Problems and solutions to chemical Engineering Principles 化工原理教研组编
Problems and Solutions to Chemical Engineering Principles 化工原理教研组编
Chapter I Fluid mechanics 1. The flame gas from burning the heavy oil is constituted of 8.5%CO2, 7.502, 76%N2 8%H2O(in volume). When the temperature and pressure are 500 Cand l atm, respectivel calculate the density of the mixed gas Solution: The molecular weight of the gaseous mixture Mn is ya2+M。2y。2+MN2y 44×0.085+32×0.075+28×0.76+18×0.08 28. 86kg/kmol Under 500C, latm, the density of the gaseous mixture is A1m1P=2886×273=045m 224*T 224273 2.The reading of vacuum gauge in the equipment is 100mmHg, try to calculate the absolute pressure and the gauge pressure, respectively. Given that the atmospheric pressure in this area is 740mmHg Solution: The absolute pressure in the equipment is equal to that atmosphere pressure minus P(absolute)=740-100 =640mmHg 1.0133×10 =640 =8.53×104N/m 760 The gauge pressure=-vacuum=-100mmHg 10133×105 =(100× )=-1.33×104N/m2 760 (100×1.33×102)=1.33×10N/m2 3. As shown in the figure, the reservoir holds the oil whose density is 960kgm. The oil level is 9.6m higher than the bottom of the reservoir. The pressure above the oil level is atmospheric pressure. There is a round hole(( 760mm )at the lower half of the sidewall, the center of which is 800mm from the bottom of the reservoir The hand-hole door is fixed by steel bolts(14mm). If the working stress of the bolts is
Chapter 1 Fluid Mechanics 1.The flame gas from burning the heavy oil is constituted of 8.5%CO2，7.5O2，76%N2， 8%H2O(in volume).When the temperature and pressure are 500℃and 1atm, respectively, calculate the density of the mixed gas . Solution: The molecular weight of the gaseous mixture Mn is Mn=M co2 y co2 + M o2 y o2 + M N 2 y N 2 + M H 2O y H 2O =44×0.085+32×0.075+28×0.76+18×0.08 =28.86kg/kmol Under 500℃,1atm,the density of the gaseous mixture is ρ= T po Mm To p 22.4* * * * = 22.4 28.86 × 273 273 =0.455kg/m³ 2.The reading of vacuum gauge in the equipment is 100mmHg,try to calculate the absolute pressure and the gauge pressure, respectively. Given that the atmospheric pressure in this area is 740mmHg. Solution: The absolute pressure in the equipment is equal to that atmosphere pressure minus vacuum P(absolute)=740―100 =640mmHg =640× 760 1.0133 105 =8.53×10 4 N/m² The gauge pressure=-vacuum =-100mmHg =-（100× 760 1.0133 105 ）=―1.33×10 4 N/m² or the gauge pressure=-（100×1.33×10 2 ）=―1.33×10 4 N/m² 3.As shown in the figure, the reservoir holds the oil whose density is 960kg/m³. The oil level is 9.6m higher than the bottom of the reservoir. The pressure above the oil level is atmospheric pressure. There is a round hole(Φ760mm )at the lower half of the sidewall, the center of which is 800mm from the bottom of the reservoir. The hand-hole door is fixed by steel bolts (14mm). If the working stress of the bolts is
400kgf/cm how many bolts should be needed Solution: Suppose the static pressure of the liquid on 0-0 level plane is p, then p is the average pressure of the liquid acting on the cover According to the basic hydrostatics equation p→pa+pgh The atmosphere pressure which acts on the outer flank of the cover is pa, then pressure difference between the inner flank and outer flank is Ap=p-p.=P, +pgh-P, =pgh △p=960×9,8196-0.8)=8.29×104Nm2 The static pressure which acts on the cover P=4p2-80076=376×0Nm The pressure on every screw is 400×9807×-×00142=6.04×103N the Number of screw=3.76x104 =6.23 6.04×10 4. There are two differential pressure meter fixed on the fluid bed reactor, as shown in the figure. It is measured that the read ing are r1=400mm, R2=500mm, respectively. Try to Iculate the pressures of points A and B Solution: There is a gaseous mixture in the U-differential pressure meter. Suppose Pg, PH2o, PHg are the densities of gas, water and mercury, respectively, then the pressure ifference could be ignored for p《pg, Then p4≈PmdB≈PD According to the basic hydrostatics equation PA≈P=Pn208R2+pmgR2 =1000×981×0.05+13600×981×0.05 716IN/m PB≈PD=P4+Pm8R1=7161+13600×981×0.4=6.05×104N/m 5. As shown in the figure, the manometric tubes are connected with the equipment a, B C, respectively. The indicating liquid in the tubes is mercury, while the top of the
