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 isChapter 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