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.

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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₯π²πΎππ²ππ π£ππ 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.

Skip to main content. Log In Sign Up. Alex Red. Noel Denevers. Many of those are discussion problems; I have tried to present enough guidance so that the instructor can lead a useful discussion of those problems. In addition I have added discussion material to many of the computation problems. I regularly assign these as computation, and then after we have agreed that the computation is correct, asked the students what this computation tells them. That leads to discussion. Wherever I can, I begin a discussion of some topic with a computation problem which introduces the students to the magnitudes of various quantities, and thus requires them to read the part of the text covering that topic.

Embed Size px. Start on. Show related SlideShares at end. WordPress Shortcode. Published in: Education , Technology , Business. Full Name Comment goes here. Are you sure you want to Yes No. DongJunPark12 Did anyone get the rest of the chapters? Ngan Hoang Hi, If possible could you please upload the solution for the rest of the chapters too.

Thank you. Show More. No Downloads. Views Total views. Actions Shares. Embeds 0 No embeds. No notes for slide. Fluid Mechanics for Chemical Engineers, 3rd Edition 1. Many of those are discussion problems; I have tried to present enough guidance so that the instructor can lead a useful discussion of those problems. In addition I have added discussion material to many of the computation problems.

I regularly assign these as computation, and then after we have agreed that the computation is correct, asked the students what this computation tells them. That leads to discussion. Wherever I can, I begin a discussion of some topic with a computation problem which introduces the students to the magnitudes of various quantities, and thus requires them to read the part of the text covering that topic.

Once we all know the magnitudes, and have all read that section of the text, we can have an interesting discussion of their meaning. In this additional discussion I have presented reference when I could. Often I relied on industry "common knowledge", folklore and gossip. I hope I got it all right. If not, I apologize for leading you astray. Where I am not sure about the folklore, I have tried to make that clear in the discussion.

Many of these problems have been class tested. Some, alas, have not. This manual, as well as the book, is certain to contain errors. I will be grateful to those who point these errors out to me, so that they can be corrected. I keep a running correction sheet, and send copies to anyone who asks for it. FAX Many of the problems go beyond what is in the text, or show derivations which I have left out of the text to make it read easier. I suggest that the instructor tell the students to at least read all the problems, so that they will know what is contained there.

Some of the problems use spreadsheets. In the individual chapters I have copied the spreadsheet solutions into the text in table format. That is easy to see, but does not let the reader modify the spreadsheets.

In the Folder labeled "Spreadsheets" I have included copies of all the spreadsheets shown in the individual chapters, and also the high velocity gas tables from the appendices. These are in Excel 4. Laws Not Used, third law of thermodynamics, all electrostatic and magnetic laws, all laws discussing the behavior of matter at the atomic or subatomic level, all relativistic laws.

The densities of other liquids with low values are: Discussion; the point of this problem is for the students to recognize that one of the principal differences between liquids and gases is the large difference in density. As a rule of thumb, the density of liquids is times that of gases.

Discussion; this assumes no volume change on mixing. That is a good assumption here, and in many other cases. This is normally ignored, but in the most careful work it must be considered.

That matches the price structure for oil, where lower density crude oils have a higher selling price, because they are more easily converted to high-priced products e. The plot covers the whole range of petroleum liquids, from propane s. Water s.

For methane and propane the values are and Propane is by far the most dangerous fuel in common use. If we have a methane leak, buoyancy will take it up and disperse it. If we have a leak of any liquid fuel, it will flow downhill on the ground, and be stopped by any ditch or other low spot. Propane, as a gas heavier than air, flows downhill, over small obstructions and depressions. It often finds an ignition source, which methane or gasoline would not find in the same situation.

Inks; spread when you are writing, don't leak when you are not. Lipstick; spread when applied, then not move. Crayons; same as lipstick. Blood; corpuscles don't settle, viscous resistance is small in large arteries and veins. Chocolate; easy to spray on warm or apply by dipping, then does not run off until it cools. Oil well drilling fluids; low viscosity when pumped up and down the well, high viscosity when leaking out of the well into porous formations.

