The firing order in a six cylinder vertical two stroke in-line engine is 1-4-2-6-3-5. The piston stroke is 100 mm and length of each connecting rod is 200 mm. The pitch distances between the cylinder centre lines are 100 mm, 100 mm, 150 mm, 100 mm and 100 mm respectively. The reciprocating mass per cylinder is 1 kg and the engine runs at 3000 rpm. Determine the out of balance primary and secondary forces and couples on this engine, taking the plane of cylinder 2 as the reference plane.

 

Six-Cylinder Two-Stroke Engine Balance Analysis

Let's analyze the out-of-balance forces and couples for this six-cylinder two-stroke engine.

Given Data:

• Engine type: 6-cylinder vertical two-stroke in-line

• Firing order: 1-4-2-6-3-5

• Piston stroke: 100 mm → Crank radius (R) = 50 mm = 0.05 m

• Connecting rod length (L) = 200 mm = 0.2 m

• Cylinder spacing: 100, 100, 150, 100, 100 mm (from cylinder 1 to 6)

• Reciprocating mass per cylinder (m) = 1 kg

• Engine speed = 3000 rpm → ω = 2π × 3000/60 = 314.16 rad/s

• Reference plane: Cylinder 2

Crank Angle Determination:

For a two-stroke engine with firing order 1-4-2-6-3-5, the crank angles are spaced at:
θ₁ = 0° (cylinder 1)
θ₄ = 60°
θ₂ = 120°
θ₆ = 180°
θ₃ = 240°
θ₅ = 300°

Primary Forces and Couples:

Primary force for each cylinder: Fₚ = mω²Rcosθᵢ

Primary Force Calculation:

Sum of primary forces:
ΣFₚ = mω²R[cos(0°) + cos(60°) + cos(120°) + cos(180°) + cos(240°) + cos(300°)] = 0

Primary Couple Calculation:

Taking cylinder 2 as reference (distance = 0):

• Distance from reference plane:
  x₁ = -100 mm = -0.1 m (cylinder 1)
  x₂ = 0 m (cylinder 2)
  x₃ = 150 mm = 0.15 m (cylinder 3)
  x₄ = -100 mm = -0.1 m (cylinder 4)
  x₅ = 250 mm = 0.25 m (cylinder 5)
  x₆ = 100 mm = 0.1 m (cylinder 6)

Primary couple components:
Cₚₓ = mω²RΣ(xᵢsinθᵢ)
Cₚᵧ = mω²RΣ(xᵢcosθᵢ)

Calculations:
Cₚₓ = 1×(314.16)²×0.05×[(-0.1)sin0° + 0sin120° + 0.15sin240° + (-0.1)sin60° + 0.25sin300° + 0.1sin180°]
= 4934.8×[0 + 0 + (-0.1299) + (-0.0866) + (-0.2165) + 0] = 4934.8×(-0.433) = -2136.8 Nm

Cₚᵧ = 1×(314.16)²×0.05×[(-0.1)cos0° + 0cos120° + 0.15cos240° + (-0.1)cos60° + 0.25cos300° + 0.1cos180°]
= 4934.8×[(-0.1) + 0 + (-0.075) + (-0.05) + 0.125 + (-0.1)] = 4934.8×(-0.2) = -986.96 Nm

Resultant primary couple:
Cₚ = √(Cₚₓ² + Cₚᵧ²) = √(2136.8² + 986.96²) = 2357.2 Nm

Secondary Forces and Couples:

Secondary force for each cylinder: Fₛ = (mω²R/4n)cos2θᵢ
where n = L/R = 200/50 = 4

Secondary Force Calculation:

Sum of secondary forces:
ΣFₛ = (mω²R/4n)[cos(0°) + cos(120°) + cos(240°) + cos(360°) + cos(480°) + cos(600°)] = 0

Secondary Couple Calculation:

Secondary couple components:
Cₛₓ = (mω²R/4n)Σ(xᵢsin2θᵢ)
Cₛᵧ = (mω²R/4n)Σ(xᵢcos2θᵢ)

Calculations:
Cₛₓ = (4934.8/16)×[(-0.1)sin0° + 0sin240° + 0.15sin480° + (-0.1)sin120° + 0.25sin600° + 0.1sin360°]
= 308.4×[0 + 0 + 0.1299 + (-0.0866) + (-0.2165) + 0] = 308.4×(-0.1732) = -53.42 Nm

Cₛᵧ = (4934.8/16)×[(-0.1)cos0° + 0cos240° + 0.15cos480° + (-0.1)cos120° + 0.25cos600° + 0.1cos360°]
= 308.4×[(-0.1) + 0 + (-0.075) + 0.05 + (-0.125) + 0.1] = 308.4×(-0.15) = -46.26 Nm

Resultant secondary couple:
Cₛ = √(Cₛₓ² + Cₛᵧ²) = √(53.42² + 46.26²) = 70.68 Nm

Summary of Results:

Parameter	Value
Primary Forces	0 N
Secondary Forces	0 N
Primary Couple	2357.2 Nm
Secondary Couple	70.68 Nm

The engine has balanced primary and secondary forces but unbalanced primary and secondary couples. The primary couple is significantly larger than the secondary couple due to the engine configuration and operating speed.