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243.627 DE6301 and 271.601/DE6301
Week 6 Part [email protected] (MIT)1Manukau Institute of technology
New Zealand Diploma in EngineeringAnd Bachelor of Engineering Technology (BEngTech) - Electrical
Fluid MechanicsGeneral Energy Equation(Mott, Chapter 7)
Contents [email protected] (MIT)2Introductory ConceptsEnergy Losses and ApplicationsNomenclature of Energy Losses and AdditionsGeneral Energy EquationPower Required by PumpsPower Delivered to Fluid Motors
Introductory [email protected] (MIT)3Think about the fluid flow system from last week. Energy can be lost from the system through friction, valves and fittingsEnergy can be added to the system through the pumpEnergy can be removed from the system by fluid motors or turbinesWe can take account of these energy gains and losses by adding terms to Bernoullis equation so that it becomes the general energy equation
[email protected] (MIT)4
Introductory concepts (cont.)[email protected] (MIT)5Bernoullis equation had a number of restrictions, however:It is valid only for incompressible fluidsThere can be no mechanical devices (e.g. pumps, fluid motors, turbines) between the two sections of interestThere can be no energy lost due to friction or turbulence (caused by valves and fittings)There can be no heat transferred into or out of the fluidWhere no fluid flow system satisfies all of these requirementsThus, when we want to consider real-life fluid flow systems, we must account for energy transfers (gains and losses)
Introductory Concepts (Cont.)[email protected] (MIT)6The diagram on the right shows a portion of an industrial fluid distribution systemFluid enters on the left, where the suction line draws fluid from a storage tank; the inline pump adds energy to the fluid and causes it to flow into the discharge line, then through the rest of the piping system. Note: the gradual reducer between the suction pipe and the pump inlet, and the gradual enlargement between the pump outlet and the discharge pipe (these size differences occur because the pipes are not the same size as the connections provided by the manufacturer a common occurrence)
Introductory Concepts (Cont.)[email protected] (MIT)7The fluid then passes through the run of a tee, where a valve in the branch line can be opened to draw some fluid off to another destination point; the fluid then passes from the tee through a valve that can be used to shut off the discharge line; there is then another tee where the fluid now takes the branch path, passes around a 90 elbow, and passes through another valve then to the discharge pipe.At each point, can you identify where energy might be gained/lost from the system, and why?
Energy Additions and Losses [email protected] (MIT)8Energy is added to the system through the pumpFriction caused by the fluid travelling along the pipe also causes energy to be lost.In addition, at each valve, tee, elbow, reducer and enlarger energy is lost from the fluid.We will now discuss these additions and losses in turn.
[email protected] (MIT)9A pump is a common example of a mechanical device that adds energy to a fluid.An electric motor or some other prime power device drives a rotating shaft in the pump; the pump then takes this kinetic energy and delivers it to the fluid, resulting in fluid flow and increased fluid pressure.
Gear pump & Piston pump.
Fluid [email protected] (MIT)10Fluid motors (and devices such as turbines, rotary actuators and linear actuators) take energy from a fluid and deliver it in the form of work (rotating a shaft or moving a piston)Most fluid motors have the same basic configuration as a pumps shown on the previous slide, however the difference between a pump and a fluid motor is that, when acting as a motor, the fluid drives the rotating elements of the device, whereas the reverse is true for pumps.
Fluid [email protected] (MIT)11A fluid in motion offers frictional resistance to flow. Part of the energy in the system is converted into thermal energy (heat), which is dissipated through the walls of the pipe in which the fluid is flowing. The magnitude of the energy loss is dependent on the properties of the fluid, the flow velocity, the pipe size, the smoothness of the pipe wall, and the length of the pipe
Valves and [email protected] (MIT)12Elements that control the direction or flow rate of a fluid in a system typically set up local turbulence in the fluid, causing energy to be dissipated as heat. Whenever there is a restriction, a change in flow velocity, or a change in the direction of flow, these energy losses occur. In a large system the magnitude of losses due to valves and fittings is usually small compared with frictional losses in the pipes. Therefore, such losses are referred to as minor losses.
