Solns to 2005级流体力学与传热考试题(双语班)(A卷)答案

1、当地大气压为745mmHg 测得一容器内的绝对压强为350mmHg ，则真空度为 mmHg 。

测得另一容器内的表压强为1360 mmHg ，则其绝对压强为 mmHg 。

2、为测量腐蚀性液体贮槽中的存液量，采用图示的装置。

测量时通入压缩空气，控制调节阀使空气缓慢地鼓泡通过观察瓶。

今测得U 形压差计读数为R=130mm ，通气管距贮槽底面h=20cm ，贮槽直径为2m ，液体密度为980kg/m 3。

试求贮槽内液体的贮存量为多少吨？解：由题意得：R=130mm ，h=20cm ，D=2m ，=ρ980kg/3m ，=Hg ρ3/13600m kg 。

（1）管道内空气缓慢鼓泡u=0，可用静力学原理求解。

（2）空气的ρ很小，忽略空气柱的影响。

g R g H Hg ρρ=∴ m R H Hg 8.113.098013600.=⨯==ρρ 吨）(15.6980)2.08.1(2785.0)(4122=⨯+⨯⨯=+=∴ρπh H D W 3、测量气体的微小压强差，可用附图所示的双液杯式微差压计。

两杯中放有密度为1ρ的液体，U 形管下部指示液密度为2ρ，管与杯的直径之比d/D 。

试证气罐中的压强B p 可用下式计算：22112)(Dd hg hg p p a B ρρρ---=证明： 作1-1等压面，由静力学方程得：g h g h P g h P B a 211ρρρ+∆+=+ （1） 22144d hD h ππ=221Dd h h =∴代入（1）式得：g h g D d h P g h P B a 21221ρρρ++=+即22112)(Dd hg g hg P P a B ρρρ---=3、查阅资料，写出比重计（密度计）的设计原理4、本题附图所示的开口容器内盛有油和水。

油层高度h 1=0.7m 、密度ρ1=800kg/m 3，水层高度h 2=0.6m 、密度ρ2=1000kg/m 3。

（1）判断下列两关系是否成立，即 p A =p'A p B =p'B（2）计算水在玻璃管内的高度h 。

05轮管流体力学试卷A答案学号姓名成绩一.选择题(共40分,每小题1分)1．下列参数属于流体的物理性质。

Ａ．表面张力，膨胀性，流量，含气量Ｂ．表面张力，含气量，压强，膨胀性2 kg/m3A．．9800 D．1000003，压力对气体的密度。

．影响很大/影响很小．影响很小/影响很小4．同种流体的密度随的变化而变化。

A．温度 B．压力．A．B．C都不对5．气体的体积膨胀系数为当不变时，增加一个单位所引起的体积相对变化量。

A．温度/压力C．温度/大气压力6．柴油机运行时，为了雾化良好，根据不同性质燃油进行适当加温，是为了调节燃油。

A．密度 B．喷油量．含气量7，与接触面上的压力。

A．有关/有关 B．无关/无关/有关8．m2C．相对粘度 D．恩氏粘度9A．增大 B．减少 C．不变10．随着温度的下降，液体的动力粘度，气体的动力粘度。

．增大/增大 C．减小/减小 D．减小/增大11。

．恩氏粘度 C．恩氏一号粘度 D．赛氏通用粘度12A．雷氏 B．恩氏 C．赛氏13．忽略粘性的流体称。

A．实际流体 B．真实流体．假想流体14．流体的运动粘度与动力粘度之比等于。

A．密度．比重 D．比重的倒数15．液体和固体相接触时，若能产生液体润湿固体的现象，这说明。

．附着力小于内聚力．不能确定16A．溶介度 B．渗气率 C．含气率17．将毛细管插入液体内，如果液体能润湿管壁，则管内液面一定。

．降低 C．不变 D．不一定18．不可压缩流体得出的规律适用于。

A．水下爆炸C．水击现象19P等于，（自由液面上的压力等于零；g一重力加速度；ρ一该液体的密度）A．h B．hg/ρ．hρ/g20．一水平放置的密闭容器盛了水，如在瓶口施加压力，则承受压强最小的内壁是 。

．底部内壁 C ．边部内壁 D ．四壁压强一样大21．A ．相同B ．不同22且其值不变，以上规律是 所表达的。

A ．牛顿内摩擦定律B ．质量守衡定律C ．伯努力方程23A ．绝对压力B ．真空24．利用连通器原理制成的仪表是。

《流体输送与传热》试卷A一、单项选择（30分，每小题2分）1. 某地大气压强为100kPa，离心泵入口处真空表读数为0.05MPa，出口处的压力表读数为0.1Mpa，则该泵进出口处的绝对压强分别为（）。

A、50 kPa，0 kPaB、150kPa，200kPaC、150kPa，0kPaD、50kPa，200kPa2. 水连续的从直径为30mm的细管流到直径为90mm的粗管，问细管中水流速是粗管中的（）倍。

A、2B、3C、6D、93. 以下对于流量计说法正确的是（）A、转子流量计能直接读出流量，孔板流量计需要计算得到；B、孔板流量计和转子流量计安装时都需要一定长度的水平直管段；C、转子流量计的转子阻碍流体流动，因此能量损失比孔板流量计大；D、孔板流量计能适用于气体、液体，而转子流量计只适用于液体。

4. 下列对气蚀产生的原因说法不正确的是（）A、进口管路阻力过大或者管路过细B、输送介质温度过高C、安装高度过高，影响泵的吸液量D、离心泵启动前没有灌泵5. 需要经常检修的管路上，要安装（）。

