ME 259 Fluid Mechanics for Electrical Students Pharos University ME 259 Fluid Mechanics for Electrical Students Application of Bernoulli Equation INTI COLLEGE MALAYSIA

Euler’s equation of motion Bernoulli equation

INTRODUCTION The three equations commonly used in fluid mechanics are: the mass, Bernoulli, and energy equations. The mass equation is an expression of the conservation of mass principle. The Bernoulli equation: Conservation of kinetic, potential, and flow energies ( viscous forces are negligible.)

Fluid Motion Two ways to describe fluid motion Lagrangian Eularian Follow particles around Eularian Watch fluid pass by a point or an entire region Flow pattern Streamlines – velocity is tangent to them

STEADY AND UNSTEADY FLOW Steady flow: the flow in which conditions at any point do not change with time is called steady flow. Then, …etc. Unsteady flow: the flow in which conditions at any point change with time, is called unsteady flow. Then, …etc.

UNIFORM AND NON-UNIFORM The flow in which the conditions at all points are the same at the same instant is uniform flow. The flow in which the conditions vary from point to point at the same instant is non-uniform flow.

ACCELERATION Acceleration = rate of change of velocity Components: Normal – changing direction Tangential – changing speed

ACCELERATION Cartesian coordinates In steady flow ∂u/∂t = 0 , local acceleration is zero. In unsteady flow ∂u/∂t ≠ 0 ; local acceleration Occurs. Other terms u ∂u/∂x, v ∂u/∂y,.. are called convective accelerations. Convective acceleration Occurs when the velocity varies with position. Uniform flow: convective acceleration = 0 Non-uniform flow: convective acceleration ≠0 Convective Local Fluid Mechanics I

Example Valve at C is opened slowly The flow at B is non uniform The flow at A is uniform Fluid Mechanics I

Laminar vs Turbulent Flow

Flow Rate Volume rate of flow Mass flow rate Constant velocity over cross-section Variable velocity Mass flow rate

Flow Rate Only x-direction component of velocity (u) contributes to flow through cross-section

CV Inflow & Outflow Area vector always points outward from CV

Examples Discharge in a 25-cm pipe is 0.03 m3/s. What is the average velocity? A pipe whose diameter is 8 cm transports air with a temp. of 20oC and pressure of 200 kPa abs. At 20 m/s. What is the mass flow rate?

Example: The velocity distribution in a circular duct is , where r is the radial location in the duct, R is the duct radius, and Vo is the velocity on the axis. Find the ratio of the mean velocity to the velocity on the axis. Find:

Example: Air (ρ =1.2 kg/m^3) enters the duct shown: Find: V/10=y/.5  V=20y dA=1*dy

Example: In this flow passage the discharge is varying with time according to the following expression: . At time t=0.5 s, it is known that at section A-A the velocity gradient in the S direction is +2m/s per meter. Given that Qo, Q1, and to are constants with values of 0.985 m^3/s, 0.5 m^3/s, and 1 s, respectively, and assuming that one-dimensional flow, answer the following questions for time t=0.5 s. a. What is the velocity a A-A? b. What is the local acceleration at A-A? c. What is the convective acceleration at A-A? Fluid Mechanics I

Example

Systems, Control Volume, and Control Surface System (sys) A fluid system: contains the same fluid particles. Mass does not cross the system boundaries. Thus the mass of the system is constant. Control Volume (C.V) A control volume is a selected volumetric region in space. It’s shape and position may change with time.(Open System) Control Surface (C.S) The surface enclosing the control volume is called the control surface.

Systems, Control Volume, and Control Surface (continued ) Consider the tank shown, assume: the control volume is defined by the tank walls and the top of the liquid. The control surface that encloses the control volume is designated by the dashed line. The liquid in the tank at time t is elected as the system and is indicated by the solid line. At this instant in time, the system completely occupies the control volume and is contained by the control surface. Thus, at this time: During the same period some liquid has entered the control volume from the left, the amount being: After a time some liquid has flowed out of the control volume to the right. The amount that flowed out is:

Systems, Control Volume, and Control Surface (continued ) Now the system has been deformed as shown in Fig. (b). Part of the system is the liquid that has flowed out across the control surface. The system remaining in the control volume has been deformed by the mass that has flowed in across the control surface. Also, the height of the control volume has changed to accommodate the net flow into the tank. The mass of the system at time t + Δt can be determined by taking the mass in the control volume, subtracting the mass that entered, and adding the mass that left. Subtracting (1) from (2), we have: Dividing by Δt and taking the limit as Δt 0 yields The equation relates the rate of change of the mass of the system to the rate of change of mass in the control volume plus the net outflow (efflux) across the control surface

Systems, Control Volume, and Control Surface (continued ) By definition, the mass of the system is constant so Lagrangian statement And Eq. (3) becomes The corresponding Eulerian statement and can be written as: This equation states that there is an increasing mass in the control volume if there is a net mass influx through the control surface and decreasing mass if there is a net mass efflux. This is identified as the continuity equation.

