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1、 Chapter2 Transportation of Fluids In order to make a fluid flow from one point to another in a pipe, it is necessary to have a driving force. Sometimes this force is supplied by gravity. Usually, the driving force is supplied by a mechanical device such as a pump or blower. Mechanical devices incre
2、ase the mechanical energy of the fluid. This energy may be used to increase the velocity , the pressure, or the elevation of the fluid,Generally, the word pump designates a machine or device for moving an incompressible liquid. Fans, blowers, and compressors are devices for moving gas (usually air).
3、 2.1Pipe, Fittings, and ValveslThe piping system including pipe, fittings and vales2.1.1 Pipe and tubinglThere is no clear-cut distinction between the terms pipe and tubing. Generally speaking,(1) pipe is heavy-walled and relatively large in diameter and comes in moderate lengths of 6 or 12 m; tubin
4、g is thin-walled and often comes in coils several hundred feet long.(2)Metallic pipe can be threaded; tubing usually cannot. (3)Pipe walls are usually slightly rough; tubing has very smooth walls.(4)Lengths of pipe are joined by screwed, flanged, or welded fittings; pieces of tubing are connected by
5、 compression fittings, flare fittings, or soldered fittings.sizelStandard nominal diameters:(1) Pipe For large pipe, more than 12 in. in diameter, the nominal diameters are the actual outside diameters; for small pipe the nominal diameter does not correspond to any actual dimension. The nominal valu
6、e is close to the actual inside diameter for 3- to 12-in. pipe, but for very small pipe this is not true.lRegardless of wall thickness, the outside diameter of all pipe of a given nominal size is the same, to ensure interchangeability of fittings. lThus the designation “2-in. nickel IPS pipe” means
7、nickel pipe having the same outside diameter as standard 2-in. steel pipe.(2) Tubing The size of tubing is indicated by the outside diameter. The normal value is the actual outer diameter, to within very close tolerances. Selection of pipe sizeslThe pipe size selected for a particular installation d
8、epends mainly on the cost of the pipe and fittings and the cost of energy needed for pumping the fluid. lThe cost of the pipe and the annual capital charges increase with about the 1.5 power of the pipe diameter, lwhile the power cost for turbulent flow varies with the -4.8 power of the diameter.lFo
9、r turbulent flow of liquids in steel pipes larger than 1 in. (25 mm) in diameter, the optimum velocity iswhere uoptoptimum velocity, m/smmass flow rate, kg/sfluid density, kg/m3lFor water and similar fluids uopt is 0.9 to 1.8 m/s; lfor air or steam at low to moderate pressures uopt is 6 to 24 m/s. l
10、For flow in heat exchanger tubes, the optimum design velocity is often higher than that given by Eq. (2.1-1) because of improved heat transfer at high fluid velocities.lWhen flow is by gravity from overhead tanks or when a viscous liquid is being pumped, low velocities are favored, in the range of 0
11、.06 to 0.24 m/s. Joints and fittingslThe methods used to join pieces of pipe depend in part on the properties of the material but primarily on the thickness of the wall. lThick-walled tubular products are usually connected by screwed fittings, by flanges, or by welding.lPieces of thin-walled tubing
12、are joined by soldering or by compression or flare fittings.lPipe made of brittle materials such as glass or carbon or cast iron is joined by flanges or bell-and-spigot joints.lFor joining pieces of large steel pipe in process piping, especially for high-pressure service, welding has become the stan
13、dard method. lWelding makes stronger joints than screwed fittings do, and since it does not weaken the pipe wall, lighter pipe can be used for a given pressure.lProperly made welded joints are leakproof, whereas other types of joints are not. lAlmost the only disadvantage of a welded joint is that i
14、t cannot be opened without destroying it.lEnvironmental protection legislation considers flanged and screwed joints to be sources of emission of volatile materials.Allowances for expansionlAlmost all pipes are subjected to expand and contract due to varying temperatureslProvision is made in all high
