1、緒論思考題與習(xí)題(89P -答案:1.冰雹落體后溶化所需熱量主要是由以下途徑得到:Q 與地面的導(dǎo)熱量f Q 與空氣的對流換熱熱量注:若直接暴露于陽光下可考慮輻射換熱,否則可忽略不計。2.略3.略4.略5.略6.夏季:在維持20的室內(nèi),人體通過與空氣的對流換熱失去熱量,但同時又與外界和內(nèi)墻面通過輻射換熱得到熱量,最終的總失熱量減少。(T T 外內(nèi)冬季:在與夏季相似的條件下,一方面人體通過對流換熱失去部分熱量,另一方面又與外界和內(nèi)墻通過輻射換熱失去部分熱量,最終的總失熱量增加。(T T 外內(nèi)掛上窗簾布阻斷了與外界的輻射換熱,減少了人體的失熱量。7.熱對流不等于對流換熱,對流換熱 = 熱對流 + 熱

2、傳導(dǎo)熱對流為基本傳熱方式,對流換熱為非基本傳熱方式8.門窗、墻壁、樓板等等。以熱傳導(dǎo)和熱對流的方式。9.因內(nèi)、外兩間為真空,故其間無導(dǎo)熱和對流傳熱,熱量僅能通過膽壁傳到外界,但夾層兩側(cè)均鍍鋅,其間的系統(tǒng)輻射系數(shù)降低,故能較長時間地保持熱水的溫度。當(dāng)真空被破壞掉后,1、2兩側(cè)將存在對流換熱,使其保溫性能變得很差。 10.t R R A = 1t R R A = 2218.331012m -= 11.q t = const =直線 const 而為=(t 時曲線12、略13.解:1211tq h h =+=18(1045.9210.361870.61124-=+2W m 111(f w q h t

3、 t =- 11137.541817.5787w f q t t h =-=-=222(w f q h t t =- 22237.54109.7124w f q t t h =+=-+=-45.92 2.83385.73q A W =14. 解:40.27.407104532t K R W A HL -= 30.2 4.4441045t R -=2m K W 3232851501030.44.44410t KW q mR -= 3428515010182.37.40710t tKW R -=15.(i w f q h t h t t =- i w f qt t h =+ 51108515573

4、=+=0.05 2.551102006.7i Aq d lq W =16.解:12441.2 1.2(100100w w t t q c =-44227350273203.96(139.2100100W m +=-=12''441.2 1.2(100100w w t t q c =-442273200273203.96(1690.3100100W m +=-='21.2 1.2 1.21690.3139.21551.1W q q q m =-=-=17.已知:224A m =、215000(W h m K =、2285(W h m K =、145t = 2500t =、

5、'2285(W k h m K =、1mm =、398=(W m K 求:k 、解:由于管壁相對直徑而言較小,故可將此圓管壁近似為平壁即:12111k h h =+=3183.5611101500039085-=+2(W m k 383.5624(5004510912.5kA t KW -=-=若k 2h '100k k k -=%8583.56 1.7283.56-=% 因為:1211h h ,21h 即:水側(cè)對流換熱熱阻及管壁導(dǎo)熱熱阻遠(yuǎn)小于燃?xì)鈧?cè)對流換熱熱阻,此時前兩個熱阻均可以忽略不記。18.略第一章導(dǎo)熱理論基礎(chǔ)思考題與習(xí)題(24P 答案:1.略2.已知:10.62(W

6、m K =、20.65(W m K =、30.024(W m K =、40.016(W m K =求:'R 、''R 解:2'3124124224259210 1.1460.620.650.016m K R W -=+=+= '"232232560.265/0.650.024R m k W =+=+= 由計算可知,雙Low-e 膜雙真空玻璃的導(dǎo)熱熱阻高于中空玻璃,也就是說雙Low-e 膜雙真空玻璃的保溫性能要優(yōu)于中空玻璃。3.4.略5.6.已知:50mm =、2t a bx =+、200a =、2000b =-/m 2、45(Wm K = 求:

