FLUID MECHANICS FOR CHEMICAL ENGINEERS NOEL DE NEVERS PDF

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Fluid Mechanics for chemical Engineers - Noel de nevers 2nd ed. Tahlia Stone. Loading Preview. Sorry, preview is currently unavailable. You can download the . Fluid Mechanics for Chemical Engineers, Third Edition Noel de Nevers Solutions Manual This manual contains solutions to all the problems in the text. Many of. Fluid Mechanics for Chemical Engineers, Third Edition. Noel de Nevers. Solutions Manual. This manual contains solutions to all the problems in the text. Many of.


Fluid Mechanics For Chemical Engineers Noel De Nevers Pdf

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Fluid Mechanics for Chemical Engineers - Noel de Nevers 2E - Ebook download as PDF File .pdf) or read book online. Fluid mechanics for chemical engineering / N. de Nevers. | π—₯π—²π—Ύπ˜‚π—²π˜€π˜ 𝗣𝗗𝗙 on ResearchGate | On Jan 1, , Noel de Nevers and others published Fluid. "Fluid Mechanics for Chemical Engineers, third edition" retains the characteristics that made this introductory text a success in prior editions.

Here for a confined fluid that is clearly not the case. If an immersed body is in a confined fluid that assumption plays no role, and the buoyant force, computed by Archimedes' principle is independent of the external pressure applied to the fluid.

M for the gauge pressure of a constant-density fluid: There, at 60 ft, the required thickness was 0. The required thickness for the assumptions in the problem is proportional to the gauge pressure at the bottom of each band of plates, which is proportional to the depth, a The values are Depth, ft Pressure, psig Design, t, in 60 It is not asked for in the problem, but the same industry source who gave me these values also told me that the maximum used is 1.

However in the middle column the lowest value, 0. The rightmost column shows the average thickness of the tapered plate except for the lowest entry, which is not tapered. So, for example, the lowest plate is 0. Thus the tapered plates reduce the weight of the plates, for equal safety factor, to 0. The same industry source who gave me the other values in the problem also told me that the tapered plate solution is quite rare.

It is apparently only done when there is an order for a substantial number of very large tanks all of one size. So it is not simply the cost of the metal, but the cost per unit stored of erection, foundations, piping, etc.

I thank Mr. Terry Gallagher of CBI who spent a few minutes on the phone and shared his insights on this problem. Compare this to Example 2. He said that they stress-relieved this tank, in the field, after it was complete, by building an insulating tent around it, and heating to a stress relieving temperature. That costs, but the economics of having this big a tank apparently made it worthwhile. For spherical tanks he and I think that the difficulties of erection become too much as the size goes over about 80 ft.

I have seen an advertisement for Hyundai, showing one of their liquid natural gas tankers, which had some huge spherical tanks in it.

LNG is transported at a low enough temperature that one must use expensive alloy steels, so the cost of those spherical tanks must have been justified. The diameter may be set by the largest diameter hemispherical heads which steel mills will regularly produce.

The heads are one-piece, hot pressed, I think. Gallagher who refers to this tank shape as "blimps" indicates that there are some field-erected tanks of this type, which are larger than the values I show. Most of the tanks of this type are factory assembled, with a significant cost saving over field erection, and then shipped to the user. Two interesting web sites to visit on this topic are www. It would probably be selected. Clearly one should not fit a single equation to that set of data. If one considers only the values for 2.

The difference is conservatism on the part of those defining the schedules. We see that below about psi the three equations give practically the same result. Above that Eq. For Eq.

From Eq. However the sausage-shaped container is small enough to be shop assembled and shipped by rail or truck, while the spherical container could be shipped by barge, but at 17 ft diameter it could probably not be shipped by truck or rail. The rule of thumb cited in App. Grooved joint connectors Perry's 7e Fig use this fact. They cut a circumferential groove half-way through the pipe to be joined. The external clamp which fits into the groove supports the pipe against external expansion, so the doubling of the hoop stress is not a hazard.

The clamp only operates on half the thickness of the wall, but that is enough to resist the axial stress, which is only half the hoop stress. If one computes the buoyant force of the air on the crown in the weighting in air, one finds 0.

