Basic chemical engineering pdf

Chemical Process Control by George stephanopoulos. Reynolds, John S. Jeris, Louis Theodore. Heat and Mass Transfer by Sudheer Siddapureddy. Missen, Charles A. Mims, Bradley A. Introduction to Fluid Mechanics by Y. Mass Transfer Operation by Robert Treybal. Petroleum Engineering Handbook by Larry W.

Reactor Design by Andrew Rosen. Branan, Editor. Semi Batch And Recycle Reactor. Statistics and Probability for Engineering Applications by W. Thermodynamics by P K Naag. Facebook Twitter. Follow Us. Contact Form. Chemical Engineering Books Pdf July 09, Alternate solution: 0. The reason for this discrepancy could be faulty flow sensor readings or possibly a leak of the condensate or steam into the process stream.

Basis: 62 kg silicon The system is the melt, and there is no generation or consumption. The balance is satisfied NaC1 balance: 0.

The closure is good for industrial practice. Propylene balance: 0. No accumulation, generation, or consumption occur. The problem cannot be solved unless one stream value is specified.

The equations have no solution — they are parallel lines. The number of unknown quantities is 3, hence a unique solution is possible. In addition three sum of mole fraction equations exist. The problem is overspecified, and will require at least squares solution. Or look at the CH4 row.

A least squares solution could be determined, or the 4 equations with the most accurate data solved. The rank of the coefficient matrix is 3 The rank of the augmented matrix is 4 Hence no unique solution exists. Basis: 2. For example, you cannot use all the flows, or all the compositions in one stream, as the resulting set of material balances will not consist of 5 independent balances, but a lesser number.

Basis: mol A d. N2: 0. The DT is the balance of each stream. Introducing the specifications and basis into the material balances: Water: 0. Unknowns: Assume all of the variables except those that are specified as zero are included in the analysis. Consult the tables at the end of the Chapter for more suggestions.

Rephrase the problem to make sure you understand it? Draw a simple diagram of what was happening? Think about what was going into the tank and what was coming out? Imagine yourself inside the tank, and ask what was going on around you?

Ask whether there were any physical laws to consider such as conservation of matter or energy? Try to imagine the answer as a number, graph, table, or whatever? Try to identify essential variables? Choose a notation? Look for a ready-made formula for the answer? Look for simplifying assumptions? Try to find an easier version of the problem? Look for bounds simple models that would definitely underestimate or overestimate the answer? Two balances can be made, N and S. Steps 8 and 9: Solve to get N: F 0.

CH4 0. Step 5: Basis: 1. Equations: 0. It is proposed to prepare the final mixture by blending four different compounds A, B, C, D ; there will still be three equations, but now there will be four unknowns. Since the rank is now less than n, there will be an infinite number of possible blends of the four mixtures. Not required: An optimization of a revenue function subject to the equations is needed. Redo the problem with a new composition for F: CH4 0. Two balances can be made, NH3 and gas.

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. For other answers, look at the answer to P You can decide how much RT accuracy is needed.

From Polymath Use Excel to solve this cubic equation to obtain a positive, real answer. Use the compressibility factor method. SRK equation: 0. Use of the Pitzer ascentric factor would require a trial and error solution. It is a liquid sp. To find the amount of gas in the cylinder, first find the approximate volume in the cylinder.

MW is Si is Let T - K. Assume the number of moles does not change. Then type IA is selected because it is the cheapest.

MW C2H4 0. Separator Bz 0. Tol 0. MW kg mass fr. Tc K pc atm Bz 50 In mm Hg kPa mm Hg 2. It is a typo. Probably the number was The weights are the respective amounts of each phase. Use semi-log paper with the horizontal axis log Obtain vertical scale rules by use of steam properties at even integers for the temperature. Or you can use Antoine Equation, Appendix G, to get 2 points and draw a line between them. Tcritical o F p critical cox chart psia pcritical psia a Acetone b Heptane c Ammonia d Ethane On logarithmic paper, set up the pressure scale on one axis.

The other axis will be a nonlinear temperature scale, which will be set up by the following method. Draw a diagonal line from the origin to the opposite corner of the graph. From the steam tables, obtain values for the vapor pressure of water at evenly spaced temperature increments. Plot these vapor pressures along the diagonal line. Now, beginning with the first point on the diagonal, draw a line extending from that point to the temperature axis and label the intersection as the corresponding temperature for that vapor pressure.

