Special Topic 1 of the Registered Chemical Engineer Professional Examination: Fluid Flow and Fluid Conveyance Machinery
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This post was last edited by zhanghp30 on 2018-10-19 at 23:15. 1. Professional knowledge: (1) Knowledge related to hydrostatic pressure and isobaric surfaces: Refer to Chapter 1 of \"Principles of Chemical Engineering\" (Volume 1) published by Tianjin University, as well as relevant sections from the Petroleum and Chemical Engineering Survey and Design Association materials. Multiple-choice questions on this topic are common, and mistakes in distinguishing between related concepts can occur, so pay attention! (2) Calculation of fluid density: It’s easy to cover this topic; the formula is p=PM/RT, or it can be converted using standard conditions: p=M*T0*P/22.4T*P0. This aspect is often tested ； (3) Concept of total potential energy/mechanical energy: The sum of potential energy and pressure energy is the total potential energy ； The sum of kinetic energy and total potential energy is mechanical energy ； (4) Pitot tubes, orifice plate flowmeters, rotameters: Pay attention to their respective characteristics; multiple-choice questions often appear ； (5) Graph of friction coefficient versus Reynolds number and relative roughness: Principles of Chemical Engineering, Tianjin University (Volume 1, pages 53–54); pay attention to the characteristics of each region, as multiple-choice questions are often asked on this topic ； (6) Friction loss: includes both straight-line pipe friction and local friction. The frictional loss in a straight pipe is the resistance generated when a fluid flows through it; it is caused by the viscous forces within the pipe, resulting in energy loss that is manifested as a decrease in the total potential energy ； Local resistance is the resistance caused when a fluid flows through local areas such as pipe fittings, valves, and sudden expansions or contractions in the pipe cross-section. The frictional loss in straight pipes is distributed evenly along the length of the pipe, while the local frictional loss is concentrated at the pipe fittings. Multiple-choice questions appear frequently, so pay attention! (7) Sudden increase/decrease in drag coefficient: In the case of a sudden increase, use 1.0 ； Set the reduction factor to 0.5. Please note that this aspect is frequently encountered in hydraulic calculations; the questions often do not mention it, and it is a common source of errors over the years. Since there are many calculation problems in exams, ignoring this detail can lead to a series of mistakes, so it must be given top priority! (8) Calculation of equivalent diameter: d = 4R, where R = A/PAI; the equivalent diameter is 4 times the hydraulic radius, and the hydraulic radius is equal to the flow cross-section divided by the wetted perimeter. Note the methods for calculating the hydraulic radius for different shapes (square, circular, etc.) ； (9) Calculation of series and parallel pipelines: includes the calculation of flow rate and resistance, with special attention paid to the method for calculating resistance in parallel pipelines, which is often tested ； (10) The characteristic curves of centrifugal pumps, the law of proportionality, and the cut-off law: these must be mastered! (11) Calculation of the installation height for centrifugal pumps: (a) Calculated from the allowable suction vacuum: Hg = Hs’ – u²/2g – Zf, where Hs’ = Pa – P1/pg. Typically, the value of Hs at 20°C is given in the problem; to find Hs’, one simply needs to write out the formulas for Hs and Hs’ based on the given information, and then subtract the two equations to obtain Hs’. This is a topic that is often tested! (b) Calculated from the net positive suction head: Hg = (P0 – Pv) / pg – NPSHa – Zf. Note that NPSHa represents the effective net positive suction head, which is determined as required by the hydraulic calculations of the process system! Generally, it is 1.2–1.3 NPSHr; NPSHr is the minimum required net positive suction head value specified by the manufacturer! Generally, if the conditions given in the problem indicate a saturated or boiling state, then P0 = Pv, and thus Hg = -NPSHa – Zf. In cases of backflow, the actual installation height must be reduced slightly (by 0.5 m); for suction-type setups as well, it should also be reduced by 0.5 m. Pay attention to the conditions provided in the problem and adjust accordingly! This topic has been a key focus in past exams! (12) Calculation of centrifugal pump power: Pe = P * H (efficiency); Pe = p * g * He * Qv; Pe = QH * p / 102 * H ； He was obtained using the white effort equation ； Pay attention to the concept of the pressure head ； (13) Effects of density and viscosity on flow rate, head, shaft power, and efficiency: The head, flow rate, and efficiency of centrifugal pumps are independent of density, whereas