1.060 Engineering Mechanics II, Problem Set 4 - Solutions

PROBLEM SET 4-SOLUTIONS PROBLEM N°1: [A diagram of a curved pipe (elbow) with fluid flowing through it. The pipe bends 45 degrees. The inlet is labeled A and the outlet is labeled B. MPA and MPB represent momentum flux at A and B, respectively. VA and VB are velocities. A control volume (C.V.) is drawn around the elbow. Forces Fx and Fz are shown acting on the fluid within the control volume. A weight W acts downwards. A coordinate system with x and z axes is shown. The text on the right says "Fx, Fz: Forces from the elbow on the fluid."] Conservation of volume between A and B: QA = QB => VA.AA = VB.AB => VB = (AA/AB)VA AA = πdA²/4 = π(0.4)²/4 = 0.1257 m² AB = πdB²/4 = π(0.2)²/4 = 0.03142 m² VB = (0.1257/0.03142)VA = 4VA Bernoulli between A and B (We neglect losses, which makes sense if the distance between A and B is small, because then we have a short transition of a converging flow). pA + ρgZA + 1/2ρVA² = pB + ρgZB + 1/2ρVB² ZA = ZB, VB = 4VA 1/2ρ(4² - 1)VA² = pA - pB 1/2 * 850 * 15 * VA² = 1.5 * 10⁵ - 1.3 * 10⁵ => VA = 1.771 m/s VB = 4VA = 7.085 m/s Thrust on A: MPA = (ρVA² + pCG,A).AA = (850 * 1.771² + 1.5 * 10⁵) * 0.1257 = 19190 N Thrust on B: MPB = (ρVB² + pCG,B).AB = (850 * 7.085² + 1.3 * 10⁵) * 0.03142 = 5425 N Weight: W = ρgV = 850 * 9.8 * 0.15 = 1250 N Conservation of momentum in x-direction: Σ Forces in x = MPA cos 45° - MPB cos 45° + Fx = 0 => Fx = -19190(√2/2) + 5425(√2/2) = -9733 N (to the left) -> The horizontal force from the fluid on the elbow is of the same magnitude acting to the right. Conservation of momentum in z-direction: Σ Forces in z = -MPA sin 45° - W - MPB sin 45° + Fz = 0 => Fz = 19190(√2/2) + 1250 + 5425(√2/2) = 18655 N (upwards) -> The vertical force from the fluid on the elbow is of the same magnitude acting downwards. PROBLEM No. 2: [A diagram showing a jet of fluid with height h0 and velocity V0 impacting a flat plate at an angle β. The plate splits the jet into two streams. The left stream has height hl and velocity Vl. The right stream has height hr and velocity Vr. The system is at atmospheric pressure (Patm). A vertical force Fv is shown acting on the plate.] a) Vl and Vr are horizontal, and therefore contribute no momentum force in vertical direction. The only vertical momentum force is from the incident jet (ρV₀²h₀ + p₀h₀)sinβ = Fv but p₀ = 0 [atmospheric pressure on all sides], so Vertical jet force is downward = ρV₀²h₀ sinβ b) Horizontal momentum balance gives, with all pressures being zero since jets are free ρVl²hl + ρV₀²h₀cosβ = ρVr²hr or since Vl = V₀ = Vr hr - hl = h₀cosβ (1) Conservation of volume (continuity) gives V₀h₀ = Qin = Vrhr + Vlhl = ΣQout or hr + hl = h₀ (2) Combining (1) and (2) gives hr = h₀(1+cosβ)/2; hl = h₀(1-cosβ)/2 c) Application of Bernoulli along streamline from O₂ to R gives V₀²/2g + p₀/ρg + z₀ = Vr²/2g + pr/ρg + zr but p₀ - pr = 0 and z₀ - zr is negligible, so V₀ = Vr Similarly, Bernoulli from O₂ to L gives V₀ = Vl (= Vr from above) q.e.d. d) From (c) we have along streamline from O₁ to R V₀²/2g + (z₀ - zr) = Vr²/2g Magnitude of V₀²/2g = 14²/(2*10) = 10m If (z₀ - zr) = lv = 1m Vr² = 14² + 2g => Vr = 14.7 m/s with gravity, 14 m/s without. Difference ~5% I would start to worry if lv > 1 to 2m PROBLEM N°3: [A diagram of a fluid flow under a sluice gate. A reservoir with surface elevation hA and fluid depth h1 and velocity V1 is on the left. The fluid flows under the gate, and downstream the depth is h2 and velocity V2. The gate has an opening d. A