Dimensions and Units

Transcription

1 Civil Engineering Hydraulics Mechanics of Fluids and Modeling Diensions and Units You already know how iportant using the correct diensions can be in the analysis of a proble in fluid echanics If you don t, I a not doing y job very well. 2 1

2 Diensions and Units Diensions are the what in a easureents l We easure length, tie, ass, weight, etc. Units are just what syste of easureent we are going to use to describe the diensions We can easure the length of soething in feet l We can easure the elapsed tie in inutes l We can easure the weight in stones l 3 Diensions and Units We define diensions Units coe with a syste Length is a diension l Feet are the unit that coes with the USCS syste l 4 2

3 Diensions and Units Based on our syste of easureent, soe diensions are chosen as fundaental to the syste They are not defined in ters of any other diension Sort of the prie nubers of the syste The syste will also give us the sybol and the coon unit for this fundaental diension 5 6 3

4 Diensions and Units Every other quantity that we work with can be defined in ters of these fundaental diensions When we are describing the diensions of soething, we use the curly braces {} around both the thing we are describing and the diensions we are using 7 Diensions and Units For exaple, if we take pressure down to fundaental diensions we would have {P} = F A {A} = {L } 2 L {F } = 2 t L 2 {P} = t 2 = 2 L Lt 8 Pressure has diensions of force per area. Area is diensioned as the square of the fundaental diension length Force is diensioned as the quotient of the product of fundaental diensions ass and length and the square of the fundaental diension tie So Pressure can be described as having fundaental diension as shown 4

5 Diensions and Units So we can describe the units of pressure in fundaental units in each syste {P} = F A {A} = {L2 } {F } = L 2 t L 2 {P} = t 2 = 2 L Lt 9 In SI, pressure would have fundaental units of kilogras per eter per second squared In USCS, pressure would have fundaental units of pound-ass per foot per second squared Diensional Hoogeneity One of the wonderful safety nets we have in engineering analysis is the ability to check out answers for the correct diensions While it is often very easy, especially with our sophisticated calculators, to just write down nubers and ake the calculation assigning the units after the calculation is ade. This can lead to soe costly and ebarrassing istakes. 10 5

6 Diensional Hoogeneity 11 It the diensions aren t the sae for each additive ter in an expression or if the diensions aren t the sae on both sides of the equal sign, you ade a istake Diensional Hoogeneity Use the in your expressions, including all expressions leading up to your calculations and you will save yourself a lot of tie and trouble Yes TIME It is a lot easier to do it correctly the first tie than to retrace your steps 12 6

7 7 13 Angular oentu, also called oent of oentu (H), is fored by the cross product of a oent ar (r) and the linear oentu (V) of a fluid particle. What are the priary diensions of angular oentu? List the units of angular oentu in priary SI units and in priary English units. { H } = {rv } = {r }{ }{v } {r } = { L } { } = { }!L$ {v } = "# t %&! L $! L2 $ { H } = { L}{ }"# t %& = " t % = L2t 1 # & { 13 } Diensional Hoogeneity You ay often see the inline for of diensions given with both positive and negative powers for the exponents This is done to siplify the presentation of units in technical papers 14 7

8 Diensional Hoogeneity Diensional hoogeneity can also help you to reeber the for of an expression For exaple, if you wanted the Bernoulli equation in ters of head which is typically given in ters of length You know that the diension of the result is in units of length, L so all the other ters would also have to work down to units of L 15 Diensional Hoogeneity 16 For the Pressure ter we know that the diensions of pressure are L-1t-2 so we would have to find a factor that that had diensions of -1L2t2 to bring the pressure ter to diensions of L! $ { p } = "# Lt 2 %&! $! L2t 2 $ " 2 %" % = {L} Lt # &# & 8

9 Diensional Hoogeneity 17 Fro previous work with the Bernoulli equation we know that weight density, ass density, and the gravitational constant are soewhere in the expression so we can see which would work here. {γ } = {ρ }{g}! $ {ρ } = 3 p = { } "# Lt 2 %& L L {g} = 2! $! L2t 2 $ t " 2 %" % = {L} L # Lt & # & {γ } = 3 2 = 2 2 L t t L Diensional Hoogeneity Fro this diensional analysis, you can develop that if you divide Pressure by weight density, you have the ter in diensions of length {γ } = {ρ }{g}! $ {ρ } = 3 { p } = "# Lt 2 %& L L {g} = 2! $! L2t 2 $ t " 2 %" % = {L} L # Lt & # & {γ } = 3 2 = 2 2 L t 18 L t 9

10 Model Developent We often have to evaluate systes and ideas in a lab scale setting because field scale is just too large and expensive to build and test For exaple, if you consider the flow under piles on a bridge, we would have to build a lot of bridges to evaluate flow patterns around different types of supports and different conditions 19 Model Developent So we build odels of the systes and test the at this uch saller scale but we have to be careful as to what characteristics we use when we ake the odels The rest of this chapter will be about first locating significant paraeters to include in the odel and then about just how to work with the paraeters on the scale odel to have the accurately represent the full scale syste

11 21 If we have a syste where we are looking to develop soe sort of predictive odel we ay want to first develop a relationship between dependent and independent paraeters We could just easure everything and use statistical ethods to develop a relationship and that is done quite often. This is known as an epirical relationship (developed fro data rather than first principles)

12 23 The Rayleigh ethod is a way to look as possible independent paraeters and see how they ight just relate to each other to describe the dependent variable. 24 You are presented a lab situation where the height h a liquid attains inside a partly suberged capillary tube is a function of surface tension σ, tube radius R, gravity g, and liquid density ρ. How are the independent variables related in describing the dependent variable? 12

