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Problems and Solutions to Chemical Engineering Principles 化工原理教研組編 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=My+ My+ My+ My =440.085+320.075+280.76+180.08 =28.86kg/kmol Under 500℃,1atm,the density of the gaseous mixture is ρ===0.455kg/m 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=8.5310N/m The gauge pressure=-vacuum =-100mmHg =-（100）=―1.3310N/m or the gauge pressure=-（1001.3310）=―1.3310N/m 3.As shown in the figure, the reservoir holds 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 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+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= p+ρgh―ρgh Δp =9609.81(9.6―0.8)=8.2910N/m The static pressure which acts on the cover is Δp=8.2910N/m2 The pressure on every screw is the Number of screw=3.7610=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 are the densities of gas, water and mercury, respectively, then the pressure difference could be ignored for 《. Then According to the basic hydrostatics equation = =10009.810.05+136009.810.05 =7161N/m =7161+136009.810.4=6.0510N/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) then =101330―（13600―1000）9.810.1 =88970N/m or =12360N/m（vacuum） and because then so =101330―（13600―1000）9.810.2 =76610N/m or 24720N/m（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/m，kgf/cm 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 of the water vapor. or or or or From the above equations, we can get so 101330+136009.81[（2.3―1.2）+（2.5―1.4）] ―10009.81[（2.5―1.2）+（3―1.4）] =364400N/m or =364400/9.80710=3.72kgf/cm 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/m and 998kg/m, 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 When the value of differential pressure meter R=300mm, the difference of liquid level between the two enlarging rooms is Δh=R Suppose are the densities of water and oil respectively, according to the basic hydrostatical principle then the pressure of gas in the pipe is p=（998―920）9.810.3+9209.810.003=257N/m 8.The tube bundle of the tubular heat exchange is constituted of 121 steel tubes(Φ252.5mm).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/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/m3 pressure in operating N/m the density of air under the operating condition =1.293 the mass flow rate of gas is 2) the volume flow rate of gas under the operating condition is 3) he volume flow rate of air under the standard condition is 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 because so then 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 connected pipe is steel pipe(Φ382.5mm),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 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 in the equation therefore we can get that 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=6.5m,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. Then we can get the equation in the equation based on the above，we can get because the pipes’ diameters are the same, and water density can be considered as constant，so the velocity at section A-A 2）volume flow rate of water 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 in the equation according to the continuity equation for the incompressible fluid，then so the pressure difference between the two sections is =（ then =868.5/9.798=88.6mmH2O the liquid level’s difference between the two pipes is 88.6mm。 because so 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=2u and Σhf?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 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 in the equation From the above，we can get the velocity of water in the pipe the water mass flow rate 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 in the equation from the data above, we get the pump’s effective power is 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 liquid supply to the reservoir, and the heigh of the liquid level h2 is 2m.The total energy loss of the liquid flowing through the pipe can be calculated by the formula Σhf=20u,where u is the velocity of liquid (in m/s).Try to calculate the required time when the liquid level of the reservoir drops 1m? Solution：The problem attributes to the instability flow，the time when the surface of the reservoir descend 1m can be calculated from material balance. Suppose F’ is instant feed into system，D’ is instant discharge out of system，dA’ is the accumulated quantity in the dθ time，then in the dθ time, the material balance equation is F’dθ―D’dθ=dA’ And then suppose in the dθ time，the surface of reservoir descend dh，the instant flow rate of liquid in the tube is u。 In the equation F’=0 D’= Then the equation reduces to or In order to work out the relation between the height of instant surface h（take the center line of drainpipe as benchmark）and the instant velocity u in equation（a），we can use the instant equation of Bernoulli. In the equation substitute the above values into formula，we have 9.81h=20 Put formula b into formula a， =―5580 integration range as flollows θ=―55802 =55802 15． A cooling cycle system of brine is shown in the figure. The density of the brine is 1100kg/m,and the circulation is 36m/h.The diameter of the pipeline is the same. The energy loss from A and two heat exchangers to B is 98.1J/kg,the loss from B TO A is 49J/kg.Try to calculate: 1) If the efficiency of the pump is 70%, what is the shaft power of it?