If one then substitutes 5. Discussion; This assumes no volume change on mixing of gold and silver. That is a very good, but not perfect assumption. As far as I know no one uses hydrogen in balloons today. One must make the distinction between kgf and kgm to solve here, finding kgm kgf 9.

Fluid Mechanics for Chemical Engineers - Noel de Nevers 2E

Whatever was pulling on it would have to have a buoyant force large enough to support its own weight, and also the weight of the battleship which it was raising. For deeper sunken vessels the fraction is higher. To the best of my knowledge the fist time this was ever tried was when the CIA attempted to raise a sunken Russian submarine for intelligence purposes during the cold war.

They lowered a clamping device on multiple cables and tried to pull it up. It broke up while they were doing it, so they only got part of it, but that was said to have revealed lots of useful technical information. In the Russians successfully raised the submarine Kursk this way, with multiple cables. However as this quantity becomes small, the two fluids tend to mix, and may emulsify. So there is a practical limit on how small it can be made.

That corresponds to To get values in terms of heights of mercury one must use the density of mercury at some temperature. This is 0. Thus the above calculation would replace the atmospheric pressure with A water barometer would need a temperature correction not only for the thermal expansion of the liquid, but also for the vapor pressure of the liquid which is about 0.

Thus it appears that the principal cause of atmospheric highs and lows is the difference in average temperature between air masses. The total depth is 4. If we use 2. One can discuss here the other ways to continually measure the level in a tank. One can use float gages, pressure difference gages, etc. If the gas flow can be tolerated, this is simpler. This system is simple and continuous.

Currently we use stacks to disperse pollutants, more than we use them for draft, but historically the height of the stack was determined by the required draft to overcome the frictional resistance to gas flow in the furnace. When the oil was pipe slowly introduced, it flowed over the water, and bypassed water, leaving some in place.

For any one such leg, the pressure difference is This low a pressure will cause a normal pressure vessel to collapse. In oil industry folklore it is reported that this has happened several times. Normally a newly installed pressure vessel is hydrostatically tested, and then the workmen told to "drain the tank". They open valve B, and soon the tank crumples. Common, cheap ones are not locally calibrated, but simply used with the factory design values.

In the factory they are not individually calibrated because each one coming down the production line is practically the same at the next. For precise work, pressure gauges are calibrated this way. This is a manometer, and if bubbles are present, they make the average density on one side of the manometer different from that on the other, leading to false readings.

If all the bubbles are excluded, then this is an excellent way to get two elevations equal to each other. The horizontal part of the device plays no role. Physically, if the two fluids have exactly the same density, then they will not form a clean and simple interface. Most likely they will emulsify. Here P is not linearly proportional to radius, so this is complicated.

If we assume that over the size of this particle we can take an average radius, and calculate the "equivalent gravity" for that location, then we can use Archimedes' principle as follows. Work in gage pressure. Two parabolas are the same. If one solves it as a manometer, from the top of the funnel to the nozzle of the jet, one will see that there is a pressure difference equal to the difference in height of the liquid in the two bottles times g times density of water - density of air.

This is clearly not a steady-state device; as it flows the levels become equal and then the flow stops. USA including money in bank vaults, in people's mattresses, etc. The boundaries are not fixed in space, nor are they fixed in size, but they are readily identifiable.

The auto including all parts, with boundaries crossing the openings in the grille and the exhaust pipe. The point is that there is one main flow out, the carbon dioxide in the exhaust, which equals the negative accumulation of carbon in the fuel, and then a lot of other minor terms. That was supposed to reshape the beaches and in other ways make the canyon more like its pristine state.

Those who organized the flood pronounced it a great success. This has the paradoxical consequence that friction, which causes the pressure to fall, causes the velocity to increase.

This type of flow is discussed in Ch. It plays no role. The answer would be the same if the initial pressure were any value, as long as the pressure was low enough for the ideal gas law to be appropriate. All real vacuum systems have leaks whose flow rate is more or less proportional to the pressure difference from outside to in assuming laminar flow in small flow passages but this comparison shows that replacing Eq.

One can look up the answers on Fig 3. Its analogs also appear in heat and mass transfer.

De Nevers, Noel 1991 Fluid Mechanics For Chemical Engineers

Then, a randomly-selected liter contains 3. The deep oceans mix very slowly. The Red Sea does not mix much with the rest of the world ocean, except by evaporation and rainfall. The perfect mixing assumption is much better for the atmosphere than for the ocean in the preceding problem.. Air resistance complicates all of this, see Prob. This appears paradoxical, but the ball is simply converting one kind of energy into another.

