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If someone could answer this in a step by step manor that would help me greatly. (I have been offered a retake for a mitigating circumstance at my University)

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- Thread starter ra180
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- #1

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If someone could answer this in a step by step manor that would help me greatly. (I have been offered a retake for a mitigating circumstance at my University)

- #2

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- #3

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Thanks Travis can i post photos of calculations i have tried?

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Of course you can do that, and we encourage you to do that.

- #5

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Sorry guys I have been really busy here is my answer.

Assuming ISA P= 101325pa ,R= 287 J/kgk and T=288.15K

Convert knots into meters per second

(1) 20 knots= 10.29/ms, 40 knots =20.58m/s 60 knots= 30.87 m/s 100 knots =51.44 m/s

(2) Using the formula R= ρux/μ where (ρ= P/RT) (u= velocity in m/s-1) (x=0.3) and (μ= 18 x10-6)

(3) Final answer at 20 knots Reynolds number = 210088, 40 knots = 420175, 60 knots = 630263, 80 knots = 8402350, 100 knots = 1050233

Assuming ISA P= 101325pa ,R= 287 J/kgk and T=288.15K

Convert knots into meters per second

(1) 20 knots= 10.29/ms, 40 knots =20.58m/s 60 knots= 30.87 m/s 100 knots =51.44 m/s

(2) Using the formula R= ρux/μ where (ρ= P/RT) (u= velocity in m/s-1) (x=0.3) and (μ= 18 x10-6)

(3) Final answer at 20 knots Reynolds number = 210088, 40 knots = 420175, 60 knots = 630263, 80 knots = 8402350, 100 knots = 1050233

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- #6

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Looks good to me, looking at the 20 knot answer. Note also that sometimes the Reynolds number over an airfoil will read as R=V*Xc / *v* where Xc is the location along the chord length and *v* is the kinematic viscosity, which is equal to μ/ρ.

Edit: though I think you have an extra 0 in the first one, that should be 210087.

What does this tell you about the boundary layer conditions at these locations?

Edit: though I think you have an extra 0 in the first one, that should be 210087.

What does this tell you about the boundary layer conditions at these locations?

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Would you believe that is the next question should I start a new thread or continue posting here?

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