(in kw) 2) If the reading of the gauge in position A is 2.5kgf/cm,what is the reading of the gauge in position B? Solution：1）The efficiency of the pump take whichever section as 1-1 and 2-2. it means liquid flow starts from section1-1 and reaches section 2-2 after a circle. For sections 1-1 and 2-2, the Bernoulli equation will be since The equation can be simplified to The mass flow rate is Then, the power of the pump is 2）The reading of the gauge in position B Set up Bernoulli equation between section A and section B，and take the center of the section A as the horizon. In the equation input the data into the Bernoulli equation （gage pressure） The reading of the gauge in position B 16.In a laboratory, the acetic acid (70%) at 20℃ is sent by the glass pipe. The inside diameter of the pipe is 1.5m, and the flow rate is 10kg/min. Calculate Reynolds number in SI system and physical unit system, respectively. What is the flow pattern? solution：1）In SI system From appendix 17 of the book, for the acetic acid (70%), (20℃)， d=1.5cm=0.015m =5657 (belong to the overfall) 2）In physical unit system 17. The liquid whose density is 850kg/m, viscosity is 8cp, flows in the steel pipe (inside diameter is 14mm). The velocity of liquid is 1m/s. Try to find : 1) The Reynolds number, and the flow pattern is pointed out. 2) The position at which the local velocity is equal to the average one. 3) Suppose the pipeline is horizontal pipe, and the pressure of the upstream is 1.5kgf/cm, How long the liquid has to flow when the pressure drop to 1.3kgf/cm? Solution：1）The Reynolds number (belong to the overfall) 2）The position at which the local velocity is equal to the average one According to the formula 1——38 and 1——39， For the position at which the local velocity is equal to the average one，then So r=0.707R=0.7077=4.95mm The position at which the local velocity is equal to the average one is 4.95mm。 3）The section of upstream is 1——1’，the section of downstream is 2——2’，because the pipeline is horizontal pipe , so According to the formula of Hagon-Poiseuille， then , 18. As shown in the figure, an inverse “U”-shape differential pressure meter is fixed between the cross sections of the two horizontal pipes with different diameters in order to measure the pressure difference of the two cross sections. When the flow rate is 10800kg/h, the reading of the “U”-shape differential meter is 100mm.The diameters of the two pipe are Φ603.5mm and Φ423mm,respectively. Calculate : 1) The energy loss of 1kg water flowing through the two cross sections; 2) The pressure drop which is equal to the energy loss (in N/m). Solution：1）The energy loss of 1kg water flowing through the two cross sections Set up Bernoulli equation between section 1——1’ and section 2——2’，and take the axis of pipe as horizon. In the equation input the data into the equation 2） 19. The liquid is flowing laminarly in the straight pipe. If the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original, how many times is the energy loss produced by resistance as much as the original? Solution：According to the formula of Hagon-Poiseuille， Use the subscript1 and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the viscosity of liquid and the length of the pipe have not changed. From the problem, we know the ratio between these two diameters Thus, if the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original，the energy loss produced by resistance has increase to 16 times as much as the original. 20. The liquid is flowing turbulently in the straight pipe. If the physical property of the liquid, the length of the pipe and the diameter are constant, but the flow rate is increased to twice of the original, how many times is the energy loss produced by resistance as much as the original? For two cases, Reynolds number is between 310~1105, the friction coefficient λ can be calculate by the formula “Palasus”. Solution: According to the frictional formula in the straight pipe, Use the subscript1 and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the diameter of the pipe and the length of the pipe have not changed. According to the formula， The coefficients of friction in these cases Because of the flow rate is 2 times as much as the original, then The viscosity and the density of the liquid are invariability，therefore =1/2 Upon that 21. The rectangular chimney (inside cross section is 10001200mm) is 30m high. The effluent gases at 400℃ is flowing from bottom to top, the average molecular weight of gases is 30kg/kmol.The bottom of the chimney keeps 5mm H2O (vacuum. The density of the air can be considered as constant in the range of the height of the chimney. The environmental temperature is 20℃, and the atmospheric pressure is 1atm .The frictional coefficient of fluid flowing through the chimney can be taken as 0.05, try to calculate the flow rate of the effluent gases (in kg/h). Solution：Suppose the bottom of the chimney is the upstream section 1——1’，and the inside cross section of the top is 2——2’, and take section 1——1’ as horizon， then set up Bernoulli equation between two sections. In this equation the sections of the chimney are the same，and the pressure of gas in the chimney does not change a lot） Because the pressure of gas in the chimney doesn’t change a lot, the temperature is 400℃, and the atmospheric pressure is 1atm. Because the pressure on the top inside of chimney is equal to the pressure of air in the same height, so we have The density of the air in the standard status is 1.293kg/m，so when the environmental temperature is 20℃, and the atmospheric pressure is 1atm，the density of the air upon that then input the data into the equation， therefore The velocity of the flue gas is The mass flow rate of the flue gas is 22. The liquid is sent to the elevated tank from the reactor by pump at 2104kg/h (as shown in the figure). The pressure keeps 20mmHg (vacuum) above the surface of the reactor, and the pressure above the surface of the elevated tank is atmosphere. The pipeline is steel pipe(Φ764mm), and its length is 50m.There are two wide open brake valves, a orifice meter (the resistance coefficient is 4), five standard elbows in the pipeline. The distance from the surface of the reactor to the exit of the pipeline is 1.5m. If the efficiency of the pump is 0.7, try to calculate the shaft power of the pump. It is given that the density of the liquid is 1073kg/m, the viscosity is 0.63cP, and the absolute roughness of pipe wallε is 0.3mm. Solution：Set up Bernoulli e