传热学课后答案方便打印版

1、绪论思考题与习题（仔_9）答案：1.冰雹落体后溶化所需热量主要是由以下途径得到:iiq=At，=constT直线CT舄手const而为舄=舄（t）时曲线a,o金/ZZ/Q、一一与地面的导热量Qf与空气的对流换热热量注：若直接暴露于阳光下可考虑辐射换热，否则可忽略不计。6.夏季：在维持20C的室内，人体通过与空气的对流换热失去热量，但同时又与外界和内墙面通过辐射换热得到热量，最终的总失热量减少。（T#汗内）冬季：在与夏季相似的条件下，一方面人体通过对流换热失去部分热量，另一方面又与外界和内墙通过辐射换热失去部分热量，最终的总失热量增加。（膈寸内）挂上窗帘布阻断了与外界的辐射换热，减少了人体的失热

2、量。7. 热对流不等于对流换热，对流换热=热对流+热传导8. 热对流为基本传热方式，对流换热为非基本传热方式门窗、墙壁、楼板等等。以热传导和热对流的方式。9. 因内、外两间为真空，故其间无导热和对流传热，热量仅能通过胆壁传到外界，但夹层两侧均镀锌，其间的系统辐射系数降低，故能较长时间地保持热水的温度。首先通过对流换热使炉子内壁温度升高，炉子内壁通过热传导，使内壁温度生高，内壁与空气夹层通过对流换热继续传递热量，空气夹层与外壁间再通过热传导，这样使热量通过空气夹层。（空气夹层的厚g壁炉的保温性能有影响，影响a。的大小。）a13.已知。=360mm、=0.6lW(mK)当真空被破坏掉后，1、2两侧

3、将存在对流换热，使其保温性能变得彳艮差。io明=咯=普吐1.2,-2=8.3310m12何卜87%*)上=一0h2=124%m2,K)1>qI耆毒=旦=285-1501082.3KWR,t7.40710£cr如切珞15已知dj=50mmI=2.5mtf=85顷求：q、tw1h=73%2K)q=5110W<2i。i+hi-h218-(-10)10.361870.61124=45.92q='(tf1t叫)-tw1=tf1q837.54=18=17.57h187q=如-心=tw=tf+，=T0+37.54=_9.7。wf2h2124=qA=45.922.83=385.7

4、3W14.已知：H=3mer=0.2mL=2m兀=45%.k)七=15。5=285。求：tw'*Iq=h：t=h(twjtf)=tw=tfh5110=85+=155c73=Aqndjq=0.05二2.55110=2006.7W16.已知:tw=50c、M=20c、寸3.96%?”)f求qi.2q.2Aq.2解:q1.2=G.2Itw4(二)4IL100=200。如4一(100)=3.96139.2%273504273204()-()IL100100cyRt=tAHL0.2v=7.407104532Kw，Lq.2=g.2w1、4CJRt=一工=4.44410，m2452732004273

5、204急96()一(F)=1690.3Wmm22285T50w10-30.4月q3R.4.44410q1.2=q'1.2q1.2=1690.3139.2=1551.1%2,K）217已知：A=24mh1=5000第一章导热理论基础h2=8t1=45。K）1思考题与习题(P24)答案:t2=500"C5=85%2.心-=1mm&=398%m.K)求：k、巾、A解：由于管壁相对直径而言较小，故可将此圆管壁近似为平壁即k=111h1h21,略2日”"°.62%nK）'2=0.65W（mK）'3=°.024W（m.K），0.016

6、W（mK）'-12L-42R,=12425920.620.650.016101_311101500039085=83.56,k）-2-32562R=0.265mk/W230.650.024=kA=83.5624>(500-45)勺"=912点KW,双Low-e膜双真空玻璃的导热热阻高于中空玻璃,也就是说双Low-e膜双真空玻璃的保温性能要优于中空玻璃。85-83.5683.5611£1因为：,hih2h2=1.72%即：水侧对流换热热阻及管壁导热热阻远小于燃气侧对流换热热前，此时前两个热阻均可以忽略不记。3.略50.118.略6已知：cr=50mmt=a+bx

