3.4 Continuity Equation The use of Eq. (3.3.1) is developed in this section. First, consider steady flow through a portion of the stream tube of Fig. 3.4. The control volume comprises the walls of the stream tube between sections 1 and 2, plus the end areas of sections 1 and 2. Because the flow is steady, the first term of Eq. (3.3.1) is zero; hence (3.4.1) which states that the net mass outflow from the control volume must be zero. Since there is no flow through the wall of the stream tube, (3.4.2) is the continuity equation applied to two sections along a stream tube in steady flow.

Figure 3.4 Steady flow through a stream tube Figure 3.5 Collection of stream tubes between fixed boundaries

For a collection of stream tubes (Fig. 3 For a collection of stream tubes (Fig. 3.5), ρ1 is the average density at section 1 and ρ2 the average density at section 2, (3.4.3) in which V1, V2 represent average velocities over the cross sections and m is the rate of mass flow. The average velocity over a cross section is given by If the discharge Q (also called volumetric flow rate, or flow) is defined as (3.4.4) the continuity equation may take the form (3.4.5) For incompressible, steady flow (3.4.6) For constant-density flow, steady or unsteady, Eq. (3.3.1) becomes (3.4.7)

Example 3.2 At section 1 of a pipe system carrying water (Fig. 3.6) the velocity is 3.0 m/s and the diameter is 2.0 m. At section 2 the diameter is 3.0 m. Find the discharge and the velocity at section 2. From Eq. (3.4.6) and

Figure 3.6 Control volume for flow through series pipes

For three-dimensional cartesian coordinates, Eq. (3. 3 For three-dimensional cartesian coordinates,  Eq. (3.3.1) is applied to the control-volume element δx δy δz (Fig. 3.7) with center at (x, y, z), where the velocity components in the x, y, z directions are u, v, w, respectively, and ρ is the density. Consider first the flux through the pair of faces normal to the x direction. On the right-hand lace the flux outward is since both ρ and u are assumed to vary continuously throughout the fluid. ρu δy δz is the mass flux through the center face normal to the x axis. The second term is the rate of increase of mass flux, with respect to x multiplied by the distance δx/2 to the right-hand face. On the left-hand lace The net flux out through these two faces is

Figure 3.7 Control volume for derivation of three-dimensional continuity equation in cartesian co-ordinates

The other two directions yield similar expressions;  the net mass outflow is which takes the place of the right-hand pan of Eq. (3.3.1). The left-hand part of Eq. (3.3.1) becomes, for an element, When these two expressions are used in Eq. (3.3.1), after dividing through by the volume element and taking the limit as δx δy δz approaches zero, the continuity equation at a point becomes (3.4.8) For incompressible flow it simplifies to (3.4.9)

In vector notation: with the velocity vector (3.4.10) (3.4.11) then Equation (3.4.8) becomes (3.4.12) Eq. (3.4.9) becomes (3.4.13) The dot product is called the divergence of the velocity vector q - it is the net volume efflux per unit volume at a point and must be zero for incompressible flow. Two-dimensional flow, generally assumed to be in planes parallel to the xy plane, w = 0, and ∂/∂z = 0, which reduces the three-dimensional equations given for continuity.

Example 3.3 The velocity distribution for a two-dimensional incompressible flow is given by Show that it satisfies continuity. In two dimensions the continuity equation is, from Eq. (3.4.9) Then and their sum does equal zero, satisfying continuity.

3.5 Euler's Equation of Motion Along a Streamline In addition to the continuity equation, other general controlling equations - Euler's equation. In this section Euler's equation is derived in differential form. The first law of thermodynamics is then developed for steady flow, and some of the interrelations of the equations are explored, including an introduction to the second law of thermodynamics. Here it is restricted to flow along a streamline. Two derivations of Euler's equation of motion are presented; The first one is developed by use of the control volume for a small cylindrical element of fluid with axis along a streamline. This approach to a differential equation usually requires both the linear-momentum and the continuity equations to be utilized. The second approach uses Eq. (2.2.5), which is Newton's second law of motion in the form force equals mass times acceleration.

