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ME3332 HOMEWORK 2 1. The diagram shows water flowing in a round pipe. In the portion of the pipe that is situated just downstream of the inlet, the flow tends to undergo some complex processes which die away with increasing distance from the inlet. When these processes have disappeared, a special type of flow regime is encountered. It is the regime of fully developed flow. Fully developed flow can be recognized by means of two characteristics: (a) the pressure decreases linearly so that dp/dx is a constant and (b) the velocity profiles at all locations are identical. These features are shown in the diagram. For fully developed flow, it is a standard design approach to predict the pressure drop by making use of a dimensionless pressure gradient called the friction factor, which is defined as: The friction factor depends on another dimensionless quantity called the Reynolds number. By any standard, the Reynolds number is the most commonly used parameter in all of fluid mechanics and heat transfer. It is defined as: In these equations, , , and (= ) are fluid properties, is the mean velocity, is the diameter of the pipe, and is the axial pressure gradient. When flow in the pipe is laminar, the equation that connects the friction factor and the Reynolds number is . Problem statement: Air flows in a round pipe at a mean velocity of . The temperature of the air is , and the pipe diameter is . (a) Find the Reynolds number of the flow. (air properties are in Table 2.1) (b) What is the magnitude of the friction factor? (c) What is the pressure drop for a length of pipe in the fully developed region? Solution: (a) From Table 2.1, and , so (b) For laminar flows, (c) From the definition of friction factor, Since is constant in the fully developed region, 2. A uniquely shaped U-tube manometer is used to measure the pressure drop between locations 1 and 2 for air flowing in a round pipe. The inner diameter of leg 1 is while that of leg 2 is . To test the accuracy of this manometer, first we have measured the pressure drop using a “normal” U-tube which has inner diameters of both legs equal to , and we have found that the pressure drop is . (a) What was the difference between the water levels in two legs for the “normal” U-tube? (the experiment was conducted at ) (b) What is the expected height difference between two legs for our newly designed U-tube. Solution: Before answering the problem, let’s look at two important force balances. Force Balance 1: Consider a cylinder of fluid of small diameter whose axis is horizontal,as shown in the diagram to the right. As the cross section of the cylinder is very small (area ) we can assume the pressure to be uniform on the two end faces, namely and , as shown in figure. As the fluid is stationary, this cylindrical mass of fluid is also stationary, so the forces acting on it should cancel out. If we focus only on the horizontal forces, the following force balance can be written. Or This result shows that the pressures at the outboard ends of an stationary cylinder of fluid with its axis horizontal must be the same. Force Balance 2: Consider the fluid mass shown in this figure, a vertical cylindrical mass of fluid. The cross-sectional area at the top and bottom faces is , and the pressures on those faces are uniform (because those faces are horizontal) and are equal to and respectfully, as shown in figure. As the fluid is stationary, this cylindrical chunk of fluid is also stationary, so the forces acting on it should cancel out. If we focus only on the vertical forces, the following force balance can be written. where is the weight of fluid inside the cylinder. So, where is the volume of the cylinder which is equal to . Rewriting the equation yields, Notice that the cross-sectional area is cancelled out from the final equation, and the pressure difference between two horizontal planes only depends on their vertical distance from each other. Now we can go back to answer the question, but let’s solve the problem for a very general case where the two legs can have any arbitrary cross section (one example is shown in this figure). Using the two force-balances mentioned above, we can write the following pressure equations. from 2nd force balance from 2nd force balance from 1st force balance Or by substituting the first two equations in the third one: Since this result is true for any manometer, we can be sure that for a given pressure drop the height difference in the two legs will be the same regardless of the legs’ cross sectional areas. So, (a) and (b) From Table 2.1, at , , therefore, As we said this result is correct for both the “normal” and “special” manometers. 3. A fluid flows from left to right through the annular space defined by . Imagine that the velocity is measured at 10 locations across the width of the annulus. One of these locations is at , and another is at . At both of these locations, the velocity is zero. At the other eight locations, numerical values of the velocity are available. The radial spacing between the points of velocity measurement are equal all across the width of the annulus. Write a formula for the mean velocity in terms of the known velocities at the ten locations and in terms of the geometry used to set up the calculation. Solution: I will try to solve this problem using four different methods, and then to see how they compare in a case for which we know the exact mean velocity. Before that, let’s see what the mean velocity is and devise a strategy to solve our problem. By definition, the mean velocity is a velocity which, when multiplied by the density and the cross-sectional area gives the same mass flow rate as the real non-uniform