冶金传输原理课后题-沈巧珍版

1、第一章1-9解11VV273V1T1800 9.87840 N/m3比重型0.810001-10 解6790 313580kg/m3500 10 3135809.8133084 N/m31-11 解27310000.279kg/m31.3-_10001 27312001.31僭27330.241kg/mRT0T01T12T20T0T11.3 273273 100030.279kg/m30丁0T21.3 273273120030.241kg/m1-12 解VtV1V14001.423273 200增大了 0.423倍。1-13 解VtV0 1 273VoVtt273273259005.818m/

2、S2731-14 解1013251.293 273287.05m2/s2 K0.082atmL/mol K 8.31J/mol K 29.27m/K1-15 解PRT旦1 RT165.1 103287273 200.774kg/m3P2RT299.2 103287273 371.115kg/m31-20 解1 dVV dPdP R P。1% 20.5 10 91071-18 解PVT1P2V2T2VlV1旦P2TlT11.032 1056.0792 105I7H00.2V2V10.2V1 V10.8V1体积缩小了1-19 解PVk CPV nRT0.8 倍。k 1TT空气kV1VTk=1.4V

3、2Vi20'4T2VlV2P21-21V21.32T11.32PP1P121.42.64P1.321521.4273380.16K107.16C2.67PP1105Pay20mm0.02m201.2P表 P大 7098168kPa68.5-98-29.5k Pa1-22解根据已知边界条件Vxy 60mm Vmax 1.08m/ s由此得抛物线方程Vx300y236yVx21.08 300 0.06 ydVxdy600 0.06 ydVxdy0.05 600 0.06 y 300.06 y01.8y 40mm0.04 m40 0.6y 60mm0.06m60 0第二章2-2解Vxx3si

4、nVy3x3 cosyVx3x2 sin yVy3x3 sin yVx土 3x2y3x3sin y 丰 0所以此平面流场不连续2-3解Vr2r s incos2r cos2Vr2 sincosVr2 sin cos4r cos sinVrVr所以此平面流场连续2-4解VxdxVyVzd dy d dzAe求解得Be当 0时（初始时刻）a, y b,zc解出A、B、C0e0,B,C则有:2V2.Rv12V1A121 -ea0e0b011ee 0z c即拉格朗日法表达式1,x1,y 2,zAeBe 1解得3e2e流线方程为3 -e ey1.5,x1,y2,z3 时51e 2e5 5- 132-5解

5、400vtqvvAqvv A12932734000.525 kg / m2738000,/ 36000.5252 0.4 0.68.82 m/s2-6解2P2V2A2d2d2V1 ALv2Al洽 0.44V2P P22V17水艮 gh 26558Pa1000 2V20.44v2 226558224qvd1222V2 65.87V2 8.12m/s2-7解方程小 8.12 a。120.064m3/sA面，取入口处为出口处为B面,以入口管中心为基准面，在A-B面间列出伯努利Pa2Va2-gzAPb2VbgzB已知Pa 68.6k Pa;Va4qv4 0.2 26.37m/s；3.14 0.22Za

6、 0Pb39.02k Pa;Vb4qv"T23.1401.59m/s;0.42ZbPa2VagzAPb2Vb2gzB6860010006.372210009.82039020 1000 1.591000 9.8 150084 38804Pa88888故由A7B2-8解 选取粗管道为1截面，喉管处为2截面，管中心为基准面。 在1-2面间列出伯努利方程2Rgz1V1P22gz2占 h失 1-2依题意可知:P1=101320+13550 X 9.8X 0.025=104639.75Pa； Z1=z2=o； P2=101320-1000 X 9.8X 0.15=9985OPaV14qvV24

