## Unit Conversion Factors

Mass 1 kg 1000 g 0.001 metric ton (tonne) 2.20461 lbm 1 lbm 453.593 g 0.453593 kg 5 x 10 ton 16 oz Force 1 N 1 kg m s2 105 dyn 105 g cm s2 0.22418 lb, 1 lb, 32.174 lbm ft s2 4.4482 N 4.4482 X 10s dyn Length 1 m 100 cm 106 p,m 1010 39.37 in. 3.2808 ft 1.0936 yd 0.0006214 mi 1 ft 12 in. 1 3 yd 0.3048 m 30.48 cm Volume 1 m3 1000 liters 106 cm3 35.3145 ft3 264.17 gal 1 ft3 1728 in.3 7.4805 gal 0.028317 m3 28.317 liters 28,317 cm3 Pressure 1 atm 1.01325 x 105 N m2 (Pa) 1.01325 bar 1.01325 x 106 dyn...

## Qkw

Figure 14-3 Two modes of settling. (a) Narrow particle size range (b) broad particle size range. 1. The large particles are hindered by the small particles, which increase the effective resistance of the suspending medium for the large particles. At the same time, however, the small particles tend to be dragged down'' by the large particles, so that all particles tend to fall at about the same rate (unless the size range is very large, i.e. greater than 6 1 or so). 2. The upward velocity of the...

## Problems

Water is flowing into the top of a tank at a rate of 200 gpm. The tank is 18 in. in diameter and has a 3 in. diameter hole in the bottom, through which the water flows out. If the inflow rate is adjusted to match the outflow rate, what will the height of the water be in the tank if friction is negligible 2. A vacuum pump operates at a constant volumetric flow rate of 10 liters min (l min) based upon pump inlet conditions. How long will it take to pump down a 100 L tank containing air from 1 atm...

## Cavitation And Net Positive Suction Head Npsh A Vapor Lock and Cavitation

As previously mentioned, a centrifugal pump increases the fluid pressure by first imparting angular momentum (or kinetic energy) to the fluid, which is converted to pressure in the diffuser or volute section. Hence, the fluid velocity in and around the impeller is much higher than that either entering or leaving the pump, and the pressure is the lowest where the velocity is highest. The minimum pressure at which a pump will operate properly must be above the vapor pressure of the fluid...

## Rv rrP r pgr559

The three components of this momentum equation, expressed in Cartesian, cylindrical, and spherical coordinates, are given in detail in Appendix E. Note that Eq. (5-59) is simply a microscopic (local) expression of the conservation of momentum, e.g., Eq. (5-40), and it applies locally at any and all points in any flowing stream. Note that there are 11 dependent variables, or unknowns in these equations (three vi's, six Tj's, P, and p), all of which may depend on space and time. (For an...

## 2

These three equations can be solved simultaneously (by iteration) for rj R, r2 R, and ri R. It is assumed that the size of the suspended drops is known as well as the density and viscosity of the liquids and the overall dimensions and speed of the centrifuge. IV. CYCLONE SEPARATIONS A. General Characteristics Centrifugal force can also be used to separate solid particles from fluids by inducing the fluid to undergo a rotating or spiraling flow pattern in a stationary vessel (e.g., a cyclone)...

## Compressors

A compressor may be thought of as a high pressure pump for a compressible fluid. By ''high pressure'' is meant conditions under which the compressible properties of the fluid (gas) must be considered, which normally occur when the pressure changes by as much as 30 or more. For ''low pressures'' (i.e., smaller pressure changes), a fan or blower may be an appropriate ''pump'' Figure 8-6 Effect of suction speed on stable operating window due to recirculation. (Numbers on the curves are the values...

## Gasliquid Twophase Pipe Flow

The two-phase flow of gases and liquids has been the subject of literally thousands of publications in the literature, and it is clear that we can provide only a brief introduction to the subject here. Although the single phase flow of liquids and gases is relatively straightforward, the two-phase combined flow is orders of magnitude more complex. Two-phase gas-liquid flows are also more complex than fluid-solid flows because of the wider variety of possible flow regimes and the possibility...

## 135

You want to set up a lab experiment to measure the velocity at which the model fluid flows down an inclined plane and scale this value to find the velocity of the glacier. (a) Determine the appropriate set of dimensionless groups. (b) Which of the above materials would be the best to use in the lab Why (c) What is the film thickness that you should use in the lab, and at what angle should the plane be inclined (d) What would be the scale factor between the measured velocity in the lab and the...