A、三通B、法兰或活接头C、弯头D、丝堵6. 起自动泄压和过压保护作用的阀门是（）。

A、安全阀B、截止阀C、闸阀D、节流阀7. 下列为间歇式操作设备的是（）A、转筒真空过滤机B、三足式离心机C、卧式刮刀卸料离心机D、活塞往复式卸料离心机8. 下列液体中（）的导热率最大A、水B、甘油C、苯D、汽油9. U形管式换热器一般让高温、高压的介质走管内，是因为（）。

A、高压的空间较小，容易密封B、克服温差应力C、管内易清洗D、结构简单、造价较低10. 反应器常采用（）换热器与夹套式换热器一起对反应介质加热或冷却。

A、套管式B、热管式C、沉浸蛇管式D、喷淋蛇管式11. 离心泵流量下降的原因不包括（）。

A、泵的出口管路有漏点B、叶轮的转速下降C、泵的吸入管路部分堵塞D、叶轮腐蚀或磨损12. 以下选项中（）不属列管换热器的振动故障产生的原因。

流体力学与传热学考试题目1-1 下图所示的两个U 形管压差计中,同一水平面上的两点A 、B 或C 、D 的压强是否相等？答：在图1—1所示的倒U 形管压差计顶部划出一微小空气柱。

空气柱静止不动，说明两侧的压强相等，设为P 。

由流体静力学基本方程式： 11gh gh p p A 水空气ρρ++=11gh gh p p B 空气空气ρρ++=空气水ρρ>∴BA p p >即A 、B 两点压强不等。

而 1gh p p C 空气ρ+=1gh p p D 空气ρ+=也就是说，Cp 、D p 都等于顶部的压强p 加上1h 高空气柱所引起的压强，所以C 、D 两点压强相等。

同理，左侧U 形管压差计中，B A p p ≠ 而DC p p =。

分析：等压面成立的条件—静止、等高、连通着的同一种流体。

两个U 形管压差计的A 、B 两点虽然在静止流体的同一水平面上，但终因不满足连通着的同一种流体的条件而非等压。

1-2 容器中的水静止不动。

为了测量A 、B 两水平面的压差，安装一U 形管压差计。

图示这种测量方法是否可行？为什么？ 答：如图1—2，取1—1/为等压面。

由1'1p p =可知：)(2H R g p O H B ++ρ=gRH h g p Hg O H A ρρ+++)(2ghp p O H A B 2ρ+=将其代入上式，整理得 0)(2=-gR O H Hg ρρ∵2≠-OHHg ρρ ∴0=RR 等于零，即压差计无读数，所以图示这种测量方法不可行。

分析：为什么压差计的读数为零？难道A 、B 两个截面间没有压差存在吗？显然这不符合事实。

A 、B 两个截面间确有压差存在，即h 高的水柱所引起的压强。

问题出在这种测量方法上，是由于导管内充满了被测流体的缘故。

连接A 平面测压口的导管中的水在下行过程中，位能不断地转化为静压能。

此时，U 型管压差计所测得的并非单独压差，而是包括位能影响在内的“虚拟压强”之差。

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+ U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2U =1m/s 2(60300.5)/g 3m Z =--=21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U U Z g+W Z g+h 22ρρP P ++=++ 10P = 5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*. 1Z 0= 2Z 15= 422.67x101.97W 15x9.81120246.88J /kg 12002=-+++= N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:h f =6.5U 2where U is the velocity in the pipe, finda. water velocity at section A-A'.b. water flow rate, in m 3 /h.22112212f U U Z g+Z g+h 22ρρP P +=++ 1U 0= 12P =P 1Z 6m = 2Z 0=2f h 6.5U = 22U 6x9.81 6.5U 2=+ U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A = 2b U 2.5*(33/47)1.23m /s == 22a a b b a b f U U Z g+Z g+h 22ρρP P +=++ a b Z =Z 22a b b a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tankand delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocal U U h h h 4f k d 22l ∑=+=+∑ 31180kg /m ρ= 300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ=k=0.025mm k/d=0.025/25=0.001c l r k =0.4 k =1 k =2x0.07=0.14el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u d Re 2.78x10ρμ== f 0.063= 2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be takenas 0.61, what is the flow rate of water?o u c =20o 0V u s 0.61x /4x0.0001π==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have 22112212f 1-2u u gZ gZ h 22ρρP P ++=+++ 2212Hg H o 0 g(R h)39630N/m ρρP =P =-= 2212f1-2c u u u 0 Z =0 h 4f +k 2.13u d 22===l We can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have 22331113f1-3u u gZ gZ h 22ρρP P ++=+++ 311Z 0 Z 6.66m u =0== 22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81u d 20.01l l ++=+== 229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++ 112120 Z 6.66 Z 0 u 0 u 3.51P ===== 22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+= 22229.81x6.66 3.15/226.2N 32970mρP =++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe line fA fB total A B22A fA A 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg = 22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /h π∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied? a. 2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+ b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d 10x0.02x1.06 Re=1.06x10100.02x10ρμ-==>12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr= ()()0.81/3081/34Nu 0027Re Pr 0.027x 1.06x10x 0.69239.66==.=. ()2i i i h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k ==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====> ()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=. (2) 12w 2w = 4421Re Re /2=2x1010=> 0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭ 0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k == (3) 44333u d 2000x0.01Re 2x10100.001ρμ===> 0.81/3Nu 0.023Re Pr = ()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease dThe first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes throughthe pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why? (a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21()20h 642.9w/m k =12t +t LMTD=702∆∆℃= Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln 14080-=--℃ 1122L t 70/86.5L t ∆==∆ 2L 0.81L1 4.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)= 0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=- C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected) ()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆()()h h 12c c 21W C T -T 'W C t 't =-2150100401515080t 15--=-- 2t 50=℃ 212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T )80/()20ln(20802---=∆h h m T T T Th=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln 10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭We can see K=0.015 qe=0.026For p=3.5Kg/cm 21-s K=2k ∆P 1-s K'=2k '∆P 1s K 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫= ⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+ q=240 V=qA=240*0.1=24(2) K 2k =∆P K'2k '=∆P'2∆P =∆P K'2K 500== ()()2q'+10500x 249.60.4=+ q ’=343.6 v=34.36。

