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The velocity profile in a pipe is AP (1)A Show that the velocity is maximal at the center of the pipe.. - (r - R2) where r is
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Play al Praungvin The Pressure force = Shear force applied Means Net force on the prod will maquial iso EF-O = TV²P - 7² (P&from en u= + V del 1 ( 2²-R².. 21dL) ZN u= -(dp) (2²_R? (IL (4) V(%) - AP (2²R2) it 4NOL well op seems the But they are Cuentsu since viscosity i.e major losses at pipes comisinto Play the pressure along pipi Starts Reducing so as to compensate the fdva so min m V Y dra So dv (%) - - OP (2+)-R² = 0 ANOL 1006 - 6 New New dv(s) = fop 4 NOL) Wholetuumfue, whallitium negetingSee first of all ∆P was not used with negative sign because value of it will be negative so to appreciate that negative value of ∆P I took minus out of it now ∆P becomes positive

Now as the rule of diffrentiation, derivative of v w.r.t r equating to 0 gives us value of r where it will be 0

Now to check whether that value of r is maximum or minimum we again diffrentiate v w.r.t r and check if it's value is positive or negative

If value is negative we call that at r here in this case r=O at the center of pipe we call it maximum velocity

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