Chapter_LC45_MKT - HES 2340: Fluid Mechanics 1 Week 4 &...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HES 2340: Fluid Mechanics 1 Week 4 & 5 Mass, Bernoulli, and Energy Equations School of Engineering, Computing & Science Sarawak Campus Introduction • The conservation laws, which are the: – Conservation of mass, – Conservation of momentum, and – Conservation of energy. • The Bernoulli equation is concerned with the conservation of kinetic, potential, and flow energies of a fluid stream and their conversion to each other. • The energy equation is a statement of the conservation of energy principle (mechanical energy balance). Conservation of mass • Conservation of mass principle is one of the most fundamental principles in nature. • Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. • Example: – when 16kg of oxygen reacts with 2 kg of H2, 18 kg of water is formed. In an electrolysis process, the water will separated back to 2 kg of H2 and 16 kg of oxygen. • The conservation of mass principle can be expressed as where and the CV, and are the total rates of mass flow into and out of / is the rate of change of mass within the CV. Conservation of momentum • The product of mass and the velocity of the body is called linear momentum. • What is conservation of momentum? • For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object. Conservation of energy • Energy can be transferred to or from a closed system by heat or work, and the conservation of energy principle requires that the net energy transfer to or from a system during a process be equal to the change in the energy content of the system. • Conservation of energy . Continuity The general form of the continuity equation is obtained by substituting the properties for mass into the Reynolds transport theorem Let and ∀ ∙ 1 , resulting in / ∀ However, ∙ 0 (conservation of mass), so the general, or integral, form of the continuity equation is ∀ ∙ 0 Continuity ∀ ∙ 0 This equation can be expressed in words as The accumulation rate of mass in the control volume The net outflow rate of mass through the control surface 0 If the mass crosses the control surface through a number of inlet and exit ports, the continuity equation simplifies to 0 Steady flow processes • For steady flow, the total amount of mass contained in CV is constant, that is, total amount of mass entering must be equal to total amount of mass leaving, 0 0 • For single-stream steady-flow systems, outflow inflow Incompressible flows • For incompressible flows ( = constant), • The is called the volume flow passing through the given cross section. The volume flow meters per second (m3/s). will have units of cubic • If the cross section is not one-dimensional, we have to integrate ∙ • The above equation allows us to define an average velocity which, when multiplied by the section area, gives the correct volume flow 1 ∙ • This could be called the volume-average velocity. • If the density varies across the section, we can define an average density in the same manner: 1 • But the mass flow would contain the product of density and velocity, and the average product would in general have a different value from the product of the averages 1 ∙ EXAMPLE Write the conservation-ofmass relation for steady flow through a streamtube (flow everywhere parallel to the walls) with a single onedimensional exit 1 and inlet 2: Solution: For single flow applies with the single inlet and exit constant Therefore, in a streamtube in steady flow, the mass flow is constant across every section of the tube. If the density is constant, then constant or The volume flow is constant in the tube in steady incompressible flow, and the velocity increases as the section area decreases. EXAMPLE The tank is being filled with water by two one-dimensional inlets. Air is trapped at the top of the tank. The water height is . Find an expression for the change in water height / . Solution 0 The flow within is unsteady, and the system has no outlets and two inlets: ∀ If 0 is the tank cross-sectional area, the unsteady term can be evaluated as follows: ∀ EXAMPLE 0 The above term vanishes because it is the rate of change of air mass and is zero because the air is trapped at the top. Therefore, ∀ For water, Euler’s Equation • The hydrostatic equations were derived by equating the sum of the forces on a fluid element equal to zero. • The same ideas are applied in this section to a moving fluid by equating the sum of the forces acting on a fluid element to the element's acceleration, according to Newton's second law. • The resulting equation is Euler's equation, which can be used to predict pressure variation in moving fluids. Euler’s Equation • Consider the cylindrical element oriented in an arbitrary direction ℓ with cross-sectional area ∆A in a flowing fluid. • The element is oriented at an angle with respect to the horizontal plane (the x-y plane). • The element has been isolated from the flow field and can be treated as a “free body” where the presence of the surrounding fluid is replaced by pressure forces acting on the