Radiator sealants; low viscosity in pumped coolant, high viscosity in potential leaks. Aircraft de-icing fluids; stick to the plane at low speeds, flow off at high speeds. Plaster, spread easily, not run once applied while it is setting. Margarine, spread easily, but remain solid when not being spread. Thus for equal volumes the sphere has the least surface, the right cylinder the next least, and the cube the most.

For the earth there is no such well-defined acceleration of gravity. For the past half century this number has remained about constant, we have found it at about the rate we used it. In the past few years this has mostly been by improved recovery methods for existing fields, rather than finding new fields. In the 's the US could produce about twice as much oil as it consumed; now we can produce about half.

This is mostly not due to a decline in production, but an increase in consumption. No one has seriously proposed this, but it makes as much sense as the slug and the poundal. Multiplying that by the acres to be irrigated gives the water demand. The south western states look to the Columbia with envy; the Northwestern states are in no hurry to give it up. So this terminology is wrong. But it is in very common usage. If they are different the common case then this value must be modified.

However European rocket engineers use the "effective exhaust velocity" which takes the pressure into account the same way US engineers use the specific impulse.

Clearly the bar is a convenient approximate atmosphere. It is very commonly used in high pressure work, as an approximate atmosphere. Meteorologists express all pressures in millibar. Currently thermodynamic tables like steam tables and chemical thermodynamic tables show pressure in bars.

The most common type of pressure gage testers, the dead-weight tester, balances a weight against the pressure in the gages. I think this is slowly losing out to the kPa, but it is not gone yet. Solutions, Chapter 2 2. This is close to One can also discuss why it hurts. So a pressure difference of about 5 psi across the eardrums is painful.

Scuba divers have the same air pressure inside their lungs as that of the outside water. The air valve that does that automatically, invented by Jaques Costeau, made scuba diving possible.

People who dive deeply wearing a face mask must equalize the pressure in their lungs with that in the face mask, or the blood vessels in their eyes will burst. This is a problem for deep free divers. Furthermore, one cannot tolerate such a high pressure in the drinking fountains on the ground floor; they would put out people's eyes.

Tall buildings have several zones in their internal water system, with suitable pressures in each, and storage tanks at the top of each zone. When I assign this problem I sketch the flow in an oil-drilling rig for the students.

High-pressure drilling fluid is pumped down the drill pipe, and flows back up the annulus between the drill pipe and the wellbore. This drilling fluid cools the drill bit and carries the rock chips up out of the well. Ask the students how one would get such dense drilling fluids? The answer is to use slurries of barite, barium sulfate s. In extreme cases powdered lead, s.

The great hazard is that high pressure gas will enter the drilling fluid, expand, lower its density, and cause it all to be blown out. Most deep drilling rigs have mechanical "blowout presenters", which can clamp down on the drill stem and stop the flow in such an emergency. Even so one of the most common and deadly drilling accidents is the blowout, caused by drilling into an unexpected zone of high-pressure gas.

If the gas is rich in H2S, the rig workers are often unable to escape the toxic cloud. The figure is sketch, roughly to scale at the right. The result is exactly one-half of that result, or 2. An internal pressure blows the vessel up like a balloon, straightening out any non-uniformities. An internal vacuum collapses the tank, starting at a non-uniformity and magnifying it.

Most students have see the demonstration in which a rectangular 5 gallon can is filled with steam and then collapsed by cooling to condense the steam. The maximum vacuum there is An internal pressure of the same amount will cause the sides and top of the can to bulge out somewhat, but the result is far less damaging than the vacuum collapse. You might suggest the analogy of pulling and pushing a rope. If you pull a rope, it straightens out. If you push one the kinks are increased and it folds up.

Then one constructs the ratio of the two pressures, and uses the spreadsheet's root finding engine to find the value of h3, which makes that ratio 1. Before we had spreadsheets we could have done the same by manual trial and error, or done it analytically. Here we see that even at the deepest point in the oceans, taking the compressibility of the water into account changes the pressure by only 2. One may do a simple plausibility check on this mathematics by computing the ratio of the density at half of the depth to the surface density.