流体力学A卷参考答案一、简要回答(共32分,每小题4分)1．定常流动答：各空间点处质点的运动要素不随时间变化的流动。

2.表面力答：表面力是作用在流体表面或截面上且与作用面的面积成正比的力，表面力包括压应力和切应力。

3. 量纲答：量纲表征物理量性质和类别的标志，是物理量的质的特征,也称为因次。

4. 流线答：流场中人为做出的光滑曲线，在同一瞬时其上每点的切线与该点的速度矢量重合。

（流线具有瞬时性）5. 缓变流答：缓变流—流线为近似平行直线的流动或曲率半径很大处的流动。

6. 层流答：各层流体质点互不干扰混杂、有秩序地一层层的流动。

这种流动称为“层流”。

7. 水力光滑管答：当δ管的粗糙度层流底层厚度时，管壁的凹凸部分完全淹没在层流中，流体的∆<紊流核区不直接与管壁接触，∆对液体紊流无影响。

由于层流边层的存在，∆对层流阻力有一定影响，这种管称水力光滑管。

8. 力学相似答：力学相似——实物流动与模型流动在对应点上的对应（同名）物理量都应该具有固定的比例关系。

力学相似的3个条件是：几何相似、运动相似、动力相似。

二、雷诺数Re有什么物理意义？它为什么能起到判别流态的作用？（10分）答：雷诺数的物理意义是流体运动时所受到的惯性力与粘性力之比。

（3分）层流和紊流的根本区别在于层流各流层间互不掺混，只存在粘性引起的摩擦阻力；紊流则有大小不等的涡流动荡于各流层之间，除了粘性阻力，还存在着由于质点掺混、互相碰撞所造成的惯性阻力。

（4分）层流受扰动后，当粘性力的稳定作用起主导作用时，扰动就受到粘性的阻滞而衰减下来，层流趋于稳定。

当扰动占上风，粘性的稳定作用无法使扰动作用衰减下来，于是流动便变为紊流。

因此，流动呈现什么状态，取决于扰动的惯性力作用和粘性的稳定作用相互制约的结果。

雷诺数之所以能判别流态，正是因为它反映了惯性力和粘性力的对比关系，是一个无量纲的纯数。

因此，当圆管中流体流动的雷诺数小于2320时，其粘性起主导作用，层流稳定。

05级《流体力学》试题及答案（2007年5月）系别： 学号： 姓名： 成绩： 注：（1）填空题答案填在试卷上，二至八题答在答卷纸上，考试时间到后一并收回。

（2）重力加速度g 取9.80；（3）旋转抛物体体积等于同高圆柱体体积的一半 一、填空：（19分） 1、（1分）流体力学中的质点应为： 体积无限小的微量流体，微观无穷大，宏观无穷小 。

2、（2分）何谓压力中心？ 总压力的作用点，为总压力作用线与受力面的交点。

何谓压力体？ 从整个受力面的最外轮廓向上引无数条垂直母线到自由液面处所包围的体积 。

3、（1分）在普通压强下，流体的粘度只随 温度 变化。

4、（1分）牛顿内摩擦定律的数学表达式为： dudyτμ= 。

5、（1分）迹线用 拉格朗日 方法进行研究。

6、（4分）何谓流线？ 流线是这样一条曲线，每一瞬时，该曲线上的每一点的速度矢量都在该点与曲线相切。

流线有什么特点？（1）流线不能相交，经过空间一点只有一条流线，除速度为零或无穷大之外；（2）流场中每一点都有流线通过，所有的流线形成流谱；（3）定常流动的流线形状和位置不随时间变化，并与迹线重合，非定常流动的流线形状和位置随时间变化 。

7、（1分）非定常流动的定义为： 流动参数随时间而变的流动 。

8、（1分）控制体定义为： 流场中某个确定的空间区域 。

9、（3分）雷诺输运方程的公式为： (),,c v c ss dB dV dA dt tβρβρ∂⎛⎫=+⎪∂⎝⎭⎰⎰V n ， 非定常可压流动连续性方程的公式为：(),,0c v c s dV dA tρρ∂+=∂⎰⎰V n 。

10、（4分）伯努力方程的使用条件：理想流体 、 质量力为重力 、 定常流动 、 无热量交换 、 无轴功 、 不可压缩 、 一维 。

二、计算（12分）：给定速度场u x t =+，v y t =--，求：过（2，2）点的流线 解： 由流线方程：dx dyu v=dx dy x t y t⇒=+-- （4）ln()ln()x t y t C ⇒+=-++ ()()x t y t C ⇒++=（4） 由于流线过（2，2）点，可得2(2)C t =+（2） ∴流线方程为：2()()(2)x t y t t ++=+（2）三、计算（15分）：已知流场的速度分布为2333(3)(2)x x y y =+-+v i j k ， 试求：（1）（1，2，1）点的加速度（2）是否有旋？若有旋，求旋转角速度矢量ω 解：318x a x = （2）225993y a x xy y =-+ （2） 25186z a xy y =- （2）在（1，2，1）点，1869120a =+-i j k （3） （2） 有旋。

全国2005年1月高等教育自学考试水力学试题课程代码：02444一、单项选择题(在每小题的四个备选答案中，选出一个正确答案，并将正确答案的序号填在题干的括号内。

每小题2分，共20分)1．牛顿流体是（）一种物质。

A．不断膨胀直到充满容器的B．实际上是不可压缩的C．不能承受剪切力的D．在任一剪切力的作用下不能保持静止的2．金属压力表的读数是（）。

A．绝对压强B．绝对压强与当地大气压的差值C．绝对压强加当地大气压D．当地大气压与绝对压强的差值3．均匀流的定义是（）。

A．断面平均流速沿流程不变的流动B．过流断面上各点流速相互平行的流动C．流线皆为平行直线的流动D．各点流速皆相等的流动4．不可压缩流体的总流连续性方程Q=υ1·A1=υ2·A2适用于（）。

