Civil Engineering Hydraulics Mechanics of

Similitude and Modeling

Dimensionless Ratios

 From using the Rayleigh method, you have seen how we can determine dimensionless ratios that are a predictive factor  Some of these dimensionless ratios are characteristic of the type of flow system we are considering

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1 Dimensionless Ratios

 In flow in a pipe or conduit, which is usually flow completely filling the pipe or conduit through the pressure drop along a length can be found to be a function of three dimensionless ratios

" !vD v L % !p = f $ , , ' # µ a D &

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Dimensionless Ratios

 The first term is known as the  The term is the and is the ratio of the actual to the of sound in the

" !vD v L % !p = f $ , , ' # µ a D &

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2 Dimensionless Ratios

 I use F for a rather than the D and L that the text uses  The first term is the Drag coefficient and the second term is the Lift coefficient

" FDrag FLift !vD % !p = f $ 2 2 , 2 2 , ' # !v D !v D µ &

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Dimensionless Ratios

 For open channel flow, the flow depth is the dependent parameter of concern  The term here is the square of the

"v2 % !z = f $ ' # gl &

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3 Similitude

 When we want to take result from a scale model to explain how a full scale system will perform, we need to maintain certain characteristics between model and full scale  We classify characteristics into two catagories  Geometric  Dynamic

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Similitude

 There are two ratios that must be maintained for geometric similarity  The ratio of the length of the model to the length of the actual in the direction of flow must be equal to the ratio of the diameter of the model to the diameter of the actual

D L model = model = ! Dactual Lactual

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4 Similitude

 Now not everything will have a diameter so you can work back from equivalent areas to see what diameter would give you this area.

D L model = model = ! Dactual Lactual

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Similitude

 In the text they show a cross sectional area with

a s by s dimension and a sm by sm dimension in the model. ! D2 A = s2 = actual 4 ! D2 A = s2 = m model m 4 4s2 4 D2 = ! D = s ! ! 4s2 4 D2 = m ! D = s m ! m ! m D s m = m D s

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5 Similitude

 For dynamic similarity, we consider force, mass, and time

Fmodel = !Factual

mmodel = "mactual

tmodel = #tactual

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Similitude

 From these ratios and from the ratios for geometric similarity other ratios can be derived

Lmodel !Lactual vmodel = = Fmodel = !Factual tmodel "tactual

mmodel = "mactual Lactual vactual = tmodel = #tactual tactual

Dmodel = $Dactual !Lactual v "t ! Lmodel = $Lactual model = actual = L vactual actual "

tactual 12 Similitude and Modeling

6 Similitude

 Utilizing these ratios and working from F=ma the text derives an expression considering the modeling of a system  The type of system under consideration will determine which we must consider

F F = !v2L2 !v2L2 model actual

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Similitude

 Considering pressure forces, the F on each side is replaced by the change in pressure acting over an area defined by L2 F F = !v2L2 !v2L2 model actual !pL2 !pL2 2 2 = 2 2 The text add an ½ to !v L !v L the bottom of each model actual fraction to make it !p !p into the common = form of the pressure !v2 !v2 coefficient. model actual 14 Similitude and Modeling

7 Similitude

 Other Relationships F F = !v2L2 !v2L2 model actual !p !p = for pressure forces !v2 !v2 model actual !vL !vL = for viscous forces µ µ model actual v2 v2 = for gravity forces gL gL model actual !v2L !v2L = for surface-tension forces " " model actual

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Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

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8 Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

Since we aren’t given any or pressure drops, that would eliminate using pressure forces. Also, since we aren’t given any changes in elevation, you can assume that gravity forces are not involved. And since the flow is contained, surface tension is probably also not involved. This leaves viscous forces.

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Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

So the relationship we can use is:

!vL !vL = for viscous forces µ µ model actual

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9 Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

L in this case isn’t the length of the pipe but the characteristic length of the system. Here that would be the pipe diameter.

!vL !vL = for viscous forces µ µ model actual

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Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

Substituting for v on each side.

Q Q ! L ! L A = A µ µ

model actual !QL !QL = Aµ Aµ model actual

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10 Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

The A and L (pipe diameter) are the same in both the actual system and the model so:

!QL !QL = Aµ Aµ model actual !Q !Q = µ µ model actual

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Dimensions and Units

 4.43 Acetone flows through a tube at a volume flow rate of 18 ft3/s. If glycerine is used in the tube instead of acetone, what volume flow rate is required for complete dynamic similarity?

Substituting known values from Table A.5

!Q !Q = µ µ model actual ! lbm$ ! lbm$ ! ft3 $ 1.263 62.4 Q 0.787 62.4 18 "# ft3 %& "# ft3 %& "# s %& = lbf lbf 1.983'105 0.659 '105 (s)( ft 2 ) (s)( ft 2 ) model actual ft3 Q = 33.75 model s 22 Similitude and Modeling

11 Dimensions and Units

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