7、qv则有:104639.7599850如2 d|16 q2104639.75 1293224 qV3.1420.0542164 q；99850 空3.1420.142 qV 0.175m3/s2-10解选取压气机前为1截面，进气管为2截面，管中心为基准面。 在1-2面间列出伯努利方程PaV2qV2-111)2V1P2Pa；PaVi0；7水艮 gh 1.25P2Pa水银gh2V2J2 13550 9.8 0.0274249.2865.19m/s1.2565.19 3.14 0.20.982.05 0.98 2m3/s220 ；h失0Pg1Pa10 ；0 ；222）热气体的伯努利方程Pg1Pa12

8、ViPg2Pa22V22Vl根据题意知2V2gggog,1327 31 2733 273 9.82.174 N/m31600aga0莎9293 273 9.811.806N /m3293当 H=2.5m273PgPa2.511.8062.17424.08 Pa当H=-0.5时PgPa0.511.8062.1744.816 Pa3）当炉门打开时，零压线以上逸气，零压线以下吸气。2-12 解P2gz2 h失Pa0.0105 6 5 -Pa 水g0.02 0301000 9.80.0105 10009.8 0.0230102.9 196 123.1 Pa第三章3-3解lv2P1P2 ； 1112 ；

9、 14qvd2111 d116qv122d14216qv222d：2d2qv1d；qv2d；qv1qv2d1df；d14AS14A27 ； d24A4ar2a254A55qv1qv22 r54A.52j r1.7721.834a3-4解对平板速度分布公式为VxVo12V0对圆管速度分布公式为R2P1LP2R2R2: 2rR2R2R23-5解1.510-6m2 /s4qV34 15 10 33.14 0.120.060.03141.91m/sRe Vd1.91 0.1-61.5 101.271052300 湍流3-6解8 10-2 2若增大Re,3-7解deReVde10-20.0114 10-

10、4则V增大；d增大;2.5 414032300 层流n减小。1 2.5 41.67 10523000.15 10-4A2.5m3-8解vd 18.87 0.75Re 一 9.415.7 104qmv -dr24 0.04 互 998.2 0.162204.43 Pa 0.05.5/602 0.086m/s960 3.14 0.02542O.。86 O.。2545.46123004 10-43-10 解v21) v徑 15.7 10 6m2/s 0.3/36000.04m/sd23.14 0.0520.04 0.051538.462300 层流0.013 106464Re1538.46O.。42

11、2 2h摩丄亠 0.042 旦 998.2 0.0413.42 Pad 20.052) v 笔心严 0.16m/s d3-11 解3.14 0.0520.16 0.056153.842300 湍流0.013 10查莫迪图0.04h摩1) v4qv4 30000/3600d23.14 0.75218.87m/s-61052300 湍流d 箫 2.67 10 4 o.000267查莫迪图0.0180.15m25 0.23-15 解l 匚 0.01822)查莫迪图1.2, c 1.6 7500.0243021.205 18.8720.7510 30.0016154.47 Pav20.024301.2

12、05 18.8723-12 解1)de2)de3-13 解1)2)3)Re3-14 解1)2)qv0.75205.96 PaS6a-d2d12S4-d2d1vd100060 /1 1vd1 0.031.56810-5vd2.5 0.0244AASReRe19001.568 10-5d1 d2)0 0.04-3400002300湍流2300 层流31902300小于4000过渡流4qvpqvAvA27004 100/360023.14 0.1521.57m/s2700/36006.25m/s0.40.325 0.23600qvd2V410 O.。823.140.05m/sV1d12V2d；828

13、.4V2V1d1210 0.0820.03271.1m/s3-16 解Pi2V12P22V2 .gZ2 h失2已知,Pi1.9621105 pa ；z120m ；P25.8861105Pa ；z10 ；v1v21.9621 105900 9.8 20-5.8861 105196210 176400-588610 -216000Pa0则知，由2向1流动。压力损失为 216000Pa。3-17 解Z10; V1V24qd2gh2V12P22V2gZ22 h失50 ； P2 1.01 10 Pa ； z1 2m ；4 15/10007.64m/s3.14 0.0521000 9.8 0.54900

14、Pa101000 10009.8 2 型g! 4900101000196000229184.8 4900154684.8 Pa3-18 解v2水V气气2 V水qv气V气 Aqv水V水 A2998.21.20528.782水V水3-19 解PP2f 2 PiP22rbb2qmvA2 Pi P22rb b23-20 解l v2v24q6168 Pa4 2.74 10 323.14 0.05261680.031010001.396m/s1.39620.0521000 1.3965846.4 947.40.3393-21 解P24qmd26000Re0.04R/d 13600 100023.14 0.