13 The first step is to identify the fundaental units for both the dependent and independent paraeters. {h} = {L} {σ } = 2 t { R} = { L } The diensions we are using here are shown in Table 4.1 in the text. We are using a M,L,T diensional syste. {g} = L 2 t {ρ} = 25 3 L Now we set up a relation between the dependent and independent variables {h} = {L} {σ } = 2 t { R} = { L } h = C σ a 1 R a 2 g a 3ρ a 4 {g} = L 2 t {ρ} = 26 3 L 13

14 Each of the independent variables has been given an exponent C is an arbitrary constant {h} = {L} {σ } = 2 t { R} = { L } h = C σ a 1 R a 2 g a 3ρ a 4 {g} = L 2 t {ρ} = 27 3 L Now we substitute the diensions on both sides of the equals sign {h} = {L} {σ } = 2 t { R} = { L } {g} = L 2 t {L} = 2 t a3 L {L}a2 2 3 t L a4 {ρ} = 28 3 L h = Cσ R a2 g a3 ρ a 4 14

15 Now we can group like fundaental units together. {h} = {L} {σ } = 2 t R = { } {L} {g} = L 2 t {L} = 2 t {L} a3 a2 L 2 3 t L a4 {L} = {}+a 4 {L}a2+a3 3a 4 {t} 2 2a3 {ρ} = 29 3 L h = Cσ R a2 g a3 ρ a 4 For diensional consistency, the exponent of like diensions ust be the sae on both sides of the expression {h} = {L} h = Cσ R a2 g a3 ρ a 4 {σ } = 2 t { R} = { L } {L} = 2 t {g} = 0 = + a4 1 = a2 + a3 3a4 L 2 t {ρ} = 30 3 L a3 L {L}a2 2 3 t t a4 {L} = {}+a 4 {L}a2+a3 3a 4 {t} 2 2a3 0 = 2 2a3 15

16 We now have three equations in four unknowns. We can t solve for the exactly but we can try and reduce the to as few unknowns as possible h = C σ a 1 R a 2 g a 3ρ a 4 0 = a 1+ a 4 a 1 = a 4 1 = a 2 + a 3 3a 4 0 = 2a 1 2a 3 a 1 = a 3 31 We now have three equations in four unknowns. We can t solve for the exactly but we can try and reduce the to as few h = Cσ R a2 g a 3 ρ a 4 unknowns as possible 0 = + a4 = a4 0 = 2 2a3 = a3 1 = a2 + a3 3a4 a2 = 1 ( a3 3a4 ) a2 = 1 ( 3( )) a2 = 1+ ( 3) =

17 So we are down to defining everything in ters of. Returning to the original expression with the new expressions h = C σ a 1R 1 2a 1 g a 1ρ a 1 a 1 = a 4 a 2 = 1 2a 1 a 1 = a 3 33 We can get it down to only one exponent because we started with four independent variables and we could generate three expressions h = C σ a 1R 1 2a 1 g a 1ρ a 1 a 1 = a 4 a 2 = 1 2a 1 a 1 = a

18 Unless we have a very unique situation (or three or less independent variables) we cannot get the expression any better than three less than the nuber of independent variables h = C σ a 1R 1 2a 1 g a 1ρ a 1 a 1 = a 4 a 2 = 1 2a 1 a 1 = a 3 35 Now we can group variables with like exponents h = C σ a 1R 1 2a 1 g a 1ρ a 1 " σ % h = CR $ 2 ' #R gρ& 36 18

19 Norally, we write the expression with any ters on the right side that we know the exponent oved to the left side. h = C σ a 1R 1 2a 1 g a 1ρ a 1 " σ % h = CR $ 2 ' #R gρ& 37 " σ % h =C $ 2 ' R #R gρ& You ay also see this written in function for h = Cσ R1 2g ρ σ h = CR 2 R gρ σ h = C 2 R R gρ σ h =f 2 R R gρ 38 19

20 Dr. Janna does not use the f notation because we will have a paraeter later that uses the f h = Cσ R1 2g ρ σ h = CR 2 R gρ σ h = C 2 R R gρ σ h =f 2 R R gρ An exaple fro the text 20

21 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 η is excluded because it is diensionless 41 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 L2 L3 3 = 2 3 t t Lt L 42 a2 a3 { L }a 4 21

22 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 L2 L3 3 = 2 3 t t Lt L a2 a3 { L }a 4 {} = {}a2 {}a3 {L}2 = {L}3 {L} a2 {L} 3a3 {L}a 4 {t} 3 = {t} {t} 2a2 43 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 L2 L3 3 = 2 3 t t Lt L a2 a3 { L }a 4 {} = {}a2 {}a3 1 = a2 + a3 {L}2 = {L}3 {L} a2 {L} 3a3 {L}a 4 2 = 3 a2 3a3 + a4 {t} 3 = {t} {t} 2a2 3 = 2a

23 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 1 = a2 + a3 a3 = 1 a2 2 = 3 a2 3a3 + a4 3 = 2a2 = 3 2a2 45 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 1 = a2 + a3 a3 = 1 a2 2 = 3(3 2a2 ) a2 3(1 a2 ) + a4 2 = 9 6a2 a a2 + a a2 = a4 3 = 2a2 = 3 2a

24 An exaple fro the text P = CQ Δp a2 ρ a3 D a 4 P = CQ 3 2a2 Δp a2 ρ1 a2 D 4+4a2 P = CQ 3Q 2a2 Δp a2 ρ1ρ a2 D 4 D +4a2 ( P = CQ 3 ρ1d 4 Q 2a2 Δp a2 ρ a2 D +4a2 P=C 47 Q 3 ρ1 ΔpD 4 D 4 Q 2 ρ ) a2 Hoework 16-1 Use a

25 Hoework 16-2 Use a2 and a