热质交换原理与设备

热质交换原理与设备习题课题目：传热学课本P305,8P305,11P286,15P286,22P287,25P305,8相对湿度为40％、温度为25℃、压力为1.0132×10^5Pa的空气，以4m/s的流速进入内径为8cm的竖直管，管内壁有25℃的薄层水不断淌下，试计算为使空气达到饱和所需的管长。P305,80.830.440.830.440.023Re0.023206050.6270.95ShSc可得3/23/220002980.22()0.25/273PTDDcmsPT622015.5310/,0.22/msDcms640.08Re206053500015.5310ud6415.53100.620.60.2510ScDDh解:先求，25℃的空气物性查得P305,840.251070.950.222/0.08DDhShmsd23,253289/3289180.0239/8314289WWPNmCkgm再求浓度℃时水壁面的水蒸气饱和压强则此时水蒸气的质量浓度2131,328940%1315.6/,1315.6180.00956/8314298PNmCkgm由已知入口处水蒸气的分压力则对应的入口处水蒸气的质量浓度:P305,832,0.0239/,:WCCkgml要使出口处空气达到饱和则出口处的水蒸气质量浓度要求所需管长则根据1221()()2DWCCmhdlCCC空气体积质量20.009560.02390.222(0.0239)(0.02390.00956)24dlu则7.2lm解得P305,11空气流入内径为25mm、长1m的湿壁管时的参数为：压力1.0132×10^5Pa、温度为25℃、含湿量3g/kg(a)。空气流量为20kg/h。由于湿壁管外表面的散热，湿表面水温为20℃，试计算空气在管子出口处的含湿量是多少。P305,112520:22.52mt解℃查得空气物性,6233/2215.5310/,1.185/295.50.220.248/273mskgmDcms3220/36000.00469/1.1850.004699.55//4VmsVumsAd空气的体积流量空气的速度P305,116649.550.025Re153733500015.531015.53100.6260.60.24810udScD0.830.440.830.440.023Re0.023153730.62655.25ShSc40.2481055.250.554/0.025DDhShmsd,(202338)PaS水面定于饱和状态故水蒸气质量浓度为℃时水蒸气PP305,1132338180.0173/8314293WCkgm211351/:/31028.91.0131048818wHOwaaawPRdPRPPa管入口处由得31488180.00354/8314298Ckgm2,C设管子出口处水蒸气质量浓度为则P305,111221()()2DWCChdlCVCC220.003540.550.0251(0.0173)0.00469(0.00354)2CC320.0122/Ckgm解得250.012283142981679.2181679.262210.3/()1.01310PPadgkga换热器热工计算的基本公式(1)mQKAt传热方程式11112222(2)(')(')QGcttGctt热平衡方程式P286,15冷却器内工作液从77℃冷却到47℃,工作液质量流量为1kg/s,比热容c=1758J/(kg.k),冷却水入口为13℃,质量流量0.63kg/s,求解在传热系数k=310W/(m^2.K)不变的条件下采用下列不同流动方式时所需传热面积(采用NTU法或LMTD法计算,均可任选)?(1)逆流(2)一壳程两管程(3)交叉流(壳侧混合,管侧为冷却水)P286,152:(1),4200/CJkg解逆流取水的定压比热℃1111(')11758(7747)52740QMCttw222252740'1332.930.634200QttMC℃12121212('')()(7732.93)(4713)38.827732.93''lnln4713mttttttttt℃2527404.3831038.83mQAmKtP286,15(2)一壳程两管程1122'77471.16'38.8213ttRtt热流体温降冷流体温升2212'38.82130.4''7713ttPtt1014,0.934.94tmtmtt逆查图2527404.8731034.94mQAmKtP286,15(3)交叉流1.16,0.4,10-14,0.8231.83tmRPt以查图℃2527405.3431031.83mQAmKtP286,22一逆流套管换热器，其中油从100℃冷却到60℃。水从20℃加热到50℃，传热量为2.5×10^4W，传热系数350W/(m^2.K)，油的比热容为2.131kJ/(kg.k)。求换热面积？如使用后产生污垢，垢阻为0.004m.K/W，流体入口温度不变，问此时换热器的传热量和两流体出口温度各为多少？P286,22解：(1)使用前面积(10050)(6020)44.810050ln6020mt℃422.5101.635044.8mQAmKt(2)'K产生污垢后的传热系数211'145.83/110.0040.004350KwmK℃P286,2212'tt(3)求产生污垢后的传热量Q及出口温度，？1122'-'tttt由于41min1142max222.510625'-100602.510833.33'-5020QCCttQCCttminminmax'145.831.66250.3730.75625833.33rCKANTUCCC，P286,221exp[(1)]1exp[0.373(10.75)]1exp[(1)]10.75exp[0.373(10.75)]rrrNTUCCNTUC0.28解得maxmin12max'('')625(10020)50000QQCttwQ又由，其中4max'0.28500001.410QQw1114111'(')'1.410'10077.6625QCttQttC又℃P286,222224222'(')'1.410'2036.8833.33QCttQttC℃P287,25质量流量为50000kg/h，入口温度为300℃的水，通过双壳程、四管程的换热器，能将另一侧质流量为10000kg/h的冷水从35℃加热到120℃。设此时传热系数为1500W/(m^2.k)，但在运行2年后，冷水只能被加热到95℃，而其他条件未变，试问产生污垢热阻为多少？P287,25111,''fRRRKKK解：原热阻运行两年后总热阻11','''fmmQQRKKKKAtAt污垢热阻，又''''''mmmmttKQQKKKQtQt2222(')(35120)77.54.193/()22fptttCKJkg℃查表得℃22221000(')4.193(12035)9900143.6pQMCttwP287,25111(300150)150225,4.6475/2ffpttCKJkg设℃，℃查表℃111119900143.6'300300153.4146.654647.5ffppQttMCC℃与假设误差不大，可以认为取值正确(300120)(146.635)143.1300120ln146.635mt逆℃设壳侧为热流体，管侧为冷流体P287,2522111222''120350.321.8''30035'ttttPRtttt300-146.6，120-3510-16()0.880.88143.1126tmtmtt逆近似查图，以单壳程代替双壳程，教材缺图℃2(3595)65'4.183/2ftCKJkg运行两年后℃，查表℃222210000'(')4.183(9535)6971663.6pffQMCttw111(300190)1902454.8/2ffPttCKJkg设℃，℃，查表℃P287,2522111222''95350.2261.743''30035'ttttPRtttt300-195.4，95-35(30095)(195.435)'181.830095ln195.435mt逆℃0.97'0.97181.8176.3mtt查图得：，℃2'697166126'1500754.9/'990014176.3mmtQKKwmQt℃42110.0013260.0006676.5910/'fRmwKK℃