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The Level 1 course is divided into two sections. Absence from any portion of the course excludes the instructor candidate from certification: An Online, day, guided section.
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You can use the total balance as a substitute. PET 0. Steps g H2O 1. Na2B4O7 0. Na2B 1 Steps 1, 2, 3, 4: F1. FeC13 6H2O kg F2. FeC13 H2O. FeC13 2. H2O FeC13 is the simplest to use. Step 5: Take as a basis g of Ba NO3 2. BaNO3 g 0. Not all will be indept balances. Mixer VM 0. Mass 1. Note that ash is a tie element to P, so that the ash balance gives 0. Choose the most accurate to use. Steps 1, 2, 3, and 4: All the known data have been placed in the figure. Solution assumes no S is in stream A and no N2 in stream B.
Steps 1, 2, 3 and 4: mass fr. NH 3 mass fr. Basis: 5. In the actual reaction the corresponding ratio is 1. The Sb2S3 required to react with the limiting reactant is 4.
We can compute from the 1. Are they independent? Might be caused by round off. Check via H: 4 The conclusion is that probably no error exists in the measurements. Additional information was the molecular weights of the compounds and the chemical reaction equations. Basis: 1 g sample d.
Species balances g. It appears to be a carbon balance h. Step 5: Basis: 1 hr 6. Ore P kg mol SO2 0. This is a steady state open process with reaction. For component i, Equation Steps 1, 2, 3, and 4 Figure P A total balance around the feed mixer. Therefore, it would be wise to condense the HCl and H2O by cooling. Perhaps the acid solution might be sold to help pay for the process.
Although the problem seems to be underspecified, when you draw a diagram of the process and place the known data on it, the situation becomes clearer. Assume the process is a steady state open one with reaction. If you use the extent of reaction, you make species balances. If the equations are independent, we can find a unique solution. After selection of a basis, the air flow is known. Similarly, the other values of the moles in P would not change.
The values of the pertinent components for NOx use NO1. The sum of the components in moles would equal the total flow, hence not all of the equations that could be written would be independent, only 2 per subsystem. Multiply Fi by the composition xij. Salt balance on the freezer: 0. No reaction occurs and the process is assumed to be in the steady state. Steps are omitted here. This is a steady state flow process without reaction. All the compositions are known except for stream C. We can pick the overall system first to get D, and then make balances on the first or second units to get C and its composition.
HC1: D 0. The solution is simplified if two balances are made first not an essential step : All material balances are in kg. Solids balance on Unit 2 and also Unit 1: 0.
Basis: mol natural gas or use mol dry gas product. The pounds of TiO2 produced P : By reaction 3 , 0. Adsorber balance of U units are U : mL 1. H2O 0. Dried cereal H2O 0. Fresh air, Basis mol Exit air, Basis 1. KNO3 0. On the basis of 1 kg of water, the saturated recycle steam contains 1.
The recycle steam composition is 0. Either a a balance around the evaporator or b a balance around the crystallizer will do. The latter is easier since only three rather than four unknown streams are involved. Single pass conversion based on H2 as the limiting reactant: 1.
Then fSP 0. The single pass conversion for Ca AC 2 is 0. Recycle Make balance on the mixing point. Balance over separator. Mole ratio of ZnBr2 to C2H2 in final products: 5.
Pick the overall process as the system. Step 5: Basis: kg mol CH4 Steps 1, 2, 3, and 4: The system is as shown in the diagram with recycle added. Recycle is not involved in the overall balance, hence the concentration of NO will not be affected because the extent of reaction with and without recycle remains the same.
The recycle does reduce the combustion temperature, which in turn will reduce the exit concentration of NO. All of it reacts. Use stochiometry mol mol fr kg mol SO3 0.
We can use a mixture of element and species balances. The bottom stream from the distillation column and the wastewater stream from the condenser are candidates. The bottoms stream from the distillation column contains no NH3 and has the highest mass fraction of AN of any waste stream so that the entire stream could be fed to the scrubber.
The change in the scrubber feed will not affect any of the stream flows or compositions upstream of the scrubber that are connected to the scrubber, namely those associated with the reactor and the subsequent condenser, nor any downstream flows not connected to the scrubber. A sequential set of material balances can be used to get the flows and concentrations in the rest of the process.
Basis: 1 second 1 The water flow rates will not change except for the stream going to treatment, which will be Let x be the kg of AN entering or leaving a particular unit. Reformer balance gives stream 3 99 kgmol CH4 conv.
Hg absolute The simplest way to proceed, now that the data are in good order, is to apply the ideal gas law. Take as a basis, Determine the initial pressure in the O2 tank alone. H2 O Hg abs. Besides inconvenience, the discharge concentration from the hood vent may be unacceptable. Comp Mol Mol.
O2: mol XS O2: 0. Also, assuming a leak would bring no additional CO2 into the system, the above CO2 balance calculations indicate the given analysis is probably correct.
You can use element balances or compound balances. The former is easier. Step 4: Calculate A given the Increase 7. Including those products will have negligible effect on the N2 and CO2 produced. Base the CO2 on the emission factor value. SO2 3. Pick the system of the reactor plus separator, and use the data for the conversion. Step 5: Basis 10, kg 1. CO2 Step 5: Basis: 1 min. Please modify your links and shortcuts to reflect this change.
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