1.3Solution ：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=％＝7.68.2812638.281-1.4Solution: 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 1.41.5Solution: 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=1.39m1.6Solution :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 ++=++ Rearranging,()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρg h d D u Hg2)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＝1.7Solution :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 integrating3rearranging 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-=232d V L p 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 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．.Solution:(1) From Fanning equationandsoFluid flows from station a to station b , mechanical energy conservation giveshence2from station c to station dhence3From 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 ，then Figure for problem 1.9 22V 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 +=ρρfcd d c h p p =-ρtherefore6Substituting equation 5 in equation 3 ，then7ThusR 1=R 21.10solution:1）1600001.01000004.04.0Re =⨯⨯==μρud from 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=⨯==fab h g l g R =+--'lg 1ρρρρ）（g R h fab ρρρ-'=1g R h fcd ρρρ-'=2Figure for problem 1.104) kerosene:427003.0800004.04.0Re =⨯⨯==μρud Pa p p 4800001.0003.01600=='∆='∆μμ m g p h 611.081.98004800=⨯=''∆='ρSolution ：(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 bFigure for problem 1.122222_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/sthe 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=2121003.51/0(page pressure Z u u m sp =≈≈=）kg J V K d l hc f /2.26251.3)5.01.015025.0(2)(222_1,=+⨯=+=∑λinput the above data into equation d ，9.81×6.66=2.261000251.322++p the pressure is: 329702=p1.14 Solution ： The variables of main pipe are denoted by a subscript1, and branch pipe by subscript2.The friction loss for parallel pipelines is 2121S S s f f V V V h h+==∑∑The energy loss in the branch pipe is22222222u d l l he f ∑∑+=λ In the equation 03.02=λsm u d ml l e /343.0053.04360072.2053.01022222=⨯⨯===+∑πinput the data into equation c kg J hf /333.02343.0053.01003.022=⨯⨯=∑The energy loss in the main pipe is333.022111121===∑∑u d l h hf f λSo s m u /36.22018.023.0333.01=⨯⨯⨯=The water discharge of main pipe ish m V h /60136.23.043600321=⨯⨯⨯=πTotal water discharge ish m V h /7.60372.26013≈+=1.16 Solution:Writing mechanical energy balance equation between the inlet 1 and throat of o o h V p V p ++=+222211ρρ 1 rearranging the equation abovef o oh V V p p +-=-22121ρ2from continuity equation11221125.625V V dd V V o o =⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛= 3 substituting equation 3 for V o into equation 2 gives()ff f f oh R h Rh V h V V p p +=+=+=+-=-94.1185.203.1903.19206.3922121211ρ4from the hydrostatic equilibriumg R p p O H Hg o )(21ρρ-=- 5substituting equation 5 for pressure difference into equation 4 obtainsf OH O H Hg h R gR +=-94.118)(22ρρρ 6rearranging equation 6kg J R R R R gR h OH O H Hg f /288.267.494.11861.12394.118)(22==-=--=ρρρ1.17.Solution: a)2.0501010==D D =⨯-=-=-81.9)1300013600(1.0)(21g R p p Hg ρρs kg V D m /187.013000183.001.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)1300013600(1.0)(21g R p p Hg ρρ588.6Pa2.1Solution:Write the mechanical energy balance equation between the suction connection and discharge connection2_1,2222121122f H gp g u Z H g p g u Z +++=+++ρρwherem Z Z 4.012=-Figure for problem 1.17()s m p p D D C V o /183.030.061.00906.061.01300081.9)1300013600(1.022.0161.021*******=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρ(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 /=η since 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 isFlow rate ，m³/h 26 Total heads ，m 18.41 Shaft