7、(10x q =、6x q = (2v q解:(100020x x x dt q bx dx =-=-=3322452(20005010910x x x dt W q bx mdx -=-=-=-= (2由220vq d t dx +=2332245(2000218010v d tW q b m dx =-=-=-=7.略8.略9.取如圖所示球坐標(biāo),其為無內(nèi)熱源一維非穩(wěn)態(tài)導(dǎo)熱故有: 22t a t r r r r = 00,t t = 0,0tr r = ,(f tr R h t t r =-=-10.解:建立如圖坐標(biāo),在x=x 位置取dx 長度微元體,根據(jù)能量守恒有: x dx x Q Q

8、Q += (1x dtQ dx =-+ (x dx d dtQ t dx dx dx +=-+4(b b Q EA E A T Udx =代入式(1,合并整理得:2420b fU d tT dx -=該問題數(shù)學(xué)描寫為:2420b fU d tT dx -=00,x t T =,0(x ldtx l dx =假設(shè)的 4(b e x ldtf T f dx =-=真實的第二章穩(wěn)態(tài)導(dǎo)熱思考題與習(xí)題(P 51-53答案1.略2.略3.解:(1溫度分布為 121w w w t t t t x -=- (設(shè)12w w t t >其與平壁的材料無關(guān)的根本原因在 coust =(即常物性假設(shè),否則t 與

9、平壁的材料有關(guān)(2由 dt q dx =- 知,q 與平壁的材料即物性有關(guān) 4.略 5.解: 2111222(0,(,w w w w d dt r dr drr r t t t t r r t t =>=設(shè)有: 12124(11w w Q t t r r =- 21214F r r R r r -=6.略7.已知:4,3,0.25l m h m = 115w t =, 25w t =-, 0.7/(W m k = 求:Q解: ,l h ,可認(rèn)為該墻為無限大平壁 15(50.7(436720.25tQ F W -= 8.已知:2220,0.14,15w F m m t =-,31.28/(

10、, 5.510W m k Q W =求:1w t解: 由 tQ F = 得一無限平壁的穩(wěn)態(tài)導(dǎo)熱312 5.510150.141520 1.28w w Q t t F =+=-+=9.已知:12240,20mm mm =,120.7/(,0.58/(W m k W m k = 3210.06/(,0.2W m k q q = 求:3解: 設(shè)兩種情況下的內(nèi)外面墻壁溫度12w w t t 和保持不變,且12w w t t >由題意知:1211212w w t t q -=+122312123w w t t q -=+再由: 210.2q q =,有121231212121230.2w w w

11、w t t t t -=+得: 123312240204(40.06(90.60.70.58mm =+=+= 10.已知:1450w t =,20.0940.000125,50w t t =+=,2340/q W m 求: 解: 412,0.094 1.25102w w t t tq m m +=+412120.094 1.25102w w w w t t t t tmq q+-=+ 445050450500.094 1.25100.14742340m +-=+= 即有 2340/147.4q W m mm 時有11.已知:11120,0.8/(mm W m k =,2250,0.12/(mm

12、 W m k =33250,0.6/(mm W m k =求:'3?= 解: '2121'3123112313,w w w w t t t t q q -=+由題意知:'q q = 即有:2121'3123112313w w w w t t t t -=+'33322=+ 0.6250505000.12mm =+= 12.已知:1600w t =,2480w t =,3200w t =,460w t = 求:123,R R R R R R 解:由題意知其為多層平壁的穩(wěn)態(tài)導(dǎo)熱 故有: w w w w w w w t t t

13、 t t t t t q R R R R -= 112146004800.2260060w w w w R t t R t t -=-223144802000.5260060w w w w R t t R t t -=- 33414200600.2660060w w w w R t t R t t -=- 13.略14.已知:111012,40/(,3,250f mm W m k mm t =,60f t = 220112,75/(,50/(h W m k h W m k = 2223,320/(mm W m k =32'23030,70/(h W m k =求:123123,q q

