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LABORATORIO DEMAQUINAS TERMICAS
Profesor: López FernándezV !en"e #$ %%er&o
Pr'!" !' (: MOTOR DIESEL
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INDICE
OB)ETIVO******************************************************
INTRODUCCION**************************************************
EQUIPO**************************************************
DESARROLLOE+PERIMENTAL*******************************************
DATOS**************************************************
MEMORIA DECALCULO*********************************************
RESULTADOS**************************************************
ANALISIS DE RESULTADOS ,CONCLUSIONES***********************************
REFERENCIAS**************************************************
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OB)ETIVO
E% des'rro%%o de %' prá!" !' !o&prende e% es"$d o de% &o"or ECOM 's0 !o&o %'de"er& n'! ón de %'s p1rd d's de ener/0'2Des!r p! ón de% &o"or d ese% de dos " e&pos !o&p'r'do !on $n &o"or EC32Re'% z'r e% 4'%'n!e de ener/0' en %' p%'n"' de e&er/en! ' de% LMT2
INTRODUCCION
EQUIPO
Pro"e!"ores '$d " 5os nd$s"r '%es6 'n'% z'dor de /'ses F r7"e de O-6 'n'% z'dor de /'ses F r7"e de CO-6 !ronó&e"ro6 ps !ró&e"ro de !'rá"$%'6 8e9ó&e"ro6 "'!ó&e"ro
DESARROLLO E+PERIMENTAL
An"es de n ! 'r %' prá!" !' 7 !o%o!'rnos %os pro"e!"ores '$d " 5os 's /n'&os %os%$/'res de "r'4' o p'r' !'d' $no2 Desp$1s en!end &os e% &o"or 7 !o&o %o ;'!e&osen "od's %'s prá!" !'s esper'&os ' ?/ 7 @n'%&en"e "o&'&os %'s &ed ! ones ne!es'r 's!o&o %'s "e&per'"$r's %' 5e%o! d'd de% &o"or !on e% "'!ó&e"ro de ;$&ed'd 7%' d s"'n! '
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opr & ó %' 5á%5$%'2 Se 4o&4eó - 5e!es p'r' HHH>> NGN -( RPMG
T /s = > JCG T '&4 -H JCGK ( G
#'s"o de !o&4$s" 4%e% -2= !&G 2 -= &G"! 6266 sGD C 2H !&G 2 H &G
BAC3ARAC3CO- H GCO 6O- 6 G
#'s"o de '/$'; 6 2> !&G 26 > &G" 3-O > sG
T 6 =( JCG T - > JCGD 3-O - !&G
MEMORIA DE C LCULO , RESULTADOS
62 Ener/0' s$& n s"r'd' ´ E1
´ E1 = ḿc ·PCA [kW ]
Donde PC ≜ Poder caloríficodel combustible
PCA Diesel ≜ 42094 [kJ kg] ρc= ρ Diesel = 800 [ kgm3 ]
#'s"o de !o&4$s" 4%e ḿc
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ḿc=(π d 24 )· l
t ⋅ ρc[kgs ]
Donded ≜ Diámetro del depósito del combustible
l ≜ Diferenciade altura en el ni el de combustible
t ≜ tiempoenel !ue se consumióel olumendel combustibleconsiderado
ḿc=(π (.094 )
2
4 )· (.023 )(61.11 ) ⋅
(800 )
ḿc= 2 "089 # 10− 3[kgs ]
´ E1 = (2.089 # 10 − 3 )· (42094 )´ E1 = 87.93 [kW ]
- Ener/0' 'pro5e!;'d' ´ E2
W = $% 1500
[& p]
´ E2 = $% 1500 ·' com[kW ]
´ E2 =W (k( )
' com[kW ]
Donde $ ≜ fuer)a aplicada conel freno& idráulicoal *E [kgf ]
% ≜
+P* del motor el,ctrico' com ≜ eficieniacombinadadelgenerador - delmotor el,ctrico = 80 [ ]
W = (11.7 )(2860 )
1500[& p]= 22.308 [& p]
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W = 16.63507 [kW ]
´ E2 =(16.635 )(0.80 )
[kW ]
´ E2 = 20 "793 [kW ]
= Ener/0' '4sor4 d' por ;$&ed'd de% !o&4$s" 4%e ´ E3
´ E3 = ḿc · . C · / & [kW ]
Donde
. C ≜& umedad delcombustible = . Diesel = .8 = 0.008 [kg 0 2 1kgcom ]/ & = cambiode entalpia = Entalpía 0 2 1lí! + Entalpía cambiode fase + Entalpía ap sobrecalen
/ & = / & 1 +/ & 2 +/& 3
Es de! r:3 = ¿c pc (2 gsc− 2 sat )
/ & 1 = c p0 2 1 (2 sat − 2 amb )3 / &2 = &fg 3 / &¿Donde
2 amb ≜2emperatura del aire [℃ ]
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2 gsc ≜2emperaturade gases decombustión [℃ ]
c pc ≜calorespecíficoa P = constantedel combustible
c p0 2 1 = 4.186
[ kJ kg℃
]c pC = 1.93 [ kJ kg℃ ]&fg(2sat = 100 C )= 2257.61
/ & = c p0 2 1 (2 sat − 2 amb )+&fg+c pc (2 gsc− 2 sat )[kJ kg]/ & = (4.186 ) (100 − 24 )+ 2257.61 + (1.93 ) (375 − 100 )
/ & = 3106 "496 [ kJ kg]En"on!es:
´ E3 = (2.089 # 10 − 3 )· (0.008 )· (3106.496 )´ E3 = 0 "0519 [kW ]
H Ener/0' '4sor4 d' por ;$&ed'd prod$!"o de %' !o&4$s" ón de ; dró/eno ´ E4
´ E4 = 9 ( 0 2 +1 28 )́mc · / & [kW ]
Co&pos ! ón de% D ese% 0 2 = 7 [ ]= .0
1 2 = 3 [ ]= .0
´ E4 = 9 (0.07 + 0.038 )(2.089 # 10 −3 )·(3 106.496 )
´ E4 = 4 "30 [kW ]
> Ener/0' '4sor4 d' por ;$&ed'd de% ' re ´ E5
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´ E5 = ḿaire · 4 · c paire ( 52 ) [kW ]Donde
4 ≜&umedad dosdelaire
ḿaire ≜ gasto másicodelaire
C paire = 1.009 [ kJ kg·6 ]4
ϕ ≜& umedadrelati adelaire
2 bs≜2 debulbo seco = 2 a
C'r"' ps !o&1"r !'
