Dimensional Analysis and Similitude

Chapter 8: Dimensional Analysis and Similitude

Dr Ali Jawarneh Department of Mechanical Engineering 8.1 The Need for Dimensional Analysis Nondimensional parameters reduces the number of necessary independent parameters in a problem. Relations of pressure, velocity, and diameter. Flow through Inviscid inverted flow nozzle

p −p = Δp = f(ρ,V ,d ,d ) Δp d1 1 2 1 1 o = ϕ( ) 1 2 d ρV o The primary purposes of dimensional analysis are: 2 1 •Trem en dous r educti on in ex perim ent al w ork • 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 (or to obtain a set of dimensionless parameters to correlate the data) • To set the data in global way Illustrative plots showing how the pressure drop per unit length in a pipe may be affected by several different factors.

ΔpL = f(ρ,V,D,μ)

DΔp ρVD An illustrative plot of pressure drop per unit L 2 = ϕ( ) length using dimensionless parameters. ρV μ Dimensionless groups 8.2 Dimensions and Equations Description Dimensions

Mass(m) M Length L BASIC Time T DIMENSIONS Temperature θ Area:...... L2 Diameter:………………… L Pressure:………………… M/LT2 Acceleration:…………….. L/T2 Work:……………………… ML2/T2 Mass flow rate:…………… M/T Volume flow rate:…………. L3/T Force:……………………… ML/T2 See Appendix A-1 Velocity:……………………. L/T Gtt(R)Gas constant(R):…………. L2/ θT2 Density(ρ):……………………. M/L3 Dynamic viscosity(μ):…….. M/LT Kinematic viscosity(ν):…… L2/T Angular speed(ω):…………. 1/T Specific weight (γ)…………M/L2T2 Surface Tension(σ)……………M/T2 8.3 The Buckingham Π Theorem

The number of independent dimensionless groups of variables (dimensionless parameters)= n - m n: number of variables Π = pi m:number of basic dimensions

Dimensional variables: p1 −p2 = Δp = f(ρ,V1,d1,do ) y1= f (y2, y3, …, yn)

Dimensionless parameters (Π-groups): Δp d Π = ϕ (Π , Π , …, Π ) = ϕ( 1 ) 1 2 3 n-m 1 2 d ρV o 2 1 8.4 Dimensional Analysis

The step-by-step method

Example: AthiA thin rec tangul ar p la te hav ing a w idth w andhihtd a height h ilis loca tdted so that it is normal to a moving stream of fluid. Assume the drag force

FD, that the fluid exerts on the plate is a function of w and h, the fluid viscosity μ and density ρ , and the velocity V of the fluid approaching the plate. Determine a suitable set of pi terms using the step-by-step method to study this problem experimentally. SltiSolution: FD =f(w, h, μ, ρ, V) 2 FD =ML/T w =L h=L μ=M/LT # of Pi terms(dimensionless grou ps)= 6-3= 3 ρ=M/L3 V=L/T The methodology: Eliminate L, then M, then T Variable [ ] Variable [ ] Variable [ ] Variable [ ]

2 2 4 2 4 2 2 FD ML/T FD /w M/T FD /ρw 1/T FD /ρw * w /V 0 2 2 =FD /ρw V w L h Lh/w 0h/w 0 h/w 0 μ M/LT μw M/T μ/ρ w2 1/T μ/ρwV 0 ρ M/L3 ρw3 M VL/TV/w 1/T V/w 1/T F h μ D = ϕ( , ) ρw2V2 w ρwV

Π1 Π2 Π3 You may choose h instead of w, if so: F w μ D = ϕ( , ) ρh2V2 h ρhV

Example: It is known that the pressure developed by a centrifugal pump, Δp, is a function of the diameter D of the impeller, the speed of rotation n, the discharge Q, and the fluid density ρ. By dimensional analysis, determine the π ggproups relatin g these variables. Use ste p-by-step method. Solution:

2 The Exponent Method

ElExample: A thin rectangular plate having a width w and a height h is located so that it is normal to a moving stream of fluid. Assume the drag force

Solution: FD =f((,w, h, μ, ρ, V)) 2 FD =ML/T w =L hLh=L μ=M/LT ρ=M/L3 # of Pi terms(dimensionless groups)= 6-3= 3 V=L/T a b c d e FD=w h μ ρ V MLT-2 = (L)a (L)b (ML-1T-1)c (ML-3)d (LT-1)e We have three equations and four unknowns. M: 1=c+d However, we can solve for three of the unknowns in terms of the fourth unknown. The solution T: -2=-c-e procedure is simplified if we select as the fourth L: 1=a+b-c-3d+e unknowns the exponent that appears most frequently in the equations. d=1-c (1) e=2-c (2) a=2-c-b (3) a b c d e 2-c-b b c 1-c 2-c FD=w h μ ρ V = w h μ ρ V If one select: b=2-c-a 2 -c -b b c 1 -c 2 -c FD = w w w h μ ρ ρ V V anddd do th e same procedures he/she will 2 2 b c get: F = (w ρV )(h/w) (μ/ρwV) 2 2 D FD/ h ρV = ϕ(w/h, μ/ρhV) 2 2 FD/ w ρV = ϕ(h/w, μ/ρwV) 8.5 Common Dimensionless Numbers