A．恒定流B．非恒定流C．恒定流非恒定流D．均不适用5．流量模数K的量纲是（）。

A．(L3T -2) B．(L3T -1)C．(L2T -2) D．(L2T -1)6．具有相同的孔口和管嘴的过水断面面积的孔口出流与管嘴出流，在相同的作用水头下，其出流流量分别为Q和Q n，则有（）。

A．Q>Q n B．Q=Q nC．Q<Q n D．以上都不是7．下面两个关于水力最优断面的定义：(1)过水断面面积一定时，过水流量最大的断面称为水力最优断面；(2)流量一定，过水断面面积最小的断面称为水力最优断面。

则（）。

A．(1)正确，(2)错B．(1)错，(2)正确C．(1)、(2)均错D．(1)、(2)均对8．在断面面积相等和其它条件相同的条件下，断面为半个正六角形的渠道的流量Q A，断面为半圆形的渠道的流量为Q B，则Q A与Q B相比，有（）。

A．Q A>Q B B．Q A=Q BC．Q A<Q B D．不一定9．图示为一水泵管路系统，断面2-3为水泵进出口断面，水泵扬程H的计算公式为（）。

1 / 52 / 5A ． H=ZB ． H=gp g p 23ρ-ρ C ． H=Z+2-0w h +5-3w hD ． H=Z+5-3w h10．宽顶堰溢流在自由出流的情况下，堰上水流的水流状态为（ ）。

【最新整理，下载后即可编辑】1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+U 1=012P =P10Z =W=60j/kg f h 30/kg = 2U =1m/s 2(60300.5)/g 3m Z =--= 21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U UZ g+W Z g+h 22ρρP P ++=++10P =5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*.1Z 0=2Z 15=422.67x101.97W 15x9.81120246.88J /kg 12002=-+++=N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation: h f =6.5U 2where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m 3 /h.22112212f U UZ g+Z g+h 22ρρP P +=++1U 0=12P =P1Z 6m =2Z 0=2f h 6.5U=22U 6x9.81 6.5U 2=+U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure thepressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A =2b U 2.5*(33/47)1.23m /s ==22aa b b a b f U U Z g+Z g+h 22ρρP P +=++a b Z =Z22abb a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocalU U h h h 4f k d 22l ∑=+=+∑31180kg /m ρ=300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ=k=0.025mm k/d=0.025/25=0.001c l r k =0.4 k =1 k =2x0.07=0.14 el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u dRe 2.78x10ρμ==f 0.063=2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be taken as 0.61, what is the flow rate of water?0o o2gR()u c ρρρ-=20o 02gx0.6x(136001000)V u s 0.61x /4x0.0001x1000π-==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have22112212f 1-2u u gZ gZ h 22ρρP P ++=+++2212Hg H o 0 g(R h)39630N/m ρρP =P =-=2212f1-2c u u u 0 Z =0 h 4f +k 2.13ud 22===lWe can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have 22331113f1-3u u gZ gZ h 22ρρP P ++=+++311Z 0 Z 6.66m u =0==22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81ud 20.01l l ++=+==229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++112120 Z 6.66 Z 0 u 0 u 3.51P =====22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+=22229.81x6.66 3.15/226.2N 32970m ρP=++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe linefA fB total A B 22A fAA 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg =22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /hπ∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied? a.2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d10x0.02x1.06 Re=1.06x10100.02x10ρμ-==> 12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr=()()0.81/3081/34Nu 0027Re Pr0.027x 1.06x10x 0.69239.66==.=.()2i ii h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1)444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====>()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=.(2)12w 2w =4421Re Re /2=2x1010=>0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k ==(3) 44333u d 2000x0.01Re 2x10100.001ρμ===>0.81/3Nu 0.023Re Pr =()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease d The first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find: (1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturatedvapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why? (a)0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 ()20h 642.9w/m k =12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca)300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m (c) 8020LMTD=86.514020ln14080-=--℃1122L t 70/86.5L t ∆==∆ 2L 0.81L1 4.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)oi h h Uo 111+= (1)o i o h h U 1'1'1+= (2) (1)-(2)=0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=-C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet andoutlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected) ()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆ ()()h h 12c c 21W C T -T 'W C t 't =- 2150100401515080t 15--=-- 2t 50=℃212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T)80/()20ln(20802---=∆h h m T T TTh=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p 1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln 10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =-()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭ We can see K=0.015 qe=0.026For p=3.5Kg/cm 21-s K=2k ∆P 1-s K'=2k '∆P1s K 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫= ⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+q=240 V=qA=240*0.1=24 (2) K2k=∆P=∆P K'2k'∆P=∆P K'2K500'2==()()2q'+10500x249.60.4=+q’=343.6 v=34.36。

流体力学与传热学试题及参考答案一、填空题：（每空1分）1、对流传热总是概括地着眼于壁面和流体主体之间的热传递，也就是将边界层的 和边界层外的 合并考虑，并命名为给热。

答案：热传导；对流传热2、在工程计算中，对两侧温度分别为t1,t2的固体，通常采用平均导热系数进行热传导计算。

平均导热系数的两种表示方法是 或 。

答案；221λλλ+=-；221t t +=-λ3、图3-2表示固定管板式换热器的两块管板。

由图可知，此换热器为 管程，管程流体的走向为 或 。

1 2 3图3-2 3-18 附图答案：4；2→4 →1→5→3；3→5→1→4→24、黑体的表面温度从300℃升至600℃，其辐射能力增大到原来的 倍. 答案: 5.39分析: 斯蒂芬-波尔兹曼定律表明黑体的辐射能力与绝对温度的4次方成正比, 而非摄氏温度,即4273300273600⎪⎭⎫⎝⎛++=5.39。

