2 CHAPTER

Basic Principles of Hydraulic Flow and Jet Theory

2.1 INTRODUCTION Hydraulic machines handle water. Machines changing fluid energy into mechanical energy are called hydraulic turbines or hydraulic motors. Machines designed to more liquids and add energy to them are called pumps. There are also special systems dealing with the transmission of hydraulic power. The hydraulic machines are classified in Fig. 2.1.

Hydraulic machines

Fluid Fluid Fluid Fluid Hydraulic Pumps Systems: Motor + P. E. turbines P. E. Pump + Piping

Work Work

1. Pelton wheel 1. Centrifugal pump 1. Hydraulic ram 2. Francis turbine 2. Reciprocating pump 2. Hydraulic accumulator 3. Propeller turbine 3. Axial flow pump 3. Hydraulic jack 4. Tubular turbine 4. Others 4. Intensifier 5. Hydraulic press

Fig. 2.1 Classification of hydraulic machines

The hydraulic mechanics are designed on the principles of fluid machanics, hydraulic flow and water jet forces.

2.2 PRINCIPLES OF FLUID MECHANICS Important principles of fluid mechanics required for the design of fluid (hydraulic) machines are summarized below. 16 Hydraulic Machines

2.2.2 Kinematics of Fluid Flow For a steady flow, the fluid properties like pressure, density, velocity, etc., do not change at a point with respect to time dV = 0, steady flow dt For uniform flow of fluid, the velocity does not change with respect to space (length of direction of flow) dV = 0, uniform flow ds The density of fluid remains constant for incompressible flow. r = constant: Incompressible flow. For laminar flow in a pipe, Reynolds number is less than 2000. For turbulent flow in a pipe, Reynolds number is more than 4000, Volumetric flow rate, & 3 Q = A1V1 = A2V2 = A3V3 [m /s] This is called continuity equation.

2.2.3 Dynamics of Fluid Flow 1. Equation of motion (Newton’s second law of motion)

Fx = m ax [N] 2. Euler’s equation of motion along a stream line, ¶p + g¶z + V¶V = 0 r 3. Bernaulli’s equation (integration of Euler’s equation of motion for a steady, ideal flow of an incompressible fluid), states that total energy consisting of pressure energy, kinetic energy and potential energy at any point of fluid is constant.

p V 2 p V 2 1 + 1 + z = 2 + 2 + z + h [m] rg 2g 1 rg 2g 2 L

p where 1 = pressure head = pressure energy per unit weight, rg V 2 1 = kinetic head = kinetic energy per unit weight 2g

z1 = datum head = datum energy per unit weight hL = loss of energy head between sections 1 and 2 Basic Principles of Hydraulic Flow and Jet Theory 17

4. Momentum equation states that net force acting on a fluid mass is equal to change in momentum per second in that direction d F = (mV) [N] dt 5. The impulse momentum equation is given as: F dt = d (mV) [N-s] 6. The force exerted by the nozzle on water

Fx = rQ(V2x – V1x) [N] 7. Moment of momentum equation states that the resultant torque on a rotating fluid is equal to the rate of change of moment of momentum

T = rQ(V2r2 – V1r1) [N-m] 8. Loss of pressure head for viscous flow through circular pipe 32muL h = [m] f rgD2 where m = coefficient of viscosity u = average fluid velocity Q = (m/s) pR2 D = diameter of pipe [m]. 9. Darcy formula. Energy loss due to friction,

4 fLV 2 h = [m] f 2gD 10. Darcy Weisbach Equation. Head loss due to friction in pipes,

4 fLV 2 h = [m] f 2gD where f = Coefficient of friction

16 = for laminar flow Re

0. 0791 = for turbulent flow. (Re)1/4 11. The velocity of water at the outlet of the nozzle, 18 Hydraulic Machines

2gH V = [m/s] 4fL a2 1+ . D A2 where H = head at inlet of pipe [m] L = length of pipe [m] D = diameter of pipe [m] a = area of nozzle outlet [m2] A = area of pipe [m2] 12. The power transmitted through nozzle,

rgQ L 4 fLV 2 O P = MH - P [kW] 1000 N 2gN Q

2.3 BASIC CONCEPTS OF HYDRAULIC FLOW The hydraulic turbines utilize the potential energy of water to produce mechanical work. The pumps are required to transfer water and other liquids. The fluid couplings, torque converters and fluid system work with special oils.