## 1

Where NRe0 dV0p and CDo 4gd(pS p) 3pVo are the Reynolds number and drag coefficient for a single particle in an infinite fluid. Data presented by Barnea and Mizrahi (1973) show that the swarm dimension-less groups NRe' and CD are related by the same expression as the corresponding groups for single particles, e.g., by the Dallavalle equation Thus, the settling velocity, or the terminal velocity of the swarm can be determined from NRe (14.42 + 1.827NA 2)1 2 - 3.798

## L

Of the conduit as was done for tube flow, a relationship can be determined between flow rate and driving force for laminar flow in a conduit with a noncircular cross section. This can also be done by application of the equivalent integral expressions analogous to Eqs. (6-6) to (6-10). The results for a few examples for Newtonian fluids will be given below. These results are the equivalent of the Hagen-Poiseuille equation for a circular tube and are given in both dimensional and dimensionless...

## 37

If the upstream pressure and flow rate are known, the downstream pressure (P2) can be found by rearranging Eq. (9-17), as follows which is implicit in P2. A first estimate for P2 can be obtained by neglecting the last term on the right (corresponding to the Weymouth approximation). This first estimate can then be inserted into the last term in Eq. (9-23) to provide a second estimate for P2, and the process can be repeated as necessary.

## 195 195 198

Non-Newtonian Fluids 205 III. FRICTION LOSS IN VALVES AND FITTINGS 206 B. Equivalent L D Method 207 D. 2-K (Hooper) Method 209 IV. NON-NEWTONIAN FLUIDS 214 V. PIPE FLOW PROBLEMS WITH FITTINGS 215 A. Unknown Driving Force 216 VII. PIPE NETWORKS 225 PROBLEMS 228 NOTATION 237 REFERENCES 238 A. Positive Displacement Pumps 239 II. PUMP CHARACTERISTICS 241 III. PUMPING REQUIREMENTS AND PUMP SELECTION 243 IV. CAVITATION AND NET POSITIVE SUCTION A. Vapor Lock and...

## 030

If the pressure drop across the valve is AP > APc, the value of APc is used as the pressure drop in the standard liquid sizing equation to determine Otherwise the value of Pj P2 is used. The notation used here is that of the Fisher Controls literature (e.g., Fisher Controls, 1990). The ANSI ISAS75.01 standard for control valves (e.g., Baumann, 1991 Hutchison, 1971) uses the same equations, except that it uses the notation FL (Km)1 2 and Ff rc in place of the factors Km and rc- For relatively...

## 4

Equation (6-11) is known as the Hagen-Poiseuille equation. This result can also be derived by equating the shear stress for a Newtonian fluid, Eq. (6-9), to the expression obtained from the momentum balance for tube flow, Eq. (6-4), and integrating to obtain the velocity profile Inserting this into Eq. (6-6) and integrating over the tube cross section gives Eq. (6-11) for the volumetric flow rate. Another approach is to use the Bernoulli equation Eq. (6-2) and Eq. (6-8) for the friction loss...

## Choked Flow

In isentropic flow (just as in isothermal flow), the mass velocity reaches a maximum when the downstream pressure drops to the point where the velocity becomes sonic at the end of the pipe (e.g., the flow is choked). This can be shown by differentiating Eq. (9-25) with respect to P2 (as before) or, alternatively, as follows For isentropic conditions, the differential form of the Bernoulli equation is Substituting this into Eq. (9-26b) gives This shows that when the mass velocity reaches a...

## Info

0 100 2 300 < 00 500 0 NO 800 500 1000 - T np a(ttt, rn O gfttJ Fahrenheit 0 100 2 300 < 00 500 0 NO 800 500 1000 - T np a(ttt, rn O gfttJ Fahrenheit I - TempcfaltKe, in Dt es FJhrenheil I - TempcfaltKe, in Dt es FJhrenheil Upper chart example The viscosity of sulfur dioxide gas (SO2) at 200 F is 0.016 cP. Lower chart example The viscosity of carbon dioxide gas (CO2) at about 80 F is 0.015 cP. Figure A-6 Viscosity of (a) gases and (b) refrigerant vapors. (From Crane Technical Paper 410,...

