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Class work 1 At normal pressure, the meters per second.. find the range of critical distance watcr at a temperature of 20 degrees passes through a smooth surface at a uniform velocity of s Xc value for laminar boundary layer to turbulent boundary layer P-998.2kg/m 100.5x10s Pa s 2. At atmospheric pressure, the air at a temperature of 30 degrees passes through the surface of a smooth plate at the rate of 10m/s.. Re, is 3.2X 10, determine by calculation that it is laminar boundary layer or turbulent boundary layer at the points 0.4m and 0.8m from the leading edge of the distance plate? Find the laminar flow boundary layer thickness. ρ-1.165Kg/mr, μ-1.86 x 10/a . s
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Answer #1

1) velocity v = 5 m/s

Turbulent flow appears at a critical NRe ( Reynolds no. )

b/w 105 and 3 x 106 if parallel flow along a plate

For NRe = 105

v v e / Mu = 105

x x 5 x 998.2 / 100.5 x 10-5 = 105

x = 0.020136 m.

For NRe = 3 x 106

x x 5 x 998.2 / 100.5 x 10-5 =  3 x 106

x = 0.60408 m.

Range of critical distance is 0.020136 to 0.60408 m.

2 ) velocity v = 10 m / s ,  NRe = 3.2 x 105

At a distance 0.4 m from the leading edge

NRe = x v e / Mu = 0.4 x 10 x 1.165 / 1.86 x 10-5

= 2.505 x 105

At 0.8 m from the leading edge

10, and-3X1 ra NRe 10 100 *10 :0.020136 m ず--:-3x1 100.5x1o 86X10 2 , 505 x 105 Ive0.8 xlox 1 1 1-86 x165 above caludated appearu nbulaut Houw n edge Boundovuy L 10x 1-165 ; 0,000 3丨勿

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