Energy Consideration in Steady Flow Lecture - 6

Introduction Up till now we have studied the motion of liquid particle without taking into consideration any force or energy causing the flow. This lecture deals with the motion of liquids and the forces causing the flow from a viewpoint of energy considerations. The first law of thermodynamics tells us that energy can neither be created nor destroyed. But it can of course, be changed from one form to other. It follows that all forms of energy are equivalent.

Energy The energy, in general, may be defined as: “The capacity to do work.” Or “Quantity that is often understood as the ability of a physical system to do work on other physical systems.” Since work is defined as a force acting through a distance (a length of space), energy is always equivalent to the ability to exert pulls or pushes against the basic forces of nature, along a path of a certain length.

Energy

Energies of a flowing fluid Though the energy exists in many forms, yet the following are important from the subject point of view. Potential Energy Kinetic Energy Pressure Energy

1. Potential Energy Energy of an object or a system due to the position of the body or the arrangement of the particles of the system.

PE in terms of Fluid “It is the energy possessed by a liquid particle by virtue of its position.” A fluid particle of weight W situated a distance z above datum possesses a potential energy of Wz. Thus its potential energy per unit weight is z, measured in units ft.lb/lb=ft or N.m/N=m. The particle's potential energy per unit mass is gz, measured in units of ft2/sec2 or m2/s2; ts potential energy per unit volume is pgz, measured in units of lb/ft2 or N/m2.

2. Kinetic Energy Anything that is moving contains kinetic energy. The kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. E.g. Wind, waves, falling rocks

Energy Conversion

KE in terms of Fluid A body of mass m when moving at a velocity V possesses a kinetic energy, KE = ½ m V 2. Thus if a fluid were flowing with all particles moving at the same velocity, its kinetic energy would also be ½ m V2; for unit weight of the fluid we can write this as: where represents the volume of the fluid mass. In BG units we express V2/2g in ft-lb/lb = ft and in SI units as N .m/N = m.

KE in terms of Fluid Similarly, The units of V2/2 of course are ft2/sec2 in BG units or m2/s2 in SI units. The units of pV2/2 are lb/ft2 or N/m2, which are units of pressure.

True KE In most situations the velocities of the different fluid particles crossing a section are not the same, so it is necessary to integrate all portions of the stream to obtain the true value of the kinetic energy. It is convenient to express the true value in terms of the mean velocity V and a factor a (alpha), known as the kinetic-energy correction factor or coriolis coefficient. Then,

Correction Factor In order to obtain an expression for alpha, consider the case where the axial components of the velocity vary across a section, as in Fig. If u is the local axial velocity component at a point, the mass flow per unit of time through an elementary area dA is: r dQ = r udA

Correction Factor Thus the true flow of kinetic energy per unit of time across area dA is: 1/2(r udA)u2 = 1/2 r u3 dA The weight rate of flow through dA is g dQ = rg udA. Thus, for the entire section, Comparing Eq. (5.3) with Eqs. (5.2) and (4.3 ), we get,

Correction Factor and we get the same result if we use True KE/Mass, where mass flow rate is: or if we use True KE/Volume, where the volume flow rate is. As the average of cubes is always greater than the cube of the average, the value of a will always be more than 1. The greater the variation in velocity across the section, the larger will be the value of a. For laminar flow in a circular pipe, a=2 (Problem 5.1), for turbulent flow in pipes, ranges from 1.01 to 1.15, but it is usually between 1.03 and 1.06.

Correction Factor In some instances it is very desirable to use the proper value of a, but in most cases the error made in neglecting its divergence from 1.0 is negligible. As precise values of a are seldom known, it is customary in the case of turbulent flow to assume that = 1, i.e., that the kinetic energy is V2/2g per unit weight of fluid, measured in units of ft.lb/lb=ft or N.m/N=m. In laminar flow the velocity is usually so small that the kinetic energy per unit weight of fluid is negligible.