## K

Is called the friction velocity, because it is a wall stress parameter with dimensions of velocity. The parameters k and A in the von Karman equation have been determined from experimental data on Newtonian fluids in smooth pipes to be k 0.4 and A 5.5. Equation (6-26) applies only within the turbulent boundary layer (outside the buffer region), which has been found empirically to correspond to y+ > 26. Within the laminar sublayer the turbulent eddies are negligible, so

## Turbulent Macroscopic Convective Transport Models

The preceding transport laws describe the rate of transfer of heat, mass, charge, or momentum from one region of a continuum to another by virtue of molecular interactions only. That is, there is no actual bulk motion of material in the transport direction (y), which means that the medium must be stationary or moving only in the direction (x) normal to the transport direction. This means that the flow (if any) must be laminar i.e., all fluid elements move in straight, smooth streamlines in the...

## 9

The cyclone efficiency increases with V up to about 1.25 Vic, after which reentrainment results in a decrease in efficiency. A similar approach to the analysis of hydrocyclones was presented by Svarovsky (1984, 1990). He deduced that the system can be described in terms of three dimensionless groups in addition to various dimensionless geometric parameters. These groups are the Stokes number, the Euler number, which is equivalent to the loss coefficient, Kf, AP it2 A PD4 In each of these...

## Temperature Dependence Of Viscosity

All fluid properties are dependent upon temperature. For most fluids the viscosity is the property that is most sensitive to temperature changes. For liquids, as the temperature increases, the degree of molecular motion increases, reducing the short-range attractive forces between molecules and lowering the viscosity. The viscosity of various liquids is shown as a function of temperature in Appendix A. For many liquids, this temperature dependence can be represented reasonably well by the...

## 8

Figure 12-3 Particle in a centrifuge. Figure 12-3 Particle in a centrifuge. When the particle reaches its terminal (radial) velocity, dVr dt 0, and Eq. (12-7) can be solved for Vrt, (the radial terminal velocity) If NRe < 1, Stokes' law holds, and CD 24 NRe, in which case Eq. (12-8) becomes This shows that the terminal velocity is not a constant but increases with r, because the (centrifugal) driving force increases with r. Assuming that all of the fluid is rotating at the same speed as the...

## Pk pj

Where k cp cv is the specific heat ratio for the gas (for an ideal gas, cp cv + P M). If the density is eliminated from Eqs. (4-14) and (4-11), the result is which relates the temperature and pressure at any two points in an isentro-pic ideal gas. If Eq. (4-15) is used to eliminate T from Eq. (4-12), the latter can be integrated to give the pressure as a function of elevation which is a nonlinear relationship between pressure and elevation. Equation (4-15) can be used to eliminate P2 P1 from...

## 120

It is important to use the hydraulic diameter substitution (D Dh) in the appropriate (original) form of the dimensionless groups e.g., NRe DVp p, f ef (2LV2 D) and not a form that has been adapted for circular tubes (e.g., NRe 4Qp nDp). That is, the proper modification of the Reynolds number for a noncircular conduit is DhVp p, not 4Qp Dh . One clue that the dimensionless group is the wrong form for a noncircular conduit is the presence of which is normally associated only with circular...

## Positive Displacement Pumps

The term positive displacement pump is quite descriptive, because such pumps are designed to displace a more or less fixed volume of fluid during each cycle of operation. They include piston, diaphragm, screw, gear, progressing cavity, and other pumps. The volumetric flow rate is determined by the displacement per cycle of the moving member (either rotating or reciprocating) times the cycle rate (e.g., rpm). The flow capacity is thus fixed by the design, size, and operating speed of the pump....

## Horizontally Accelerating Free Surface

Consider a pool of water in the bed of your pickup truck. If you accelerate from rest, the water will slosh toward the rear, and you want to know how fast you can accelerate (ax) without spilling the water out of the back of the truck (see Fig. 4-4). That is, you must determine the slope (tan 6) of the water surface as a function of the rate of acceleration (ax). Now at any point within the liquid there is a vertical pressure gradient due to gravity Eq. (4-5) and a horizontal pressure gradient...

## The Venturi And Nozzle

There are other devices, however, that can be used to determine the flow rate from a single measurement. These are sometimes referred to as obstruction meters, because the basic principle involves introducing an obstruction (e.g., a constriction) into the flow channel and then measuring the pressure drop across the obstruction that is related to the flow rate. Two such devices are the venturi meter and the nozzle, illustrated in Figs. 10-2 and 10-3 respectively. In both cases the fluid flows...