5、3-24 用0.1Mpa 的饱和水蒸气在套管换热器中加热空气。

空气走管内，由20℃升至60℃，则管内壁的温度约为 。

答案：100℃6、热油和水在一套管换热器中换热，水由20℃升至75℃。

若冷流体为最小值流体，传热效率0.65，则油的入口温度为 。

答案：104℃ 分析： ε=2020751--T =0.65 ∴1T =104℃1 2 37、因次分析法的基础是 ，又称因次的和谐性。

答案：因次的一致性8、粘度的物理意义是促使流体产生单位速度梯度的_____________。

答案：剪应力9、如果管内流体流量增大1倍以后，仍处于滞流状态，则流动阻力增大到原来的 倍。

答案：210、在滞流区，若总流量不变，规格相同的两根管子串联时的压降为并联时的 倍。

答案：411、流体沿壁面流动时，在边界层内垂直于流动方向上存在着显著的_______________，即使____________很小，____________仍然很大，不容忽视。

答案：速度梯度；粘度；内摩擦应力 12、雷诺数的物理意义实际上就是与阻力有关的两个作用力的比值，即流体流动时的______ 与__ ____ 之比。

,考试作弊将带来严重后果！华南理工大学期末考试《 流体力学与传热 》试卷1. 考前请将密封线内填写清楚；所有答案请直接答在试卷上(或答题纸上)； ．考试形式：闭卷；、 流体在圆形管道中作完全湍流流动，如果只将流速增加一倍，阻力损失为原来的 4倍；如果只将管径增加一倍而流速不变，则阻力损失为原来的 1/2 倍。

、离心泵的特性曲线通常包括 H-Q 曲线、 N-Q 和 η-Q 曲线等。

、气体的粘度随温度升高而 增加 ，水的粘度随温度升高而 降低 。

、测量流体体积流量的流量计有 转子流量计 、 孔板流量计 和 涡轮流量计。

、(1)离心泵最常用的调节方法是 B（A ） 改变吸入管路中阀门开度 （B ） 改变压出管路中阀门的开度（C ） 安置回流支路，改变循环时的大小 （D ） 车削离心泵的叶轮2）漩涡泵常用的调节方法是 B（A ） 改变吸入管路中阀门开度（B ） 安置回流支流，改变循环量的大小 （C ） 改变压出管路中阀门的开度 （D ） 改变电源的电压。

6、沉降操作是指在某种 力场 中利用分散相和连续相之间的 密度 差异，、最常见的间歇式过滤机有 板框过滤机和叶滤机 连续式过滤机有 真空转筒过滤机 。

、在一卧式加热器中，利用水蒸汽冷凝来加热某种液体，应让加热蒸汽在 壳程流动，加热器顶部 排放不凝气，防止壳程α值大辐度下降。

、（1）为了减少室外设备的热损失，保温层外所包的一层金属皮应该是 A（A ）表面光滑，颜色较浅； （B ）表面粗糙，颜色较深 （C ）表面粗糙，颜色较浅（2）某一套管换热器用管间饱和蒸汽加热管内空气，设饱和蒸汽温度为C ︒100，空气进口温度为C ︒，出口温度为C ︒80，问此套管换热器内管壁温应是_C__。

（A ）接近空气平均温度 （B ）接近饱和蒸汽和空气的平均温度 （C ）接近饱和蒸汽温度、举出三种间壁式换热器 套管 、 夹套换热器 、 蛇管换热器 。

二、问答题：（每题5分，共10分） 1．离心泵发生“汽蚀”的主要原因是什么？离心泵工作时，在叶轮中心区域产生真空形成低压而将液体吸上。

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2U =1m/s 2(60300.5)/g 3m Z =--=21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U UZ g+W Z g+h 22ρρP P ++=++10P = 5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*. 1Z 0= 2Z 15=422.67x101.97W 15x9.81120246.88J /kg 12002=-+++=N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation: h f =6.5U 2where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m 3 /h.22112212f U UZ g+Z g+h 22ρρP P +=++ 1U0= 12P =P 1Z 6m = 2Z 0=2f h 6.5U = 22U 6x9.81 6.5U 2=+U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A = 2b U 2.5*(33/47)1.23m /s ==22aa b b a b f U U Z g+Z g+h 22ρρP P +=++ a b Z =Z22abb a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocal U U h h h 4f k d 22l ∑=+=+∑31180kg /m ρ= 300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ= k=0.025mm k/d=0.025/25=0.001 c l r k =0.4 k =1 k =2x0.07=0.14 el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u d Re 2.78x10ρμ== f 0.063=2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meterand orifice coefficient can be taken as 0.61, what is the flow rate of water?o u c =20o 0V u s 0.61x /4x0.0001π==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalentlength of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3,ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have22112212f 1-2u u gZ gZ h 22ρρP P ++=+++ 2212Hg H o 0 g(R h)39630N/m ρρP =P =-= 2212f1-2c u u u 0 Z =0 h 4f +k 2.13ud 22===lWe can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have22331113f1-3u u gZ gZ h 22ρρP P ++=+++ 311Z 0 Z 6.66m u =0==22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81u d 20.01l l ++=+== 229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++112120 Z 6.66 Z 0 u 0 u 3.51P =====22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+= 22229.81x6.66 3.15/226.2N32970mρP=++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe linefA fB total A B22A fAA 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg =22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /hπ∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C. a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied?a. 2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1kJ/kg-K443u d 10x0.02x1.06 Re=1.06x10100.02x10ρμ-==> 12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr=()()0.81/3081/34Nu 0027Re Pr 0.027x 1.06x10x 0.69239.66==.=.()2i i i h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find: a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain? Hint: for laminar flow, Nu=1.86[Re Pr]1/3 for turbulent flow Nu=0.023Re 0.8 Pr 1/3(1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====>()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=.(2) 12w 2w = 4421Re Re /2=2x1010=>0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k ==(3) 44333u d 2000x0.01Re 2x10100.001ρμ===>0.81/3Nu 0.023RePr = ()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease d The first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ? (2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged? (4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?(a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 ()20h 642.9w/m k = 12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln14080-=--℃1122L t70/86.5L t ∆==∆ 2L 0.81L1 4.4m ==(d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface areaof pipe) is 2115 w/(m 2o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uois 2660 w/( m 2o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored) o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)=0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=-C=2859io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected)()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆()()h h 12c c 21W C T -T 'W C t 't =-2150100401515080t 15--=-- 2t 50=℃212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturatedvapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T)80/()20ln(20802---=∆h h m T T TTh=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p 1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatureschange from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2o C) and 1.7kw/(m 2o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -==()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1) 10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3 660 9.10⨯10-33.50 17.1 2.27⨯10-3 233 9.10⨯10-3 Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭We can see K=0.015 qe=0.026For p=3.5Kg/cm21-s K=2k ∆P 1-s K'=2k '∆P1sK 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E eV KA 2V+V d d θ⎛⎫=⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows: (q+10)2 = 250(t+ 0.4) where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+ q=240 V=qA=240*0.1=24 (2) K 2k =∆P K'2k '=∆P '2∆P =∆P K'2K 500== ()()2q'+10500x 249.60.4=+ q ’=343.6 v=34.36。

