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 betweerh3=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