312CET05 Mechanics of FluidsMaximum: - 100 Marks.PART-A(10X2=20)Answer all Questions1. What is continuum of a fluid?

Continuum is the distribution of molecules of fluid uniformly in the fluid mass assuming that there is no void (or) space between two molecules of fluid.

2. Define control volume.

The open system (or) control volume is the one which has a fixed volume in space. Here the boundaries do not change their shape but the mass and energy can cross its boundaries.Example: Car engine.

3. What do you understand by "Total Pressure"?

The total pressure on a immersed surface may be defined as the total pressure exerted by the liquid on it and is given by,P =

4. How does pressure variation in a static fluid is expressed?

The rate of increase of pressure in a vertically downward direction must be equal to the specific weight of the fluid at that point,P =

5. Define "Vena contracta".

The stream lines are straight and parallel to each other and perpendicular to the plane of the orific. This section is called Vena-Contracta from the surface. This phenomenon is called the boundary layer separation.

6. Express the momentum equation for frictionless flow.

It is based on the law of conservation of momentum (or) on the momentum principle, which states that the net force acting on a fluid mass is equal to the change in momentum of flow per unit time in that direction,F =

7. Define Hydraulic gradient line.

It is defined as the line which gives the sum of pressure head, datum head of a flowing fluid in a pipe with respect to some reference lines. Sometimes the hydraulic grade line is also known as piezometric head,

8. Define Boundary layer separation?

Along the length of the solid body at a certain point a stage may come when the boundary layer may not be able to keep sticking to the solid body if it cannot provide kinetic energy to overcome the resistance offered by the solid body. In other words, the boundary layer will be separated.

9. What is dynamic similarity in dimensional analysis?

Dynamic similarity means the similarity of forces between the model and prototype.

10. What is meant by geometric, kinemetic and dynamic similarities?

The ratio of all linear dimensions of the model and of the prototype should be equal. It is called geometric similarity. Kinematic similarity means the similarity of motion between model and prototype. Dynamic similarity means the similarity of forces between the model and prototype.

PART-B(5X16=80)11. a) State and explain the three basic systems of dimensions.

The various physical quantities used in fluid phenomenon can be expressed in terms of fundamental quantities (or) primary quantities. The fundamental quantities are mass, length; time designated by the letters M, L, T respectively. The quantities which are expressed in terms of fundamental (or) primary qualities are called derived (or) secondary qualities. The expression for a primary and derived quantity in terms of M, L, T is called three basic system (or) we can expressed as F, L, Q.

b) Find the capillary rise in a tube of 5mm diameter, when immersed in (i) Water and (ii) Mercury. The temperature of water and mercury is 200C and the surface tension of water and mercury at 200C in contact with air are 0.07358 N/m and 0.51 N/m and the density of water and mercury at 200C are 998 kg/m3 and 13550 kg/m3. The contact angle for water is 00 and for mercury is 1300.

Given: d = 5 mm = 5 x 10-3m.Capillary effect for water:h = = = 6.01 x 10-3 m = 601 mmCapillary effect for mercury:h = = = -1.97 x 10-3 m = -1.97 mmNegative sign indicates capillary depression.

12. a) Derive the equation for variable of pressure due to gravity from point to point in a static fluid

The pressure at any point in a fluid at rest is obtained by the hydrostatic law which states that the rate of increase of pressure in a vertically downward direction must be equal to the specific weight of the fluid at that point.

Consider a small fluid element as shown in figure.Let A = cross section area of elementZ = height of fluid elementp = pressure on face ABz = Distance of fluid element from free surface1) Pressure force on AB = p x AA and acting perpendicular to face AB in the downward direction.2) Pressure force on CD =, acting perpendicular to face CD, vertically upward direction.3) Weight of fluid element = Density X g X volume = x g x (A x Z).4) Pressure forces on surfaces BC and AD are equal and opposite. For equilibrium,p A - A + x g x (A x Z) = 0p A - p A - Z A + (A x Z) = 0- Z A + (A x Z) = 0 Z A = A x Z = = W Weight density of fluid = p = z = p is the atm. pressurez pressure head.

b) Explain how the resultant pressure on a curved surface immersed in a liquid can be found out.

Consider a curved surface AB, sub-merged in a static fluid as a shown in figure. Let dA is the area of a small strip at a depth of h from water surface.