2.3.1 Fluid Characteristics The following fluid characteristics are assumed for the working substance used in hydraulic machines and systems.

1. Ideal fluid An ideal fluid is an imaginary fluid which is both incompressible and non-viscous. Such liquids do not exist in nature. However, water is assumed to be incompressible and has very low value of viscosity. Therefore, it is nearly an ideal fluid. For incompressible fluid, the water density is constant. r = const.

2. Newtonian fluid The oils used in power transmission machines and fluid systems are also incompressible. These oils are assumed to be Newtonian fluids whose viscosity is independent of velocity gradient, du t = m dy

2.3.2 Closed and Open Systems A system is a quantity of matter in space upon which attention is made in the study of changes of properties and analysis of a problem. Everything external to the Basic Principles of Hydraulic Flow and Jet Theory 19 system is called the surrounding. The boundary separating the system from the surrounding may be real (solid) or imaginary. There may be energy transfer into or out of the system. 1. Closed system. A closed system has fixed identity with fixed mass. There is no mass (fluid) transfer across the system boundary. All positive displacement machines such as reciprocating pumps, gear pumps, etc. torque converters, fluid couplings, various fluid power systems and fluid control systems will be analyzed as constant mass closed systems. 2. Open system. In an open system, the fluid crosses the boundary of the system in addition to interaction of energy between the system and the surrounding. The mass of an open system may or may not change. The identity of the fluid changes continuously. The boundary of the open system is kept fixed without any change in its volume. An open system is also referred to as control volume system. The closed boundary of a control volume is called the control surface. There is a transfer of both mass and energy across the control surface. All types of hydraulic turbines, centrifugal pumps, axial flow pumps, slurry pumps, jet pumps, pneumatic lift pumps, various supply and disposal fluid systems will be analyzed and studied as control volume systems.

2.3.3 Macroscopic Properties Matter is composed of several molecules and description of the motion of a fluid will consider the behaviour of discrete molecules which constitute the fluid. The microscopic properties can be found out by statistical summation of properties of individual molecules. Liquids have extra-strong intermolecular attractive forces. Therefore, molecular description is not required. The entire liquid mass behaves as a continuous mass. The time-averaged values of properties are sufficiently accurate and valid. The matter of the system is assumed as a continuous distribution of mass with no empty space and no conglomeration of separate molecules. The fluid machines and system will be designed and analyzed on the basis of time-averaged macroscopic properties measured with the help of instruments. This is called the concept of continuum model.

2.3.4 Conservation of Mass In non-nuclear processes, the matter can neither be created nor destroyed. The conservation of mass is inherent to the concept of a closed system. For a control volume, the rate of mass entering the system must be equal to the rate of mass leaving the system plus the rate of storage of mass in the system. If the fluid flow is steady, the rate of liquid stored is zero. Basic Principles of Hydraulic Flow and Jet Theory 21

2.3.6 Governing Equation The flow of liquid through hydraulic machines is divided in imaginary stream tubes and the behaviour of the hydraulic flow is studied with the help of the following governing equations. 1. Continuity Equation This is based on the principle of conservation of mass flow (m& ). 2. Energy Equation This is derived from the principle of conservation of energy (m& V2) 3. Equation of Motion This is based on the principle of conservation of momentum (m& V).

2.4 CONTINUITY EQUATION D Vt2 The continuity equation of flow is derived on V 2 the principle of conservation of mass. For an incompressible, uniform and steady state flow, the quantity of fluid entering at one end of stream tube and leaving at the other must D Vt1 be same provided there is no sink or source in A 2 the tube. Consider a stream tube with two arbitrary V 1 A 1 cross-sections A1 and A2, the mass of fluid enters the tube through A1 with a velocity V1 Fig. 2.2 Flow through a and leaves through A2 with velocity V2. In an stream tube increment of time Dt, the fluid particles entering A1 have moved an infinitesimal distance V1Dt. The particle at A2 have moved a distance V2Dt. By the definition of a stream tube no fluid can cross the stream tube. The mass of fluid entering the stream tube at A1 in time Dt must be exactly equal to the mass of the fluid leaving the stream tube at A2 during the same interval of time. For a steady flow, the rate of fluid stored is zero. Mathematically, the statement can be expressed as follows:

m& 1 = m& 2 & & rQ1 = rQ2

rA1 V1 Dt = rA2 V2 Dt For incompressible flow, density r is constant. & 3 \ V1 A1 = V2 A2 = VA = Constant = Q [m /s] This is called equation of continuity and the constant Q& represents the volumetric flow rate. Therefore, the volume of fluid which passes through each cross-section of the stream tube per unit time remains constant for a steady flow and incompressible fluid. 22 Hydraulic Machines

