A mono-cylinder engine comprises a non-offset slider crank mechanism as shown in the figure. The input crank pivoted at A rotates with a constant angular velocity co = 200 rad/s in counter-clockwise sense (refer to figure below). In the position shown, the input angle 0 is 45°, B=169.82°, and AC = 464.4 mm. a) Draw the vector loop for the given mechanism. b) Find the velocity of the piston. c) Find the acceleration of the piston. d) Using energy method, determine the output torque obtained from the combustion of air-fuel mixture. Note: The pressure in the combustion chamber is 6 MPa, and the piston bore diameter is 6.5 cm. The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in mm. 100 mm B 400 mm B M F Position velocity and acceleration analysis: Slider-crank linkage α3 ABA 003 AB R3 ΑΒΑ AA A b R2 AA 002 α2 d R1 (0) 4 C R4 x AB AA ΑΒΑ ABA AA 04 X (b) Position analysis (asino2 031 = arcsin or b asin02 d = acos02-bcos03 032 = arcsin +π Velocity analysis a cos02 d=-aw2 sin 02 +bw3 sin 03 03 = b cos03 α3= Acceleration Analysis aα2 cose₂-awsin 02 +bw sin 03 bcos03 =-aα2 sin 02-aw²¾½ cos 02 +bα3 sin 03 + bw cos03 A₁ = (A+A)=(aα₂ je₂-awe₂) ABA = (A'BA + ABA) = (bα3 je j°³ - bw³½³ej³°³) AB = AB = ä
A mono-cylinder engine comprises a non-offset slider crank mechanism as shown in the figure. The input crank pivoted at A rotates with a constant angular velocity co = 200 rad/s in counter-clockwise sense (refer to figure below). In the position shown, the input angle 0 is 45°, B=169.82°, and AC = 464.4 mm. a) Draw the vector loop for the given mechanism. b) Find the velocity of the piston. c) Find the acceleration of the piston. d) Using energy method, determine the output torque obtained from the combustion of air-fuel mixture. Note: The pressure in the combustion chamber is 6 MPa, and the piston bore diameter is 6.5 cm. The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in mm. 100 mm B 400 mm B M F Position velocity and acceleration analysis: Slider-crank linkage α3 ABA 003 AB R3 ΑΒΑ AA A b R2 AA 002 α2 d R1 (0) 4 C R4 x AB AA ΑΒΑ ABA AA 04 X (b) Position analysis (asino2 031 = arcsin or b asin02 d = acos02-bcos03 032 = arcsin +π Velocity analysis a cos02 d=-aw2 sin 02 +bw3 sin 03 03 = b cos03 α3= Acceleration Analysis aα2 cose₂-awsin 02 +bw sin 03 bcos03 =-aα2 sin 02-aw²¾½ cos 02 +bα3 sin 03 + bw cos03 A₁ = (A+A)=(aα₂ je₂-awe₂) ABA = (A'BA + ABA) = (bα3 je j°³ - bw³½³ej³°³) AB = AB = ä
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
I want to solve it based on the slider crank linkage rule, which is given in an image. Please, I do not want to solve it at all using the geometric method.
Transcribed Image Text:A mono-cylinder engine comprises a non-offset slider crank mechanism as shown in the figure. The input
crank pivoted at A rotates with a constant angular velocity co = 200 rad/s in counter-clockwise sense (refer
to figure below). In the position shown, the input angle 0 is 45°, B=169.82°, and AC = 464.4 mm.
a) Draw the vector loop for the given mechanism.
b) Find the velocity of the piston.
c) Find the acceleration of the piston.
d) Using energy method, determine the output torque obtained from the combustion of air-fuel
mixture.
Note: The pressure in the combustion chamber is 6 MPa, and the piston bore diameter is 6.5 cm.
The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in mm.
100 mm
B
400 mm
B
M
F
Transcribed Image Text:Position velocity and acceleration analysis: Slider-crank linkage
α3
ABA
003
AB
R3
ΑΒΑ
AA
A
b
R2
AA
002
α2
d
R1
(0)
4
C
R4
x
AB
AA
ΑΒΑ
ABA
AA
04
X
(b)
Position analysis
(asino2
031
= arcsin
or
b
asin02
d = acos02-bcos03
032
= arcsin
+π
Velocity analysis
a cos02
d=-aw2 sin 02 +bw3 sin 03
03 =
b cos03
α3=
Acceleration Analysis
aα2 cose₂-awsin 02 +bw sin 03
bcos03
=-aα2 sin 02-aw²¾½ cos 02 +bα3 sin 03 + bw cos03
A₁ = (A+A)=(aα₂ je₂-awe₂)
ABA = (A'BA + ABA) = (bα3 je j°³ - bw³½³ej³°³)
AB = AB = ä
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