Incompressible Viscous Flows Q. Choose the crect answer (i) The maximum velocit of a one-dimensional incompressible full developed viscous flow between two fixed parallel plates is 6m/s. The mean velocit of the flow is (a) m/s (b) m/s (c) m/s (d) 5 m/s [Ans.(c)] (ii) Loss of head due to fluid friction f stead, full developed, incompressible laminar flow through circular pipe is (a) QL hf gd (b) 8QL hf gd (c) 8QL hf gd (d) 6QL hf gd [Ans.(c)] (iii) The velocit profile of a full developed laminar flow in a straight circular pipe r is given b, where dp is the constant, R is the radius of R the pipe and r is the radial distance from the centre of the pipe. The average velocit of fluid in the pipe is (a) (b) (c) (d) 8 [Ans.(d)] (iv) F full developed, incompressible, laminar flow through circular pipes, shear stress varies (a) linearl with the radial distance from the axis (b) parabolicall with the radial distance from the axis (c) exponentiall with the radial distance from the axis (d) linearl with the radial distance from the pipe wall [Ans.(a)]
Q. Water at 0 C flows between two large stationar parallel plates which are cm apart. If the maximum velocit is m/s, determine (a) average velocit, (b) the velocit gradients at the plates and (c) the difference in pressure between two points 0 m apart. Viscosit of water at 0 C is 0.00 Pa-s. Consider the flow to be a full developed one. The flow geometr is shown in the figure below. x p p cm 0 m (a) The velocit distribution f full developed flow between two stationar hizontal parallel plates can be expressed as dp u where is channel half height and is measured from the centerline. The maximum velocit is obtained as dp umax The mean velocit is obtained from the above velocit field as uda ud 0 dp u A From the above expressions f mean velocit and maximum velocit, one can write u umax Substituting umax m/s in the above equation, we get u u 0.667 m/s max (b) Substituting the values of u and in the velocit profile, one can write 0.667 u 0000 0.0 The velocit gradients at the plates are u 0000 0000
0000 0.0 00 /s (c) The mean velocit is given b dp u dp u The pressure gradient can be expressed as dp p p L where p and p are the pressures at point and respectivel, and L is the distance between the two points. Then, the difference in pressure between two points L distance apart can be expressed as ul p p Substituting the respective values in the above equation, we have 0.000.6670 p p 00 N/m 0.0 Q. An oil of viscosit 0. Ns/m and densit 900 kg/m is flowing between two stationar hizontal parallel plates 0 mm apart. The maximum velocit is m/s. Find the mean velocit and the location at which this occurs. Also find the velocit at mm from the wall of the plates. Consider the flow to be a full developed one. The velocit distribution f full developed flow between two stationar hizontal parallel plates can be expressed as dp u where is channel half height and is measured from the centerline. The maximum velocit is obtained as dp umax The mean velocit is obtained from the above velocit field as uda ud 0 dp u A From the above expressions f mean velocit and maximum velocit, one can write u umax Substituting umax m/s in the above equation, we get u m/s Now, it is given that u u
dp dp 0.577 Substituting the value of, we get 0.5770.005 0.00885 m.885 mm The velocit distribution can be expressed in terms average velocit as u u Substituting the respective values in the above equation, we have u 0000 0.005 The velocit at mm from the plates equivalentl 5 mm 0.00 m from the centerline is u 00000.00.88 m/s Q. An oil with densit and viscosit flows between two hizontal parallel plates apart. The upper plate is moving with a unifm speed U, while the lower one is kept stationar. A constant pressure gradient of dp drives the flow in such a wa that the net flow rate across an section is zero. Find out the point where maximum velocit occurs. Also find the magnitude of maximum velocit. The flow geometr and the velocit profile across a section are shown in the figure below. U dp x The governing differential equation is du dp d Integrating the above equation twice with respect to, we get
du dp C d dp u CC where C and C are constants of integration. To evaluate the constants, we appl the boundar conditions. The boundar conditions are: at 0, u 0 at, u U. Appling the boundar conditions, we get U dp C and C 0 Therefe, the velocit distribution becomes dp U u The volume flow rate is given b Q uwd w width of the plate 0 dp w U d 0 dp U w F zero flow rate ( Q 0 ), we have dp U w 0 dp 6U u F maximum velocit, 0 dp U 0 dp dp U U dp U 6U 6 5
The maximum velocit is dp U u 6U U U U U U U 9 Q5. Find the radial location in a stead, full developed, laminar flow in a circular hizontal pipe where the local velocit is equal to the average velocit The velocit distribution f stead, full developed, laminar flow in a circular pipe is given b R dp r R where R is the radius of the pipe, r is the radial distance measured from center of the pipe and dp is the constant pressure gradient that drives the flow. Average velocit is found to be R rdr Q o A R R R dp rdr R 8 Now, it is given that R dp r R 8 o r R R r R 0.707R Q6. An oil with densit 900 kg/m and viscosit 0.6 Ns/m is flowing through a 0 cm diameter pipe. The maximum shear stress at the pipe wall is.5 N/m. Determine (a) the pressure gradient, (b) the average velocit of flow and (b) the maximum velocit of flow. (a) 6
The velocit distribution f stead, full developed, laminar flow in a circular pipe is given b R dp r R Shear stress at an point of the pipe flow is given b vr dp rz r z r z dp rz r The wall shear stress is given b dp w rz wall R Substituting the respective values, we obtain dp.5 0. dp 50 Pa/m (b) The velocit distribution f stead, full developed, laminar flow in a circular pipe can be expressed in terms of average velocit as r R The wall shear stress can also be expressed in terms of average velocit as vr r w rz wall z r R rr rr w R Substituting the respective values, we obtain 0.6v.5 z 0. 0.9 m/s (c) The maximum velocit of flow is found to be,max 0.9 0.78 m/s Q7. Consider stead, incompressible full developed flow through a hizontal concentric pipes of radii R and R R respectivel. Find an expression f the velocit distribution. Also show that the velocit distribution approaches that of pipe flow as R 0. 7
A small element of thickness dr at a radius r from the centre is considered as shown in the figure below. r dr r z R R dr r p r dp p F stead, incompressible, full developed flow in the annular space between two concentric pipes, the governing differential equation becomes d d dp r rdr dr d d r dp, r dr dr Integrating with respect to r, we get d dp r r C dr where C is a constant of integration. d dp C Now, r dr r Integrating once me with respect to r, we have dp r Cln r C where C is another constant of integration. To evaluate the constants, we appl the boundar conditions. The boundar conditions are: at r R, 0 At r R, 0 Appling the boundar conditions, we get dp R R dp R R C and C ln RR lnr R lnr R Substituting the values of C and C in the expression of velocit, we obtain dp R R r R r ln lnr R R This is the required expression f the velocit distribution f stead, incompressible, full developed flow in the annular space between two concentric pipes. 8
As R 0, the above expression f the velocit becomes dp R r This implies that the velocit distribution approaches that of pipe flow. 9