In Fig. 3.8 a prismatic control volume of very small size, with cross-sectional area δA and length δs, is selected. Fluid velocity is along the streamline s. By assuming that the viscosity is zero (the flow is frictionless), the only forces acting on the control volume in the x direction are the end forces and the gravity force. The momentum equation [Eq․(3.3.8)] is applied to the control volume for the s component. (3.5.1) The forces acting are as follows, since as s increases, the vertical coordinate increases in such a manner that cos θ = ∂z/∂s. (3.5.2) The net efflux of s momentum must consider flow through the cylindrical surface mt, as well as flow through the end faces (Fig. 3.8c). (3.5.3)

Figure 3.8 Application of continuity and momentum to flow through a control volume in the S direction

To determine the value of mt, the continuity equation (3. 3 To determine the value of mt, the continuity equation (3.3.1) is applied to the control volume (Fig. 3.8d). (3.5.4) (3.5.5) Substituting Eqs. (3.5.2) and Eq. (3.5.5) into equation (3.5.1) (3.5.6) Two assumptions have been made: (1) that the flow is along a streamline and (2) that the flow is frictionless. If the flow is also steady, Eq․(3.5.6) (3.5.7) Now s is the only independent variable, and total differentials may replace the partials, (3.5.8)

3.6 The Bernoulli Equation Integration of equation (3.5.8) for constant density yields the Bernoulli equation (3.6.1) The constant of integration (the Bernoulli constant) varies from one streamline to another but remains constant along a streamline in steady, frictionless, incompressible flow. Each term has the dimensions of the units metre-newtons per kilogram: Therefore, Eq. (3.6.1) is energy per unit mass. When it is divided by g, (3.6.2) Multiplying equation (3.6.1) by ρ gives (3.6.3)

Each of the terms of Bernoulli's equation may be interpreted as a form of energy. In Eq. (3.6.1) the first term is potential energy per unit mass. With refence to Fig. 3.9, the work needed to lift W newtons a distance z metres is Wz. The mass of W newtons is W/g kg  the potential energy, in metre-newtons per kilogram, is The next term, v2/2, is interpreted as follows. Kinetic energy of a particle of mass is δm v2/2; to place this on a unit mass basis, divide by δm  v2/2 is metre-newtons per kilogram kinetic energy.

The last term, p/ρ is the flow work or flow energy per unit mass The last term, p/ρ is the flow work or flow energy per unit mass. Flow work is net work done by the fluid element on its surroundings while it is flowing. In Fig. 3.10, imagine a turbine consisting of a vaned unit that rotates as fluid passes through it, exerting a torque on its shaft. For a small rotation the pressure drop across a vane times the exposed area of vane is a force on the rotor. When multiplied by the distance from center of force to axis of the rotor, a torque is obtained. Elemental work done is p δA ds by ρ δA ds units of mass of flowing fluid  the work per unit mass is p/ρ. The three energy terms in Eq (3.6.1) are referred to as available energy. By applying Eq. (3.6.2) to two points on a streamline, (3.6.4)

Figure 3.9 Potential energy Figure 3.10 Work done by sustained pressure

Example 3.4 Water is flowing in an open channel (Fig. 3.11) at a depth of 2 m and a velocity of 3 m/s. It then flows down a chute into another channel where the depth is 1 m and the velocity is 10 m/s. Assuming frictionless flow, determine the difference in elevation of the channel floors. The velocities are assumed to be uniform over the cross sections, and the pressures hydrostatic. The points 1 and 2 may be selected on the free surface, as shown, or they could be selected at other depths. If the difference in elevation of floors is y, Bernoulli's equation is Thus and y = 3.64 m.

Figure 3.11 Open-channel flow

Modification of Assumptions Underlying Bernoulli's Equation Under special conditions each of the four assumptions underlying Bernoulli's equation may be waived. When all streamlines originate from a reservoir, where the energy content is everywhere the same, the constant of integration does not change from one streamline to another and points 1 and 2 for application of Bernoulli's equation may be selected arbitrarily, i.e., not necessarily on the same streamline. In the flow of a gas, as in a ventilation system, where the change in pressure is only a small fraction (a few percent) of the absolute pressure, the gas may be considered incompressible. Equation (3.6.4) may be applied, with an average unit gravity force γ.

