流体力学与传热 ：1-5 Pipe Flow Systems
管道输送中的流体力学与热传递
管道输送中的流体力学与热传递管道输送是现代工业生产中最为常见的方式之一，利用管道输送各种流体物质，包括气体、液体等等。
而在管道输送中，除了要考虑到工艺条件、设备选型、管道大小等因素之外，还需要考虑到流体力学与热传递等问题。
因此，本文将重点介绍管道输送中的流体力学与热传递问题，并对其重要性和应用进行探讨。
一、流体力学在管道输送中的重要性流体力学是研究流体运动规律的学科，它可以很好地解释管道输送中液体或气体的流动情况。
在管道输送中，流体力学可以帮助我们分析和判断液体或气体在管道内的流速、阻力等因素，并据此确定管道设计方案和流体工艺参数等。
因此，流体力学在管道输送中具有重要的作用。
1、流速的分析管道输送中，流体的流速直接关系到物质输送的效率和速度。
流速过快会增加管道损耗和能耗，从而影响经济效益；流速过缓则会影响工艺效果和产量。
因此，需要根据工艺要求和管道特性等因素来确定合适的流速。
流体力学分析可以帮助我们精确地计算得出流速大小，从而达到最佳的输送效果。
2、阻力的分析管道输送中，管道内壁和流体之间存在着摩擦力，这种摩擦将会对流体的流动产生阻力。
阻力大小直接关系到能耗大小，因此需要对阻力进行分析，找出减小阻力的方法。
而流体力学可以帮助我们对管道内流体的速度、黏度等因素进行分析，计算得出精确的阻力大小，为管道设计和优化提供科学依据。
3、管道结构的优化流体力学分析还可以帮助我们优化管道结构和设计。
在管道输送中，管道的长度、直径、弯头角度、内壁光滑度等因素都会影响流体流动情况。
因此，需要将这些因素纳入考虑范围，采取相应措施，优化管道结构和设计，以提高管道输送效率和经济性。
二、热传递在管道输送中的应用除了流体力学之外，热传递也是管道输送中不可缺少的一环，其重要性不亚于流体力学。
管道输送中的热传递主要表现为两种形式：传导和对流。
在管道输送过程中，热传递可以帮助我们解决流体的温度失控、物料结冻、物料沉淀、物料变质等问题。
流体力学与传热 ：1-3 Basic Equations of Fluid Flow
It is usual to take α to 1 in the calculation.
1.3.4 Overall Mechanical Energy Balance for Steady-state Flow System
The total energy balance, Eq. (4) is not often used when appreciable enthalpy changes occur or appreciable heat is added (or subtracted), since the kinetic- and potentialenergy terms are usually small and can be neglected.
a
b
At steady state the mass flow in equals the mass flow out,
m Q VaSaa VbSbb const
(1.3-6 )
The equation is also called the equation of continuity.
