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1 Dynamics of Viscous Flows (Lectures 8 to ) Q. Choose the correct answer (i) The average velocity of a one-dimensional incompressible fully developed viscous flow between two fixed parallel plates is m/s. The maximum velocity of the flow is (a) m/s (b) m/s (c) m/s (d) 5 m/s [Ans.(b)] (ii) Which of the following statements is correct for a fully developed pipe flow? (a) pressure gradient is greater than the wall shear stress (b) the velocity profile is changing continuously (c) inertia force balances the wall shear stress (d) pressure gradient balances the wall shear stress only and has a constant value. [Ans.(d)] (iii) The velocity profile of a fully developed laminar flow in a straight circular pipe r is given by vz, where dp is the constant, R is the radius of dz R dz the pipe and r is the radial distance from the centre of the pipe. The maximum velocity of fluid in the pipe is (a) dz (iv) (b) (c) dz dz (d) 8 dz [Ans.(c)] For fully developed, laminar flow through circular pipes Darcy friction factor (f) is given by (a) 6 f (b) 6 f (c) f (d) f [Ans.(b)]

2 Q. Water at C flows between two large parallel plates which are mm apart. If the maximum velocity is.5 m/s, determine the average velocity, pressure drop per unit length and the shear stress at walls of the plate. Viscosity of water at C is. Pa-s. Solution The average velocity is given by u u.5 m/s max The pressure drop in a two dimensional straight channel is given by dp umax, where half the channel height Putting the respective values, we get dp..5. N/m kn/m per m The wall shearing stress in a channel flow is given by u dp τ w. ( ) N/m y wall Q. Two viscous, incompressible, immiscible fluids of same density ( ρ ) but different viscosities (viscosity of the lower fluid layer and that of the upper fluid layer < ) flow in separate layers between parallel boundaries located at y ± as shown in the figure below. Thickness of each fluid layer is identical and their interface is flat. The flow is driven by a constant favourable pressure gradient of dp. Derive expressions for the velocity profiles in the fluid layers. Assume the flow to be stea and the plates to be of infinitely large width. y Stationary plate ρ, x ρ, Stationary plate Fig.

3 Solution Large lateral width of the plate renders the basic flow consideration to be two dimensional, for which the continuity equation under incompressible flow conditions reads: u v + x y u For fully developed flow, x v ence, y v vy ( ) Since v at y ± as a result of the no-penetration at the walls, v is identically equal to zero for all y, i.e., v at y Now, considering x-momentum equation, we have, for stea flow, u u dp u u ρ u + v x y + + x y Under assumptions mentioned as above, the above equation becomes d u dp constant Function of y only Function of x only For y du dp Integrating the above equation, we have du dp y C + dp y u Cy C + + () where C are constants of integration. For y du dp Integrating the above equation, we have du dp y C + dp y u Cy C + + () where C are constants of integration.

4 Eqs () and () are subjected to the following boundary conditions At y, u, At y, u, At y, u u(continuity of flow velocity) du du At y, (continuity of shear stress) From the boundary conditions at y, u, and at y, u, we get C dp C dp C C du du From the boundary condition at y,, we have C C C C From the boundary condition at y, u u, one can write C C dp dp C C ( ) dp + dp + C dp ( ) C + dp C ( ) ( ) ( + ) Substituting the values of C, C, C in the Eqs () and (), we have dp ( ) ( + ) ( ) ( + ) ( ) u y + y+ dp ( ) u y + y

5 Q. Consider stea, incompressible, fully developed flow between parallel plates distance apart. The upper plate moves to the right at a uniform speed ofu, while the lower plate is stationary. Determine the pressure gradient required to produce zero net flow at a section. Solution U y x The governing equation for the motion of flow can be written as d u dp Integrating the above equation twice with respect to y, we obtain du dp y + C dp u y + Cy + C where C are constants of integration. To evaluate the constants, we apply the boundary conditions. The boundary conditions are: at y, u at y, u U Applying the boundary conditions, we get U dp C Therefore, we obtain dp U u ( y y ) y + The volume flow rate is given by Q uw [ w width of the plate] 5

6 dp y w ( y y) + U dp U w + For zero flow rate ( Q ), we have dp U w + dp 6 U or Q5. An oil of density 8 kg/m and viscosity. Ns/m is flowing through a pipe of diameter cm at the rate of 5 litre/s. Find the head lost due to friction and power required to maintain the flow for a length of 5m. Solution Average velocity of flow is Q Q.5 u.78 m/s A π π D. ynolds number is ρud The flow is laminar since <. ead lost due to friction is given by QL hf πρ gd m π The power required to maintain the flow is P ρ gqh f W 6

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