Using the data here, the density at a pressure of psi is about 1. For the depths of ordinary industrial equipment this ratio is almost exactly 1. The fact that the atmosphere extends above this height shows that there is heat transfer in the atmosphere, which contradicts the adiabatic assumption.

At this elevation the predicted pressure from equation 2. These are normally "corrected to sea level" so that, for example in Salt Lake City the reported barometric pressures are normally about Those values are used to set altimeters in airplanes, which have a dial-in for the sea level barometric pressure, which they compare to the observed pressure to compute the altitude.

These are called lapse rates and the minus sign is normally dropped, so that the standard lapse rate is 3. This small value shows how all hydraulic systems work. A small pressure, acting over a modest area can produce an impressive force. The small air compressor in the service station easily lifts the heavy car, fairly rapidly. You can ask the students whether this is psig or psia. You can also ask them about buoyancy.

Is the piston a floating body? Answer; not in the common sense of that term. If you look into Archimedes principle for floating bodies, there is an unstated assumption that the top of the floating body is exposed to the same atmospheric pressure as the fluid is. 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. 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. 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. If one re-plotted it on normal log-log paper, one would see that it has three families of parallel straight lines, the laminar flow lines with slope 1, the turbulent flow lines with slope about 0.

The from Fig. We draw a line of proper slope through it completes the zero viscosity boundary. We draw a line with slope 1 through it, for the laminar region, and a line with slope 0. Then we complete the laminar region, by drawing lines parallel to the first line, for various viscosities. In the laminar region, at constant flow rate the pressure drop is proportional to the viscosity, so, for example, the 20 cs line is parallel to the 40 cs line, but shifted to the left by a factor of two, and the 80 cs line is parallel to the 40 cs line, but shifted to the right by a factor of two.

We must guess a value of the kinematic viscosity for which the line will fall between the transition and zero viscosity lines. From Fig. Then for s. We could compute similar lines for other kinematic viscosities if we wished.

The problem does not ask the students to compute the lower section which corrects for differences in density. You might bring that up in class discussion. The lines there would have slope 1 on an ordinary piece of log-log paper. The values calculated above are all for s. If, for example, the true s. This is obvious for turbulent flow, where we set the density to For laminar flow it is a bit more subtle.

In making up the laminar part of the plot we used the kinematic viscosity, with an assumed s. Thus, in this formulation the laminar pressure gradient is also proportional to the specific gravity. Here I have used as the transition Reynolds number. The authors of Fig. But this problem shows that one could.

For this value we can compute the pressure gradients for laminar and turbulent equations, finding 0. The value in the table is 0. A few of the other smallest velocity values are also laminar. All the other values are for turbulent flow. These match Fig. Comparing them to the vales from Eq. For increasing flow rates this ratio diminished. For velocities greater than 9.

As a further comparison I ran the friction factors from Eq. Clearly even at this size pipe, we do not have smooth tube behavior. This is tolerable agreement.

The main reason for this is that in turbulent flow, the volumetric flow rate is proportional to D5. By visual interpolation the required pipe diameter is about 0. If the friction factors are the same, we would expect the pressure gradients to be proportional to the densities.

Thus, this is only an approximate way of estimating the behavior of hydrogen. For gases with properties more like those of air it works better. All of these give practically the same answer. To find the slopes I read the 4 inch line as beginning at 15 and ending at The curvature seems to increase with increasing pipe size. These do not flatten out until higher values of the velocity, so the curves correspond to a value of f which decreases slowly with increasing Q, giving the curves shown.

From the caption for Fig. So this matches Table 6. The computations are easy on a spreadsheet. The plot, covering the range from 3 to 12 ft is shown below. I don't know which is best, other than to follow Lapple's suggestion in the text. Another peculiarity of this comparison is that the estimate by the "K-factor" method is uninfluenced by a change in fluid viscosity, as long as the flow is turbulent.