7、2a=200c、b=2000C/m2、=42)qx£qv代入式（1）,合并整理得:K)2-dtp4cI0dxfqx=/*x="2bXx=0该问题数学描写为：.2,dt_二bUT4_0dx2fdt/udx-2bx(2)由qv略略史Jdxx=0,t=T°=_45乂2乂(2000)x5010里.dtxT。dxxmd2t%-=赤2b=T5x(2000)x2=1801照10dx9.取如图所示球坐标，其为无内热源一维非稳态导热故有:.:taf2ft一="r:r:r.:r-=0,t=t°r=0,=0frr=R"=h(t-tf):r10.解：建立如图

8、坐标，在x=x位置取dx长度微元体，根据能量守恒有：QxQxdxdt一赤+dxd=九十dx(tdt-dx)dxQ；=EA=；EbA=-bT4(Udx)=9103W2=0(假设"也=8bTe4f(真实的)dxx4第二章稳态导热思考题与习题（P51.53）答案33,解：（1）温度分布为tw1'tw2)t=tw1tw1一tw2x(设其与平壁的材料无关的根本原因在？=coust即xx4-tixIxdt2)由q=&知，q与平壁的材料即物性有关4.略、=0.7W/(mk),2=0.58W/(mk)"b05.解：r辛ri,t=twi(设twi>tw2)r=r2，t=

9、tw2-4K,、有'Q=(tw1-'tw2)1&3=0.06W/(mk),q2=0.2qtw26.略2-14二，r2r17.已知：I=4m,h=3m,、=0.25twi=15c,tw2=-5c'如=0.7W/(mk)求：Q解：二I,h=6,可认为该墙为无限大平壁8.已知F=20m2,5=0.14m,tw2=15"=1.28W/(mk),Q=5.5103Wtw1情1r况下的内外面12X1I2-2-32315-(-5)=0.7(43)0.25tw20tw1tw2q2'1.'般Ftw1=T50.14=15求：tw1社解：由Q=F得一无限平壁的

10、稳态导热535.510201.28再由q2=0.2q1,有tw1-tw2.tw1-tw2-_0.27；二一2W12nnnrtrt1'2'31'2得224020，彳项一1圣)=40.06()=90.6mm9.已知q=240mm,心2=20mm7勺姻0.70.5810,已知twi=4500.6=25050=500mm0.1212,已知twi=600c,tw2=480c,tw3=200c,=0.0940.000125t,tw2=502<340W/m2求：解°Ctw4=60C求:R1R2R3R，R，R.=呵迪,m=0.094i.25i04七产解:由题意知其为多层

11、平壁的稳态导=m-=0.094i.25i04twi2侦2q=1一Lw223R2445050450-50=0.094i.25i02340twi-tw2=0.R474rtWi-tw4R:iR=R+R+R600-480】23=0.22600-60R3即有q_340W/m时有i47.4mm11.已知§=i20mm,=0.8W/(mk),2=50mm,2=0.i2W/(mk)R2_w2-tw3t,一t,kwiw4480-200600-60=0.52tw3tw4解:i一tw4200-60=0.26600-60、3=250mm,3=0.6W/(mk)求:略已知、=2mm,豚=40W/(mk),、0

12、=3mm,tfi=250tw2Twi，_tw2-tw-,q=fYn'i2'3'i'3twi32itw2由题意知：q=q即有:tw2-】wi、3'i3：3.2一iItfqtwi3矿3tw?ctf=60ctf2a2.2.、.2.、0=、,'=75W/(mk),h2=50W/(mk)3)求:q0kk31'0q12)62=3mm,兀=320W/(mk)'2)&=80,A3=由h2=70W/(m叫，.归2,q,k1,k2,k3k)tf1f2250-601h1E1'0h2175310；401+501、11-rrh'1h

13、212101TT7540501h2_1310-17540702=36.11W/(mk)2q3=k3.t=36.11(250-60)=6860.7W/m2=5687貌昂q3q。=6860.7-5687.2=1173.5W/m2q3=AqAq,第三种方案的强化换热效果最好=29.96阴儒.雄=35m*B=130mm，其余尺2=k1t=29.96(250-60)=5692.4W/m22q=q1-q°=5692.4-5687.2=5.2W/m2寸如下图所示,A=C求R.k2q2=1.53W/(mk),b=0.742W/(mk)解：该空斗墙由对称性可取虚线部分，成为三个并联的部分l_-=-=.

