Problems and Solutions to Chemical Engineering Principles 化工原理教研组编 Chapter 1 Fluid Mechanics 1.The flame gas from burning the heavy oil is constituted of 8.5%COz,7.502. 76%N2.8%HzO(in volume).When the temperature and pressure are 500'Cand 1atm. respectively.calculate the density of the mixed gas Solution:The molecular weight of the gaseous mixture Mn is Mn=M yoo+My:+M way w:+M woy w2o =44x0.085+32×0.075+28x0.76+18x0.08 -28.86kg/kmol Under 500C.1atm the density of the gaseous mocture is .m*70+卫.28.86,273 -0.455kgmm 22.4*T◆022.4273 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 n the equipment is equal to that atmosphere pressure minus vacuum Pab50hute)=740-100 =640mmHg 640xL0133x10 =8.53x10Nm 760 The gauge pressure-vacuum =-100mmHg 1.0133×103 =.(100× )=-1.33×10'wm 760 or the gauge pressure=-(100x1.33x10)=-1.33x10 N/m 3.As shown in the figure,the reservoir hoids the oil whose density is 960kg/m'.The oil level is 9.6m higher than the bottom of the reservoir.The pressure above the oil level
Problems and Solutions to Chemical Engineering Principles 化工原理教研组编 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(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 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。+pgh The atmosphere pressure which acts on the outer flank of the cover is p.,then pressure difference between the inner flank and outer flank is Ap"p-P,"p.+pgh-P.pgh △p=960✉9.81(9.6-0.8)=8.29x10NW㎡ The static pressure which acts on the cover is p=Apxd2=829x10'xIx0.762-3.76x10Nm 4 4 The pressure on every screw is 400x9.807xZx0.0142-6.04x10N the Number of screw=3.76x10 76.04x102623 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 PP.P are the densities of gas,water and meroury.respectively.then the pressure difference could be ignored for P (PeThen Ps P,cndpa Po According to the basic hydrostatics equation Pa P.=Pm3ogR2 PnegR =1000x9.81￥0.05+13600✉981×0.05
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/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㎡ Pg元Pe=P4+PmgR,=7181+13800*9.810,4=6.05x10,Nm 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? 20 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 drferent among 1.2 and 3.They are at the same level plane of static liquld,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)P=Ps then P+Puaogh:Pa+Piao8(hh)+Puegh Pa P-(Pug-Pm2o)gh =101330-(13600-1000).9.8101 =88970Nh or Pa=12360N/m (vacuum) and because P=P then P+Pirsogh:=Pe+prgh s0P。=P,-(P-Pogh =101330-(13600-1000)x9.81.02 =76610N
=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 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 B p =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 p.=24720N/m (vacuum degree) 6.As shown in the figure,measure the steam pressure above the boiler by the series 'U"-shape dfferential pressure meter.the indicating liquid of the dfferential pressure meter is meraury.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 respedtively .The distance between the level of the water in the boier 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/m.kgf/cm? respectively.) Solution:Choose 2.3 and 4 in the U-pipe of series connection as referring level. According he basic hydrostatical principle.start from 2.we can get the equation of every basic level.Then we can get the pressure p of the water vapor. 乃■P■P+Pg(%-)arP:-P。-PgA-A)】 月=P=P。-Pwog6-h)arB-乃=-Pm2ogh,-A） P.=p =p.+Png(h,-h)or p.-P:Pug(h-h) P=P,-Pw3o8(h-h）oNP6-P,=-Pmgh-A】 From the的ove equ减ions,we can get P%=P,+Pm8h-A)+(h,-h】-Pmo8h-h)+(h-h,刀 50 745 p%三760 101330+13600*9.81(2.3-1.2)+(2.5-1.4)1 -1000×9.81(2.5-12)+(3-1.4)1 =364400NP crP=3644009.807%10'=3.72 kgf/cm 7.Based on the reading of the dfferential 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
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/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
920kgim and 998kg/m.respectively.The distance of the water and ol 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 he lquid levels in the two enlarging room are at the same level.Then the relation between the enlarging room and differential pressure meter is D2h=d产R 4 4 When the value of drferential pressure meter R300mm the difference of liquid level between the two enlarging rooms is ah=R(y-0.3产-0.003m 60 Suppose P.P,are the densities of water and oil respectively,according to the basic hydrostatical principle p-P.=(p-P:)gR+P:gN then the pressure of gas in the pipe is P=(998-920》×9.81x0.3+920x9.81×0.003=257NWm 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 9ms.The average temperature of the air in the tube is 50C.he pressure is 2kgf/cm'(gauge pressure).The local atmospheric pressure is 740mmHg.Try to calculate 1)The mass velocty of the ar. 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 x1.0133×103+2×9807×10=2.95×10Nm the density of air under the operating condition
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 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