streamline S4 to A is shown.] a) Continuity between inflow (1) and outflow (2): Q = h₁V₁ = h₂V₂ => V₁ = (h₂/h₁)V₂ This is a short transition of a converging flow, so ΔH~0 and Bernoulli between sections 1 and 2 yields p₁ + ρgz₁ + 1/2ρV₁² = p₂ + ρgz₂ + 1/2ρV₂² Since flow in sections 1 and 2 is well-behaved (straight parallel streamlines), p+ρgz is constant along the depth, i.e., p₁ + ρgz₁ = ρgh₁ (anywhere along section 1) p₂ + ρgz₂ = ρgh₂ ( " " " 2) Thus, gh₁ + 1/2(h₂/h₁)²V₂² = gh₂ + 1/2V₂² h₂ = 0.61d = 0.61m V₂ = √[2g(h₁-h₂)/(1-(h₂/h₁)²)] = √[2*9.8(3-0.61)/(1-(0.61/3)²)] = 6.990 m/s Q = V₂h₂ = 6.990 * 0.61 = 4.26 m³/s V₁ = Q/h₁ = 4.26/3 = 1.421 m/s b) Now we apply Bernoulli along the streamline from S4 to A: pS4 + ρgh₁ + 1/2ρV₁² = pA + ρghA + 1/2ρVA² pS4=0, pA=0 VA=0 since A is a convex corner hA = h₁ + V₁²/2g = 3 + 1.421²/ (2 * 9.8) = 3.10 m c) To calculate the horizontal force on the gate, we'll apply conservation of linear momentum in x-direction to the control volume between sections 1 and 2: (Bottom friction is neglected since it is a "short" transition) [Diagram of the control volume showing momentum fluxes MP1 and MP2 and a force F on the gate.] MP₁ = (pCG1 + ρV₁²)h₁ = (ρgh₁/2 + ρV₁²)h₁ = (98003/2 + 10001.421²) * 3 = 50158 N/m MP₂ = (pCG2 + ρV₂²)h₂ = (ρgh₂/2 + ρV₂²)h₂ = (98000.61/2 + 10006.990²) * 0.61 = 31628 N/m ΣFx = 0 (steady flow) => MP₁ - F - MP₂ = 0 => F = MP₁ - MP₂ = 18530 N/m (force from fluid on gate acts to the right). d) Assuming hydrostatic pressure: [A diagram of the hydrostatic pressure distribution on the gate, forming a triangle.] F = area of the triangle of pressure = 1/2ρg(hA-d)² = (9800/2) * 2.40² = 21609 N/m e) When flow is moving past the gate, there is velocity near the bottom of the gate. Therefore, the pressure will be reduced near the bottom of the gate with respect to hydrostatic pressure (this is suggested by Bernoulli equation). Note also that point B is in contact with the atmosphere, so pB=0. Thus, the total force on the gate (18530 N/m) is smaller than in the hydrostatic case (21609 N/m). [A diagram comparing hydrostatic pressure distribution with pressure distribution with flowing fluid, showing lower pressure at the bottom for the latter.] PROBLEM N°4: a) [A diagram of a hydraulic jump. A shallow, fast flow (depth h1, velocity V1) transitions to a deep, slow flow (depth h2, velocity V2). A control volume is shown with momentum fluxes MP1 and MP2.] Continuity: V₁h₁ = V₂h₂ => V₂ = V₁(h₁/h₂) (1) Momentum: MP₁ = MP₂ NOTE: Short transition => Friction is negligible (But ΔH≠0 since flow is expanding). (1/2ρgh₁ + ρV₁²)h₁ = (1/2ρgh₂ + ρV₂²)h₂ (2) Substituting (1) into (2): 1/2ρgh₁² + ρV₁²h₁ = 1/2ρgh₂² + ρ(V₁h₁/h₂)²h₂ g/2 h₁² + V₁²h₁ = g/2 h₂² + V₁²h₁²/h₂ g/2 h₂³ - (g/2 h₁² + V₁²h₁)h₂ + V₁²h₁² = 0 h₂³ - h₁²h₂ - 2V₁²/g h₁h₂ + 2V₁²/g h₁² = 0 h₂(h₂² - h₁²) - 2V₁²/g h₁(h₂ - h₁) = 0 h₂(h₂ + h₁)(h₂ - h₁) - 2V₁²/g h₁(h₂ - h₁) = 0 h₂ > h₁ => h₂ - h₁ ≠ 0, therefore h₂(h₂ + h₁)h₂ - 2V₁²/g h₁ = 0 => h₂ = (-h₁ + √(h₁² + 8V₁²h₁/g))/2 h₂ = (-1 + √(1 + 8*10²/9.8))/2 = 4.05 m; V₂ = V₁(h₁/h₂) = 2.47 m/s b) Energy conservation: HEAD LOSS V₁²/2g + p₁/ρg + z₁ = V₂²/2g + p₂/ρg + z₂ + ΔH₁->₂ Well-behaved flow in sections 1 and 2 => p₁/ρg + z₁ = h₁; p₂/ρg + z₂ = h₂ Thus, ΔH₁->₂ = (V₁²-V₂²)/2g + (h₁-h₂) = (10²-2.47²)/(29.8) + (1-4.05) = 1.75 m Rate of energy dissipation per unit width of the channel: Ė₁->₂ = ρgQΔH₁->₂ = 9800 * (101) * 1.75 = 1.72 * 10⁵ J/ms c) To obtain h₁ and V₁ from h₂ and V₂, we need to apply continuity and momentum, as we did in (a). The two equations