14、J-J1A1B1C11JA2C9.99W/(m2k)1工3勺0'275320_-_5A)r_C3_L111.'2.10h1'2h2=k2t=29.99(250-60)=5698.4W/m2q2=q2-q0=5698.4-5687.2=11.2W/m2R1R1R1l1L_-.111_rI1R2R2R211»-11iI1R3R3R31IIIRR1=Ra1Rb1Rc1=R3R=Ra2Rb2Rc2c3510-=21.5313010RR3O30?(mL-W/V17=1.66410-(mk)/W1.532"di58160-、A2、B2、C3c3510；R=/_B

15、2_C3=2'A2'B2'C31.532；2心R.20.拙柿华0.7422二0.09316.In17060=0.517(mI170R1R22工0.13070.2212、22-3=5.0R,3102'22R，1R"：R，22二0.17,1706080°In=017060d1=160mmd2EZamm,、=58W/(mk)2)、2=30mm,2=0.093W/(mk)R2R3300-50314.1W/m0.5170.2793=40mm,3=0.17W/(mk),=3003)由qitw1-tw2R1C，tw4=50C求:1)R丸，%，Rm:2)qi

16、:3)tw2,tw3.tw2qR,1=300-314.11.66410=299.95解:c同理:tw3=tw4qR3=5。314.1o.279=137.631，17.已知：&=&2,7吃=匚兀，d2m=2d1m2求:qiqi解:忽略管壁热阻indo2、1do变）qiqiq=0.783qn由2'1222；2dmd2m1R)=in2'2do2、.ido1in2"qi12do：tR.JU15'in3in5R=23=1.277R-in3in-2q-q=21.7%-qdo*箱如损失比原来减小21.7%。18.B知do2：do2、1（管内外壁温twi,tw

17、3不d=1mm,dodim2tW212tw1R=2.221o"/m,=0.15W/(mk)tw1max=65C，tw2=40c2qido2、11.do2、12、2nin2：*2do25do2、”、=o.5mm,1.do2、11.do2、12、.inin2fdo1in2do-2do2；2do解:21maxind°4'1do2、1ind°jdo2、1max(do2)=d°1d1m(d1m2、2)=do3、12即19.d1d2m=2dm=do3、"=2(d°)=d°（代入上式）word文档可自由编辑"Cctt2w

18、1maxw2-1maR-1ind2ttw1maxlw2Rd2、din2皿65-4o2.221o项，12o.5n2二o.15=1=85mm,d2=1oomm,4oW/(mk),tw1=180tw1tW2tw3!I1'-ind2+d22=0.053W/(mk),tw3£40qi=52.3W/m1/11、1,(-)(4"114：w11d1(d121)(d121)2(25273.15-77.4)：tqlR1R'2tw1-tw31,d21,d22、2ln工ln22"d12-2d21乂16.31/11()0.030.3560.416或:=102.7W整理得:L

19、lndZ)(e2睫.0.053(180"0蝴翳tw12二(25273.15-77.4)0.031：1)=72mm11=102,7W(r2一r3)(0.356一0.416)或：R?=R角-故有_：t_tw1Iw3一I顽1耕2'22*2d2t12ft一=2=也伦ql-1)=72mmtf,hik20.已知：；t,1t>t2fd1c1=KM7*kg/hr199.6!ik_=0.35mm,S=3mm,%=30mm,r=199.6kJ/kQtw1m=21.略22.略"Cd-m2。=0,8=ttfX=0尸-，"tftw3=25=0.03W/(mk),1=16.3W

20、/(mk),=1h3解Xl广-弓=t2Tf求:解:tw3一版Q=RFiRf2解微分方程可得其通解：Qmx二GeRR.tw1入F1tw2入F2tw3I1>IlI_mx1111c2e4兀(r-习顽以为由此得温度分布（略）F=-=ch(ml)ch(ml)%-tf24.、=3mm,l=25mm,=140W/(mk),h=75W/(m2k),t0=80tf向伽1。84ch(2)-4°弗93ch(ml)-1ch(2)-1°Cch(2)=3.7622解：ml=hUl=&L2hl=0.025,一275=04725100%=99.93一84100%=15.9%"A、L

