# HOW TO GET A GOOD GRADE ON THE MME 2273B FLUID MECHANICS 1 EXAM. Common mistakes made on the final exam and how to avoid them

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1 HOW TO GET A GOOD GRADE ON THE MME 2273B FLUID MECHANICS 1 EXAM Common mistakes made on the final exam and how to avoid them

4 Manometry Manometry problems are generally straightforward and you should get the right answer if you start at one end of the manometer (either end) and then apply the manometry rules until you get to the other end of the manometer system: (1) on moving down a distance z in a fluid of density ρ the pressure increases by ρgz, (2) on moving up a distance z in a fluid of density ρ the pressure decreases by ρgz, (3) the pressure is the same at the same height in a continuous mass of the same fluid, which allows us to jump across from one arm of a manometer to the next one within the same fluid. Remember that the shape of the manometer pipes and containers is completely irrelevant all that matters are the heights of the fluid levels, and the interfaces between the fluids, above datum. The only time the shape and size matters is when you want to use the pressures to work out the forces on the wall of container that forms part of the manometer, for example, as in hydrostatic forces problems where you need to know the surface area of the container. If you think I am labouring the point by going over the manometry rules again here, generally the majority of students in each year problems have been set on manometers do not get the right answer. Typically, there are errors in mixing up gauge and absolute pressures, getting the signs the wrong way round when moving up and down in the manometer or using the wrong distances from the diagram. Hydrostatic forces on plane and curved surfaces In the case of a flat plate the force acting normally to the surface is the pressure at the centroid of the plate (= pressure at the fluid surface + pressure due to vertical depth of liquid from the surface to the centroid) multiplied by the plate area. The force acts at the centre of pressure (CP) which we find by computing y CP and, for asymmetrically shaped bodies, x CP. Remember we must find where CP is in order to compute the moment due to the normal force acting on the plate, for example to determine an opening or closing torque on a shaft attached to a plate gate. In the case of curved surfaces we need to split the hydrostatic force into a horizontal and a vertical component, compute them separately and then use vector addition to get the resultant force. For some reason a number of students each year on the exam also split the force up this way to tackle flat plate problems of the type mentioned in the previous paragraph. Of course, you can do this but there is no need to go through the additional calculations required and those students who do it this way tend to get the wrong result as there are more calculations and so more possibilities for a calculation error. To compute the forces on curved surfaces it is necessary to know some details about the shape of the surface. If a mathematical equation is available to describe its shape (e.g. y = Ax 2 ) you ll need to use integration in order to compute the area (and, hence, volume and weight) of fluid above the gate (i.e. above the equation) to then get the vertical force on the surface. Determining where that force acts requires computation of the centroid of this area and this

5 needs an integration of the moments of small vertical slices of area about a vertical axis (e.g. the y axis) over the whole area and then division by the value of the area. A special case is where the surface is curved at a constant radius, for example part of a circle or of a sphere. Here, the areas or volumes above the surface (needed to get the vertical component of the force) can be computed from the fraction of the area of a circle or the fraction of the volume of a sphere and, hopefully, you know the equations for the area of a circle and the volume of a sphere! Since the fluid pressure acts normally to the surface at all points and, for these special cases the normal is always along a radius, the resultant force (that is the vector sum of the horizontal and vertical force components, F R = (F H 2 + F V 2 ) 0.5 ) must act through the centre of the radius of curvature (that is, through the centre of the circle or sphere). When the fluid is below the curved surface, we compute the horizontal component the same way as for when it is above (i.e. it is the force on the plane vertical projection of the surface) but the vertical component now acts upward and (if the gauge pressure at the fluid surface is zero i.e. open to atmosphere) is equal to the weight of the fluid removed above the outside of the curved surface, using the principles of buoyancy. Buoyancy When dealing with a fully or partially submerged body there may be up to 3 forces acting on it; (1) the weight that acts vertically downwards through the centroid (the centre of mass of the body), (2) the buoyancy force that acts vertically upwards through the centre of buoyancy (the centre of mass of the displaced fluid volume), (3) an externally applied mechanical force on the body, such as a tethering cable or rope. Make sure you consider all three as a common mistake is to forget one. If, and only if, the body is freely floating without a tether attached, the force in (3) is zero and so we have equilibrium when the buoyancy force (which is always equal to the weight of the fluid volume displaced) equals the weight of the body. For a balloon or other body that is made from material that can expand or contract, where the body is filled with gas, you will need to take the weight of the gas into account as well as the weight of the body material. The size of the body will be determined by the external pressure, which, in the case of a balloon moving up or down in the atmosphere, will vary with height. Fluid dynamics All of the problems we have studied relating to fluid dynamics may be solved by using two or three of the following equations; (1) the Bernoulli (energy) equation, (2) the continuity (volume or mass flow rate) equation, (3) the momentum equation, and (4) the modified Bernoulli equation that takes into account energy losses and is used only in pipe network problems. In most cases you will need to use the continuity equation with one or more of the other equations in order to have enough equations to find the unknowns. The Bernoulli equation relates pressure, velocity and elevation above datum and allows energy to be transferred between

