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GivenTemperature of Water T_(oo)=15^(@)CTemperature of the wall of the plate T_(s)=25^(@)CUniform Velocity of Water U_(oo)=1.5m//sDistance between parallel plates d=1mmSolutionFrom properties of water,Prandtl Number of water Pr=6.13Kinematic Viscosity of Water v=0.8937 xx10^(-6)m^(2)//sWater properties are taken at 25^(@)C as the properties would change as it get heated due to wall temperature - as worst case scenario.lAs the distance between two parallel plates d=1mm, the maximum boundary layer thickness can bedelta=(d)/(2)=(0.001)/(2)=0.0005mBoundary layer thickness for laminar flow over flat plate is given by(delta )/(x)=(5)/(sqrt(Re_(x)))=(5)/(sqrt((U_(oo)x)/(v))){:[delta=(5x)/(sqrt((U_(oo)x)/(v)))=(5sqrtx)/(sqrt((U_(oo))/(v)))],[=>quad x=((delta)/(5))^(2)(U_(oo))/(v)]:}Substituting the values, we getx=((0.0005)/(5))^(2)xx(1.5)/(0.8937 xx10^(-6))=0.01678mFrom this, Reynolds Number isRe_(x)=(U_(oo)x)/(v)=(1.5 xx0.01678)/(0.8937 xx10^(-6))=2.8164 xx10^(4)This Reynolds number is Re_(x) < 10^(6) in the laminar region only. So, our assumption of laminar region is correct.Next let us find where (at what ... See the full answer