21、"、'14031。tf99.93m=18.926.已知6=0.8mm,l=160mm,t0=60c,chm(l-x)830)ch0.4725-18.9对6.3W/(mk)，其他条件同25题'0顶""30)ch(0.4725)求：.Lt=44.91ch(0.4725-18.9x)t=3044.91ch(0.4725-18.9x)mlE“c20人”,l=160，'i36.2716.30.810QhU2hql=Wh(ml)=u°th(ml)LmLm275=(80-30)th(0.4725)=174.7W/m18.925.已知：t_tlc

22、h(ml)-t。84ch(6.27)-608409fch(ml)-1ch(6.27)-1"Cch(6.27)=264.24=15mm,l=20mm,，=48.5W/(mk),tl=84七tf-tltf100%=84.09-84X.84.09100%=0.11%c，t0=40ch=20W/(m2k)27.已知：&=3mm,l=16mm，=140W/(mk),h=80W/(m2k)ml"Xlhdl=,二dl=0.1220一248度)1'.54卵/k),h=125W/(m2k)求：f解ml=hUl-Al=1610；1"2h'旱fF33"

23、0)1140310=0.312501.3410=0.82128.th(ml)th(0.312)72ml0.312=0.9733.5=2.15fml=th(ml)ml已查图得：f=0.781610"40123：名戚印煮Q=0.853Q1=fQ0=fhF(t0-tf)0.73知22二2二(房-12)fh(t0-tf)d1=77mm,d2=140mm,、=4mm,P=25mm,，=50W/(mk)=2二(726lc=l=33.52-38.52)10”0.7860(320-75)=266.7Wtf2h=60W/(m2k),t0=320c每米肋片管的散热量为:=75。q=nQ(n-1)Q2ql

24、n=四00+1=41片/米25=41266.7401.48=11kWQ2为两肋片间的表面的散热量Q=二dFOtf)r2c=1L=72f=、(r2c一)=410侦(7238.5)10-17.3410034r2510(320-75)=1.48W.已!七=3x2.2m2,5=0.3m,%=0.56W/(mk),tw0tw2=30CI求：ql解：l2llSiAiI1L63L10L0.3cII2CI1H,解：S1=,S2=1=I2QQS2A2I2L2.2L0.3=7.33LS3=0.54H,S4=0.54S3qI=0.54L1I1,I2'5-twi-tw2(2S12S24S3).：tQ=(S14

25、S24&4S4).：tI1I24IiH0.520.5240.5204:Lt_4&_4S434-40.540.42-40.023(3014)=(21027.3340.54)0.56(30-0)=618.6W/m=3.62102m=36.m2n31.已知d=165mmtw1=90c33.、=5mm/=2.54m,P=2MPa,.：t=80=1.5m,舄=1.05W/(mk),tw2=6。2h=20W/(m2k)tW2解Iar,H>3r入12叫otw1c"=180W/jd，(mk)In(2H)rQs档2皿=地II.,2H、In()r求：itc解：由=2.54Pm,P求

26、:qiqi2二1.05=2MPa,查表得，21.5(90-6)=154.2W/mRc=0.8810顼(m2k)/WIn0.165/232.I1l2=0.520.52m2,Ht1tw2=-14c,Qc-0.42m/=0.023W/(mk),tw1=30=34Wt1AIt2B1I2t2ARct2Bt3Q=-，AL=t2A'UbRc-t-tf解：由一=1%-0'o-仔R0.8810t=tf+0.01x(t°tf)=120+0.01x(25120)=119.05匚tc=4=380495104r2+Rc2乂+0.881七c180"C第三章非稳态导热hVBiv=FhR9

27、50.510'2322=3.6100.1M=131.略2.略3.略4.略FOv5.已知：,故满足集总参数法的求幡条件，有:d=0.15mmcp=420J/(kgk),8400kg/m3,h1=58W/(m2k)01h2=126W/(m2k)求：-01,-02hF七pV=T=hF8.In。0国840030啷03h1138930400-0.51039522358h=39W/(m2k)"=48.5W/(mk)_2ln(110:t0=300ctf=20c02ps%p房a=12.710-62m/s,t=50c6.略7.3h284004200.15'危勺0=0.7(s)t解231

