Dimensional Analysis ME 305 Fluid Mechanics I • Consider that we are interested in determining how the drag force acting on a smooth sphere immersed in a uniform flow depends on other fluid and flow variables. Part 7 • Important varibles of the problem are shown below (How did we decide on these?). 푉 Dimensional Analysis and Similitude 휇 , 퐹퐷 퐷 These presentations are prepared by Dr. Cüneyt Sert Department of Mechanical Engineering • Drag force 퐹퐷 is thought to depend on the following variables. Middle East Technical University 퐹퐷 = 푓(퐷, 푉, 휇, ) Ankara, Turkey [email protected] • In order to find the actual functional relation we need to perform a set of experiments. You can get the most recent version of this document from Dr. Sert’s web site. • Dimensional analysis helps us to design and perform these experiments in a Please ask for permission before using them to teach. You are NOT allowed to modify them. systematic way. 7-1 7-2 Dimensional Analysis (cont’d) Dimensional Analysis (cont’d) • The following set of controlled experiments should be done. • It is possible to simplify the dependency of drag force on other variables by using nondimensional (unitless) parameters. • Fix 퐷, 휇 and . Change 푉 and measure 퐹퐷. 퐹퐷 푉퐷 • Fix 푉, 휇 and . Change 퐷 and measure 퐹 . = 푓 퐷 푉2퐷2 1 휇 • Fix 퐷, 푉 and 휇 . Change and measure 퐹퐷. Note : These are only illustrative Nondimensional drag figures. They do not correspond to Nondimensional • Fix 퐷, 푉 and . Change 휇 and measure 퐹퐷. any actual experimentation. force (Drag coefficient) Reynolds number (푅푒) 퐹퐷 Constant 퐹퐷 Constant 퐹퐷 Constant 퐹퐷 Constant 퐷, 휇, 푉, 휇, 퐷, 푉, 휇 퐷, 푉, 퐹퐷 푉2퐷2 Illustrative figure 푉 퐷 휇 • We need to perform too many experiments. • Also there are major difficulties such as finding fluids with different densities, but 푅푒 same viscosities. Flow over a sphere at 푅푒 = 15000 7-3 7-4 1 Dimensional Analysis (cont’d) Buckingham Pi Theorem • To find this new relation, we only need to change the Reynolds number. • Buckingham Pi theorem can be used to determine the nondimensional groups of variables (Pi groups) for a given set of dimensional variables. • We can do it in any way we want, e.g. the simplest way is to change the speed of air flow in a wind tunnel. • For the flow over a sphere problem studied previously, dimensional parameter set is (퐹 , 퐷, 푉, 휇, ) and this theorem helps us to find two Pi groups as • All 푅푒 = 15000 flows around a sphere will look like the same and they all provide 퐷 the same nondimensional drag force. It does not matter what fluid we use or how 퐹 푉퐷 Π = 퐷 and Π = big the sphere is (be aware of very extreme cases). 1 푉2퐷2 2 휇 • Dimensional analysis is used to formulate a physical phenomenon as a relation • Let’s explain how this works using “the drag force acting on a sphere” problem. between a set of nondimensional (unitless) groups of variables such that the number of these groups is less than the number of dimensional variables. • Step 1 : List all the dimensional variables involved in the problem. • 푛 푛 = 5 • It is important to develop a systematic and meaningful way to perform experiments. is the number of dimensional variables. for our example. • These variables should be independent of each other. For example if the diameter of • Nature of the experiments are simplified and the number of required experiments is a sphere is in the list, frontal area of the sphere can not be included. reduced. • If body forces are important in a problem, gravitational acceleration should be in the list, although it is a constant. 7-5 7-6 Buckingham Pi Theorem (cont’d) Buckingham Pi Theorem (cont’d) • Step 2 : Express each of the variables in tems of basic dimensions, which are • Step 3: Determine the repeating variables that are allowed to appear in more than one Pi group. 퐿 : length , 푇 : time , 푀 : mass • There should be 푟 many repeating variables. • For problems involving heat transfer Θ (temperature) can also be a basic dimension. • If 퐿 is a primary dimension of the problem, we should select one geometric variable • For the example we are studying basic dimensions of variables are as a repeating variable. 