element. • Assume that the viscous forces are zero. Euler’s Equation • Here the element is being accelerated in the ℓ-direction. • Note that the coordinate axis z is vertically upward and that the pressure varies along the length of the element. • Applying Newton's second law in the ℓ-direction results in ℓ ℓ ℓ • The mass of the fluid element is ∆ ∆ℓ Euler’s Equation • The net force due to pressure in the ℓ-direction is ∆ ∆ ∆ ∆∆ • Note that any pressure forces acting on the side of the cylincer element will not contribute to a force in the ℓ-direction. Euler’s Equation • The force due to gravity is the component of weight in the ℓ-direction ∆ℓ ∆ sin where the minus sign occurs because the component of weight is in the negative ℓ-direction. • From the diagram, it showing the relationship for angle α with respect to ∆ℓ, and ∆z, one notes that sin ∆ /∆ℓ, so the force due to gravity can be expressed as ∆ ∆ ∆ ∆ℓ∆ ∆ℓ ∆ℓ • Note that the weight of the element is ∆ ∆ℓ∆ . ℓ ∆ ∆∆ ∆ℓ∆ ∆ℓ Dividing through by ∆ ∆ℓ results in ∆ ∆ℓ ℓ ∆ ∆ ℓ ∆ℓ ∆ℓ Taking the limit as ∆ℓ approaches zero (element shrinks to a point) leads to the differential equation for acceleration in the ℓ-direction ℓ For an incompressible flow, ℓ ℓ is constant and thus ℓ ℓ The above equation is Euler's equation for motion of a fluid. ℓ ℓ • Euler’s equation shows that the acceleration is equal to the change in piezometric pressure with distance, and the minus sign means that the acceleration is in the direction of decreasing piezometric pressure. • In a static body of fluid, Euler's equation reduces to the hydrostatic differential equation. – In a static fluid, there are no viscous stresses, which is a condition required in the derivation of Euler's equation. – Also there is no motion, so the acceleration is zero in all directions. – Thus, Euler's equation reduces to ℓ = 0. EXAMPLE A column water in a vertical tube is being accelerated by a piston in the vertical direction at 100 m/s2. The depth of the water column is 10 cm. Find the gage pressure on the piston. The water density is 103 kg/m3. Solution Because the acceleration is constant there is no dependence on time so the partial derivative in Euler’s equation can be replaced by an ordinary derivative. Euler’s equation in z-direction: EXAMPLE Integrate between sections 1 and 2: ∆ Gage pressure ∆ Gage pressure 10.9 kPa Gage pressure The Bernoulli Equation • From the dynamics of particles in solid‐body mechanics, one knows that integrating Newton's second law for particle motion along a path provides a relationship between the change in kinetic energy and the work done on the particle. • Integrating Euler's equation along a pathline in the steady flow of an incompressible fluid yields an equivalent relationship called the Bernoulli equation. Frictionless flow: the Bernoulli equation • The Bernoulli equation is an approximate relation between pressure, velocity, and elevation and is valid in regions of steady, incompressible flow where net frictional forces are negligible. • Equation is useful in flow regions outside of boundary layers and wakes, where the fluid motion is governed by the combined effects of pressure and gravity forces. The Bernoulli equation • The Bernoulli equation is developed by applying Euler's equation along a pathline with the direction , ℓ replaced by s, the distance along the pathline, and the acceleration , the ℓ replaced by direction tangent to the pathline. • Euler's equation becomes: The tangential component of acceleration is The Bernoulli equation 0 Since the properties along a streamline depend only on the distance s Steady flow • Euler’s equation now becomes 2 • Moving all the terms to one side yields 2 0 ⇒ 2 constant Bernoulli equation states that the sum of the piezometric pressure and kinetic pressure /2 is constant along a streamline for the steady flow of an incompressible, inviscid fluid. Illustration of the concept underlying the Bernoulli equation • The constant in the Bernoulli equation is the same at all three locations 2 2 2 • Even though the elevation, pressure head, and velocity head vary through the venturi section, the sum of the three heads is the same. Assumptions of the Bernoulli equation • Steady flow ( / =0) – It should not be used during the transient start-up and shut-down periods, or during periods of change in the flow conditions. • Incompressible flow ( ∙ 0) – Acceptable if the Mach number is less than 0.3. • Frictionless flow – Very restrictive, solid walls introduce friction effects. Assumptions of the Bernoulli equation • Flow along a single streamline – Different streamlines may have different “Bernoulli constant” , depending on the flow conditions. • No shaft work ( 0) – The Bernoulli equation can still be applied to a flow section prior to or past a machine (with different Bernoulli constants) • No heat transfer (qnet,in 0) Illustration of the regions of validity and invalidity of the Bernoulli equation For wind-tunnel model test, the Bernoulli equation is only valid in the core flow of the tunnel but not in the tunnel-wall boundary layers, the model surface boundary layers, or the wake of the model, all of which are regions with high friction. Illustration of the regions of validity and invalidity of the Bernoulli equation In propeller flow, Bernoulli’s equation is valid both upstream and downstream, but with a different constant , cause by the work addition of the propeller. Illustration of the regions of validity and invalidity of the Bernoulli equation For wind-tunnel model test, the Bernoulli equation is only valid in the core flow of the tunnel but not in the tunnel-wall boundary layers, the model surface boundary layers, or the wake of the model, all of which are regions with high friction. Static, dynamic, and stagnation pressures 2 • constant along a streamline is the static pressure; it represents the actual thermodynamic pressure of the fluid. • This is the same as the pressure used in thermodynamics and property tables. • is the dynamic pressure; it represents the pressure rise when the fluid in motion. • is the hydrostatic pressure, depends on the reference level selected. • • The sum of the static, dynamic, and hydrostatic pressures is called the total pressure (a constant along a streamline). The sum of the static and dynamic pressures is called the stagnation pressure, kPa • 2 Therefore, the fluid velocity at that location can be calculated from 2 Pitot‐static probe 2 , , The Bernoulli equation 2 constant Dividing the equation by specific gravity 2 where and 2 is the piezometric head and /2 is the velocity head, is a constant. Pressure head Elevation head 2 Velocity head Constant along streamline It is often convenient to plot mechanical energy graphically using heights. constant 2 Dividing the equation by specific gravity , constant 2 Constant along Velocity Pressure Elevation head head head streamline • / is the pressure head; it represents the height of a fluid column that produces the static pressure . • /2 is the velocity head; it represents the elevation needed for a fluid to reach the velocity V during frictionless free fall. • is the elevation head; it represents the potential energy of the fluid. • is the total head. HGL and EGL • Hydraulic Grade Line (HGL) • Energy Grade Line (EGL) (or total head) Tips for Drawing HGLs and EGLs • For stationary bodies such as reservoirs or lakes, the EGL and HGL coincide with the free surface of the liquid, since the velocity is zero and the static pressure (gage) is zero. • • For steady flow in a Pipe of constant diameter and wall roughness, the slope Δ /Δ for the EGL and the HGL will be constant. The EGL is always a distance /2 above the HGL. • A pump causes an abrupt rise in the EGL and HGL by adding energy to the flow. • Height of the EGL decreases in the flow direction unless a pump is present. • A turbine causes an abrupt drop in the EGL and HGL by removing energy from the flow. • Power generated by a turbine can be increased by using a gradual expansion at the turbine outlet. The expansion converts kinetic energy to pressure • • • If the outlet to a reservoir is an abrupt expansion, the kinetic energy is lost. When a flow passage changes diameter, the distance between the EGL and the HGL will change because velocity changes. In addition, the slope on the EGL will change because the head loss per length will be larger in the conduit with the larger velocity. • When a pipe discharges into the atmosphere the HGL is coincident with the system because / = 0 at these points. • For example, the HGL in the liquid jet is drawn through the jet itself. • If the HGL falls below the pipe, then / is negative, indicating subatmospheric pressure and a potential location of cavitation EXAMPLE Water is flowing from a hose attached to a water main at 400 kPa gage. A child places his thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. If the hose is held upward, what is the maximum height that the jet could achieve? Assumptions: The flow exiting into the air is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). The velocity inside the hose is relatively low (V1 = 0) and we take the hose outlet as the reference level (z1 = 0). At the top of the water trajectory V2 = 0, and atmospheric pressure applied. 40.8 m EXAMPLE A piezometer and a Pitot tube are tapped into a horizontal water pipe to measure static and stagnation pressures. For the indicated water column heights, determine the velocity at the center of the pipe. EXAMPLE Find a relation between nozzle discharge velocity and tank free-surface height . Assume steady frictionless flow. Solution: • Note that mass conservation is usually a vital part of Bernoulli analyses. • If is the tank cross section and the nozzle area, this approximation a onedimensional flow with constant density 1 Bernoulli’s equation gives 2 2 But since sections 1 and 2 are both exposed to atmospheric pressure , the pressure terms cancel, leaving 2 2 2 Combining Eqs. (1) and (2) 2 1 / Generally the nozzle area is very much smaller than the tank area , so that the ratio / is negligible, and an accurate approximation for the outlet velocity is 2 The above equation