《水力学》李炜、徐孝平 习题解答4

'第四章4-1解：4Q40.004vd0.510.1v0.51m/s，Re5080322水6d3.140.11.00310水Re508032000为紊流。水0.510.1Re3402000，为层流。重油41.5104-2解：2AA0.52m，2.14m，R0.243m。RevR51020.2431211450061.0031065005001.003106保持层流V206410m/s0.21cm/smaxR0.2434-3解：vd120.45Re3.4102000为紊流。51.41105Re1.41102000ve0.282m/s对管流而言，此流速是很小的。cd0.14-4解：uu1hmaxln，断面（宽矩形）平均流速vuky*1hu*hhu*yyu*vulndyulndu，0max0maxmaxhkykhhku*hhhhuuln要uv则ln1，得y0.368hmaxkyye2.7184-5解：101T1100.5（1）uxuxdtuxdtuxt/100.5T0100.5011.882.052.342.32.171.741.621.931.982.190.5100.52.02m/s101T1100.5uyuydtuydtuyt/100.5T0100.50110.10.060.210.190.120.180.210.060.040.1100.007m/s0"2ux（2）（A点在x方向相对紊动强度）v26 "2uy（A点在y方向相对紊动强度）v"uuuxxx-0.140.030.320.280.15-0.28-0.4-0.09-0.040.17(m/s)相应的"2ux0.01960.00090.10240.07840.02250.07840.160.00810.00160.028922（m/s）"2"2"2"2220.5008/10uxux将ux相加得：ux0.5008(m/s)0.111(相对紊动强度为无2.02vux量纲量)"uuuyyy0.093-0.067-0.217-0.1970.1130.1730.203-0.053-0.047-0.107(m/s)"2相应的uy0.0086490.0044890.0470890.038810.0127690.0299290.0412090.0028090.0022090.01144922（m/s）"2"2"2"20.019941uxux将uy相加平均得：uy0.0677(相对紊动强度为无量纲量)2.02vux(3)A点的紊动附加切应力"""uu998.2yxxy0.140.0930.030.0670.320.2170.280.1970.150.1130.280.1730.40.2030.090.0530.040.0470.170.107/102998.20.02738/1027.33N/m。2""2dux22（4）uul27.33l0.26xydy27.33掺长：l0.6364m2998.20.26du2x22l998.20.63640.26105Ns/m(Pas)dydu2x22l0.63640.260.1053m/sdy已知水温，查得分子粘度321.00210Ns/m(Pas)621.00310m/s计算紊动动力粘性系数（又叫涡粘度）105Pas计算得紊动运动粘性系数（又叫涡运动粘度）0.1053(Pas)10550.10535两种相比1.0510;1.0510361.002101.00310可见紊动粘性远大于分子粘度，在均质流有4-6解：27 5vd2104Re,断面平均流速v0.113m/s263.1415104℃时水的运动粘度261.5722m/s100.1130.015Re10782000为层流61.572210222lv64lv6440.113h0.0103mfd2gRd2g10780.01519.6e2644v当水流从紊流转为层流时，Re2000,hf20000.0152g3这时流速及流量就不是0.113m/s6200020001.572210v0.2096m/scd0.01526440.2096h0.0191mf20000.01519.64-7解:3重量流量G(m/s)2373237333体积流量Q245.7m/h0.0683m/s4.662220.0683Ad0.7850.30.0707m，vv0.966m/s1240.07070.9660.3t1=10℃时，Re115.92000为层流。1425100.9660.3当t2=40℃时，Re19322000也为层流。241.510432vl3225100.9665000h438mf122gd9.80.34321.5100.9665000h26.3mf229.80.3NQh96600.0683438288982w289KW1f1NQh96600.068326.317352w17.35KW2f24-8解：t=15℃的水的621.13910m/svd3.00.35R7.9102000为紊流e61.13910h22fv0.0153水力坡度J0.022960.023ld2g0.319.6（1）r0处切应力0；0.5处切应力RJ0r11∵R0.5r处的R0.5rr000224r0.150RJ98000.02398000.0238.45Pa44du63（2）1.1391010004.344.9410Pa1dy28 22du22y2l1000ky14.342dyr00.5r222010000.40.50.15118.84r05000.000918.848.478Pa8.450.004948.445Pa2总122du8.445Ll0.021m2dudy10004.34dyyl0.21ly1，0.395970.4r0y0.50.150.5y1r04-9解：2JJr220urr，而v048uv得：r0.5r0.707r处可量得圆管层流的平均流速004-10解：2222ada26，得：d1.128a;b0.707a4dabJ；J；J。