shaft power changes with density (Pe=QHp/102H); as the viscosity of the fluid being transported increases, the flow rate, head, and efficiency decrease, while the shaft power increases. This point must be memorized, as multiple-choice questions will be set on it! (14) Characteristic equations of pipelines and pumps/Parallel and series connection of centrifugal pumps: When two centrifugal pumps of the same model are connected in series, the flow rate is greater than that of a single pump, but the head is less than twice that of a single pump ； The parallel head pressure increases, but the flow rate is less than twice that of a single unit. Generally, in systems with low resistance, parallel connection is preferable to series connection, while in systems with high resistance, series connection is better than parallel connection. Note: This is a multiple-choice question ； The intersection point of the pipeline curve and the pump curve represents the operating point of the centrifugal pump; at this point, Q=Qe and H=He. In parallel operation, He=H and Qe=Q/2; in series operation, Qe=Q and He=2H. This topic is frequently tested! Must master it! (15) Reciprocating pumps (positive displacement pumps): The head is limited by the carrying capacity of the piping, and has nothing to do with the pump’s geometric dimensions. The discharge capacity is related to the pump’s geometric dimensions and piston displacement, and is independent of the head pressure and pipeline conditions ； Adjustment method: Bypass adjustment ； Change the piston stroke and number of reciprocations. Pay attention to the applicable scenarios of different types of positive displacement pumps (such as gear pumps, screw pumps, etc.) and the methods for calculating head pressure; in particular, note that screw pumps are not suitable for the Whitehead equation! (16) Centrifugal fan: N = Ht * Q / 1000H, where Ht is the wind pressure; Ht = (p2 – p1) + p * u2 / 2. Note that when selecting a fan, the values are usually converted to those at 20°C, with Ht = Ht’ * 1.2 / p’. Please be aware that Ht’ and p’ represent the actual wind pressure and density under real conditions. [Exam Prediction]1. Water flows steadily in a circular pipe under steady-state conditions. If the mass flow rate of water remains constant, compared to winter, what will happen to the Re value in summer?
A. It will increase.
B. It will decrease.
C. It will remain unchanged.
D. It cannot be determined.
2. During an experiment to determine the characteristic curve of a centrifugal pump, Xiao Wang noticed that no water came out from the discharge pipe after the pump was started. The vacuum gauge at the pump inlet indicated a very high level of vacuum. Unsure of the cause, he asked Li, a top-performing student in his group, for help. Li identified the problem and resolved it. As someone who is about to become a licensed chemical engineer, can you figure out what caused this issue? A. High water temperature
B. Suction pipe is blocked
C. Vacuum gauge is damaged
D. Discharge pipe is blocked
3. Regarding the series and parallel connections of centrifugal pumps, which of the following statements is incorrect?
A. When two identical pumps are connected in parallel, the flow rate in the pipeline increases; correspondingly, the fluid resistance also increases.
B. When two identical pumps are connected in series, each pump operates under conditions of higher flow rates and lower head.
C. When there are significant variations in the liquid level at the suction point, two or more pumps can be connected in series based on the required head.
D. The greater the number of pumps connected in parallel, the greater the increase in flow rate.
4. To prevent cavitation in centrifugal pumps, which of the following measures is incorrect?
A. Place the pump as close to the liquid source as possible.
B. Make the diameter of the suction pipe slightly larger; note that the suction pipe diameter should be smaller than that of the discharge pipe.
C. Minimize the number of bends in the suction pipe and eliminate any unnecessary fittings.
D. Install a flow control valve on the suction pipe.
5. In a piping system using a centrifugal pump, when only one pump is operating, the flow rate is qV and the head is He. Now, if another pump of the same model is added to the same system, with all other conditions remaining unchanged: when the pumps are connected in series, the flow rate becomes qv and the head becomes He'; when they are connected in parallel, the flow rate becomes qv'' and the head becomes He''. Then:
A. qv' = qv, He' = He; qv'' = 2qv, He'' = He
B. qv' > qv, He' > 2He; qv'' > 2qv, He'' > He
C. qv' = 2qv, He' = He; qv'' = qv, He'' = 2He
D. It depends on the specific conditions of the piping system.
The above represents some professional knowledge I’ve summarized personally. I hope everyone can review these key points—they may prove useful to you.
