LECTURE #11 1.060 ENGINEERING MECHANICS II THE MOMENTUM PRINCIPLE ∑(∂/∂t) ∫ρqdV = ∫ρqdA + ∑MP + ∑ other forces acting on fluid MP = Thrust (Momentum M + Pressure P) at inflow/outflow section of Ain & Aout If flow is well behaved at Ain & Aout: Straight parallel streamlines // A Then MP = ∫(ρq₁ + p)dA(-nₚ) = (KmρV² + PcA)f(-nₚ) Km = ∫q₁²dA / (V²)A = 1 (very well behaved) V = Q/Af; Pc = pressure at Center of Gravity of Af FOR STEADY FLOW: ∂/∂t = 0 ∫ρqdA + ∑MP + ∑ other forces on fluid in V = 0 |MP| = ρV²Af + PcAf and it acts INWARDS towards t₀ in direction L. Af ∑ Forces on t₀ = -∑ Forces from t₀ EXAMPLE I: The Firehose Patm MP₁ MP₂ U₁ U₂ Patm F x Gravity acts in z-direction Plate wide enough to completely deflect the jet Jet is freely falling In x-direction MP₁x + MP₂x + Fx = 0 -MP₂ = -p₂U₂²A Fx = force on fluid from plate = -MP₂ = p₂U₂²A -Fx = force from fluid on plate = -p₂U₂²A Firehose acts as a pressure force (p₂U₂²)A In y-direction MP₁y + MP₂y + MP₃y = 0 → MP₁ = MP₂ P₁ = P₂ = 0 since jets in atmosphere R₁ = R₂ because of symmetry (P₁ + p₁U₁²)A₁ = (P₂ + p₂U₂²)A₂ → U₁ = U₂ But what is U₁, U₂? Bernoulli along surface streamlines from "0" to "1" and "0" to "2", with same z and p₀ = p₁ = p₂: z₀ρg + (1/2)ρU₀² = z₁ρg + (1/2)ρU₁² and z₀ρg + (1/2)ρU₀² = z₂ρg + (1/2)ρU₂² Note: If flow in x₂-plane gravity enters the problem and we get from Bernoulli: z₀ρg + (1/2)ρU₀² = z₁ρg + (1/2)ρU₁² = z₂ρg + (1/2)ρU₂² or U₁² = U₀² + 2g(z₀ - z₁); U₂² = U₀² + 2g(z₀ - z₂) Thus, U₀ ≈ U₁ ≈ U₂ still hold if U₀² / 2g >> |z₀ - z₁| = elevation difference In problem where the characteristic velocity head V²/2g is much larger than elevation differences, these may be neglected. In momentum principle applications this translates into a neglect of gravity forces. is acceptable when V²/2g >> ΔZ. EXAMPLE E: Flow around a bend flanged connection Gravity acts in z (Neglect here) Find forces & moment (in xy-plane) on flange at 1-1 M N ② MP₂ \[Q_{2}\] Flow well behaved at 1-1 & 2-2 (far enough from bend) Volume Conservation Q₁ = Q₂ = Q; A₁ = A₂ = A (always!) → V₁ = V₂ = Q/V Momentum in X-direction: MP₁ + Fx = 0 → Fx = -MP₁ = -(ρV² + p₀₁)A Force on Pipe in y-direction: MP₂ + Fy = 0 → Fy = -MP₂ = -(ρV² + p₀₂)A Force on pipe = Fy = -(ρV² + p₀₂)A - T Moment of forces around flanged connection: MP is acting as a total force - so just use it as such to get moments around CG of 1-1. MP₁ has no moment arm and MP₂ has no contribution MP₂ has moment arm = Xb M = MP₂ · Xb = (ρV² + p₀₂)AXb (counter clockwise)