Massachusetts Institute of Technology Department of Civil and Environmental Engineering 1.060 Fluid Mechanics. In-class Examination March 9, 2001 Problem 1 (35%) Air 1=6.0m S 1,000 h-3m h=1.5m Air G 0-37 h=3m b=10m (Into paper) The sketch above shows a plane gate that backs up water (p = 1000 kg/m³) in a 10-m- wide rectangular channel to a depth of h 3.0 m. The gate itself is inclined at an angle of 037° to horizontal and is supported by a shaft, S, spanning the width of the channel. The shaft is located h 3.0 m above the bottom at a distance I= 6.0 m "downstream" of the gate's intersection with the still water level in the channel. The total weight of the gate is W, and is assurned to act through the mid-point of the submerged portion of the gate (point G in sketch). a) Determine the total horizontal pressure force acting on the gate, Fr, and its line of action. b) Determine the total vertical pressure force acting on the gate, Fv, and its line of action. c) Determine the necessary weight of the gate, W, for the gate to remain closed. d) Corresponding to the weight of the gate obtained in (c) determine the total force acting on the shaft, S. The sketch above shows a precision manometer used to measure small pressure differences in gaseous systems. It consists of two circular cylindrical chambers, of diameters de, partially filled with a chamber fluid, of density pe. The bottoms of the two chambers are connected by manometer tubes, of diameter dm, containing a manometer fluid, of density P. When the two chambers are open to the atmosphere, ie. when p₁ = P2. the manometer is in equilibrium as shown in the sketch. When one chamber is connected to a pressure p, and the other to a pressure p₂ (in a gas) the manometer fluid in the right leg (connected to the chamber with pressure p2) is "deflected" a vertical distance, hm, from its equilibrium position. a) Derive the general expression for the pressure difference pi-pa corresponding to a manometer reading of hm b) Evaluate the expression obtained in (a) for de = 10 cm, dm = 1 cm, pc = 970 kg/m², Pm = 1000 kg/m³, and hm = 5.0 cm. The sketch above shows a horizontal pipe of diameter d,= 10 cm, carrying a discharge Q = 0.02 m³/s of water, p= 1000 kg/m². The discharge is supplied from a very large container in which the water level is a vertical distance he above the pipe elevation. The pipe discharges into the atmosphere through a nozzle with a circular outflow of diameter d = 5 cm. a) Determine the average velocity in the 10-cm-diameter pipe, V. b) Determine the elevation of the free surface in the container, he, corresponding to the given problem specifications. In-class Examination No: 1, Spring 2001 Problem No: 1 Given: Air Water Gate with dimensions: Length (l) = 6.0 m Height (h) = 3 m Width (b) = 10 m (into the paper) Angle (θ) = 37° Calculations: a) Total horizontal pressure force on gate: Horizontal force on gate's projection on vertical plane = ρghb F_H = 1000 kg/m³ * 9.81 m/s² * 3 m * 10 m = 294,300 N = 294.3 kN Acting towards the right (on the gate) Q_F = h/3 = 1.0 m above bottom b) Total vertical pressure force on gate: Weight of fluid above gate's translation into channel = ρgh(hcotθ)b F_V = 1000 kg/m³ * 9.81 m/s² * (1.33h) * 6 m * 10 m = 588,000 N = 588 kN Acting upwards (on the gate) Q_V = (2/3)hcotθ = 2.67 m to the right of the gate's intersection with still water level.