Dear in DAP (NH4)2HPO4 formula Nitrogen 2 molecule. 👉molecule of nitrogen contains two nitrogen atoms so 1 nitrogen molecule -14 2 nitrogen molecule -28 If any other query feels free to contact me.
Sir humko ye samjh nahi aa raha hai ki 5ppm dose karna hai chemical water humko itna lena hai kaise pata kr rahe hai hum ki stock 6l/hrs rakhna hai to hamara chemical 5 ppm chala jayega please reply sir
Here are a few types of calandria 👇👇 1. Vertical Calandria 2. Horizontal Calandria 3. Natural Circulation Calandria 4. Forced Circulation Calandria 5. External Calandria 6. Internal Calandria If you were referring to a different type of "calandria," feel free to contact me !
5-micron cartridge filter helps improve the efficiency, lifespan, and cost-effectiveness of the RO system by protecting the membrane and other components from larger particles. Reason behind 👇 Filters with a 5 micron rating remove large amounts of debris from liquids. While there are filters with other, smaller micron ratings, filters below 5 microns are prone to intense blockage or quick debris build-up. The smaller the micron rating, the faster the filter will clog.
1. TMP (Transmembrane Pressure): In an RO plant, TMP refers to the pressure difference across the membrane. It is the force that drives water through the RO membrane, leaving behind impurities. TMP is a key parameter in determining the efficiency and performance of the membrane. 2. DP (Differential Pressure): DP measures the difference in pressure between two points in the RO system, typically the inlet and the outlet. It helps monitor any blockage or fouling in the system. Both TMP and DP are important indicators for the operation and health of the RO system.
डिगैसर को SAC और SBA दोनों के बाद लगाने से प्लांट की दक्षता कम हो जाती है, पानी की गुणवत्ता गिर जाती है, और लागत बढ़ जाती है। सही तरीका है कि डिगैसर को SAC के बाद और SBA के पहले इंस्टॉल किया जाए ताकि प्रक्रिया को ऑप्टिमाइज़ किया जा सके। If a degasser is installed after both the SAC and SBA exchangers, several issues can arise: 1. Higher CO₂ load on SBA 2. Reduced efficiency of ion exchange 3. Increased operational costs 4. Decreased water quality
Maximum kitina %doshing rokhna chahiye har ro me alag alaga dosing pump ki set kiya jata hai 1 hrs me sabhi ro ka 50%ka dosing hota hai kiya sir or sabhi ro mein 1 hrs 5 ltr ka dosing set kiya jata hai
RO प्लांट में एंटीस्केलेंट की डोजिंग रेंज आमतौर पर 2 से 6 ppm के बीच होती है। हालांकि, यह कई कारकों पर निर्भर करती है, जैसे: 1. Feed water Quality: यदि पानी में घुले हुए लवण, कैल्शियम, मैग्नीशियम, सल्फेट या सिलिका की मात्रा अधिक होती है, तो उच्च मात्रा में एंटीस्केलेंट की आवश्यकता हो सकती है। 2. Water Recovery Rate: उच्च रिकवरी रेट के लिए अधिक मात्रा में एंटीस्केलेंट की जरूरत हो सकती है। 3. Antiscalant types : विभिन्न प्रकार के एंटीस्केलेंट के लिए अलग-अलग डोज़ अनुशंसित होते हैं, इसलिए हमेशा निर्माता के दिशा-निर्देशों का पालन करना चाहिए।
Sir Aeration me bacteria 🦠 nhi ban pa Raha hai daily uria Dap jigri dal rhe hai COD 9000ppm hai 1liter me 500g mlss sattling time 10 minutes tha kuchh din pahle or ab mlss 1liter me 10 rhegya hai out let ki COD bahut adhik aa rhi hai to sir kya karna chahiye
To determine the normality of a ferrous ammonium sulfate (FAS) solution, you need to know the following: 👉The molar mass of ferrous ammonium sulfate [Fe(NH4)2(SO4)2·6H2O] is approximately 392.14 g/mol. 👉The equivalent weight of FAS depends on the reaction it is involved in. In redox reactions where iron (Fe²⁺) is oxidized to Fe³⁺, FAS has an equivalent weight equal to its molar mass because 1 mole of Fe²⁺ loses 1 mole of electrons. ✅Steps to Calculate Normality: 1. Weigh the FAS: Accurately weigh a certain mass of ferrous ammonium sulfate. 2. Dissolve in Water: Dissolve the weighed FAS in distilled water and make up to a known volume (typically in liters or milliliters). 3. Determine Equivalent Weight: For ferrous ammonium sulfate in redox reactions, the equivalent weight is 392.14 g/equivalent. 4. Formula for Normality: Normality (N)= Mass of FAS (g) \Equivalent weight (g/equivalent) x Volume of solution (L) ✅Example Calculation: 👉Suppose you dissolve 19.6 g of FAS in 1 liter of solution. 👉The equivalent weight of FAS is 392.14 g/equivalent. Now, apply the formula: N = 19.6 / {392.14 x 1} = 0.05 N So, the normality of the solution is 0.05 N. ✅Key Points: 👉Use the molar mass of FAS (392.14 g/mol) for determining the equivalent weight. 👉The volume of the solution should be in liters to get the normality in equivalents per liter (N). 👉The normality is a measure of the concentration of reactive equivalents in a solution.