For an incompressible fluids
a b
If the fluid flows through a channels of circular cross section ,then the volumetric flow rate is
流体力学与传热习题：problems and solutions forchapter1,2
1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressuredifference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B ismethane （甲烷）, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and thatliquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure differenceover the instrument In meters of water, (a) when the change in the level in the reservoirs isneglected, (b) when the change in the levels in the reservoirs is taken into account? What is thepercent error in the answer to the part (a)?Solution ：p a =1000kg/m 3 p c =815kg/m 3 p b =0.77kg/m 3 D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubesR d x D 2244ππ= (1) so R D d x 2⎪⎭⎫ ⎝⎛= (2) and hydrostatic equilibrium gives following relationshipg R g x p g R p A c c ρρρ++=+21 (3)sog R g x p p c A c )(21ρρρ-+=- (4)substituting the equation (2) for x into equation (4) givesg R g R D d p p c A c )(221ρρρ-+⎪⎭⎫ ⎝⎛=- (5) （a ）when the change in the level in the reservoirs is neglected,()Pa g R g R g R D d p p c A c A c 26381.98151000145.0)()(221=⨯-=-≈-+⎪⎭⎫ ⎝⎛=-ρρρρρ（b ）when the change in the levels in the reservoirs is taken into account()Pa g R g R D d g R g R D d p p c A c c A c 8.28181.98151000145.081.9815145.0515.6)()(22221=⨯-+⨯⨯⨯⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-ρρρρρρ error=％＝7.68.2812638.281-1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R 1=400mm ，R 2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R 3=50mm. Try to calculate the pressure at point A and B .Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by Hg O H g ρρρ,,2, respectively. The pressure at point A is given by hydrostatic equilibriumg R R g R g R p g Hg O H A )(32232+-+=ρρρg ρis small and negligible in comparison with Hg ρand ρH2O , equation above can be simplifiedc A p p ≈=232gR gR Hg O H ρρ+=1000×9.81×0.05+13600×9.81×0.05=7161N/m²1gR p p p Hg A D B ρ+=≈=7161+13600×9.81×0.4=60527N/mFigure for problem 1.41.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit ofdrainpipe are all open to air.Solution: Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: 2222222111u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation1The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:22⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛D d u u o 22⎪⎭⎫ ⎝⎛=d D u u o 2 Bernoulli equation is written between the throat and the station 2-23 Combining equation 1,2,and 3 gives222u Hg =222200u u p =+ρSolving for HH=1.39m1.6 A liquid with a constant density ρ kg/m 3 is flowing at an unknown velocity V 1 m/s through a horizontal pipe of cross-sectional area A 1 m 2 at a pressure p 1 N/m 2, and then it passes to a section of the pipe in which the area is reduced gradually to A 2 m 2 and the pressure is p2. Assuming no friction losses, calculate the velocities V 1 and V 2 if the pressure difference (p 1 - p 2) is measured. Solution :In Fig1.6, the flow diagram is shown with pressure taps to measure p 1 and p 