Problems and solutions to chemical Engineering Principles 化工原理教研组编
Problems and Solutions to Chemical Engineering Principles 化工原理教研组编
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 is
Chapter 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
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 the
400kgf/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 6.0410 =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
tube is water, water surfaces of the three equipments are at the same level. Try to determine 1)Are the pressures equal to each other at the points of 1, 2, 3? 2)Are the pressures equal to each other at the points of 4, 5, 6? 3)If h1=100mm, h2=200mm, and the equipment A is open to the atmosphere(the tmospheric pressure is 760mmHg), try to calculate the pressures above the water in the equipment B and C Solution: I) The pressure is different among 1, 2 and 3. They are at the same level plane of static liquid, but they don t connect the same liquid 2)The pressure is the same among 4, 5 and 6. They are at the same level plane of atic liquid, and they connect the same liquid 3)∵P4=P +Puzogh,=PB+PRog(,-h,)+pu Pe=P-(pHe-Puzo)gh, =101330—(13600-1000)×981×0.1 88970N/m2 or P:=12360N/m2(vacuum se P4= p6 then PA+ PH2ogh2=Pc+ PHggh2 P=Pu-(PHg-PH2o)gh2 =101330—(13600-1000)×9.81×0.2 =76610N/m or P=24720N/m2(vacuum degree 6. As shown in the figure, measure the steam pressure above the boiler by the series U-shape differential pressure meter, the indicating liquid of the differential pressure meter is mercury, the connected tube between the two"U -shape meters is full of water. Given that the distance between the mercury levels and the referring level are h1=2.3m, h2=1.2m
tube is water, water surfaces of the three equipments are at the same level. Try to determine: 1) Are the pressures equal to each other at the points of 1,2,3? 2) Are the pressures equal to each other at the points of 4,5,6? 3) If h1=100mm，h2=200mm，and the equipment A is open to the atmosphere(the atmospheric pressure is 760mmHg),try to calculate the pressures above the water in the equipment B and C. Solution: 1) The pressure is different among 1, 2 and 3. They are at the same level plane of static liquid, but they don’t connect the same liquid. 2) The pressure is the same among 4, 5 and 6. They are at the same level plane of static liquid, and they connect the same liquid. 3) p4 = p5 then 2 2 2 2 1 1 pA + pH O gh = pB + H O g(h − h ) + Hg gh 2 1 pB = pA − ( Hg − H O )gh =101330―（13600―1000）×9.81×0.1 =88970N/m² or pB =12360N/m²（vacuum） and because p4 = p6 then pA + H 2O gh2 = pC + Hg gh2 so pc = 2 2 p A − ( Hg − H O )gh =101330―（13600―1000）×9.81×0.2 =76610N/m² or pc = 24720N/m²（vacuum degree） 6. As shown in the figure, measure the steam pressure above the boiler by the series “U”-shape differential pressure meter, the indicating liquid of the differential pressure meter is mercury, the connected tube between the two “U ”-shape meters is full of water. Given that the distance between the mercury levels and the referring level are h1=2.3m，h2=1.2m
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 betweer
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/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
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