Then pressure intensity on the area dA = ghPressure force dF = p x Area= gh x dAThis force dF acts normal to the surface.Hence total pressure force on the curved surface,F =. The total force in the x and y directions (i.e.) Fx and Fy are obtained by integrating dFx and dFy. Then total force on the curved surface is, F = inclination of resultant with horizontal is tan = Resolving the force dF given by equation (1) in x and y directions:dFx = dF sin = sin dFy = dF cos = cos Total forces in the x and y directions are,Fx = x = sin = dA sin (2)Fy = y = cos = dA cos (3)Figure (b) shows the enlarged area dA, from this figure (i.e.) EFGEF = dAFG = dA sin EG = dA cos Thus in equation (2), dA sin = FG = Vertical projection of the area dA and hence the expression.dA sin Represents total pressure force on the projected area of the curved surface on the vertical plane. Thus Fx = Total pressure force on the projected area of the curved surface on vertical plane dA cos = EG = horizontal projection of dA. dA cos is the volume of the liquid contained in the elementary area dA upto free surface of the liquid. dA cos is the total volume contained between the curved surface extended upto free surface. dA cos is the total weight supported by the curved surfaceFy = dA cos = Weight of liquid supported by the curved surface upto free surface of liquid. The curved surface AB is not supporting any fluid. In such cases Fy = Weight of the imaginary liquid supported by AB upto free surface of liquid. The direction of Fy will be taken in upward direction.

13. a) Derive the Bernoulli's equation with correction for pump work and friction losses. Give some applications of Bernoulli's equation.

Consider an ideal in compressible fluid through a non-uniform pipe as shown:

Consider two sections LL and MM assume that the pipe is running full and there is continuity of flow between the two sections.Let P1 Pressure at LLV1 Velocity of liquid at LLZ1 Height of LL above the datumA1 Area of pipe at LL andP2, V2, Z2, A2 corresponding values atm.Let the liquid between two sections CC and MM move the LL and MM through very small lengths dl1 and dl2.

b) State the momentum equation. How will you apply momentum equation for determining the force exerted by a flowing liquid on a pipe bend?

It is based on the law of conservation of momentum (or) on the momentum principle, which states that the not force acting on a fluid mass is equal to the change in momentum of flow per unit time in that direction. The force acting on a fluid mass m is given by the Newtons second law of motion.F = m x aWhere a is the acceleration acting in the same direction as force Fa = F = m = This is known as the momentum principle.

F dt = d(mv)Which is known as the impulse-momentum equation.Force exerted by a flowing fluid on a pipe bend:The impulse-momentum equation is used to determine the resultant force exerted by a flowing fluid on a pipe bend.

Net resultant force (FR) acting on bend = F2x + F2yAngle made by resultant force with horizontal direction is given by,tan =

14. a) Water flows at a rate of 0.0084 m3/s through 75 mm diameter smooth pipe with L = 100 m, attached to a constant level reservior, with a square edged inlet. The density and viscosity of water may be taken at standard temperature and pressure. The friction may be assumed as 0.017 and loss coefficient to be 05. The water discharge to the atmosphere.

L = 100 md = 0.075mf = 0.017Loss L/2 = 0.5

Applying Bernoullis equation, + + Z1 = + + Z2 + losses0 + 0 + H = 0 + + 0 + hf1 + hf2V2 = = = 1.9 m/shf1 = = = 16.68 mH = + 16.68 + 0.5H = 17.36 mReservoir depth = 17.36 m

b) Derive an expression for the displacement thickness in boundary layer with the necessary assumption.

Displacement Thickness (*) is defined as the distance, measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in flow rate on account of boundary layer information. It is denoted by *. It is also denoted as: The distance perpendicular to the boundary, by which the free-stream is displaced due to the formation of boundary layer.