## Bingham Plastics

The Bingham plastic model usually provides a good representation for the viscosity of concentrated slurries, suspensions, emulsions, foams, etc. Such materials often exhibit a yield stress that must be exceeded before the material will flow at a significant rate. Other examples include paint, shaving cream, and mayonnaise. There are also many fluids, such as blood, that may have a yield stress that is not as pronounced. It is recalled that a ''plastic'' is really two materials. At low stresses...

## Chemical Engineering Fluid Mechanics

Second Edition, Revised and Expended This book is printed on acid-free paper. 270 Madison Avenue, New York, NY 10016 tel 212-696-9000 fax 212-685-4540 Hutgasse 4, Postfach 812, CH-4001 Basel, Switzerland tel 41-61-261-8482 fax 41-61-261-8896 The publisher offers discounts on this book when ordered in bulk quantities. For more information, write to Special Sales Professional Marketing at the headquarters address above. Copyright 2001 by Marcel Dekker, Inc. All Rights Reserved. Neither this book...

## Fundamentals A Basic Laws

The fundamental principles that apply to the analysis of fluid flows are few and can be described by the conservation laws'' 2. Conservation of energy (first law of thermodynamics) 3. Conservation of momentum (Newton's second law) 4. The second law of thermodynamics 5. Conservation of dimensions (fruit salad'' law) 6. Conservation of dollars (economics) These conservation laws are basic and, along with appropriate rate or transport models (discussed below), are the starting point for the...

## 12PV12PV2NReQ145

T Mi(12V d) + t0 1 + NBi 12 1 Equation (11-45) could be used in place of the traditional Reynolds number for correlating the drag coefficient. Another approach is to consider the effective shear rate over the sphere to be V d, as was done in Eq. (11-25) for the power law fluid. If this approach is applied to a sphere in a Bingham plastic, the result is This is similar to the analysis obtained by Ainsley and Smith (see Chhabra, 1992) using the slip line theory from soil mechanics, which results...

## 51

You want to use a plate and frame filter to filter an aqueous slurry at a rate of 1.8 m3 per 8 hr day. The filter frames are square, with a length on each side of 0.45 m. The ''down time'' for the filter press is 300 s plus an additional 100 s per frame for cleaning. The filter operates with a positive displacement pump, and the maximum operation pressure differential for the filter is 45 psi, which is reached after 200 s of operation. (a) How many frames must be used in this filter to...

## The Speed of Sound

Sound is a small-amplitude compression pressure wave, and the speed of sound is the velocity at which this wave will travel through a medium. An expression for the speed of sound can be derived as follows. With reference Figure 9-1 Sound wave moving at velocity c. Figure 9-1 Sound wave moving at velocity c. to Fig. 9-1, we consider a sound wave moving from left to right with velocity c. If we take the wave as our reference, this is equivalent to considering a standing wave with the medium...

## Gravity Settling

Solid particles can be removed from a dilute suspension by passing the suspension through a vessel that is large enough that the vertical component of the fluid velocity is lower than the terminal velocity of the particles and the residence time is sufficiently long to allow the particles to settle out. A typical gravity settler is illustrated in Fig. 12-2. If the upward velocity of the liquid (Q A) is less than the terminal velocity of the particles (Vt), the particles will settle to the...

## Problem Solution

It should be emphasized that the specific relationship between the variables or groups that is implied in the foregoing discussion is not determined by dimensional analysis. It must be determined from theoretical or experimental analysis. Dimensional analysis gives only an appropriate set of dimen-sionless groups that can be used as generalized variables in these relationships. However, because of the universal generality of the dimen-sionless groups, any functional relationship between them...

## Qmr

Using the values of NRe,pl from step 2 and D from step 3, calculate the value of f and Kpipe) from Eq. (6-44), and the Kfit values from the 3-K equation. 5. Insert the K values into Eq. (7-50) to find a new value of D. 6. If the value of D from step 5 does not agree with that from step 3, use the value from step 5 to calculate a revised NRe,pl, and repeat steps 4-6 until agreement is attained. The procedure for a Bingham plastic is similar to the foregoing, using Eq. (6-62) for the pipe...