2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,2112A A V V = For the items in the Bernoulli equation , for a horizontal pipe,z 1=z 2=0Then Bernoulli equation becomes, after substituting 2112A A V V = for V 2, ρρ22121211212020p A A V p V ++=++ ()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρg h d D u HgRearranging,2)1(21212121-=-A A V p p ρ ⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-12221211A A p p V ρ＝Performing the same derivation but in terms of V 2,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--21221212A A p p V ρ＝1.7 A liquid whose coefficient of viscosity is µ flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity V . Show that the pressure loss in a length of pipe L p ∆ is 232dV μ. Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.Solution :The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2 substituting equation 2 for u into equation 1 and integrating3 rearranging equation 3 gives ⎰⎰==R R rdr u R udA A V 020211ππ⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--=22014R r R L p p u L μ2032D L p p V L μ-=1.8. In a vertical pipe carrying water, pressure gauges areinserted at points A and B where the pipe diameters are0.15m and 0.075m respectively. The point B is 2.5m belowA and when the flow rate down the pipe is 0.02 m 3/s, thepressure at B is 14715 N/m 2 greater than that at A.Assuming the losses in the pipe between A and B can beexpressed as g V k 22where V is the velocity at A, find the value of k . If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres.Solution:d A =0.15m; d B =0.075mz A -z B =l =2.5mQ =0.02 m 3/s,p B -p A =14715 N/m 2s m d QV V d Q A A AA /132.115.0785.002.044222=⨯===ππs m d QV V d Q B B BB /529.4075.0785.002.044222=⨯===ππWhen the fluid flows down, writing mechanical balance equation222222A B B B A A A V k V g z p V g z p +++=++ρρ 213.1253.4100014715213.181.95.2222k ++=+⨯ Figure for problem 1.8 232d V L p μ=∆Pa d VL p 115201.01206.005.0323222=⨯⨯⨯==∆μk 638.0260.10715.14638.0525.24++=+=k 0.295making the static equilibriumgR g x g l p g R g x p Hg A B ρρρρρ+∆++=+∆+()()mm g g l p p R g H A B 7981.91260081.910005.214715-=⨯⨯⨯-=---=ρρρ1.9．The liquid vertically flows down through the tube from thestation a to the station b , then horizontally through the tube fromthe station c to the station d , as shown in figure. Two segments ofthe tube, both ab and cd ，have the same length, the diameter androughness.Find:（1）the expressions of g p ab ρ∆, h fab , g p cd ρ∆ and h fcd , respectively. （2）the relationship between readings R 1and R 2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b , mechanical energy conservation giveshence2from station c to station dFigure for problem 1.922V d l h fab λ=22V d l h fcd λ=fcdfab h h =fab b a h p p +=+ρρlg fab b a h p p =+-lg ρfcd d c h p p +=ρρhence3From static equationp a -p b =R 1（ρˊ-ρ）g -l ρg 4p c -p d =R 2（ρˊ-ρ）g 5Substituting equation 4 in equation 2 ，thentherefore6Substituting equation 5 in equation 3 ，then7ThusR 1=R 21.10 Water passes through a pipe of diameter d i=0.004 m with the average velocity 0.4 m/s, as shown in Figure.1) What is the pressure drop –∆P when water flows through the pipe length L =2 m, in m H 2O column?2) Find the maximum velocity and point r at which it occurs.3) Find the point r at which the average velocityequals the local velocity. 