Question

4. Consider fully developed Couette flow-flow between two infinite parallel plates separated by distance h, with the top plate moving and the bottom plate stationary. The flow is steady, incompressible, and two-dimensional in the xy- plane. Use the method of repeating variables to generate a dimensionless relationship for the x component of fluid velocity u as a function of fluid viscosity , top plate speed V, distance h, fluid density p, and distance y Show all your work. Hint: u = f (u, V, h, p, y) Choose repeating parameters u, h, and y PVh = Reynolds number = Re

Answer 1

Answer:

Given data

There are spre parameters in this problem Therefore, n=6 The five parameters are, . a fluid velocity : y distance ry -Fluid density & P Fluid viscosity - u Top plate speed - u Distance =h Here the parameter u is a function of other five parameters 13 Piu fig List of relevant parameters uefCP, M, Vi hiy) Nowe we have to write the primary dimension of each paramedlem primary dimension of velocity, duy = 47

primary dimension of density, [p] = kwotame] primany dimension of viscosity ; {u} = (m? primary dimension of distance y, ty) = {L'} primary .dimension of distance hid =d17 primary dimension of top plate spoed, du? -457 The number of primary dirrensions represented in the problem is 3, (milit) first guess, j is first set equal to three j=3 of this value of j is correct the number of ti's predicted by Buckingham pon theoren is, Number of expected ti's es ken-j 63 : k=3 Sinco here j = 3 we have to choose three repeating parameters. Based on the guidelines for seleching repeating choose hip anda. parometers we

Now, we combine 3 ? these repeating parameters with the othoy remaining parameters to create ni's First of all a dependent T is generated using the dependent variable u. Dependent Ti = uhalph va u Now, primary dimension of It, is, 2013 = dmzoly Now, primary dimension whapbiver is, zuhai poluon} = (2'4 ")43°(me)"Çl'&"*} Now , equation becomes dºlº20} = f('t+1(0)61(m2)*(e't"))} Now, equalling the exponents of each primary dimension independently for solving an be and a mass : domo = {mbiz ... Ozbi . . bizo Mass}

Time: 100} = {€47-} O- -|- → 43-4 Length: {}=i'e' +361, 017 -.0 14-36179 o=xta-ot -> 01= 0. . Now, substituing the values of ar bi and ca in equation ④ Ti = uhopoul Therefore, th=4 - a) Now, a independent t is generated using the independent variable ii . independent It, Tauhan ba (2) . Now primary dimension of the is, {12}=almolot? Now, pinany dimension of wharbor ya is,

a whopbwcy} - {(m*') (L')"(m13) by L'{-1967 now, equation (2) becomes, {m^cºte} = come!c*)__)©4m2:362(t2-99} now, equate the exponent of each primary dimensian independenty to solve for azı bi and G. mass: {no} = {m' mb] Ost be - bazwi 1to3 = +45c72 Os --G (93-1 Length: 42°3=4142024622603 o=-Haz-3bztca 0:-/+azt3-1 = -2+3 + Oy = -1 Now, substitute the values of aribe and as in equation ③ Tg = whiptul We have

The above ratio is the rechbrocal of Reynolds number hyper Therefore, Itaa hev The = Re - Now , a independent. I is generakd using the independent variable y independent Ti. Tz = ghaspbyes - 6 Now , Primany dimension of 113 is, fitzy - Homologo3 Now, primary dimension of ghasphzig is, yhazpbywe}} ={[[C]ja: (ml") by 14433} now, equaton (3) becomes, 20194079) 14)')"}(m12) 63c0'+4393 Now, equating the exporents of each primary dimension independently for solving azı b3 and 3. mass: {m} = (mb)} osib - bzzo

Length: Time! {to } = {1-3} o = -63 350 4203 = {1!2*31.36197 o-l-taz -3b₃ + c3 o - Hazlo to az:-) Now substituting the value of azı by and a in equ ③ Tz = ght povo 7 - Ylh , Tig = Y/- © from equations @ 6 and @ we can make a relation Yv-f{re, din]