15、-temperature lines for taking up expansion, so that the fittings and valves are not put under strain. This is done by bends or loops in the pipe, by packed expansion joints, by bellows or packless joints, and sometimes by flexible metal hose.2.2. Pumps lThis section deals with the transportation of
16、liquids through channels.l Liquid are sometimes moved by gravity from elevated tanks, or from a “blowcase”, but by far the most common devices for the purpose are pumps. Pumps increase the mechanical energy of the liquid. The two major classes are positive-displacement pumps and centrifugal pumps. P
17、ositive-displacement units apply pressure directly to the liquid by the reciprocating piston, or by rotating members. Centrifugal pumps generate high rotational velocities, then, convert the resulting kinetic energy of the liquid to pressure energy.2.2.1 Developed head A typical pump application is
18、shown diagrammatically in figure. The pump is installed in a pipeline to provide the energy needed to draw liquid from a reservoir and discharge at the exit of the pipeline.z2z2 above the level of the liquid. At pump itself, the liquid enters the suction connection at station a and leaves the discha
19、rge connection at station b. Since the only friction is that occurring in the pump itself and is accounted for by mechanical efficiency , hf = 0. Then equation (1.3-25) can be written 2.2-1 The equation(2.2-1) can be divide by g, gives(2.2-2)The quantity H is called total head, which each term has t
20、he dimension of length. Often the head developed by a pump is expressed in meters of fluid.Power requirement2.2-3 Brake power is denoted by Ne as mass flow rate m is transported Using the total mechanical-energy-balance equation (2.2-1) on a pump and piping system, the theoretical mechanical energy
21、W added to the fluid by the pump can be calculated. 2.2-5 The power delivered to the fluid is also calculated from the mass flow rate and the head developed by pump. If is the fractional efficiency and N the shaft work delivered to the pump, Eq. (2.2-5)gives Electric motor efficiency lSince most pum
22、ps are driven by electric motors, the efficiency of the motor must be taken into account to determine the total electric power to the motor. lHence, the total electric power input equals the shaft work divided by the electric motor drive efficiency e: 2.2-6 2.2.2. Suction lift of pumps (NPSH)If the
23、suction pressure is only slightly greater than the vapor pressure, some liquid may be flash to vapor inside pump , a process called cavitation , which greatly reduces the pump capacity and causes severe erosion.If the suction pressure is actually less than thevapor pressure, cavitation will occur in
24、 the suction line, and no liquid can be drawn intothe pump. To avoid cavitation, the pressure at the pump inlet must exceed the vapor pressure by a certain value. Called the net positive suction head(NPSH).Suction liftHg12To avoid cavitation, the pressure in station 2 must exceed the vapor pressure
25、by a certain value NPSH 2.2-7 From an energy balance, the least height of the suction station 2 of pump above the liquid level may be found asTherefore, suction lift isFor a pump taking suction from a reservoir, like that shown in figure, the available NPSH is customarily calculated as2.2-9 But this
26、 term is usually small and is accounted for in the recommended value for NPSH furnished by pump suppliers. the velocity head at the pump inlet u2/2 could be subtracted from the result given by equation to give more theoretically correct value of the available NPSH Required value of NPSH is: For smal
27、l centrifugal pump(100gal/min), 2 to 3m For very large pumps, value up to 15m.The NPSH is calculated asThe suction lift is calculated by 2.2-9 For the special situation where the liquid is practically nonvolatile (Pv=0), the friction is negligible (hf=0), and pressure at station a atmospheric, the m
28、aximum possible suction lift can be obtained by subtracted the required NPSH from the barometric head.Rise in temperature and fall in pressure induces vaporization lAny decrease in external pressure or rise in operating temperature can induce vaporization and the pump stops pumping.lThus, the pump a