velocity profile. In mathematical terms, the mean velocity is the area-weighted average (if density is constant) of the velocity profile, or: so, As we do not have the algebraic equation for the velocity profile and only know the velocity magnitude at certain given locations, we cannot integrate the equation given above. Therefore, we have to approximate the integral in the numerator by a summation over some finite areas. (Integrating is summing over infinite number of infinitesimal areas, and unlike finite areas, the velocity is constant at infinitesimal areas because of their being very small. Thus, after substitution of infinitesimal areas by finite areas and replacing integration with summation, some error is introduced into our calculation.) where is the number of small finite areas. Within each such element, the velocity is assumed to be uniform. The following four methods use the above-mentioned summation to calculate the mean velocity, and they only differ in the way we choose the finite areas ( ’s) for which we are going to apply the summation and/or the magnitude of the mean velocity in those areas. Method I: In this method we divide the cross sectional area as it has been shown in figure. This is very similar to what we saw in class where we applied this method to flow inside a circular pipe. The measurement points in this method lie halfway between the two lines that define each . Exceptions to this designation of the velocity must be made for the first and last points at which velocity is measured. The first point lies on the inner surface of its element, and the last point lies on the outer boundary of its element. For each of the rings: For the first ring: For the second to the ninth rings: And finally for the tenth ring: Also, As , so Method II: This method, and the next two methods, uses another choice of the finite areas. The areas of the elements used in these three methods are defined by circles passing through the measurement points. The only difference between the methods is the choice of constant velocity over these areas. The first method among these three methods uses the velocity at the inner boundary of every finite area as the constant velocity in that area. For this method and the next two methods, So, Method III: In this method, we use the velocity at the outer edge of each element, so that. Method IV: For method IV, we will use the average of the velocities at the inner and the outer boundaries of each element. The resulting equation for follows as, Comparison: In order to compare the three methods introduced above, we are going to apply them to a situation where we know the exact mean velocity. We have measured the local velocities at the 10 points mentioned above for a laminar flow in an annular space (outer pipe’s inner radius being and inner pipe’s outer radius being ). The true mean velocity for this flow is . The following table lists the measured velocities: Measurement point Velocity ( ) Measurement point Velocity ( ) 1 0 6 1.455 2 0.654 7 1.279 3 1.107 8 0.974 4 1.380 9 0.546 5 1.492 10 0 Now if we apply the four methods, we will get the following results: Method Calculated mean velocity ( ) Relative error (%) I 0.98 2 II 1.02 2 III 0.94 6 IV 0.98 2 4. A dam creates a lake that is upstream of it. The diagram shows the situation in which the lake water height is below the top of the dam, while a receiving body of water below the dam is at a substantially lower level. The respective levels are at and . The water is at a temperature that is approximately . Find the net horizontal force on the dam. Solution: Again we are going to try two different methods in solving this problem. Before we present the solution, it is worth noting that the first method is more general and in addition to the horizontal force, it can be used to calculate the vertical force too, but the second method can be used for calculating the horizontal force only. Method I: In this method, we are going to sum all the horizontal forces on the dam. For the purposes of this problem, the only force acting on the dam is the pressure forces. But, as we know, pressure acts normal to the surface, so for the curved face of the dam we need to calculate the horizontal component of the pressure force exerted on the dam. If we enlarge a portion of dam and look at the forces acting on a very small part of dam lying at a depth equal to , we will see a picture very similar to one shown here. From this picture: where is the width of the dam. Also, from the figure, , so that Note that this result can be applied to any surface, no matter if the surface is curved or straight (it only has to be an extrusion of a two-dimensional curve). Therefore, if we had neglected the curvature of the surface and considered it as a vertical line, we would have arrived at the same result. The force exerted by the upstream lake and ambient air is: and that exerted by the downstream lake and the ambient air is: Therefore the net horizontal force can be calculated as: Note that the ambient pressure has cancelled out, and we could have got to the same result if we had neglected the air pressure (or in other words, if we had used the gauge pressure instead of absolute pressure) Performing the integration yields, Method II: In this method instead of summing the forces acting on the dam, we will consider the illustrated control volume which encompasses the dam and cuts its base, so that the external horizontal force acting on the assumed control volume is equal to the net horizontal force acting on the dam. Using this method we do not need to project the pressure force on the horizontal plane, as the faces of the control volume are vertical. The force balance is exactly the same as above: