Chemical Engineering Laboratory i(1)

8/3/2019 Chemical Engineering Laboratory i(1)

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CHEMICAL ENGINEERING LABORATORY ILABORATORIUM TEKNIK KIMIA I


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KIND OF EXPERIMENTS

1. Energy Losses In Pipes And Bends (FP1)

2. Friction Losses In Small Bore Pipe (FP2)

3. Piping System With Centrifugal Pump (KP1)

4. Centrifugal Pump Characteristic (KP2)

5. Fluid Measuring System (M)

6. Liquid-Solid Fluidized Bed (FZ1)

7. Gas-Solid Fluidized Bed And Heat Transfer (FZ2)

8. Drag Coefficient In Air Flow (DG)

9. Natural Convection Heat Transfer (NC)

10. Natural And Forced Convection And Radiation Heat Transfer (NF)11. Double Pipe Heat Exchanger (HE1)

12. Shell And Tube Heat Transfer (HE2)

13. Condensing Vapor Heat Transfer (CV)

14. Liquid-Liquid Mixing (MX1)

15. Solid-Solid Mixing (MX2)

16. Compressible Flow Through Constant Area Conduit (CF1)

17. Compressible Flow Through Convergent-Divergent Nozzle (CF2)


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1. ENERGY LOSSES IN PIPES AND BENDS

ObjectiveTo study the variation in friction factor, f, used in the Darcy Formula with the Reynolds number in

both laminar and turbulent flow. The friction factor will be measured as a function of Reynolds

number and the roughness will be calculated using the Colebrook equation.

TheoryThe loss of head resulting from the flow of a fluid through a pipeline is expressed by the Darcy

Formula

2

f2

l

L Vh

D g 0.1

where hf is the loss of head (units of length) and the average velocity is V. The friction factor, f,

varies with Reynolds number and a roughness factor.

Laminar flow

The Hagen-Poiseuille equation for laminar flow indicates that the head loss is independent ofsurface roughness.

2

32l

LVh

gD

0.2

Thus in laminar flow the head loss varies as V and inversely as D2. Comparing equation 0.1 and

equation 0.2 it can be shown that

64 64f

VD R

0.3

indicating that the friction factor is proportional to viscosity and inversely proportional to the velocity,pipe diameter, and fluid density under laminar flow conditions. The friction factor is independent ofpipe roughness in laminar flow because the disturbances caused by surface roughness are quicklydamped by viscosity.

Equation 0.2 can be solved for the pressure drop as a function of total discharge to obtain

4

128 LQp

D

0.4

Turbulent flow

When the flow is turbulent the relationship becomes more complex and is best shown by means ofa graph since the friction factor is a function of both Reynolds number and roughness. Nikuradse

showed the dependence on roughness by using pipes artificially roughened by fixing a coating ofuniform sand grains to the pipe walls. The degree of roughness was designated as the ratio of the sand

grain diameter to the pipe diameter (/D).The relationship between the friction factor and Reynolds number can be determined for every

relative roughness. From these relationships, it is apparent that for rough pipes the roughness is moreimportant than the Reynolds number in determining the magnitude of the friction factor. At high

Reynolds numbers (complete turbulence, rough pipes) the friction factor depends entirely onroughness and the friction factor can be obtained from the rough pipe law.

1 3.72log

f

D

0.5

For smooth pipes the friction factor is independent of roughness and is given by the smooth pipe law.


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1 Re f 2log

2.51f

0.6

The smooth and the rough pipe laws were developed by von Karman in 1930.Many pipe flow problems are in the regime designated transition zone that is between the smooth

and rough pipe laws. In the transition zone head loss is a function of both Reynolds number androughness. Colebrook developed an empirical transition function for commercial pipes. The Moodydiagram is based on the Colebrook equation in the turbulent regime.

1 2.512log

3.7f Re f

D

0.7

The Colebrook equation can be used to determine the absolute roughness, , by experimentallymeasuring the friction factor and Reynolds number.

1

2 f2.51

3.7 10 Re fD

0.8

Alternatively the explicit equation for the friction factor derived by Swamee and Jain can be solvedfor the absolute roughness.

2

0.9

0.25f

5.74log

3.7 ReD

0.9

When solving for the roughness it is important to note that the quantity in equation 0.9 that is

squared is negative!

-1

2 f

0.9

5.743.7 10

ReD

0.10

Equations 0.8 and 0.10 are not equivalent and will yield slightly different results with the

error a function of the Reynolds number.

Other correlation for estimating Darcy fare following.

Serghides Equation (for Re>2100 and any e/D)

f= [A [(B-A)2/(C-2B+A)]]-2

A = -2.0 log((e/D)/3.7 + 12/Re)B = -2.0 log((e/D)/3.7 + 2.51A/Re)

C = -2.0 log((e/D)/3.7 + 2.51B/Re)

Moody Equation (4000


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Churchill Equation (for all values of Re and e/D)

f= 8((8/Re)12 + 1/(A+B)1.5)1/12

A = (-2.457ln((7/Re)0.9 + 0.27e/D))16

B = (37530/Re)16

Chen Equation (for all values of Re and e/D)

1/(f)1/2 = -2.0log((e/D)/3.7065 5.0452A/Re)

A = log((e/D)1.1098/2.8257 + (5.8506/Re0.8981))

Zigrang and Sylvester Equation (for 4000


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4. CENTRIFUGAL PUMP CHARACTERISTIC

A centrifugal pump converts the input power to kinetic energy in the liquid by accelerating the liquid

by a revolving device - an impeller. The most common type is the volute pump. Fluid enters the pumpthrough the eye of the impeller which rotates at high speed. The fluid is accelerated radially outwardfrom the pump chasing. A vacuum is created at the impellers eye that continuously draws more fluidinto the pump.

The energy created by the pump is kinetic energy according the Bernoulli Equation. The energytransferred to the liquid corresponds to the velocity at the edge or vane tip of the impeller. The fasterthe impeller revolves or the bigger the impeller is, the higher will the velocity of the liquid energytransferred to the liquid be. This is described by the Affinity Laws.

Pressure and HeadThe kinetic energy of a liquid coming out of an impeller is obstructed by the pump casing whichcatches the liquid and slows it down. When the liquid slows down the kinetic energy is converted topressure energy. In Newtonian fluids the term head is used to measure the kinetic energy which apump creates. Head is a measurement of the height of the liquid column the pump creates from thekinetic energy the pump gives to the liquid.

The main reason for using head instead of pressure to measure a centrifugal pump's energy is that the

pressure from a pump will change if the specific gravity of the liquid changes, but the head will not.

The maximum head of a pump is mainly determined by the outside diameter of the pump's impellerand the speed of the rotating shaft. The head will change as the capacity of the pump is altered.

The pump's performance on any Newtonian fluid can always be described by using the term head.There are different term associated to pump head, such as: Total Static Head, Total Dynamic Head(Total System Head), Static Suction Head, Static Suction Lift, Static Discharge Head, Dynamic

Suction Head/Lift, and Dynamic Discharge Head.

Suction Head

Low pressure at the suction side of a pump can encounter the fluid to start boiling with reducedefficiency, cavitations, and even damage of the pump as a result. Boiling starts when the pressure inthe liquid is reduced to the vapor pressure of the fluid at the actual temperature.

Based on the Energy Equation - the suction head in the fluid close to the impeller can be expressed as

the sum of the static and the velocity head:


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hs = ps / + vs2

/ 2 g (1)

where hs = suction head close to the impeller, ps = static pressure in the fluid close to the impeller, =specific weight of the fluid, vs = velocity of fluid, and g = acceleration of gravity.

Liquids Vapor Head

The liquids vapor head at the actual temperature can be expressed as:

hv= pv / (2)

where hv= vapor head, pv= vapor pressure

Net Positive Suction Head - NPSH

The Net Positive Suction Head - NPSH - can be expressed as the difference between the Suction Headand the Liquids Vapor Head and expressed like

NPSH = hs - hv (3)

or NPSH = ps / + vs2

/ 2 g - pv / (3a)

Available NPSH - NPSHa

The Net Positive Suction Head made available at the suction system for the pump is often

named NPSHa. The NPSHa can be determined during design and construction, or determined

experimentally from the actual physical system.

For a common application - where the pump lifts a fluid from an open tank at one level to an other,the energy or head at the surface of the tank can be expressed as:

h0= hs + hl (4)

where h0 = head at surface, hs = head before the impeller, hl= head loss from the surface to impeller.

In an open tank, the head at the surface can be expressed as:

patm / = ps / + vs2

/ 2 g + he + hl (4a)

where he = elevation from surface to pump - positive if pump is above the tank, negative if the pump isbelow the tank.

The head available before the impeller can be expressed as:

ps / + vs2

/ 2 g = patm / - he - hl (4b)

or as the available NPSHa:

NPSHa = patm / - he - hl - pv / (4c)

Required NPSH - NPSHr

The NPSHr, called as the Net Suction Head as required by the pump in order to prevent

cavitation for safe and reliable operation of the pump. The required NPSHrfor a particular

pump is in general determined experimentally by the pump manufacturer and a part of thedocumentation of the pump.

The available NPSHa of the system should always exceeded the required NPSHrof the pump to avoidvaporization and cavitation of the impellers eye. The available NPSHa should in general be significanthigher than the required NPSHrto avoid that head loss in the suction pipe and in the pump casing,local velocity accelerations and pressure decreases, start boiling the fluid on the impeller surface.Note that the required NPSHrincreases with the square capacity.


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Regulating discharge flow capacityThe capacity of a centrifugal pump can be regulated either at constant speed, or varying speed.

Capacity Regulating by Varying Speed

Speed regulating is energy efficient since the energy to thepump is reduced with the decrease of speed. The change inpower consumption, head and volume rate can beestimated with the affinity laws.

Capacity Regulating by Constant SpeedCapacity can be regulated at constant speed by throttling, bypassing flow, changing impeller diameter,

or modifying the impeller.

ThrottlingThrottling can be carried out by opening and closing a

discharge valve. Throttling is energy inefficient since the

energy to the pump is not reduced. Energy is wasted by

increasing the dynamic loss.

Bypassing Flow

The discharge capacity can be regulated by leading a part ofthe discharge flow back to the suction side of the pump.

Bypassing the flow is energy inefficient since the energy tothe pump is not reduced.

Changing the Impeller DiameterReducing the impellers diameter is a permanent change and the method can be used where the change

in process demand is temporary. The method may be energy efficient if the motor is changed and theenergy consumption reduced.

Modifying the Impeller

The flow rate and the head can be modulated by changing the pitch of the blades. Complicated andseldom used.


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For non-cavitating flow of an incompressible fluid through a pump of specified geometry, thepressure rise can be expressed as

DNQfP ,,,, (5)

Where P = pressure rise across the pump (N m-2)Q = volumetric flow rate (m

3s

-1)

= fluid density (kg m-3)

= fluid viscosity (N s m-2

)N = rotational speed (radians/s)D = a typical dimension such as diameter.

Using dimensional analysis one can obtain

),(2

3122

ND

ND

Qf

DN

P

(6)

The dimensionless pressure rise is thus related to the dimensionless flow rate and a Reynolds number,Re, defined as

DND..Re

For a turbo machine, it is found that if Reynolds number is greater than some critical value, then thedimensionless pressure rise becomes independent of the Reynolds number.

The apparatus consists of a simple flow loop as shown in the Figure ?.1. The pump is of turbineimpeller type with directly coupled to the motor. The rotational speed of the motor is 2880 rpm. Theflow rate could be controlled by valve A, valve B, or both.

The mass flow rate can measured from the time required to collect a certain volume of liquid in the

top tank or by reading the liquid level in the top tank (for higher flow rate).

Static pressure tapings are placed at the suction side, P1, and discharge side, P2, of the pump. Thepresence of bubbles in the suction flow can be observed through the sight glass C.

Procedure.

Valve B

Valve A

P1

P2

Sight Glass C

Bottom Tank

Top Tank

Pump

Figure ?.1.


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1. Before starting the pump, make sure that the bottom tank has enough water. The pH of the wateris adjusted to slightly alkaline to reduce corrosion and a dye is added so the water color turns red.

2. Open fully both valve A and B. Turns on the pump and wait for a few minutes to get a steadyflow. Adjust the discharge valve B to avoid overfilling the top tank.

Data collection.You are required to collect data of1. discharge pressure rise against flow rate at three different values of suction pressure.

2. flow rate against decreasing suction pressure at some constant discharge pressure. Adjust bothvalve A and B to control.

Do not forget to note the sound of the pump, the presence of bubble in the flow, thestability of pressure readings, the difficulty of controlling the pressures, and so on.

Data analysis1. Confirm that eq.6 is valid for flow in the absence of cavitations and for Reynolds number above a

critical value its value become unimportant.Using SI unit calculate

2

322and,,

ND

ND

Q

DN

P

Plot the first and the second dimensionless groups against the last for your experimental data. Canyou observe the effect of cavitations?

Theory suggests that a simplification will roughly yield parabolic curve with an intercept of a

little bit higher than 0.125 on the22DN

P

axis. Does your experimental data confirm this?

2. Calculate Suction Specific Speed,

43

21

a

SSNPSH

QN

against suction pressure at some constant discharge pressure. A rule of thumb says that to avoidcavitations, the suction specific speed should be below 5200 (Metric Unit). Does yourexperimental data confirm this?


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5. FLUID MEASURING SYSTEM

THE VENTURI METER

OBJECTIVE:

1. Experimental verification of the Bernoulli equation.

2. To measure the discharge and to investigate the characteristic of a Venturi Meter, Orifice,and Rotameter.

Description of the main component of the set-up:

The Venturi meter used in this experiment consists of successive converging, uniform and

diverging sections equipped with pressure taps at selected locations See Fig. 1 for a

schematic diagram.

Experimental method:

1 2 3 4 5, 6 7

8

9


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A flow rate is supplied through the test section. After steady state flow is achieved the flow

rate is measured by a flow meter. The pressure variation along the test section is measured by

the use of piezometer tubes. The experiment is repeated for two more flow rates.

THEORY

(a). Bernoulli Equation

Assume that the following apply:

Steady flow Incompressible flow Flow along a streamline No frictional forces

The Bernoulli equation, which expresses the principle of conservation of linear momentum

under special circumstances, can written as :

2

2

221

2

11

22gz

VPgz

VP

(1)

In addition, let assume 21 zz . Therefore, we have

22

2

22

2

11 VPVP

(2)

Let us consider the continuity equation (conservation of mass) and limit it to the same

assumption as above. Then we have

2211 AVAV

where

4

2

11

DA

4

2

2

2

DA (3)

2

2

1

1

2

1

2

D

D

A

A

V

V(4)

Solving forP2

from equation (2)

2

1

2

1

2

2

112

V

V

VPP

Substituting forV2

/ V1

from equation (4)


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2

1

4

2

1

2

112

D

D

VPP

Noting that

11 AVm

11

A

mV

21

2

1A

mV

4

2

1

2

1

2

12 1

2 D

D

A

mPP

(5)

orpressure at any location x is given by

4

1

2

1

2

1 12 xD

D

A

mPxP

(6)

or, since

g

xPxh

)()(

one can write

4

1

2

1

2

2

1)(

12

)(xD

D

Ag

mhxh

(7)

Hence the pressure head at any location along the test section can be expressed in term of the

pressure at a reference location (inlet).

(b). Venturi meter

Assuming that there is no loss of energy along the pipe, and that the velocity and


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piezometric heads are constant across each of the sections considered, then Bernoullis

theorem states that

(8)

Where u1, u2 and un are the velocities of flow through section 1, 2, and n. The equation of

continuity is

(9)

Q denotes the volume flow or discharge rate.

Substituting in equation (8) for u1 from equation (9)

(10)

And solving this equation for u2 leads to

(11)

So that the discharge rate, from equation becomes:

(12)

In practice, there is some loss of energy between sections 1 and 2, and the velocity is not

constant across either of these sections. Consequently, measured values of Q usually fall

a little short of those calculated from equation (3) and it is customary to allow for this

discrepancy by writing:

(13)

C is known as the coefficient of the meter, which may be established by experiment. Its

value varies slightly from one meter to another and even for a give meter it may vary

slightly with the discharge, but usually lies within the range of 0.92 to 0.99.

(c). Orifice meter


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The governing equation is similar to eq.(13) of the venturi meter, but with different values of

coefficient and characteristic.

(14)

Experimental Procedure

1. Connect the Venturi Meter apparatus to the water supply line.

2. Adjust the water flow rate by the flow control valve until a steady reading of 10

liters/min is observed by the use of the flow meter.

3. Ensure that the water heights in the piezometer tubes become stabilized.

4. Read and record these heights in mm.

5. Change the flow rate to 12 liters/min and repeat steps 3, 4.

6. Change the flow rate to 14 liters/min and repeat steps 3, 4.

Lab Report Requirements

1. Determine the mass flow rate for each run.

2. Perform the calculations for each flow-rate and compare with the measured values.

a. Evaluate xh for each pressure location using the corresponding diameter xD .b. Record the xh values in Table 2.

3. Calculate % error between theoretical and experimental results and list it in the

appropriate row in Table 2.

5 6 7


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Note: %100%exp

exp

h

hhERROR

calor %100%

cal

cal

h

hhERROR

exp

depending on which of and you consider to be more accurate. You must give a

justifying statement as to which one you consider to be more accurate.

4. Plot xh vs. x for both theoretical and experimental data for the three different flowrates separately, i.e. each graph should include a set of theoretical and a set of

experimental data points.

5. Considering the assumptions given in the beginning of this write-up, is it all proofed

to be true? Explain.

6. For Venturi meter and Orifice meter, plot Cvs. NRe and compare with data on the

textbook. Give comments.

7. Plot the head loss for the rotameter vs. fluid velocity.

Note: For theoretical pressure calculations for stations 2, 3, 4, 5, 6, 7, 8, and 9 the

distance is measured from station 1, and the measured pressure at station 1 must

be assumed to be the inlet pressure.

Table 1

Experimental values of pressure as a function of flow-rate: m in (kg/sec) and h in

(mm of water).

Q

mh

1

h2

h3

h4

h5

h6

h7 8h 9h

Table 1a

Corrected experimental values of pressure as a function of flow-rate: m in (kg/sec)

and h in (mm of water).

Q m h1 h2 h3 h4 h5 h6 h7 8h 9h

Table 2

Theoretical values of pressure as a function of flow-rate: m in (kg/sec) and h in (mmof water).


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Q m h1 h2 h3 h4 h5 h6 h7 8h 9h

%error

in h

%

error

in h

%

error

in h

Figure1

Schematic diagram of the Venturi meter

Table 3Pressure tap locations and diameters at pressure tap locations

Pressure

location

A B C D

Diameter

(mm)

26 16 26

x (mm) 0 46 156

A

BC

D


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6. LIQUID-SOLID FLUIDIZED BED


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7. GAS-SOLID FLUIDIZED BED AND HEAT TRANSFER

OBJECTIVES Observe and measure fluidization of a solid bed.

Measure the pressure loss across the fluidized bed

Study heat transfer in a fluidized bed.

INTRODUCTIONPacked and fluidized beds play a major role in many chemical engineering processes.

Packed-bed situations include such diverse processes as filtration, wastewater treatment, and

the flow of crude oil in a petroleum reservoir. In these cases, the interest centers on the

pressure drop through the bed as a function the volumetric flow rate or superficial velocity.

If the particles in the bed are loose and there is sufficient volume in the device containing the

particles, the particles may fluidize at high flow rates. Such beds inherently possess excellent

heat transfer and mixing characteristics. In the study of the fluid-mechanical behavior of

these beds, the focus is on the incipient fluidization velocity and the dependence of bedexpansion on the superficial velocity.

The term fluidized bed describes a finely granulated layer of solid material (referred to as the mass)that is loosened by fluid flowing through to such an extent that the particles of solid material are freeto move to a certain degree. It is called fluidized because the solid material takes on propertiessimilar to those of a fluid. Fluidized beds are used widely in engineering for applications such as

combustion, the cracking of high-molecular-weight petroleum fractions, drying plants, andpowder coating.

To characterize a fluid bed, the pressure loss of the fluid flowing through the bed can be used. As the

fluid flows through the solid material, the pressure below the mass initially rises with increasing airspeed. This occurs until the pressure forces match the weight of the mass and the material becomes

suspended. At this point, the layer reaches a fluid state. With further increasing flow rate, the pressureloss is almost constant. After a certain flow rate, the top particles no longer fall back into the fluidized

bed; they are drawn along with the fluid flow and removed.

The characteristics of the transfer of heat from a heated body to the surroundings also change on the

formation of a fluidized bed. In the solid bed, the transfer of heat is determined largely by the verylow conductivity of the mass of particles. Part of the heat is removed by the fluid flow; therefore, the

heat transfer slowly increases with fluid flow. However, once the particles are in motion, the heattransfer is defined by the moving particles. Due to the higher specify heat capacity of the particlematerial, the heat transfer increases significantly. This allows for an extremely even temperature in

the fluidized bed.

THEORY

The theory for this experiment is covered in Chapter 7 of the 4th Edition of McCabe, Smith,

and Harriott (M,S,&H). The following material is a condensation of that chapter as it relates

to the experiment at hand. As an aid to you, some specific equations in M,W,&H are referred

to. The 5th Edition was recently published and is considerably revised and where possible,

pertinent equations from that edition are given as well. However, generally speaking, the 4th

Edition will probably be more helpful.

There are three areas of interest to us: (1) Relationship between the pressure drop and the

flow rate; (2) Minimum fluidization velocity, and; (3) Behavior of the expanded bed.

(1) Relationship between the pressure drop and the flow rate


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The flow of a fluid, either liquid or gas, through a static packed bed can be described in a

quantitative manner by defining a bed friction factor, fp, and a particle Reynolds number,

NRe,p, as follows:

(1.1)

(1.2)

where

P = pressure drop across the bedL = bed depth or length

gc = conversion constant (= unity if SI units are used)

Dp = particle diameter

= fluid density= bed porosity or void fractionVo = superficial fluid velocity

= fluid viscositys = sphericity

The friction factor and the Reynolds number are unitless. Some typical sphericity factors are

given in McCabe, Smith and Harriott (4th Ed.: p. 750, Table 26-1; 5th Ed.: p. 928, Table

28.1).

For laminar flow, where only viscous drag forces come into play, NRe,p 20, experimentaldata may be correlated by means of the Kozeny-Carman equation:

(1.3)

(From combining Eqns. (7 - 21) and (7 - 23) MS& H with Eq.(2) above)

Note: According to Yates ("Fundamentals of Fluidized-bed Chemical Processes," by J. G.

Yates, Published by Butterworths, 1983, p. 7-8) the factor of 150 was originally given by

Carman as 180 for the case of laminar flow. Ergun later suggested a better value was 150

when the particles are greater than about 150 m in diameter.

For highly turbulent flow where inertial forces predominate, NRe,p 1000, experimentalresults may instead be correlated in terms of the Blake-Plummer equation:

(1.4)

(From combining Eqs. (7 - 24) and (7 - 21). Also, related to Eq. (7- 20) in the 5th Ed.)

While both equations (3) and (4) have a sound theoretical basis, Ergun empirically found that

the friction factor could be described for all values of the Reynolds number by simply adding

the right-hand sides of equations (3) and (4). Thus:

(1.5)

(2) Minimum fluidization velocity


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At a sufficiently high flow rate, the total drag force on the solid particles constituting the bed

becomes equal to the net gravitational force and the bed becomes fluidized. For this situation

a force balance yields:

(1.6)

whereM = void fraction at the minimum fluidization velocityA = cross-sectional area of the bed

p = particle densityg = gravitational constant

M = total mass of packing.

This is Eq. (7-50) on Page 149 of the 4th Ed. and Eq. (7.48) of the 5th. The superficial fluid

velocity at which the fluidization of the bed commences is called the incipient or minimum

fluidization velocity, V0M . The incipient fluidization velocity may be determined by

combining equations (1), (3), and (6) with the following result [Eq. (7-54), Page 149 of the4th Ed. and Eq. (7.52) of the 5th]:

(1.7)

This equation is the basis for some empirical equations found in the literature. The terms can

be grouped as follows:

(1.8)

The first factor contains the sphericity of the particles and the bed porosity at the point of

incipient fluidization. Neither of these factors is usually known with a high degree of

accuracy. If spheres are assumed s 1and a reasonable value of voidage, say M 0.4 , thenthe first factor is 0.00071. The factor is quite sensitive to M . For example, ifM 0.413, thenthe factor is 0.0008.

One investigator, [D. Geldhart, "Types of Fluidization," Powder Technology, 7 (1973), 285-

292; Geldhart and Abrahamsen, Powder Technology, 19 (1978), 133-136] simply determined

the first factor from his data and actually found 0.0008 to be the best value; that is, he

reported the following correlation:

(1.9)

Behavior of the expanded bed

The expansion of fluidized beds is discussed in the text on Pages 152-156 of the 4th Ed. and

Pages 170-173 of the 5th Ed. The treatment to be used here is slightly different. For fluid

velocities exceeding the incipient fluidization velocity, the bed expands. The porosity, , of

an expanded bed may be related to the superficial fluid velocity, , by means of an empiricalrelation suggested by Richardson and Zaki (1,2):


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(1.10)

where ut is the terminal velocity of a spherical particle in a fluidizing medium (3). The

exponent, n, depends on the flow conditions -- that is, on the Reynolds number. Thus:

NRe,p 0.2 (11)0.2 NRe,p 1.0 (12)

1NRe,p 500 (13)NRe,p 500 (14)

Because the terminal velocity, ut, is a constant for a given particle, it can be seen that

Equation (10) above is essentially the same as the empirical equation in the text; namely Eq.

(7-61), p. 152 of the 4th Ed. and Eq. (7.59), p. 171 of the 5th Ed.

The void fraction of the expanded bed, , is related to that at incipient fluidization by thefollowing equation:

This is Eq. (7 - 60) of the 4th Ed. and Eq. (7.58) of the 5th. (7-60) where LM and M are thebed height and void fraction at incipient fluidization, and L is the measured height of the

expanded bed. Therefore, since LM and M are known, can be calculated from the measuredheight, L, of the expanded bed.

In Equations (11)-(14) the Reynolds number is based on the particle diameter, Dp, and theterminal velocity, ut. Therefore it is necessary to know the terminal velocity. By means of a

force balance it is shown that the terminal velocity for spherical particles is:

(15)

(Eq. (7 -39), p. 142 of the 4th Ed. and Eq. (7.37), p. 159 of the 5th.

where CD denotes the drag coefficient.

A graph of CD versus NRe,p is shown in the text (Figure 7-3, p. 131 of the 4th Ed. andFigure 7.6, p. 158 of the 5th Ed.).

Thus, to find CD, you need to know ut in order to calculate NRe,p. One could do this by

trial-and-error. Thus, you could guess ut, calculate NRe,p, look up CD on the graph, and put the

resulting value in Eq. (15). If the calculated value of u t did not match the guess, you would

guess again.

However, the trial-and-error can be avoided. Square both sides of Eq. (15) and utilize the

definition of NRe,p (Eq. (2)). One obtains:

(16)


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All parameters on the right are known. This suggests that a plot of versus NRe,p can

be constructed and used to avoid the trial-and-error procedure.

The plot is prepared in the following way. Pick a series of point coordinates off the plot

shown above. Some examples for spheres are:

NRe,p CD

.001 22,000 22

.01 2,200 22

.1 220 22

1,000 0.48 480

Pick off a dozen similar pairs. Then plot CDNRe,p as the ordinate against corresponding NRe,p as

the abscissa. For each bed, calculate from Eq. (16). From your plot read the

corresponding NRe,p . Then use Eq. (2) to calculate ut.

Heat Transfer Coefficient

The heat transfer coefficient can be calculated from:

where Pis the heater power, Tis the difference between the heater and bed temperatures, and is

the surface area of the heater. The heater surface area is the area of the cylindrical surface and theopen face.

EQUIPMENT AND PROCEDUREIn this experiment the friction factor will be measured as a function of Reynolds number for

the flow of air through a bed of solid particles. Experimental results will be compared with

theoretical predictions for the appropriate flow regimes. A flowsheet of the experimental set-

up is depicted schematically in Figure 1 at the end of this handout. The equipment includes

two transparent beds, rotameters, manometers, a source of low pressure air, and appropriate

valves and fittings.

Measure the bed height after tapping the bed gently until no further change is observed.

Close Valves B and C. Open Valve A. Control the flow of air through the system by

manipulating Valve B or C depending on the rotameter used. Open Valve D for pressure drop

measurements. Increase the flow rate of air in small steps noting the rotameter and

manometer readings until the bed is fluidized and the pressure drop does not change

appreciably. Also, record the corresponding bed height at each flow rate. Continue themeasurements until the bed is appreciably fluidized and obtain at least fifteen readings in the

packed-bed region and ten readings in the fluidized-bed region. Decrease the flow rate noting

the flow rate and pressure drop values.