28、26BiV303103248.5=0.9810侦故有:=0.5mmP=893。原/m3,C400J/佝给"25解""Ctf=120"ChF了CpV口02口c,=95W/(mk),=1%,=22W/(mk)%求:,t（康铜）?CpVQInhF气VIn一haF气c/348.51：1026ln3912.710*150-20300-20二328saA-A9.略、.2Ai0.已知t0=80c,d=20mm,tf12.u=12m/s,t=5min,t=34c.3-8954kg/m求：h解：假设可使用集总参数法,查表得：FoA=0.24=FoBABaB=.A=-B、B

29、2A©BJ2=12x2=48min3一axbkc=0.5x0.5x0.5m,t0=30ctf=800J九=52W/(mk),cp=383.1J/(kgk),，=386W/(mk)22a=0.063m/h,h=80W/(mk),=30min故有:elnF%hF二-尸%求：tm解:Bi-112038954383.1102252-c=2.6h'8010.52560,34-20Foq?07、.2对2、26尹睇蹭0尊5k)36000.252一.于hVhRBiv=上-室83.220102386tff_mtf-t0%=2.161(3：0M1=re)3m扑J平板0.12B=2.6Fo0.佳图

30、有：二满足集总参数法的计算，上述假设成立。ii.<6m10.砰板=2'b,A=，B,CpA=CpB,tA=tB,tf,hA=hB:,-B=0.9=12mintm=tffe_mChI0J平板3=800-(800-30)0.9=239"C13.求：.A=40mm,a=510=m2/s,，=4W/(mk),t0=25Bi4mB=°"。C,tf=1260c缺少2hh=40W/(mk),1655.8X4=0.45m14.已知：tf=40ctw=20c535.5%=。.8叹l=0.45m5Ree=5105e口=2ax图略=0.1mx2=0.2mx3=0.3mX4

31、=l15.已知：l=0.3m、u=0.9S求：:xa解：tm_1-2按tmXif+tw)=30c=30c查表得Pr=5.42、tf=25c求'maxU|(y)由tf=25c查=0.618%mk)寸8.05'10'v=9.05510由Rexu：xx1=0.1muJRe|0.30.9Rex=9.94104v9.05510*2.98105：ReceRex=1.99105Rex5=2.98105Rex5=4.47105x1=0.1m803.9X2X3-0.2mRe均为层流=0.3m=0.45m=0.3321136.32k)4.6414.643一0.3=2.5510mRei2.9

32、8105u"2(丁max、1,y、3一2、u：:Jmax30.9322.551013y0.9(3y)32.5510Pr13Rex'2x2=0.2m16.17.18.x3=0.3mUi(y)=529.4y2.71107y3已知U8=10%tf=80"求:解:tw=30cl=0.8mRe。=5>d05b=1m1，、一tm=(tf+tw)=55c2tm=55c=1.846X10直mZPr=0.697、k)=2.865102u一xcRecvRe。=xvl:xc.全板长均为层流dx2即:：d§)dx二U.u5U.一-const即有:dx.51051.86410

33、10两、d、.a=2%=0.664土Re2Pr13U一一'-AJ-0.923m、.2危：6“U一一2.8651。'=0.664,100.822.0.81.84610*)20.697U(1u一一'-AJdx=13.92k)=Ft=13.910.8(80-30)556Wxv6dx=)UX.22=6Vxu:：.12Rex1Rex22,23.已知t=aby+cy,twtf求:由边界y=ot=tw19.略一,2dx20.略x心uy521.已知：=层流求：Ug&x一得:c解：由动量积分方程有：0dt2=0t=tfa=tw、b=(twtftw一tfdyE(-b2cy)y。bt

34、w一tf24.略1r。25.略a=c(rdrQ=:Ft有：Rei=口项1=5012.31106v12.161011,-L26.略a=-faxdx,Q=aFAt27.略adtdyRe2U：2l2V28.0851.74610*=2.31106228.已知F=3mI1=1m知Re1=Re2Pr1P&，且几何相似求:tm1"Ctw1=140cI2=5m捉=7021,=2(折tw1tf1=30c、Q1=15000w、tw2=20c°C1)"(14030)=85得:而:11扁"(捉"以070)=45按稣tm2查表得:Nu1=Nu22I2NU1I1I