4）if kerosene flows through this pipe ，how do thevariables above change ？（the viscosity and density of Water are 0.001 Pasand 1000 kg/m 3，respectively ；and the viscosityand density of kerosene are 0.003 Pas and 800kg/m 3，respectively ）solution:1）1600001.01000004.04.0Re =⨯⨯==μρud fcd d c h p p =-ρfab h g l g R =+--'lg 1ρρρρ）（g R h fab ρρρ-'=1g R h fcd ρρρ-'=2Figure for problem 1.10from Hagen-Poiseuille equation1600004.0001.024.0323222=⨯⨯⨯==∆d uL P μ m g p h 163.081.910001600=⨯=∆=ρ 2）maximum velocity occurs at the center of pipe, from equation 1.4-19max 0.5V u = so u max =0.4×2=0.8m3）when u=V=0.4m/s Eq. 1.4-172max 1⎪⎪⎭⎫ ⎝⎛-=wr r u u 5.0004.01max2=⎪⎭⎫ ⎝⎛-u V r ＝ m r 00284.071.0004.05.0004.0=⨯== 4) kerosene:427003.0800004.04.0Re =⨯⨯==μρud Pa p p 4800001.0003.01600=='∆='∆μμ m g p h 611.081.98004800=⨯=''∆='ρ1.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the loss coefficient of the entranceis 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m³/h)b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m²). l e /d ≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.Solution ：(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 1—1’ and the section of pressure point 2—2’，and take the center of section 2—2’ as the referring plane, then ∑+++=++21,2222121122—f h p u gZ p u gZ ρρ （a ） In the equation 01=p (the gauge pressure)222/396304.181.910004.081.913600m N gh gR p O H Hg =⨯⨯-⨯⨯=-=ρρ0021=≈Z uWhen the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).gR h Z g Hg O H ρρ=+)(12 （b ）where h=1.5mR=0.6mSubstitute the known variables into equation b 2222_1,113.22)5.01.015025.0(2)(66.65.110006.013600V V V K d l h m Z c f =+⨯=+==-⨯=∑λ Substitute the known variables equation a9.81×6.66=2213.21000396302V V ++ the velocity is V =3.13m/s Figure for problem 1.12the flow rate of water is h m V d V h /5.8813.312.0436004360032=⨯⨯⨯=⨯=ππ2） the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 1—1’ and 3-3´，then∑+++=++31,3233121122—f h p V gZ p V gZ ρρ （c ） since m Z 66.61=311300p p u Z =≈=2223_1,81.4 2]5.0)151.035(025.0[ 2)(V V V K d l l h c e f =++=++=∑λ input the above data into equation c ，9.8122V 81.4266.6+=⨯V the velocity is: V =3.51 m/sWrite mechanical energy balance equation between thestations 1—1’ and 2——2’, for the same situation of water level ∑+++=++21,2222121122—f h p V gZ p V gZ ρρ （d ）since m Z 66.61=212103.51/0(page pressure Z u u m s p =≈≈=）kg J V K d l hc f /2.26251.3)5.01.015025.0(2)(222_1,=+⨯=+=∑λ input the above data into equationd ， 9.81×6.66=2.261000251.322++pthe pressure is: 329702=p1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe (Φ60⨯3.