Systems Conservation of Mass Momentum Energy Laws of Mechanics Written for systems System = arbitrary quantity of mass of fixed identity Fixed quantity of mass, m Conservation of Mass Mass is conserved and does not change Momentum If surroundings exert force on system, mass will accelerate Energy If heat is added to system or work is done by system, energy will change

Control Volumes Solid Mechanics Fluid Mechanics Follow the system, determine what happens to it Fluid Mechanics Consider the behavior in a specific region or Control Volume Convert System approach to CV approach Look at specific regions, rather than specific masses Reynolds Transport Theorem Relates time derivative of system properties to rate of change of property in CV

CV Inflow & Outflow

Continuity Equation In the case of the continuity equation, the extensive property in the control volume equation is the mass of the system, Msys, and the corresponding intensive variable, b, is the mass per unit mass, or Substituting b equal to unity in the control volume equation yields the general form of the continuity equation. This is sometimes called the integral form of the continuity equation.

Continuity Equation The term on the left is the rate of change of the mass of the system. However, by definition, the mass of a system is constant. Therefore the left-hand side of the equation is zero, and the equation can be written as This is the general form of the continuity equation. It states that the net rate of the outflow of mass from the control volume is equal to the rate of decrease of mass within the control volume. The continuity equation involving flow streams having a uniform velocity across the flow section is given as

Example: at a certain time rate of rising is 0.1 cm/s: Continuity equation

Select a CV that moves up and down with the water surface Example: Both pistons are moving to the left, but piston A has a speed twice as great as that of piston B. Then the water level in the tank is: a) rising, b) not moving up or down, c) falling. Select a CV that moves up and down with the water surface Continuity Equation CS VA=2VB VB h Fluid Mechanics I

Euler Equation Fluid element accelerating in l direction & acted on by pressure and weight forces only (no friction) Newton’s 2nd Law

Example 1: Given: Steady flow. Liquid is decelerating at a rate of 0.3g. Find: Pressure gradient in flow direction in terms of specific weight. Flow 30o l

Example 2: Given: g = 10 kN/m3, pB-pA=12 kPa. vertical Given: g = 10 kN/m3, pB-pA=12 kPa. Find: Direction of fluid acceleration.

dV/dx = (80-30) ft/s /1 ft = 50 ft/s/ft Example 3: Given: Steady flow. Velocity varies linearly with distance through the nozzle. Find: Pressure gradient ½-way through the nozzle V1/2=(80+30)/2 ft/s = 55 ft/s dV/dx = (80-30) ft/s /1 ft = 50 ft/s/ft

Bernoulli Equation Consider steady flow along streamline s is along streamline, and t is tangent to streamline

The Bernoulli equation states that the sum of the kinetic, potential, and flow energies of a fluid particle is constant along a streamline during steady flow. An alternative form of the Bernoulli equation is expressed in terms of heads as: The sum of the pressure, velocity, and elevation heads is constant along a streamline.

Example 4: Given: Velocity in outlet pipe from reservoir is 6 m/s and h = 15 m. Find: Pressure at A. Solution: Bernoulli equation Point 1 Point A

Example 5: Given: D=30 in, d=1 in, h=4 ft Find: VA Point A Point 1 Given: D=30 in, d=1 in, h=4 ft Find: VA Solution: Bernoulli equation

Static, Stagnation, Dynamic, and Total Pressure: Bernoulli Equation Hydrostatic Pressure Static Pressure: moves along the fluid “static” to the motion. Dynamic Pressure: due to the mean flow going to forced stagnation. Hydrostatic Pressure: potential energy due to elevation changes. Following a streamline: Follow a Streamline from point 1 to 2 0, no elevation 0, no elevation Note: “Total Pressure = Dynamic Pressure + Static Pressure” H > h In this way we obtain a measurement of the centerline flow with piezometer tube.

Bernoulli equation: The sum of flow, kinetic, and potential energies of a fluid particle along a streamline is constant. Each term in this equation has pressure units, and thus each term represents Energy per unit volune P is the static pressure. ϱV2/2 is the dynamic pressure. ϱ gz is the hydrostatic pressure The sum of the static, dynamic, and hydrostatic pressures is called the total pressure. Bernoulli equation states that the total pressure along a streamline is constant. The sum of the static and dynamic pressures is called the stagnation pressure, and it is expressed as

The static, dynamic, and stagnation pressures are shown. When static and stagnation pressures are measured at a specified location, the fluid velocity at that location can be calculated from

Stagnation Tube

Stagnation Tube in a Pipe Flow Pipe 1 2

Pitot Tube

Example – Venturi Tube Given: Water 20oC, V1=2 m/s, p1=50 kPa, D=6 cm, d=3 cm Find: p2 and p3 Solution: Continuity Eq. Bernoulli Eq. D d 1 2 3 Nozzle: velocity increases, pressure decreases Diffuser: velocity decreases, pressure increases Similarly for 2  3, or 1  3 Pressure drop is fully recovered, since we assumed no frictional losses Knowing the pressure drop 1  2 and d/D, we can calculate the velocity and flow rate Fluid Mechanics I

Air conditioning (~ 60 oF) Ex Given: Velocity in circular duct = 30 m/s, air density = 1.2 kg/m3. Find: Pressure change between circular and square section. Solution: Continuity equation Bernoulli equation Air conditioning (~ 60 oF)

Ex Given: r = 1000 kg/m3 V1= 30 m/s, and A2/A1=0.5, gm=20000 N/m3 Find: Dh Solution: Continuity equation Bernoulli equation Heating (~ 170 oF) Manometer equation

Pitot Tube Application 1 2 V y l z1-z2