1、Chapter 9 Utilizing Experience-based Principles to Confirm the Suitability of a Process Design.利用经验准则确认过程设计的适用性,王 卓,9.1 The role of experience in the design process,Checking new process designProviding equipment size and performance estimatesHelping to troubleshoot problems with operating systemsVer
2、ifying（核实） the reasonableness of results of computer calculations and Simulations Providing reasonable initial values for the input to a process simulator required to achieve program convergenceObtaining approximate costs for process unitsDeveloping preliminary process layouts.,Experienced chemical
3、engineers have formulated a number of experienced-based short-cut methods and guidelines that are used for:,1/19,（a）正三角形排列,（b）正方形排列,换热管的排列：,9.1.1 Introduction to Technical Heuristics and Shot-Cut Methods,排列紧凑，管外流体湍动程度高，给热系数大。,排列比较松散，给热效果较差，但管外清洗方便，对易结垢液体更为适用。,2/19,多组分精馏的FUG简捷算法：,用芬斯克（Fenske）公式估算最少理论
4、板数用恩特伍德（Underwood）公式估算最小回流比用吉利兰（Gilliland）图估算实际回流比下的理论板数用Kirkbride公式近似确定进料位置,3/19,化工常用金属材料特性：,4/19,9.1.2 Maximizing the benefits obtained from experience,An engineer must be capable of transferring knowledge gained from one or more experiences to resolve future problems successfully. Experienced eng
5、ineers retains a body of information, made up largely of heuristics and short-cut calculation methods, that is available to help solve new problemsThe process by which an engineer uses information and creates new heuristics consists of three steps: Predict、 Authenticate(验证)、Re-evaluate，they form the
6、 basis of the PAR Process.,5/19,Example 9.1估测一下温度为93，流速3.05m/s的水在直径为38mm管内流动的传热系数。根据以往经验，我们知道21流速1.83m/s的水在相同管内的传热系数为5250W/m2。 下面用PAR过程分析：Step 1predict:假设流速和温度没有影响，预测传热系数为5250W/m2Step 2Authenticate/Analyze:利用下表物性数据计算Reynold Number(雷诺数):,6/19,利用 Sieder-Tate 方程验证假设: 两个条件下的传热系数之比： 显然，最初的假设，流速和温度没有影响是不对
7、的。Step3Re-evaluate/Rethink: 对假设进行改进：1.温度对黏度的影响的影响必须考虑2.温度对Cp,和k的影响可以忽略3.管径对h的影响不大4.结果仅限于Sieder-Tate方程的适用范围内,7/19,根据上述假设，把21下的数值代入方程（1），即得到估测水在38mm管内流动传热系数的经验式：for u(m/s), (kg/ms)The heuristics or rules are contained in a number of tables(9.2-9.18) and apply to operating conditions that are most ofte
8、n encountered. The information provided used in the following example and should be used to check any information on any PFD.,9.2 Technical heuristics in the application of chemical equipment,8/19,Example 9.2参考第一章甲苯加氢工艺图1.3，表1.5和表1.7所给数据，估算下面设备尺寸和其它一些操作参数：V-101E-105C-101T-101下面分别对这几个设备进行估算：,V-101Hig
9、h-Pressure Phase Separator,Vapor flow =stream 8=9200kg/h, P=23.9bar, T=38, Liquid flow =stream 17+18=11570kg/h, P=2.8bar, T=38 v=8kg/m3 and l=8kg/m3 (表1.7),Question:怎样估算V-101的尺寸（直径和高度）？,9/19,We use the following heuristics:,Rule 1 Gas/liquid phase separators are usually vertical.Rule 2 Optimum lengt
10、h/diameter =3, but the range 2.5 to 5 is common.Rule 3 Hold time is 5 min for half-full gas/liquid separators,5-10 min for a product feeding another tower.Rule 4 Gas velocity in gas/liquid separators , u=k(l/v-1)0.5 m/sK=0.0305 for vessels without mesh entrainers.Rule5 Good performance can be expect
11、ed at velocities of 30-100% of those calculated with the given k; 75% is popular.,10/19,From Rule 4，u=0.0305(850/8-1)0.5=0.313m/sFrom Rule 5, uact=0.75u=0.75*0.313=0.23m/sMass flowrate of vapor=9200/3600=2.56kg/s =uvD2/4 D=1.33mFrom Rule 3,volume of liquid=0.5LD2/4=0.726L m3 5 minutes of liquid flow
12、=5*60*11570/850/3600=1.13m3 0.726L=1.13L=1.56mFrom Rule2, L/D should be in range 2.5 to 5. For our case L/D=1.56/1.33=1.17,is out of range.We should change to L=2.5D=3.3m,11/19,Conclusion: Heuristics above suggest that V-101 should be a vertical vessel with D=1.33m,L=3.3mFrom Table 1.7,we see that t
