Introducing
Your new presentation assistant.
Refine, enhance, and tailor your content, source relevant images, and edit visuals quicker than ever before.
Trending searches
14
MEC 3101 DYNAMICS OF MACHINERY
DATE
MODULE II
ENGINE DYNAMICS / PART 1
Inertia Force Analysis
Dr. S. RASOOL MOHIDEEN, Dean, SMS
THANK YOU!
Velocity
&
Acceleration
of
Piston
Velocity
&
Acceleration
of
Piston
31
Forces Acting in Engine
Forces in Engine
Force acting along the connecting rod:
Thrust on the sides of the cylinder walls or normal reaction on the guide bars:
Thrust on crank shaft bearings:
Crank-pin effort:
Crank effort or turning moment or torque on the crank shaft:
PROBLEM 2
The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 60° from I.D.C., the difference between the driving and the back pressures is 0.35 N/sq.mm. The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate : (i). Piston Effort , (ii). Thrust on slide bars, (iii). Thrust in the connecting rod, (iv). Tangential force on the crank-pin, and (v). Turning moment on the crank shaft.
Problem 2
Given: r = 300 mm = 0.3 m ; mR = 250 kg; θ = 60°; p1 – p2 = 0.35 N/mm.sq; l = 1.2 m ; D = 0.5 m = 500 mm ; N = 250 r.p.m. or ω = 2 π × 250/60 = 26.2 rad/s
(i) To find the piston effort (FP):
We know that net load on the piston,
Ratio of length of connecting rod and crank,
n = l / r = 1.2 / 0.3 = 4
and accelerating or inertia force on reciprocating parts,
Piston effort, FP = FL – FI = 68 730 – 19 306 = 49 424 N = 49.424 kN
Solution
(ii) Thrust on slide bars:
Let φ phi = Angle of inclination of the connecting rod to the line of stroke
φHence Phi = 12.5°
We know that thrust on the slide bars,
FN = FP tan (phi)φ = 49.424 × tan 12.5° = 10.96 kN
(iii) Thrust in the connecting rod:
thrust in the connecting rod,
Solution (Contd.)
(iv) Tangential force on the crank-pin:
We know that tangential force on the crank pin,
FT = FQ sin (θ theta + phi φ) = 50.62 sin (60° + 12.5°)
= 48.28 kN Ans.
(v) Turning moment on the crank shaft:
We know that turning moment on the crank shaft,
T = FT × r = 48.28 × 0.3 = 14.484 kN-m Ans.
Solution (Contd.)
Dynamically Equivalent System
Dynamically Equivalent System
Inertia Force Analysis
Inertia Force Analysis
N
M
A
P
B
O
Q
K
Fc
P
A
D
d
G
B
g
O
L
Q
Fp
I
Y
FP
WC
FC
X
FT
FQ
FN
WC
I
Problem 4
Problem 4
The connecting rod of an internal combustion engine is 225 mm long and has a mass 1.6 kg. The mass of the piston and gudgeon pin is 2.4 kg and the stroke is 150 mm. The cylinder bore is 112.5 mm. The centre of gravity of the connection rod is 150 mm from the small end.
Its radius of gyration about the centre of gravity for oscillations in the plane of swing of the connecting rod is 87.5 mm. Determine the magnitude and direction of the resultant force on the crank pin when the crank is at 40° and the piston is moving away from inner dead centre under an effective gas pressure of 1.8 MN/sq.m. The engine speed is 1200 r.p.m.
(Courtesy: TOM, R.S. Khurmi)
Solution
Given : l = PC = 225 mm = 0.225 m;
mC = 1.6 kg;
mR = 2.4 kg;
L = 150 mm or r = L/2 = 75 mm = 0.075 m ;
D = 112.5 mm = 0.1125 m ;
PG = 150 mm ;
kG = 87.5 mm = 0.0875 m ; θ
teta= 40° ;
N = 1200 r.p.m. or ω = 2π × 1200/60 = 125.7 rad/s
p = 1.8 MN/sq.m = 1.8 × 10E6 N/sq.m ;
Solution
Inertia Force Diagram
Y
Step 1: Draw the mechanism
Step 2: Draw the acceleration diagram
I
Step 3: Fix the point D, the position of dynamically equivalent mass from G
Step 4: Find the position of Inertia force of connecting rod , Fc
Step 5: Fix the instantaneous centre , I
X
K
Step 6: Take moments of all the forces about
FC
FT
S
M
C
d
D
Q
g
P
FP
G
O
N
40
L
WC
Solution (Contd.)
Inertia Force Diagram (Contd.)
Y
Step 6: Take moments of all the forces about I
I
Step 7: By measurement,
NO = 0.0625 m ; gO = 0.0685 m ; IC = 0.29 m ; IP = 0.24 m ; I Y = 0.148 m ; and IX = 0.08 m
Step 8: To find FP
Force due to gas pressure,
X
K
Inertia force due to mass of the reciprocating parts,
FC
FT
S
M
C
d
D
Q
g
P
FP
G
N
O
40
L
WC
Solution (Contd.)
Inertia Force Diagram (Contd.)
∴Net force on the piston,
FP = FL − FI = 17 895 − 2370 = 15 525 N
FC
FP
Step 5: Inertia force due to mass of the connecting rod,
FN
FQ
FT
Step 6: Taking moments about point I, to find FT
FR
FP x IP= (WC x IY) + (FC x IX) + (FT x IC)
FORCE POLYGON
15 525 x 0.24 =(1.6 x 9.81 x0.148)+ ( 1732x 0.08)+ (FT x 0.29)
FT = 12 362 N
By measurement, FN = 3550 N; and FR = 7550 N
FQ ,the resultant of FR and FT
By measurement, FQ = 13 750 N Ans.