Chapter 8 Dimensional Analysis and Similitude Ahmad Sana Department of Civil and Architectural Engineering Sultan Qaboos University Sultanate of Oman Email: [email protected] Webpage: http://ahmadsana.tripod.com Significant learning outcomes Conceptual Knowledge State the Buckingham Π theorem. Identify and explain the significance of the common π-groups. Distinguish between model and prototype. Explain the concepts of dynamic and geometric similitude. Procedural Knowledge Apply the Buckingham Π theorem to determine number of dimensionless variables. Apply the step-by-step procedure to determine the dimensionless π- groups. Apply the exponent method to determine the dimensionless π-groups. Distinguish the significant π-groups for a given a flow problem. Applications (typical) Drag force on a blimp from model testing. Ship model tests to evaluate wave and friction drag. Pressure drop in a prototype nozzle from model measurements. CIVL 4046 Fluid Mechanics 2 8.1 Need for dimensional analysis • Experimental studies in fluid problems • Model and prototype • Example: Flow through inverted nozzle CIVL 4046 Fluid Mechanics 3 Pressure drop through the nozzle can shown as: p p d V d 1 2 f 0 , 1 0 2 V / 2 d1 p p d For higher Reynolds numbers 1 2 f 0 V 2 / 2 d 1 CIVL 4046 Fluid Mechanics 4 8.2 Buckingham pi theorem In 1915 Buckingham showed that the number of independent dimensionless groups of variables (dimensionless parameters) needed to correlate the variables in a given process is equal to n - m, where n is the number of variables involved and m is the number of basic dimensions included in the variables. Buckingham referred to the dimensionless groups as Π, which is the reason the theorem is called the Π theorem. y1 f (y2 , y3, y4 ,.....yn ) 1 ( 2 , 3,...... nm ) CIVL 4046 Fluid Mechanics 5 For example, the drag force F of a fluid past a sphere would be a function of the velocity V, mass density , viscosity , and diameter D, a total of five variables (F, V, , , D) and three basic dimensions (L, M and T) are involved. By Buckingham pi theorem there will be 5-3=2 groups. So, these two groups can be correlated as: 1 ( 2 ) CIVL 4046 Fluid Mechanics 6 8.3 Dimensional analysis Purposes of dimensional analysis ►To generate nondimensional parameters that help in the design of experiments (physical and/or numerical) and in the reporting of experimental results ►To obtain scaling laws so that prototype performance can be predicted from model performance ►To (sometimes) predict trends in the relationship between parameters CIVL 4046 Fluid Mechanics 7 8.3 Dimensional analysis Step by step method 1. Identify the significant dimensional variables and write out the primary dimensions of each. 2. Apply the Buckingham Π theorem to find the number of π- groups. 3. Set up table with the number of rows equal to the number of dimensional variables and the number of columns equal to the number of basic dimensions plus one (m + 1). 4. List all the dimensional variables in the first column with primary dimensions. CIVL 4046 Fluid Mechanics 8 8.3 Dimensional analysis Step by step method 5. Select a dimension to be eliminated, choose a variable with that dimension in the first column, and combine with remaining variables to eliminate the dimension. List combined variables in the second column with remaining primary dimensions. 6. Select another dimension to be eliminated, choose from variables in the second column that has that dimension, and combine with the remaining variables. List the new combinations with remaining primary dimensions in the third column 7. Repeat Step 6 until all dimensions are eliminated. The remaining dimensionless groups are the π-groups. List the π-groups in the last column CIVL 4046 Fluid Mechanics 9 CIVL 4046 Fluid Mechanics 10 CIVL 4046 Fluid Mechanics 11 CIVL 4046 Fluid Mechanics 12 CIVL 4046 Fluid Mechanics 13 CIVL 4046 Fluid Mechanics 14 CIVL 4046 Fluid Mechanics 15 CIVL 4046 Fluid Mechanics 16 CIVL 4046 Fluid Mechanics 17 CIVL 4046 Fluid Mechanics 18 8.4 Common groups CIVL 4046 Fluid Mechanics 19 CIVL 4046 Fluid Mechanics 20 8.5 Similitude Similarity between model and prototype Geometric similarity: the model must be the same shape as the prototype, but may be scaled by some constant scale factor Kinematic similarity: velocity at any point in the model flow must be proportional (by a constant scale factor) to the velocity at the corresponding point in the prototype flow Dynamic similarity: all forces in the model flow scale by a constant factor to corresponding forces in the prototype flow CIVL 4046 Fluid Mechanics 21 Geometric similarity: the model must be the same shape as the prototype, but may be scaled by some constant scale factor lm wm cm 2 3 Lr Ar Lr Vr Lr l p wp cp 22 Kinematic similarity: velocity at any point in the model flow must be proportional (by a constant scale factor) to the velocity at the corresponding point in the prototype flow Vm L / tm Lr Lr Vr tr Vp L / tp tr Vr a V / t V V V 2 m m r r r a p V / tp tr Lr /Vr Lr CIVL 4046 Fluid Mechanics 23 Dynamic similarity: all forces in the model flow scale by a constant factor to corresponding forces in the prototype flow 3 Fm mam VV / tm L V / tm 3 Fp map VV / tp L V / tp 3 3 Fm L m gL m 3 3 Fp L p gL p 3 gL m VV / tm tm L /V m 3 , Also gL p VV / tp t p L /V p V 2 V 2 V V Fr Fr gL gL gL gL m p m p m p CIVL 4046 Fluid Mechanics 24 CIVL 4046 Fluid Mechanics 25 CIVL 4046 Fluid Mechanics 26 3m CIVL 4046 Fluid Mechanics 27 CIVL 4046 Fluid Mechanics 28 CIVL 4046 Fluid Mechanics 29 CIVL 4046 Fluid Mechanics 30 CIVL 4046 Fluid Mechanics 31 .