states that the discharge velocity equals the speed which a frictionless particle would attain if it fall freely from point 1 to point 2; that is the potential energy of the surface fluid is entirely converted to kinetic energy. Euler’s equation ℓ ℓ The Bernoulli equation constant along a streamline 2 2 constant along a streamline The Energy Equation The energy equation involves energy, work, and power as well as machines that interact with flowing fluids. Energy, Work and Power Energy • When matter has energy, the matter can be used to do work. • A fluid can have several forms of energy. For example a fluid jet has kinetic energy, water behind a dam has gravitational potential energy, and hot steam has thermal energy. Work • An energy interaction is work if it is associated with a force acting through a distance. • For example wind passing over the blades of a wind turbine. • The wind exerts a force on the blades; this force produces a torque and work is given by Work = force X distance = torque X angular displacement Energy, Work and Power • A system may involve numerous forms of work, and the total work can be expressed as When air passes across the rotor of a wind turbine, the air exerts forces that result in a net torque. This torque does work on the blades Energy, Work and Power • Work and energy both have the same primary dimensions, and the same units, and both characterize an amount or quantity. – For example, 1 calorie is the amount of thermal energy needed to raise the temperature of 1 gram of water by 1°C. Power • The time rate of doing work is called power, quantity of work or energy ≡ interval of time , which is defined by: ∆ lim ∆→ ∆ • A derivative is used because power can vary with each instant in time. • To derive an equation, let the amount of work be given by the product of force and displacement ∆ where ∆: ∆ lim ∆→ ∆ is the velocity of a moving body. • When a shaft is rotating, the amount of work is given by the product of torque and angular displacement ∆ ∆ . In this case, the power equation is ∆ lim ∆→ ∆ • Work Done by Pressure Forces: the work done by the pressure forces on the control surface The associated power is Turbine and Pump • In fluid mechanics, a turbine is a machine that is used to extract energy from a flowing fluid. • Similarly, a pump is a machine that is used to provide energy to a flowing fluid. Energy transfer/change • One of the most fundamental laws in nature is the 1st law of thermodynamics, which is also known as the conservation of energy principle. • It states that energy can be neither created nor destroyed during a process; it can only change forms • Falling rock, picks up speed as PE is converted to KE. • If air resistance is neglected, PE + KE = constant • The conservation of energy principle: • We frequently refer to the sensible and latent forms of internal energy as heat, or thermal energy. • For single phase substances, a change in the thermal energy is equivalent to a change in temperature. • The transfer of thermal energy as a result of a temperature difference is called heat transfer. • A process during which there is no heat transfer is called an adiabatic process: insulated or same temperature • An adiabatic process an isothermal process. Energy transfer General energy equation • The energy equation for a system is: • The above equation is also called the first law of thermodynamics, can be stated in words: net rate of thermal energy entering system • • • net rate at which system does work on environment rate of change of energy of the mater within the system A system is a body of matter that is under consideration. A system always contains the same matter. An imaginary boundary separates the system from all other matter, which is called the environment. General energy equation • The energy equation involves sign conventions: • Thermal energy is positive when there is an addition of thermal energy to the system and negative when there is a removal. • Work is positive when the system is doing work on the environment and negative when work is done on the system. • To extend the energy equation to a control volume, apply the Reynolds Transport Theorem ∀ ∙ • Let the extensive property be energy and let energy per unit mass, we obtain following for a fixed control volume: ∀ The net rate of energy transfer into a CV by heat and work transfer The time rate of change of the energy content in the CV ∙ The net flow rate of energy out of the control surface by mass flow ∀ The ∙ term: • In the field of heat transfer, / can be broken down into conduction, convection, and radiation. In this fluid mechanics course, we consider it only occasionally. The term: • The work term can be divided into three parts: • The shaft work isolates that portion of the work which is deliberately done by a machine (pump impeller, fan blade, piston, etc) protruding through the control surface into the control volume. • According to the sign convention , pump work is negative. Similarly, turbine work is positive, thus • The rate of work done on pressure forces occurs at the