8.6 Similitude Non-dimensionalization of an equation by inspectional analys is is useful only when one knows the equation to begin with. However, in many cases in real-life engineering, the equations are either not known or too difficult to solve; oftentimes experimentation is the only method of obtaining reliable information. In most experiments, to save time and money, tests are performed on a geometrically scaled model, rather than on the full-scale prototype.

Model: the replica of the structure on which the tests are made. EiExperimenta liifl testing is often per forme dihllld with a small scale rep lica

prototype: Full-scale structure employed in the actual engineering design

The model is usually made such smaller than the prototype for economic reasons Model: Π 1m= Π 1p Π 1m= ϕ (Π 2m, Π 3m, …, Π nm) Π2m= Π 2p Π3m= Π 3p Prototype: . . Π 1p= ϕ (Π 2p, Π 3p, …, Π np) .

Π nm= Π np

Example: FD =f(w, h , μ, ρ, V) Solution:

FD h μ Geomet r y Simili tude 2 2 = ϕ( , ) ρw V w ρwV h h ( )m = ( )p F F w w ( D ) = ( D ) ρw2V2 m ρw2V2 p μ μ Dynamic Similitude ( ) = ( ) ρwV m ρwV p Model Scales Example: length scale =1/10 scale Length scale or scale model: Lm/Lp = λL mode or 1:10 scale model Velocity scale: Vm/Vp = λv Lm/Lp =1/10 Density scale: ρm/ ρ p = λρ

Viscosity scale: μm/ μp = λ μ

Temperature scale: Tm/Tp = λT

Notes:

1- sometime as an example: ρm= ρ pp, , or μm = μp, or gm=gp or Tm=Tp, ………. Q V A V d V L t 2- p = p p = ( p )( p )2 p = p m Qm Vm Am Vm dm Vm Lm tp

p ρ T p = p p (for ideal gas) pm ρm Tm Example: The drag on a submarine moving below the free surface is to be determined by a test on a 1/20 scale model in a water tunnel. The velitlocityof protttotype inseawater (ρ=1015 k/kg/m3, ν=1.4 x 10-6 m2/)/s) is 2m/s. The test is done in pure water at 20 0C. Determine the speed of the water in the water tunnel for dynamic similitude and the ratio of drag force on the model to the drag force on the prototype.

Solution: FD =f(L, D , μ, ρ, V)

F D μ D = ϕ( , ) ρL2V2 L ρLV

ρmLmVm ρpLpVp Rem = Rep = = μm μp From table at T=20 0C −6 LmVm LpVp Lp νm 1×10 = ⇒ Vm = ( )( )Vp = (20)( −6 )2 = 28.6 m/ s νm νp Lm νp 1.4×10 F F Dm Dp 2 2 = 2 2 ρmLmVm ρpLpVp

From table at T=20 0C

F ρ V L 998 28.6 1 Dm = ( m )( m )2( m ) = ( )( )2( )2 = 0.503 F ρ V L 1015 2 20 Dp p p p Example: A large venturi meter is calibrated by means of a 1/10 scale model using the prototype liquid. What is the discharge ratio Qm/Qp for didynamicsiilit?imilarity? If a pressure difference of 300 kpa ismeasured across ports in the model for a given discharge, what pressure difference will occur between similar ports in the prototype for dynamically similar conditions?

V d d0

Solution:

Δp=f(d, d0, μ, ρ, V)

Δp = f(d,d0 ,μ,ρ,V)

Δp d ρVd 2 = ϕ( , ) ρV d0 μ Vmdm Vpdp Vm dp νm Rem = Rep ⇒ = ⇒ = ( )( ) νm νp Vp dm νp Q V A V d d ν d d ν m = m m = m ( m )2 = ( p )( m )( m )2 = ( m )( m ) Qp VpAp Vp dp dm νp dp dp νp

Q 1 1 ∴ m = ( )(1) = Qp 10 10

Δpm Δpp ρp Vp 2 ρp dm 2 νp 2 2 = 2 ⇒ Δpp = Δpm ( )( ) = Δpm ( )( ) ( ) ρmVm ρpVp ρm Vm ρm dp νm

1 ∴Δp = 300(1)( )2(1)2 = 3 kpa p 10