山东大学工程流体力学2005.2006试题参考答案及评分细则2005年工程流体力学（A卷）参考答案及评分细则一、选择题：1B 2D 3B 4C 5B 6C 7C 8D 9B 10D11A 12D 13C 14 A 15D 16B 17B 18C 19D 20A二、计算题1. 圆柱体表面的粘性切应力为。

粘性力与重力在斜面上的分量相等。

即（3分）且（1分）则（4分）2. 设平衡时杯子液面与圆管内水、油界面的高度差为。

取时圆管右边的水、油界面作为等压面，可得如下方程：（4分）由图中可以看出，（1分）将上式代入等压面方程，化简后得（3分）3. 利用连续性方程可以求出面积比。

（1分）（4分）（2分）（1分）三、计算题1. 取如图所示的控制体，截面1-1和2-2的平均流速分别为（2分）弯管水平放置，则根据伯努利方程，可得（3分）列方向的动量方程：得（3分）得（3分）则合力大小为，与y方向的夹角为。

（1分）2. 因为，满足连续性方程，故存在流函数。

（1分）因为，为无旋流动，存在速度势。

（1分），故有，则，所以，C为积分常数。

故有（5分），故有，则，所以，C为积分常数。

故有（5分）3. 对水泵的出口截面2-2和高水池液面3-3应用伯努利方程，有（2分）整理得先令，可得出（2分）验证：，所以成立。

（1分）（1分）对低水池液面0-0和水泵的进水口截面1-1应用伯努利方程，有（2分）（1分），所以成立。

（1分）代入数据，可得，即（2分）2006年工程流体力学（A卷）参考答案及评分细则一、选择题： 1C. 2B. 3A 4D 5A 6C 7A 8A 9C 10B 11D 12C 13A 14C 15A 16D 17B 18D 19A 20B二、计算题（每题8分，共24分）1. 间隙中的油液的速度呈线性分布，活塞表面的粘性切应力为（1分）粘性力与外力F相等，有下式成立：（4分）则可求得（3分）2. 选取等压面，列方程如下：（5分）代入数据，整理化简后得（3分）3. 空气（2分）（2分）（1分）（2分）（1分）三、计算题（每题12分，共36分）1. 对截面0-0和2-2应用伯努利方程：可得（3分）（2分）假设截面1-1的相对压强为，则对截面1-1和2-2列伯努利方程，有（3分）取截面1-1和2-2之间的流体体积为控制体，应用动量方程：（2分）求得（2分）2. 因为，满足连续性方程，故存在流函数。