Work was done on the ball as the airplane lifted it from the ground. This seems odd, but the water is merely transferring work from the pump to the piston, rack and car. Here the dWn.

Solving for vin dmin dWn. If we had worked it in absolute, then we would have needed to include a term for driving back the atmosphere as the car, rack and piston were driven up. But it shows how one would proceed if one did not have access to a suitable steam table. Here the students must use the fact, stated at the beginning of this set of problems, that for a perfect gas u and h are functions of T alone, and not on the pressure.

Unfortunately the adiabatic assumption cannot be realized, even approximately if one tries this in the laboratory. Even in the few seconds it takes to fill such a container the amount of heat transferred from the gas makes the observed temperature much less than one calculates this way.

The amount of heat transferred is small, but the mass of gas from which it is transferred is also small so their ratio is substantial. In spite of that this is a classic textbook exercise, which is repeated here. Using the solution to problem 4. Ask them why. Some will figure out that the heating value is for the fuel plus the oxygen needed to burn it, taken from the air, and not included in the weight of the fuel.

High explosives include their oxidizer in their weight. They do not release large amounts of energy per pound, compared to hydrocarbons. What they do is release it very quickly, much faster than ordinary combustion reactions.

You can estimate the pressure for this from the ideal gas law, finding amazing values. Fats are roughly CH2 n. Carbohydrates are roughly CH2O n. This simplifies the chemistry a little, but not much. You might ask your students what parts of plants have fats. Some will know that the seeds have fats, the leaves and stems practically zero.

Then some will figure out why. The assignment of a seed is to find a suitable place, put down roots and put up leaves before it can begin to make its own food. It is easier to store the energy for that as a fat than as a carbohydrate. Our bodies make fats out of carbohydrates so that we can store them for future use in case of famine. Thermodynamics explains a lot of basic biology!

Note that the metal walls of the calorimeter are outside the system. The value here is between that for fats and for carbohydrates. This latter is not quite correct, because particles are blown off the sun by solar storms, but their contribution to the energy balance of the sun is small compared to the outward energy flux due to radiation.

What nature does is independent of how we think about it. We may show that they do not violate the first law by making an energy balance for each.

These all violate the second law. They all violate common sense. If you doubt that, put a baseball on a table and watch it, waiting to see it spontaneously jump to a higher elevation and cool.

Be patient! For 1 psig, this factor is P1 If one is not going to use the approach in Ch. The ask about a can of gasoline. Torricelli's Eq. See Chapter 6. Apparently the safety doors divide the boat's hull into some number 7 on the Titanic?

So if the hull is punctured, as in this problem, and the doors close properly, one compartment will fill with water up to the level of the water outside, but the other compartments will remain dry. Apparently the impact with the iceberg opened the hull to more compartments than the ship could survive, even with the safety doors closed. Presumably if the compartments were sealed at the top, then even that accident would have been survivable, but it is much easier to provide closed doors on passageways parallel to the axis of the boat, in which there is little traffic, than on vertical passageways stairs, elevators, etc.

However it is correct. The reason is the very low density of the helium, which makes the difference in atmospheric pressure seem like a large driving force. These devices most often send a signal which is proportional to the pressure difference. If this is shown directly on an indicator or on a chart, the desired information, the flow rate, is proportional to the square of the signal.

One solution to this problem is to have chart paper with the markings corresponding to the square of the signal. The chart manufacturers refer to these as "square root" charts, meaning that in reading them one is automatically extracting the square root of the signal. Moving blade or moving cup anemometers are most often used for low-speed gas flows, e. The students have certainly seen these in weather stations.

On the dashboard of an airplane private or commercial is an "indicated air speed" dial. The lift is directly proportional to the indicated air speed, so this is also effectively a lift indicator, see Secs. You can ask the students why commercial airliners fly as high as they can. For a given weight, there is only one indicated air speed at which they can fly steadily in level flight.

As the air density goes down, the corresponding absolute velocity goes up. So by going high, they go faster. This means fewer hours between takeoff and landing. That means less fuel used, fewer hours to pay the pilots and crews for, and happier customers who do not like to sit too long in an aircraft.

