# Formulae that you may or may not find useful. E v = V. dy dx = v u. y cp y = I xc/a y. Volume of an entire sphere = 4πr3 = πd3

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2 CE30 Test 1 Date: 26 Sept points) The following are multiple-choice questions with possibly multiple correct answers. Circle clearly all responses that you consider correct; if there is any ambiguity in the responses, the responses graded will be decided by the grader. Note that incorrect answers will be penalized but the minimum score for each question is zero. a) The mechanical properties of the fluid determine how the fluid will respond to mechanical forces. i. Waterhammer is associated with the bulk modulus of elasticity. ii. Cavitation occurs when the gage pressure in a fluid becomes negative. iii. An ideal fluid is characterized by zero vapor pressure. iv. A fluid is defined as a material that deforms continuously when a normal force is applied. v. The force due to surface tension is equal to the product of the surface tension coefficient and the area of the gas-liquid interface. vi. Capillary effects, such as the rise in tubes, become more important, i.e., are larger, when the tube diameter or its equivalent) increases. Solution: Waterhammer is associated with the bulk modulus of elasticity E v ), as high pressures develop due to the very fast closure of valves, so that response i) is correct. Cavitation occurs when the fluid pressure approaches the saturated) vapor pressure, not when the gage pressure becomes negative, so response ii) is incorrect. An ideal fluid is characterized by zero viscosity, so response iii) is incorrect. A fluid is defined as a material that deforms continuously when a shear force is applied, so response iv) is incorrect. The force due to surface tension is equal to the product of the surface tension coefficient, σ, and the length of the relevant gas-liquid interface, so response v) is incorrect. Capillary effects become more important at smaller scales, such as when the tube diameter or its equivalent) decreases, e.g., h = 2σ cos θ)/γr) from which it is seen that the capillary rise h increases as r decreases. In summary, only the response i) is correct. 2

3 CE30 Test 1 Date: 26 Sept b) A pressurized closed container contains a liquid of specific weight, γ. Above the liquid is a gas at a pressure, p 0. A cylinder of height L and cross-sectional area, A cyl, of specific weight, γ cyl < γ, floats vertically in the liquid, such that a portion of height h of the cylinder is submerged as shown). i. The magnitude of the force on the bottom of the cylinder due to hydrostatic pressure is γh. ii. The magnitude of the force on the bottom of the cylinder due to hydrostatic pressure is γha cyl. iii. The magnitude of the force on the bottom of the cylinder due to hydrostatic pressure is p 0 + γh)a cyl. iv. The magnitude of the force on the bottom of the cylinder due to hydrostatic pressure is p 0 + γl)a cyl. v. The fraction of cylinder submerged, h/l, is equal to γ cyl /γ. vi. The fraction of cylinder submerged, h/l, is equal to γ cyl /γ) p 0 /γ)/l). Solution: The bottom of the cylinder is a horizontal plane surface, and so the pressure, p bot, is constant over this surface. The pressure can be found by applying the hydrostatic equation between the liquid surface and the cylinder bottom, namely, p bot p 0 = γz bot z 0 ) = p bot = p 0 + γh. The magnitude of the force on the cylinder bottom is therefore F bot = p bot A cyl = p 0 + γh)a cyl and so responses i), ii), and iv) are incorrect, while response iii) is correct. The vertical force balance on the cylinder will determine the fraction of cylinder submerged, and can be expressed as + F y = 0 = F bot W cyl F top = p 0 + γh)a cyl γ cyl A cyl L) p 0 A cyl = h/l. = γ cyl /γ where the force, F top, on the top of cylinder is due to the constant) gas pressure, p 0. Thus, the effect of the gas on the submergence cancels out. Hence, response v) is correct, while response vi) is incorrect. In summary, only responses iii) and v) are correct. 3

