![]() ![]() SolutionThe velocity profile of a fluid flowing though aĬircular pipe is given. Discussion Note that the drag force acting on the pipe in this case is independent of the pipe diameter. R Then the friction drag force exerted by the fluid on the inner surface of the r pipe becomes 0 F = τ w A w n μ u = max (2 π R ) L = 2 n πμ u R max L u max Therefore, the drag force per unit length of the pipe is F / L = 2 n πμ u max. (Or, du /dr = - du /dy since y = R – r ). Analysis The wall shear stress is determined from its definition to be du τ w = − μ = − μ u max d ⎛ ⎜ 1 − r n ⎞ ⎟ n ⎟ = − μ u max − nr n − 1 n = n μ u max dr r = R dr ⎝ R ⎠ r = R R r = R R u ( r ) = u max (1- r n / R n ) Note that the quantity du /dr is negative in pipe flow, and the negative sign is added to the τ w relation for pipes to make shear stress in the positive (flow) direction a positive quantity. Assumptions The viscosity of the fluid is constant. The friction drag force exerted on the pipe by the fluid in the flow direction per unit length of the pipe is to be determined. Limited distribution permitted only t o ⎜ Solution The velocity profile of a fluid flowing though a circular pipe is given. Chapter 2 Properties of Fluids 2- 44 2- 18 PROPRIE T ARY MATERIAL. ![]()
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