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 the400kgf/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 6.0410 =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