The region where fluidization just begins -- namely, at the "minimum fluidization velocity" --

is of special interest. In Figure 2, it occurs at Point B. [This plot is based upon one in "Design

for Fluidization, Part 1," by J. F. Frantz, Chemical Engineering, September 17, 1962, pp. 161-


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178.] At this point, where the pressure drop through the fixed bed becomes equal to the

weight of the bed per unit area, a slight rearrangement of particles occurs and the particles

shift position so as to present maximum flow area to the gas. Quoting from the article above:

"This causes a slight decrease in pressure drop, and channeling occurs. Only at

a higher gas velocity does the entire bed become fully supported by the gas

stream. Leva, Shirai and Wen realized this phenomenon took place, and thusthe defined minimum fluidization velocity as a gas velocity 10% greater than

the point at which, with increasing velocity, the pressure drop through the fixed

bed first equals the weight of the bed per unit area."

In view of this, repeat the run to check for reproducibility, making sure that you get sufficient

data in the somewhat ill-defined region between Points B and D.

The theory of this experiment is built around the assumption that at "steady state," the

particles are uniformly distributed in the bed. As you increase the flow rate of air, take some

notes concerning what the bed looks like at various stages. This may help to explain some

discrepancies between measured and theoretical values.

Repeat the measurements on the other bed by reconnecting the appropriate lines.In a separate experiment, the porosity of a container of solid particles was measured using the

method of water displacement. This involves first weighting a measured volume of dry

particles. Water is then slowly added to the particles until the upper surface is wet. The

weight of the water added can then be used to calculate the porosity of the bed. This porosity

corresponds to the value, M , in Equation (6).

EXPERIMENTAL PROCEDURE

A. Setup

1. Check that the air compressor valve is closed.

2. Switch on the fluidized bed equipment.3. Close the valve V1 (which controls flowrate).4. Open the valve on the air compressor.

B. Experiment 1. Measuring the pressure loss.1. Slide the pressure sensor all the way to the bottom of the bed.

2. Adjust the flow rate to 10 L/min. using valve V1.3. Record the flow rate and pressure.4. Increase flow rate by steps of 2 L/min.

5. Continue to record flow rate and pressure.6. As soon as the first particle movements are seen, the loosening speed has been reached. Note

the flow rate at this point.7. Continue increasing the flow rate by steps of 5 L/min. until a flow rate of 70 L/min. isreached.

8. Now reduce the flow rate by steps of 5 L/min. until a flow rate of 10 L/min. is reached.9. Plot the pressure loss versus fluid velocity.

C. Experiment 2. Measuring the pressure distribution in the fluid bed.1. Adjust the flow rate to 40 L/min. using valve V1.2. Record the pressure.3. Raise the pressure sensor 10 mm and record the pressure.

4. Repeat measurements until the pressure sensor has reached the surface.5. Graph pressure loss versus height and compare with theoretical.

D. Measuring Heat transfer


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1. Adjust the heater such that it is around 30 mm deep in the mass.2. Adjust the heater power to 30 W. (Note: adjust by reading the ampere and the voltage; the

scale around the control knob is not in units of W)3. Adjust flow rate to 10 L/min.4. Wait about 5 minutes until the heater has stabilized.5. Read and note the heater temperature T1 and the fluid bed temperature T2.6. Increase flow rate by 10 L/min.7. Wait about 2 minutes. Then read the temperature three times at one-minute intervals. Record

the average of the three temperatures.8. Repeat steps 6 and 7 until the flow rate is 70 L/min. (Note: you may have to adjust the heater

power)

9. Calculate the heat transfer coefficient and graph versus fluid velocity.

E. Heat transfer as function of immersion depth.1. Adjust flow rate to 90 L/min.

2. Adjust the height of the heater such that it is just above the surface.3. Wait about 5 minutes until the heater has stabilized.

4. Read and note the heater temperature T1 and the fluid bed temperature T2.5. Lower the heater in 10 mm steps.6. Wait about 2 minutes. Then read the temperature three times at one-minute intervals. Record

the average of the three temperatures.7. Repeat steps 6 and 7. (Note you may have to adjust the heater power)

8. Calculate the heat transfer coefficient and graph versus depth.

Calculations1. For each bed, plot the measured pressure drop (in cm of H2O) versus the volumetric

flow rate in liters/min. There will likely be a "hysteresis effect," in that the pressure

drop curve for increasing flow rate will differ from that for decreasing flow rate.

Therefore, use different symbols for increasing and decreasing flow rates. Explain thelikely cause of this effect.

2. For each bed, calculate the friction factor and corresponding Reynolds number for

each data point using Equations (1) and (2). Then prepare a single plot of fp versus

NRe,p which combines the results for the two beds. Use a different symbol for each bed

and show symbols only (no lines). On this plot, also show the predicted values from

Equations (3), (4), and (5). Show these as solid or dashed lines, and do not show the

points used for determining these plots.

3. From your plots in Part 1 above, determine the pressure drop at the point where

fluidization begins in cm H2O. Using Equation (6), calculate the predicted value of the

pressure drop at this point in cm H2O. Note that you have to know M. Leva [MaxLeva, "Chemical Engineering, November, 1957, pp. 266-270] gives a correlation thatcan be used. Alternatively, you could assume M as provided to you by the TA forthe packed bed before fluidization commences. Comment on your findings.

4. As noted earlier, the minimum fluidization velocity, V0M , is of considerable interest.

From your results in Part 1, calculate V0M in m/s and ft/s for each of your runs and

beds. Using the theoretical Equation (7), calculate V0M in m/s and ft/s. Use a value of

0.86 fors since your particles seem to be "sand-like," and the value ofM you gotfrom the plot in step 3. Several literature equations are available based on

experimental data. Three of these will be used here for comparison with your

experimental results.

(a) Leva (discussed in the two references in "Chemical Engineering")


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Leva gives the following equation for the mass velocity at the point at which

fluidization begins:

where: Dp = particle diameter, inches

= fluid density, lbm/ft3

s = particle density, lbm/ft3

= fluid viscosity, centipoisesGmfV0M = mass velocity, lbm/ft2-hr

The value of 688 was chosen by Leva as best based on 223 experimental points.