35、2F1“t13.810_23.091029.30.I1l2已知:Ga母00=8.24w(m2.Q53乂(140-30)da=16mmdb=30mm=2Gbta=tb求：（1）是否相似（2）如何相似Pr=0.691V1k)=3.0910°=2.1610*解：（i）ta=tb，且为同类流体二Va=*Pa=PbP2=0.6985k)=2.810V2=1.746X10'mZ有:UmaumbRea2d即:Rea知:Ga:a-d"aa4Gb二.2bb7dumadaVaUmbdbVb=1-Reb两者流态不相似(2)若要相似，需Rea即:Reaa=1>Reb而:UmaGadb

36、2umaUmbGbdadadb2dbda2da今=2Umbdbdbdad；db=RebUmadaUmbdb,带入上式有:1683015即：要使两者相似，两者的质量流量之比应为31.32.33.34.略略略略=1Gad；d2815第六章单相流体对流换热及准则关联式1、(1)、不同。夏季一一热面朝下，冬季一一冷面朝下(相当于热面朝上)。(2) 、不同。流动情况及物性不同。(3) 、有影响，高度为其定型尺寸。(4)、在相同流速下,(ReUd在相同流量下，d大tRe小t0小(5)、略2、略3、不可以，其不满足边界层类型换热问题所具备的4个特征。4-15、略16,已知tf1=10C,G=0.045kg/

37、s,dl/d10,满足管长条件-t'tf2=110CHmtf1:tft=51mm,l=2m,tw=200Ctw1t''tf2.=tIn：t190-90133.8C,190In90tf=twtm=200133.8=66.2C按tf查表有:f=19.621m2/s,Prf=0.695,f=2.93710袖Cpf=1.0075kJ/(kgK),七=1.041kg/m3,f=20.4G4d"f,RefUmdG一30.045二.饵一.-511020.414f0.80.40.023RePrfdff.管内流动处于旺盛紊流按迪图斯-贝尔特公式计算4080NUf=U.U乙UQD

38、f30.023(5.5104)(0.695)=71W/(m12K)5110=2.937102=0.023Ref8Pr0.4,twtf,l/d10f4、0.80.4Nuf=0.023(2.8710)(5.31)d0.62312.610Q=GCpf(tf2tf1)=Qdl(twtf)tf2-tf1'一汶dl(2)按西得-塔特公式计算:GCpf(tw-tf)NUf=0.027Re.8Pr；30.14Pf)w3.7151102'：.-=10(200133.8)=43.22C一一.一一30.0451.007510-Nu0.027(2.87104)0.8(5.31d0.6212.610；相

39、比较知：tf2#tf2，故原假设不合理，重新假设:2=9545W/(mK)18.tf=70C，重复上述步骤，直至tf2tf2,符合计算精度要求，结果略。(tf=85C)tw=250C,tf1=160C,tf2=240C,Um=1m/s,q=3.841(17.=280,tlf2=34C,l/d20d=12.6mm,um=1.8m/s,tw=80C,tf1由q=coust知求:tf解:：t:t虬羹=1.13280-341(tf1tf2)=200C2按tf查表得:-tf=0.15810°m2/s,%=1.3610，Ns/m2,Prf=0.93按tf=31C查表得:=0.62W/(mK),f

40、=7.910m2/s,pf6巍3110业可/(mK),w=1.。910'Ns/m2RefUmd-f1.812.610曰7.910=假设管内流动为紊流,_44=2.8710110RefUmdfq=:(twf)而另Nuf解设tf2=40C有：竺=4<2,即.：t90-400.1419.求:0.8=0.027RefLlVwJtf=(tf1tf2)=30C2按tf=30'C查表有"=0.02亡R0e8d3_=995.7kg/m3,Cpf=4.174kJ/(kgK),，f=0.618W(u0.027也y0.8Prf0.027=0.114mRefumdif原假设成立0.14

41、f=0.80510-6m2/s,Pr5.42旦(tw-tf)qRef-fUmd1210室1.74,“46=2.53101100.8051050.8-0.45"f200)3.84105d3；R=110.3-R0.840.140.80ccc；1.36湖0，取Nuf066®23RejPf代I(0.93Y3乂号50二0.158100.14口.0910有0.114=7.21051040.15810再由Q=dlq=GCpf(tf2tf1)_二2、G=d'fum4且有:tf2:=Nud408040.628f=0.023(2.5310).(5.42).10.012=7774W/(m

