 0.) η #e2+ 0.)* λ +0.2* PL9/ 0 1 2 3 4 5 % * ) P1= difusor
0.)*
0.)%
#ah+ - √ 401.1o + 0.130
P8pa 101.325 102.4)0* 13.)*52 401.%3% 301%.124* 2*.2*5 11).))05 55%.0)%* 2)).0500 101.325
8- 2**.15 2*).131) 333.024) 434.*02 %)2.2%% 15%3.15 1253.*1%0 103%.0%5 *)0.1150 321.%)
Pi+Po 1 ; 0.)44.4444<22010 o <3.5 + 102.4)0*8Pa 1+ o 1 ; 0.20.130<2 + 2*).132) 8 π "if+ 1.0115
P2= FAN
0.))
π fan+ πstgf ¿ n=( 1.6 )1=1.6 P2+ P1 πfan ¿=163.9852 KPa
2
¿ T 1 ( πfan )
0.3007
=333.0246 K
P3= Booster πstg =
√
13
13 OPR = √ 18.4132 =1.2511 πbooster =( 1.2511 )4 =2.4500 πdif .πfan
P3+401.%3%8Pa 3+22.4500<0.2)% + 434.*03 8
P4 !c πMC =( 1.2511 )
9
=7.5097
P4+P3 πMC ¿=3017.1248 KPa
4+3 πMC ¿0.2976=792.2776 K
P" ## 5+15%3.15 8 P5+P4 1( ⋀ Pcc + 2*.2*58Pa P.=t +5 >
1005 ( t 4 − t 3 )
(1148 ) ( πmec 2 )
( )
πMt =
P6=
T 5 T 6
P5
πHr P7 . LT
3.9496
❑
=2.4500=
( T −T ) 3
1
(1148 ) ( nmeci )
( )
P7=
T 6 T 7
P6
P5 P6
=1169.9055 kpc
T 7 =T 6 1005 πLT =
¿ + 1253.*1%0 8
=1037.0675 k
❑3.9093=2.1000 =
P6 P7
= 557.0978 kpc
πLT
P$ %&BE'A #&'E P 7 1 1 4.0303 ❑ = .5368 =1.8628 X = =5.4981 = Po 0.1430 0
%$= "()$3)$ P)= %&BE'A FAN %a= t2 =t2*+32,32,3,41-+..344Pa=Po =1.132" /a= 1".4.()
!r = !fan + !core 4.4482 # =111,205 # 25,000 " bt 1 " bf
$=
mfan % mfan=5.4 mcore mcore
6fan + 5.4 #ore a(o
Efan = 572.2029 mcore
mg= mcore + mfre "
mfre" =
( mcore ) (ℷcc ) Li
=0.0190 mcore
mg= mcore + 0.0190 mcore
mg= mcore ( 0.0190 ) =1.0190 mcore
&sa" =
mg 1.0190 mcore = =0.01485 mcore " 8 ' ( 1.1829 ) ( 579.8398 ) 8
"
p 8 kg =1.1829 3 R g m
8=
0
P8−¿ P¿( 0.01485 mcore) !core =( 1.0190 mcore ) ( ' 8 )−( mcore ) ( 'o ) +¿ !core=( 590.8567 mcore − 44.4444 mcore ) + 2936.2162 mcore !core=3481.6285 mcore !T = 572.2029 mcore + 3282.6285 mcore 111,205= 4054.8314 mcore kg "b 1 "b =60.4081 mcore =27.4253 s 0.454 kg s kg "b mfan=148.0966 = 326.2040 s s kg "b mg= 27.9463 =61.5559 s s kg "b mfre" =0.521 =1.1477 s s kg "b mT = mfan + mcore =353.6293 =778.9191 s s 1 "bf =3527.9070 "bf !fan=15692.8361 ∈ 4.4482 # 1 "bf =21472.0856 "bf !core =95512.1314 # 4.4482 # !tota" =111,204.9675 # = 24,999.9926 "bf =25,000 "bf
(
)
(
)
(
)
PE'F&'!AN#E PA'A!E%E'0
!) mfan mfre" "b "b =0.1652 = 0.1652 T(*C = !T +r "bf . +r "bf (=
1
1
2
2
2
, !c = mg' 8− mfan' 8