6 these three terms along a streamline through a system, whilst the total energy stays constant. For open channel flows it is always sensible to follow the surface streamline through the system as the pressure there is always P ATM and so cancels out on both sides of the equation that represents two different points on the same streamline. Also, remember that if you see the phrase very large reservoir you can assume that at the surface the fluid velocity is zero. The Bernoulli equation is an energy equation and so, by itself, it cannot be used to calculate any forces that arise due to the fluid motion (e.g. impact of jets, force on a reducing nozzle). To find those forces we must use the momentum equation (and usually continuity as well) since the force is related to rate of change of momentum. When using the modified Bernoulli equation for pipe network problems remember to break the calculation of losses down into major (friction) and minor (all other losses), compute them one by one and then add them all up. You can use the Moody diagram or the Colebrook equation (or the approximate versions of the Colebrook equations) but when you find the value of the friction factor (f) just stop for a moment and see if it makes sense. It should fit within the range of values given on the Moody chart and so if you find that f = , for example, it is clearly nonsense and you should go back and check your calculations! Remember that even if the question says that the pipe is smooth the flow can still be turbulent. The key is to check the Reynolds number if it is less than 2,000 the flow is laminar and you can use f = 64 / Re, otherwise you may assume it is turbulent and so you have to use one of the equations or the Moody chart. When using the Bernoulli or Modified Bernoulli equation, decide at the start which form you are going to use; Work (or energy) per unit mass (units will be m 2 /s 2, with losses expressed as φ), Work per unit weight (units will be m, with losses expressed as head loss, h loss ) or Work per unit volume (units will be N/m 2, with losses expressed as a pressure loss ΔP loss ), and then be consistent throughout the entire question. Flows through orifices The Toricelli equation, u = (2gh) 0.5, is a result of applying the Bernoulli equation and gives the velocity (u) of the flow coming out of an orifice in a tank that is open to atmosphere, where the fluid surface in the tank is at h above the orifice. A common error from this is in computing the volume flow rate (Q) using this velocity value and the cross-sectional area A of the orifice. The coefficient of velocity for the orifice (C v ) modifies the velocity only, not the volume flow rate. The cross section of the jet often reduces just after the orifice, due to the vena contracta and so we use the contraction coefficient (C c ) that modifies the cross-sectional area of the jet only, not the velocity or the volume flow rate. It is only when these are all put together as Q = C c C v Au, that we get the volume flow rate. If, instead we are only given a discharge coefficient (C d ) for the orifice, we use Q = C d Au and assume that A is modified by C d since the modification of u by any coefficient is usually very small for an orifice (C v is typically 0.97, for example). When we consider the jet that comes out of an orifice and we see the term free jet we can apply the Bernoulli equation at a point in the jet just downstream of the vena contracta where the streamlines are now all straight. Because the streamlines are straight there is no pressure gradient

7 across the jet and so the pressure in the jet will be the same as that in the fluid outside of the jet. So, for example, a water jet that issues into air will be at the pressure of the air (atmospheric pressure). Dimensional analysis and similarity Whilst most topics on the course require the appropriate tools to be used in slightly different ways for different problems, the process of dimensional analysis using the Buckingham π- theorem and the use of the resulting π parameters to compare the performance of geometrically experimental models and prototypes, is well-defined and followed exactly the same way for every problem. Thus, if you know the procedure you should get the right answers every time. However, sadly, most students tend to lose a lot of marks when working on these problems. First, remember to list all the variables and then write down their dimensions. If you do not know the dimensions of a variable try to derive them from parameters that you know. For example the units of pressure = units of (mass x acceleration / area). If you do the π theorem analysis and you do not get the groupings asked for there are two possible reasons; (1) you got the dimensions of one or more of the variables wrong at the beginning, (2) you chose a different set of (valid) repeating parameters. If it is (2) then you should be able to get the parameters asked for in the question by combining the ones you found from your analysis, using multiplication, division or raising to a power. You can do this because if your parameters are dimensionless then so will, for example, two parameters multiplied together or divided into another. If you cannot get the parameters asked for in the question using this approach then you have probably made an error in the units of one or more of the variables at the outset so go back and check. Once you have established the π parameters you can then say that, for example if there are three parameters, π 1 = f (π 2, π 3 ), where f denotes function of and π 1 contains the dependent parameter (e.g. force, pressure, flow rate, power), in other words the parameter you want to find out how it is influenced by the other parameters. Many mistakes are made when using this expression to assess how a prototype (subscript p) may perform given the performance of a geometrically similar laboratory model (subscript m). The first thing to do is list everything you know about the prototype in one column, leaving blanks or question marks alongside those that are unknown at the outset, and then the same for the model in the next column. Sometimes you might not know the values of the parameters but you might know that, for example, the full-scale diameter of the test component is ten times the model value of the diameter so D p = 10 D m. A common error is to try to write that π 1p = π 1m straightaway in order to find the value of the dependent parameter quickly. Remember, though that we can only say that π 1p = π 1m if the model and prototype flows are similar. In order to ensure similarity we first must have variable values that ensure that both π 2p = π 2m and π 3p = π 3m. Only once we have achieved this (and we may need to solve π 3 before π 2 ), can we then solve for π 1p = π 1m and get the value of the dependent parameter. Eric Savory March good luck on the exam!

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