Forces on the Reciprocating parts of an engine, Neglecting the weight of the connecting rod.
1 Answer

[1] Piston Effort:

$\rightarrow$ $F_p$ = Net load on the piston $\mp$ inertia force.

$\rightarrow$ $F_p = F_L \mp F_I$ - - - - - Neglecting frictional resistance.

$\rightarrow$ $F_p = F_L \mp F_I – R_F$ - - - - Considering frictional resistance.

$F_I = m_R. w^2.r.$ $[cos \ \theta + \frac{cos \ 2 \theta}{n}]$ {Inertia force of the reciprocatary parts.}


[1] Negative sign is used when the piston is accelerated, and positive sign is used when the piston is retarded.

[2] In a double acting reciprocating steam engine, net load on the piston,

$F_L = P_1 A_1 – P_2 A_2 = P_1 A_1 – P_2 [A_1 – a]$

Where, $P_1 A_1$ = Pressure and cross sectional area on the back end side of the piston.

$P_2 A_2$ = Pressure and cross sectional area on the crank end side of the piston.

$a =$ c/s area of the piston rod.

Reaction between the piston and cylinder.

$F_N = F_P \ tan \ \phi$

[3] If ‘P’ is the net pressure of steam or gas on the piston and ‘D’ is the diameter of the piston, the.

$F_L \ = \ Pressure \ \times \ Area$

$= p \times \frac{\pi}{4} (D^2)$

[4] Force acting along the connecting rod (OR) Thrust in the connecting rod.

$F_Q = \frac{F_p}{cos \ \phi}$

[5] Thrust on the sides of the cylinder walls OR normal reaction on the guide bars OR pressure on slide bars.

$F_N = F_Q \ sin \ \phi$

[6] Crank pin effort OR Tangential force on crank.

$F_T = F_Q \ sin \ (\theta + \phi)$

[7] Thrust on crankshaft bearings.

$F_B = F_Q . \ cos \ (\theta + \phi)$

[8] Turning moment OR torque on the crankshaft OR crank effort.

$T = F_T . r$

[9] To get $\phi$ $\rightarrow$ $sin \ \phi = \frac{sin \ \theta}{n}$

Please log in to add an answer.