the two enlarging rooms is (b32=0360)2=00 Suppose P1, p2 are the densities of water and oil respectively, according to the basic hydrostatical principle p-pa=(p,-p2)gR+p2gAh then the pressure of gas in the pip (998—920)×981×0.3+920×9.81×0.003=257N/r 8. The tube bundle of the tubular heat exchange is constituted of 121 steel tubes( ( 25x2. 5mm). The air flows in the tube bundle at 9m/s. The average temperature of the air in the tube is 50C, the pressure is 2kgf/cm(gauge pressure). The local atmospheric pressure is 740mmHg. Try to calculate 1)The mass velocity of the air; 2) The volume flow rate of the air in the operating condition 3) The volume flow rate of the air in the standard condition Solution: 1)the density of air is 1. 293kg/m pressure in operating 740 10133×103+2×9.807×104=2.95×103N/m the density of air under the operating condition 273×295×10 =3.18kg/m3 (273+50)1.0133×10 flow rate of w,=u4p=9×121××0.02×3.18=1.09kg/s 2)the volume flow rate of gas under the operating condition V=,/p=1.09/3.18=0.343m3/s 3)he volume flow rate of air under the standard condition i V=w,/p’=1.09/1293=0.843m3/s 9. The gas at the average pressure of latm flows in the pipe((76x3mm) When the averagethe two enlarging rooms is Δh=R m D d ) 0.003 60 6 ( ) 0.3( 2 2 = = Suppose 1 2  ,  are the densities of water and oil respectively, according to the basic hydrostatical principle p − pa = (1 −  2 )gR +  2 gh then the pressure of gas in the pipe is p=（998―920）×9.81×0.3+920×9.81×0.003=257N/m² 8.The tube bundle of the tubular heat exchange is constituted of 121 steel tubes(Φ25×2.5mm).The air flows in the tube bundle at 9m/s.The average temperature of the air in the tube is 50℃,the pressure is 2kgf/cm²(gauge pressure).The local atmospheric pressure is 740mmHg. Try to calculate : 1) The mass velocity of the air; 2) The volume flow rate of the air in the operating condition; 3) The volume flow rate of the air in the standard condition. Solution: 1) the density of air is 1.293kg/m3 pressure in operating 5 4 5 1.0133 10 2 9.807 10 2.95 10 760 740 p =   +   =  N/m² the density of air under the operating condition  =  =   Tp T p 1.293× 3 5 5 3.18 / (273 50)1.0133 10 273 2.95 10 = kg m +    the mass flow rate of gas is w uA k g s s 0.02 3.18 1.09 / 4 9 121 2 = =     =   2) the volume flow rate of gas under the operating condition is V w m s s s / 1.09 / 3.18 0.343 / 3 =  = = 3) he volume flow rate of air under the standard condition is V w m s s s / 1.09/1.293 0.843 / 3 =  = =   9. The gas at the average pressure of 1atm flows in the pipe (Φ763mm).When the average