power ，kW 2.45 Efficiency ，%53.12.2Solution:Equation(1.6-9)sm Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫ ⎝⎛-=-⎪⎭⎫⎝⎛-=ρρρ）（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 .. Solution :From an energy balance,WhereP 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 ?s m D d V V /12112.42200=⎪⎭⎫⎝⎛⨯=⎪⎭⎫ ⎝⎛=m g u d l fH f 1.581.92105.0200025.02422=⨯⨯==NPSH H gp p H f vog ---=ρmNPSH H gp p H f vo g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=（ρ• 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=⨯⨯⨯==-μρudfrom 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=⨯==∆=ρkPa p 5.8271840450=⨯=∆or friction factor is calculated from equation1.4-25kgJ 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=391.92kPa8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u u u dl u d l p p =⎪⎭⎫ ⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯==∆='∆----ρμρρ＝sos 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 32.0Re 500.00056.0+=f for hydraulically smooth pipeSolution:Write energy balance equation:f h gu 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 λρΔp=46.92×1840×9.81=847.0kpa2.6solution ：∑+++=+++f AA A A AA h u p g z w u p g z 2222ρρρλρ022p 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⨯⨯3.2Solution:The gravity settling is followed Stocks ’ law, so maximum diameter of particle settled can be calculated from Re that is set to 11Re ==μρt c t u d , thenρμc t d u =equation 3.2-16 for the terminal velocityμρρρμ18)(2g d d S c c -=solving for critical diameter32)(224.1gd S c ρρμ-=Check up the appendixThe density of 20℃ air ρ=1.205 kg/m³ and viscosity µ=1.81×10-3N ·s/m 2mm d c μ3.571073.5205.1)205.12650()1081.1(224.15323=⨯=⨯-⨯=--when Reynolds number ≥1000, the flow pattern follows Newton ’s law and terminal velocity can be calculated by equation 3.2-19()ρρρ-=p p t gd u 75.1 1critical Reynolds number is 1000Re ='=μρt ct u d ， 2rearranging the equation 2 givesρμct d u '=1000 3 combination of equation 1 with equation 3ρρρρμg d d S c c )(74.11000-'=' solving for critical diameter32)(3.32μρρρ-='S cdumm d c151210512.1205.1)205.12650()1081.1(3.323323=⨯=⨯-⨯='--3.3solution:to calculate terminal velocity from the equation 3.2-16()μρρ182g d u p pt -=The density of 21℃ air ρ=1.205 kg/m³ and viscosity µ=1.81×10-5N ·s/m 2()s m g d u p pt /181.01081.11881.9)205.12403()1050(185262=⨯⨯⨯-⨯-=--＝μρρt BLu Q = so286.20181.0605.226m u Q BL t =⨯==1 from equation3.3-4tu H u L = the maximum permissible velocity of the air is 3m/s181.03HL =H L 58.16= 2set B to be 3m , then from equation 1 L =7m And H =0.42m3.4 Solution: D =0.4m B =D /4=0.1m h =D /2=0.2ms m hB Q u i /9.131.02.036001000=⨯⨯==According to the equation3.3-12 for N=5：m m u N B d i p c μπρρπμ81089.13)02300(5)1.0)(106.3(9)(965=⨯=⨯-⨯≈-=--3.6Solution:The equation for the constant-pressure filtrationt KA VV V m 222=+5min .1l 51.02122⨯=+K V m 10min .6.1l 101.06.126.122⨯⨯=⨯+K V msolving the equations above for V m and K7.0=m V and K =48For min 15=t 151.0487.0222⨯⨯=⨯⨯+V VSolving for V =2.073 l3.7 The following data are obtained for a filter press of 0.0093 m 2 filtering area in the test(1) filtration constant K , V m at the pressure difference of 1.05(2) if the frame of the filter is filled with the cake at 660s, what is the final rate of filtration Edt dV ⎪⎭⎫ ⎝⎛ (3) and what is the compressible constant of cake n ?solution:①from equation 3.4-19aKt qq q m =+22 For pressure difference 05.1=p ㎏/㎝2500093.01027.220093.01027.2323⨯=⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 1 6600093.0101.920093.0101.9323⨯=⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 2 solving the equations 1 and 2 gives s m K mm q m /1056.10379.02323-⨯==② )(22m E V V KA dt dV +=⎪⎭⎫ ⎝⎛=s m /1014.736-⨯For pressure difference 5.3=p ㎏/㎝271.10093.01027.220093.01027.2323⨯'='⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 3 2330093.0101.920093.0101.9323⨯'='⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 4 solving the equations 3 and 4 gives s m K m m q m /1037.40309.02323-⨯='='n p p -⎪⎪⎭⎫ ⎝⎛∆'∆=K K '1 then n ---⎪⎭⎫ ⎝⎛=⨯⨯13305.15.31056.11037.433.3ln 8.2ln 1=-n solving for n =0.1423.8solution:(1)22250)4.06.249(250)4.0(250)10(=+=+=+t qq=240 l/m 2V=24 l (2) the pressure difference is double2=∆'∆='pp K K soK ´=500210102=⎪⎪⎭⎫ ⎝⎛++'q q 2/5.3421025041.110)10(2m l q q =-⨯=-+='V=34.25 lSolution:Area of filtration: A =2×0.32×20=3.6m 2Δp =248.7kN/m 2t 1=300s, V 1=1/4×0.7=0.175m 3t 2-t 1=1800s, ΔV=0.7-0.175=0.525m 3or t 2 =2100s, V 2=0.7m 3KV andKt KA V m m 21006.37.027.03006.3175.02175.022222⨯=⨯+⨯==⨯+V m =0.2627m 3 And 522101517.3210096.122627.04.149.021006.34.17.0-⨯=⨯⨯+=⨯+=m V K capacitys m Q /10692.250021007.034-⨯=+=for the rotary drum filterArea of filtration: A =πdL =3.14×2.2×1.5=10.362m 2Operating pressure:Δp =70kN/m 2Andthe drum submerged : φ=25%For rotary drum filter the filtration coefficient: keep V m unchanged, K changes with changing in pressure difference6510871.87.24870101517.3--⨯=⨯=∆'∆='p p K K the capacity of rotary drum filter is equal to that of press filter⎪⎪⎭⎫ ⎝⎛-+⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛-+'==⨯--2627.02627.025.0362.1010871.810692.2226224n n V V n A K n nV m m ϕSolving for n from the equation above by trial and error⎪⎪⎭⎫ ⎝⎛-+=⎪⎪⎭⎫ ⎝⎛-+'==⨯-2627.0069.00002381.010692.2224n n V V n A K n nV m m ϕ n=0.0088=0.048rpm(转/min)Solution:From equation4.2-11 (a) heat loss:22211/89.68838.1229.0138.0114.06.76760m W k B k B T R T A q i i =+-=+∆=∆=∑∑ (b) the temperature at interface:211/89.688138.0114.0760m W t k B T A q =-=∆= t=191℃(c) contact resistance is 0.088°C·m 2/W, heat loss:22211/4.683088.038.1229.0138.0114.06.76760088.0m W k B k B T R T A q i i =++-=++∆=∆=∑∑4.4.Solution ：Similar to compound resistances in series through the flat wall, the total resistance across insulating layer is as followsLL kA B R m 2357.0)426.0/278.1ln(/)426.0278.1(615.0426.0=-⨯==π Heat loss per meter tube through the wall ism W R T L q /7.5892357.038177=-=∆= and temperature distribution is()m W r r t R T L q /7.589213.0ln 213.02615.0177=-⨯-=∆=π()213.0ln213.05.2277177r r t --=4.7. Solution:106001002.006.11002.0Re 3=⨯⨯⨯==-μρdu 692.00289.01002.01000Pr 3=⨯⨯==-k c p μ 6.47863.05.166003324.0692.01060002.00289.0023.0Pr Re 023.04.08.04.08.0=⨯⨯=⨯⨯==i i d k h 4.8. Methyl alcohol flowing in the inner pipe of a double-pipe exchanger is cooled with water flowing in the jacket. The inner pipe is made from 25-mm Schedule 40 steel pipe. The thermalconductivity of steel is 45 W/(m·o C). The individual coefficients and fouling factors are given in the following Table. What is the overall coefficient, based on the outside area of the inner pipe? TABLE Data for Problem 8Coefficient/ W/(m 2·°C)Alcohol coefficient h i 1020Water coefficient h o 1700Inside fouling factor h di 5680Outside fouling factor h do 2840Solution: From equation (4.3-37)od o m o i i o di i o o h h d d k b h d d h d d U 111++++= from appendix 3, the wall thickness is 3.25mm and outside diameter is 33.5mm for nominal diameter 25mm Schedule 40 standard steel pipemm d m 25.302275.33=+=Cm W U o o 244534/4.40710521.310882.510998.710216.110184.21284011700125.305.334500325.01020275.335680275.331=⨯+⨯+⨯+⨯+⨯=+++⨯+⨯=-----4.9.Solution:Outside diameter of inner tube d o =19mm, inside diameter of outer tube Di=32mmThe equivalent diameter of annular space of double tube de ismm d D d D d D d o i o i o i e 131932)()(4422=-=-=+-=ππfrom equation4.08.0023.0⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=k c u d d k h p e e μμρ 1 C t o m 5022080=+= the properties of benzene: µ=0.45×10-3Pas ，c p =1.81kJ/kg.