TD3
Correction
Exercice 1
This exercise considers a thin dam with a wedge-shaped opening. The flow through the openingQ[m3·s−1] depends on the height H[m] of the opening, the speed of the water flowing out V[m·s−1], the gravitational constantg[m·s−2] and the angleφ[−] of the wedge-shaped opening.
The aim is to determine the non-dimensional numbers Πi=QaHbVcgdφe.
Variable Q H V g φ
Unit m3·s−1 m m·s−1 m·s−2 -
Coefficient a b c d e
The dimensional functions are : m : 3a+b+c+d= 0
s :−a−c−2d= 0
The first non-dimensional number is easy to spot : Π1 = φ[−]. There remain 4 variables and 2 independent equations, thus 4-2=2 non-dimensional numbers. Therefore, we pick 2 free variables. We chooseQ, because it is the most interesting variable. The second variable preferably has all the units, but the remaining non-free variables should also contain all of the units. We chooseV. The coefficients areaandc.
— For Π2 we choosea= 1 andc= 0.
3 +b+d= 0 ⇒ b=−5/2
−1−2d= 0 ⇒ d=−1/2 Thus we have :
Π2=Q H−5/2g−1/2= Q H−5/2√g
— For Π3 we choosea= 0 andc= 1.
b+ 1 +d= 0 ⇒ b=−1/2
−1−2d= 0 ⇒ d=−1/2 Thus we have :
Π3=H−1/2 V g−1/2= V
√Hg =F r
Exercice 2
This exercise considers a jetty protecting a harbor. Waves are hitting the jetty with a periodT = 7.5 s. We have a scale model of the jetty and harbor at a 1/20 scale. The question is : what is the period of the wavesTm
hitting the jetty in the model ?
H jet´ee
λ
Figure1 – The jetty problem
The 1/20 scale gives us a condition on the wave length and the depth : λm
λr
=Hm
Hr
= 1
20 (1)
We can consider that the depth close to the jetty is small and write the wave phase velocity :c=√
gH =Tλ. Then we can write the following equations for the model (m) and the reality (r) :
λm=Tm
pgHm (2) λr=Tr
pgHr (3)
Combining (2) and (3) we get :
λm
λr = Tm
Tr
rHm
Hr
With the scale condition (1) :
Tm= Tm
√20 = 1.67s
Exercice 3
This exercise considers the drag coefficient Cx on a boat, which depends on the Reynolds number (this is an empirical assumption). The question is : if the cruising speed of a real boat is given byUc, what is the speed Um in a model at scale 1/10 ?
1. Using the equality of the Reynolds number :
Rem=Rec ⇒ ρUmLm
µ =ρUcLc
µ ⇒ Um=Uc Lc
Lm = 10·Uc
Then the scale model need to be 10 times spider than the real boat. That can be a problem because we can create other physical effects...
2. The ratio of the forces is given by : FFm = Um2 ·Sm
U2·S . With S = W·L and Sm = Wm·Lm. we get
Fm
F = L·Wm
Lm·W = 1. The drag force in the model is equal to the force in the reality. To build a model with a more correct similitude (reasonable speed) we could use a fluid with a higher viscosity (eg : Glycerol).
Another example, for this problem of similitude, is the plane model. This plane models are immersed in a water tunnel to get a higher viscosity than air (otherwise the speed is higher than the noise and that induces shock waves).
Exercice 4
This exercise considers a model of a levee made off concrete blocks with mass m = 1 kg. In the model a block can be lifted by wave that has a height larger than 30cm. The question is : how heavy should a real block be in the case of a real levee when it should withstand 6mhigh waves ?
The lifting force working on the block Fa ∝U2L2 and the gravitational force on the blockFp =mascale linearly :Fp/Fa =ǫ. We will use this, and also the fact thatǫin the model and in reality should be equal :
Fp,m
Fa,m
= Fp,r
Fa,r
.