R:R:R0.282:0.25:0.236y圆方矩123443：：0.282:0.25:0.2361.0:0.886:0.837园方矩22222lvvu*2v2u*hJ;又8uv8f2*4R2g4R2gv822uu8RgJ8RgJ**2J8，也有vv8RgRg由于J互等，也互等vR，而A又互等，Q:Q:QR:R:R123123Q:Q:Q0.282:0.25:0.2360.531:0.5:0.4861.0:0.9416:0.91512314-11解："12udAv2AA2"1r0J228r0224对圆管层流：rr2rdrrrrdr1.33200600Jr24r3002r0829 21hJ对宽矩形明槽："2y2hydy02J2hh39h29h243222hyydyy4hy4hydy4h504h20h59y4423914hyhy154h5345309861.241554-12解：nyuu，mrn0r0yu222maxrv2rydyu2rydyr0r000maxr00n1n202umaxv2umaxn1n2v,或n1n2un1n2v2maxn1n2uvmax23n33r0n1n23yv2rydy00138r0udA3232vrvr0A033n1n243n13n22n22r0n1n22yv2rydy00"124r0同理：udA2222vrvr0A02n1n242n14-13解：nnyhyuumaxh，单宽流量q0umaxhdynq1yyy断面平均流速vhumaxhdh,令h01u则有：nmaxvudmax0n133n131h3n1h2y313n1n1udAudyn13020maxvAumaxhh3n103n130 22n12"1h2n1h2y212n1n1udAudyn12020maxvAumaxhh3n102n14-14解:是一道绘制水头线的绘图题，数据来源例题4-12（p.164）4-15解：30.864Q40.9410甲管：0.0043,v0.03m/s22d200d3.140.20.030.20.3164Revd5268,10.0371(假设的）甲61.13910R4e32.8d32.80.26.56粘性底层厚度0.00646mlR52680.03711014.8e0.860.1330.4校核为光滑区。l6.4622v0.03710.030.00003339J0.00000852d2g0.219.63.9230.194Q43.510乙管：0.00457,v2.787m/s22d40d3.140.042.7870.04Rvd97876,先假设处于紊流过渡区e甲61.13910用Jain公式（应为Swamee-Jain公式）121.2521.251.142lg1.142lg0.004755.670.90.9dRe978760.250.25680.19680.25用0.11计算过渡区0.110.110.004750.000695dR0.0497876e过渡区0.03与0.0311的误差为3.54%10.0311,25.6732.8d32.80.04粘性底层厚度0.000076m0.076mmlR978760.0311e0.192.56,但大于0.4校核确为紊流过度区0,076l2v0.03112.7870.2416J0.308d2g0.0419.60.7844-16解：10.120.4330.001;0.004;0.03100d100d100624Q40.015t=20℃运动粘度1.00310m/s，平均流速v1.91m/s22d3.140.131 vd65R1.910.1/1.003101904301.910，查moody图（p.148）e得：0.057阻力平方区，0.03过度区，0.0208过度区32122lv10001.91h0.02080.0208186138.7mf11d2g0.119.6h18610.03186155.83m；h0.0571861106.1mf22f322v1.911，J0.02080.02081.8610.03871d2g0.119.62v12J0.031.8610.02081.8610.0558，J0.0571.8610.106。23d2gd2ugRJ,R0.025m,g9.8m/s所以有：*4ugRJ9.80.0250.03870.0974m/s,*11u0.495J0.4950.05580.117m/s,u0.161m/s*22*3损失功率：N14738.75689W5.689kW1N14755.838207W8.207kW2N147106.115597W15.597kW34-17解：在层流内，圆管64，与R数线性反比。hR1；Ref1ee在光滑区，0.3164，与R的1次方成反比。hv1.75；1R4e4f2e2在紊流粗糙区只与相对粗糙度有关，而与R数无关。hv；ef34-18解：梯形断面面积2Ahbmh3101339m断面平均流速vQ391m/s。