3. Pressure Energy It is the energy, possessed by a liquid particle, by virtue of its existing pressure. A particle of fluid has energy due to its pressure above datum, most usually its pressure above atmospheric, although we normally do not refer to this as pressure energy. From Eq. (3.4) this pressure is p = gh, and so the depth of liquid that would produce this pressure, or the "pressure head", is h=p/g.

4. Internal Energy Internal energy is stored energy that is associated with the molecular, or internal state of matter; it may be stored in many forms, including thermal, nuclear, chemical, and electrostatic.

Problem: In Laminar flow through a circular pipe the velocity profile is a parabola (Fig), the equation of which is: Where U is the velocity at any radius r, and Um is the maximum velocity in the center of the pipe where r = 0 and ro is the radius to the wall of the pipe. Find a ?

Solution:

EQUATION FOR STEADY MOTION OF AN IDEAL FLUID ALONG A STREAMLINE, AND BERNOULLI'S THEOREM Referring to Fig., let us consider frictionless steady flow of an ideal fluid along the streamline. We shall consider the forces acting in the direction of the streamline on a small element of the fluid in the stream tube, and we shall apply Newton's second law, that is F = ma. The cross-sectional area of the element at right angles to the streamline may have any shape and varies from A to A + dA. Recalling that in steady flow the velocity does not vary at a point (local acceleration = 0), but that it may vary with position (convective acceleration 0).

Bernoulli's Theorem: The mass of the fluid element is m = r ds(A + 1/2dA) = r dsA when we neglect second order terms. The forces tending to accelerate or decelerate this mass along s are: (a) the pressure forces: (b) the weight component in the direction of motion: Applying ƐF = ma along the streamline, we get,

Bernoulli's Theorem: Dividing by the volume dsA, This states that the pressure gradient along the streamline combined with the weight component in that direction causes the acceleration a of the element. Recalling that a = V(dV/ds) for steady flow (Equation 4.24), we get: Multiplying by ds/p and rearranging,

Bernoulli's Theorem: We commonly refer to this equation as the one-dimensional Euler3 equation, because Leonhard Euler (1707-1783), a Swiss mathematician, first derived it in about 1750. It applies to both compressible and incompressible flow, since the variation of p over the elemental length ds is small. Dividing through by g, we can also express Eq. (5.5) as:

Compressible Fluid: For the case of a compressible fluid, since Sp. weight (g) is not constant, we must introduce an equation relating g (or r) to p and T before integrating Eq. (5.5) or (5.6).

Incompressible Fluid: For the case of an incompressible fluid (g=constant), we can integrate Eq. (5.6) to give: Energy per unit weight: We know this famous equation as Bernoulli's theorem, in honor of Daniel Bernoulli (1700-1782), the Swiss physicist who presented this theorem in 1738. If we multiply each term first by g and then by r, we obtain the alternate forms:

Incompressible Fluid: Energy per unit mass: Energy per unit volume: The constant (of integration) is known as the Bernoulli constant.

Assumptions: 1. It assumes viscous (friction) effects are negligible 2. It assumes the flow is steady 3. The equation applies along a streamline 4. It assumes the fluid to be incompressible 5. It assumes no energy is added to or removed from the fluid along the streamline

Problem: Glycerin (specific gravity 1.26) in a processing plant flows in a pipe at a rate of 700 L/s. At a point where the pipe diameter is 600 mm, the pressure is 300 kPa. Find the pressure at a second point where the pipe diameter is 300 mm if the second point is 1.0 m lower than the first point, neglect the head loss.

Solution:

Assignment: Assume frictionless flow in a long, horizontal, conical pipe, which has a diameter of 3.6 ft at entrance and 2.4 ft at exit. The pressure head at the smaller end is 15 ft of water. If water flows through this cone at the rate of 95 cfs, find the velocities at the two ends and the pressure head at the larger end. Assume the flow to be frictionless in the siphon shown in Fig., where a = 3 ft, b = 12 ft. Find the rate of discharge in cfs and the pressure head at B if the pipe has a uniform diameter of 3 in. Application of Bernoulli’s Theorem in real world…