0"pT2-1293 273×2.95×10 Tp' (273+501.0133×10-318g/m the mass flow rate of gas is W,=4p=9×121×T×0.022×3.18=1.09kg1s 2)the volume flow rate of gas under the operating condition is '.=p,/p=1.09/3.18=0.343m3/s 3)he volume flow rate of air under the standard condition is =w,/p'-1.09/1.293=0.843m3/s 9.The gas at the average pressure of 1atm flows in the pipe (76x3mm).When the average pressure changes to be 5atmif it is required that the gas flows in te 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 stale under 5atm. In the two casesw=w,=w, T=T=T 4=出= because ,=4A月=AP A-d TPu =P:TP: 50 丹丹 then d:=di B=0.07 =00313mm P: 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 0.1kgf/cm'.and the feed rate is 5m/h.The corected
= = Tp T p 1.293× 3 5 5 3.18 / (273 50)1.0133 10 273 2.95 10 = k g 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 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 f eed rate is 5m³/h. The connected
pipe is steel pipe(38x2.5mm),the energy loss in the connected pipe of the feed liquid is 0.1kgf/cm(the energy loss in the exit is not included).What is the distance between the liquid level of the clevated tank and the feed inlet? Solution:Suppose the liquid level of header tanker as the upper reaches,and the inner side of the comecting pipe as the lower reaches.And suppose section1-1'as basic level.We can get the equation 2 P in the equationZ=0 车0 P=0表压) 5 3=',/A= =1.62m/s 3600×x00332 4 P:=0.1×9.807×10'=9807N/m2 ∑h,=3U/kg therefore we can get that 1.6229807 Z2= +30)/9.81=-4.37m 2850 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 te 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 byEhf6.5m,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
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 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 2+号+A=2,+号+2+∑， 2 P 2 P in the equation Z=8m Z=2m 高1名0 P=P: ∑h,=6.5m2=65u based on the above.we can get 42=√981×6/7=2.9m/s because the pipes'diameters are the same,and water density can be considered as constant.so the velocity at sectionA-A ar =2.9m/s 2)volume flow rate of water K=3600Am=3600×x0.12×2.9=82m1h 4 12.The water at 20C flows through the horizontal pipe(38x25mm)at 2.5m's.The ppe is connected with the another horizontal (53x3mmythrough 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 he glass tube(in mm)?Draw the relative postion of the liquid level in the figure. Solution:From section A-A and section B-B',then the Bernoulli equation is g2,++P=g,+ 2 P ++∑如 2 P Z,=Z。=0 in the equation M2.5m/s ∑Aa=15UIkg according to the continuity equation for the incompressible fluid,then
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 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 =
33 50 Ma M(- 2-25 =1.23m/s d。 47 the pressure difference between the two sections is p,-p,=巴，g-∑p 2 =（25-1.23 -1.5)x1000=868.5N1m3 then Pg-P,=868.59.798=88.6mH0 the liquid level's difference between the two pipes is 88.6mm. because pa =886+p S0 Pa>P 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 sarubber from the reservoir by the centrfugal pump.The liquid level of the reservor keeps constant,as shown in the figure. The diameter of the pipeline is 76x2.5mm.In the operating condition ,the reading of the vacuum gauge in the pump entrance is 185mmHg.The energy loss of te suction tube (The entrance of te pipe is not included)and the discharge tube can be calculated according toformulaghf.1=2u and Ehf2=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 1kgf/cm.Try to calculate the effective power of the pump. Solution:Suppose liquid level of storing tank as section 1--1.the connedtion of vacuum meter as section 2--2.And suppose section 1--1 as basic level.Then the Bernoulli equation is g肠+号+A=g+号+A+∑ 2 P 2 P in the equation Z =0 Z=1.5m
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 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
A=0 ∑nu=2w2 码0 185 P乃=- ×1.0133×103=-247×10N/m2 760 From the above.we can get the velocity of water in the pipe 247×10 -9.81×1.5)2.5=2m/s 1000 the water mass flow rate W,=4p-2×Tx0.072×1000-7.92kg/3 4 2)pump's effective powe时 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 g,+生+A+所=g昭+兰++∑+∑4： 2 P 2 P in the equation Z=0 乙3=14m 4=0 2=2m/s 片=0 P3=9.807×10°N/m2 Ehr +hy:=2+10u =12u from the data above.we get 形，=981×14+9807x10+125x22-2854J1g 1000 the pump's effective power is N。=Wm,=2854×7.92=2260W-226kf 14.As shown in the figure,the inside diameter D of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm.There is no
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 0 0 0 1 1 1 = = p u Z 4 2 2 2 2 9.807 10 / 2 / 14 p N m u m s Z m = = = 2 2 2 hf ,1 +hf ,2 = 2u +10u =12u from the data above, we get We 12.5 2 285.4J / k g 1000 9.807 10 9.81 14 2 4 + = = + the pump’s effective power is Ne = Wews = 285.47.92 = 2260W = 2.26kW 14. As shown in the figure, the inside diameter D of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm.There is no