Frantz noted that the standard deviation was 33% and the average deviation was 22%.

The equation is valid for Reynolds numbers, GmfDp/, less than 5. Above 5, the valueof Gmfmust be multiplied by the correction factor, Fg, given in the following plot.

(b) Perry's Chemical Engineering Handbook, 6th Edition, p. 20-59.

Baeyens and Geldhart ["Fluidization and Its Applications," Proc. Int. Symp.

Toulouse, 253 (1973)] gives "one of the better correlations:"

where V0M = minimum fluidization velocity, m/s

s = particle density, kg/m3

= fluid density, kg/m3g = 9.81 m/s2

DP = particle diameter, m

= viscosity, kg/m,s

(c) Equation of Geldhart in "Powder Technology."

Using Equation (9).

For each bed, calculate V0M from the equations of Leva and Perry's in m/s and ft/s.

Compare the results with your experimental values.

5. Finally investigate the behavior of your beds when they are expanded. At each data

point in the expanded regime, you can calculate from Equation (7-60). ForM,assume that it has the value given to you by the TA for the packed bed. Then tabulate

corresponding values of V0 on the abscissa. Note that taking the logarithm of both

sides of Equation (7) gives:

Hence the slope of each log-log plot will give n and the "intercept" should be ut. Whatvalues of ut do you find by this method?


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Now prepare a log-log plot of versus NRe,p . From this plot, Equation (16),

and Equation (2), determine ut. Compare its value with that from the log-log plot of

versus .

6. Calculate the heat transfer coefficient and graph versus fluid velocity, the heat transfercoefficient and graph versus depth.

7. Discuss your results critically.

REFERENCES1. J.F. Richardson and W.N. Zaki, Trans. Inst. Chem. Engrs., 32, 35 (1954).

2. J.M. Coulson and J.F. Richardson, "Chemical Engineering," Vol. II, p. 510-527,

Pergamon Press, Oxford (1960).

3. R.B. Bird, W.E. Stewart, and E.N. Lightfoot, "Transport Phenomena," p. 190-194,

John Wiley and Sons, New York (1960).

4. S. Ergun, Chem. Eng. Prog., 48, 89 (1952).

5. W.L. McCabe and J.C. Smith, "Unit Operations of Chemical Engineering," 3rd

Edition, p.146-150, 159-160, McGraw-Hill, New York (1976).

6. A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus, and L.B. Anderson, "Principles of

Unit Operations," 2nd edition, p. 637-547, 699-714, John Wiley and Sons, New York

(1980).

7. Kunii, D. and Levenspiel, O.: Fluidization Engineering, Butterworth-Heinemann,Boston, 1991

8. McCabe, W.L., Smith, J.C., and Harriott, P: Unit Operations of Chemical

Engineering, McGraw-Hill, 1985


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8. DRAG COEFFICIENT IN AIR FLOW

Drag on a Circular Cylinder

Aim:

The aim of the experiment is to investigate the pressure distribution in an air flow

around a cylinder in order to determine the drag forces and drag coefficient on thecylinder.

Apparatus

The apparatus consists of a small cylinder of 1/2 outside diameter mounted

diametrically across a 8 diameter wind tunnel. The cylinder can be rotated 360 and

a small hole in the center of the cylinder is provided for pressure tapping. The hole

points directly into the air stream when the degree scale reads zero. The pressure

differences between the cylinder and the tunnel wall are measured by 10 inclined

manometer.

The wind tunnel is driven by a turbo fan which its rotation can be varied, so its air

velocity.

Theory

When a body is in a fluid stream, the fluid exerts a resultant force on the body. The

components of the resultant force parallel and normal to the direction of the flow are

called respectively drag and lift.

Figure E?.1

The flow pattern around a cylinder in a fluid stream is illustrated in Figure E?.1 for

NRe > 20. The flow separates at points S, about 80 to the front stagnation point.

Vortices are formed in the behind of the cylinder (in the wake region). The drag on

the cylinder results partly from the skin friction and partly from the fact that the

pressure on the rear side is lower than on the front.

The drag resulting from the pressure differences around the surfaces of an object is

called the form drag.

This experiment trying to investigate and calculate the form drag of a cylinder object

by measuring the pressure distribution around the surfaces of a cylinder.

s

s

Wide eddywake

d

P

d

Figure E?.2Flow


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Page 30 of 54

If the flow is symmetrical and by resolving the force exerted by the pressure in the

direction of the approaching flow, the form drag FD per unit length of cylinder is

given by

0 0

cos2.cos2 dPdd

d

PFD

Where Pis the pressure around the surface of the cylinder and a function of.

During the experiment, the manometer does not read the Pvalues, but the differences

between Pand Pstatic at the tunnel wall. Therefore the expression forFD becomes

0

cos dPPdFD s

where Ps is the static force at the wall tunnel and taken as constant.

The drag coefficient for a body in a fluid is defined as:

)(221 areaV

dragtotalCD

Where 221 V is the dynamic pressure of the fluid stream, and the area is the projected

area of the body in the direction of the approaching stream. For a circular cylinder,

dV

DCD 2

21

For a form drag of the cylinder, the form drag coefficient can be calculated using the

following relation

0

2

21 cos dV

PPC sD

Applying the Bernoullis theorem between the point at the wall and the 0 point at the

cylinder gives

sPPV 02

21

Finally, the form drag coefficient can be written as

0 0

cos dPP

PPC

s

sD

Experimental Procedure1. Before starting the flow in the tunnel, connects the pressure tap on the left side of the

manometer to the pressure tap of the cylinder and the other side to the pressure tap on

the tunnel wall. Make sure that all connections are tight!

2. Adjust the scale of the manometer to get the zero reading of the scale match with the

meniscus.