2. Case knowledge: (1) There are two parallel pipelines, ADB and ACB. The length of pipeline ADB is 50 m, while the length of pipeline ACB is 5 m (including pipe fittings but excluding the equivalent length of valves). The diameter of both pipelines is DN80. The resistance coefficient of the gate valves is 0.17, and that of the straight sections is 0.03. Each pipeline is equipped with one gate valve and one heat exchanger; the local resistance coefficient for each heat exchanger is 3. Determine the flow rate ratio between the two pipelines when both valves are fully open. 【Analysis】 Let 1 represent ACB and 2 represent ADB. We examine the calculation of resistance in parallel pipelines. Since the resistances of the branches are equal, we have hf1 = hf2. Also, u1²/u2² = (0.03×50/0.08 + 0.17 + 3) / (0.03×5/0.08 + 0.17 + 3) = 4.34. The ratio of flow rates is q1/q2 = 0.785d1²u1 / 0.785d2²u2 = u1/u2 = 4.34^(1/2) = 2.08. It’s important to utilize the characteristics of parallel pipelines to efficiently calculate flow rates and velocities; pay attention to these techniques! (2) Water is pumped to the high-level tank using a centrifugal pump; the height difference is 12 m, and the vapor pressure in the high-level tank is 118 KPaG. At a specific rotational speed, the pump’s performance curve is given by H=42-7.56*10^4*Q^2 (with units of m3/s). When the water flow rate is 0.01 m3/s, the flow enters the fully turbulent regime. Now, a solution with a density of 1200 kg/m3 is to be transported, with all other conditions remaining unchanged. Determine the flow rate and effective power of the centrifugal pump for transporting this solution. 【Analysis】 Applying the Bernoulli equation to the suction liquid level and the liquid level in the surge tank: He = 12 + 118*10^3 / (1000*9.81) + BQ^2 = 24 + BQ^2. Given that H = 42 – 7.56*10^4Q^2, Q = Qe = 0.01, and B = 1.044*10^5, the pipe characteristic equation becomes: He = 24 + 1.044*10^5Qe^2. For transporting the solution: He = 12 + 118*10^3 / (1200*9.81) + 1.044*10^5Qe^2 = 22 + 1.044*10^5Qe^2. By combining this equation with the pump’s characteristic equation, we obtain Q = 0.01054 m³/s, H = 33.6 m, and Ne = QHp/102 = 4.17 kW
(3) Technical renovation of a domestic ten-million-ton refinery. The current installation height of the centrifugal pump is 3.5 m. The liquid at the bottom of the tower needs to be pumped out using this centrifugal pump; the vacuum level in the tower bottom is 67 KPa, meaning that the tower is in a saturated state at this time. The pump is located on the ground surface. The total head loss in the inlet pipeline is 1.2 m, and NPSHr = 3.7. Therefore, the correct answer is: A The base of the tower should be raised by 1.9 m; B The base of the tower should be lowered by 1.9 m ； The base of Tower C is increased by 1.4m ； The design of the high pump is reasonable. 【Explanation】 A: Hg = (P0 - Pv) / (ρg) + NPSHa + Zf. Here, P0 = Pv (in a saturated state). Normally, NPSHa should be at least 0.5 m greater than NPSHr. Therefore, Hg = (3.7 + 0.5) + 1.3 = 5.4 m. This indicates that the installation height has increased; the increase in height is 5.4 - 3.5 = 1.9 m. Please note that it’s easy to calculate this value as 1.4 m by using NPSHr in the calculation; however, this is incorrect! Pay special attention to this part! (4) A fine chemical company uses DN100 steel pipes to transport 19 tons of oil per hour. The oil has a specific gravity of 0.9 and a viscosity of 72 cP. The total length of the pipes is 160 km, and the total pressure in the pipes is 60 barg. The pipes are installed horizontally, with local resistance being negligible. Determine quantitatively whether it is necessary to add more pressure stations during transportation, and if so, how many such stations should be added A does not need B needs it ； Add 3; C is needed, add 4 ； It is necessary to add 5 more units. 【Explanation】 Cu = w/Ap = 19000/0.785*0.01/900/3600 = 0.747 m/s. Re = dpu/miu = 0.1*0.747*900/72/1000 = 934; thus, the flow is laminar. The friction coefficient s = 64/Re = 64/934 = 0.06852. The pressure drop deltaP = p*Hf = 900*0.06852*160000/0.1*0.747^2/2 = 275.29 bar, which is greater than the allowable value. Therefore, additional pressure stations are needed. The number of additional stations required, N, is calculated as DeltaP/60 = 275.29/60 = 4.6; rounding up gives 5. Since one pump station has already been installed for transportation, 4 more pressure stations need to be added along the way.
