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# 硅酸盐工业热工基础习题.doc

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《硅酸盐工业热工基础》课后习题参考答案【 1-5 】 P64 解:因为孔径比较大,所以应该使用气体通过炉门的流出和吸入公式。)/(283 .01000 273 273 32 .1 )/(205 .120 273 273 293 .1 3 00 3 00,mkg T T mkg T T a aa??????????????由 P14 ( 1-46 )式????)(2 0?? agZ FV 流出见 P14 ?可取 0.52 ~ 0.62 。此处取?=0.62 )/( 11.0)/(396 3600 283 .0 )283 .0205 .1(5.081 .924 2.062 .0 )(2 33 2 0smhm gZ FV a????????????????流出)/(053 .0)/(192 3600 205 .1 )205 .1283 .0()5.0(81 .924 2.062 .0 )(2 33 2 0smhm gZ FV a a??????????????????吸入【 1-6 】 P64 解:若空气从孔口流入大气, 由 P= ρ RT ( 1-2 ) P1)/(177 .2)273 15 (1.287 10 8.1 3 5mkg RT P???????由于不考虑阻力损失, ξ=0 P11101 11 1??????? P1264 .0164 .0???????由 P11 ( 1-41 )????)/(1.271 177 .2 10 110 8.121 2 55 12sm PPw a????????????由 P12 ( 1-42 ) ??????)/(6.306 )/(085 .0 177 .2 10 110 8.12025 .04 64 .0 2 33 552 1hmsm PPFV a???????????????)/(667 177 .26.306 hkg vm??????若空气通过渐缩喷嘴喷入大气中。由上可知, 64 .0 1 )/(177 .2?????????skg s由 P18 ( 1-57 ))/(1.299 10 8.1 10 11177 .2 10 8.114.1 4.1211 2 4.1 14.15 55 12 2sm P P Pw ss s????????????????????????????????????????????????????????由 P18 ( 1-58 )??)/(2.756 10 8.1 10 110 8.1 10 1177 .210 8.114.1 4.12025 .04 1 2 4.1 14.15 54.1 25 55 2 12 222hkg P PP PPFm ss ss???????????????????????????????????????????????????????????????????????????????? P18 ( 1-60 ))/(484 2.756 64 .0hkg mm????????【 1-8 】 P64 解:(1 )管嘴为收缩管已知528 .0148 .010 09 .7 10 05 .1 5 5< ???? s aP P ,表明喷嘴出口截面已达到临界状态。压缩空气的滞止密度为: ??)/(43 .8273 20 287 10 09 .7 3 5mkg RT P s ss???????喷嘴出口处气体的临界参数由表 1-4 中的简化公式可得: P21 速度:)/(313 293 287 08 .108 .1sm RT w s cr????实际流出速度: )/(282 313 9.0sm ww cr cr??????(2 )管嘴为拉伐尔管由上可知:喉部已达临界状态。由 P24 ( 1-73b )式,扩张段出口马赫数: 90 .1110 05 .1 10 09 .714.1 211 2 4.1 14.15 5 12?????????????????????????????????????????????????? a sP P Ma 由 P24 ( 1-75 )喷嘴出口处温度: )(170 90 .12 14.11293 2 11 12 1222K Ma TT s???????????????????????由 P17 ( 1-50 ) 出口处断面当地音速: )/(4.261 170 287 4.1 22sm RT a??????由 P17 马赫数定义: )/(497 90 .14.261 222sm Ma aw????实际喷嘴出口速度: )/(447 497 9.0 22sm ww??????【 1-15 】 P65 解:由 P30 式( 1-83 )2198 .007 .014 .314 .3 tan ????a?由 P32 式( 1-93 )a RL 0676 .0?1036 .0676 .0 107 .0676 .0 0???? aL R 由 P30 式( 1-84 )8636 .41036 .020 07 .04.34.3 0???????Rax R x由 P29 式( 1-80 )22 31???????????????????? xcR rw w)/(223 .98636 .4 21 51 22 3 22 3sm R r ww x c?????????????????????????????????????由 P31 式( 1-87 )x cR Rw w 003.3?)/(21 .131 1036 .03.3 8636 .4223 .93.3 0 0sm R Rww xc?????由)/(42 .4)/(15927 3600 21 .131 1036 .0 3600 3 2 0 200smhm wRV??????????【 1-17 】 P65 解:)/(205 .120 273 273 293 .1293 .1 3 0mkg T T a a?????假设烟囱每米温降 1.5 ℃/m。总温降: CH?????5.52 5.135 5.1)(5.252 5.52 200 Ct B????)(25 .226 2 5.252 200 2 C ttt BTm??????)/(722 .025 .226 273 273 32 .1 3 00mkg T T m m??????假设)/(0.2s Nm w T?)(4.22 94 4 0m w Vd T T???????由25 .1/? TBdd)(34.225 .1m d B???)/(273 .13 94 4 22 0s Nm d Vw B B???????)(7.22 34.22 m ddd TBav?????)/(572 .17.2 94 4 22 0s Nm d Vw av av???????)(4.164 722 .02 572 .17.2 35 05 .0722 .02 273 .12)722 .0205 .1(81 .935 22 )( 2 22 222 Pa wd H ww Hg P m av av m BT 内容来自淘豆网www.taodocs.com转载请标明出处.