Repeat step five for every point on the diagonal line. This process will establish the temperature scale. Using the established temperature scale, plot the data for benzene and draw a line through the points. The result from the Cox chart is that the vapor pressure for aniline at o is 22 atm.

CCl3NO2 0. That means their breath can hold a bit more moisture than ours, and on that day, that extra moisture could have been just enough for condensation to occur. It could be that it was a few degrees colder on the beach than where the observer was sited. And, as you know if you have ever experimented with seeing your breath, the bigger the volume of air, the more moisture available to condense out and so the condensed moisture was more visible.

Assume the barometer is mm Hg abs. The above greatly exceeds this limit. Hg total lb mol hr Alternate solution: make air and water balances. A mercury spill can be cleaned up reasonably effectively by dusting with sulfur, then vacuuming with a vacuum cleaner specifically designed for mercury pick-up, which prevents the escape of mercury vapor.

H2O sat 2. The 2 answer is close to correct because of the very small amount of water. The process is open, steady state with reaction. The first objective is to make material balances. F Coal dry P flue gas CO2 0. Hg equivalent to oF The total pressure is 1 atm.

Assuming a basis of g of solution, we can construct the following: Molecular Mol g weight Mol fraction Water 25 18 1. Use the Antoine equation to get T Assume ideal liquid and vapor. Basis: Data given in problem statement P 20 Calculate mole fractions in the liquid. Dew point temperature y 0. Enriched gas 0. Material balance on Comp. Get the mole fraction of the compound in the liquid phase. Basis: g water 5. K y mol fr. H 2O Glycerol 5. Glycerol and phenol have the lowest concentrations.

If volatization is proportional to the vapor phase concentration, only MEK might be a problem. Intensive b. Intensive c. Intensive d. Extensive 9. Total energy consumed is the same. The higher power consumption is compensated for in the shorter travel time. Also, the location would have to be in the far western united so power would have to be transmitted over long distances.

Watts are power, and if the energy is transferred for a very short period to time, the power can be quite large. For example, 1J a very small amount of energy transferred in 10—6s is 1 MW! And sometimes that just creates confusion. What the manufacturer is doing is basing the advertisement on the weight of the hot dog. A hot dog weights about 43 grams and has 8 grams of fat in it.

They divided 8 by 43 and get 19 percent fat. But you are concerned with the percentage of fat in the calories. You want 30 percent or less to come from fat. Your figures are right. But for a , liquid to solid, there are probably no examples. If any, they would be pathological cases. Ignore the volume of the liquid water at the initial conditions, and assume the initial amount of vapor is also negligible. Heat is energy transferred from one system to another, and from the viewpoint of either system for which an energy balance is made , heat is not conserved.

Thus a is wrong, heat is part of a balance; b same as a ; c heat is not conserved—it is transferred and is part of the energy balance; d the answer should be no with the explanation offered. Heat is energy, not material. Heat is energy by definition. Heat is energy; cold is related to temperature. See c.

Heat is a transfer of energy. See a. Heat transfer can result from the change in state. Heat transfer can be terminated. Data here taken from the steam tables. State 1 assumed saturated State 2 superheated Heat is energy that flows between the system and the surroundings because of a temperature gradient.

Therefore they are not the same class of energy. The initial and final states for U and H are the same. Since the enthalpy of liquid water changes negligibly with pressure, no loss of accuracy is encountered for engineering purposes if the initial pressure on the water is not stated.

Enthalpy is a function of the temperature and pressure, but the effect of pressure on liquid water under these conditions is negligible. Therefore the enthalpy of water at K and The final conditions are presumably a state in which water is all vapor. A check of the steam tables shows this assumption to be true. Only enthalpy changes can be computed using the chart; the enthalpy itself cannot be determined.

Benzene properties: Mol. Enthalpy decrease 7 psia The CD gives — J. The corresponding pressure is the saturation pressure, namely about 2. Another temperature would correspond to another pressure. U should have the same reference temperature. At kPa saturated Open, steady state. Closed, unsteady state. The process is closed, unsteady state. A similar analysis applies to the second equation in the problem.

The cooling at the maximum efficiency is 0. The main question with the advertisement is: where does the energy go that is removed from the room if the unit does not require outside venting?

At the end of the process the system contains nothing. No reaction occurs. The process is clearly a flow process open system. Assume that the entering velocity of the air is zero.