14、q k k k 解: 未變前的 122030102250605687.2/1113101754050f f t t q W m h h -=+ 121311121129.96/(1112101754050k W m k h h -=+ 21129.96(250605692.4/q k t W m =-= 21105692.45687.2 5.2/q q q W m =-=-= 222321221129.99/(11131017532050k W m k h h -=+ 22229.99(250605698.4/q k t W m =-= 22205698.45687.211.2/q q q W

15、 m =-=-= 3 22330'101136.11/(11131*k W m k h h -=+ 23336.11(250606860.7/q k t W m =-= 23306860.75687.21173.5/q q q W m =-=-=321q q q > ,第三種方案的強(qiáng)化換熱效果最好 15.已知:35,130A C B mm mm =,其余尺寸如下圖所示, 1.53/(,0.742/(A C B W m k W m k = 求:R 解:該空斗墻由對稱性可取虛線部分,成為三個并聯(lián)的部分111132222,A B C A B C R R R R R R R R R =+

16、=+3321111311135101301020.1307(/1.53 1.53C A B A B C R R m k W -=+=+= 332322222335101301020.221(/1.530.742C A B A B C R m k W -=+=+= 2212115.0410(/1111220.13070.221R m k W R R -=+16.已知:121160,170,58/(d mm d mm W m k =,2230,0.093/(mm W m k =33140,0.17/(,300w mm W m k t =,450w t = 求:1123,R R R ; 2 l q

17、: 3 23,w w t t . 解nln 1.66410(/2258160d R m k W d -= 2222221117060lnln 0.517(/220.093170d R m k W d +=223332222111706080lnln 0.279(/2220.1717060d R m k W d +=+132R R R < 2 2330050314.1/0.5170.279l i t t q W m R R R -=+ 3由 121w w l t t q R -=得 4211300314.1 1.66410299.95w w l t t q R -

18、=-=-= 同理:34350314.10.279137.63w w l t t q R =+=+=17.已知:1221211,22m m d d =求:'ll q q 解:忽略管壁熱阻010121020122211lnln 222d d R d d +=+ '010122010122211lnln 222d d R d d +=+ '',l l t tq q R R = (管內(nèi)外壁溫13,w w t t 不變 01012'20101'010121020122211lnln 22222211ln ln 222l l d d q R d d d d

19、q R d d +=+01010010101001241lnln 22241ln ln 22d d d d d d d d +=+由題意知: 1001011(22m d d d d =+=+2112011(232m m m d d d d =+=+即:21010101232(m m d d d d d =+=+= (代入上式''15ln 3ln23 1.27715ln 3ln 23l l q R q R +=+ 即: '0.783l l q q ='21.7%l llq q q -=即熱損失比原來減小21.7%。 18.已知:1,d mm =32.2210/,l

20、 R m -=0.15/(W m k =1max 65w t =,240w t =,0.5,mm =求:max I解: 21max 2max 12ln 2w w l l t t q I R d d-=+11221max 2max36540123.7(2 2.2210120.5ln ln 220.151w w l t t I A R d d - -= + 19.已知:121185,100,40/(,180w d mm d mm W m k t = 230.053/(,40w W m k t =,52.3/l q W m = 求:2 解: 13222121122211ln ln 22w w l t

21、 t tq d d R R d d -=+整理得:221111804011002(ln 20.053(ln 2252.3240852100(1(17222l d tq d d ee mm -=-=-= 或:21R R ,故有 132222221ln 2w w l t t tq d R d -=+ 2222(1722lt q demm =-=20.已知:4.7715.273(,/6.199,30,3,35.01211+-=w t kg kJ r mm mm mm d 325w t =,210.03/(,16.3/(,1W m k W m k h =求:m 解: 1231w w F F t t Q

22、 R R -=+311211111112111111(111144(2(2(222222w w t t d d d d -=-+-+2(25273.1577.4111111(16.30.350.3560.030.3560.416+-=-+-102.7W = 或:12F F R R ,故有: 312232(25273.1577.40.03102.711111(40.3560.416w w t t Q W r r -+-=-102.7 3.6 1.85/199.6Q m kg h r = 21.略 22.略23. 解: ff f t t l x t t x t t m dxd -=-=-=-221