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is2.72m 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.Solution ： The variables of main pipe are denoted by a subscript 1, and branch pipe by subscript 2.The friction loss for parallel pipelines is2121S S s f f V V V h h +==∑∑The energy loss in the branch pipe is 22222222u d l l h e f ∑∑+=λ In the equation 03.02=λs m u d ml l e /343.0053.04360072.2053.01022222=⨯⨯===+∑πinput the data into equation ckg J h f /333.02343.0053.01003.022=⨯⨯=∑The energy loss in the main pipe is 333.022111121===∑∑u d l h h f f λ So s m u /36.22018.023.0333.01=⨯⨯⨯= The water discharge of main pipe is h m V h /60136.23.043600321=⨯⨯⨯=π Total water discharge ish m V h /7.60372.26013≈+=1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to be R 5.2 m/s, where R is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution: Writing mechanical energy balance equation between the inlet 1 and throat o for Venturi meterf o o hg z V p g z V p +++=++22121122ρρ 1 rearranging the equation above, and set (z 2-z 1)=xf o oh xg V V p p ++-=-22121ρ 2 from continuity equation 11221125.625V V d d V V o o =⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛= 3 substituting equation 3 for V o into equation 2 gives()f f f f oh xg R h xg R h V h xg V V p p ++=++=+=++-=-94.1185.203.1903.19206.3922121211ρ 4from the hydrostatic equilibrium for manometerg x g R p p Hg o ρρρ+-=-)(1 5substituting equation 5 for pressure difference into equation 4 obtainsf Hgh xg R gx g R ++=+-94.118)(ρρρρ 6 rearranging equation 6 kg J R R R R g R h Hg f /288.267.494.11861.12394.118)(==-=--=ρρρFigure for problem 1.161.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid,calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m 3Solution: a)2.0501010==D D =⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3pas kg V D m /268.0130063.201.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3Pa pressure difference due to increase of velocity in passing through the orificePa D D V V V V p p o 8.44882)2.01(63.213002242412222212221=-=⎪⎪⎭⎫ ⎝⎛-=-=-ρρ pressure drop caused by friction lossPa p f 5.75778.44883.12066=-=∆2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suction connection of the pump is 24.7 kPa as the flow rate is 26m 3/h. The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition. ()s m p p D D C V o /63.231.461.056.1861.0130081.9)130013600(1.022.0161.021*******=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρSolution:Write the mechanical energy balance equation between the suction connection and discharge connection 2_1,2222121122f H gp g u Z H g p g u Z +++=+++ρρ wherem Z Z 4.012=-(Pa 1052.1(Pa 1047.22_1,215241≈=⨯=⨯-=f H u u pressure gauge p pressure gauge p ））total heads of pump is m H 41.1881.9100010247.01052.14.055=⨯⨯+⨯+= efficiency of pump is N N e /=ηsince kW g QH N e 3.1360081.9100041.18263600=⨯⨯⨯==ρ N=2.45kWThen mechanical efficiency %1.53%10045.23.1=⨯=η The performance of pump is Flow rate ，m³/h26 Total heads ，m18.41 Shaft power ，kW2.45 Efficiency ，%53.12.2 Water is transported by apump from reactor, which has200 mm Hg vacuum, to thetank, in which the gaugepressure is 0.5 kgf/cm 2, asshown in Fig. The totalequivalent length of pipe is200 m including all localfrictional loss. The pipeline isφ57×3.5 mm , the orificecoefficient of C o and orificediameter d o are 0.62 and 25mm, respectively. Frictionalcoefficient λ is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)Solution:Equation(1.6-9)Mass flow rates kg S V m o o /02.21000025.0414.312.42=⨯⨯⨯==ρ 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V 1=V 2f H z gp p H +∆+-=ρ12 ∆z=10mPa p 7570710013.17602001081.95.054=⨯⨯+⨯⨯=∆ ∆p/ρg=7.7mThe relation between the hole velocity and velocity of pipeFriction losssoH=7.7+10+5.1=22.8m2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the netpositive suction head must be at least 3m above the cavitation vaporpressure of 710mm mercury vacuum. If losses in the suction pipeaccounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet?Solution :From an energy balance,s m Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫ ⎝⎛-=-⎪⎭⎫ ⎝⎛-=ρρρ）（s m D d V V /12112.42200=⎪⎭⎫ ⎝⎛⨯=⎪⎭⎫ ⎝⎛=m g u d l f H f 1.581.92105.0200025.02422=⨯⨯==NPSH H gp p H f v o g ---=ρWhereP o =760-640=120mmHgP v =760-710=50mmHgUse of the equation will give the minimum height H g as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?• Density of acid 1840kg/m 3• Viscosity of acid 25×10-3 PasSolution: Velocity of acid in the pipe:s m d m d mpipe of area tional cross flowrate volumetric u /32.3025.01840785.03785.04sec 222=⨯⨯===-=ρπρReynolds number:6109102532.31840025.0Re 3=⨯⨯⨯==-μρud from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9kg J u d l f ph f /450232.3025.0600085.042422=⨯==∆=ρ kPa p 5.8271840450=⨯=∆ or friction factor is calculated from equation1.4-25kg J u d l u d l f ph f /426232.3025.0606109046.042Re 046.042422.022.02=⨯⨯⨯==∆=--＝ρkPa p 84.7831840426=⨯=∆ if the pressure drop falls to 783.84/2=391.92kPam NPSH H g p p H f v o g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=（ρ8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u u u d l u d l p p =⎪⎭⎫ ⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯==∆='∆----ρμρρ＝ so s m u /27.236.489..1079012.03919208.18.1==⨯= new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?Density of acid 1840kg/m 3Viscosity of acid 25×10-3 Pa Friction factor 32.0Re500.00056.0+=f for hydraulically smooth pipe Solution: Write energy balance equation:f hg u z g p H g u z g p +++=+++2222222111ρρ gu d l g p h H f 22λρ=∆== 342=ρπu ds m d u /32.31840025.014.3124322=⨯⨯=⨯=ρπ 611510251840025.032.3Re 3=⨯⨯⨯=- 0087.061155.00056.0Re 500.00056.032.032.0=+=+=f 92.4681.9232.3025.0600087.04222=⨯⨯==∆==g u d l g p h H f λρ Δp=46.92×1840×9.81=847.0kpa2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38⨯2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient C o ＝0.63. the permanent loss in pressure is3.5×104N/m 2, the friction coefficient λ=0.024. find:(1) What is the pressure drop along the pipe AB?(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when the shaft work is 500W, 60%efficiency? (The density of fluid is 870kg/m 3 )solution ：∑+++=+++f A A A A AA h u p g z w u p g z 2222ρρ ρλρ022p u d l h p p f BA ∆+==-∑ 247.0334.162=⎪⎭⎫ ⎝⎛=A A o ()()s m gR C u /5.8870870136006.081.9297.063.02247.01200=-⨯⨯=''--=ρρρ ∴u = (16.4/33)2×8.5=2.1m/s∴242/76855105.321.2033.030870024.0m N h p p f B A =⨯+⨯==-∑ρ (2)W u d p Wm 1381.2033.0785.0768554Ne 22=⨯⨯⨯=∆==ρπρ sothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid ％％＝461006.0500138⨯⨯。
