205 kgm3 since by definition u 0062 kgm s2 0062 pa ans

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Unformatted text preview: pe). The average velocity V is found by integrating the logarithmic-law velocity distribution Part (c) Q A V 1 R2 R u 2 r dr (2) 0 Introducing u u*[(1/ ) ln (yu*/ ) B] from Eq. (6.21) and noting that y R carry out the integration of Eq. (2), which is rather laborious. The final result is V 0.835u0 4.17 m/s r, we can Ans. (c) We shall not bother showing the integration here because it is all worked out and a very neat formula is given in Eqs. (6.49) and (6.59). Notice that we started from almost nothing (the pipe diameter and the centerline velocity) and found the answers without solving the differential equations of continuity and momentum. We just used the logarithmic law, Eq. (6.21), which makes the differential equations unnecessary for pipe flow. This is a powerful technique, but you should remember that all we are doing is using an experimental velocity correlation to approximate the actual solution to the problem. We should check the Reynolds number to ensure turbulent flow Red Vd (4.17 m/s)(0.14 m) 1.51 10 5 m2/s 38,700 Since this is greater than 4000, the flow is definitely turbulent. 6.4 Flow in a Circular Pipe As our first example...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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