流体力学泵与风机课后题答案详解中国建筑工业出版社

流体力学泵与风机课后题答案详解中国建筑工业出版社流体力学泵与风机部分习题答案 2-15解：（1）当1γ为空气 21p p =()ABp h z p =++γ()h z p p p BA+=-=∆γ 3.010008.9⨯⨯= kpa pa 94.22940==（2）当1γ为油 31p p =()z H h p p A+++=γ1()H h p p B γγ++=13Hh z H h p p p p p B A γγγγγ--+++-=-=∆131 hz h 1γγγ-+=1.090002.010008.91.010008.9⨯-⨯⨯+⨯⨯=kpapa 04.22040==2-16 解：21p p=()211h h H p p M +++=水γ212h h p p a 汞油γγ++=()2121h h p h h H p a M 汞油水γγγ++=+++()2.010008.96.1378502.05.110008.998011⨯⨯⨯+⨯=++⨯⨯+-h h 26656785098002.098005.1980098011+=+⨯+⨯+-h h1960147009802665619501--+=hmh 63.51=2-28 解：()21h h p -=γ()()()b h h h b h h h h P 02210212145sin 45sin 21-+--=γγ()()145sin 22310008.9145sin 232310008.92100⨯-⨯⨯+⨯-⨯-⨯⨯⨯=kNN 65.343465022510008.9==⨯⨯=()()()P bl h h h bl h h h h l D D D 2022110212145sin 45sin 21-+--=γγm45.222510008.9222210008.92322210008.9=⨯⨯⨯⨯⨯+⨯⨯⨯=2-32 解：b h h b h h P 02202145sin 2145sin γγ+=2222210008.9212222110008.9⨯⨯⨯⨯⨯+⨯⨯⨯⨯=kNN 8576.1106.1108572810008.9==⨯⨯=Ph h b h h h h b h h l D 02102202102145sin 3245sin 2145sin 245sin ⎪⎭⎫ ⎝⎛++⎪⎭⎫ ⎝⎛+=γγ2810008.92372410008.9222410008.9⨯⨯⨯⨯⨯+⨯⨯⨯=2613=26722613=-=p lTP G l T l P l G ⨯=⨯+⨯22672810008.9162.19⨯=⨯⨯⨯+⨯TkN T 31.10134.27481.9=+=2-41解：245sin 0=⨯=r h b h h P x⨯⨯⋅⋅=21γ 4212210008.9⨯⨯⨯⨯⨯=kN N 2.3939200== V P zγ=br r r ⎪⎭⎫⎝⎛⨯⨯⨯-=00245cos 45sin 2136045πγ 4212281214.310008.92⨯⎪⎭⎫ ⎝⎛⨯⨯-⨯⨯⨯⨯=kNN 344.2222344== kNP 1.45344.222.3922=+=03057.0arctan 2.39344.22arctan arctan≈===x z P P α3-3解：（1）sm v d Q /0049.010025.04432323=⋅⋅=⋅=ππ s kg Q /9.4=ρ（2）sm v d d v /625.032131=⎪⎪⎭⎫ ⎝⎛=sm v d d v /5.232232=⎪⎪⎭⎫ ⎝⎛=3-5解：sm h mQ /778.2/1000033==sm d Qv /2042≤=π所以，177.04=≥πv Qd所以，mm m d 45045.0== 此时，s m d Q d Qv /4.1763585.0112.114422====ππ3-6解：22543212054d d A A A AA ππ======22114012021d d A A ππ=⋅="=' 22224012021d d A A ππ=⋅="=' 22334012021d d A A ππ=⋅="=' 22444012021d d A A ππ=⋅="=' 22554012021d d A A ππ=⋅="=' 2214014d d ππ=d d 1011=d r 10211=2224034d d ππ=d d 1032=d r 10232=2234054d d ππ=d d 1053= d r 10253=2244074d d ππ=d d 1074=d r 10274=2254094d d ππ=d d 1035=d r 10235=()()54321254321220240u u u u u d u u u u u d Q G ++++=++++==πρπρρ3-7解：干管前端的质量流量为：42562.2211111d A v Q πρρ⨯⨯==()skg /128544.005.042562.22=⨯⨯⨯=πs kg Q Q Q /064272.02132===ρρρ()sm A Q v /247.2204.043.2064272.022222=⋅⋅==πρρ ()sm A Q v /05.18045.0424.2064272.023333=⋅⋅==πρρ3-10解：将基准面建立在B 点经过的水平面上，列能量方程：gv p z gv p z 222222221111αγαγ++=++其中，mz 2.11=mp 5.11=γsm v /21=sm v d d v /5.4122212==121==ααgp g 25.40225.12.1222++=++γ871.125.4225.12.1222=-++=gg p γ3-11解：将2点所在的水平面作为基准面，列能量方程：gv p z gv