## Solution of High Speed Gas Problems

We will illustrate the procedure for solving the three types of pipe flow problems for high-speed gas flows unknown driving force, unknown flow rate, and unknown diameter. The unknown driving force could be either the upstream pressure, Pj, or the downstream pressure, P2. However, one of these must be known, and the other can be determined as follows. j. Calculate NRe DG pj and use this to findfj from the Moody diagram or the Churchill equation. 2. Calculate NMaj (G PJ)(RTj kM)j 2. Use this...

## Control Valves

Flow control is achieved by a control valve, which is automatically adjusted (opened or closed) continuously to achieve a desired flow rate. The valve is controlled by a computer that senses the output signal from a flow meter and adjusts the control valve by pneumatic or electrical signals in response to deviations of the measured flow rate from a desired set point. The control valve acts as a variable resistance in the flow line, because closing down on the valve is equivalent to increasing...

## Loss Coefficient

The total friction loss in an orifice meter, after all pressure recovery has occurred, can be expressed in terms of a loss coefficient, Kf, as follows. With reference to Fig. 10-12, the total friction loss is P1 P3. By taking the system to be the fluid in the region from a point just upstream of the orifice plate (P1) to a downstream position where the stream has filled the pipe (P3), the momentum balance becomes F m (V Vi ) 0 PjAo + P2(AI A0) P3Aj (10-19) k 1.3 approximately (for CO,. SO,....

## 40

(a) Determine the viscosity of this sample. (b) How would you describe the viscosity of this material (c) What model would be the most appropriate to represent this viscosity (d) Determine the values of the parameters in the model that fit the model to the data. 27. Consider each of the fluids for which the viscosity is shown in Fig. 3-7, all of which exhibit a typical ''structural viscosity'' characteristic. Explain why this is a logical consequence of the composition or ''structural makeup''...

## 23432 047569 028866

Downhill Stratified -1.3303 4.8081 4.17584 56.262 0.07995 0.50489 15.519 0.37177 0.39395 to a number of other relations that were checked against a variety of data. The Hughmark correlation is equivalent to the following expression for slip 1 - K + (1 - x) x)pff pL K(1 - x) x)pG Pl where the parameter K was found to correlate well with the diensionless parameter Z

## Centrifugal Pumps

The term centrifugal pumps'' is also very descriptive, because these pumps operate by the transfer of energy (or angular momentum) from a rotating impeller to the fluid, which is normally inside a casing. A sectional view of a typical centrifugal pump is shown in Fig. 8-1. The fluid enters at the axis or eye of the impeller (which may be open or closed and usually contains radial curved vanes) and is discharged from the impeller periphery. The kinetic energy and momentum of the fluid are...

## Momentum Equation In Spherical Coordinates

9vr v + v 9V.L+ v0 9vr v2 + v0 pl 9t vr 9r r 90 r sin 0 90 r I T + IF (r Trr) +-(Tr0 sin 0) 1 9tr0 T00 + T00 . + . a ,-- + pgr + v + v 9v + v0 9ve v v0cot0 9t vr 9r r 90 r sin 0 90 r r I -7F77+ 139(r Tre) + TTTTT ee sin0) + r 90 r2 9r r sin 0 90 r sin 0 9v0 9v0 ve 9v0 v0 9v0 v0vr , v0v0 + p - + vr - +---- +--r---- + ---1--- COt 0 ryz rzy HM rzx rxz HM

## Classification Of Materials And Fluid Properties

What is a fluid It isn't a solid, but what is a solid Perhaps it is easier to define these materials in terms of how they respond (i.e., deform or flow) when subjected to an applied force in a specific situation such as the simple shear situation illustrated in Fig. 3-1 (which is virtually identical to Fig. 1-1). We envision the material contained between two infinite parallel plates, the bottom one being fixed and the top one subject to an applied force parallel to the plate, which is free to...

## Pipe Flow Problems With Fittings

The inclusion of significant fitting friction loss in piping systems requires a somewhat different procedure for the solution of flow problems than that which was used in the absence of fitting losses in Chapter 6. We will consider the same classes of problems as before, i.e. unknown driving force, unknown flow rate, and unknown diameter for Newtonian, power law, and Bingham plastics. The governing equation, as before, is the Bernoulli equation, written in the form DF - + J2 ef +1 A( V2) (7-42)...