23、1222,0,0解微分方程可得其通解: 12mx mx c e c e -=+ 由此得溫度分布(略24.已知:25,l mm =3,mm =20140/(,75/(,80W m k h W m k t = 30f t =,0x l q = 求:,l q 解: 3222750.0250.4725140310hU h L h ml l l l A L -= 18.9m = 0(0.472518.9(8030(0.4725ch m l x ch x ch ml ch -=-44.91(0.472518.9ch x =- 3044.91(0.472518.9t ch x =+-002(l Q hU h

24、 q th ml th ml L mL m = 275(8030(0.4725174.7/18.9th W m =-=25.已知:15,20,48.5/(,84l mm l mm W m k t =,040t = 220/(h W m k =求:t 解: 3200.12248.5 1.510hU h d h ml l l l A d -= 00(f l l ft t ch ml ch ml t t -=-0(84(24099.93(1(21l f t ch ml t ch t ch ml ch -=-(2 3.7622ch =99.9384100%100%15.9%99.93f l ft t

25、t t -=26.已知:00.8,160,60mm l mm t =,16.3/(W m k =,其他條件同25題求:t 解: 320160 6.2716.30.810hml l -= 0(84(6.276084.09(1(6.271l f t ch ml t ch t ch ml ch -=-(6.27264.24ch = 84.0984100%100%0.11%84.09f l ft t t t -=27.已知:3,16mm l mm = 2(1140/(,80/(W m k h W m k = 2(240/(,125/(W m k h W m k = 求:f 解:/p>

26、00.312140310hU h L h ml l l l A L -= (0.3120.970.312f th ml th ml = (23322212516100.7340310hU h L h ml l l l A L -= (0.730.8530.73f th ml th ml = 28.已知:1277,140,4,25,50/(d mm d mm mm P mm W m k = 2060/(,320h W m k t =,75f t = 求:l q解: 211(31.52l d d =-=33.52c l l =+=2172c c r r l =+=334221(410(7238.5

27、10 1.3410c f r r m -=-=-=P (1133223224226033.5100.82150 1.3410c h l f -= 2172 2.1533.5c r r = 查圖得: 0.78f = 每片肋片的散熱量為1Q 100(f f f Q Q hF t t =-222102(c f f r r h t t =-2262(7238.5100.7860(32075266.7W -=-= 每米肋片管的散熱量為:12(1l q nQ n Q =+- 100014125n =+=片/米 41266.740 1.4811kW =+= 2Q 為兩肋片間的表面的散熱量 210(f Q d

28、 P t t =-3377102510(32075 1.48W -=-= 29.略30.已知:21213 2.2,0.3,0.56/(,0w l l m m W m k t =,230w t =求:l q解: 1113100.3A l LLS L =222 2.27.330.3A l L LS L =30.54S L = 121,5l l >123(224l S S S tQ q L L+= ,12w w t t t =-(21027.3340.540.56(300=+- 618.6/W m =31.已知:1165,90w d mm t =,21.5, 1.05/(,6w H m W m

29、 k t = 220/(h W m k = 求:l q解: ,3l r H r >> 22ln(ls H r=l 1l 2Hdtw 1tw 222ln(l Q s t q t H l l r=2 1.05(906154.2/2 1.5ln 0.165/2W m =-= 32.已知:21210.520.52,0.42,0.023/(,30w l l m H m W m k t = 214w t =-, 34Q W = 求: 解: 1211212,l l l HS S l l =3410.54,0.54S H S l = 1234(444Q S S S S t =+1213440.52

30、0.5240.520.42344440.540.4240.540.520.023(3014l l l H Q S S t +=-+ 23.621036.2m mm -=33.已知:5, 2.54,2,80mm m P MPa t =,180/(W m k = 求:c t 解:由 2.54,2m P MPa =,查表得,420.8810(/c R m k W -= c tQ R =+ 31t t t =-再由 c ctQ R =,22c A B t t t =-得4340.88108049510220.8810180cc c R t t R -=+第三章 非穩(wěn)態(tài)導(dǎo)熱l 1l 2Q 底=0HRct