Nomenclature of Energy Losses and [email protected] (MIT)13We will account for energy losses and additions in a system in terms of energy per unit weight of fluid flowing in the system. This is also known as head.Specifically, we will use the following terms
The magnitude of energy losses produced by fluid friction, valves, and fittings is directly proportional to the velocity head of the fluid. This can be expressed mathematically asThe term K is the resistance coefficient.
General Energy [email protected] (MIT)14The general energy equation as used in this text is an expansion of Bernoullis equation, which makes it possible to solve problems in which energy losses and additions occur.Fluid flow system illustrating the general energy equation.For such a system the expression of the principle of conservation of energy is
General Energy Equation (cont.)[email protected] (MIT)15The energy possessed by the fluid per unit weight is
It is essential that the general energy equation be written in the direction of flow, that is, from the reference point on the left side of the equation to that on the right side.
Example [email protected] (MIT)16Water flows from a large reservoir at the rate of 0.034 m3/s through a pipe system. Calculate the total amount of energy lost from the system because of the valve, the elbows, the pipe entrance, and fluid friction.
Then the total amount of energy lost from the system isExample 1 [email protected] (MIT)17Water flows from a large reservoir at the rate of 0.034 m3/s through a pipe system. Calculate the total amount of energy lost from the system because of the valve, the elbows, the pipe entrance, and fluid friction.
The answer is hL=4.8 m. Here is how it is found. Because Q was given as 0.034 m3/s and the area of a 7.6-cm-diameter jet is 0.0045 m2
Then the total amount of energy lost from the system isExample [email protected] (MIT)18The volume flow rate through the pump is 0.014m3/s. The fluid being pumped is oil with a specific gravity of 0.86. Calculate the energy delivered by the pump to the oil per unit weight of oil flowing in the system. Energy losses in the system are caused by the check valve and friction losses as the fluid flows through the piping. The magnitude of such losses has been determined to be 1.86Nm/N.
Example 2 [email protected] (MIT)19The volume flow rate through the pump is 0.014m3/s. The fluid being pumped is oil with a specific gravity of 0.86. Calculate the energy delivered by the pump to the oil per unit weight of oil flowing in the system. Energy losses in the system are caused by the check valve and friction losses as the fluid flows through the piping. The magnitude of such losses has been determined to be 1.86Nm/N.
Using the sections where the pressure gages are located as the sections of interest, write the energy equation for the system, including only the necessary terms.
Example 2 Solution (cont.)[email protected] (MIT)20Because pB = 296kPa and pA = -28kPa, we have
Notice that point B is at a higher elevation than point A and, therefore the result of zB zA is that is a positive number.
We can use the definition of volume flow rate and the continuity equation to determine each velocity:
Example 2 Solution (cont.)[email protected] (MIT)21Then, solving for the velocities and using the flow areas for the suction and discharge pipes from Appendix F gives
Finally,
The only remaining term is the energy loss hL which is given to be 1.86 Nm/N or 1.86m. We can now combine all of these terms and complete the calculation of hA.The energy added to the system is
That is, the pump delivers 42.9Nm of energy to each newton of oil flowing through it. This completes the programmed instruction.
Power Required by [email protected] (MIT)22Power Required by [email protected] (MIT)23Mechanical Efficiency of [email protected] (MIT)24Mechanical Efficiency of Pumps (Cont.)[email protected] (MIT)25In general the overall efficiency is analogous to the mechanical efficiency discussed for other types of pumps in this section.Volumetric efficiency is a measure of the actual delivery from the pump compared with the ideal delivery found from the displacement per revolution times the rotational speed of the pump.Torsional efficiency is a measure of the ratio of the ideal torque required to drive the pump against the pressure it is developing to the actual torque.