Consider the flow of a fluid having free-stream velocity equal to U over a thin smooth plate as shown in the above diagram. At a distance x from the leading edge consider a section 1 -1. The velocity of fluid at B is zero and at C, which lies on the boundary layer is U. Thus velocity varies from zero at B to U at C. Where BC is equal to the thickness of boundary layer.Distance BC = .At section 1 -1, consider an elemental stripLet y = distance of elemental strip from the platedy = thickness of the elemental stripU = velocity of fluid at the elemental stripb = width of plateThen area of element strip,dA = b x dyMass of fluid per second flowing through elementalstrip = x velocity x area of elemental strip = U dA = u x b x dy(A) If there had been no plate then the fluid would have been flowing with a constant velocity equal to free-stream velocity (V) at the section 1 -1. Then mass of fluid per second flowing through elemental strip would have been= x velocity x area = x U x d x dy(B)As U is more than u, hence due to the presence of the plate and consequently due to the formation of the boundary layer, there will be a reduction in mass flowing per second through the element strip.This reduction in mass/sec flowing through elemental strip= mass/sec given by equation (B)mass/sec given by equation (A)= Ubdy Ubdy = b(U-u)dy Total reduction in mass of fluids flowing through BC due to plate= dy = b dy(C){if fluid is compressible}Let the plate is displaced by a distance * and velocity of flow for the distance * is equal to the free-stream velocity (i.e., U) Loss of the mass of the fluid/sec flowing through the distance *.= x Velocity x Frequence= x U x * x bEquating equation (C) and (D) we getbdy = x U x *bCancelling b from both sides, we havedy = U x *(or) * =dy = ( U is constant and can be taken inside the integral)* = .

15. a) The frictional pressure drop p for the flow of a fluid through a long straight rough pipe depends on:1) Length of pipe (L)2) Diameter of pipe (D)3) Roughness of pipe (e)4) Average fluid velocity (V)5) Density of fluid ()6) Viscosity of fluid ()Using Dimensional analysis, bring out various dimensionless groups involving the above variables.

p = f(l, d, , , e, v)The repeated variables may be choose as , v and dRoughness of pipe may be taken as average height of pipe and hence dimension of [L]i = aVb dcp = [ML-3]a [LT-1]b [Lc] [L] = [M0 L0 T0] M a = 0L -3a + b + c + 1 = 0T -b = 01 = Thus a velocity gradient is set up in the fluid near the surface of the plate. This velocity gradient develop shear resistance, which retards the fluid Thus the fluid wth a uniform free stream velocity (U) is retarded in the vicinity of the solid surface of the plate and the boundary layer region begins at the sharp leading edge. At subsequent points downsteam the leading edge, the boundary larger region increases because the retarded fluid is further retarded. This is also referred as the growth of boundary layer. Near the leading edge of the surface of the plate, where the thickness is small, the flow in the boundary layer is laminar through the main flow is turbulent This layer of fluid is said to be laminar boundary layer The length of the plate from the leading edge upto which laminar boundary layer exists is called laminar zone.

b) State and explain the conditions under which prototype behavior can be predicted from model tests.

The model is the small scale replica of the actual structure (or) Machine is called prototype. It is not necessary that the models should be similar than the prototypes, they may be larger than the prototype. Similitude is defined as the similarity between the model and its prototype in every respect, which means that the model and prototype have similar properties (or) Model and prototype are completely similar. Three types of similarities must exist between the model and prototype. They are:1) Geometric similarity2) Kinematic similarity3) Dynamic similarityGeometric similarity: Geometric similarity is said to be exist between the model and the prototype. The ratio of all corresponding linear dimension in the model and prototype are equal.Let, Lm = Length of Modelbm = Breath of ModelDm = Diameter of Model Am = Area of ModelVm = Volume of ModelLp, bp, Dp, Ap, Vp = Corresponding values of prototype = = = LWhere L is called scale ratio = = L x L = L2and = 3 = 3 = 3

Automatic Similarity: It means the similarity between model and prototype. This kinematic similarity is said to exist between the model and the prototype if the ratios of the velocity and acceleration at the corresponding points in the model and at the corresponding points in the prototype are the same.LetVp1 = Velocity of fluid at point 1 in prototypeVp2 = Velocity of fluid at point 2 in prototypeap1 = Acceleration of fluid at point 1 in prototypeap2 = Acceleration of fluid at point 2 in prototypeVm1, Vm2, am1, am2 Corresponding values at the corresponding points of fluid velocity and acceleration in the model = = VV = Velocity ratio = = aa is called as acceleration ratioAlso the direction and velocities in these model and prototype should be sameDynamic Similarity: It means the similarity of forces between the model and prototype. Thus dynamic similarity is said to exist between the model and the prototype if the ratios of the corresponding forces acting at the corresponding points are equal. Also the directions of the corresponding forces at the corresponding points should be same.Let (Fi)p = Inertia force at a point in prototype(Fv)p = Viscous force at the point in prototype(Fg)p = Gravity force at the point in prototype(Fi)m, (Fv)m, (Fg)m Corresponding values of forces at the corresponding point in model = = Also the directions of the corresponding forces at the corresponding points in the model and prototype should be same.