For unsteady flow with gradually changing conditions, e. g For unsteady flow with gradually changing conditions, e.g., emptying a reservoir, Bernoulli's equation may be applied without appreciable error. Bernoulli's equation is of use in analyzing real-fluid cases by first neglecting viscous shear to obtain theoretical results. The resulting equation may then be modified by a coefficient, determined by experiment, which corrects the theoretical equation so that it conforms to the actual physical case.

Example 3.5 (a) Determine the velocity or efflux from the nozzle in the wall of the reservoir of Fig. 3.12. (b) Find the discharge through the nozzle. Solution: (a) The jet issues as a cylinder with atmospheric pressure around its periphery. The pressure along its centerline is at atmospheric pressure for all practical purposes. Bernoulli's equation is applied between a point on the water surface and a point downstream from the nozzle, With the pressure datum as local atmospheric pressure, p1 = p2 = 0; with the elevation datum through point 2, z2 = 0, z1 = H.

The velocity on the surface of the reservoir is zero (practically); hence. and which states that the speed of efflux is equal to the speed of free fall from the surface of the reservoir. This is known as Torricelli's theorem. (b) The discharge Q is the product of velocity of efflux and area of stream,

Figure 3.12 Flow through nozzle from reservoir

3.7 Reversibibility, Irreversibility, and Losses A process may be defined as the path of the succession of states through which the system passes, such as the changes in velocity, elevation, pressure, density, temperature, etc. Example of a process is the expansion of air in a cylinder as the piston moves out and heat is transferred through the walls. Normally, the process causes some change in the surroundings, e.g., displacing it or transferring heat to or from its boundaries. When a process can be made to take place in such a manner that it can be reversed, i.e., made to return to its original state without a final change in either the system or its surroundings – reversible process. Actual flow of a real fluid: viscous friction, coulomb friction, unrestrained expansion, hysteresis, etc. prohibit the process from being reversible. However, it is an ideal to be strived for in design processes, and their efficiency is usually defined in terms of their nearness to reversibility.

When a certain process has a sole effect upon its surroundings that is equivalent to the raising of  mass, it is said to have done work on its surroundings. Any actual process is irreversible. The difference between the amount of work a substance can do by changing from one state to another state along a path reversibly and the actual work it produces for the same path is the irreversibility of the process - may be defined in terms of work per unit mass or work per unit time. Under certain conditions the irreversibility of a process is referred to as its lost work, i.e., the loss of ability to do work because of friction and other causes. In the Bernoulli equation (3.6.4), in which all losses are neglected, all terms are available-energy term, or mechanical-energy terms, in that they are directly able to do work by virtue of potential energy, kinetic energy, or sustained pressure.

Example 3.6 A hydroelectric plant (Fig. 3.14) has a difference in elevation from head water to tail water of H = 50 m and a flow Q = 5 m3/s of water through the turbine. The turbine shaft rotates at 180 rpm, and the torque in the shaft is measured to be T = 1.16×105 N · m. Output of the generator is 2100 kW. Determine (a) the reversible power for the system, (b) the irreversibility, or losses, in the system, (c) the losses and the efficiency in the turbine and in the generator. Solution (a) The potential energy of the water is 50 m · N/N. Hence, for perfect conversion the reversible power is (b) The irreversibility, or lost power, in the system is the difference between the power into and out of the system, or

(c) The rate of work by the turbine is the product of the shaft torque and the rotational speed: The irreversibility through the turbine is then 2451.5 – 2186.5 = 265.0 kW, or, when expressed as lost work per unit weight of fluid flowing, The generator power loss is 2186.5 – 2100 = 86.5 kW, or Efficiency of the turbine η1 is And efficiency of the generator ηg is

Figure 3.14 Irreversibility in hydroelectric plant