湿周b2h1m2A393910231118.5m,水力半径RA2.11m18.5431212R22用曼宁公式vR3J2vJJvn4nn2R322J10.0240.000148，这表示每公里长渠道上损失沿程水头0.148m,2.113ugRJ9.82.110.0001480.0553m/s,*28u80.05532利用书上公式(4-4-14,p.136)*220.0245v14-19解:列上下游水池的能量方程,并忽略水池的流速水头,32 4Q40.5有:Hhwhjhf,已知条件中有v222.548m/sd3.140.522vlvh2h,j进弯90阀门出f2gd2g先计算hj;进0.5,13.53.5r20.25029020.1311.84620.1311.84610.291R90122.548阀门5.52,出1.0,hj0.50.2915.521.019.6vd2.5480.56hj7.3110.3312.42m;再计算hf:Re61.27101.003100.60.0012查Moody图,得0.021,故h0.0211000.3311.39md500f0.5hwhjhf2.421.393.81m4-20解:2lv按长管计算,H管系统两端列能量方程得v2gHd/ld2g0.3160.316(1)光滑管区时,10.250.25R4vde0.250.250.250.250.250.3160.316有;I0.250.25II0.250.25vdvdIII在光滑管区中,与R有,当d,H,一定时,e10.125l0.25QIIVIIIIIIIlIVIIlIVIIVIIlI即:0.25QIVI1IIlIIVIlIIVIVIlIIlII0.8754V1V17QIIIIII故;1.1791.179;V0.75V0.75QIII(2)完全粗糙管：当,d一定，即就一定，这两个方案，V1IIH,d,g,一定，V1，1.155lV0.75I4-21解：列河流至抽水机进口之间的能量方程：p222ap泵进口vvlv00h2g2gd2gpp22a泵进口lv2012vh1210.2160.3d2g0.152g33 0.6求：0.004先假定位于阻力平方区，查Moody图（不必先知道R）ed150得0.004时的0.0275d2pap泵进口32v21.00.21.06.00.30.02756mHO20.152g2v419.6414.37,v2.336m/s2g14.373Qd2v0.7850.1522.3360.04126m41.26升4s秒校核流区：Revd2.3360.1563493502000紊流，1.0031032.8d32.80.150.0275,0.000085m0.085mmlR3493500.0275e0.67.066,校核确处于紊流粗糙区（完全粗糙区，阻力平方区）。0.085l4-22解：用曼宁公式，借用书中例4-11（p.164）的成果21d78.4nR3，衬砌方案：d7.0m,R21.75mn0.0142242不衬砌方案：d7.8m,R1.95m,n0.033,0.0683411112178.40.0141.7530.015370.830.01275;24Q4700m22v14.66lvlv122s12d3.147.8hf11,hf22;d2gd2g470012v18.2m23.147.02s220014.66h0.0683419.21mf17.819.6220018.2h0.012756.16mf27.019.6可见减小糙率将使水头损失有较大的下降。4-23解：22v1vvv2dhjh,02vv2vvj122g2gdv22得：vv1v2,hv10.5v10.5v20.5v10.5v2v22j改进2g2g34 1212vvvv1221h144j改进hj改进2g2gh2j原4-24解：2v∵d和Q相同，所以相等，局部损失系数决定了局部水头损失值2g24Q40.06vv1.91m,0.1863m22sd3.140.22g（1）圆管90°折角管，1.1,h1.10.18630.205m1j1两个45°折角管，0.35,h20.350.18630.13m2j23.5120.190一个圆角缓弯管，30.1311.8460.1316190h0.13160.18630.0245mj3可见做成缓弯段能减少不少水头损失。1.910.2（2）R3808572000，紊流e61.003100.30.0015，查Moody图得0.046d200d折算成等值沿程水头损失的等值长度用公式1d1.10.2l1.14.3484.783m10.046d2l0.74.3483.044m1d3l0.13164.3480.5722m14-25解：vd0.40.0140010Re39880.5，紊流61.003101.003246C0.40.40.0060.0940.5DR1Ree2u0210000.42FCA0.5d0.50.7850.0052000.001963ND2424下沉速度vs1gd，s为小球的比重3CD2223vC3vC30.40.5DDs1s1110.6121.6124gd4gd49.80.014-26解：35 uu1ruymax0，*lnuulnmaxukykr*01r0u*yr0u*r0rrrvumaxln2rdr2umaxlndr0kr0krrr200000r1u令*,v2uln1d0maxr0k下：当0时t1令1t,1t,ddt,故有上：当1时t00u*1u*v2ulnt1tdt2ulnt1tdt1max0maxkk1u2u22umax*tt3*2utlnttlntumaxmax2k242k0证毕！36'