stst10-2 - Date: 27 Oct. 2010 CE340 Test 2 Last Name: First...

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Unformatted text preview: Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: COVER PAGE Write your name on each sheet of paper that you hand in. Read all questions very carefully. If the problem statement is not clear, you should ask the proctors present to clarify if possible the question. Pay particular attention to consistent units. Please attempt to do all work in an organized manner, so that the papers can be graded generously. Draw and label all diagrams neatly if they are to be considered in the grading. Circle all main results including the final result, which you wish to be considered in the grading. If you write on the back of a sheet, or on additional paper, please note on your first sheet that you have done work on additional sheets and where the additional work is to be found. ALL RELEVANT WORK SHOULD BE SHOWN. MARKS MAY BE DEDUCTED IF THE GRADER IS NOT ABLE TO UNDERSTAND HOW AN ANSWER WAS OBTAINED. Unless otherwise specified, the fluid is water, and room temperature and standard atmospheric pressure apply. There are four (4) problems in total. It is highly recommended that all problems be attempted, so budget your time accordingly. Formulae that you may or may not find useful 4σ cos θ dp h= Ev = − = ρc2 γD d V /V I xc ycp − y = yA ￿ ￿ ∂p al − +z = ∂l γ g ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ p a0 l p a0 l p ω 2 r2 p ω 2 r2 +z+ = +z+ , +z− = +z− γ gA γ gB γ 2g A γ 2g B ￿ ￿ 2 ∂ Vs ∂ Vs V ∂u ∂u ∂u ∂u a= + Vs es + s er , a= +u +v +w ∂t ∂s r ∂t ∂x ∂y ∂z dy v = dx u for orifices and Venturi meters, for rectangular weirs, for triangular weirs, ￿ 2gh† 1 − (d/D)4 ￿ ￿ 2￿ 3, Q = Cwr Lw 2gH Cwr = 0.611 + 0.075(H/P ) 3 ￿ ￿ ￿￿ ￿ 8 θ 5 Q = Cwt tan 2gH 15 2 V2 hL = K 2g Q = Cd A 1 Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: 1. (20 points) For each of the following multiple choice questions, choose one and only one response, which is the most correct answer. In addition to completing the bubble form, it is recommended that you also indicate clearly on the test paper the answer that you have selected. (a) Three cases of water flow (from left to right in all cases) in an open channel are shown in profile view, with gravity acting vertically downwards as shown. The bottom boundary and the free surface can be taken as streamlines. In cases I and II, these are curved, while in case III, the streamlines are straight and parallel everywhere. The depths in all three cases are equal (namely, h), at the vertical section containing the point A shown. ❧ 1 In all three cases, the pressure at the bottom boundary (at point A) is γ h, where γ is the specific weight of water. ❧The pressure at A in case I is larger than the pressure at A in case II. 2 ❧ 3 Of all cases, the pressure at A in case III is least. ⑤ 4 The pressure at A in case II is larger than the pressure at A in case III. ❧ 5 The pressure at A in case I is larger than the pressure in case III. ❧ 6 The piezometric head is constant throughout the vertical section containing point A in all three cases. ❧1 and 6 7 ❧ 8 2 and 5 (b) For a given steady flow over a sharp-crested weir, ❧ 1 the magnitude of the flow velocity at a point over the weir crest section increases in the vertical direction. ❧ 2 the pressure distribution at the weir crest section is assumed to be hydrostatic in the derivation of the standard weir formulae. ❧ 3 the nappe refers to the flow upstream of the weir section. ❧ 4 for accurate evaluation of the weir discharge using the standard weir formulae, it is required that the water surface elevation at the weir crest section be measured accurately. ❧the discharge coefficient of the weir accounts only for the frictional effects neglected 5 in the theoretical analysis. ⑤for accurate evaluation