31、 1t 2At 2Bt 3t 1t 1A t 2Bt 3x1.略2.略3.略4.略5.已知:3210.15,420/(,8400/,58/(p d mm c J kg k kg m h W m k = 22126/(h W m k = 求:0102,解:33012111484004200.1532210 1.52(3235842p p p d d c c c V s h F h d h - =同理:302284004200.152100.7(323126p d c s h -= 6.略7.已知:300.5,8930/,400/(,25p d mm kg m c J kg k t =,120f

32、t =2095/(,1%,22/(h W m k W m k =(康銅 求:,t 解:由 001%f ft t t t -=- 00.01(1200.01(25120119.05f f t t t t =+-=+-=34950.51013 3.6100.10.123223R V hhF Biv M -=<=故滿足集總參數(shù)法的求解條件,有:V VBi Fo e -=320189304000.5103ln ln(11014.43952p c V s hF -=-=-= 8.已知:23,11,mm F m =239/(,48.5/(h W m k W m k =,0300t =,20f t =

33、,6212.710/,a m s -=50t =求:解: 33330101220.98100.10.148.53V hBi M -=<= 滿足集總參數(shù)法的求解條件,故有:phFc Ve -=00lnln p c VV hFhaF =-=- 36348.511050202ln 3283912.710130020s -=-=-9.略10.已知:080t =,20,20f d mm t =,12/,5min,34u m s t = 38954/,383.1/(,386/(p kg m c J kg k W m k = 求:h解:假設(shè)可使用集總參數(shù)法,故有: 0phFc Ve -=3201208

34、954383.110342022ln ln 83.2/(5608020p c V h W m k F -=-=-=-由 3383.2201012 2.16100.10.123862V Rhh F Biv M -=<= 滿足集總參數(shù)法的計算,上述假設(shè)成立。11.已知:2,12min A B A B pA pB A B f A B B c c t t t h h =0050%mA mB = 求:A 解: 11000,0.52AmAmB A A A ABi Bi h -= 查表得:0.24A B Fo Fo =即: 222212248min A A A B BA AB B B a a = 12

35、.已知:300.50.50.5,30a b c m t =,800f t =,52/(W m k =220.063/,80/(,30min a m h h W m k =求:m t 解: 1522.61800.52Bi h -=220.06330600.536000.25a Fo =對于正六面體有:3000f mm m f t t t t -= -平板由 12.60.5Bi Fo -=, 查圖有:00.9m = 平板3300800(800300.9239m m f t t =-=-= 平板13.已知:72040,510/,4/(,25mm a m s W m k t -=,1260f t =

36、240/(,1hh W m k =缺少40.14=f t 已知:、20=w t 、sm u 8.0=、m l 45.0=、5105Re =c 、m x 1.01=、m x 2.02=、m x 3.03=l x =4求:x 解:30(21=+=w f m t t t 按30=m t 查表得:42.5Pr =、(618.0k m w =、smv 271005.8-=由 vxu x =Re 得m x 1.01= 41094.9Re =x m x 2.02= 51099.1Re =x < ec R 均為層流m x 3.03= 51098.2Re =x m x 45.04= 51047.4Re =

37、x =2131Re Pr 332.0x x m x 1.01= 3.1136mx 2.02= 9.803(2k m wm x 3.03= 8.655m x 45.04= 5.535 x 2= 圖略15. 已知:m l 3.0=、s m u 9.0=、25=f t 求:max 、(y u l解:由25=f t 查表得 smv 2710055.9-= c l v l u Re 1098.210055.99.03.0Re 57<=- m l l35max 1055.23.01098.264.4Re 64.4-=-=u y y y u l 3max max (21(23(=3331055.21(

