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 average
the 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 gh 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 (Φ763mm).When the average
pressure changes to be Satm, if it is required that the gas flows in the tube at the same temperature rate and mass velocity, what's the inside diameter of the tube? Solution: Suppose the subscribe l as the state under latm and subscribe 2 as the state under 5atm In the two cases ws=w2=w T=T,=T l because W,=MAP=u,A,P2 A=-d P,=p, 2P1 TP2 PI pI P2 p2 temd2=d1P=0071=0013m 10. As shown in the figure, the feed liquid whose density is 850kg/m is sent into the tower from the elevated tank The liquid level of the elevated tank keeps constant. The gauge pressure in the tower is 0. lkgf/cm2, and the feed rate is 5m/h. The connected pipe is steel pipe( 38x2. 5mm), the energy loss in the connected pipe of the feed liquid is 0. lkgf/cm(the energy loss in the exit is not included). what is the distance between the liquid level of the elevated tank and the feed inlet? Solution: Suppose the liquid level of header tanker as the upper reaches, and the inner side of the connecting pipe as the lower reaches. And suppose sectionl-l' as basic level. We can get the quation u pI 822 P in the equation Z1=0
pressure changes to be 5atm,if it is required that the gas flows in the tube at the same temperature , rate and mass velocity, what’s the inside diameter of the tube? Solution: Suppose the subscribe 1 as the state under 1atm and subscribe 2 as the state under 5atm. In the two cases ws1 = ws2 = ws T1 = T2 = T u1 = u2 = u because ws = u1A11 = u2A2 2 1 2 2 1 1 2 2 4 T P T p A d = = so 2 1 2 2 1 1 2 ( ) p p d d = = then mm p p d d 0.0313 5 1 0.07 2 1 2 = 1 = = 10. As shown in the figure, the feed liquid whose density is 850kg/m³ is sent into the tower from the elevated tank .The liquid level of the elevated tank keeps constant. The gauge pressure in the tower is 0.1kgf/cm²,and the feed rate is 5m³/h. The connected pipe is steel pipe(Φ382.5mm),the energy loss in the connected pipe of the feed liquid is 0.1kgf/cm²(the energy loss in the exit is not included).What is the distance between the liquid level of the elevated tank and the feed inlet? Solution: Suppose the liquid level of header tanker as the upper reaches, and the inner side of the connecting pipe as the lower reaches. And suppose section1-1’ as basic level. We can get the equation + + = + + +hf u p gZ u p gZ 2 2 2 2 1 2 1 1 2 2 in the equation Z1 = 0
P1≈0(表压) u,=V/A= =162m/s 3600×-×0033 P2=0.×9.807×10=9807N/m2 ∑h=30J/kg therefore we can get that 6229807 2a5o+30)/981=-437m the liquid level of header tanker should be 4. 37m higher than the orifice for raw stuff. 11. The liquid level of the elevated tank is &m higher than the floor. The water flows out of the pipeline((108x4mm) The exit of the pipeline is 2m higher than the floor. In the given condition, the energy loss of the water flowing through system (the energy loss of the exit is not included )can be calculated by 2hf=6.5m2, where u is water velocity (in m/s).Try to calculate 1)The velocity of the water at the"A--A "cross section; 2)The flow rate of water(in m/h) Solution: 1)Suppose the liquid level of header tanker as the upper reaches, and the inner side of the pipe's exit as the lower reaches. Suppose the ground as basic level. Then we can get the quation g g ∑b in the equation Z1=8m Z2 0 P1=P2 ∑h=6.52=65h2 based on the above, we can get
u V A m s p u s 1.62 / 0.033 4 3600 5 / 0( 0 2 2 1 1 = = = 表压） 4 2 2 p = 0.19.80710 = 9807N / m hf = 30J / kg therefore we can get that Z 30)/ 9.81 4.37m 850 9807 2 1.62 ( 2 2 = − + + = − the liquid level of header tanker should be 4.37m higher than the orifice for raw stuff. 11. The liquid level of the elevated tank is 8m higher than the floor. The water flows out of the pipeline(Φ1084mm).The exit of the pipeline is 2m higher than the floor. In the given condition, the energy loss of the water flowing through system (the energy loss of the exit is not included )can be calculated byΣhf=6.5m²,where u is water velocity(in m/s).Try to calculate : 1) The velocity of the water at the “A--A” cross section; 2) The flow rate of water(in m²/h) Solution: 1) Suppose the liquid level of header tanker as the upper reaches, and the inner side of the pipe’s exit as the lower reaches. Suppose the ground as basic level. Then we can get the equation + + = + + +hf u p gZ u p gZ 2 2 2 2 1 2 1 1 2 2 in the equation Z1 = 8m Z2 = 2m 2 2 2 1 2 1 6.5 6.5 0 h u u p p u f = = = based on the above，we can get u 9.81 6 / 7 2.9m /s 2 = =