p z 222222221111αγαγ++=++31=z2=zγγ21p p =sm v /31=gv p g p 2023322221++=++γγsm gh v /2.83222=+=32.822112=⎪⎪⎭⎫ ⎝⎛=d d v v 所以，md12.02=3-14解：以水面为基准面，列0-0和D-D 的能量方程：g v p z gv p z DD DD 22220000αγαγ++=++0=z=γp2200=gv α 4-=D z=γDpgv DD 2040002α++-=++ 所以，422=gv DD α，即，sm v D /85.88.924=⋅⋅=所以，sm v d Q D/017368.085.805.044322=⋅⋅==ππ81:1:2:24422==A D DD A A d d gv gv αα列0-0和A-A 断面的能量方程：gv p z gv p z AA AA 22220000αγαγ++=++8147000++-=++γAp所以，8147-=γAp所以，kpapA1.68=列0-0和B-B 断面的能量方程：gv p z gv p z BB BB 22220000αγαγ++=++kpa p B 484.08.9814-=⋅-=列0-0和C-C 断面的能量方程：gv p z gv p z CC CC 22220000αγαγ++=++kpap C 1.208.98142-=⨯⎪⎭⎫ ⎝⎛+-==D p3-18解：将基准面建在管道所在的水平面上，列能量方程：21222222111122-+++=++l h gv p z gv p z αγαγ128.998.0008.9490222+++=++g v α9.3222=gvsm v /74.82=3-19解：（1）（a ）将基准面建在A 所在的水平面上，列0-0和C-C 断面的能量方程：gv p z gv p z CC CC 2222000αγαγ++=++gv CC 2000042α++=++422=gv CC αsm v C /85.88.98=⨯=1:4:2:22222==B C CC B B s s gv gv αα122=gv BB αsm v /43.48.921=⨯= 且BA v v =（b ） （c ）gv p z gv p zAA AA 222200αγαγ++=++10004++=++γAp3=γApkpap A 4.29=（2）（a ）21220022-+++=++l CC CC h gv p z gv p z αγαγ其中，gv g v hl 2324222121+=- g v g v g v 223200004222222++++=++54222=g v 所以，sm v/96.32=sm v v /96.12121==（b ） （c ）gv g v p z g v p z 222221211112000+++=++αγαγ5300041++=++γp5341-=γpkpap 32.331= gv g v g v p z g v p z 22324222222222222000++++=++αγαγ5423545400042⋅++++=++γpkpap 76.112=3-20 解：()()212221221122-++=--++l a p v p z z v p ργγρs m d Qv /38.2005.014.34202.042221=⨯⨯⨯==πsm dQv /19.1005.014.3402.04222=⨯⨯==π2423222121v v p l ρρ+=-()()242322222122212211v v v p z z v p a ρρργγρ+++=--++22214v v =()()8.930306.02.1224232300212221221⨯+---+++=v v v v p ρρρρ ()()8.930306.02.12424212230022222222⨯+---+++=v v v v ρρρρ8.9606.019.1026.0133002⨯⨯-⨯⨯+=pa 16.352=mm p h 6.449.716.3521===γ3-22解：s kN h kN G /048944.0/2.176==sm GQ /1347.77.08.910048944.033=⨯⨯==γs m d Q dQv /09.914.31347.7444222=⨯===ππ()2122221122-++=-++l a p v p H v p ργγρ其中，01≈v，pah p988.9101010331=⨯⨯⨯==-γ ()γgv d H H 2035.0209.97.008.97.02.1098222+⨯+=⨯⨯-++-()8.97.08.9209.9035.0209.97.008.97.02.109822⨯⨯⨯+⨯+=⨯⨯-++-H HHH 0122.19.289.498+=+-所以，m H 64.32=()212211212212-++=-++l M M a p v p H v p ργγρ()8.97.08.9209.9164.322035.0209.97.064.328.97.02.12109822⨯⨯⨯+⨯+=⨯⨯-++-M p 52.169.28968.7998++=+-M p所以，pap M45.63-=3-263-28解：列连续性方程：s m DQ v /18.34.014.344.04221=⨯⨯==π sm dQv /96.501.014.344.04222=⨯⨯==π列能量方程：gv p z g v p z 222222221111αγαγ++=++gv gv p 222112221ααγ-=m98.1318.9218.396.5022=⨯-=kpap 404.12938.998.1311=⨯=列动量方程： ()12v v Q F -=∑ρ()12222144v