OPC PPC typically refers to a type of composite material made from Ordinary Portland Cement (OPC) and Pulverized Fuel Ash (PPC). The specific percentage of each ingredient can vary depending on the formulation, but a common mix might consist of around 70-80% OPC and 20-30% PPC. For precise percentages, it's best to refer to specific product guidelines or standards set by manufacturers.
What???
polarity order of ehanol methanol n-hexane and distilled water
its used a fuel , but has less calorific value as the carbon monoxide percentage is less which is the main combustible gas in the b.f gas. its used in heating the blast of hot air in the blast furnance itself and also can be used in rolling mill, sinter plant, and many more where you need heating
In a thermal power plant, the carbon content in ash typically ranges from 1% to 10%, depending on the type of fuel used and the combustion efficiency. Fly ash, a byproduct of combustion, may have lower carbon content, while bottom ash can contain slightly higher levels. The specific percentage can vary widely based on operational conditions and the nature of the coal or biomass burned.
What You Should Know About Flash Tanksby J. E. TROCOLLI • Sarco Co., Inc. • Allentown, PA (Actual Specifying Engineer/71)High-pressure steam systemsrequire flash tanks -here is a guide to determining sizesCondensate temperatures in high-pressure steam systemsgenerally are only slightly less than the saturatedtemperature of the steam. When hot condensates are dischargedinto lower-pressure areas, condensate temperatureimmediately drops to the saturated temperature ofthe low-pressure area. As the result of the drop in temperature,heat released evaporates a portion of the condensate,generating flash steam.To return condensate to the boiler or to discharge it tothe sewer, it is necessary to separate flash steam fromthe condensate. This is accomplished by discharging condensatethrough steam traps into a vented tank, referredto as a flash tank.Flash steam produced in the flash tank may be ventedto the atmosphere or piped to a low-pressure main. Condensateremaining may then be returned to the boiler ordischarged to drain.Tanks must be large enough to ensure dryness of thereleased steam and to avoid carryover by the steam ofwater in droplet form. When using horizontal flash tanks,the required area is found by multiplying the diameter ofthe tank by its length. This measures the tank's capacityto handle condensate. Table 1 illustrates the required areafor each 1,000 lb. of condensate/hour with varying steamand flash pressures.How big a tank?Problem:An absorption machine condenses 12,000lb. of steam/hr. Assuming the flash tank is vented to theatmosphere, determine the size of the flash tank requiredwith a steam pressure of 12 PSIG.Solution:Enter table 1 at 12 PSIG steam pressure,moving horizontally to 0 PSIG. Find .75, which is the numberof square feet required for each 1,000 lb. of condensate.Since 12,000 lb. of steam are generated, it may bedetermined that by multiplying 12 x .75, 9 sq. ft. of surfacewill be required, or that the diameter of the tank infeet times its length in feet must equal 9. Thus, a tank 2ft. by 4.5 ft. may be used.Problem:A dryer operating at 100 PSIG condenses18,000 lb. of steam/hr. The flash tank is to discharge itsflash into a 5-PSIG heating main. Determine the size ofthe flash tank required.Solution:Again, using table 1, enter at 100 PSIGinitial pressure. Move horizontally to the 5-PSIG columnand find 1.92.Then:18 x 1.92 = 34.6 sq. ft.A flash tank 4 ft. by 9 ft. will be satisfactory for theapplication.Sizing vent linesIf flash steam is to be discharged to the atmosphere,a properly sized vent line must be provided. To determinethe proper size, first find the area of the flash tank,using the method described above.Problem:Determine the size of the vent line usingtable 2. If, as in the first problem, tank size is 2 ft. by 4.5ft. (9 sq. ft.), refer again to table 2, where 9 sq. ft. falls on7.4-to-12 line. For this range, a 2-in. vent line would besatisfactory.Problem:Determine the size of the vent line if theflash tank size is 4 ft. by 9 ft. or 36 sq. ft. In table 2, 36 isin the 27-to-36 line, and, in this case, a 31/2 in. vent wouldbe needed.Figure 1 shows