2.5 THREE-DIMENSIONAL FLOW In a three-dimensional fluid flow, the fluid properties (velocity, pressure, density, viscosity) may vary in all directions. Examples are: flow in a river, flow within a fluid machine. However, for simplicity, the fluid machines are analyzed as one-dimensional fluid flow.

Z Z X w y D C DY B u DZ v X X O Y D Y

Fig. 2.3 Three-dimensional flow

Consider an elementary cube Dx, Dy, Dz in the fluid body. The difference between the amounts of fluid which flows into and out of its faces during time Dt, must be equal to the increase in the mass which the edges enclosed. The mass of fluid of density r entering the cross-section across the face OB in time Dt = ru(Dy.Dz)Dt. The mass of fluid leaving across the face CD in time Dt = ru(Dy.Dz)Dt + ¶ (ruDyDzDt)Dx ¶x ¶ \ Gain of mass across the above faces = – ru(Dx.Dy.Dz)Dt ¶x ¶ Similarly, gain of mass across the faces BD and OC = – rv(Dx.Dy.Dz)Dt ¶y ¶ And, gain of mass across the faces BC and OD = – rw(Dx.Dy.Dz)Dt ¶z F ¶ ¶ ¶ I Total gain = – G rrru + v + wJ (Dx.Dy.Dz)Dt. ...(1) H ¶x ¶y ¶z K Mass of fluid contained at time t = r(Dx.Dy.Dz) ¶ Mass of fluid contained at time t + Dt = r(Dx.Dy.Dz) + (r.Dx.Dy.Dz)Dt ¶t 24 Hydraulic Machines

2.6 MOMENTUM EQUATION The momentum equation is based on Newton’s second law of motion which states that the time rate of change of momentum is proportional to the applied force and takes place in the direction of force. Assume a resultant force F acts on a mass m Y x v along x-axis. The change in velocity of mass m in time V dt be dV. du u \ F = m. ...(3) X dt w X or FX dt = mdu ...(4) Equation (3) is called linear momentum equation. Equation (4) is called impulse-momentum equation Z which shows that impulse of applied force (FX.dt) is Fig. 2.6(a) Velocity equal to change of momentum (m.du). components Consider a stream tube. For steady state flow, mass entering the tube is equal to mass leaving the tube.

r1 A1V1Dt = r2 A2V2Dt Momentum of fluid entering

= (r1A1V1Dt)V1 Momentum of fluid leaving

= (r2A2V2Dt)V2

D Vt2 V 2 q2 x A 2

Vt1D q V 1 1 x

A 1

Fig. 2.6(b) Momentum equation

\ Change of momentum in x-direction,

= [(r2A2V2Dt)V2 cos q2] – [(r1A1V1Dt)V1 cos q1] Basic Principles of Hydraulic Flow and Jet Theory 25

But for a steady state, equation of continuity:

(r1A1V1Dt)= (r2A2V2Dt) = rQDt \ Change of momentum in x-direction,

= rQDt (V2 cos q2 – V1 cos q1) From impulse-momentum equation,

åFx Dt = årQ Dt(V2 cos q2 – V1 cos q1) F = rQ(V cos q – V cos q ) = rQ(v – v ] x 2 2 1 1 x2 x1 Similarly, F = rQ(V sin q – V sin q ) = rQ(v – v ) y 2 2 1 1 y2 y1 Resultant force acting on the flowing fluid,

22 F = FFxy+

The angle between F and Fx can be found out as: F tan q = y Fx The flowing fluid will also exert an equal and opposite force on the boundary of the stream tube. The forces may include dynamic force due to change of momentum, static pressure, weight of fluid, drag force due to friction, gravity and inertia forces due to centrifugal force.