(5) Water at 20 ℃ is pumped out of the storage tank using a pump, with vacuum and pressure gauges installed before and after the pump. It is known that the sum of the total resistance and velocity head in the pump’s suction pipeline is 2 mH2O; the permissible suction lift is 5 m; the atmospheric pressure is 1.013×10^5 Pa; and the level of the liquid in the storage tank is 2 m below the pump’s suction point. When the water temperature reaches 60 °C, the readings of the vacuum gauge and pressure gauge will ( ). It is known that the saturated vapor pressures of water at 20 ℃ and 60 ℃ are 2335 Pa and 19923 Pa, respectively. A. The vacuum gauge rises, and the pressure gauge falls. B. The vacuum gauge rises, and the pressure gauge rises. C. The vacuum gauge falls, and the pressure gauge falls. D. The vacuum gauge falls, and the pressure gauge rises. 【Analysis】C When the water temperature is 60 °C, Ps = 19923 kPa and p = 983.2 kg/m3; at 20 °C, Ps = 2335 Pa and ρ = 998.2 kg/m3. Then, the allowable vacuum level at 60 °C is given by: Hs’ = Hs + (P0 – Ps’) / p’g – (P0 – Ps) / pg = 5 + (101325 – 19923) / 983.2 / 9.81 – (101325 – 2335) / 998.2 / 9.81 = 3.33 m. The allowable installation height for the pump is: Hg = Hs’ – u1² / 2g – Zf = 3.33 – 2 = 1.33 m. The actual installation height, however, is 2 m, which is too high. Cavitation occurs inside the pump, causing both the pressure gauge and the vacuum gauge readings to drop suddenly. A, B, D are incorrect answers. (6) A natural gas compression station transports natural gas at 25°C through steel pipes with an inner diameter of 500 mm to a gas storage tank located 22 km away. The gas delivery rate is 5×104 Nm3/h. The gauge pressure of the gas storage tank is 165 kPa. When the height difference between the two ends of the delivery pipeline is small and the flow can be considered isothermal, the relationship between the volumetric flow rate of natural gas and its pressure can be expressed using the Weymouth equation: Vh = 2.538*10^(-5)*d^2.667*((p1^2 - p2^2/rL)*(273/T))^1/2. Here, Vh represents the volumetric flow rate of the gas, in Nm3/h (standard m3/h) ； d is the inner diameter of the pipe, mm ； l is the length of the pipe, m ； p1 and p2 are the pressure at the start and end of the pipeline, in kPa ； r is the relative density of the gas, a dimensionless value; it represents the ratio of the gas density ρ at 0°C and 101.3 kPa to the density of air, which is 1.294 kg/m3. In this case, its value is 0.552. Then the initial pressure p1 of the natural gas at the outlet of the compression station should be ( ) kPa. A. 483.5 B. 454.5 C. 14370 D. 542.6 [Explanation] A: 5*10^4 = 2.538*10^(-5)*500^2.667*((p1^2 - 165^2)/(0.552*22))*(273/298)^(1/2). Solving this equation gives p1 = 483.5 kPa
This post was last edited by zhanghp30 on 2015-10-13 at 15:29 (7 times). The liquid phase at the bottom of Tower A is transported to Tower B via the pressure difference between the two towers; the flow rate is 12 m3/h, the fluid density is 616 kg/m3, the viscosity is 0.15 mPa•s. The total length of the pipeline is 55 m. The pressure at the starting point of the pipeline is 350 kPa at a height of 2 m, while the pressure at the ending point is 150 kPa at a height of 18 m. The sum of the local resistance coefficients along the pipeline is 10.07. Use a pipe diameter of 0.05 m with an absolute roughness of 5×104. Then the pressure drop of the piping system (excluding the static pressure drop) is ( ). A. 72.61 kPa B. 55.37 kPa C. 60.56 kPa D. 64.57 kPa Hot water (70 ℃) from a tank is pumped into an open container. It is known that the inner diameter of the water delivery pipeline is 50 mm, the flow velocity of the water is 1.5 m/s, and the flow resistances in the pump’s suction and discharge pipelines are 9.81 kPa and 39 kPa respectively. The atmospheric pressure in this area is 90.66 kPa, and the outlet of the water pipe is 5 m above the liquid level of the hot water tank. Given that the permissible suction lift HS of the pump is 7.6 m (for clear water at 101.33 kPa and 20 °C), then the permissible installation height HS of the pump is ( ). A. 2.58 m B. 3.69 m C. 2.69 m D. 3.58 m【Solution】A