surface only; all work on internal portions of the material in the control volume is by equal and opposite forces and is self-cancelling. • The pressure work equals the pressure force on a small surface element times the normal velocity component into the control volume ∙ , • The total pressure work is the integral over the control surface ∙ If part of the control surface is the surface of a machine part, we prefer to delegate that portion of the pressure to the shaft work not to • The shear due to viscous stresses occurs at the control surface and consists of the product of each viscous stress and the respectively velocity component ∙ ⇒ ∙ where is the stress vector on the elemental surface . • This term may vanish or be negligible according to the particular type of surface at that part of the control volume: • Solid surface: for all parts of the control surface which are solid 0. confining walls, 0 from the viscous no-slip condition; hence • Surface of a machine: here the viscous work is contributed by the machine, and so we absorb this work in the term . • An inlet or outlet: at an inlet or outlet, the flow is approximately normal to the element ; hence the only viscous-work term comes . Since viscous normal stress are from the normal stress extremely small in all cases, it is customary to neglect viscous work in inlets and outlets of the control volume. • • This term may vanish or be negligible according to the particular type of surface at that part of the control volume: • Streamline surface (SS): if the control surface is a streamline such as the upper curve in the boundary-layer, the viscous-work term must be evaluated and retained if shear stresses are significant along this line. In summary, the net result of the rate-of-work term consists essentially of The • • ∙ ∙ term: The system energy per unit mass may be of several types: where could encompass chemical reactions, electrostatic or magnetic field effects. If we neglect here and consider only the first three terms, with defined as “up” 1 2 Recalling the Reynolds transport theorem for a fixed control volume: ∀ ∙ Substituting ∙ ∀ ∙ ∙ Rearrange the equation ∀ Next, substitute 1 2 Using the definition of enthalpy 1 2 ∀ 1 2 / , we obtain ∀ 1 2 ∙ ∙ • 1 1 ∀ ∙ 2 2 If the control volume has a series of one-dimensional inlets and outlets, the surface integral reduces to a summation of outlet fluxes minus inlet fluxes: 1 2 ∙ 1 2 Finally, the simplified form for the energy equation for a fixed control volume: 1 2 1 2 1 2 ∀ 1 2 The steady-flow energy equation • For steady flow, time rate of change of the energy content of the CV is zero. 1 2 1 2 • This equation states that the net rate of energy transfer to a CV by heat and work transfers during steady flow is equal to the difference between the rates of outgoing and incoming energy flows with mass. • For steady-flow with one inlet and one outlet, the equation reduces to a relation used in many engineering analyses. • Let section 1 be the inlet and section 2 the outlet, then 1 2 1 2 The net rate of energy transfer into a CV by heat and work transfer The time rate of change of the energy content in the CV 1 2 1 2 ∀ 1 2 The net flow rate of energy out of the control surface by mass flow The net rate of energy transfer into a CV by heat and work transfer The time rate of change of the energy content in the CV 1 2 1 2 ∀ Steady‐flow 1 2 The net flow rate of energy out of the control surface by mass flow Steady‐flow 1 inlet and 1 outlet The net rate of energy transfer into a CV by heat and work transfer 1 2 Flow rate of energy in to the control surface by mass flow 1 2 Flow rate of energy out of the control surface by mass flow The steady-flow energy equation 1 2 1 2 , the equation can rearrange as follows: • From continuity, 1 2 1 2 where , , Recall that is positive if heat is added to the control volume and that and are positive if work is done by the fluid on the surroundings. 1 1 2 2 • Note that each term in the equation has the dimensions of energy per unit mass, which is a form commonly used by mechanical engineers. • If we divide through by , each term becomes a length, or head, which is a form preferred by civil engineers. • The traditional symbol for heard is , in order to avoid the confusion / ), we use internal energy in rewriting the with enthalpy ( head form of the energy relation: 2 2 where , , • A very common application of the steady-flow energy equation is for low-speed flow with negligible viscous work ( → 0). 