15、0421.327m/s5.3 1042300 湍流0.3 30.9vd1.327 0.04110 6开启角40°的旋塞阀2 17.32P0.04 匹 0.9 17.3 1000 1.3270.0428.2 880.5 24829Pa24.829k Pa若加大管径，则速度减小，压差减小，压强减小。第六章6-1解1)cVkRT(1.4 327345 J1850696.41360.4m/s2) cJkRTJ1.4 8314_273 56.6V 86855.5V 29294.7m/ sMa2400000/3600 2.26294.76-2解在1-2截面间列出连续性方程、能量方程、状态方程和绝

16、热方程qmL R1P11V1A1 2V2 A2 22RT1k P2 V；k 12RT22Pk2V22k2JRT1 T2 v1P2P1V1AT2A2T1J2LRTT2V12V k 16-3解CP4詈273 15辰3莎340m/sMa3401.47Pokk 12Ma2Po101325ToToIV2k 1M2k 1212k 12Ma26-4解PeePeR V12VeqmeVe AeqmeVeAedede4Aekk""1.4 11.4721.41.4 11013257356211.74 Pa1.43218 22731511.4 11.47412.47K13.21'4 9.8

17、1 1041.08 106kPe2Ve2P1Pe2.742.38 5331142.38kg/m22 1.4 1.08 106*1.42.16j4 2.16 10 39.81 104 屮.42.381013.23m20.0525m240m/ s2.3843.147Ma912孺隘愷2.22Ce 240L73MaAe2.06Q qF9-3孺0.02 05m2J4。|5 36.6mmN 3厶40.032 LI 6.4w、m20.025F 1.5 20 10 s 150CW032s1503024si、m2osb?1CDOl耳Qmmmq T|ogoCD5Th5OOoo5 oo004oco55O*K4565

18、32COk o00g242CO42COO52 ai2o2wm23k) ai322CO4s o3wm29-4孺9-6孺Q EFTF0.95567273 847 4 34 0.S25 7.98 S2W917孺0.7 567 s 8 2505504w、m29-8孺5677.06TJ FS 8 273 800 4 273 250 4 1=9-9解Qi2h t2 F 丄_丄，- hT302010cc c,10.003 r100.8400.240.1 0.00375 0.0257.255.9W0.128759-107000kcal/k g0.02kcal/h m7000 4.2kJ/kg 29400kJ/

19、kgC 0.02 1.163W/ m C 0.02326W/ m C第十章10-1r1h1900.01m2 C/W0.03456.6710 5m21h260001.6710 4m21021 88.15W/m2h1h210-2 解根据tw11)2)3)t w1tw2320270/0.03/236t w1tw2320270/0.03/43.2t w1tw2320270/0.03/0.9711/393333.33qqq21618.5 W/m72000 W/m2W/m2其中,3202700.8 0.00058 0.97114) qt w1tw2/320 270263.33W/m20.03/0.158其

20、中,0.099 0.0002 320 2700.15810-3解设0 1 bttw2根据Q如/ Ftw1 tw2 F14.98 0.02180301.31 10 415.25W/m°C20.01 0.02250 501.31 10 415.28 W/mC211000 1 105b15 251105b1 150b 一-一 1 150b15.281105b15.28 1604.4b15.252287.5b1.434520.290.26 10-3 800 1500.4315q tw1 tw3M1250-1500.345_0.1150.2405 0.27812121.1 曲b 4.39 10