29、lways needs to have a sufficient amount of suction head present to prevent this vaporization at the lowest pressure point in the pump. NPSH increases as capacity increases lThe NPSH required varies with speed and capacity within any particular pump. The NPSH required increase as the capacity is incr
30、easing. lPump manufacturers curves normally provide this information. lThe NPSH is independent of the fluid density as are all head terms. NPSH as a measure to prevent liquid vaporization lThe manufacturer usually tests the pump with water at different capacities, created by throttling the suction s
31、ide. When the first signs of vaporization induced cavitation occur, the suction pressure is noted. This pressure is converted into the head. This head number is published on the pump curve and is referred as the net positive suction head required (NPSHr) or sometimes in short as the NPSH. Thus the N
32、et Positive Suction Head (NPSH) is the total head at the suction flange of the pump less the vapor pressure converted to fluid column height of the liquid. NPSH is a function of pump design NPSH required is a function of the pump design and is determined based on actual pump test by the vendor. As t
33、he liquid passes from the pump suction to the eye of the impeller, the velocity increases and the pressure decreases. There are also pressure losses . The NPSH required is the positive head absolute required at the pump suction to overcome these pressure drops in the pump and maintain the majority o
34、f the liquid above its vapor pressure. The available NPSH for a given pump should be at least 1 m more than that required by the manufacturer. Note: It is to be noted that the net positive suction head required NPSH number shown on the pump curves is for fresh water at 20 and not for the fluid or co
35、mbinations of fluids being pumped. problemlWhen the temperature of fluid to be transported increases, the maximum lift of suction will ( ). If the velocity of suction pipe decreases, the maximum suction lift will ( ). If the flow rate remains constant but the diameter of suction pipe decreases, the
36、maximum suction lift will ( ) A） increase; B ）decrease; C）keep the same D） be uncertain lWhen the suction pressure is ( ) to the vapor pressure, the cavitation will occurproblemlIf a centrifugal pump operates on air, then, liquid from an initially empty suction line。lPositive-displacement pumps air
37、binding.lBefore a centrifugal pump starts up, close the valve in discharge in other to ( ) the start-up power requiredproblemlThe difference between the theoretical and actual curves results from: ; ; .lWhen higher developed head is required, the best selection is .A) operation in series B) multista
38、ge centrifugal pump C) throttled by valve probleml ( )One pump can develop more H than that two same pumps which work in series. l The methods of adjusting volume flow rate of centrifugal pump are (1) ; (2) ; (3) ; The commonest way in practice is ( ). A centrifugal pump is to be used to extract wat
39、er from a condenser in which the vacuum is 640 mm of mercury. At the rated discharge, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the suction pipe accounted for a head of 1.5m. What must be the least height of the liquid
40、 level in the condenser above the pump inlet?Hg problem A petroleum fraction is pumped 2 km from a distillation plant to storage tanks through a mild steel pipeline, 150 mm in diameter, at the rate of 0.04m3/s. what is the pressure drop along the pipe and the power supplied to the pumping unit if it
41、 has an efficiency of 50%? The pump impeller is eroded and the pressure at its delivery falls to one half. By how much is the flow rate reduced? ( is not changed) specific gravity of the liquid =0.705 viscosity of the liquid =0.5mPa.s roughness of pipe surface=0.004mmWater is transported by a pump (
42、efficiency of pump 65%) from the reactor to tank 2, as shown in Figure. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is 894.5 mm , the orifice coefficient of Co and orifice diameter do are 0.61 and 20 mm, respectively. Frictional coefficient is 0.025
43、 and the readings of vacuum gauge in reactor and pressure gauge in tank 2 are 200 mm Hg and 49000N/m2, respectively. Find:1. Water mass flow rate, in kg/s when the reading R of U in orifice meter is 600 mm Hg? (H2O =1000 kg/m3, Hg =13600 kg/m3)2. Developed head H of pump, in m H2O?3. Practical power W of pump, in kW?
