1.
1. A fve-cyliA fve-cylindernder, our, our-str-stroke cycoke cycle SI engine has a comprle SI engine has a compressiession ratio rc = 11:1,on ratio rc = 11:1, o
orre e ! ! = = "."."#"#cmcm, , ststrrokoke e S S = = ".".$# $# cmcm, , anand d coconnnnecectiting ng rorod d lelengngth th % % == 11.&
11.&&cm. 'ylinder inlet &cm. 'ylinder inlet condconditionitions s are ()*' are ()*' and and +#k+#ka. a. %he intake valve%he intake valve closes at 1*a!' and the spark plug is
closes at 1*a!' and the spark plug is fred at/1:"*%'.fred at/1:"*%'.
'alculate: 0a %emperature and pressure in the cylinder at ignition, assuming 'alculate: 0a %emperature and pressure in the cylinder at ignition, assuming 2tto cycle analysis 0i.e., assume the intake valve closes at !' 2tto cycle analysis 0i.e., assume the intake valve closes at !' and ignition is at %'. 34,
and ignition is at %'. 34, ka5ka5
0 67ective compression ratio 0i.e., actual compression o the air-uel 0 67ective compression ratio 0i.e., actual compression o the air-uel
mi8ture eore ignition. mi8ture eore ignition.
0c Actual temperature and pressure in the cylinder at ignition. 34, 0c Actual temperature and pressure in the cylinder at ignition. 34,
ka5 ka5 Solution:
Solution:
00aa %%## = % = %110r0rcck-1k-1 = 0))( 4011 = 0))( 4011&.)"&.)" = $$9 4 = "&" = $$9 4 = "&" °°''
## = = 110r0rcckk = 0+# ka011 = 0+# ka0111.)"1.)" = #)# ka = #)# ka
0
0 'ranksha'rankshat o7set = a = S# = "t o7set = a = S# = ".$# # = #.9( cm.$# # = #.9( cm ; = ra =
; = ra = 11.&#.9( = ).9"11.&#.9( = ).9"
'rank angle <hen intake valve closes and actual compression starts 'rank angle <hen intake valve closes and actual compression starts
θ
θ = 19& = 19&°° 1 1°° = ##1 = ##1°°
'rank angle <hen ignition occurs
'rank angle <hen ignition occurs θθ = )" = )"°°
comustion chamer volume <hen intake valve closes comustion chamer volume <hen intake valve closes >
>I>'I>'>>cc = 1 ?0r = 1 ?0rcc @ 1 3; 1 - cos @ 1 3; 1 - cosθθ - 0; - 0;## @ sin @ sin##θθ1#1#55
>
>I>'I>'>>cc = 1 ?011-1 0).9" 01 @ cos0##1 = 1 ?011-1 0).9" 01 @ cos0##1°° @ 30).9" @ 30).9"## @ sin @ sin##0##10##1°°551#1#B B ==
1&.&" 1&.&"
'omustion chamer volume <hen ignition occurs 'omustion chamer volume <hen ignition occurs >
>igig>>cc = 1 ?011-1 0).9" 01 @ cos0)" = 1 ?011-1 0).9" 01 @ cos0)"°° @ 30).9" @ 30).9"## @ sin @ sin##0)"0)"°°551#1#B B ==
1.#1 1.#1
67ective compression ratio 67ective compression ratio >
>I>'I>'>>igig = 0> = 0>I>'I>'>>cc 0> 0>igig>>cc = 01&.&" 01.#1 = 9.#9 = 01&.&" 01.#1 = 9.#9
00cc %%## = 0))( 4 09.#9 = 0))( 4 09.#91.)"-11.)"-1 = $& 4 = )1 = $& 4 = )1 °°''
#. A ).(-liter, >( SI engine is designed to have a ma8imum speed o $&&& ;C. %here are t<o intake valves per cylinder, and valve lit eDuals one-ourth valve diameter. !ore and stroke are related as S = 1.&( !. esign temperature o the air-uel mi8ture entering the cylinders is (&*'.
'alculate: 0a Ideal theoretical valve diameter. 3cm5
0 Ca8imum Eo< velocity through intake valve. 3msec5 0c o the valve diameters and ore siFe seem compatileG Solution:
Hor 1 cylinder
>d = ).(( = &.( = &.&&&( m) = 0J01.&(!)
! = &.&9+$ m = 9.+$ cm
S = 01.&(09.+$ cm = +."& cm 0a Sonic velocity at inlet conditions
' = 3k;%51# = 301.&0#9$ Kkg.40)))451# = )(( msec
Average piston velocity at ma8imum speed
0Lpma8 = #SM = 0# stokescycle0&.&+"& mstroke0$&&&(& revsec = ##.1$
msec
Area or the # valves
Ai = '!#30Lpma8c15 = 01.+09.+$ cm#0##.1$ msec0)(( msec = (.) cm#
Hor 1 valve
A1 = (.)# = ).1$ cm# = Jdvl = Jdv0dv = 0Jdv#
dv = #.&1 cm
0 Ca8imum Eo< velocity <ill e sonic >ma8 = c= )(( msec
0c Nith proper design valves could e ft into comustion chamer <ith diOculty
). A si8-cylinder, our-stroke cycle SI engine <ith multipoint uel inPection has a displacement o # liters and a volumetric eOciency o 9$Q at )&&& ;C, and operates on ethyl alcohol <ith an eDuivalence ratio o 1.&(. 6ach cylinder
has one port inPector <hich delivers uel at a rate o &.&# kgsec. %he engine also has an au8iliary inPector upstream in the intake maniold <hich delivers uel at a rate o &.&&) kgsec to change the air-uel ratio and give a richer mi8ture <hen needed. Nhen in use, the au8iliary inPector operates continuously and supplies all cylinders.