remain unchanged when we reverse flow direction. Therefore, if h₂ and V₂ have the same values as in (a), h₁ and V₁ will also have the same values, that is: h₁=1m and V₁=10m/s. d) Now energy conservation yields: Compare with (b) V₂²/2g + p₂/ρg + z₂ = V₁²/2g + p₁/ρg + z₁ + ΔH₂->₁ => ΔH₂->₁ = -ΔH₁->₂ = -1.75m < 0!! and Ė₂->₁ = ρgQΔH₂->₁ = -1.72*10⁵ J/ms < 0!! Obviously, energy dissipation cannot be negative, since this would mean creation of energy. Therefore, we conclude that this hypothetical flow from right to left is impossible. This conclusion is true for any values of h₁, V₁ such that h₁ n/(n+1)ũs = U => n/(n+1) * 0.881 = 0.780 => n = 7.72 On next page, I have represented the log profile vs. the power-law approximation, as well as the relative error of the power-law (defined as |ũ(y) - u(y)| / u(y)). As seen in the figure, the agreement is good. The power-law profile slightly underpredicts the log profile (error ~ 1%) in the upper region, while it yields a 25% overprediction of the near-bottom velocity. Other than in the near-bottom (relevant, e.g., for sediment transport calculations), the agreement is very good. e) Km = (7.72+1)² / (7.72(7.72+2)) = 1.013; Ke = (7.72+1)³ / (7.72²(7.72+3)) = 1.038 Both have values close to 1, so the momentum and energy of the flow are well represented by the average velocity values. These results are also consistent with what we saw in recitation 5: δ² ~ 0.013 => Km = 1 + δ², Ke = 1 + 3δ². [Top graph: Log profile vs. power-law. y (m) vs. u (m/s). The log profile (solid line) and power-law, n=7.72 (dashed line) are very close. Bottom graph: Relative error of the power-law. y (m) vs. relative error. The error is large near y=0 and decreases rapidly.] -PROBLEM N°6: a) [A diagram of flow through a sudden contraction in a pipe. The upstream pipe has area ap² and the downstream pipe has area aN². A manometer measures the pressure difference.] At 1-1 and 2-2 we have well-behaved flow and pressure is hydrostatic. With this, from the manometer reading, we can obtain the pressure difference between 1 and 2: p₁,CG - ρgh₀ - ρmghm + ρg(hm+h₀) = p₂,CG => p₁,CG - p₂,CG = (ρm - ρ)ghm = (2600 - 9.8) * 0.061 = 7532 Pa The CG of sections 1 and 2 are on the same streamline. Neglecting wall friction effects (since transition is short), we apply Bernoulli between 1-1 and 2-2 considering the center streamline: p₁,CG + 1/2ρV₁² = p₂,CG + 1/2ρV₂² (1) where V₁ = Q/A₁ and V₂ = Q/A₂ (Km=Ke=1 assumed) and we have applied that Z₁,CG = Z₂,CG. Continuity dictates: Q = V₁ap² = V₂aN² => V₂ = (ap/aN)²V₁ = 4V₁ (2) Plugging (2) into (1): p₁,CG + 1/2ρV₁² = p₂,CG + 8ρV₁² => V₁ = √[2(p₁,CG - p₂,CG) / (15ρ)] = √[27532 / (151000)] = 1.00 m/s V₂ = 4.00 m/s, Q = V₁ap² = 1.0 * (1)² = 0.01 m³/s b) As seen in Lecture 13, the headloss in an abrupt expansion is ΔHexp = (V₂ - V₃)² / 2g By continuity, Q = V₁ap² = V₃ap² => V₃ = V₁. Therefore ΔHexp = (4-1)² / (2 * 9.8) = 0.459 m c) [A diagram showing the flow expanding from section 1 to 3, with a force F on the nozzle.] Conservation of energy between 1-1 and 3-3 dictates p₁,CG/ρg + Z₁,CG + V₁²/2g = p₃,CG/ρg + Z₃,CG + V₃²/2g + ΔHexp (Ke≈1 assumed) Since Z₁,CG = Z₃,CG; V₁ = V₃; ΔHf=0 (wall friction is neglected since transition is short). p₁,CG - p₃,CG = ρgΔHexp = 9800 * 0.459 = 4498 Pa Momentum conservation in the C.V. represented above yields: 0 = MP₁ - MP₃ - F => F = (ρV₁² + p₁,CG)A₁ - (ρV₃² + p₃,CG)A₃ = (Km≈1 assumed) = (p₁,CG - p₃,CG)ap² = 4498 * (0.1)² ≈ 45 N (to the left) The force from the fluid on the nozzle is 45 N to the right.