2.7 APPLICATIONS OF MOMENTUM EQUATION There are two types of applications of Impulse-Momentum equation. 1. Determination of forces exerted by the flowing fluid on the boundaries of flow passage of hydraulic machines due to change of momentum. The examples are: (i) Forces caused by a fluid jet striking a surface, i.e., fixed and moving blades of hydraulic machines. (ii) Jet propulsion. Propulsion of ships, boats, rockets, turbojet, ramjet, etc. (iii) Propellers. Marine propellers, helicopters. (iv) Pipe bends and reducers 2. Determination of flow characteristics due to energy loss in the flow systems. The examples are: (i) Sudden enlargement or constriction in a pipe such as orifices and mouth pieces. (ii) Hydraulic jump in open channel. Basic Principles of Hydraulic Flow and Jet Theory 27

But rate of change of momentum is equal to torque. m \ T = (V cos a r – V cos a r ) t 2 2 2 1 1 1

= rQ(V2 cos a2r2 – V1 cos a1r1)

For a circular path, r1 = r2 = r3 and a1 = a2 = a3 = 0

\ T = rQr(V2 – V1)

Power P = Tw = rQr(V2 – V1)w where w = angular velocity.

Blade velocity, Vb = wr \ P = rQ(V V – V + V ) w2 b2 w1 b1 The moment of momentum equation is applied for: 1. Analysis of flow problems in turbines and centrifugal pumps. 2. Finding torque exerted by water on sprinkler.

2.9 EULER’S FUNDAMENTAL EQUATION Euler’s fundamental equation is the equation of motion for a fluid with the following assumptions.

Assumptions 1. Fluid is non-viscous and frictional losses are zero. 2. Fluid is homogeneous and incompressible. r = constant 3. Flow is steady. 4. Flow is one-dimensional along the streamline. 5. Velocity is uniform over the section 6. Flow is continuous 7. Except gravity and pressure forces, no other forces are involved. Other forces like forces due to viscosity, turbulence and compressibility are neglected. Energy is defined as ability to do work. It manifests in various forms and can change from one form to another. 1. Gravitational potential energy or elevation energy Z in metres of liquid column. V 2 2. Kinetic energy due to mass and velocity of fluid in N-m. 2 3. Pressure energy required to move the fluid against its pressure. Pressure energy = p [N/m2]. Basic Principles of Hydraulic Flow and Jet Theory 29

From equation (5), ¶p VV¶ – dAdS – rgdAdS cos q = rdAdS . ¶s ¶S Dividing throughout by rdS dA ¶p ¶V \ – – g cos q + V = 0 r¶S ¶S dZ But cos q = dS 1 ¶p dZ dV \ + g + V = 0 r ¶S dS dS ¶p or + gdZ + VdV = 0 r ¶p or + gdZ + VdV = 0 ...(6) r Equation (6) is called Euler’s equation of motion.

2.9.1 Bernoulli’s Equation Bernoulli’s equation is also called energy equation and is obtained by integration of Euler’s equation of motion dp z + z gdZ + z VdV = constant r For incompressible flow, r = constant p V 2 \ + gZ + = constant r 2

p V 2 or + Z + = constant ...(7) rg 2g Equation (7) is called Bernoulli’s equation. p = Pressure head = Pressure energy per unit weight of fluid. rg

V 2 = Kinetic head = Kinetic energy per unit weight 2g Z = Potential head = Potential energy per unit weight. 30 Hydraulic Machines

The Bernoulli’s equation can also be written as p V 2 p V 2 1 + z + 1 = 2 + z + 2 . rg 1 2g rg 2 2g

2.10 JET THEORY When a nozzle is fitted at the end of a pipe, a jet of liquid comes out with high velocity utilizing the pressure of liquid in the pipe. Applying continuity equation to the sections 1 and 2 of the nozzle,

A1V1 = A2V2 Pipe 1 Nozzle The velocity of liquid jet, V 2 Liquid jet 1 V 2 A1 V2 = V1 A2 Fig. 2.9 Liquid jet The kinetic energy of the jet, 1 KE = m& V 2 j 2 2

Let jet velocity = Vj = V2 1 \ KE = m& V 2 j 2 j 1 1 = rA V ´ V 2 = rA V 3 [ j/s] 2 j j j 2 j j The jet power,

1 3 rAVjj P = 2 [kW] j 1000 The efficiency of nozzle, Jet power at outlet of nozzle h = n Liquid power at inlet of nozzle