v Q R d p D p -=-⨯-⨯ρππ()18.396.504.04.04404.12932-⨯=-⨯⨯R πkNR 339.14378.474.04.04404.12932=⨯-⨯⨯=πkN R94.1112=3-33解：列能量方程：gv p z gv p z 222222221111αγαγ++=++其中，5321=vv2221259v v =gv gv 209.0205.1222211αα++=++gv g v 225926.02222-=sm v /3.42= sm v /58.21=()12v vQ F -=∑ρ()1222212121v v Q R b h b h -=--ργγ其中，s m Q /644.45.12.158.23=⨯⨯=72.1644.410009.0108.9215.1108.9212323⨯⨯=-⨯⨯⨯-⨯⨯⨯RNR 2.480=4-2 （1）mmm d 1.0100==skg Q /10=ρsm Q Q /01.03==ρρs m d Q v /274.11.014.301.04422=⨯⨯==πs m /10519.126-⨯=ν8387110519.11.0274.1Re 6=⨯⨯==-νvd（紊流）（2）skg Q /10=ρ s m Q Q /011765.0850103===ρρs m d Q v /4987.11.014.3011765.04422=⨯⨯==π sm /1014.124-⨯=ν13151014.11.04987.1Re 4=⨯⨯==-νvd4-3 解：m d 3.0= C T 020= sm /107.1526-⨯=νsm d v /1067.1043.0107.152000Re 36max --⨯=⨯⋅=⋅=νsm A v Q /103947.743.014.31067.1043323max max --⨯=⨯⨯⨯=⋅=hkg Q /9.3136002.1103947.73=⨯⨯⨯=-ρ4-4 解：212=d d4212221==d d v v222111Re 2214Re ===ννd v d v 所以，2ReRe21=4-12 紊流粗糙区，5106Re ⨯>νvd=Re ，所以，sm d v /14.325.010308.1106Re 65=⨯⨯⨯==-νsm d v Q /154.0425.014.314.34322=⨯==π4-13sm s L Q /2.0/20031==s m dQ v /076433.44211==π661107791.010308.125.0076433.4Re ⨯=⨯⨯==-νvdsL Q /202= sm v /4076433.02=4210791.7Re ⨯=sL Q /53=sm v /1019.03=43109478.1Re ⨯=查尼氏图，得到，5106Re ⨯=u4104Re ⨯=l123Re Re Re Re Re <<<<u l ，所以，1Q 属于紊流粗糙区，2Q 属于紊流过渡区，3Q 属于紊流光滑区，（1） 对于1Q ，采用希弗林松公式，02326.025.0105.011.011.025.0325.01=⎪⎪⎭⎫ ⎝⎛⨯=⎪⎭⎫⎝⎛=-d K λmg v d l h f 888.78.92076433.425.010002326.0222111=⨯⨯⨯==λ（2） 对于2Q ，采用阿公式，02547.010791.76825.0105.011.0Re 6811.025.04325.02=⎪⎪⎭⎫ ⎝⎛⨯+⨯=⎪⎭⎫⎝⎛+=-d K λmg v d l h f 086.08.924076433.025.010002547.0222222=⨯⨯⨯==λ（3） 对于3Q ，采用布公式 02678.05.194773164.0Re 3164.025.025.03===λm g v d l h f 005676.08.9244076433.025.010002678.0222333=⨯⎪⎭⎫⎝⎛⨯⨯==λ4-15 5102Re ⨯=u4000Re =lm d 05.0= m K 31025.0-⨯=sm d v u /028.405.010007.1102Re 65max =⨯⨯⨯==-νs L d v Q /905.7405.014.3028.4422maxmax =⨯==π 26min10056.805.010007.14000Re --⨯=⨯⨯==d v l νs L s m d v Q /1581.0/1001581.0405.014.310056.8432222minmin =⨯=⨯⨯==--π4-21 （1）a d d =212211av v =g v d l d v g v d l g v d l h f 2642Re 64221111211121111νλ===4212221211a d d v v h h f f ==19.1=a（2）75.425.12275.12122225.0225.0225.021125.0125.0125.021123164.023164.0a d d v v gv d l d v g v d ld v hhf f =⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛==νν16.1=a（3）25.525.11222122225.0221125.01211211.0211.0a d d v v gv d l d K g v d l d K hhf f =⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=14.1=a4-24 解：s m Q /002742.0602329.03=⨯= s m d Q v /3972.105.014.3002742.04422=⨯⨯==π629.022=⎪⎭⎫ ⎝⎛+gvd l ζλ()629.08.923972.162=⨯+ζ3151.0=ζ4-26 解：（1） 突然缩小375.03145.7815.015.0121=⎪⎭⎫⎝⎛-=⎪⎪⎭⎫ ⎝⎛-=A A ζ mmm g v h j 5.760765.08.922375.022211==⨯==ζ（2）5.02=ζmmm g v h j 102102.08.9225.022222==⨯==ζ （3）1693145.781122213=⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛-=A A ζmmm h j 115115.08.92216923==⨯=（4）14=ζmmm h j 204204.08.922124==⨯=4-27解：()()gv v gv v h h m m jj222121-+-=''+' ()()()()02212221=-+--=''+'gv v g v v h h m m vmj j所以，221v v v m+=此时，()jj j h gv v g v v v g v v v h h 2221222222121212211=-=⎪⎭⎫ ⎝⎛-++⎪⎭⎫ ⎝⎛+-=''+'4-29 解：sm h m Q /1044.4/16333-⨯==sm d Q v /2624.205.014.31044.44423211=⨯⨯⨯==-πs m d Q v /5656.01.014.31044.44423222=⨯⨯⨯==-πmg v v p p h j 140674.08.925656.02624.28.910001739.522222121=⨯-+⨯⨯-=-+-=γgv h j 2211ζ= 5387.01=ζ gv h j 2222ζ= 619.82=ζ5-17 解：5.6082.014.32.12.01002.08842412111=⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛=d d l S p πρλ7.30422.014.32.12.05002.08842422222=⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛=d d l S p πρλ973671.014.32.11.05002.08842432333=⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛=d d l S p πρλ3.101018973677.30425.608321=++=++=p p p p S S S S22211/91.227215.03.101018m N Q S p p =⨯==22222/1.258616.03.101018m N Q S p p =⨯==5-25 解：()()⎪⎩⎪⎨⎧=++=++=1021520232322223221SQ Q Q S SQ Q Q S SQ 610=S解得，sm Q /10472.4331-⨯= sm Q /1041.2332-⨯=sm Q /1063.0333-⨯=5-27 解：94.10348.92.014.32.020002.08842412111=⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛=g d d l S πλ8.206988.91.014.31.0100025.08842422222=⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛=g d d l S πλ 78.37258.92.014.32.072002.08842432333=⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛=g d d l S πλ038035.087.14311705.321111211=+=+='S S S 所以，25.6911='S1）()HQS S =+'231s m S S H Q /10186.604417163331-⨯==+'=2）HSQ=2H Q S =⎪⎭⎫⎝⎛'2211325133831432=+'=-'=S S S S gd πζ()1.25688.92.014.31325142=⨯⨯⨯=ζ5-28 解：286.1368.93.014.383.020002.084242=⨯⨯⨯⨯==g d d l SAB AB AB ABπλ029.1098.93.014.383.016002.084242=⨯⨯⨯⨯==g d d l S AC AC AC ACπλ34.328.94.014.384.020002.084242=⨯⨯⨯⨯==g d d l S AD AD AD ADπλ772.818.93.014.383.012002.084242=⨯⨯⨯⨯===g d d l S S BC BC BC CDBC πλ5108.2⨯=A p 2ABAB A QS p γ= s m S p Q AB A AB /457868.08.91000286.136108.235=⨯⨯⨯==γ2ADAD A Q S p γ=s m S p Q AD AAD /93993.08.9100034.32108.235=⨯⨯⨯==γ()()222BC BC BC AC A Q S Q S p +=()sm S S p Q Q BC AC ACD BC /23488.043=+==γs m Q Q Q BC AB /69275.022=+=sm Q Q Q CD AD /17481.123=+= sm Q Q Q /86756.13321=+=22/2.44m kN Q S p BC BC C ==γ。