## Problems Pumps

The pressure developed by a centrifugal pump for Newtonian liquids that are not highly viscous depends upon the liquid density, the impeller diameter, the rotational speed, and the volumetric flow rate. (a) Determine a suitable set of dimensionless groups that should be adequate to relate all of these variables. You want to know what pressure a pump will develop with a liquid having an SG of 1.4 at a flow rate of 300 gpm using an impeller with a diameter of 12 in. driven by a motor running...

## Pipe Flow Problems

There are three typical problems encountered in pipe flows, depending upon what is known and what is to be found. These are the ''unknown driving force,'' ''unknown flow rate,'' and ''unknown diameter'' problems, and we will outline here the procedure for the solution of each of these for both Newtonian and non-Newtonian (power law and Bingham plastic) fluids. A fourth problem, perhaps of even more practical interest for piping system design, is the ''most economical diameter'' problem, which...

## Fluidsolid Twophase Pipe Flows

The conveying of solids by a fluid in a pipe can involve a wide range of flow conditions and phase distributions, depending on the density, viscosity, and velocity of the fluid and the density, size, shape, and concentration of the solid particles. The flow regime can vary from essentially uniformly distributed solids in a pseudohomogeneous (symmetrical) flow regime for sufficiently small and or light particles above a minimum concentration to an almost completely segregated or stratified...

## Dimensional Analysis and ScaleUp

In the steady flow of a Newtonian fluid through a long uniform circular tube, if NRe < 2000 the flow is laminar and the fluid elements move in smooth straight parallel lines. Under these conditions, it is known that the relationship between the flow rate and the pressure drop in the pipe does not depend upon the fluid density or the pipe wall material. (a) Perform a dimensional analysis of this system to determine the dimension-less groups that apply. Express your result in a form in...

## Problems Compressible Flow

ID gas pipeline carries methane (MW 16) at a rate of 20,000 scfm. The gas enters the line at a pressure of 500psia, and a compressor station is located every 100 mi to boost the pressure back up to 500psia. The pipeline is isothermal at 70 F, and the compressors are adiabatic with an efficiency of 65 . What is the required horsepower for each compressor Assume ideal gas. 2. Natural gas (CH4) is transported through a 6 in. ID pipeline at a rate of 10,000 scfm. The compressor stations...

## Turbulent Drag Reduction

A very remarkable effect was observed by Toms during World War II when pumping Napalm (a jellied solution of a polymer in gasoline). He found that the polymer solution could be pumped through pipes in turbulent flow with considerably lower friction loss than exhibited by the gasoline at the same flow rate in the same pipe without the polymer. This phenomenon, 3 fB(x) fB(x) 9 B A (x, y) dy dy + (x, B) (x, A) 3x)a(x) Ja(x) x x 3x Figure 6-5 Drag reduction data for polyacrylamide solutions (NRe,s...

## Newtons Law of Viscosity

Momentum is also a conserved quantity, and we can write an equivalent expression for the transport of momentum. We must be careful here, however, because velocity and momentum are vectors, in contrast to mass, energy, and charge, which are scalars. Hence, even though we may draw some analogies between the one-dimensional transport of these quantities, these analogies do not generally hold in multidimensional systems or for complex geometries. Here we consider the top plate to be subject to a...

## Cupand Bob Couette Viscometer

As the name implies, the cup-and-bob viscometer consists of two concentric cylinders, the outer cup and the inner bob, with the test fluid in the annular gap (see Fig. 3-2). One cylinder (preferably the cup) is rotated at a fixed angular velocity (Q). The force is transmitted to the sample, causing it to deform, and is then transferred by the fluid to the other cylinder (i.e., the bob). This force results in a torque (T) that can be measured by a torsion spring, for example. Thus, the known...

## Problems Flow Measurement

An orifice meter with a hole of 1 in. diameter is inserted into a 12 in. sch 40 line carrying SAE 10 lube oil at 70 F (SG 0.93). A manometer using water as the manometer fluid is used to measure the orifice pressure drop and reads 8 in. What is the flow rate of the oil, in gpm 2. An orifice with a 3 in. diameter hole is mounted in a 4 in. diameter pipeline carrying water. A manometer containing a fluid with an SG of 1.2 connected across the orifice reads 0.25 in. What is the flow rate in the...