of the weir discharge using the standard weir formulae, the 6 nappe should be well aerated, i.e., surrounded on all sides by the atmosphere. ❧2 and 4 7 ❧ 8 3 and 6 2 CE340 Test 2 Date: 27 Oct. 2010 Last Name: First Name: (c) The analysis of practical flows is generally based on a control volume, to which the concepts of mass, momentum, or energy conservation are applied. For steady flows, ❧ 1 the flows of mass, momentum, and energy into a control volume must be equal to the corresponding flows out of the control volume. ❧the control volume always contains the same physical material. 2 ❧ 3 for ideal fluids, the flow of momentum into a control volume may differ from the flow of momentum out of the control volume only due to the action of pressure forces. ⑤the energy equation, applied in its usual form, may be interpreted in terms of the 4 difference in the flows of a quantity, (p/γ ) + z + αV 2 /2g , out of and into a control volume and relating this difference to the rate of frictional loss and energy input or loss to shaft devices. ❧1 and 2 5 ❧ 6 2 and 3 ❧ 7 3 and 4 ❧ 8 1, 3, and 4 (d) Four pipe systems with real fluids are shown in the figure, together with two piezometers in each system. Systems I and II are constant-diameter systems, but in systems III and IV, the cross-sectional area at A is smaller than the cross-sectional area at B. Streamlines may be assumed straight and parallel where the piezometers are installed. In systems I and III, the levels in the two piezometers at A and B are the same, while in systems II and IV, the level in the piezometer at B is higher than the level in the piezometer at A. Velocites may be assumed to be constant over all cross-sections (i.e., α = 1). ❧ 1 In system III, there cannot be any flow. ❧ 2 In system I, there cannot be any flow. ❧ 3 In system III, if there is flow, it must be from B to A. ❧ 4 In system II, the flow must be from B to A. ❧ 5 In system IV, the flow may be from A to B or from B to A depending on the magnitude of the difference in piezometric heads, and on the change in cross-sectional areas, between A and B. ❧In system III, the flow may be from A to B or from B to A depending on the 6 magnitude of the difference in piezometric heads, and on the change in cross-sectional areas, between A and B. ❧2 and 3 7 ❧ 8 2 and 4 ⑤ 9 2, 4 and 5 3 Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: 2. (20 points) A sharp-edged orifice of diameter 1.4 cm is installed in a pipe of diameter 2 cm, inclined with respect to the horizontal, as shown in the figure. A differential manometer, with a manometer fluid of specific gravity of 0.75, is applied across the orifice, and its deflection is observed to be 0.2 m for a given water flow. Determine the water discharge in the pipe. Solution: This is a straightforward application of the orifice equation for a pipe, namely ￿ 2gh† Q = Cd Ao 1 − (d/D)4 where Ao = π d2 /4 is the area of the orifice (d is the orifice diameter), and not the area of the pipe, and h† = ∆[(p/γ ) + z ], the change in piezometric head across the orifice. The usual analysis of the manometer (see end of solution) leads to h† = [1 − (γm /γ )]∆h = (1 − 0.75)(0.2 m) = 0.05 m where γm /γ = sm = 0.75, the specific gravity of the manometer fluid, and ∆h = 0.2 m is the manometer deflection. The discharge coefficient, Cd , is correlated (see inset in correlation figure) with the Reynolds number, Re = V D/ν , where V = Q/Apipe is the average velocity in the pipe, D is the pipe diameter, and ν is the fluid kinematic viscosity. Because Q is unknown, we cannot evaluate directly Re, we guess an initial value of Cd based on the known value of d/D =1.4 cm/2 cm = 0.7. From the given graphical correlation, Cd = 0.61 for large Re (where the curve flattens out, which is usually a good initial guess), so we now evaluate Q with Cd = 0.61, ￿ 2(9.8 