21、 5015.1815.181 4.39 10 5t10-4 解 设 tw2 800 C1.4345 0.4135根据tw1 tw2 q -77tw2 tw1 q125012121.1 朦 zc800-739.87 100%800误差过大，需要重新计算。 假设7.5%4%tw2740 C0.840.5810-312507400.290.2610"3tw1 tw3tw2tw110-6 解qLt21.41710.3d22007401500.40571250-1500.3450.11511000.24340.28352087-68 W/m21.41710.405712502087.680.3

22、451.4171741.74 CJi1 , d2 In 2 d1137.77465In0.20m1504001250I d2 Inc c 一CCLC ccc " 3 400 2500.152 3.14 0.052 0.29 10465d 20.15200mm25mm2最小厚度为25mm。10.7 解t1 t2163qL400 5012d132 3.140.033 0.00023 10ln生400 500.11.143 ln0.1d2314mm314 100107mm所以厚度为107mm。10.8 解qLt1t21. d21. d3ln一 ln一2 1 d1 2 2 d2,130 In

23、2 3.14 0.1350350400 50_142 3.14 0.121301.17 0.698 187.37W/mqLt1 t2ln d1t2 t14001d 2qLln400 187.372d1219.2180 8C2 3.14 0.1350矿渣棉tmax550-600 C ；煤灰泡沫砖tmax 300 C ；增加煤灰泡沫砖的厚度t2增加。10-9 解1)t1 t22 1*2 2 d?,250 in2 3.14 0.06100tin型2 3.14 0.182502)2.43 0.422.85qL2t1 t21 d21 d3In丄In2 1 d122 d 2t,2501 cz 400 in

24、 06 in 2 3.14 0.181002 3.14 0.18250tt0.81 1.252.06 qL1 qL2所以第一种好（热导率小的在内） 对平壁，不影响。10-10 解qF50003.14 0.06 32826Wh tftwtwtf14-2 解Q12EbiEb2第十四章4T21001005.67111048301001000.0465.6715180.74745.80.0462721.6W14-3 解F1R2F2R21112F112F2212112222114-4 解1)3(1,2)zr van UESEIl »弓题图14-43132313(1,2) -320.25-0.1

25、80.07由相对性:II 11IIflur IPVBV raiSiiturESIIEHI：.3 71+237 2z/x0.80.4Y/X0.60.60.250.181F3 机1=Fi 如313F3ft313 50.0750.1052)(3 4)( 1 2)0.242-0.1670.0753+471+23+4 7 2471+247 2z/x0.40.20.40.2Y/X0.40.40.20.20.2420.1670.3380.2684 2F3(34)1F14(12)41F4411413F3410F4F114-5 解F141F10.075 0.15411342421414140.3380.2680

26、.070.070.150.070.070.08Eb2C0T114-6 解E11Eb10.8E22 Eb20.8Eb15.674096Eb25.671004T21000.025.67 409618579 W/m25.67 81367 W/m223224 W/m281459 W/m2b2J2Eb1TEb1T 1J11J1J2Eba1223224 45922765 15176 W/m21.50.82120 2J1Eb123224 15176 23224 3794 19430 W/m20 . 8J2Eb24590 2一 15176 459 3794 4253 W/m20.8G14253 W/m2R1G

27、10.24253851 W/m2R2G20.2194303886 W/m2A1G10.842533402 W/m2A2G20.81943015544 W/m214-7 解1) qiQ1Eb1Eb2LL 11 0.8 1 2F11 F1 12 F2 2其中：F1 2rLdLF210.815.67G2J119430 W/m25421qi3412)0.080.32-7711.22 rLdLF21.8120.81213.14 1 L 0.83.14W/m4 L 7.2LF11qi L3.14 1 L 0.85.67625 811 0.80.0814-8解设板R3R5R75.6754 341 0.83.