1 rAV3 2´ 1000 jj V 2 = = j rgQ&H 2gH 1000 where H = total head of liquid at the inlet of nozzle. The liquid jet exerts a force on a plate placed in front of it. 32 Hydraulic Machines

p Solution F = rQ& V = rA V V = r D2 V 2 j j j j j 4 j j p F 75 I 2 = 1000 ´ G J ´ (20)2 = 1766.8 N 4 H 1000K = 1.767 kN

2.11.2 Inclined Plate

A fluid jet with a velocity Vj strikes a fixed plate inclined at an angle q with the direction of jet. The components of jet velocity,

Vjx¢ = Vj sin q

Vjy¢ = Vj cos q = 0 The jet force, & Fj = –rQVjx¢ & = –rQVj sin q. The jet force has components in X-direction and Y-direction. 2 Fjx = Fj sin q = –rQVj sin q & Fiy = –rQVj sin q cos q = 0

Plate Y

Q 1 F Nozzle V j sin q j X q q Q Fj cos q 2 Fj

Fig. 2.11 Jet force on inclined fixed plate

Y Y¢

Vj sin q

Vj cos q q X V j q X¢

Fig. 2.12 Components of jet velocity Basic Principles of Hydraulic Flow and Jet Theory 33

The volumetric discharge Q is divided into Q1 and Q2.

Q = Q1 + Q2 Q Q = (1 + cos q) 1 2 Q Q = (1 – cos q) 2 2

2.12 JET FORCE ON MOVING FLAT PLATE

The absolute velocity of jet issuing from the nozzle, = Vj The velocity of moving plate in the direction of jet = U. The jet force & Fj = rQ(Vj – U) The quantity of fluid mass striking the plate per second, Q = A(V – U) j Nozzle Plate 2 A \ Fj = rA(Vj – U) V U The work done by the jet on the plate j & Wj = Fj×U = rQ(Vj – U)U.

The kinetic energy of jet Fig. 2.13 Jet force on moving flat plate 1 1 1 KE = m& V 2 = rQ& V 2 = rAV 3 j 2 j 2 j 2 j The efficiency of the system,

Work done on the plate W h = = j Kinetic energy of jet KE j

rAV().- U2 U 2 = j = [V 2U + U3 – 2U U2] 1 V 3 j j rAV 3 j 2 j

For a given Vj, maximum efficiency. dh 2 = (V 2 + 3U2 – 4V U) = 0 3 j j dU Vj 2 ¹ 0 3 Vj Basic Principles of Hydraulic Flow and Jet Theory 35

F VVI 2GV - jjJ H j 22K h = = 0.5 or 50%. max 2 Vj

2.13 JET FORCE ON CURVED PLATE WHEN JET STRIKES TANGENTIALLY 2.13.1 Stationary Vane

The fluid jet enters the curved vane with absolute velocity V1 glides along the smooth inner surface and leaves with absolute velocity, V2, making angles a1 and a2 with the horizontal direction.

V V V 1 1 1

a1 a1 a1 Y

X

a2 a2 a2 V V V 2 2 2

(i)a2 < 90° (ii)a2 > 90° (iii)a1 = a2

Fig. 2.15 Jet force on stationary vane

The velocity components in the X-direction are: V = V cos a X1 1 1 V = V cos a . X2 2 2 The force exerted by the jet on the curved vane,

FX = rQ(V1 cos a1 – V2 cos a2)

But Q = AV1 where A = cross-section area of jet.

\ FX = rAV1(V1 cos a1 – V2 cos a2)

(i) If a2 > 90°

FX = rAV1(V1 cos a1 + V2 cos a2)

In order to get more power, a2 > 90°. 36 Hydraulic Machines

(ii) For symmetrical vane, a1 = a2 = a and V1 = V2 = V 2 2 FX = rAV (cos a + cos a) = 2rAV cos a. The hydraulic thrust in Y-direction 2 Fy = rAV (sin a – sin a) = 0

(iii) For semicircular vane, a1 = a2 = 0. 2 2 \ FX = rAV (cos 0° + cos 0°) = 2rAV 2 Fy = rAV (sin 0° – sin 0°) = 0 Semicircular vanes give maximum hydraulic force and zero hydraulic thrust.

