Question

Air at a temperature of 25°C flows over a smooth flat plate with a uniform velocity of 10 m/s along the plate. Find at what distance from the entrance of the plate will the boundary layer begin to transition to turbulence? (for air at 25°C: \(\rho= 1.184 \text{ kg/m}^3\) and \(\mu=1.849\times10^{-5} \text{ kg/m.s}\) A 0.854 m B 1.261 m. C 0.156 m D 0.074 m

          Air at a temperature of 25°C flows over a smooth flat plate with a uniform velocity of 10 m/s along the plate. Find at what distance from the entrance of the plate will the boundary layer begin to transition to turbulence? (for air at 25°C: \(\rho= 1.184 \text{ kg/m}^3\) and \(\mu=1.849\times10^{-5} \text{ kg/m.s}\)
A
0.854 m
B
1.261 m.
C
0.156 m
D
0.074 m
        
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Air at a temperature of 25°C flows over a smooth flat plate with a uniform velocity of 10 m/s along the plate. Find at what distance from the entrance of the plate will the boundary layer begin to transition to turbulence? (for air at 25°C: ρ= 1.184  kg/m^3 and μ=1.849×10^-5 kg/m.s
A
0.854 m
B
1.261 m.
C
0.156 m
D
0.074 m

Added by Linda B.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Air at a temperature of 25°C flows over a smooth flat plate with a uniform velocity of 10 m/s along the plate. Find at what distance from the entrance of the plate will the boundary layer begin to transition to turbulence? For air at 25°C: p = 1.184 kg/m³ and μ = 1.849x10^-5 kg/(m·s). 0.854 m 1.261 m 0.156 m 0.074 m
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Transcript

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00:01 Hello students here we have a thin plate that has the temperature inlet as 34 degrees celsius and the temperature outlet as 120 degrees celsius.
00:26 So here the velocity of the inlet is given as v equal to 20 meter per second and the length of the plate l equal to 0 .9 meter and the kinematic viscosity nu equal to 20 .92 into 10 power minus 6 meter square per second and the pr value is 0 .7.
00:53 So here we have to calculate the heat flux.
00:59 Heat flux that is q equal to k t2 minus t1 divided by l.
01:07 This is our equation number one.
01:11 So we will get t1 equal to 34 plus 273 that is 307 kelvin and t2 equal to 120 plus 273 that is 393 kelvin.
01:28 Substitute the value in the q dash equation we will get q double dash equal to 30 into 10 power minus 3 into 393 minus 307 divided by 0 .9.
01:45 So we will get q double dash equal to 2866 .67 watt per meter square.
01:53 Next we have to calculate the temperature at which the air reaches a free stream velocity...
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