13、he actual V-101 is vertical vessel with D=1.1m,L=3.5mWe should conclude that the design of V-101 in Chapter 1 is consistent with the heuristics.,b. E-105 Product cooler,热物流：105 38 Q=1085MJ/h=301kw (表1.7),Question: 怎样估算换热器E-105的传热面积？,12/19,We use the following heuristics:,Q=KAFTm,Rule 1 For conservat
14、ive estimate set F=0.9 for shell and tube exchangers with no phase Changes.Rule2 Cooling water inlet is 30,maximum outlet 45。Rule3 Heat transfer coefficients for estimating purposes,W/m2：Water to liquid,850; condensers,850; liquid to liquid,280; liquid to gas,60; gas to gas,30; reboiler 1140.Rule4 D
15、ouble-pipe exchanger is competitive at duties requiring 9.3-18.6m2,From the Rule we can set:F=0.9，K=850W/m2Water enters at 30 and leaves at 40,13/19,Tm=(105-40)-(38-30)/ln(105-40)/(38-30)=27.2A=Q/K tm F=301000/850/27.2/0.9=14.46m2From Rule 4, this exchanger should be a double-pipe or multiple-pipe d
16、esign.Comparing the results:Heuristic:Double-pipe design,Area=14.46m2Table 1.7:Multiple-pipe design,Area=12m2The heuristic analysis is close to the actual design.,c. C-101,Question:怎样估算压缩机的功率？,From Table1.7, flow=6770kg/h, T1=38=311K,P1=23.9bar, P2=25.5bar,14/19,We use the following heuristics:,Rule
17、: Theoretical reversible adiabatic power (理论绝热可逆功率) Wrev,adiab=mz1RT1(P2/P1)a-1/a. where T1=inlet temperature, R=Gas Constant=8.314J/mol K, z1=compressibility(压缩系数), m=molar flow rate, a=(k-1)/k and k=Cp/Cv,m=molar flow rate(Sream5+Stream7) =(758.8+42.6)/3600=0.223kmol/sk=1.41（assume）and a=(k-1)/k=0
18、.2908Wrev,adiab=mz1RT1(P2/P1)a-1/a =0.223*1.0*8.314*311*(25.5/23.9)0.2908-1 =37.7kWUsing a compressor efficiency of 75% Wactual=37.7/0.75=50.3KWFrom Table 1.7 , W=49.1KW ,the results are very close.,15/19,d. T-101,xD=0.9962,xW=0.0308, D=2.44,W=2.13, F=142.2kmol/h,D=105.6kmol/h ，=（ DW ）0.5=2.28,Quest
19、ion:怎样估算塔径、回流比、板数、全塔压降？,We use the following heuristics:,Rule1 ：Optimum reflux in the range of 1.2-1.5RminRule2 : Optimum number of stages approximately 2NminRule3: Nmin=lnxD/(1-xD)/xw/(1-xw)/ln (Fenske方程)Rule4: Rmin=(F/D)/(-1), when feed is at the bubble point.Rule5: Use a safety factor of 10% on n
20、umber of trays,16/19,Rule 6: Lmax=53m (wind load and foundation considerations) and L/D30Rule 7: Peak efficiency of trays is at values of the vapor factor Fs=uv0.5=1.2 1.5m/s(kg/m3)0.5Rule 8: Pressure drop per tray Ptray=0.007barRule 9: Tray efficiencies for distillation of light hydrocarbon and aqu
21、eous solutions(水溶液) are 60-90%.,Use the Rules above ,we get:Nmin=ln0.9962/(1-0.9962)/0.0308/(1-0.0308)/ln2.28=10.9Rmin=F/D/(-1)=(142.2/105.6)/(2.28-1)=1.05Range of R=(1.2 1.5 )Rmin=1.26 1.58Ntheoretical2*10.9=21.8 tray=0.6Nactual (21.6/0.6)*1.1=40 traysv=6.1kg/m3u=(1.2 1.5)/6.10.5=0.49 0.60m/s,17/19
22、,Vapor flow rate (stream 13)=22700kg/h体积流率v=1.03m3/sDtower=4v/ u0.5=4*1.03/3.14/(0.490.60)0.5 =1.641.48mPtower =Nactual ptray=40*0.007=0.28barA comparision of the actual equipment design and the predictions of the heuristic methods are give below:,18/19,9.3 Summary:In this chapter, we have introduce
23、d a number of heuristics for typical chemical equipment. From the worked example above, it is clear that the sizing of the equipment in Table 1.7 agrees well with the predictions of the heuristics presented in this chapter. They are useful guides that allow the engineer to flag out possible errors and help focus attention on areas of the process that may require special attention.,19/19,Thank you for listening!,返回,