38、9.0211055.219.023y y - 371071.24.529(y y y u l -= sm 圖略16. 略 17. 略18. 已知:s m u 10=、80=f t 、30=w t 、m l 8.0=、5105Re =c 、m b 1=求:c x 、Q解:55(21=+=w f m t t t 按55=m t 查表得:smv 2510846.1-=、697.0Pr =、(10865.22k m w =-由 m u v x v x u c c c c 923.01010864.1105Re Re 55=- c x l < 全板長均為層流3121Pr Re 664.02ll l

39、=(9.13697.010846.18.0108.010865.2664.02312152k m w =-(w t F Q 5563080(8.019.13=-=19. 略 20. 略 21. 已知:y u u = 層流 求:x 解:由動量積分方程有:(1(1(202020=-=-=-=u dx d dy y y u dx d dy u u u u u dx d dy u u u dx d 式左式右=U udydU uy 0即: =U u U dx d (2即 const U = 即有:dx U v d U v dx d =6 兩邊積分有:=u v dx U v d x x62162021Re

40、 12x x=,vxU x =Re 22. 略23. 已知:w t cy by a t ,2+-= f t 求:x 解:由邊界層特點知 0=y w t t = 0022=y xt d d=y f t t = 得:0=c 、 w t a = 、(f w t t b -=-=+-=-=b t t cy b t t d d t t fw y fw y yt fw x 002( 24. 略25. 略 dr r r r =001,t F Q =26. 略 dx l lx =01,t F Q =27. 略 0=-=y f w dydtt t 28. 已知:23m F =、m l 11=、1401=w t

41、、301=f t 、s m u 501=、w Q 150001=、m l 52=、202=w t 、702=f t 、s mu 82=求:2 解:8530140(21(21111=+=+=w f m t t t 457020(21(21222=+=+=w f m t t t 按 1m t 2m t 查表得:691.0Pr 1=、(1009.321k m w=-、sm v 2511016.2-=6985.0Pr 2=、(108.222k m w=-、smv 25210746.1-=有:6511111031.21016.2150Re =-v l u6522221031.210746.1508.8R

42、e =-v l u 知:1Re =2Re ,1Pr 2Pr ,且?guī)缀蜗嗨?得:21Nu Nu = 121121222=l l Nu l 而:1111t =F Q =22112l l 111t F Q (24.830140315000511009.3108.3222k m w =-=-( 29. 略30. 已知:mm d a 16=,mm d b 30=,b a G G 2=,b a t t =求:(1是否相似(2如何相似解:(1b a t t = ,且為同類流體 b a v v =,b a p p = 再:24d GFGu m =有:2222244ab b b baa ambB.C.D. A.

43、 從甲國販賣布匹到乙國B. 從甲國販賣木材到乙國C. 從乙國販賣木材到甲國D. 從乙國販賣布匹到甲國26.黨的十八大報告提出,在發(fā)展平衡性、協(xié)調(diào)性、可持續(xù)性明顯增強(qiáng)的基礎(chǔ)上,2020年實現(xiàn)國內(nèi)生產(chǎn)總值和城鄉(xiāng)居民人均收入比2010年翻一番的目標(biāo)。實現(xiàn)這一目標(biāo),應(yīng)推進(jìn)經(jīng)濟(jì)結(jié)構(gòu)戰(zhàn)略性調(diào)整,轉(zhuǎn)變經(jīng)濟(jì)發(fā)展方式ma d u d u v d u v d u 22a b d d b ad d 12=a b d d 即:b a Re Re A.B.知:兩者流態(tài)不相似(2若要相似,需b a Re Re = 即:11Re Re =bmb a ma b a d u du 而:22a b b a mb ma d G d

44、 G u u =,帶入上式有:122=baa b b a d d d d G G 1583016=b a b a d d G G 即:要使兩者相似,兩者的質(zhì)量流量之比應(yīng)為15831. 略 32. 略 33. 略 34.略第六章 單相流體對流換熱及準(zhǔn)則關(guān)聯(lián)式30.公安部提出一系列措施,進(jìn)一步簡化辦事程序,方便群眾辦證、辦戶口。下列做法與材料體現(xiàn)的政府職能一致的是A.全國法院建立“失信者黑名單制度”,向社會公布失信被執(zhí)行人名單B.中國人民銀行授權(quán)中國外匯交易中心公布匯率中間價C.國家工商總局決定組織開展專項整治“霸王條款”活動D.某氣象局利用自身的網(wǎng)絡(luò)技術(shù)建立“農(nóng)網(wǎng)”,為農(nóng)民免費(fèi)提供信息咨詢31.