4 c) In a region of flow, the velocity is everywhere the same from t = 0 to t = T, whereas from t = T to t = 2T, the velocity is again everywhere the same, but differs in direction from the earlier velocity. A particle released at point A at t = 0 travels to point B at t = T and to point C at t = 2T along the path shown. CE30 Test 1 Date: 26 Sept i. Over the period from t = 0 to t = T, the flow is unsteady. ii. Over the period from t = 0 to t = 2T, the flow is uniform. iii. If a streakline is started at t = 0 by releasing particles at A, the resulting streakline at t = 2T would be as shown in Figure a. iv. If a streakline is started at t = 0 by releasing particles at A, the resulting streakline at t = 2T would be as shown in Figure b. v. The streamline passing through A immediately before t = 2T would be as shown in Figure c. vi. The streamline passing through A immediately after t = 0 would be as shown in Figure c. Solution: Over the period from t = 0 to t = T, the flow is steady as the velocity remains the same everywhere, so response i) is incorrect. On the other hand, in each interval from t = 0 to t = T, and from t = T to t = 2T ), the velocity is everywhere the same, and so in each time interval, and so for both intervals, the flow is uniform. Response ii) is therefore correct. The streakline started at t = 0 and observed at t = 2T corresponding to the given pathline is shown and hence the responses iii) and iv) are incorrect. Similarly the streamline passing through A immediately after t = 0 or t = 0 + ) and immediately before t = 2T or t = 2T ) are also shown. Thus, response v) is correct, while response vi) is incorrect. In summary, only responses ii) and v) are correct.

5 2. 20 points) SAE 30 oil at 20 C flows between two stationary parallel plates, each 500 mm in length, and 00 mm in width into the page), and separated by a distance of 2H = 200 mm as shown. The fluid velocity varies as uy) = u max 1 y2 H 2 CE30 Test 1 Date: 26 Sept where the maximum velocity, u max = 15 cm/s, y being measured positive upwards from the midpoint between the two plates. ), a) Determine the fluid shear stress at the bottom plate, and at the midpoint where the maximum velocity occurs. b) Determine the net force magnitude and direction) that the flowing oil exerts on both plates over the streamwise length of 500 mm. Solution: i) For the given simple flow, the shear stress is related to the velocity by the relation, τ = µdu/dy), where µ = 0. N s/m is the dynamic or absolute) viscosity of SAE30 oil read from the table of fluid properties. Thus, we need only to take the derivative of the given velocity profile to obtain the solution, namely, τy) = µ du dy = µ d dy u max { 1 y H ) 2 } = µu max 2y H 2 ) = 2µu maxy H 2. We can then evaluate the value of the shear stress for any given y. At the bottom plate, y = H = 0.1 m, so that τy = 0.1 m) = 2µu max H) H 2 = 2µu max H Similarly, at the midpoint, y = 0, so that τy = 0) = 0. = 20. N s/m2 )0.15 m/s) 0.1 m = 1.32 Pa. ii) The magnitude of the force on the fluid is obtained as τy = ±H) A plate. If the shear stress is evaluated at the top plate, it is in fact found to be negative, i.e., τy = H) = τy = H), but the force on the fluid at both the top and the bottom plate is in the same direction, namely resisting the flow, and so in upstream direction. The direction of the force of the fluid flow on the plate is opposite, namely in the flow or downstream direction. Thus, F plates = 2τy = H)L plate W plate = N/m 2 )0.5 m 0. m) = 0.53 N, where the area on which the shear stress is being exerted is A plate = L plate W plate, with the length and width of each plate being L plate 0.5 m and W plate = 0. m. 5

6 CE30Test 1 Date: 26 Sept points) A compound manometer, consisting of two different manometer fluids I and II), connects a water reservoir and a pipe at A) as shown. Manometer fluid I has a specific gravity of 1.6 while manometer fluid II has a specific gravity of 1.3. If the observed manometer deflections are as shown, determine the difference in piezometric heads between the pipe at A and the tank, indicating clearly which has the higher piezometric head. Solution: In a fluid in static equilibrium, such as in the tank, the piezometric head, defined by p/γ) + z, is the same everywhere, and as such any arbritrary point in the tank can be chosen here point E is chosen). On the other hand, the pressure, p, or pressure head, p/γ), is not the same everywhere in a fluid in static equilibrium. We want p/ ) + z A p/ ) + z E. Despite the fact that the elevations of neither point A nor point E are specified, these elevations are not needed for the solution. The solution starts from repeatedly writing the hydrostatic equation, starting from either A or E and working towards the other point. Here we start from A and work sequentially towards E: p A p B = z A z B ) p B p C = γ I z B z C ) p C p D = γ II z C z D ) p D p E = z D z E ) adding = p A p E = z A z B ) γ I z B z C ) γ II z C z D ) z D z E ) where, γ I, and γ II are the specific weights of the water, and manometer fliuds I and II, respectively. We now divide through by and transfer terms, z A and z E, from the right-hand side to the left-hand side to express the left-hand side in terms of piezometric heads: pa + z A ) pe + z E ) = z B z D γ I z B z C ) γ II z C z D ) = z B z C ) + z C z D ) γ I z B z C ) γ II z C z D ) γ w = z B z C ) 1 γ ) I + z C z D ) 1 γ ) II = 60 mm) ) mm) ) = 81.6 mm where we use z B z D = z B z C ) + z C z D ). Note neither z A nor z E appears on the right-hand size, and so it is not necessary to know these elevations. As the result is positive, we conclude that the piezometric head at A is higher than the piezometric head at E in the tank). 6