m/s2 )(0.05 m) Q = 0.61(π (0.014 m2 /4) = 9.2 × 10−5 m3 /s 1 − 0.54 We now check with our Cd correlation. We evaluate Re = (Q/Apipe )D/ν = 4Q/(νπ D) = 4(9.2 × 10−5 m3 /s/10−6 m2 /s = 5857 From our graphical correlation, this value of Re is unfortunately not within the range, but we choose closest available value, namely Cd ≈= 0.645, so we make another iteration, with Cd = 0.645 as the new guess. We find Q = 9.8 × 10−5 m3 /s, Re = 6240, which is closer to our desired value. Manometer analysis: pA − p1 = −γ (zA − z1 ) p1 − p2 = −γm (z1 − z2 ) p2 − pB = −γ (z2 − zB ) =⇒ ￿ p +z γ ￿ A − ￿ p +z γ ￿ =h = † B ￿ γm 1− γ ￿ ∆h Note that if we performed the manometer analysis from B to A, i.e., assuming a downwards flow, we would have obtained a negative value, which would indicate that the assumed downwards flow direction was not correct. 4 Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: 3. (25 points) The pipe fixture shown in the figure connects two inflow pipes, and combines them into a single outflow pipe. All pipes of different diameters are in a horizontal plane. The fixture is isolated mechanically by flexible joints such that the net forces on the pipe fixture due to the flow are sustained entirely by the pipe support shown. Pressure gages at the inlets (A and B) and the outlet (C) of the fixture record pressures of 725 lbf/ft2 , 721 lbf/ft2 , and 720 lbf/ft2 . A discharge of 0.06 cfs flows in the inflow (2-in diameter) pipe, while a discharge of 0.09 cfs flows in the other (3-in diameter) inflow pipe. The flow in the outlet (4-in diameter) pipe is in a direction directly opposite to that in the 2-in diameter inflow pipe, while the other (3-in diameter) inflow pipe lies at 50◦ angle to the other two pipes as shown in the figure. Determine the magnitude and direction of the two components of the net force acting on the pipe supports due to the flow. Sketch and label carefully all control volumes as well as any coordinate system used in your analysis. Solution: We draw our control volume and coordinate system as shown. Note that pressure forces are always normal and compressive. We assume a net external force on the flow due to the pipe (or pipe support) in the negative x- and positive y - directions, and write the x- and y - component equations, with forces on the left hand side, and momentum flows (analagous to ma in Newton’s second law) on the right hand side, as : −Fx + (pA AA )x + (pB AB )x + (pC AC )x = (ρQV )Cx − [(ρQV )Ax + ρQV )Bx ] Fy + (pA AA )y + (pB AB )y + (pC AC )y = (ρQV )Cy − [(ρQV )Ay + ρQV )By ] We now simplify where possible (e.g, (pB AB )x = −pB AB sin θ, (ρQV )Bx = ρQB VB (− sin θ), (pA AA )y = 0, (pC AC )y = 0, (ρQV )Cy = 0, (ρQV )Ay = 0, (ρQV )By = ρQB VB (− cos θ), choosing carefully the signs and the appropriate components (θ = 50◦ ) according to our coordinate system, Fx = pA AA − pB AB sin θ + pC AC − ρQC (−VC )+ [ρQA VA + ρQB VB (− sin θ)] Fy = −pB AB (− cos θ) − ρQB VB (− cos θ) Because QC = QA + QB = 0.15 cfs, and in each pipe V = Q/A, the terms on the right hand side of the equations can be evaluated, and hence Fx = [725 lbf /ft2 × π (1/6 ft)2 /4] − [721 lbf /ft2 × π (1/4 ft)2 /4] sin(50◦ ) + [720 lbf /ft2 × π (1/3 ft)2 /4] + 1.94 slugs/ft3 × [(0.06 ft3 /s)2 /{π (1/6 ft)2 /4}] − 1.94 slugs/ft3 × [(0.09 ft3 /s)2 /{π (1/4 ft)2 /4}] sin(50◦ ) = 52.1 lbf Fy = (721 lbf /ft2 )[π (1/4 ft)2 /4] cos(50◦ ) + 1.94 slugs/ft3 × [(0.09 ft3 /s)2 /{π (1/4 ft)2 /4}] cos(50◦ ) = 23.0 lbf 5 Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: As the results for Fx and Fy are positive, the assumed directions are correct, namely that the force components on the flow due to the pipe (or pipe support) are in the negative x- and positive y -directions. The force components on the pipe support are equal in magnitude but opposite in direction to that shown in the control volume diagram (the force in the x-direction is to the right, and that in the y -direction is downwards). 