化工原理计算题解答

目 录

第一章 流体流动与输送设备·····················································(2) 第二章 非均相物系分离·························································(26) 第三章 传热···································································(32) 第四章 蒸发···································································(44) 第五章 气体吸收·······························································(48) 第六章 蒸馏···································································(68) 第七章 干燥···································································(84) 第八章 萃 取··································································(92)

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第一章 流体流动与输送机械

1. 燃烧重油所得的燃烧气,经分析知其中含CO28.5%,O27.5%,N276%,H2O8%(体积%),试求此混合气体在温度500℃、压力101.3kPa时的密度。

解:混合气体平均摩尔质量

Mm??yiMi?0.085?44?0.075?32?0.76?28?0.08?18?28.86?10?3kg/mol∴ 混合密度

pMm101.3?103?28.86?10?3??0.455kg/m3 ?m?RT8.31?(273?500)

2.已知20℃下水和乙醇的密度分别为998.2 kg/m3和789kg/m3,试计算50%(质量%)乙醇水溶液的密度。又知其实测值为935 kg/m3,计算相对误差。

解:乙醇水溶液的混合密度

1?m?a1?1?a2?2?0.50.5? 998.27893 ??m?881.36kg/m

相对误差:

?m实??m?881.36? ?100%??1???100%?5.74%

?m实935??

3.在大气压力为101.3kPa的地区,某真空蒸馏塔塔顶的真空表读数为85kPa。若在大气压力为90 kPa的地区,仍使该塔塔顶在相同的绝压下操作,则此时真空表的读数应为多少?

解:p绝?pa?p真?pa?p真

''?p真?pa?(pa?p真)?90?(101.3?85)?73.7kPa

4.如附图所示,密闭容器中存有密度为900 kg/m3的液体。容器上方的压力表读数为42kPa,又在液面下装一压力表,表中心线在测压口以上0.55m,其读数为58 kPa。试计算液面到下方测压口的距离。

解:液面下测压口处压力

p?p0??g?z?p1??gh

题4 附图

''p1??gh?p0p1?p0(58?42)?103??z???h??0.55?2.36m

?g?g900?9.81

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5. 如附图所示,敞口容器内盛有不互溶的油和水,油层和水层的厚度分别为700mm和600mm。在容器底部开孔与玻璃管相连。已知油与水的密度分别为800 kg/m3和1000 kg/m3。

(1)计算玻璃管内水柱的高度;

(2)判断A与B、C与D点的压力是否相等。 解:(1)容器底部压力 p?pa??油gh1??水gh2?pa??水gh

h1 h2 A B D

C 题5 附图

?油h1??水h2?油800 ?h??h1?h2??0.7?0.6?1.16m

?水?水1000 (2)pA?pB pC?pD

6.水平管道中两点间连接一U形压差计,指示液为汞。已知压差计的读数为30mm,试分别计

算管内流体为(1)水;(2)压力为101.3kPa、温度为20℃的空气时压力差。

解:(1)?p?(?0??)Rg?(13600?1000)?0.03?9.81?3708.2Pa (2)空气密度

pM101.3?103?29?10?3??1.206kg/m3 ??RT8.31?(273?20)''' ?p?(?0??)Rg?(13600?1.206)?0.03?9.81?4002.1Pa ' ∵ 空气密度较小,∴?p??0Rg

7.用一复式U形压差计测量水流过管路中A、B两点的压力差。指示液为汞,两U形管之间充满水,已知h1=1.2m,h2=0.4m,h4=1.4m,h3=0.25m,试计算A、B两点的压力差。

解:图中1、2为等压面,即p1?p2

p1?pA??gh1 p2?p3??0gR1

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?pA??gh1?p3??0gR1 (1)

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3 4

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又 p5?p6?pB??gh3??0gR2 p3?p4?p5??g(h4?h2)

?pB??gh3??0gR2??g(h4?h2) (2) 将(2)代入(1)中:

pA??gh1?pB??gh3??0gR2??g(h4?h2)??0gR1

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2 5 6

题7 附图

??pAB?pA?pB

??gh3??0gR2??g(h4?h2)??0gR1??gh1 ??g(h2?h1?h3?h4)??0g(R1?R2) ?(R1?R2)(?0??)g

??pAB?(1.2?1.4?1.4?0.25)?(13600?1000)?9.81

?241031.7Pa?241.0kPa

8.根据附图所示的双液体U管压差计的读数,计算设备中气体的压力,并注明是表压还是绝压。已知压差计中的两种指示液为油和水,其密度分别为920 kg/m3和998 kg/m3,压差计的读数R=300mm。两扩大室的内径D为60mm,U管的内径d为6mm。

解: 1. 2 为等压面, p1?p2 p1?p??1gh1

p2?pa??(h1?R)??Z??1g?R?2g ?p??1gh1?pa?(h1?R??Z)?1g?R?2g p?pa?Rg(?2??1)??Z?1g 又 R1 h1 2

题8 附图

?4d2??Z?4D2

d262 ??Z?2?R?()?0.3?0.003m

60D ?p?pa?0.3?9.81?(998?920)?0.003?920?9.81 ?256.6Pa (表压)

9. 为了排出煤气管中的少量积水,用附图所示的水封装置,水由煤气管道中的垂直支管排出。已知煤气压力为10kPa(表压),试求水封管插入液面下的深度h。

解: 煤气表压 p??gh

p10?103?3?1.02m h??g10?9.81题9 附图

10.绝对压力为540kPa、温度为30℃的空气,在φ108×4mm的钢管内流动,流量为1500m3/h

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