Using the methods in this book, we find for that pressure difference lbf 0. We can get other values by ratio, e. The upper one, on log-log coordinates is probably the more useful, because the fractional uncertainty in the Q, cfs reading is practically the same over the whole range.

The lower one, on arithmetic coordinates is probably easier for non-technical people to use, and would probably be selected if Delta P, psig 10 Here I have chosen the pressure drop as the independent variable, 8 because that is the observational instrument reading. From Fig 5. See Eq 5. This explains why this type of carburetor was the practically-exclusive choice of automobile engine designers for about 80 years. With a very simple device, one gets a practically constant air-fuel ratio, independent of the throttle setting.

The rest of the carburetor was devoted to those situations in which one wanted some other air-fuel ratio, mostly cold starting and acceleration, for which one wants a lower air fuel ratio " rich ratio". The air density falls while the fuel density does not, so the air fuel ratio would become 0. High altitude conversion kits smaller diameter jets are available to deal with this problem.

Rich combustion leads to increased emissions of CO and hydrocarbons; there are special air pollution rules for autos at high elevations. Demands for higher fuel economy and lower emissions are causing the carburetor to be replaced by the fuel injector. In a way it is sad; the basic carburetor is a really clever, simple, self-regulating device.

The density is proportional to the molecular weight. For the values shown, the predicted jet velocity in the burners will be the same for propane and natural gas. Because of the higher density of propane, the diameter of the jets will normally be reduced, to maintain a constant heat input.

But the velocities of the individual gas jets are held the same, to get comparable burner aerodynamics. For 0. This is a simple scale change for each curve on Fig. Solving Eq. Either gives the same result. Clearly, the greater the depth, the higher the speed at which the propeller can turn without cavitation, so this is not as severe a problem for submarines submerged as it is for surface ships. However the noise from propeller cavitation is a serious problem for submarines, because it reveals the position of the submarine to acoustic detectors.

Submarines which do not wish to be detected operate with their propellers turning slower than their minimum cavitation speed. As in that example take 1 at the upper fluid gasoline surface and 2 at the outlet jet.

Take 2a at the interface between the gasoline-water interface. If we replaced the 20 m of gasoline with So this is really the same as Ex.

Then one repeats the whole derivation in Ex 5. One can also compute the time from when the interface has passed the exit to when the surface is 1 m above the exit, finding the same time as for water, because in this period the density of the flowing fluid is the same as that of gasoline.

See Prob. I have the device described in that problem. I regularly assign the problem, then run the demonstration.

One can estimate well, down to an interface one or two diameters above the nozzle, but not lower. The easiest way to work the problem is to conceptually convert the 10 ft of gasoline to 7.

Then this is the same as Ex. The final pressure is This gives the answer to part c ; the 5-fold expansion of the gas lowers its absolute pressure by a factor of 5, which would produce a vacuum and stop the flow if the initial pressure were 20 psig. Instead we proceed by a spreadsheet numerical integration. The original spreadsheet carries more digits than will fit on this table. One may test the stability of this solution, by rerunning it with smaller height increments.

For increments of 0. Both round to 6. Then this becomes the same as Ex. Following Ex. Much of the time is spent in the last inch, where the assumption that the diameter of the nozzle is negligible compared to the height of the fluid above it becomes poor see Sec 5. Some students like that, others don't. We then square both sides and differentiate w. Then using the same ideas as in Ex 5. In this problem two Vs appear, the volume of the can and the instantaneous velocity.

Following the nomenclature, both are italic. When a V is the volume of the can, it has a "can" subscript. Try it, you'll be impressed! However it is easy to solve this on a spreadsheet. The solution is shown below. Most of the spreadsheet carries more significant figures than are shown here. For the plug flow model the agreement is V 1. We solve V0 3. That happens here. In making up this solution I found two errors in [7]. I hope the version here is error free.

As of spring you could watch a film clip of a version of this demonstration at http: For me it is a fine demonstration of unsteady-state flow. They see all sorts of interesting combustion-related issues in it.

I give this problem as an introduction to a safety lecture. Propane is by far the most dangerous of the commonly-used fuels. If there is a large leak of natural gas, buoyancy will take it up away from people and away from ignition sources. If there is leak of any liquid fuel gasoline, diesel, heating oil it will fall on the ground and flow downhill. But the ground will absorb some, and ditches, dikes or low spots will trap some or all of it.