【流体力学与传热】4.3 对流传热
圆缺形：Nu
0.36Pr 1/ 3 Re 0.55
W
0.14
适用范围：2000 Re 106
定性温度：tm 特征尺寸：de
正方形
de
4(t 2
0.78
5d
2 0
)
πd0
正三角形
4( de
3 2
t
2
0.785d02
)
πd0
d0 t
t
u按流过管间最大截面积计算：
Smax
1/3
例【4-5】 在外径为152mm的水平管内通有171oC 的饱和水蒸气，管外空气温度为21oC，试求：
（1）管道不保温，每米管长因空气自然对流 引起的热损失？ （2）管外包有一层50mm厚的保温层，使保温层 的表面温度降为59oC，此时每米管长因空气自然 对流造成的热损失又为多少？保温层的导热系数 是多少？
h u0.8
h2 20.8 h1 1.7411271.77 2214.28w m-2 0 C-1
（4）如苯走壳程，水蒸气走管内
de
4A P
4
D
4
2
nd 2
4
D nd
0.452 38 0.0252
0.45 38 0.025
0.128m
u
qv A
8.32
0.069m s1
hC
gtl 3 2
(
cP )n
do
2
=
2.754102 0.54
(4.63107 )1/ 4
4.9
Wm-2K-1
0.252
包保温层后的热损失
Q/L=hdo∆t=4.90.252(59-21)=147.3 Wm-1
热传递与流体力学中的数值计算
热传递与流体力学中的数值计算一、简介热传递和流体力学是两个紧密相关的领域，都涉及物质的运动和转换，成为热力学体系中不可或缺的一部分。
数值计算则是解决热传递和流体力学问题的重要方法。
今天我们将从数值计算的角度出发，探讨热传递和流体力学的数值计算方法，分析其应用和局限性。
二、热传递中的数值计算热传递包括传导、对流和辐射，其中最为重要的是传导。
传导热量-流量的表达式是 Fourier 定律，它指出了热流的大小和热梯度的相关性。
传导热量的数值计算方法包括：1. 显式方法显式法是一种直接求解离散方程形式的传统计算方法，它的计算精度较低，但现在已经逐渐淘汰。
例如，TFLUIDS 软件提供了一种标准的显式方法，用于传导问题的数值计算。
2. 隐式方法隐式法是一种求解离散方程变量的计算方法，它的计算精度较高，但需要更多的计算量。
在隐式方法中，计算可以逐步迭代，直到满足预设的精确性要求。
为了获得高精度的计算结果，通常使用数值计算软件，例如 CFD 和 ANSYS。
3. 软件仿真软件仿真是一种基于多物理场和多机构模型的高级计算方法。
它是一种计算大型和复杂热传递问题的高精度方法，可以处理各种传导模型，包括两相流、相变和复杂结构材料。
此类方法已经被广泛应用于汽车、航空航天、能源和建筑等领域的规划和设计，并得到了广泛的认可。
三、流体力学中的数值计算流体力学是液体和气体力学的研究领域，其主要研究对象是流体的运动和转换。
流体力学的主要模拟对象是流体场中的速度和压力，因此流体力学的核心是 Navier-Stokes 方程组，其中包括质量、动量和能量守恒方程。
流体力学的数值计算方法包括：1. 有限体积方法有限体积方法是一种离散流体力学方程的高精度方法，它考虑了流体的受力、耗散和粘度等因素。
有限体积方法的最大优点是可以处理高速和复杂的流体场问题，例如，超音速飞行器、汽车和火箭引擎等问题。
2. 有限元方法有限元方法是一种更为通用的计算方法，它不仅可以应用于流体力学问题，还可以应用于结构力学、热传递等其他力学问题。
流体力学研究生涉及传热
流体力学研究生涉及传热摘要：一、流体力学基本概念与研究内容1.流体力学的定义与作用2.流体力学的研究范畴二、传热基本原理与过程1.热传导2.对流传热3.辐射传热三、流体力学与传热的关系1.流体流动与热传递的相互影响2.工程应用中的流体力学与传热问题四、流体力学与传热在工程中的应用1.流体输送2.非均相机械分离3.传热及换热设备4.蒸发五、流体力学与传热的研究方法与挑战1.研究方法与技术发展2.工程实际问题处理3.学术研究与应用的结合正文：流体力学作为一门研究流体在不同条件下运动和变形的学科，广泛应用于工程技术领域。
研究生阶段对流体力学的研究，涉及到多个方向，其中传热是一个重要的研究领域。
本文将简要介绍流体力学基本概念、传热原理，以及流体力学与传热在工程中的应用和研究方法。
一、流体力学基本概念与研究内容流体力学是研究流体在不同条件下运动和变形的学科，涉及的范围包括流体静力学、流体动力学、湍流理论等。
流体力学在工程技术中具有广泛的应用，如航空航天、建筑、水利、化学工程等领域。
1.流体力学的定义与作用流体力学主要研究流体在静止和运动状态下的力学性质，包括流体的内部结构和宏观运动规律。
流体力学的作用在于揭示流体的基本规律，为工程设计和实际应用提供理论依据。
2.流体力学的研究范畴流体力学的研究范畴包括流体静力学、流体动力学、湍流理论、边界层理论、流体机械等。
在这些领域中，研究者需要探讨流体的基本性质、流动规律、力矩传递、能量转换等问题。
二、传热基本原理与过程传热是指热量从高温物体传递到低温物体的过程。
在自然界和工程应用中，传热过程普遍存在。
传热主要包括三种方式：热传导、对流传热和辐射传热。
1.热传导热传导是指热量通过固体、液体或气体等物质直接传递的过程。
热传导的速率取决于物质的导热性能和物体的厚度。
2.对流传热对流传热是指热量通过流体（如气体和液体）的运动而传递的过程。
对流传热的速率取决于流体的速度、温度梯度和流体的热导率。
流体力学与传热(英文版)chapter1b
dy
(the proportionality constant μ is called the viscosity, unit of μ is kg/(m.s) ) Eq. (1.2-3) is called Newton’s law of viscosity. It is good for Newton’s fluids and laminar flow. Gases and most liquids belong to Newtonian fluids
Chapter 1
1.2 Fluid-flow phenomena
• Ideal fluid
which has zero viscosity. Without friction or resistance when fluid flows through the channel.
• Potential flow
The velocity field
• If the wall is at rest in the reference frame, the velocity of the fluid at the interface is zero, but there’re variations in velocity from point to point in the flowing stream at distances away from the solid . • The velocity field exists in the space occu pied by the fluid.
