传热习题

Q?WnCpn?t?14.65?1.842?103?(80?35)?1214?103w (3). 计算平均温差，先按逆流计算：

?tm'?(80?37)?(35?30)?17.6?C80?37ln35?3080?35 R??6.4337?3037?30P??0.1480?30查图得：??t?0.85.可用单壳程 所以：?tm???t??tm'?0.85?17.6?15?C (4). 初选换热器型号： 选k??400w/(m2??C),所以

Q1214?103S????200m2

k??tm400?15因为：Tm?tm?57.5?33.5?24?C?50?C，选用固定管板式换热器 由附录查得换热器规格如下：

D/mm 800

S/m2 200 L/m 6

管数n 444 程数 6

管子尺寸/mm ?25?2. 5管子排列 ?(正三角形排列) 2. 计算压强降：

(1). 管程压强降.苯为易燃液体 u气?1m/s

444???0.022?0.0232m2464V60ui?s??0.718m/s

A3600?0.0232du?0.02?0.718?879Rei?i??30790?0.41?10?3Ai?n?di2??湍流

取：?/d?0.1/20?0.005.查第一章??Re图得??0.032

L?ui26879?(0.718)2?Pi????0.032???2175Pad20.022Pu2879?0.7182 ?P2?3?()?3?()?680Pa22??Pi?(?P1??P2)FtNpNs其中：Np?6,Ns?1,Ft?1.4则：

??P?(2175?680)?1.4?6?1?23980P

ia(2).管程压强降：

??P?(?P??P)NF.其中：N?'1'2sss?1,Fs?1.15

则：?P?Ff?nc(NB?1)'1?u22

管子为正三角形排列.取：Fs?0.5

nc?1.1n?1.1444?23 取折流板间距h?0.3m则：NB?水的流量为：

L6?1??1?19 n0.31214?104Wc?4.186?103?(37?30)?41.43kg/sA??h(D?ncd?)?0.3?(0.8?23?0.025)?0.0675m2?Vs??A?41.430.0675?0.617m/s?960?R?u??e?????0.025?0.617?9950.745?10?3?20600

f?0.228??5Re??5?(20600)?0.228?0.519?P'1?0.5?0.519?23?20?995?0.6172?2?22610Pa?P'2h?u22?0.3995?0.61722?NB(3.5?D)2?19?(3.5?0.6)?2?8996Pa??P??(22610?8996)?1.15?1?36350Pa管程和壳程压降均小于50kPa，可视为合适. 3. 核算总传热系数 (1). 管程?i

?0.3i?0.023??d?R0.8eiPri0.1371.842?103?0.41?10?3 ?0.023?0.02?(30790)0.8?(0.137)0.3

?1025(w/m2??C)(2). 壳程??

??13??0.36d(deu??)0.55(Cp?)(??)0.14 e??w正

三角形排列4?(3t2??d23?2de?24?)4?(2?0.0322?4?0.025)?d取管心距t?0.032m?0.025??0.0202m.Ad?)?0.3?0.8?(1?25)?0.0525m2??hD(1?流通面积：t32.

u?41.43?995?0.0525?0.793m/s：

0.6220.0202?0.793?9950.554.186?103?0.745?10?30.33???0.36??()?()?1.05?3 0.02020.745?100.622?4770w/(m2??C)(3). 污垢热阻 取

Rs??0.00035(m2??C)/w.Rsi?0.0002(m?C)/w2?

(4). 总传热系数

dbd11???Rsi???k??idi?dm??0.0250.0250.00025?0.0251?0.0002???0.00035?1025?0.020.0245?0.022547701 ?0.0208k?478w/m2??Ck计478??100???120??k选400?所选换热器合适.