2 2 This is the difference between the available head upstream and downstream and is normally positive, representing the loss in head due to friction, denoted as . • Therefore, in low-speed (nearly incompressible) flow with one inlet and one exit, we may write 2 2 • The terms are all positive; that is, friction loss is always positive in real flows, a pump adds energy (increases the left-hand side), and a turbine extracts energy from the flow. , are included, the pump and/or turbine must lie between • If points 1 and 2. Kinetic-energy correction factor • Often the flow entering or leaving a port is not strictly one-dimensional, the velocity may vary over the cross section. • In this case the kinetic-energy term for a given port should be modified by a dimensionless correction factor so that the integral can be proportional to the square of the average velocity through the port 1 2 • By letting 1 2 ∙ 1 ≡ 2 where 1 be the velocity normal to the port, for incompressible flow, 1 1 or 2 • The term is the kinetic-energy correction factor, having a value of about 2.0 for fully developed laminar pipe flow and from 1.04 to 1.11 for turbulent pipe flow. Incompressible steady-flow energy equation • The complete incompressible steady-flow energy equation, including pumps, turbines, and losses, would generalize to: 2 2 • The terms on the right ( , , numerically positive. • All additive terms have dimensions of length ) are all Power equation • Here we discuss how to relate head to power and efficiency. • These parameters are used for applications such as selecting a motor for operating a centrifugal pump, calculating the amount of power that can supplied by a proposed hydroelectric plant, and estimating the pump size for a piping system. • Using the definition of shaft head: ⇒ ⇒ Eq. 1 Eq. 2 • From Eq. (1) and Eq. (2), the generalized power equation is: • Both pumps and turbines lose energy due to factors such as mechanical friction, viscous dissipation, and leakage. These losses are accounted for by the efficiency , which is defined as the ratio of power output to power input: power output from a machine or system power input to a machine or system • If the mechanical efficiency of the pump is , the power output delivered by the pump to the flow is: where is the power supplied to the pump, usually by a rotating shaft that is connected to a motor. • If the mechanical efficiency of the turbine is the output power supplied by the turbines is: where is the power input to the turbine from the flow. EXAMPLE A hydrostatic power plant takes in 30 m3/s of water through its turbine and discharge it to the atmosphere at 2 m/s. The head loss in the turbine and penstock system is 20 m. Assuming turbulent flow, 1.06, estimate the power in MW extracted by the turbine. We neglect the viscous and heat transfer and take section 1 at the reservoir surface where 0, , and 100 m. Section 2 is at the turbine outlet. The steady-flow equation becomes 2 0 2 100 m 1.06 2 m/s 2 9.81 m/s 0 20 m 79.8 m The turbine extracts about 79.8 percent of the 100-m head available from the dam. The total power extracted may be evaluated from the water mass flow: 23.4 MW EXAMPLE A pump delivers water ( 1000 kg/m3) at 0.085 m3/s to a machine at section 2, which is 6.096 m higher than the reservoir surface. The / 2 , where 7.5 losses between 1 and 2 are given by is a dimensionless loss coefficient. Take 1.07. Find the power required for the pump if it is 80 person efficient. If the reservoir is large, the flow is steady, with 0. 0.085 /4 0.0762 18.64 m/s The steady-flow energy equation becomes 2 2 2 101350 68950 1000 9.81 6.096 1.07 7.5 18.64 2 9.81 154 m The pump head is negative, indicating work done on the fluid. The power delivered is computed from 1000 0.085 9.81 154 128 kW We drop the negative sign when merely referring to the “power” required. If the pump is 80 percent efficient, the input power required to drive it is efficiency 128 kW 0.8 160 kW Incompressible steady‐flow energy equation 1 inlet and 1 outlet 2 2 Power Efficiency Pump Turbine Contrasting the Bernoulli Equation and the Energy Equation The Bernoulli Equation The Energy Equation • Was derived by applying • Was derived by starting with the Newton’s second law to a first law of thermodynamics and particle and then integrating the then using the Reynolds transport resulting equation along a theorem. streamline. • The Bernoulli equation involves • The energy equation includes both only mechanical energy. mechanical and thermal energy. • The Bernoulli equation is applied by selecting two points on a streamline then equating terms at these points: 2 2 • The energy equation is applied by selecting an inlet section and an outlet section in a pipe and then equating terms as they apply to the pipe: 2 2 The Bernoulli Equation The Energy Equation • Applies to steady, incompressible, and inviscid flow. • Applies to steady, viscous, incompressible flow in a pipe with additional energy being added through a pump or extracted through a turbine. • Under special circumstances the energy equation can be reduced to the Bernoulli equation. • If the flow is inviscid, there is no head loss. • If the “pipe” is regarded as a small stream tube enclosing a streamline, then 1. • There is no pump or turbine along a streamline, so • In this case the energy equation is identical to the Bernoulli equation. • Note that the energy equation cannot be developed starting with the Bernoulli equation. 0. ...
View Full Document

This note was uploaded on 11/19/2011 for the course HES 2340 taught by Professor Tomedwards during the Three '09 term at Swinburne.

Ask a homework question - tutors are online