(9) When the compression ratio of a reciprocating compressor remains constant, an increase in the clearance coefficient will result in an increase in the compressor’s suction volume. A. Increase B. Decrease C. Remain unchanged D. Uncertain (10) Clean air at 30°C and 101.3 kPa is pumped by a centrifugal fan at a flow rate of 26,000 m3/h; it is heated to 90°C using an air heater before being sent to the dryer. Under average operating conditions (60°C, 101.3 kPa, ρ=1.06 kg/m3), the total wind pressure required for the conveying system is HT=2600 Pa. The flow rate and total pressure on which the fan is selected are ( ). A. 26,000 m3/h and 2,600 Pa B. 26,000 m3/h and 2,943 Pa C. 23,430 m3/h and 2,943 Pa D. 28,570 m3/h and 2,600 Pa 【Explanation】B: The flow rate is based on the actual inlet air volume, while the total wind pressure is calculated using the experimental conditions; that is, Ht = 2,600 * 1.2 / 1.06 = 2,943 KPa. Therefore, the parameters for selecting the fan are: Q = 26,000 m3/h, HT = 2,943 Pa. (11) To control the flow rate of the absorbent entering the absorption tower using a control valve and ensure efficient and stable operation, it is advisable to choose ( ). A. Closed-centrifugal pump B. Reciprocating pump C. Vortex pump D. Open-centrifugal pump [Explanation] A. Reciprocating pumps cannot be adjusted; vortex pumps are not suitable; open and semi-open types are appropriate for applications with particles ； Leading to another question: if feeding in proportion is required, then choose a metering pump! (12) A liquid is pumped from an open storage tank to a distillation tower. The vertical distance between the pipe inlet to the tower and the liquid level in the storage tank is 12 m ； The pressure loss of the liquid as it flows through the heat exchanger is 0.3 kgf/cm2 (29.4 kPa). The pressure in the distillation tower (gauge pressure) is 1 kgf/cm2 (98.1 kPa). The discharge pipeline is made of steel pipe with dimensions of ?114×4 mm, and its length is 120 m (including the equivalent length due to local resistances). The flow velocity of the liquid is 1.5 m/s, and its relative density is 0.96. All other physical properties are very similar to those of water; the friction coefficient λ = 0.03, the pressure loss in the pump suction pipeline is 1 m of liquid column, and the diameter of the suction pipe is. Try to determine, through calculation, which of the following centrifugal pumps is the most suitable: A. n=22 m3/h, H=1.6 m, n=2900 r/min, η=66%, HS=6.0 m; B. n=50 m3/h, H=37.5 m, n=2900 r/min, η=64%, HS=6.4 m; C. n=90 m3/h, H=91 m, n=2900 r/min, η=68%, HS=6.2 m; D. n=87 m3/h, H=20 m, n=2900 r/min, η=65%, HS=6.2 m. 【Analysis】 For option B, the flow rate Q=0.785d2u=47.6 m3/h. The pressure head required in the pipeline is given by He=(Z2-Z1)+DetaP/pg+deta u^2/2g+Hf. The resistance losses in the pipeline system include those in the suction line, discharge line, and heat exchanger. Hf=Hf,in+Hf,out+Hf,heat exchanger=8.02m. Therefore, he＝12+10.4+8.02+0.11＝30.53 m. Based on these results, it is more appropriate to use centrifugal pump 2#. (13) A certain centrifugal pump is used to transfer the liquid in a storage tank to a high-level tank; now that the liquid level in the storage tank has risen, and assuming the characteristics of the other pipelines remain unchanged, the flow rate will: A Increase B Decrease C Remain unchanged D Uncertain. [Explanation] A Using graphical methods, the pipeline characteristic equation is: He=(Z2-Z1)+(p2-P1)/pg+lanbuda*(L/D+S)*u^2/2g=(Z2-Z1)+(p2-P1)/pg+K*q^2. The characteristic curve of the pump is: Hg=A-B*q^2. Since this value remains unchanged, but (Z2-Z1) has decreased, the pipeline curve shifts downward, resulting in an increase in flow rate. One must learn to use curves to qualitatively determine the relationship between pipeline characteristics and centrifugal pumps.