Step 1 is solving this :
Fp,r=Fp,m
Fa,r
Fa,m =Fp,m
Ur2L2r
Um2L2m (4)
Step 2 is using the equality of the Froude number :
F rm=F rr ⇒ Um2 gHm
= Ur2
gHr ⇒ Um2 Ur2 =Hm
Hr
(5)
Step 3 is rewriting the gravitational force using the volume of a blockV =L3 :
Fp =mg=ρV g=ρL3g ⇒ L= Fp
ρg 1/3
(6)
Combining (4), (5) and (6) we find :
Fp,r =Fp,m
Hr
Hm
L2r
L2m =Fp,m
Hr
Hm
(Fp,r/ρrg)2/3 (Fp,m/ρmg)2/3 The blocks are from concrete in the scale model as well as in realityρp =ρr :
Fp,r=Fp,mHr
Hm
Fp,r2/3
Fp,m2/3
⇒ Fp,r1/3=Fp,m1/3Hr
Hm ⇒ Fp,r=Fp,m
Hr
Hm
3
With Fp=mg we get :
mr=mm
Hr
Hm
3
= 1 6
0.3 3
= 8000 kg
Exercice 5
This exercise considers the pressure drop ∆p along a circular tube. The aim is to determine how many experiments need to be performed to characterize the configuration.
1. The different parameter areD [m] is the tube diameter,V[m·s−1] the flow speed,ρ[kg·m−3] the fluid density,µ[kg·m−1·s−1] the fluid viscosity and ∆p[kg·m−2·s−2] the pressure drop.
2. We can build the non-dimensional numbers Πi=DaVbρcµd∆pe.
Variable D V ρ µ ∆p
Unit m m·s−1 kg·m−3 kg·m−1·s−1 kg·m−2·s−2
Coefficient a b c d e
The dimensional functions are : m :a+b−3c−d−2e= 0 s :−b−d−22 = 0
kg :c+d+e= 0
There are 5 variables and 3 independent equations, thus 5-3=2 non-dimensional numbers. Therefore, we pick 2 free variables. We chooseδp, because it is the most interesting variable. The second variable preferably has all the units, but the remaining non-free variables should also contain all of the units. We chooseµ. The coefficients aredande.
— For Π1 we choosed= 0 and e= 1.
a+b−3c−2 = 0 ⇒ a= 1
−b−2 = 0 ⇒ b=−2 c+ 1 = 0 ⇒ c=−1 Thus we have :
Π1=DV−2ρ−1∆p=∆p D V2ρ
— For Π2 we choosed= 1 and e= 0.
a+b−3c−1 = 0 ⇒ a=−1
−b−1 = 0 ⇒ b=−1 c+ 1 = 0 ⇒ c=−1 Thus we have :
Π2=D−1V−1ρ−1µ= µ ρDV = 1
Re
∆p D
V2ρ =f(Re)
3. To understand the pipe problem we need to know the evolution of ∆Vp D2ρ withRe, thanks to experiences.
In 1944, Lewis Ferry Moody performed this experiences (the number of experiences depends of the evolution) and plotted the Darcy-Weisbach friction factor into what is now known as the Moody chart :
Figure2 – Moody chart
We can see that another hidden parameter is the relative pipe roughness which allow the evolution to vary. For a given pipe with a given roughness and for a high Reynolds number we see that the evolution is constant. Then, if we already knew this, we could say that, in this condition and only in this condition, (high Reynolds, and high roughness) that we can have the friction factor with only one experience and keep it in a large domain of Reynolds number (without forget it becomes not truth for a not so low Reynolds). For instance, with a roughness of 0.03 the friction factor start to be constant atRe= 104.
Exercice 6
This exercise considers the drag force on a plate that is positioned in a flowing liquid. The aim is to determine the non-dimensional numbers Πi=FDawbhcµdρevf, whereFD[kg·m·s−2] is the drag force,w[m] the width of the plate,h[m] the height of the plate submerged,µ[kg·m−1·s−1] the viscosity of the liquid,ρ[kg·m−3] the fluid density andv[m·s−1] the speed of the fluid.
Variable FD w h µ ρ v
Unit kg·m·s−2 m m kg·m−1·s−1 kg·m−3 m·s−1
Coefficient a b c d e f
The dimensional functions are : kg :a+d+e= 0
m :a+b+c−d−3e+f = 0 s :−2a−d−f = 0
There are 6 variables and 3 independent equations, thus 6-3=3 non-dimensional numbers. Therefore, we pick 3 free variables. We choose FD, because it is the most interesting variable. The second and third variables preferably have all the units, but the remaining non-free variables should also contain all of the units. We chooseµ andh. The coefficients area, candd.