6 Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: 4. (25 points) A 5-cm diameter steady round water jet discharges into the atmosphere, impinging on a stationary 15-cm diameter circular disk, and so is deflected to flow radially (see figure). The distance between the jet exit and the circular disk is 20 cm, and the thickness of the flow at the edge of the disk is observed to be 0.5 cm. The total volume of water from the jet exit to the edges of the disk is estimated to be 600 cm3 . It may be assumed that the velocity magnitude at the jet exit is constant over the jet exit section and similarly the velocity magnitude at the disk edge is constant vertically, though the velocity magnitude at the jet exit is not necessarily the same as the velocity magnitude at the disk edge. The fluid can also be assumed to be ideal (i.e., frictionless). Sketch and label carefully any control volume, as well as any coordinate system or streamline used in your analysis. (a) Determine the jet discharge. (b) If the circular disk is maintained in place entirely due to its weight (i.e., no other external force is applied), what must be the weight of the circular disk? Solution: a) This is a combination of various types of problems, and involves applying the Bernoulli equation and the continuity (i.e., mass conservation) equation. We apply the Bernoulli equation along the streamline between the points A and B shown, ￿ ￿ ￿ ￿ p V2 p V2 +z+ = +z+ γ 2g A γ 2g B At both points, the pressure is zero because the jet is in the atmosphere, and so rearranging, we find that ￿ 2￿￿ ￿ ￿2 ￿ 2 2 VA − V B AB − A2 Q 1 1 Q2 A = zB − zA = ∆ z =⇒ −2= = ∆z 2g 2g A2 AB 2gA2 A2 A A B If a different streamline had been taken, e.g., say, along the edge of the jet from jet exit to disk edge, from A￿ to B￿ , a slight difference in resulting exit velocity at the disk edge would have arisen due to the slight difference in elevation (between B and B￿ ), but we are assuming that the velocity is constant over the control surface at the disk edge. After some algebra, we can solve for Q in terms of the other variables as ￿ 2g ∆z Q = AA 1 − (AA /AB )2 which is very similar to the orifice equation (which should not be too surprising). It is clear that AA = π d2 /4, where d is jet diameter, but AB = π Dh, where D is the disk diameter, and 7 Date: 27 Oct. 2010 CE340 Test 2 Last Name: First Name: h is thickness of the radial jet at the disk edge (recall that the flow is radially outwards at the disk edge, and so the appropriate area is the area of the side of cylinder of diameter D and height h). So eventually with AA /AB = (π d2 /4)/(π Dh) = d2 /4Dh, we can evaluate the discharge as ￿ ￿￿ π (0.05 m)2 2g (0.2 m) Q= = 7.04 × 10−3 m3 /s 2 /(4 × 0.15 m × 0.005 m)}2 4 1 − {(0.05 m) b) To determine the weight, Wdisk , of the circular disk, we simply need to consider the z -momentum or force balance (because of radial symmetry, no other force components is relevant). With the control volume and the coordinate system shown, we write, forces (with the proper signs) on the left hand side, and momentum flows on the right hand side, −Wdisk − Wfluid = [(ρQV )out ]z − [(ρQV )in ]z Because the jet is in the atmosphere, pressures are all zero, and so pressure forces do not arise, but the weight of the fluid, Wfluid , within the control volume should be taken into consideration (especially since the volume of fluid, Vfluid = 600 cm3 , in the control volume is given). Because the z -component of the velocity at the outlet is zero, i.e., (Vout )z = 0, the balance is very simple, namely Wdisk = ρQVin − Wfluid = 103 kg/m3 × [(7.04 × 10−3 m3 /s)2 /{π (0.05 m)2 /4}] − (9.8 × 103 kN/m3 )(600 × 10−6 m3 ) = 19.5 N 8 ...
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