But released propane forms a gas heavier than air, which flows downhill, and flows over ditches, dikes and low spots, looking for an ignition source, and then burning at the elevation where people are. Thus by BE, the pressure must be rising steadily in the radial direction.

In the center hole the pressure must be higher than atmospheric, in order to give the gas its initial velocity and to overcome the frictional effect of the entrance into the channel between the cardboard and the spool.

Thus the figure is as sketched below. This is a very simple, portable, cheap, dramatic demonstration to use in class. Since the pressure increases by 1. It must be clear that at low values of the pressure ratio the three curves are practically identical.

As far as I know there are still ferry boats that use this system to stay above the water, regularly crossing English Channel. There are two put up each winter within five miles of my house. I wish I remembered where. As the following calculation shows, by simple B. However choosing 1.

It is often in laminar flow in laboratory or analytical sized equipment. The corresponding viscometer for very high viscosity fluids replaces gravity with a pump, and attempts with cooling to hold the whole apparatus and fluid isothermal. All serious viscometry is done inside constant temperature baths, see Fig. A1 shows that this low a viscosity is rarely encountered in liquids, so that this kind of viscometer can be applied to most liquids.

Cryogenic liquids, however, have very low viscosities, so there might be a laminar-turbulent transition problem using this particular viscometer on them. Saybolt Seconds Universal, SSU which is the standard unit of viscosity for high boiling petroleum fractions.

The x component force depends on the speed of the train, and how often the balls are thrown back and forth, but not on their velocity. This shows how the 4f appears naturally in the Fanning friction factor.

They don't. Assign this problem, after you have discussed laminar and turbulent flow, and you will be appalled at how few of the students can solve it. After they have struggled with it, they will be embarrassed when you show them how trivial it is. Maybe that way they will learn about the different relation between pressure drop and velocity in laminar and turbulent flow!

This is very similar to that shown in Table 6. One enters at gpm, reads horizontally to the zero viscosity boundary which corresponds to the flat part of the curves on Fig. This is less than the available 28, so a 3 inch pipe would be satisfactory. The most likely reason for the lower roughness in Fig 6. The value for the roughness of steel pipe Table 6. The first three columns are the same as those in Table 6. The bottom 4 rows are in addition to what is shown in Table 6.

Variable First guess Solution Prob. Looking at the original spreadsheet we see it went from 0. The bottom 2 rows are in addition to what is shown in Table 6.

Fluid Mechanics for Chemical Engineers

If there were no change in f then quadrupling the volumetric flow rate would cause the required diameter to double. As shown here it increases by a factor of 1. With the same pressure drop 1.

This is only an 0. The result is in the laminar flow region, so we could solve directly from Pouisueille's equation, lbf 5. The density is easily measured in simple pyncnometers, to much greater accuracy than it could possibly be measured by any kind of flow experiment. The viscosity is measured in laminar flow experiments, as shown in Ex.

Furthermore, in a flow experiment we would presumably choose as independent variables the choice of fluid, the pipe diameter and roughness and the velocity. We would measure the pressure drop, and compute the friction factor.

If we were at a high Reynolds number, i.

The friction factor would also be independent of the density which is part of the Reynolds number , although the calculation of the friction factor from the observed pressure drop would require us to know the density in advance.

Q and 6. I doubt that this has an analytical solution; it would be tolerably easy numerically, but the procedure shown in Ex. For Ex. X and 6. Again, I doubt that this has an analytical solution; it would be tolerably easy numerically, but the procedure shown in Ex. First you should point out that Fig. It has a scale ratio of about 1. Three brand new chapters are included. Get A Copy. Hardcover , pages. More Details Original Title. Other Editions 1.

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The air valve that does that automatically, invented by Jaques Costeau, made scuba diving possible. They open valve B, and soon the tank crumples. Much of the time is spent in the last inch, where the assumption that the diameter of the nozzle is negligible compared to the height of the fluid above it becomes poor see Sec 5.

If they are different the common case then this value must be modified. That matches the price structure for oil, where lower density crude oils have a higher selling price, because they are deg API more easily converted to high-priced products e. This shows that the friction effect of a square duct is more than that of a circular one, and we need a little more than equal area.

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