'alculate: 0a %ime o one inPection pulse or one cylinder or one cycle. 3sec5
0 AH i the au8iliary inPector is not eing used. 0c AH i the au8iliary inPector is eing used. Solution:
actual AH ratio
0AHact = 0AHstoichR = +.& 1.&( = 9.+
Hor 1 cylinder >d = 0#. ( = &. = &.&&& m)
0a mass o air into one cylinder or one cycle
ma = a>dTv = 01.191 kgm) 0&.&&& m) 0&.9$ = &.&&&11 kg
mass o uel into 1 cylinder or 1 cycle
m = ma0AHa = 0&.&&&11 kg 9.+ = &.&&&&9 kg
time o inPection = 0&.&&&&9 kg 0&.&# kgsec = &.&&## sec 0 Actual AH rom aove 0AHa = 9.+
0c Nhen au8iliary inPector is used at )&&& ;C %ime o 1 cycle
% = 0# revcycle0)&&&(& revsec = &.& seccycle Cass o uel rom port inPector or 1 cylinder or 1 cycle m = &.&&&&9 kg
mass o uel rom au8iliary inPector or 1 cylinder or 1 cycle m = 0&.&&) kgsec0&.& sec0( cylinders = &.&&&&#& kg
1 cylinder or 1 cycle
AH = mam = &.&&&11 0&.&&&&9 &.&&&&#& = (.&1
. A (.#-liter, >9, our-stroke cycle SI engine is designed to have a ma8imum speed o ("&& ;C. At this speed, volumetric eOciency is 99Q. %he engine is eDuipped <ith a our-arrel caruretor, each arrel having a discharge
coeOcient o ' t = &.+". %he uel used is gasoline at AH = 1":1 0density o gasoline g = $"& kgm).
'alculate: 0a Cinimum throat diameter needed in each caruretor venturi. 3cm5
0 Huel capillary tue diameter needed or each venturi throat i tue discharge coeOcient ' c = &.9" and the capillary tue height di7erential is small. 3mm5
Solution:
air Eo< rate needed at ma8 speed ´
ma= ρ
aV d N ηv/n = 01.1910&.&&(#0("&&(&0&.99# = &.)+ kgsec
Hor each o arrels ´
ma = 0&.)+ kgsec = &.&9$) kgsec
0a throat area o 1 arrel
0 m´a ma8 = #)(." 'IU3At5 = &.9$) = 0#)(."0&.+"3At5
At = &.&&&)99( m# = ).99( cm#= 0Jdt#
dt= #.## cm
0 uel Eo< rate needed or 1 arrel ´
mf = m´
a 0AH = 0&.&9$) kgsec1" = &.&&"9# kgsec
pressure in caruretor throat
t = 0&."#9)& = 0&."#9)01&1 ka = "). ka
pressure di7erential in uel capillary
∆t = ∆a = & @ t = 1&1 -"). = $.( ka
uel capillary tue Eo< area ´
mf = '
cAc3# ∆t51#
0.00582kg/sec¿
0&.9"Ac30#0$"& kgm)0$.( kMm#01 kg-mM-sec#51#
Ac = &.&&&&&&91 m# = &.91 mm# = 0Jdc#
". A >9 engine <ith $."-cm ores is redesigned rom t<o valves per cylinder to our valves per cylinder. %he old design had one inlet valve o ) mm diameter and one e8haust valve o #+ mm diameter per cylinder. %his is replaced <ith t<o inlet valves o #$ mm diameter and t<o e8haust valves o #) mm diameter. Ca8imum valve lit eDuals ##Q o the valve diameter or all valves.
'alculate: 0a Increase o inlet Eo< area per cylinder <hen the valves are ully open.3cm#5
0 Vive advantages and disadvantages o the ne< system. Solution:
0a <ith 1 intake valve
0At1 = Jdvl = J0). cm 30&.##0).5 = $.++ cm#
Nith t<o intake valves
0At# = 0# valves J 0#.$ cm 30&.##0#.$ cm5 = 1&.&9 cm#
Increase in Eo< area
∆A = 01&.&9 cm# @ 0$.++ cm# = #.&+ cm#
0 Advantages: Vreater intake valve Eo< area <hich improves volumetric eOciency.
Vreater e8haust valve Eo< area <hich allo<s or a shorter e8haust lo<do<n process.
Vreater Ee8iility in intake e allo<ing variation in valve timing and lit et<een the t<o valves.
isadvantages: Meed or greater numer andor more comple8 camshats. Wigher cost in manuacturing.