45、具有157年歷史的東莞道生園是廣東四大名園之一,日前被拆,引發(fā)廣泛關(guān)注。不久,東莞市政府決定在原址復(fù)建。保護(hù)好文物古跡主要是因為它是民族歷史文化的物質(zhì)載體作為文化遺產(chǎn)是一個國家和民族歷史文化成就的重要標(biāo)志具有很高的經(jīng)濟(jì)價值是民族文化的集中展示和民族情感的集中表達(dá)A.B.C.D.32.2013年第二屆嶺南民俗文化節(jié)在廣州舉行,來自廣東省內(nèi)12個地市的30多支民俗隊伍參加表演和巡游。舉辦民俗文化節(jié)是尊重文化多樣性的表現(xiàn)促進(jìn)了中華民族精神的豐富和發(fā)展有利于在文化的包容性中形成和保持文化特色是人們以自省、自律的方式,提高思想道德修養(yǎng)A.B.C.D.33.右邊漫畫給我們的哲學(xué)啟示是A.關(guān)鍵部分對整體起決

46、定作用B.量變必然引起質(zhì)變C.事物發(fā)展是前進(jìn)性和曲折性的統(tǒng)一D.實踐是認(rèn)識的基礎(chǔ)34.世界各國對海洋經(jīng)濟(jì)可持續(xù)發(fā)展理念,雖有不同的理解,但都認(rèn)為是經(jīng)濟(jì)發(fā)展、環(huán)境以及社會之間的平衡。材料體現(xiàn)的唯物辯證法的哲理有矛盾的普遍性與特殊性相互聯(lián)結(jié)辯證否定的實質(zhì)是揚(yáng)棄事物是普遍聯(lián)系的認(rèn)識事物的角度不同,其價值評價A.B.按f t 查表有:62219.6210/,Pr 0.695, 2.93710/(f f f m s W m K -=3621.0075/(, 1.041/,20.4110/f f f Cp kJ kg K kg m N s m -=42360.045,Re 5.510511020.4110

47、444m m f ffu dGGd d 材料三:十八屆三中全會立足黨和國家全局,提出堅持和完善基本經(jīng)濟(jì)制度、加快完善現(xiàn)代市場體系、構(gòu)建開放型經(jīng)濟(jì)新體制等一系列新部署。全會要求全面深化改革必需堅持以馬列主義、毛澤東思想、中國特色社會主義理論體系為指導(dǎo),必須牢牢把握正確方向、掌握科學(xué)有效的改革方法,才能不斷推動改革開放的偉大事業(yè)向前發(fā)展。(1從材料一反映了什么經(jīng)濟(jì)態(tài)勢?(4分(2運(yùn)用經(jīng)濟(jì)生活的知識,結(jié)合材料一、二,為我國全面提高開放型經(jīng)濟(jì)水平出謀劃策。(12分(3分析材料三所體現(xiàn)的辯證唯物論道理。(15分37.(21分閱讀下列材料,結(jié)合所學(xué)知識回答問題。材料一:廣東省惠州市在“崇文厚德,包容四海,敬業(yè)樂群”的惠州精神指導(dǎo)下,深度激活文化生產(chǎn)力,大力發(fā)展文化產(chǎn)業(yè),促進(jìn)惠州經(jīng)濟(jì)蓬勃發(fā)展;扎實開展各項文化惠民工程,相繼建成一大批公共文化場館,充分尊重人民群眾的主體地位,推動不同人群共享文化發(fā)展成果;以熱情開放的客家文化為起點,大力