Midterm Examination for Fluid Mechanics and Heat Transfer Grade 2004Name Student No. Class Score1.(10 points) True or false(1)( ) When a fluid passes through a pipe, the total mechanical energy in section1 (inlet) is always equal to that in section2 (outlet).(2)( ) When the developed head H is the same, one pump can not develop morevolume flow rate than that two same pumps which work in parallel.(3)( ) In a laminar flow, the frictional loss is proportional to velocity square.(4) ( ) A reciprocating pump is not subject to air binding.(5) ( ) There are three types of flow, that is, laminar, transitional and turbulent flow.2.(30 points) Fill in the blanks or choose the correct answer(1) The orifice meter belongs to , flow rate meter; while the rotameter is , flow rate meter.A) constant pressure drop; B) constant area; C) variable pressure drop; D) variable area(2) ( ) In order to avoid the cavitation, which of the following methods is feasible?A)increase the operating temperature;B)decrease the diameter of suction pipe;C)install the valve or fittings in the suction pipeline;D)reduce the frictional loss in the suction pipeline(3) The orifice meter is used to measure ; the rotameter is utilized tomeasure ; the Pitot tube is used to measure .A.average velocity in the orifice;B.volume flow rate in the orifice;C.average velocity in annular space between the float and tube wall;D.volume flow rate in annular space between the float and tube wall;E.local velocity in the pipe;F.average velocity in the pipe(4) The methods of adjusting volume flow rate of centrifugal pump are1) ; 2) ;(3) .The most common used way in practice is ( ). (5) To avoid cavitation, the pressure at the centrifugal pump inlet must exceed the vapor pressure by a certain value called the .(6) 1atm= mH2O= N/m2；1cp= P= N·s/m2.(7) When the temperature of fluid to be transported increases, the maximum suction lift will ( ). If the velocity of suction pipe decreases, the maximum suction lift will ( ).(8) The viscosity of a gas with temperature increasing, and viscosity of a liquid usually significantly when the temperature is raised.(9) In the boundary layer, there exist . Outside the boundary layer, potential flow survives.(10) When the pipe installation changes from horizontal to vertical position and the velocity keeps unchanged, the frictional loss will be .A. increased;B. decreased;C. the same;D. uncertain3. (30 points) A liquid with density 900 kg/m3 and viscosity μ=30 cp flows from tank 1 to tank 2 through the pipeline of inner diameter 40 mm. The levels of liquid surfaces in two tanks keep constant. When the valve is closed, the readings of gauges A and B are 8.82⨯104 N/m2 and4.41⨯104 N/m2, respectively. When the valve is open, the total length of pipeline is 100 m including all local frictional equivalent length, find:(1)the volumetric flow rate in the pipe, in m3/h?(2)when the valve is open, what happens to the readings of gauges A and B,explain simply why?Hint: for laminar flow, f=16/Re or λ=4f=64/ReFor turbulent flow, λ=0.3164/Re0.254. (30 points)The characteristics of a centrifugal pump are measured under certain rotational speed. The inside diameters of the suction and discharge pipelines are 70mm and 50mm respectively. When the volumetric flow rate of water is 30m3/h, the readings of the vacuum gauge at the suction station and the pressure gauge at the discharge station of the pump are 40kPa and 215kpa respectively. The vertical distance between the pressure tap (pressure measurement point) at the suction station and the pressure tap at the discharge station is 0.4m. The shaft power is 3.45kW. Calculate the developed head and efficiency of the pump.。

《流体力学与流体机械》试题A卷答案一、填空（20分）1、液体气体平衡运动2、总是沿其作用面的内法线方向大小作用的方位3、速度、加速度空间位置时间的4、毕托管文德里流量计孔板流量计喷嘴流量计集流器5、几何相似运动相似动力相似6、叶片式容积式其他型式7、前盖板叶片后盖板轮毂8、平衡孔平衡（背）叶片双吸式叶轮叶轮对称布置9、叶轮导叶外壳集流器疏流罩扩散器10、闸门节流法风门调节法二、选择（10分）1、B2、A3、B4、C5、C6、D7、C8、A9、C 10、A三、判断（10分）1、χ2、√3、χ4、√5、χ6、√7、√8、χ9、χ 10、√四、名词解释（15分）1、流体的粘性分层流动的流体，层与层之间因速度不等而产生相对运动。

带动与阻力是一对大小相等，方向相反的作用力，流层之间的内摩擦力，流体的粘性就是阻止发生剪切变形的一种特性。

粘性大小的物理量有动力粘度和运动粘度。

2、音速和马赫数压力扰动在可压缩流体中的传播速度称为音速。

音速是气体动力学中非常重要的参数。

当扰动源与气体存在相对运动时，相对运动速度V与当地音速a的比值称为马赫数。

其物理意义是惯性力与弹性力之比。

3、允许吸上真空度允许吸上真空度是某一流量下工作时，根据压头特性曲线，功率特性曲线和效率曲线就可以确定相应的压头，功率和效率。

这一组流量，压头，功率和效率在特性曲线上的对应点称为泵或风机的工况点。

五、简答题（10分）1、伯努力方程应用的限制条件是什么？（1）理想不可压缩流体。

（2）作定常流动（3）作用于流体上的质量力只有重力（4）沿同一条流线（或微小流束）2、离心式风机集流器蜗壳的作用是什么？集流器的作用是合格证气流均匀的充满叶轮进口，减少流动损失和降低进口涡流噪声。

有筒形，锥形，弧形和组合形等几种。

蜗壳的作用是将叶轮出口的气体汇集起来，导至风机的出口，并将气的分动能转变为静压。

六、问答题（20分）1、流体运动总流及其包括的三个内容是什么？流动边界内所有流束的总和称为总流。

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:h f =6.5U 2where U is the velocity in the pipe, finda. water velocity at section A-A'.b. water flow rate, in m 3 /h.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be taken as 0.61, what is the flow rate of water?7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we haveWe can get Z1 from the valve closed(2) when the valve opens fully, for section 1-1’ and 3-3’, we haveFor section 1-1’ and 2-2’8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe line10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied?a.2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+ b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====> (2) 12w 2w = 4421Re Re /2=2x1010=> (3) 44333u d 2000x0.01Re 2x10100.001ρμ===> (4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )π (5) there methods : increase u or hi or decrease dThe first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?(a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln 14080-=--℃ (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)= 0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=- C=2859 io h Uo h 111-= ho=8127W/(m2K) 17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected)18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored)Th=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h 22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 2We can see K=0.015 qe=0.026For p=3.5Kg/cm 2 23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+q=240 V=qA=240*0.1=24(2) K 2k =∆P K'2k '=∆Pq ’=343.6 v=34.36。