a typical flash-tank piping diagramin which the flash is discharged to the atmosphere.If it is desired to utilize the flash steam by dischargingit into a low-pressure main, refer to figure 2. In thisinstance, it will be necessary to properly size the lineconnecting the flash tank to the low-pressure main.Using table 3, determine the percent of flash. Multiplythis percentage by the condensate load in lb./hr. todetermine the number of pounds of steam that are flashed.Determine what steam velocity will be acceptable in theline. If a low noise level is desired, a relatively low velocitymust be selected - 4,000 to 6,000 FPM.Where noise is not a factor, a velocity of 12,000 FPMor higher may be used. After velocity has been determined,the required pipe size can be found in table 4.Problem:10,000 lb./hr. of condensate is dischargedinto a flash tank from a 125-PSI steam system. Flashsteam is to be piped into a 10-PSIG low-pressure heatingmain. Determine the size pipe required for connectingthe flash tank to the steam main.Enter table 3 at 125-PSI initial pressure. Move horizontallyto the 10-PSIG column and find 12.2 percentflash. Then the amount of flash steam/hr. is found thisway:10,000 x 12.2 percent = 1,220 lb./hr.SECTION ENG - PAGE 0100FLASH TANKSTABLE 1 FLASH TANK IN SQ. FT. = DIAMETER x LENGTH OF HORIZONTALTANK FOR 1,000 LB. CONDENSATE PER HOUR BEING DISCHARGEDTABLE 2 VENT LINE SIZE TABLE 3 PERCENT FLASHFOR HORIZONTALFLASH TANKS½"FLASH TANKSTABLE 4 STEAM VELOCITY CHARTFORM FLASH TANKSRevised 4/04 SHIPCO® IS A REGISTERED TRADEMARK OF SHIPPENSBURG PUMP CO., INC.SHIPPENSBURG PUMP COMPANY, INC.,P.O. BOX 279, SHIPPENSBURG, PA 17257 • PHONE 717-532-7321 • FAX 717-532-7704 • WWW.SHIPCOPUMPS.COMPRINTED IN THE U.S.A. • BEIDEL PRINTING HOUSE, INC., 717-532-5063 PERMISSION TO REPRINT BY SPIRAX SARCO INC.Since low noise level is important, a velocity in the4,000 to 6,000 FPM range must be selected. Enter table4 at 1,220 lb./hr., moving horizontally to a flash-steampressure of 10 PSIG. Then move up to 4,000 to 6,000FPM velocity. Here, the chart shows that a 3-in. pipe willhandle about 6,000 FPM, or a 4-in. line would handleabout 3,500 FPM.When vent lines cannot be extended to dischargeoutside the buildings, it is important that the condensatebe cooled below the dewpoint to prevent the exhaustfrom condensing and wetting walls, machinery, floors andso on.Estimating temperatureSince the dewpoint depends on several factors, includingrelative humidity and temperature (which are variable),the temperature to which the condensate must becooled should be estimated for individual cases.The following is the recommended procedure:Calculate the size of the flash tank in the methoddescribed above:Estimate the dewpoint, assuming unfavorable conditions,and, when making the estimate, take ventilation intoaccount - it is a factor in determining relative humidity:Once the dewpoint is known, estimate the quantityof cooling water that will be required and finally:Determine the pipe size and size of temperatureregulator valve required. A self-contained regulator witha normally closed valve to open when the temperaturerises is recommended. All such controls have an amplerange over and under the calibration point so that settingmay be adjusted after installation.Problem:Calculate the quantity of cooling water requiredto cool condensate in a flash tank vented to anenclosed space, assuming the following data:Steam pressure is 100 PSIG;Condensate is entering the flash tank at 1,550 lb./hr.at 335°F;Ambient temperature of the space into which ventdischarges is 75°F; andCold water temperature is 50°F.Assuming that ventilation at the above temperaturewill be sufficient to have not more than 70 percent relativehumidity, the dewpoint will be 64.5°F (determinedfrom psychrometric chart).To allow 1°F for safety, condensate should be cooledfrom 338°F to 63.5°F. When the installation is completed,further adjustment can be made by resetting the regulator.The heat to be extracted from the condensate is equalto:1,500 lb./hr. x (338-63.5)°F = 410,000 BTUHThe quantity of cooling water required:410,000 BTUH(63.5 -50)°F= 30,500 lb./hr.or 3,670 GPH = 61 GPMA 1.5-in. pipe to supply the water and a 1.25-in. temperatureregulator are recommended for this