℃，k=0.138W/m ℃，ρ= 850kg/m 3 at average temperature of 50℃andaverage velocity of benzene flowing through the annular space of double tube is calculated as followu d D mo i )(422-=πρ()s m d D m u o i /13.1019.0032.0850360018004)(42222=-⨯⨯⨯=-=ππρ Substituting the variables into equation 1 givesC)·W/(m 17809.527748244.0138.01045.01081.11045.085013.1013.0013.0138.0023.0023.024.08.04.0338.034.08.0︒⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯⎪⎭⎫ ⎝⎛⨯⨯⨯⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=--＝＝k c u d d kh p e e μμρ4.13Solution:From the appendix 6, we obtain the latent heat of the saturated vapor, 2256.7kJ/kg at 100 o C.The average temperature of the condensate film=(100+92)/2=96℃The properties of the saturated water at 96 o C are as follows:µ=0.282×10-3Pas(from appendix 8); ρ=958kg/m 3; k=0.68W/m.K(from appendix5)from equation (4.5-12)()())/(5471101327.1943.010282.05.292100107.225681.995868.0943.0943.0241154133234123C m W L T g k h o f o f f ⋅=⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯-⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛∆=-μλρ()h kg s kg t t hA m w s /8.20/005785.0107.2256921005.2038.05471)(3==⨯-⨯⨯⨯=-=πλ Check the flow pattern:)/(04848.0038.014.3005785.0ms kg d m o =⨯==Γπ 68810282.004848.044Re 3=⨯⨯=Γ=-μ(laminar flow) If the tube is placed horizontally, then from equation (4.5-14)4123729.0⎪⎪⎭⎫ ⎝⎛∆='f o o f f d T g k h μλρ then2.2848.2773.0038.05.2773.0943.0729.04141=⨯=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛='od L h h h´=2.2h=2.2×5471=12036.2W/m 2℃hence,hh m m '=' m´=2.2×20.8=45.8kg/h4.14.Solution: (1)From equation (4.3-37)3001212345002.070011923111=++=++=o o m o i i o o h h d d k b h d d U 300110868.4001729.015=+⨯+-ohh o =642.8 W/m 2℃(2) write energy balance equationm o o m o o p t L d U t A U t t mc ∆=∆=-π)(12 1logarithmic mean temperature difference for countercurrent flow∆t 1=150-80=70 o C ; ∆t 2=90-20=70 o C so ∆t m =70 o Ccalculate the length of double tube from equation 1m t d U t t mc L m o o p 5.570023.03003600)2080(1000500)(12=⨯⨯⨯-⨯=∆-=ππ (3)∆t 1=140-80=60 o C ; ∆t 2=140-20=120 o C so2121ln t t t t t m ∆∆-=∆ m t t d U mc t t t t d U t t mc t d U t t mc L o o p o o p m o o p 4.412060ln 023.030036001000500ln ln )()(2121211212=⨯⨯⨯-=∆∆=∆∆--=∆-='ππππ(4) the scale depositing on the surfaces of the tube to give additional thermal resistances4.15.Solution:From equation (4.3-37), neglecting thermal resistance of tube wall giveso i o h h U 111+=when the velocity of water is 1m/s, the individual coefficient of water to wall of tubes is h i , overall coefficient2115111=+=o i o h h U 1when the velocity of water is 1.5m/s, the individual coefficient of water to wall of tubes is h ´i , and h ´i related to h i by equation(4.4-25)383.15.18.08.0==⎪⎭⎫ ⎝⎛'='u u h h i iand overall coefficient is26601383.111111=+=+'=o i o i o h h h h U 2solving equation 1 and 2 for h i and h o h i =2858 W/( m 2⋅ o C ); h o =8135 W/( m 2⋅ o C )4.22Solution :(1)Write energy balance equation211212ln )(t T t T t t A U t A U t t mc o o m o o p ---=∆=- sopo o mc A U t T t T =--21ln set U o =h i because the thermal resistance from saturated vapor to wall of tube and resistance of wall are very small and negligible 608.0100025.1203817030170ln 2=⨯⨯=--t 837.1100025.12038170301702=⨯⨯=--t t 2=93.8 o Cthe rate of condensate for one condenser ()s kg t t mc m p h /0388.010002054308.93100025.1)(121=⨯-⨯=-=λ sum of condensate for two condensers in parallel m h =2m h1=2×0.0388=0.0776kg/s(2) two condensers operate in series heat transfer coefficient between the air and wall of tubes 74.128.08.0==⎪⎭⎫ ⎝⎛'=⎪⎪⎭⎫ ⎝⎛'='u u h h U U i i o oU ´o =1.74U o =1.74×38=66.12 W/m 2⋅K. po o mc A U t T t T 2ln21⨯'='-- so 058.110005.24012.6617030170ln 2=⨯⨯='--t 88.2170301702='--t t´2=121.4 o C()s kg t t c m m p h /111.010*********.12110005.2)(12=⨯-⨯=-''=λ。