热传递与流体力学的理论研究
热传递与流体力学的理论研究一、热传递基本理论热传递是指物质内部或外部因温度差异而产生的热量流动现象，是自然界不可缺少的物理现象之一。
热传递包括传热的过程、机理和规律，热传递与其他物理现象之间存在着密切的联系。
1. 热传递模式热传递过程中，热量的传输方式分为三种：传导、对流和辐射。
传导是指物体内部由高温区到低温区热量的传递。
常用的传导计算公式是傅里叶传导定律。
对流是指物体表面的传热。
常见的对流模式有自然对流和强制对流。
辐射是指通过电磁波传递的热量，无需介质参与传递，例如太阳辐射地球的热量。
2. 热传递机理传热的机理主要取决于物质状态和性质。
固体的热传递主要通过传导实现，液体和气体的热传递则主要通过对流实现。
3. 热传递规律热传递中常用的物理量有流量、传热系数和传热阻力。
流量是指单位时间内热量的传递量，传热系数则是单位时间内单位温差下热源的热流密度，传热阻力是指传递热量时所经过的障碍和阻力。
4. 热传递应用热传递理论广泛应用于工业生产、温度测量和环境工程等领域。
例如，气体的热传递常常影响飞行器的设计、航空气象预报以及设备冷却等。
二、流体力学基础流体力学是一门研究流体的运动规律和性质的学科，包括流体的流动、动力学、稳定性和湍流等领域，也是应用数学的一部分。
1. 流体的基本性质流体的基本性质包括密度、压力、温度、速度和流量等。
密度时不同物质的质量与体积的比值，压力则是单位面积上垂直的力的大小。
2. 流体在静力学下的运动流体在静力学下的运动主要涉及莫门定理、贝努利定理和亚松森定理等。
通过这些定理，可以描述流体在管道、水渠和水坝等建筑物中的运动规律。
3. 流体在动力学下的运动流体在动力学下的运动主要涉及连续性方程、动量方程和能量方程等。
通过这些方程，可以描述流体在航空、航天和海洋工程等领域的运动规律。
4. 流体力学应用流体力学的应用范围非常广泛，包括航空、航天、海洋、地质、环境和工业生产等领域。
例如，在飞行器的设计和制造中，流体力学可以用来研究飞行器的空气动力学性质和热力性质。
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1.5 Pipe Flow Systems
• In the previous sections, we discussed concepts concerning the energy balance and the flow in pipes.
• The purposes of this section is to apply these ideas to the solutions of pipe systems.
Re
1.12 103
45000
Thus ,the flow is turbulent flow
Governing equation for either case (a), (b), or (c) is the mechanical balance equation given by
p1
V12 2
Solution: Since the velocity in the pipe is given by
V1
Q A1
0.756 103
0.01912
2.64m / s
4
And the fluid properties are ρ=999.9kg/m3 and μ=1.12×10-3Pas
999.9 2.64 0.0191
1.5.1 Single Pipes
For flowing through a pipe three most common types of problems are shown in table in terms of the parameters involved.
Variable
type I type II type III
z1g
p2
V22 2
z2g hf
Where z1=0, z2=6.1m, p2=0, and outlet velocity is
V2
Q A2
0.756 103
0.01272z2 g
V22
V12 2
h f
1
Where the friction loss is different for each of three cases
With thisε/D and Reynolds number (Re=45000), the value of friction coefficient is obtained from Moody chart as λ=0.0215.
the potential and kinetic energy are the same as part (a). Eq.1 gives
p1
z2 g
V22
V12 2
L D
V12 2
74133 999.9 0.0215 18.29 2.642 0.0191 2
145872Pa
Of this pressure drop, the amount due to skin friction is 71739 Pa
a. Fluid
Density Viscosity
b. Pipe
Diameter Length Roughness
c. Flow
Flowrate
d. Pressure
Pressure drop
given given given given
given determine
given given given given
due to the increase in kinetic energy
(b)
If only losses included are the skin friction losses , the friction loss is
hf
L V12 D2
Relative roughness ε/D=8×10-5
determine given
given given determine given
given given
Example 1
Water at 15.6ºC flows from the basement to the second floor through the 1.91cm diameter copper pipe (ε = 0.001524mm ) at a rate of Q=0.756 l/s and exits through a faucet of diameter 1.27cm. As shown in Fig.
• We will concern two classes of pipe systems: those containing a single pipe and those containing multiple pipes in parallel, series.
Pipe system
Some of the basic components of a typical pipe system shown in Fig.
(a) If all losses are neglected, Eq.1 gives
p1
6.1 999.9 9.81 999.9 5.972
2.642 2
74133Pa
Note that for this pressure difference, the
amount due to elevation change and amount
They include : • the pipes themselves, • the various fittings used to connect the
individual pipes to form the desired system, • metering devices, and • the pumps or turbines that add energy to or remove energy from the fluid.