application.Flash tanks separate flash steam from the condensateby venting the flash steam to the atmosphere orpiping it to a low-pressure main, while returning the remainingcondensate to the boiler or discharging it to thedrain. If flash steam is discharged to the atmosphere, aflash tank and a properly sized vent line must be determined;if discharged to low-pressure mains it is necessaryto calculate the correctly sized line connecting theflash tank to the low-pressure main. Also, the proper temperaturefor cooling the condensate must be determinedfor projects in which vent lines cannot be extended todischarge outside the buildings.Different situations require individual solutions to determinethe correctly sized flash tank, connecting pipeand cooling temperature needed, but the calculation examplesoffered here provide the means to determine thenecessary installations and accessories required.Figure 1A typical flash tank pipingdiagram discharging to atmosphere.NOTE: Omit trap if condensate isdischarged into vented pump receiver.Figure 2A typical flash tank pipingdiagram with flash discharging to lowpressuresteam system.Figure 3This diagram depicts acombination flash tank installation withsubcooling condensate.VENTPRESSURERELIEF VALVEVENT IFDESIREDTO LOWPRESSURE MAINHIGH PRESSUREFROMCONDENSATEHIGH PRESSURESYSTEMFLASH TANKSTRAINERFLOAT ANDTHERMOSTATICTRAPFIN TUBECOOLING LEGREQUIREDTO LOW PRESSURERETURN LINE ORDRAINVENT TOATMOSPHEREHIGH PRESSUREFROMCONDENSATEHIGH PRESSURESYSTEMFLASH TANKSTRAINERFLOAT ANDTHERMOSTATICTRAPFIN TUBECOOLING LEGREQUIREDTO DRAINCONDENSATEFROM HIGHPRESSURESTEAM SYSTEMTEMPERATUREREGULATORSTRAINERCOLD WATERSUPPLYVENT TOT ATMOSPHEREWATER LEVELCOILDRAINTO SEWERFLASH TANKS
OPC is ordinary Portland cement and PPC is Portland pozolona cement. OPC is obtained by grinding clinker and gypsum whereas PPC is obtained by grinding clinker gypsum and high reactive silica. High reactive silica improves the quality of cement by reducing liberated lime.
Well, both types of cements have got merits of its own. OPC is a pure type cement releasing heat energy at a faster rate & the chances of shrinkage cracks are higher. PPC is a low heat cement in comparison to OPC requiring lesser degree of immediate post concrete setting curing. OPC is not suitable for hot weather conditions. PPC is suitable for adverse weather conditions. OPC cannot be used for thick & massive elements. PPC is to some extent suitable for thicker elements. Temperature controls are better conducted on PPC.
Ordinary Portland Cement, also known as OPC, is a type I cement that does not only harden, but also becomes water-resistant once it cures. Pozzolanic Portland Cement, or PPC, is a type IV cement that includes a mixture of a pozzuolanic material, which can increase the strength of the concrete and reduce the amount of OPC used.
there are 3type of cement1)opc (ordnary portland cement) 2)PPC (pozolan portland cement) 3) special cement
there are 3type of cement1)opc (ordnary portland cement) 2)PPC (pozolan portland cement) 3) special cement
uhm, and EGG!
the percentage varies from 15% to 35%.
opc-dart
OPC DA will access the real time value from control system. OPC HDA will access the real time value of historian processed. Hence the historian will save the value in their database as well as it feed to OPC server also.
Ordinary Portland Cement sets more quickly, and attains 85% of its maximum strength within 7 days. OPC is most valuable in large construction projects Portland Pozzolana Cement (OPC mixed with fly ash) takes longer to get to full strength, but it is more eco friendly, because it neutralizes otherwise toxic fly ash, and it is more resistant to corrosive environments. PPC can be used in most situations that OPC can.
Portland Pozzolana cement (PPC) is manufactured by a combination of pozzolanic materials. Pozzolana is an artificial or natural material that has silica in it in a reactive form. Along with pozzolanic materials in specific proportions, PPC also contains OPC clinker and gypsum. These pozzolanic materials include volcanic ash, calcined clay or silica fumes, and fly ash which makes around 15% to 35% of cement weight. To know more visit Banshiwala Cement- Leading Cement Company in Rajasthan India
1.44 ton/m3