This post was last edited by zhanghp30 on 2018-10-19 at 23:18. (14) Among the forms of energy involved in fluid flow, which ones are mechanical energy? A Potential energy B Kinetic energy C Work D Heat [Explanation] ABC. Here, option C is often overlooked; work is a form of energy conversion, as described in Bernoulli’s equation. (15) In flowmeters designed based on fluid mechanics principles, the one used to measure the velocity distribution across the cross-section of large-diameter gas pipelines is: A. Orifice flowmeter B. Venturi flowmeter C. Anemometer D. Rotameter 【Explanation】 C. Pay attention to the measurement principles of various components; multiple-choice questions involving these topics are common! (16) The conservation laws relevant to fluid mechanics include the following: A. Mass conservation B. Energy conservation C. Momentum conservation D. Heat conservation. [Explanation] Options ABC are correct; momentum conservation is often overlooked. Heat conservation is related to heat balance calculations. For additional questions on this topic, such as the significance of mass balance and energy balance, please refer to the examination guidelines provided by Tianjin University and those issued by the Survey and Design Association. Multiple-choice questions may be asked. (17) In the experiment to determine the characteristic curve of a centrifugal pump, the pressure reading at the outlet is P2 = 0.15 MPa, while the vacuum reading at the inlet is P1 = 0.082 MPa. Assuming that the diameters of the suction pipe and the discharge pipe are equal, and that the fluid used in the experiment is water, what is the head of the pump? 【Analysis】The energy equation is applied at the inlet and outlet of the pump: Hg = (Z2 – Z1) + (P2 – P1)/pg + (u2² – u1²)/2g + Hf. Since u1 = u2, and the distance between the inlet and outlet pipes is short, Z2 – Z1 = 0; therefore, Hf can be ignored. Thus, Hg = (P2 – P1)/pg = (P2 – Pa)/pg + (Pa – P1)/pg = 0.15*10^6/1000/9.81 + 0.082*10^6/1000/9.81 = 23.65 m. Special attention should be paid to the concepts of gauge pressure and vacuum pressure, as these are often tested!
This post was last edited by zhanghp30 on 2020-1-16 at 10:39. [Summary] The key points of this chapter are as follows: 1. Comprehensive application of the hydrostatic equation and the Blasius equation; pay attention to the calculation of friction forces, as well as the values applicable when the local friction coefficient increases or decreases suddenly! The effective power and useful work required by the transmission system are obtained from the formula. Pay attention to the selection of isobaric surfaces, etc ； ‘ 2. Calculation of fluid transfer machinery (pumps, compressors, fans): Pay attention to the characteristics of different types of machinery and the methods for calculating power, with special emphasis on the method for calculating the power of screw pumps! 3. Calculation of resistance in series and parallel pipelines as well as branch pipelines, application of the Fanning formula! 4. Calculation of the installation height for centrifugal pumps, calculation of the net positive suction head, calculation of the allowable suction vacuum level, characteristic curves and the three laws ； 5. Characteristics of pitot tubes, orifice plates, and rotameters ； Combining the above content with the summary tables in the guidelines for the course \"Principles of Chemical Engineering\" at Tianjin University will yield excellent results!
Very good, thank you for the guidance from the experienced expert!
Thank you, I’ll review it one more time before the exam