because the pipes'diameters are the same, and water density can be considered as constant, so the elocity at section A-A u, =2.9m/s 2) volume flow rate of water V=3600A=3600×2×0.12×29=82m/h 12. The water at 20C flows through the horizontal pipe((38x2. 5mm)at 2.5m/s The pipe is onnected with the another horizontal( 53 x3mm )through a conical pipe. As shown in the figure, two vertical glass tubes are inserted in either side of the conical pipe. If the energy loss through sections A and B is 1.5J/kg, try to calculate the distance between the two liquid level of the glass be(in mm)? Draw the relative position of the liquid level in the figure Solution: From section A-A and section B-B, then the Bernoulli equation is p+∑b Z=Z=O ∑ according to the continuity equation for the incompressible fluid, then )2=2.5()2=1.23m/ the pressure difference between the two sections is P8-PA 2.52-1.232 1.5)×1000=8685N/m then PB-PA=868.5/9.798=88.6mmH20 the liquid level's difference between the two pipes is 88.6mm 886 so P8> PA the pipe's liquid level B is higher than the liquid level-A 13. The water at 20C is sent to the top of the scrubber from the reservoir by the centrifugal
because the pipes’ diameters are the same, and water density can be considered as constant，so the velocity at section A-A u m s A = 2.9 / 2）volume flow rate of water 2 3 3600 3600 0.1 2.9 82 / 4 V Au m h h = = = 12. The water at 20℃ flows through the horizontal pipe(Φ382.5mm) at 2.5m/s.The pipe is connected with the another horizontal (Φ533mm)through a conical pipe. As shown in the figure, two vertical glass tubes are inserted in either side of the conical pipe. If the energy loss through sections A and B is 1.5J/kg,try to calculate the distance between the two liquid level of the glass tube(in mm)? Draw the relative position of the liquid level in the figure. Solution：From section A——A’ and section B——B’，then the Bernoulli equation is + + = + + + f AB B B B A A A h u p gZ u p gZ , 2 2 2 2 in the equation = = = = h J kg u m s Z Z f AB A A B 1.5 / 2.5 / 0 , according to the continuity equation for the incompressible fluid，then 2 2 4 4 u A d A uB d B = so m s d d u u B A B A ) 1.23 / 47 33 ( ) 2.5( 2 2 = = = the pressure difference between the two sections is ) 2 ( , 2 2 − − − = f AB A B B A h u u p p =（ 2 2 2 1.5) 1000 868.5 / 2 2.5 1.23 − = N m − then pB − pA =868.5/9.798=88.6mmH2O the liquid level’s difference between the two pipes is 88.6mm。 because pB = 6 + pA 88. so pB pA the pipe’s liquid level B is higher than the liquid level-A. 13. The water at 20℃ is sent to the top of the scrubber from the reservoir by the centrifugal
pump. The liquid level of the reservoir keeps constant, as shown in the figure. The diameter of the pipeline is (762.5mm. In the operating condition, the reading of the vacuum gauge in the pump entrance is 185mmHg The energy loss of the suction tube (The entrance of the pipe is not included )and the discharge tube can be calculated according to formulachf, 1=2u2 and Ehf,2=10u?, where u is water velocity in the suction and discharged tubes. The gauge pressure of the joint of the discharged tube and the spray nozzle is lkgf/cm2. Try to calculate the effective power of the pump Solution: Suppose liquid level of storing tank as section 1---l, the connection of vacuum meter as section 2---2. And suppose section 1---I as basic level. Then the Bernoulli equation Z 82+Y3 +2+∑h in the equation Z=0 Z2 1.5m PI ∑h/:=2 2s、l85 ×10133×105=-247×104N/ From the above, we can get the velocity of water in the pipe 2.47×10 9.81×1.5)2.5=2m/s 1000 w,=uAp=2××0.0712×1000=7.92kg/s 2) Suppose the liquid level of storing tank as upper reaches section 1--l; suppose the connection point of drain pipe and blow head as lower reaches section 3---3; Suppose section I---1 as basic level Then we get the equation pI +w=gZ +∑b+∑ in the equation
pump. The liquid level of the reservoir keeps constant, as shown in the figure. The diameter of the pipeline is Φ762.5mm.In the operating condition ,the reading of the vacuum gauge in the pump entrance is 185mmHg.The energy loss of the suction tube (The entrance of the pipe is not included)and the discharge tube can be calculated according to formulaΣhf，1=2u² and Σhf‚2=10u²,where u is water velocity in the suction and discharged tubes. The gauge pressure of the joint of the discharged tube and the spray nozzle is 1kgf/cm2 . Try to calculate the effective power of the pump. Solution: Suppose liquid level of storing tank as section 1---1, the connection of vacuum meter as section 2---2. And suppose section 1---1 as basic level. Then the Bernoulli equation is + + = + + + ,1 2 2 2 2 1 2 1 1 2 2 hf u p gZ u p gZ in the equation Z1 = 0 Z2 =1.5m 0 2 0 2 2 ,1 1 = = u h u p f 5 4 2 2 1.0133 10 2.47 10 / 760 185 p = − = − N m From the above，we can get the velocity of water in the pipe u 9.81 1.5)2.5 2m/s 1000 2.47 10 ( 4 − = = the water mass flow rate w uA k g s s 0.071 1000 7.92 / 4 2 2 = = = 2）pump’s effective power Suppose the liquid level of storing tank as upper reaches section 1---1; suppose the connection point of drain pipe and blow head as lower reaches section 3---3; Suppose section 1---1 as basic level. Then we get the equation + + + = + + + ,1 + ,2 2 2 2 2 1 2 1 1 2 2 e hf hf u p W gZ u p gZ in the equation