— For Π1 we choosea= 1,c= 0 andd= 0.
e=−1 ⇒ e=−1
1 +b−3e+f = 0 ⇒ b=−2
−2−f = 0 ⇒ f =−2 Thus we have :
Π1=FDw−2ρ−1v−2= FD
w2ρv2
— For Π2 we choosea= 0,c= 1 andd= 0.
e= 0 ⇒ e= 0 b+ 1 = 0 ⇒ b=−1 f = 0 ⇒ f = 0 Thus we have :
Π2=h w−1= h w
— For Π3 we choosea= 0,c= 0 andd= 1.
1 +e= 0 ⇒ e=−1
b−1−3e+f = 0 ⇒ b=−1 1 +f = 0 ⇒ f =−1
Thus we have :
Π3=w−1µρ−1v−1= µ wρv = 1
Re
Exercice 7
[7.1]To calculate the time it takes to cook a turkey of 6 kg we use the fact that it takes 50 minutes to cook one kilogram of turkey.
For a turkey of 6 kg we get : tc= 50·6 = 300 min.
If you use this linear relation, you will probably get a burned turkey...[7.2]
We can understand that the solution can’t be a simple linear relation and we can attempt to reach a more realistic solution with the Π theorem. First of all, we need to think about the physical problem.
L
Tf
T(t), χ
Figure 3 – The turkey problem
It’s also good to remind the heat equation to understand which parameters are important :
∂T
∂t =χ∂2T
∂x2 (7)
Then the parameters to solve this heat diffusivity problem are :T, Tf, tc, χ, L. whereT[K] is the temperature of the turkey (in the middle), Tf [K] the temperature of the oven,χ [m2·s−1] the diffusivity,L[m] the size of the turkey andtc [s] the cooking time.
One is tempted to take the mass and the density as parameter, but change this parameters doesn’t change the heat diffusivity problem if the diffusivityχ stay the same.
We have 5 parameters for 3 units.
Next we want to determine the non-dimensional numbers Πi=TaTfbχcLdtec,
Variable T Tf χ L tc
Unit K K m2·s−1 m s
Coefficient a b c d e
The dimensional functions are : K :a+b= 0
m : 2c+d= 0 s :−c+e= 0
There are 5 variables and 3 independent equations, thus 7-4=2 non-dimensional numbers. The first non dimen- sional number is build with the two parameter which have only the K unit a or b, and for the the second one we choose arbitrary c, d or e. At the end we get the same non dimensional number.
— For Π1 we choosea= 1 andc= 0.
a=−b, c= 0, d= 0, e= 0 Thus we have :
Π1=T Tf−1= T Tf
— For Π3 we choosea= 0 andc= 1.
b= 0, d= 0
2 +d= 0 ⇒ d=−2
−1 +e= 0 ⇒ e= 1 Thus we have :
Π2=χL−2tc=tc χ L2 This gives us the relation :
T Tf
=f tc χ
L2
[7.3]We assume that when they grow, turkeys are geometrically similar and keep the same values ofρand χ. We want to find the relationship between mass M and cooking timetc in an oven at temperatureTf.
We use the non-dimensional number 2 to donate two different turkeys :
tc χ L2
1
= tc χ
L2
2
Assumingχ1=χ2 we have :
tc1
L21
= tc2
L22 ⇒ L1
L2
= tc1
tc2
1/2
(8) Just with a measure of the turkey we can already have the time of cooking. Finally we can assume that M ∝ L3 because the density are the same and we get :
M1
M2
=L31
L32
(9)
Substituting (9) in (8) we find :
tc1
tc2
= M1
M2
2/3
(10) [7.4]Using the result of7.3and the cooking time found in 7.1we want to calculate the cooking timetc for a turkey of 6 kg. From7.1we have for a turkey of 3 kilo :tc= 50·3 = 150 min. Using (10) :
tc2=tc1
M1
M2
2/3
= 150· 6
3 2/3
= 238min
This value is much smaller than the time find with the linear law. Let’s eat an unburned turkey !