42、2K)由di，fCpf(tf2tf)4qQ=：dl：(twtf)=GCpf0.1148634.505103(240-160)l/d=23.1/0.114X0，满足Nuf计算关联式的要求。d=12mm,D=180mm,n=4,tf1(tf2-tfJ状一12气8第X4心pfdlc兀G+=23.1mw-tf)dn：D：，,4nD-(tw-tf)=tf1"=d2CpffCpfum4=20C,um=1.7m,2p=90C180995.74.1741031.71210遛._3777410(90-30)=70Ctf2#tf2,与假设不符，重新假设tf2的值，重复上述的步骤，直至计算得满足要求的值。

43、结果略！'f5、0.80.40.644(tf=65C):-=Nuf=0.023(1.3610)(3.77)d0.1620.略21.=1.8Cd=0.16m,l=2.5m,U=5V,I=911.1A,tf1=47宅,5=0.5/孚06qw=022.求:，讨=tw-tfG=2.5kg/s,tf1=40C,d=50mmtw=85C"：=0.0002,1解:Q=IU=5911.1=4555.5W求tf2,QQ=:Ft=GCpf(tf2-tf1)tf2=70C,tfCPf=4.18103J/(kgK),*-980kg/12(tf15皿按tf=55C查表得:,.Qtf2-tf1777GC

44、Pf-tf1'一.2r-d:fUmCPf4Cpf1.005kJ/(kgK),，f=2.8710W/(mK),.f=18Prf=0.697,，：f=1.077kg/m3=47=47.11C土23-0.1629800.54.1810344555.5Um=:fG"=2戒"d2可按tf=tf1=47C查取物性:St-Pr23由:CL6米5,E(St=)0.644W/(mK),f=0.58710*m2/s,Prf=3.778:fCpfumRef逐二=1.361051104；?fCPfUmf0.587101.0771.005103'1.0778Pr31.742lg!18

45、0.697'2.31.742lg-且1/d10,twtf取Nuf=0.023Re8P0r4.一2=2437W/(mK)-_'一一.一Q=GCpf(tf2Tf)=:F(twTQtf2=tf1(twf)GCpfUmdRef1.30.01944=4.781041040.51710-F="l=402437100.05二3(85-55)=85C2.51.00510知Nuf-0.023Re0.8Pr0.4'tf2"tf2，与假设不符，重新假设tf2的值，重复上f0.80.40.023RfePr=d0.6530.00.019(4.784f082=7030W/(m2

46、K)述的步骤，直至计算得满足要求的值。结果略！(tf=75C,Q=3.67105W=367kW)23.略24.25.Um1.3m/s,d=19mm,l=5.5m,F=42mmUg,tw1钮mC$tff=5®怎C,tw=57.9C,d=22mm,l=2.5m求：解按tf=55C查表得:解按tf=38.5C查表得:69W=985.6kg/m3,Cpf=4.177kJ/(kgK),ff65)3)269Wv(mmKK)f=16.7410m2/s,Prf=0.7=0.51710"m2/s,Prf=3.265得:由:1 P-2fUmStfPrf=:fCpfUm42133.325.50.

47、019212985.61.32f二(St=)8;?fCPfUm985.64.1771031.32.32310望283.2653UmdRef=f1.272210；3=1.6710316.7410l/d60NUf=0.0214任号8-100)f°.|=0.0214宣(Re*8100)Pr0.41+【a立djJw27061W/(m2K)0.0269=°.°214磁一0.45-(1.67乂103广-100*(0.711326.Um2=6.24W/(m2K)=0.023-067(2.4104)0.8(2.445)°.4=8897.2W/(m2I0.004=3.5m

48、/s,tf=58.lC,tw=90C,其他同2驰28.d=50mm,G=0.0125kg/s,l=6m,tf1=73.1C,tf2=62C解:按tf=58.1C查表得求tw,a,Qf=0.02836W/(mK),.f=18.7210*m2/s,Prf=0.7092解按tf=(tf1tf2)=42.75C查表得:2f=0.0275W/(mK),.f=17.24106m2/s,Prf=0RefUmd3.50.022q七=1.175kg/m3=4.11乂10，过f18.7210渡流080/d0.0214(R§-100)Pr0|1+，d0.80.4日J)j<TwUmd有0.4网27.d

