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CHEME 3230, Spring 2013 Problem Set 2 Problem 1. 1a) If the coecient of interfacial tension between air and water is = 74 dynes/cm, what diameter of

I need concepts to solve problem 1. Its a submerged gate hinge problem. I don't know how to do the momentum balance in a static fluid

CHEME 3230, Spring 2013 Problem Set 2 Problem 1 . 1a) If the coefficient of interfacial tension between air and water is γ = 74 dynes/cm , what diam- eter of the capillary is needed to raise water to a height 100 m ? Does this seem reasonable for, say, redwood trees? If not, what other mechanism might be at work? 1b) Two fluids are held back by a hinged gate as illustrated in figure 1. The lower fluid is of depth h 1 and density ρ 1 and the upper fluid is of depth h 2 and density ρ 2 , with ρ 2 ρ 1 . Determine the moment per unit width about the base of the hinge. Recall that the moment (torque), M , is defined as the cross product of the force f and the moment arm—the position vector measured from the point of interest, r . Hint: compute the moment about the hinge—that is, perform an angular momentum balance: ∂t Z V ρ r × u dV = Z V r × f g dV + Z S r × t dV - M hinge , (1) where dS = dw dz ( w is the width of the gate into the page and z is the height of the gate / depth of the fluid). Note that the left-hand side is zero for a static fluid, and the first term on the right-hand size is zero because the gravitational force passes through the gate. The traction vector t is given by the Cauchy relation discussed in class; here, the stress is the isotropic hydrostatic pressure acting normally on the gate. Recall that the cross product between two vectors a and b is given by a × b = | a || b | sin θ . Figure 1: Figure for problem 1(b). 1
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1c) The gate shown in the figure 2 is rectangular and has dimensions of 9 m by 9 m . Neglecting the weight of the gate, determine the force acting on the hinge of the gate. Figure 2: Figure for problem 1(c). Problem 2 . 2a) The gate shown in figure 3 is rectangular and has dimensions of 6 m by 4 m . What is the reaction at point A? Neglect the weight of the gate. Figure 3: Figure for problem 2(a). 2
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Dear Oscar, Now... View the full answer

Solution for 1(a) -Oskar.pdf

The formulae is,
γ (2πr) = ρh(πr2)g
74 (2πr) = 990 * 100*102*(πr2)*9.81
R = 7.33 * 10-5 cm The result we get is not reasonable as trees shows negative pressure which means suction

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