Combustion Solved Problems

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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+#ka. 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, ka5ka5

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 eore ignition. mi8ture eore ignition.

0c Actual temperature and pressure in the cylinder at ignition. 34, 0c Actual temperature and pressure in the cylinder at ignition. 34,

ka5 ka5 Solution:

Solution:

00aa %%## = % = %110r0rcck-1k-1 = 0))( 4011 = 0))( 4011&.)"&.)" = $$9 4 = "&" = $$9 4 = "&" °°''

## =  = 110r0rcckk = 0+# ka011 = 0+# ka0111.)"1.)" = #)# ka = #)# ka

0

0 'ranksha'rankshat o7set = a = S# = "t o7set = a = S# = ".$# # = #.9( cm.$# # = #.9( cm ; = ra =

; = ra = 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 = )" = )"°°

comustion chamer volume <hen intake valve closes comustion chamer volume <hen intake valve closes >

>I>'I>'>>cc = 1  ?0r = 1  ?0rcc @ 1 3;  1 - cos @ 1 3;  1 - cosθθ - 0; - 0;## @ sin @ sin##θθ1#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°°551#1#B B ==

1&.&" 1&.&"

'omustion chamer volume <hen ignition occurs 'omustion chamer 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)"°°551#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.#91.)"-11.)"-1 = $& 4 = )1 = $& 4 = )1 °°''

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#. A ).(-liter, >( SI engine is designed to have a ma8imum speed o $&&& ;C.  %here are t<o intake valves per cylinder, and valve lit 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. 3msec5 0c o the valve diameters and ore siFe seem compatileG Solution:

Hor 1 cylinder

>d = ).(( = &.(  = &.&&&( m) = 0J01.&(!)

! = &.&9+$ m = 9.+$ cm

S = 01.&(09.+$ cm = +."& cm 0a Sonic velocity at inlet conditions

' = 3k;%51# = 301.&0#9$ Kkg.40)))451# = )(( msec

Average piston velocity at ma8imum speed

0Lpma8 = #SM = 0# stokescycle0&.&+"& mstroke0$&&&(& revsec = ##.1$

msec

Area or the # valves

Ai = '!#30Lpma8c15 = 01.+09.+$ cm#0##.1$ msec0)(( msec = (.) cm#

Hor 1 valve

A1 = (.)# = ).1$ cm# = Jdvl = Jdv0dv = 0Jdv#

dv = #.&1 cm

0 Ca8imum Eo< velocity <ill e sonic >ma8 = c= )(( msec

0c Nith proper design valves could e ft into comustion chamer <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

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has one port inPector <hich delivers uel at a rate o &.&# kgsec. %he engine also has an au8iliary inPector upstream in the intake maniold <hich delivers uel at a rate o &.&&) kgsec 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

0AHact = 0AHstoichR = +.&  1.&( = 9.+

Hor 1 cylinder >d = 0#.   ( = &.  = &.&&& m)

0a mass o air into one cylinder or one cycle

ma = a>dTv = 01.191 kgm) 0&.&&& m) 0&.9$ = &.&&&11 kg

mass o uel into 1 cylinder or 1 cycle

m  = ma0AHa = 0&.&&&11 kg  9.+ = &.&&&&9 kg

time o inPection = 0&.&&&&9 kg  0&.&# kgsec = &.&&## sec 0 Actual AH rom aove 0AHa = 9.+

0c Nhen au8iliary inPector is used at )&&& ;C  %ime o 1 cycle

 % = 0# revcycle0)&&&(& revsec = &.& seccycle 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&.&&) kgsec0&.& sec0( cylinders = &.&&&&#& kg

1 cylinder or 1 cycle

AH = mam  = &.&&&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 caruretor, each arrel having a discharge

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coeOcient o ' t = &.+". %he uel used is gasoline at AH = 1":1 0density o  gasoline g = $"& kgm).

'alculate: 0a Cinimum throat diameter needed in each caruretor venturi. 3cm5

0 Huel capillary tue diameter needed or each venturi throat i  tue discharge coeOcient ' c = &.9" and the capillary tue height di7erential is small. 3mm5

Solution:

air Eo< rate needed at ma8 speed ´

ma= ρ

aV d N ηv/n  = 01.1910&.&&(#0("&&(&0&.99# = &.)+ kgsec

Hor each o  arrels ´

ma  = 0&.)+ kgsec = &.&9$) kgsec

0a throat area o 1 arrel

0 m´a ma8 = #)(." 'IU3At5 = &.9$) = 0#)(."0&.+"3At5

At = &.&&&)99( m# = ).99( cm#= 0Jdt#

dt= #.## cm

0 uel Eo< rate needed or 1 arrel ´

m  = m´

a 0AH = 0&.&9$) kgsec1" = &.&&"9# kgsec

pressure in caruretor throat

t = 0&."#9)& = 0&."#9)01&1 ka = "). ka

pressure di7erential in uel capillary

t =a = & @ t = 1&1 -"). = $.( ka

uel capillary tue Eo< area ´

m  = '

cAc3# ∆t51#

0.00582kg/sec¿

 0&.9"Ac30#0$"& kgm)0$.( kMm#01 kg-mM-sec#51#

Ac = &.&&&&&&91 m# = &.91 mm# = 0Jdc#

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". 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 lit 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

0At1 = Jdvl = J0). cm 30&.##0).5 = $.++ cm#

Nith t<o intake valves

0At# = 0# valves J 0#.$ cm 30&.##0#.$ cm5 = 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 Ee8iility in intake e allo<ing variation in valve timing and lit et<een the t<o valves.

isadvantages: Meed or greater numer andor more comple8 camshats. Wigher cost in manuacturing.

Figure

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References

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