49、1,Um:f一2-15.78W/(m2二2d2K)4Gd-二】2=12mm.d2=16mm.l=400mm.um=2.4m/Retf±73.14C,tw=卵65乂104>104f-f解按tf=73.1C查表得:f0.8-=0.023Refd=0.67W/(mK),f=0.399510”m2/s,Prf=2.445_042Prf."6.11W/(mK)Q=GCpf(tf2tf1)=amdl(twtf)得de.22d22-d124*2d)=d2一d=4mmtw=tfGCPf(tf2tf1)=62.5C:二dlRefUmdef410'2.40.399510*=2.4

50、104104tf一上tm盘t-tb=53.66C查In,:t故可有:fct=xde0.023Re0.8Pr0.4表得:f=0.02805W/(mK),f=18.2910m2/s,F再由Q=口<n：dl，(twtf)得七=1.0808kg/m3,Cpf=1.009kJ/(kgK)tw=tfQ=159.5Cdl计算得(步骤同上)：32.已知um=5.89m/s,Ref=1.61104104d=14mm,l=1.5m,u：=3m/s,tf=55C,tw=95C求：Q.升加5cccr兀f0.8C13Tfa=0.027RefPrf3dJw,-2=30W/(m解：按tf=55'C查表得：K)

51、f=0.51710-6m2/s,f=66.3510=W/(mK),PrQ=GCpf(tf2tf1)=485.6W按tw=95'C查表得P麟=1.85Qtw=tf=70.82C右ajrdl有29.略30.略cud30.014431.已知：Ref=8.12尺10f0.51710um=25.5m/s,d=35mm,l=0.5m,Q=900W,tf=25.3C查表得：c=0.26,n=0.6，即求：tw解按tf=25.3C查表得:f=15.610m2/s,frP展.25Nuf=0.26Re0.6Pr0.37M<PrwJ0.250.0261W/(mK),时骄澎房"7*；=19.4

52、kW/(m2K)Ref=25A|5=5.72104f15.610查表知c=0.26,n=0.6，取Pw=Pf即:Nuf0.60.37=0.26RefPrf斗.25PR<Prw/Qmmdl(匕-tf)=51.2kW33.已知：d=12mm,u二=14m/s,tf=30.1C,tw=12C求：解按tf=55C查表得：-f-16.0410*m2/s,f=0.0264W/(mK)0.3740.6-0.260.71255.721040.02612-122W/(m2K)0.035有Refud140.0124=-=1.05104.f16.0410-Refmaxd-f4.870.0191.00210项_

53、4=9.2310按Ref查表得：c=0.26,n=0.6，即查表有：&z=0.92Nuf-0.88cRef,=0.880.26(1.05104)0.6=130.2W/(m2NCf)=0.35Re0.6Pr0.36PrfPrw0.250.2I34.P=6,umax已知：=15.5m/s,tf=19.4C,tw=67.8C36.已知求：解按tf-19.4C查表得:Pr,=0.709,；z=0.95解按tf=60C查表得:P=12,d=25mm,S=50mm,S2=45mm,umax=5m/s,tf=.f=15.0610£m2/s,f=0.02567W/(mK),Prf=0.713

54、求：叉,ct顺f=18.910"m2/s,，f=0.0285W/(mK)Refumaxd15.50.019“4四6=1.9610f15.0610Nuf0.630.36=0.027RefPrfPrf0.25c"maxd50.0253Re=6.6110-f18.91035.0.25:=0.027里Re0.63d0.36PrfPrwJ对于叉排查表得：；z=0.98,另:鱼=0.92&=155.5W/(mK)0.6Nuf=0.31Ref寸2.JP=5,umax287m/s,tf-SiS20.2=20.2C,tw=25.2C,d=0.3.弩辞胃2z=66.4W/(mK)解按tf=20.2C查表得:对于顺排有：z=0.98f=1.00210§m2/s,，f=59.9410W/(mK),Prf(6.99?Prw-6.1420.24Re0；z=68.4W/(m2K)d37-39.略40.已知：d=30mmtf=37.1C,tw=64.5C9.81(90-20)(0.5)30.71_8=5.4810(27355)(18.42106)2求：1解：按tm=(tf+tw)=50.8C查表得:2查表得：c=0.59,n=14.m=0.5510"m2/s,Prm=3.495"m=64.890.59W(G(rnPrJ°.59l£