400kgf/cm2 , how many bolts should be needed ? Solution: Suppose the static pressure of the liquid on 0-0 level plane is p, then p is the average pressure of the liquid acting on the cover. According to the basic hydrostatics equation p=p a +ρg h The atmosphere pressure which acts on the outer flank of the cover is pa, then pressure difference between the inner flank and outer flank is Δp=p―p a = p a +ρgh― pa = ρgh Δp =960×9.81(9.6―0.8)=8.29×10 4 N/m² The static pressure which acts on the cover is p = Δp× 2 4 d =8.29×10 4 2 4 0.76 3.76 10 4 = N/m2 The pressure on every screw is N 2 3 0.014 6.04 10 4 400 9.807 = the Number of screw=3.76×10 3 4 6.0410 =6.23 4. There are two differential pressure meter fixed on the fluid bed reactor, as shown in the figure. It is measured that the reading are R1=400mm，R2=500mm, respectively. Try to calculate the pressures of points A and B. Solution: There is a gaseous mixture in the U-differential pressure meter. Suppose g H O Hg , , 2 are the densities of gas, water and mercury, respectively, then the pressure difference could be ignored for g 《 Hg . Then pA pcandpB pD According to the basic hydrostatics equation pA pc = H 2O gR2 + Hg gR2 =1000×9.81×0.05+13600×9.81×0.05 =7161N/m² pB pD = pA + Hg gR1 =7161+13600×9.81×0.4=6.05×10 4 N/m 5. As shown in the figure, the manometric tubes are connected with the equipment A , B, C, respectively. The indicating liquid in the tubes is mercury, while the top of the
tube is water, water surfaces of the three equipments are at the same level. Try to determine 1)Are the pressures equal to each other at the points of 1, 2, 3? 2)Are the pressures equal to each other at the points of 4, 5, 6? 3)If h1=100mm, h2=200mm, and the equipment A is open to the atmosphere(the tmospheric pressure is 760mmHg), try to calculate the pressures above the water in the equipment B and C Solution: I) The pressure is different among 1, 2 and 3. They are at the same level plane of static liquid, but they don t connect the same liquid 2)The pressure is the same among 4, 5 and 6. They are at the same level plane of atic liquid, and they connect the same liquid 3)∵P4=P +Puzogh,=PB+PRog(,-h,)+pu Pe=P-(pHe-Puzo)gh, =101330—(13600-1000)×981×0.1 88970N/m2 or P:=12360N/m2(vacuum se P4= p6 then PA+ PH2ogh2=Pc+ PHggh2 P=Pu-(PHg-PH2o)gh2 =101330—(13600-1000)×9.81×0.2 =76610N/m or P=24720N/m2(vacuum degree 6. As shown in the figure, measure the steam pressure above the boiler by the series U-shape differential pressure meter, the indicating liquid of the differential pressure meter is mercury, the connected tube between the two"U -shape meters is full of water. Given that the distance between the mercury levels and the referring level are h1=2.3m, h2=1.2m
tube is water, water surfaces of the three equipments are at the same level. Try to determine: 1) Are the pressures equal to each other at the points of 1,2,3? 2) Are the pressures equal to each other at the points of 4,5,6? 3) If h1=100mm，h2=200mm，and the equipment A is open to the atmosphere(the atmospheric pressure is 760mmHg),try to calculate the pressures above the water in the equipment B and C. Solution: 1) The pressure is different among 1, 2 and 3. They are at the same level plane of static liquid, but they don’t connect the same liquid. 2) The pressure is the same among 4, 5 and 6. They are at the same level plane of static liquid, and they connect the same liquid. 3) p4 = p5 then 2 2 2 2 1 1 pA + pH O gh = pB + H O g(h − h ) + Hg gh 2 1 pB = pA − ( Hg − H O )gh =101330―（13600―1000）×9.81×0.1 =88970N/m² or pB =12360N/m²（vacuum） and because p4 = p6 then pA + H 2O gh2 = pC + Hg gh2 so pc = 2 2 p A − ( Hg − H O )gh =101330―（13600―1000）×9.81×0.2 =76610N/m² or pc = 24720N/m²（vacuum degree） 6. As shown in the figure, measure the steam pressure above the boiler by the series “U”-shape differential pressure meter, the indicating liquid of the differential pressure meter is mercury, the connected tube between the two “U ”-shape meters is full of water. Given that the distance between the mercury levels and the referring level are h1=2.3m，h2=1.2m