7 CE30 Test 1 Date: 26 Sept points) A solid cylinder of radius, 0.8 ft, has a specific weight of 300 lb/ft 3 and acts as a plug for an opening in the corner of the water tank. Determine the reactions magnitude and direction) per unit width into the page) at A and B that the tank exerts on the cylinder, assuming that the tank wall surfaces are smooth so that they support only normal reactions. Solution: To find the reactions at A and B, it is sufficient to consider the horizontal and vertical force balances moment balances could also be applied but is not necessary the solutions to the force balances are consistent with the momentum balances). Based on the free-body diagram shown, we have + F x = 0 = F h R B = R B = F h, + F y = 0 = W cyl F v + R A = R A = W cyl + F v. where we have assumed that the net horizontal force component of the hydrostatic force is directed to the right, and the net vertical force component is directed downward. We also use that the resultant hydrostatic force must pass through the center of the cylinder because all hydrostatic force elements act normal to the cylinder surface, and so act through the center). Thus, if the horizontal and vertical components, F h and F v, of the hydrostatic pressure force are obtained, then the desired reaction forces can be found. Horizontal component: The magnitude of the horizontal component of the hydrostatic pressure forces can be found by identifying the relevant vertical plane projected surfaces), and determining the magnitude of the force acting on these projected surfaces. Hydrostatic forces in the horizontal direction act on both the left and right sides of the cylinder in opposite directions. We could consider two separate vertical plane projected surfaces, one for the left and one for the right side of the cylinder see figure), but it is quicker to realize that the hydrostatic pressure force on the right-hand side cancels out exactly the hydrostatic pressure force on the upper half of the left-hand side recall lecture example and homework problem). Thus, it is sufficient to consider a single projected surface, namely for the lower half of the cylinder. Further, the vertical plane projected surface is a rectangle, so that, per unit width, F h = pa) proj = h proj A proj H R/2)R) = 62.3 lb/ft 3 ft 0.8 ft/2)0.8 ft) = 179. lb/ft Vertical component: The magnitude of the vertical component of the hydrostatic pressure forces is found by identifying the relevant volumes), the combined weight of which is equal to the vertical component magnitude. We distinguish between the vertical components of pressure force elements that act in the downward or in the upward direction, as the signs are different, and must be considered. Thus, we divide the cylinder surface into an upper half 7

8 where the vertical component of the pressure force elements is directed downwards, and the lower half where the vertical component is directed upwards. On the upper half, the vertical component, denoted as F v, is the weight of the volume of fluid above the curved surface to the zero-pressure level and is directed downwards, while on the lower half the vertical component, denoted by F + v, is again the weight of the volume of fluid above the curved surface to the zero-pressure level and is directed upwards: F v = H R)2R) πr2 2, F + v = H R)R + πr2. The magnitude of the net vertical component is the algebraic sum of the these two contributions: F v = F + v + F v = H R)R + πr2 + H R)2R) πr2 2 = H R)R 3πR2 = 62.3 lb/ft 3 ft 0.8 ft)0.8 ft) 3π0.8 ft)2 It is also possible to combine the volumes so that the relevant volumes become as shown in the figure to the side, but there is no substantial simplification, and no large advantage to combining volumes. Reactions: With both horizontal and vertical components of the hydrostatic pressure force found, we find the reactions from the above force balances as R B = F h = 179. lb/ft R A = W cyl + F v ) = lb/ft lb/ft) = lb/ft, = 65.5 lb/ft where W cyl = γ cyl πr 2 ) = 300 lb/ft 3 π)0.8 ft) 2 = lb/ft. Some students also tried to use the line of action of the horizontal component in a moment balance. This is correct but this requires that the separate line of action of the vertical component also must be used in the moment balance. The determination of the line of action of the vertical components), basically the centroids of the relevant volumes, requires substantial effort however, and so the above procedure is much to be preferred. 8

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