Download DYNAMICS OF RECIPROCATING ENGINES

DYNAMICS OF
RECIPROCATING ENGINES
Indicator Diagrams
In-Line Engines
V Engines
GAS FORCES
Binomial Theorem:
So,
Since in most engines,
Therefore,
We can write
is so small, and only the first two terms of the series
may be taken with sufficient accuracy.
and since
and finally
It is important to note that always angular velocity () is assumed constant.
Therefore, acceleration of the piston may be obtained as,
-------------------------------------------------------------------------------------------------------
Using Binomial theorem for the denominator
of this ratio and taking only first two terms of
the series,
Therefore,
Similarly,
T12k + xi X (F14j)=0
T12
T12=-F14 x k =-P tan x k
T21=P tan x k
P

x
F12
F14
Making appropriate substitutions and neglecting higher order terms,
torque delivered by the crank to the shaft is obtained
EQUIVALENT MASSES
• A
A
•
m3
=
G3
•
G3
•
B
•
IG3
•
B
• 1) Total masses should be the same:
• 2) The mass center positions should be the same:
• (If
,
)
In practice,
for a connecting
rod
EQUIVALENT MASSES
• The first two conditions is called STATIC
EQIVALENCE.
• 3) The mass moment of inertias should be the
same:
• The third condition is called DYNAMIC
EQIVALENCE.
• For a connecting rod of a slider-crank mechanism, our aim is to divide the
total mass into two masses; one, at point A (rotating) and the other one,
at point B (reciprocating). The third condition does not conform our aim ,
therefore it is not satisfied throughout the analysis of connecting rod.
Dividing the crank mass into two parts, at O 2 and A, regarding the static
equivalence conditions.
+0
INERTIA FORCES
Constant angular velocity assumption for crank gives the acceleration of point A as,
Inertia force at A
Acceleration of the piston has been found as,
Inertia force of the reciprocating parts
The components of total inertia force,
Primary
İnertia
force
Secondary inertia force
Inertia Torque
T12k + xi X (- F14j)=0
T12=F14 x k =-mBaB tan x k
-mAaA
T12

-mBaB
x
Making appropriate substitutions, neglecting higher
order terms and using trigonometric identities, inertia
torque exerted by the engine on the shaft is obtained ,
F12
Inertia torque is a periodic function, including the first three harmonics.
F14=-mBaB tan 
BEARING LOADS IN THE
SINGLE-CYLINDER ENGINE
• The resultant bearing loads in a single-cylinder
engine are made up of due to the following four
forces:
• 1) Gas force (examined at the beginning).
• 2) Inertia force due to the mass of piston m4.
• 3) Inertia force due to the mass of connecting
rod m3B at piston-pin end.
• 4) Inertia force due to the mass of connecting
rod m3A at crank-pin end.
• (It is assumed that crank has been balanced.)
2) Inertia force due to the mass of piston m4
F34’’
F14’’
3) Inertia force due to the mass of connecting
rod m3B at piston-pin end
R
F43’’’

R
4) Inertia force due to the mass of connecting
rod m3A at crank-pin end
A
r
If a counterweight of
m3A is added at point C,
O2
r
r
C
r O2
FBD
C
A
no bearing force
will act at O2.
Therefore shaking
force due to m3A
İs eliminated
Superposition
CRANKSHAFT TORQUE
•
O2
1
•
•
T21
F21
B
MO2=0
F41
The torque delivered by the crankshaft to the load.
ENGINE SHAKING FORCES
(due to only reciprocating masses)
for
Graphically, F21x = OA’+OB’
Shaking Force:
(Linear vibration in x direction)
Shaking Couple:
(Torsional vibration about crank center)
PROBLEM
Stroke=2r =90 mm
d=100 mm
n=4400 rpm
m 3A=0.8 kg
Obtain cylinder pressure for the 30% expansion case.
p= 1.857 Mpa=1857000 Pa(N/m 2)
P=1857000(7.854x10-3 ) = 14584.878 N
r=45 mm
A=d2/4=1002/4=7853.98 mm 2=7.854x10-3 m 2
=4400(2)/60=460.767 r/s
m 3B=0.38 kg
m 4=1.64 kg
Force Calculations
F41=-[(0.38+1.64)(-3578.663)+14584.878](0.1155)j
F41= -849.616j N
F41
F34=[1.64(-3578.663)+14584.878]i – 849.616j = 8715.87i-849.616j N