ḿaire =( 28 % 2
(12 ) (1 2 +C1 ) (0.7685 ) )·C · ḿc[kg
aires ]
DondeN- O- 7 CO son prod$!"os de %' !o&4$s" ón en 2Se !ons der' CO 6
% 2 = 100 − (C1 2 +C1 +1 2 ) [ ]
% 2 = 100 − (4 +1 +16 ) [ ]
% 2 = 79 [ ]
C ≜ contenidodecarbonoenlacomposición = 0.8 = 80 [ ]
ḿaire =( 28 (79 )(12 ) (16 +1 ) (0.7685 ) )· (0.8 )· (2.089 # 10 −3 )ḿaire = "0235 [kg
aire
s ]O4"en endo ';or' 4 &ed 'n"e e% $so de %' !'r"' ps !o&1"r !':
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P'r' K (= G 7 T' -H JC o4"ene&os 5 2 6>> [kg 0 2 1kgaire ]A;or':
/ 2 = (2 gs− 2 a )´ E5 = 4 · ḿaire · c pa (2 gs− 2 a ) [kW ]´ E5 = (0.01559 )· (0.0235 )·(1.009 ) (375 − 24 )
´ E5 = 0 "1297 [kW ]
Ener/0' '4sor4 d' por /'ses se!os ´ E6
´ E6 = ḿgs · c pgs ·(2 gs− 2 a ) [kW ]Donde
ḿgs ≜ gastomásico de gasesde combustión
c pgs ≜calor específico aP = constante delosgases decombustión
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c pgs= 1.0048 [ kJ kg·6 ]ḿgs=(700 +1 2 +4 C1 23 (C1 2 +C1 ) )· C · ḿs[ kggcs ]ḿgs=(700 +(16 )+4 (4 )3 (4 +1 ) )· (0.8 )·(2.089 # 10 − 3 )ḿgs= 0.08155 [kggcs ]
Por %o "'n"o:´ E6 = (0.08155 )· (1.0048 )·(375 − 24 )´ E6 = 28.76 [kW ]
Ener/0' por $n d'd de " e&po perd do por e% '/$' de enfr '& en"o ´ E7
´ E7 = ḿ 0 2 1 · c p 0 2 1 (2 2 &2 o− 2 1 &2 o) [kW ]Donde
ḿ 0 2 1 ≜ gastom á sico deaguade enfriamiento
c p0 2 1 ≜ calor espec í fico a P = constantedel agua de enfriemiento
c p0 2 1 = 4.186 [ kJ kg·6 ]
ḿ 0 2 1 =( π ·D 0 2 1
2
4 )·& 0 2 1t 0 2 1
⋅ ρ 0 2 1 [kgs ] ρ 0 2 1 = 1 000 [ kgm3 ]
Donde& 0 2 1 ≜ alturaenel tan!ue deaguade enfriamiento
t 0 2 1 ≜ tiempo enel !ue&a- cambiode ni el eneltan!ue de aguadeenfriamiento
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ḿ 0 2 1 =( π · 0.2 24 )· (0.105 )
(5 ) ⋅ (1000 )
´
m 0 2 1= 0
"6597
[kgs ]
´ E7 = (0 "6597 )· (4.186 ) (50 − 38 )´ E7 = 33 "138 [kW ]
( Ener/0' perd d' por !o&4$s" 4%e no
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´ E9 = 87.93 − (20.793 +.0519 +4.30 +.1297 +28.76 +33.138 +7.98 )
´ E9 =− 7 "222 [kW ]
6 E@! en! ' '
' =´ E2´ E1
=20.793
87.93 8100
' = 23 "64
RESULTADOS
E6 ( 2 = ?E- - 2 = ?E= 2 >6 ?EH H2= ?E> 26- ?E -(2 ?E ==26=( ?E( 2( ( ?E 2-- ?
'-=2 H
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-
H
-(=-
(
EnergiaE- E= EH E> E E E( E
Serie 1
Ser e 6
ANALISIS DE RESULTADOS , CONCLUSIONES
6= . P á / n '
E6
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