(2) There are many industrial filters, give three of them:1) Cake filters (pressure filters, vacuum filters) ; 2) Clarifying filters ;3) crossflow filters (ultrafilter) .(3) The orifice meter belongs to B , C flow meter; while the rotameter is A , D flow meter.A) constant pressure drop; B) constant area; C) variable pressure drop; D) variable area(4) When the filter cake is incompressible, a cake resistance is not influenced by ( D ).A) the cake layer thicknessB) the specific cake resistanceC) the porosity (void fraction)D) the filtration rate(5) Air flows along the tube and saturated vapor passes through the shell in a shell-tube exchanger. In order to enhance heat transfer, which way is feasible in practice as follows? ( C )A) increase vapor velocity; B) employ superheated vapor;C) increase air velocity; D) set up the baffles in the shell.(6) The removal of non-condensable gas in the condensing vapor will ( A )A) increase condensing film coefficient.B) decrease condensing film coefficient.C) not change condensing film coefficient.D) be uncertain.(7) Based on the temperature distribution as shown in theattached figure, which layer has the largest heat resistance?A , and which has the largest thermal conductivity?B , if the thickness of each layer is the same.A) layer A; B) layer B; C) layer C(8) When fouling (scale) and pipe wall resistances are ignored and the difference between h i and h o is very huge, overall heat transfer coefficient U is close to ( B ); and wall temperature is close to ( A ) side.A) larger h ; B) smaller h ; C) average value between h i and h o ; D) uncertain (9) For laminar flow, the average velocity in a pipe is 0.5 times the maximum velocity, for turbulent flow average velocity is 0.82~0.87 times the maximum velocity. (10) For laminar flow in a pipe, if the flow rate is constant, when diameter D i increases, the friction coefficient will increase , and friction loss will decrease .3. (15 points) It is proposed to pump 10000 kg/h of toluene at 114℃and 1.1 atm abs pressure from the reboiler of a distillation tower to a second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the reboiler and pump is 7 kN/m 2 and the density of toluene is 866 kg/m 3, how far above the pump must the liquid level in the reboiler be maintained to give a net positive suction head of 2.5 m? Solution.Use the equation for the suction lift H g .m NPSH g h g p p H fs v a g 32.35.280665.9803.8-=--=---'=ρ4. (20 points) A centrifugal pump takes brine from the bottom of a supply tank and delivers it into the bottom of another tank. The brine level in the discharge tank is 45 m above that in the supply tank, and both brine levels are open to the atmosphere. The line between the tanks is 180 m long of 100 mm inner diameter pipe. The flow rate is 90 m 3/h. In the line are 2 gate valves, 4 standard tees, and 4 elbows. What is the power of the motor required for running this pump? The density of brine is 1180 kg/m 3, and the viscosity of brine is 1.2 cP. The overall efficiency of pump and motor is 60 percent.Hint: Loss coefficient for gate valve is 0.17, for elbow is 0.75, for tee is 1.0 For laminar flow, f=16/Re or λ=4f=64/Re For turbulent flow, λ=0.3145/Re 0.25Solution.Bernoulli equation between station a and b becomesb fa b b b p a a ah Vgz p W V gz p -+++=+++2222ρηρ According to the given conditions,2)4()81.945(6.01)(1%60,,0452V K K K D L f h h h g z W p p V V m z z z f e c bfa b fa b fa p b a b a a b ⨯∑+++⨯=+⨯=+⋅∆=∴=====-=∆---ηη Here, from the given conditions,53221013.3102.11180183.31.0Re /183.336001.04904,1.0,180⨯=⨯⨯⨯==∴=⨯⨯⨯====-μρππV D s m D q V m D m LSo it is turbulent flow, and0133.0)1013.3(3145.0425.05=⨯==λf34.70.1475.0417.020.1,4.0=⨯+⨯+⨯=∑==f e c K K KkgJ V K K K D L f h f e c bfa /5.1652183.3)34.70.14.01.01800133.0(2)4(22=⨯+++⨯=⨯∑+++⨯=- kg J h h g z W b fa b fa p /1012)5.16581.945(6.01)81.945(6.01)(1=+⨯=+⨯=+⋅∆=∴--ηPower of motor W W mP p B 3109.2910123600118090⨯=⨯⨯==5. (20 points) In the shell-and-tube heat exchanger with the parallel-current flow, water is used to cool the oil. The temperature of entering water and leaving water are 15 o C and 40 o C, respectively. While the temperature of entering and leaving oil are 150 o C and 100 o C, respectively. Because of production condition changes, the leaving temperature of the oil is to be decreased to 80 o C. Suppose that the mass flow rate, the entering temperatures and the physical properties of fluids are not changed, if the tube length of the original heat exchanger is 1m, how long has the tube to be increased to meet the production requirement? The heat loss of the heat exchanger can be ignored. Solution.The logarithmic mean temperature difference is(15015)(10040)92.513560m t Cn---∆==︒ From the heat balance2112()40150.5()150100h Ph C Pc W C t t W C T T --===-- When the temperature of the oil drops to 80℃, from the heat balance2(15080)(15)h Ph c pc Q W C W C t '=-=-or2150.515080h PhC Pct W C W C '-==-From which 250t C '=︒And (15015)(8050)7013530m t C n ---∆==︒From the heat transfer rate equation, The former heat exchanger:(150100)(92.5)h ph m W C KS t Kn dL π-=∆= The new heat exchanger:(15080)(70)h ph m W C KS t Kn dL π'''-=∆=7092.5, 1.855070hence L m '=⨯=6. (15 points) A slurry is filtered by a filter press of 0.1m 2 filtering area at constant pressure, the equation for a constant pressure filtration is as follows)4.0(250)10(2+=+t qwhere q=filtrate volume per unit filtering area in l/m 2, t= filtering time, in min calculate:(1) how much filtrate will be gotten after 249.6min?(2) If the pressure difference is doubled and both the resistances of the filtration medium and cake are constant, how much filtrate will be obtained after 249.6min?solution:(1) From 22250)4.06.249(250)4.0(250)10(=+=+=+t q We can get 2/240m l q = So filtrate volumel qA V 241.0240=⨯==(2) ∵P rcPK ∆∝∆=μ2 ∴5002==∆'∆='K K pp K And t cons t K Kt q mm m tan 1004.02502=''==⨯== ∴[]min 2.0500100100=='='K t m∴()()m mm t t K t t K q t K q +'≈'+'=+'=+'22)10( ∴222502)10(⨯≈+'q∴2/6.343102250m l q =-=' ∴l A q V 36.341.06.343=⨯='='。