h3=2.5m, h4=1. 4m respectively. The distance between the level of the water in the boiler and the base level is h5=3m. The atmospheric pressure Pa is 745mmHg. Try to calculate the vapor pressure P above the boiler. (in N/m2, kgf/cm2respectively. Solution: Choose 2, 3 and 4 in the U-pipe of series connection as referring level. According the basic hydrostatical principle, start from 2, we can get the equation of every basic level. Then we can get the pressure Po of the water vapor. p2=p,=Pa+ Pug(h,-h,) or p2-P=Pueg(h-h,) P3=p3=Pa-PHzog(h, -h,)or P3-P2=-PH2og(h,-h2) P4=P4=Pa+ PHgg(h, -h,) or P4-P3=PHgg(h3-h4) P0=P4-pn08(h5-h4)orp0-p4=-p28(h-h) From the above equations, we can get Po=pa+Ph[(h-h2)+(h1-h4)-P20g{(h3-h2)+(h-h2) 745 pm01330+136009811(23-1.2)+(25-14) 1000×9.81[(2.5-1.2)+(3-14)] 364400N/m orPo=3644009807×10=3.72 kgf/cm2 7. Based on the reading of the differential pressure meter as shown in the figure, calculate the gauge pressure of the the tube line. The indicating liquids in the differential pressure meter are oil and water, respectively, the density of which are 920kg/m and 998kg/m, respectively. The distance of the water and oil interfaces in the" Ushape tube R is 300mm. The inner diameter of the reservoir is 60mm, and the inside diameter of the"U-shape tube is 6mm. When the gas pressure in the tube line is equal to the atmospheric pressure, the liquid level is flush with each Solution: If the barometric pressure in the pipe equals the atmosphere pressure, then the liquid levels in the two enlarging room are at the same level. Then the relation between the enlarging room and differential pressure meter is z When the value of differential pressure meter R=300mm, the difference of liquid level betweer
h3=2.5m，h4=1.4m respectively .The distance between the level of the water in the boiler and the base level is h5=3m. The atmospheric pressure Pa is 745mmHg.Try to calculate the vapor pressure P above the boiler.(in N/m²，kgf/cm² respectively.) Solution: Choose 2, 3 and 4 in the U-pipe of series connection as referring level. According the basic hydrostatical principle, start from 2, we can get the equation of every basic level. Then we can get the pressure 0 p of the water vapor. ( ) p2 p2 = pa + Hg g h1 − h2 = or ( ) p2 − pa = Hg g h1 − h2 ( ) p3 p3 = pa − H 2O g h3 − h2 = or ( ) p3 − p2 = − H 2O g h3 − h2 ( ) p4 p4 = pa + Hg g h3 − h4 = or ( ) p4 − p3 = Hg g h3 − h4 ( ) p0 = p4 − H 2O g h5 − h4 or ( ) p0 − p4 = − H 2o g h5 − h4 From the above equations, we can get [( ) ( )] [( ) ( )] p0 = pa + Hg g h1 − h2 + h3 − h4 − H 2O g h3 − h2 + h5 − h4 so 760 745 p0 = ×101330+13600×9.81[（2.3―1.2）+（2.5―1.4）] ―1000×9.81[（2.5―1.2）+（3―1.4）] =364400N/m² or 0 p =364400/9.807×10 4 =3.72kgf/cm² 7. Based on the reading of the differential pressure meter as shown in the figure, calculate the gauge pressure of the gas in the tube line. The indicating liquids in the differential pressure meter are oil and water, respectively. the density of which are 920kg/m³ and 998kg/m³, respectively. The distance of the water and oil interfaces in the “U”shape tube R is 300mm.The inner diameter of the reservoir is 60mm,and the inside diameter of the “U”-shape tube is 6mm.When the gas pressure in the tube line is equal to the atmospheric pressure, the liquid level is flush with each other. Solution: If the barometric pressure in the pipe equals the atmosphere pressure, then the liquid levels in the two enlarging room are at the same level. Then the relation between the enlarging room and differential pressure meter is D h d R 2 2 4 4 = When the value of differential pressure meter R=300mm, the difference of liquid level between