But the equivalent length method is sensitive to changes in viscosity. I reran the spreadsheet that generated the values for the above plot, taking the fluid viscosity as 1 cP instead of the 50 cP in those examples.

That raised the Reynolds numbers by a factor of 50, reducing the computed f's by almost a factor of 2. That reduced the estimated delta P by roughly a factor of 2, going from In this case the curve on the above plot ends up nearly horizontal; the computed value for 24 inches is You might ask your students what a check valve is.

Normally there will be one visible in the vicinity of your building, as part of the lawn sprinklers, which you can point out. We can simply look up the two pipe terms, in Table A. For the expansion term D 2. Then we read Fig. From Table A. The solution is a trial and error, in which one guesses f and then does a suitable trial and error. The three terms in the denominator under the radical have values 1, 0.

Each contributes to the answer. In a lecture room with a sink one can fill the tank with a piece of tygon tubing, and let the students watch the results. Then we do the calculations, then run the test. The above value is for galvanized pipe, as shown in the problem. They use the value for steel pipe, getting a much lower value. As it was used more and more and the pipe rusted the time climbed slowly to 85 sec. Cleaning it with a wire brush got it back to its original value.

The fundamental cussedness of inanimate objects This works well, one can see several oscillations in the flow rate. One can sketch Fig. True leakage paths are certainly more complex than the uniform annulus assumed here, but this description of the leakage path is generally correct. You might ask bright students how much error is introduced that way? I checked by comparing the slit solution to that for an annulus with inside diameter 0. The computed flow for the slit is 1.

The other values are taken from Ex. Making those substitutions in Eq. If one of you will work that out satisfactorily and send me a copy, I will be grateful. A gap of 0. The flow is laminar. All real gaskets are the equivalent of this; the flow rate through them is not zero, it is simply too small to detect. We first assume that a square duct with 8 inch sides will be used.

In this problem the computed 7. 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. Vactual design 3. We assume that the stack is made of concrete a common choice , so from table 6. We estimated the viscosity of the stack gas as 0. Students are generally uncomfortable with the idea that the pressure drop though the furnace is estimated by a constant times the kinetic energy based on the flue gas temperature.

For the highest-quality estimate we would follow the flow through the furnace, taking into account the changes in density, viscosity and velocity from point to point in the furnace. That is now possible with CFD programs.

But the simple estimating method here is widely used, and reasonably reliable. The required head will be highest for the lowest specific gravity 0. So we would specify a pump with a design point of gpm, and ft of head. This is a high head for a centrifugal pump at this volumetric flow rate; we would probably have to specify some other type of pump.

The minimum corresponds to the opposite of those conditions. Here the open globe valve adds x 3. The values shown are those at the end of that trial and error. Then we solve for the velocity applying BE. Here the adjusted length of the pipe is At that time high quality plastic pipe had not developed to the extent it has now, so the pipe was made of aluminum.

If you walk down the trail from the North Rim to the river, you can see the occasional valve boxes, which allow the pipe, which is buried by the trail, to be shut down and drained for maintenance and repair when needed.

I don't know if they had a special pipe made, or used an 8 inch pipe. The final flow rates are A gpm, B 60 gpm, C gpm. The actual spreadsheet carries and displays more significant figures than are shown in this abbreviated table.

For each of the three pipe sections we have a trial and error for the velocity, done using the spreadsheet's numerical engine. These are somewhat higher than those shown in the example in the first edition, but within the accuracy of interpolations in Table A. This is a somewhat clumsy way to solve this class of problems. Then in the sum of flows into point 2, we must take flow B as positive rather than negative.

The three Q's are For multiple tubes one rewrites these equations for successive pairs of tubes. In all cases the velocity is declining in the 2 flow direction, because each tube bleeds off some of the flow. One can build the device to the dimensions show in the article, and see that these predictions are observed.

Second, if the experienced workers in some facility you are new at are goading you into betting on something, don't bet very much. For gpm in a 3.