1 / 12

Problem 1: The concentration of MLVSS is equal to X V —i.e. X V = MLVSS

1. A four-pass step-feed activated sludge system, shown in the figure below, has equal aeration tank volumes in each pass of 240m 3 , Using the design parameters given below, determine the MLVSS concentration in each tank. Problem 1:

dyllis
Download Presentation

Problem 1: The concentration of MLVSS is equal to X V —i.e. X V = MLVSS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1. A four-pass step-feed activated sludge system, shown in the figure below, has equal aeration tank volumes in each pass of 240m3, Using the design parameters given below, determine the MLVSS concentration in each tank.

  2. Problem 1: The concentration of MLVSS is equal to XV—i.e. XV= MLVSS From the mass equation , we can get : XR QR =X1 (QR +Q1)=X2 (QR + Q1 +Q2 )=…… • 1m3 =1000L。 For wastewater A, there are : Tank 1: • X1 = (QR +Q1 )/(XR QR )=(2000+800)/(10,000x2,000)=1.4x10-4 mg/d • MLVSS= X1 V= 1.4x10-4x240=0.0336 mg/d • Tank 2: X2= (QR + Q1 +Q2 )/ (XR QR )=(2,000+800+1,200)/(10,000x2,000) =2x10-4mg/d MLVSS= X2 V=2x10-4 x240=0.0480 mg/d Tank3: X3= (QR + Q1 +Q2 +Q3)/( XR QR ) =(2000+800+1,200+1,000)/ (10,000x2,000)=2.5x10-4 mg/d MLVSS= X3 V=0.0600mg/d Tank 4: X4= (QR + Q1 +Q2 +Q3+Q4)/( XR QR ) =(2,000+800+1,200+1,000+1,000)/ (10,000x2,000)=3x10-4 MLVSS= X4V=0.0720mg/d

  3. 2. 已知一个连续流反应器，体积为1000L，进水流量为100L/h。80％的出流通过微生物分离器后，分离出来的所有颗粒物质都回流到反应器。其余的出流直接从反应器中排出。求HRT和SRT。

  4. Problem2: • HRT:Hydraulic Retention Time水力停留时间 • SRT:生物固体平均停留时间 • HRT=V/Q=1000L/(100L/h)=10h • SRT=VX/△X=VX/(100X20%X)=50h

  5. 3. A 2-L settlemeter is used to perform an SVI test. The MLSS Concentration for the test is 3500mg/L and the settled sludge volume after 30 min is 840mL. what is the sludge volume index?

  6. Problem3: • Known Values: MLSS=3500mg/L • =3.5g/L • SV=804ml/2L=402mL/L • SVI=SV(mL/L)/MLSS(g/L)=402/3.5 • =114.85 • 由于SVI以介于70~100之间为宜，而现在的SVI为114.85，过高，说明沉降性能不是很好，还有可能产生膨胀现象。

  7. 4、某城镇污水量Q=30000m3/d，原污水经初次沉淀池处理后BOD5值Sa＝200mg/L，要求处理水 BOD5为Se＝20mg/L，去除率90％，求定鼓风曝气时的供气量。    有关参数为：混合液活性污泥( 挥发性 ) Xv＝2000mg/L；曝气池出口处溶解氧浓度C=2mg/L；计算水温25度，有关设计的各项系数：a'＝0.5；b'＝0.1；α＝0.8；β＝0.9；ρ＝1； (EA)＝18％。Cs(20)=9.17mg/L; Cs(25)=8.4mg/L。经计算曝气池有效容积V=10000m3，空气扩散装置安设在水下4米(H=4m)。

  8. （1求定需氧量 • 按公式 • R=Q2= a'Q(Sa-Se)+ b'XvV • 代入各值 • R=Q2=(30000*0.5(200-20))/1000+(2000*10000)/1000 • =4700KgQ2/d • （2）计算曝气池内平均溶解氧饱和度，按公式 • Csb=Cs（Pb/(2.026*105)+Qt/42） • 计算，为此，确定式中各参数值： • 求定空气扩散装置出口处的绝对压力，按公式： • Pb=1.013*105+9.8*4*103=1.405Pa

  9. 2) 求定气泡离开池表面时，氧的百分比Qt值，按公式： • Qt=21*（1-0.18）/(79+21*(1-0.18))*100%=17.9% • 由题得： • CS(20)=9.17mg/L; CS(25)=8.4mg/L • 代入各值，得： • Csb（25）=8.4*（1.405/2.026+17.9/42）=9.4mg/L • Csb（20）=9.17*（1.405/2.026+17.9/42）=10.27 mg/L • (3)计算20摄氏度时脱氧清水的需氧量，按公式，代入各值，则： • Ro=（4700*9.17）/(0.8*(0.9*1*9.4-2)*1.024^(25-20)) • =7407 KgQ2/d=308.6Kg/h • (4)计算供气量，按公式 • Gs=7407/（0.3*18）*100=177166.7m3/d=95.3 m3/min

  10. 5.污泥膨胀的原因有哪些？如何防治污泥膨胀？若已产生污泥膨胀，可采用什么方法补救？举具体实例说明。5.污泥膨胀的原因有哪些？如何防治污泥膨胀？若已产生污泥膨胀，可采用什么方法补救？举具体实例说明。

  11. 丝状菌膨胀原因： • ①pH、水温 • ②运行Ns、DO • ③生产装置和其他运行方面的问题。 • 非丝状菌膨胀原因： • 非丝状菌性膨胀污泥含有大量的表面附着水，细菌外面包有粘度极高的粘性物质。这种粘性物质是由葡萄糖、甘露糖、阿拉伯糖。鼠李糖。脱氧核糖等形成的多糖类。 • 原因： 水温低、Ns高

  12. 防止：加强操作管理、经常检测污水水质、曝气池内DO、污泥沉降比、SVI、镜检等。 • 控制措施： • ①控制曝气量； • ②调整pH值； • ③营养比例合适； • ④投加化学药剂；（解决一时问题） • 参考： • 《排水工程》（下册）第四版

More Related