2.13.2 Moving Vane A fluid jet of cross-sectional area A strikes a curved vane with an absolute velocity

V1. The curved vane moves with a velocity U in X-direction. The tip angles of vane are a1 and a2 at inlet and exit respectively. The velocity triangles at inlet and outlet of vane are drawn.

U 1 V W 1 1 b1 a1 U Direction of motion of vane

b2 W a2 1 V 2 U 2

Fig. 2.16 Jet force on moving vane.

For frictionless vane, the relative velocity does not change,

W1 = W2

Q = A(V1 – U)

FX = rA(V1 – U)(V1 cos a1 – V2 cos a2)

If a2 > 90°

FX = rA(V1 – U)(V1 cos a1 + V2 cos a2) In actual hydraulic machines, a series of vanes is mounted on the periphery of a wheel. 38 Hydraulic Machines

The efficiency of the system, WD.. rQ().WUU- 2(WUU- ) h = = = 1 2 E rQW2 W 2 For maximum efficiency of the system dh = 0 dU d 2()WU- U 2 \ = 0 dU W2 W 2 But ¹ 0 2 W \ U = 2 F WWI 2GW - J H 22K 1 h = = = 50% max W 2 2

QUESTION BANK NO. 2 1. How are the hydraulic machines classified? 2. Derive Euler’s equation applied to fluid machines. 3. Derive the continuity equation in Cartesian coordinates. 4. Discuss the following aspects of hydraulic flow: (a) Closed and open systems. (b) Concept of continuum. (c) Lagrangian and Eulerian methods. (d) Streamline and stream tube 5. Derive momentum and angular momentum equation. 6. What are the applications of momentum and angular momentum equations in fluid machines? 7. Establish the Bernoulli’s equation from Euler’s equation of motion. 8. Explain water jet theory and how are the following principles established. (a) Impulse principle (b) Reaction principle (c) Propulsion principle 9. Derive equation of a jet force for moving flat plate and moving curved vane. Basic Principles of Hydraulic Flow and Jet Theory 39

TUTORIAL SHEET NO. 2 1. A 25 cm diameter pipe carries lubrication oil of 0.9 specific gravity at a velocity of 3 m/s. Find the mass flow rate of oil. What will be the oil velocity at another section where diameter is reduced to 20 cm. Solution Section 1:

D1 = 25 cm = 0.25 m p p A = D2 = (0.25)2 1 4 1 4 = 0.049 m2.

V1 = 3 m/s. r = 0.9 ´ 1000 = 900 kg/m3.

m& = rA1V1 = 900 ´ 0.049 ´ 3 = 132.23 kg/s Ans. Section 2

D2 = 20 cm = 0.20 m p A = (0.20)2 = 0.314 m2 2 4 Applying equation of continuity at sections 1 and 2,

A1V1 = A2V2

AV11 00. 49 ´ 3 \ V2 = = = 4.68 m/s Ans. A2 00. 314 2. Water under a pressure of 29.43 N/cm2 (g) and velocity of 2 m/s is flowing through a 5 cm diameter pipe. Find the total head of water at a section 5 m above datum line. p 29. 43´ 104 Solution Pressure head = = = 30 m. rg 1000´ 9.81

V 2 22´ Kinetic head = = = 0.204 m 2g 29´ .81 Datum head = Z = 5 m. \ Total head = 30 + 0.204 + 5 = 35.204 m Ans 3. A jet of water 60 mm in diameter, having a velocity of 20 m/s, strikes a flat plate inclined at an angle of 30° to the axis of the jet. The plate moves at 5 m/s in the direction of the jet. Basic Principles of Hydraulic Flow and Jet Theory 41

A W

G q G C Nozzle a D V F j jn F jn B

B

Vj. The plate swings through an angle q. The plate experiences the following forces 1. Weight W of plate acting vertically through G. 2. Normal water jet force, F , jn F = rAV 2 sin a jn j where a = angle of jet with the centre line of plate. \ F = rAV 2 sin (90° – q) jn j 2 = rAVj cos q. The plate is in equilibrium under the two forces. Taking moments about point A. F (AD)= W(CG) jn AG AD = cos q CG = AG sin q. ()AG \ F = W(AG) sin q jn cosq

()AG rAV 2 cos q = W(AG) sin q. j cos q

rAV 2 \ sin q = j Proved. W