Chapter 3 FLUID-FLOW CONCEPTS AND BASIC EQUATIONS. The statics of fluids is almost an exact science, unit gravity force (or density) being the only quantity that must be determined experimentally. On the other hand, the nature of flow of a real fluid is very complex.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Chapter 3FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
η is not a fluid property alone; depends upon the fluid motion and the density - the eddy viscosity.
Figure 3.1 Velocity at a point in steady turbulent flow
produces the differential equations of a streamline; any continuous line that satisfies them is a streamline.
Figure 3.2 Streamlines for steady flow around a cylinder between parallel walls
In two-dimensional, incompressible steady flow around an airfoil the streamlines are drawn so that they are 10 mm apart at a great distance from the airfoil, where the velocity is 40 m/s. What is the velocity near the airfoil, where the streamlines are 7.5 mm apart?
wherem is the total mass.
m is the constant mass of the system.
∑F refers to the resultant of all external forces acting on the system, including body forces such as gravity; v is the velocity of the center of mass of the system.
Figure 3.3 System with identical control volume at time t in a velocity field
where dA is the vector representing an area element of the outflow area (Fig 3.3c).
The minus sign is needed as v · dA (or cos α) is negative for inflow, Fig. 3.3d.
This eq`n states that time rate of increase of N within a system is just equal to the time rate of increase of the property N within the control volume (fixed relative to xyz) plus the net rate of efflux of N across the control-volume boundary.
In words, the continuity equation for a control volume : the time rate of increase of mass within a control volume is just equal to the net rate of mass inflow to the control volume.
or by use of Eq. (3.3.2)
In words, the resultant force acting on a control volume is equal to the time rate of increase of linear momentum within the control volume plus the net efflux of linear momentum from the control volume.
which states that the net mass outflow from the control volume must be zero.
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
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
the continuity equation may take the form
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)
Figure 3.6Control volume for flow through series pipes
since both ρ and u are assumed to vary continuously throughout the fluid. ρuδy 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
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,
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.
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)
and their sum does equal zero, satisfying continuity.
Figure 3.8 Application of continuity and momentum to flow through a control volume in the S direction
Figure 3.10 Work done by sustained pressure
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
and y = 3.64 m.
Figure 3.11 Open-channel flow
Figure 3.12Flow through nozzle from reservoir
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.
Figure 3.14 Irreversibility in hydroelectric plant
qH is the heat added per unit mass of fluid flowing, and ws is the shaft work per unit mass of fluid flowing.
This is the energy equation for steady flow through a control volume.
Figure 3.15 Control volume with flow across control surface normal to surface
Figure 3.16 Cooling-water system
in whichT is the absolute temperature.
which is a very important thermodynamic relation; one form of the second law of thermodynamics. Although it was derived for a reversible process, since all terms are thermodynamic properties, it must also hold for irreversible-flow cases as well.
Figure 3.17 Steady-stream tube as control volume
d (losses) ≡Tds – dqH(3.9.4)
d (losses) is positive in irreversible flow, is zero in reversible flow, and can never be negative.
This is a most important form of the energy equation. In general, the losses must be determined by experimentation. It implies that some of the available energy is converted into intrinsic energy during all irreversible process.
If work is done on the fluid in the control volume, as with a pump, then ws is negative. Section 1 is upstream, and section 2 is downstream
in which each term now is energy in metre-newtons per newton, including the loss term.
The work term has been omitted but may be inserted if needed.
in which ρvδA is the mass per unit time passing δA and v2/2ρ is the kinetic energy per unit mass. By equating this to the kinetic energy per unit time passing the section, in terms of αV2/2,
By solving for α, the kinetic-energy correction factor,
Figure 3.18 Velocity distribution and average velocity
as mass per second entering and leaving is ρQ = ρ1Q1 = ρ2Q2.
in whichβ is dimensionless. Solving for β yields
In applying Eq. (3.11.1-2) care should be taken to define the control volume and the forces acting on it clearly.
Figure 3.21 Control volume with uniform inflow and outflow normal to control surface
Figure 3.23 Jet impact on pipe flowing full over partial length
Figure 3.26 Solution of linear-momentum problem by addition of vectors
(1) stationary in a flow as indicated;(2) moving to the left with velocity V1 through a stationary fluid, since the relative picture is the same.
Figure 3.28 Propeller in a fluid stream
in which A is the area swept over by the propeller blades. The propeller thrust must be equal and opposite to the force on the fluid. After substituting and simplifying,
since z1 = z2 = z3 = z4. In solving for p3 - p2, with p1 = p4,
Vabs = ∆V (Fig. 3.29) is the absolute velocity of fluid in the jet and is the mass per unit time being discharged.
Figure 3.29 Walls of flow passages through jet engines taken as inpenetrable part of control surface for plane when viewed as a steady-state problem
Figure 3.30 Steady-state flow around a boat
and by Eq. (3.11.15)
Figure 3.31 Control surface for analysis of rocket acceleration. Frame of reference has the velocity V1 of the rocket.
Figure 3.34 (a) Moving vane. (b) Vane flow viewed as stead-state problem by superposition of velocity u to the left. (c) Polar vector diagram
Figure 3.37 Vector diagram for vane doing work on a jet
Figure 3.37 Jet acting on a moving vane
Figure 3.39 Hydraulic jump in a rectangular channel
Figure 3.40 Standing wave
The relations between the variables for the hydraulic jump in a horizontal rectangular channel:
hj - losses due to the jump. Eliminating V2 in the first two equations
the plus sign has been taken before the radical (a negative y2 has no physical significance). The depths y1 and y2 : conjugate depths.
which is the cross (or vector), product of F and the position vector r of a point on the line of action of the vector from O.
Fr sin θ
which is the product of F and the shortest distance from O to the line of action of F. The sense of the final vector follows the right-hand rule.
Figure 3.42 Notation for moment of a vector
On figure the force tends to cause a counterclockwise rotation around O. If this were a right-hand screw thread turning in this direction, it would tend to come up, and so the vector is likewise directed up put of the paper. If one curls the fingers of the right hand in the direction the force tends to cause rotation, the thumb yields the direction, or sense, of the vector.
This is general moment-of-momentum equation for a control volume.
When applied to an annular control volume, in steady flow (Fig. 3.43b)
Figure 3.43 Two-dimensional flow in a centrifugal pump impeller
Figure 3.47 Plan view of sprinkler and control surface