## The Expansion Factor

The adiabatic flow equation Eq. 9-25 can be represented in a more convenient form as where pj PjM RTj, AP Pj P2, and Y is the expansion factor. Note that Eq. 9-37 without the Y term is the Bernoulli equation for an incompressible fluid of density pj. Thus, the expansion factor Y Gadiabatic Gincompressible is simply the ratio of the adiabatic mass flux Eq. 9-25 to the corresponding incompressible mass flux and is a unique function of P2 Pj, k, and Kf. For convenience, values of Y are shown in...

## Units and Dimensions

Determine the weight of 1 g mass at sea level in units of a dynes b lbf c gf d poundals. 2. One cubic foot of water weighs 62.4 lbf under conditions of standard gravity. a What is its weight in dynes, poundals, and gf b What is its density in lbm ft3 and slugs ft3 c What is its weight on the moon g 5.4 ft2 in lbf d What is its density on the moon 3. The acceleration due to gravity on the moon is about 5.4 ft s2. If your weight is 150 lbf on earth a What is your mass on the moon, in slugs b...

## Small Sample Of Ground Coal Is Introduced Into The Top Of A Coloumn Of Water

By careful streamlining, it is possible to reduce the drag coefficient of an automobile from 0.4 to 0.25. How much power would this save at a 40 mph and a b 60 mph, assuming that the effective projected area of the car is 25 ft2 2. If your pickup truck has a drag coefficient equivalent to a 5 ft diameter disk and the same projected frontal area, how much horsepower is required to overcome wind drag at 40 mph What horsepower is required at 70 mph 3. You take a tumble while water skiing. The...

## Darby Melson Equation

Note The ANSI pipe grades correspond approximately to Sched 20, 30, 40, 80, and 120 for commercial steel pipe. Note The ANSI pipe grades correspond approximately to Sched 20, 30, 40, 80, and 120 for commercial steel pipe. Figure 7-4 Cost of pump stations 1980 . Pump station cost CCPS A B hp e where A 172,800 and B 451 hp for stations of 500 hp or more. Figure 7-4 Cost of pump stations 1980 . Pump station cost CCPS A B hp e where A 172,800 and B 451 hp for stations of 500 hp or more. and J L D...

## Structural Viscosity Models

The typical viscous behavior for many non-Newtonian fluids e.g., polymeric fluids, flocculated suspensions, colloids, foams, gels is illustrated by the curves labeled structural in Figs. 3-5 and 3-6. These fluids exhibit Newtonian behavior at very low and very high shear rates, with shear thinning or pseudoplastic behavior at intermediate shear rates. In some materials this can be attributed to a reversible structure or network that forms in the rest or equilibrium state. When the material is...

## D

Where both ts and Do the pipe outside diameter are measured in inches. This relation between schedule number and pipe dimensions can be compared with the actual dimensions of commercial pipe for various schedule pipe sizes, as tabulated in Appendix F. 1. The manometer equation is A Apg Ah, where A is the difference in the total pressure plus static head P pgz between the two points to which the manometer is connected, Ap is the difference in the densities of the two fluids in the manometer, Ah...

## An Inclined Manometer Is Used To Measure The Pressure Drop Between Two Taps

The three-fluid manometer illustrated in Fig. 4-P11 is used to measure a very small pressure difference P1 P2 . The cross-sectional area of each of the reservoirs is A, and that of the manometer legs is a. The three fluids have densities pa, pb, and pc, and the difference in elevation of the interfaces in the reservoir is x. Derive the equation that relates the manometer reading h to the pressure difference P1 P2. How would the relation be simplified if A a 12. A tank that is vented to the...

## 000524

The most economical diameter is 1.5 m, or 59.2 in. The standard pipe size'' closest to this value on the high side or the closest size that can readily be manufactured would be used. A procedure analogous to the one followed can be used for non-Newtonian fluids that follow the power law or Bingham plastic models Darby and Melson, 1981 . For power law fluids, the basic dimensionless variables are the Reynolds number, the friction factor, and the flow index n . If...

## Conservation Of Dimensions

Physical laws, theories, empirical relations, etc., are normally expressed by equations relating the significant variables and parameters. These equations usually contain a number of terms. For example, the relation between the vertical elevation z and the horizontal distance x at any time for a projectile fired from a gun can be expressed in the form This equation can be derived from the laws of physics, in which case the parameteers a and b can be related to such factors as the muzzle...