푀퐿 퐿 푀 푀 [퐹 ] = , 퐷 = 퐿 , [푉] = , [] = , [휇] = 퐷 푇2 푇 퐿3 퐿푇 • If 푇 is a primary dimension of the problem, we should select one kinematic variable as a repeating variable. • Our example involves 푟 = 3 primary dimensions. For most fluid mechanics problems • If 푀 is a primary dimension of the problem, we should select one dynamic variable 푟 will be 3. as a repeating variable. • Variables having only 퐿 in their dimension are called geometric variables. • Note that this selection is not unique and the resulting Pi groups will depend on our selection. Certain selections are ‘‘better’’ than others. • Variables having only 푇 or both 퐿 and 푇 are called kinematic variables. • For the problem of interest we can select 퐷 , 푉 and as repeating variables. • Variables having 푀 in their dimension are called dynamic variables. • If there is an obvious dependent variable in the problem, do not select it as a • For our example 퐷 is a geometric, 푉 is a kinematic and 퐹 , 휇 and are dynamic 퐷 repeating variable. In our example 퐹 is a dependent variable. We are trying to variables. 퐷 understand how it depends on other variables. 7-7 7-8 2 Buckingham Pi Theorem (cont’d) Buckingham Pi Theorem (cont’d) • Step 4: Determine (푛 − 푟) many Pi groups by combining repeating variables with • Now determine the second Pi group which has 휇 as the nonrepeating variable. nonrepeating variables and using the fact that Pi groups should be nondimensional. 푎 푏 푐 Π2 = 휇 퐷 푉 • For our example we need to find 5 − 3 = 2 Pi groups. Each Pi group will include only 푀 퐿 푀 푐 one of the nonrepating variables. • Π should be unitless : − = [퐿]푎 [ ]푏 2 퐿푇 푇 퐿3 푎 푏 푐 Π1 = 퐹퐷 퐷 푉 We need to determine Π2 should have no 퐿 dimension : 0 = −1 + 푎 + 푏 − 3푐 푎, 푏 and 푐. 푎 = −1 휇 A nonrepeating Π should have no 푇 dimension : 0 = −1 − 푏 푏 = −1 Π = Unknown combination 2 2 퐷푉 parameter 푐 = −1 of repeating parameters Π2 should have no 푀 dimension : 0 = 1 + 푐 푀퐿 퐿 푀 푐 • Π should be unitless : − = [퐿]푎 [ ]푏 • Therefore the relation of nondimensional groups that we are after is 1 푇2 푇 퐿3 퐹 휇 Π = 푓 Π → 퐷 = 푓 Π should have no 퐿 dimension : 0 = 1 + 푎 + 푏 − 3푐 1 1 2 푉2퐷2 1 푉퐷 1 푎 = −2 퐹퐷 Π should have no 푇 dimension : 0 = −2 − 푏 푏 = −2 Π = 푉퐷 1 1 퐷2푉2 • It is better to write the second Pi group as because it is the well known 푐 = −1 휇 Π1 should have no 푀 dimension : 0 = 1 + 푐 Reynolds number. 7-9 7-10 Exercises for Buckingham Pi Theorem Important Nondimensional Numbers of Fluid Mechanics Exercise : Consider the flow of an incompressible fluid through a long, smooth- • Following nondimensional numbers frequently appear as a Pi group. walled horizontal, circular pipe. We are interested in analyzing the pressure drop, 푉퐿 푉퐿 • Reynolds number : 푅푒 = = . Ratio of inertia forces to viscous forces. ∆푝, over a pipe length of 퐿. Other variables of the problem are pipe diameter (퐷), 휇 휈 average velocity (푉) and fluid properties ( and 휇). Determine the Pi groups by a) Δ푝 • Euler number : 퐸푢 = 1 . Ratio of pressure forces to inertia forces. 푉2 selecting as a repeating parameter, b) selecting 휇 as a repeating parameter. 2 푉 • Froude number : 퐹푟 = . Squareroot of the ratio of inertia forces to gravitational Exercise : In a laboratory experiment a tank is drained through an orifice from initial 푔퐿 liquid level ℎ0. The time, 휏, to drain the tank depends on tank diameter, 퐷, orifice forces. diameter, 푑, gravitational acceleration, 푔, liquid properties, and 휇. Determine the 푉 푉 • Mach number : 푀푎 = = . Squareroot of the ratio of inertia forces to Pi groups. 퐸푣/ 푐 compressibility forces. Exercise : The diameter, 푑, of the dots made by an ink jet printer depends on the ink 푉2퐿 휇 퐷 퐿 • Weber number : 푊푒 = . Squareroot of the ratio of inertia forces to surface properties, and , surface tension, , nozzle diameter, , the distance, , of the nozzle from the paper and the ink jet velocity, 푉. Determine the Pi groups. tension forces. 휔퐿 • Strouhal number : 푆푡 = . Used for flows with oscillatory (periodic) behavior. Exercise : The power, 풫, required to drive a propeller is known to depend on the 푉 following variables: freestream speed, 푉, propeller diameter, 퐷, angular speed, 휔, 푝−푝푣 • Cavitation number : 퐶푎 = 1 . Used for possibly cavitating flows. 휇 푐 푉2 fluid properties, and , and the speed of sound . Determine the Pi groups. 2 7-11 7-12 3 Model and Prototype Three Basic Laws of Similitude • In experimental fluid mechanics we sometimes can not work with real sized objects, • A similitude analysis is done to make sure that the results obtained from an known as prototypes. experiment can correctly be transferred to the real flow field. • Instead we use scaled down (or up) versions of them, called models. • Three basic laws of similitude must be satisfied in order to achieve complete • Also sometimes in experiments we use fluids that are different than actual working similarity between prototype and model flow fields. fluids, e.g. we use regular tap water instead of salty sea water to test the 1.