Question

2. A 4.8 L, SI, four-stroke, V-8 stationary engine operates at 2000 RPM using gasoline with air/fuel ratio of 14.7. The engine has a volumetric efficiency of 90%, with a bore (B) and stroke (S) related as B:S=1.1:1. You can assume the density of fresh air is 1.16 kg/mº. Please calculate: (30 Points) a) Displacement of each cylinder, and length of stroke; b) Average velocity of piston; (This is a very simple question. You need find the distance piston moved and the time needed for this movement). c) The mass of air and fuel consumed by each cylinder in each cycle; d) Mass flow rate of air consumed by this engine in kg/min; e) Mass flow rate of fuel consumed by engine in kg/hour;

Answer 1

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 2

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 3

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 4

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 5

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 6

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 7

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =

Answer 8

Ans - Stivar 4.82 vs Engince displaj comat vi Volume - 4.8L=48X10m :?v=907 N = 2000 RPM 3. Boc. LS: Stroke s No.of cylinders, kus - S = 116 kg / m3 volt Air fuel Ratio 2 21407 aj Disp loucemiant volume of each cylindor = Total dis placemat volume No of cylindov s hy of x 10-3 - 0.6 X 10 m3 = 0.66 Displacement volume of each cylindor a = 1 / 2 B² xs 6-67 va (1.1%) vs
​​​​​​
S = 0.08678 85.7844 So stroke length is 95.73 MM b) Aug velacity of piston i = 2 SN - 2 x85.75 X 2000 X1000 5.7126 MLS vol of . aur intake c] volumetric officiency on y displacerat volume 0.9. Vas Vas 0.97006xió? Ya 0154 x 1ő red As in one cycle cach Cylinder is fixed once. So in tuke of avv occurs only once in ele cycle. Moves of intake ain, ma forva = 106 Xously x103 : 662 CM X 102K9

in Mass of earch fuel consumed in each cylinder cycle, me = mars of aire O Gair I fuel ratio) = 0.6264 14.7 = 0.04 26X163 kg. flow rate of air consumed in d ] Mass enginee. ma ma x k X N. k = no. of cylindor Es N = Enginee Revolution = 2,000 RPM ma = mass of air used by each Cylinder in each cele = 0.62614 lry n e (for four stroke anginee) 6.6GHX8 X Y DOG mov = 50" Kg/min

9 Mass flow vite of fuel Consumed by cininke. . Mass-fireng nie of air Air fiel vatic - 5 ° 11 kg Hàn E 14.7 = 0.3409 lig (min 2o. 445-38' lag Ihr =