28、14 1 L 17.2L 0.80.32 0.047010.18 W/m3的两个表面分别为 3L及3R,房间表面为4,则辐射网络图如下Ri10.05133 F323 F20.05 119R4190.415 12.41R6122F 21 0.60.6 10.6671.709R8F 3 341.709R9R101.709Eb15.677.73420244W/m2Eb25.676.2348542W/m2Eb45.672.934417.9W/m20.58513对J11 1F11420244 - J1J 3LJ1417.9 J11.52.411.7098542-J2J3RJ2417.9 J20.6672

29、.411.709J1J3LEb3J3L417.9 J32.41191.70900对J2对J3L0对J 3REb3 J 3R19_417.9 J 3R J2 J 3R 02.411.709则:J191962W/m2 J256332W/m2 J3L23823 W/mJ3R28472W/mEb33155W/m2QiEb1J120244 9196Q21.57365WEb2J212F2 28542 56330.6674361W平板3的温度T3紅3155/5.67 100 486K14-9解设热电偶指示值为h1 tft1C0 T1遮热罩平均温度为t3,则有10041002h2 tft3C0T34Tw100

30、1002 25180273T30.65.67T3490 27350453 T33.402T3100100100173.6则T3439.5K40180 273T13.402T1100373.1贝y T1=450Kt1450273177Ch tft1C0Tw14-10 解T11001007.8%Tit1 V 100Tw100504.434 3.634170 14.4 184.4C测温误差：184.4-170 100%184.414-12 解Eb1 R1 J1R2J3LR3Eb3 R4J3RR5J2R6Eb2Eb1 Eb2 q1 =Eb1 Eb2q2-4-2445.676310.8445.67634

31、 20.84592W/m22296W /m214-13 解(1) h t f t1C0Ti4T210010058.2 tf 8000.45.674410.738.73t f 1090C1090-8001090100%26.6%174tf8000.45.67410.738.73 4tf897C897-800897100%10.8%174tf8000.45.67410.73410.23tf8300830-800830100%3.6%14-15 解3/ 1F1=F2=O.2 X 0.2=0.04F3=4 X 0.2 X 0.2=0.16查图(12=0.2 ,(1=0，贝U (3=1-0.2=0.8根

32、据相对性，F1(13=F3(31 , (31=(3F1/F3=0.2=(32Q12Eb1Eb21 1F1 15.671 1227311F2231F1 13427 44273 271001002R11R313.8912F 15.670.04 0.22401 810.04 0.80.04 0.86.250.008 0.01613154.414-16 解(1)把四周绝热面作为一个表面来处理时，辐射网络图如下:F1=F2=0.3 X 0.3=0.09F3=4 X 0.3 X 0.3=0.366.25 0.00831.25 31.2513154.416.25 41.67274.51W也2=0.2,如3=

33、0.8 ,()23=0.81 11F11 0.80?右 2.778R222 F20.70.7* 1一55.560.2 0.32R4R523F210.20.321 1 1 1RR3R4R555.56R3一1一0.0542 13.891.2表面间的净辐射换热量Eb1 Eb25.676444RR22.778 18.52 4.762226.3WEb1 J1R1J 2 Eb2R2226.3J16719.6W/m2J2=2529.16 W/m 2J3=4264.38 W/m 2=Eb3T3=534K14-17 解（1）表面3壁温已知,而且三个面面积相等,J3E3R3GEb3 (1A3) J1 13 J 2

34、 230T341J113 J223表面1、2已知辐射换热热流密度，根据EbiJi1i则EbiJ1E1Eb1R1G1(1A1) J33121J1J11q1 J3q31 J21q1J111J 331J 4 21111J331J22323同理J2 q2J 331 J112且3112230 .5由J1、0T14J2、J3的联立方程可解的j1、1JlqiJ2、J3,0T24J2 -q2(3)解出J3后，即可确定q3q3Eb3 J 3"118解查表，s 0.6d ,Ph2oSO.lbar 0.6m0.06 barpco2s0.2bar 0.6m0.12 bar mT 1400K，由图得H2O0.065CO20.10 Cco2H2O1.06由图近似的查得0.017,H2OCH2OCO2CCO20.0651.06 0