3. Run the turbofan at predetermined RPM, and wait for a while to attain stable air

flow.4. Rotate the cylinder to the zero position and read the manometer reading. Now, rotate

clockwise in 10 intervals from 0 until 360 and note the manometer readings at

each interval stops.5. Plot the pressure reading vs. the degree of rotation ().


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6. Measure the temperature of the air leaving the wind tunnel and hence look up for the

kinematics viscosity value of the air.

Analysis: Data

1. Identify the separation point on your plot of(P Ps) vs. . In the front part of thecylinder, the pressure falls rapidly as the fluid accelerates. After the separation point a

wake is formed and the pressure is more or less constant.

2. Find the form drag coefficient by considering the area between the curve and the axis,

taking account of the sign and do not forget to convert degrees to radians. Discuss how

you obtained the error estimate for coefficient.

3. Calculate the Reynolds number. The contribution of skin friction to the total drag

coefficient can be estimated from the correlation5.0

Re4fC

Compare the values of the form and skin friction drag coefficient with one another and

compare the total drag coefficients with the values in the textbooks.4. Comment on the accuracy of the experiment.


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9. NATURAL CONVECTION HEAT TRANSFER

Natural Convection Experiment

Todemonstratethebasicprinciplesofnaturalconvectionheattransferincluding

determinationoftheconvectiveheattransfercoefficient.

Naturalconvectionoccurswhentemperaturegradientwithinafluidinducesitsmotion.Ingeneralthedensityofafluiddecreaseswithincreasingtemperatureandtherefore creatingapressuregradienttowardcoolerpartofthefluid.Thispressuregradientistheresponsiblecause forthefluidmotion.Thehotterfluidmovesupwardandthecoolerfluidmovesdownward.Asthefluidmovesthedragforceson the fluidwillbalancethepressuregradientandsolimitingthefluidvelocity.

Tosustainthe natural motion,heatmustbetransferredtothefluidcontinuously.AccordingtoNewtons LawofCooling,therateofheattransfer couldbeexpressedas

)( fs TThq

Where h=heattransfercoefficient, Ts=hotsurfacetemperature,and Tf=coldfluidtemperaturefarawayfromthesurface.Tomaketheexpressionusable, hshouldknownorcanbepredictedeasily.Sinceanalyticalapproachofpredicting hisdifficult,anexperimentalapproach(orempirical)isoftenchosenthroughtheuseofdynamicsimilarity.Itisfoundthatcertainsystemsinfluidmechanicsorheattransferarefoundtohavesimilarbehaviorseventhoughthephysicalsituationsmaybequitedifferent.Inconvectiveheattransferwemayapplydynamicsscalingtomakeatransformation.

IthasbeendefinedadimensionlessconvectiveheattransfercoefficientcalledtheNusseltnumberas

k

LhNu c (1)

wherehc:convectiveheattransfercoefficient

L:characteristiclengthk:thermalconductivityofthefluid.

L1

L2

L3


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Thecharacteristiclengthischosenasthesystemlengththatmostaffectsthefluidflow.Forflow inapipethecharacteristiclengthisthe pipediameter, D,and theexpressionbecomes

k

DhNu c (2)

TheRayleighnumberisindicativeofthebuoyancyforcethatisdrivingtheflowandisgivenby

3)( LTTgRa

fs (3)

whereg:accelerationduetogravity:fluidthermalexpansioncoefficientTs:surfacetemperature

Tf:fluidtemperature

L:characteristiclength:fluidkinematicviscosity:fluidthermaldiffusivity.

ThedimensionlessparameterwhichisusedtorepresenttheaffectoffluidpropertiesisthePrandtlnumber

Pr (5)

Theinfluenceofgeometrymaybeseeninacoupleofways.First,forthoseconfigurationsthathavetwolengthdimensions,suchasacylinder, weintroducea

dimensionlessgeometricparameter(6)

ThesecondwayinwhichweseegeometricalinfluencesisthroughthefunctionalformoftheNusseltnumbercorrelation.Ingeneralwemaywrite

(7)

Forsimplesituationsthesemayoftenbewrittenaspowerlawrelationships

(8)

wheretheconstants a,m,and nwillchangefordifferentgeometries.


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10. NATURAL AND FORCED CONVECTION HEAT TRANSFER

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35/54

Page 35 of 54

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36/54


37/54

Page 37 of 54

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38/54

Page 38 of 54

MMeeaassuurree tthhee rreeaaddiinngg ffoorr tthhee ssuurrffaaccee tteemmppeerraattuurree ooff tthhee ccyylliinnddeerr,, tthhee tteemmppeerraattuurree

aanndd vveelloocciittyy ooff tthhee aaiirr ffllooww aanndd tthhee ppoowweerr ssuupppplliieedd bbyy tthhee hheeaatteerr..

RReeppeeaatt sstteeppss 11 aanndd 22 ffoorr ddiiffffeerreenntt vveelloocciittiieess tthhee aaiirr ffllooww aanndd ppoowweerr iinnppuutt..

QQiinnppuutt hhrr hhCC11tthh hhCC22tthhSSeett

WW WW//mm22KK WW//mm

22KK WW//mm

22KK

11 44

22 88

33 1122

44 1166

TThhee ttoottaall hheeaatt iinnppuutt iiss:: QQiinnppuutt == VVII

TThhee hheeaatt ttrraannssffeerr rraattee bbyy rraaddiiaattiioonn iiss::

QQrraadd == AA ((TTss44 TTaa

44)) == hhrr AA ((TTss TTaa))

SSoo,,as

asr

TT

TTh

)( 44

TThhee hheeaatt ttrraannssffeerr rraattee bbyy ccoonnvveeccttiioonn iiss::

QQccoonnvv == QQiinnppuutt -- QQrraadd

FFrroomm NNeewwttoonnss llaaww ooff ccoooolliinngg

)( ascconv TTAhQ

AAnndd

)( as

convc

TTA

Qh


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FFoorr aann iissootthheerrmmaall lloonngg hhoorriizzoonnttaall ccyylliinnddeerr,, MMoorrggaann ssuuggggeessttss aa ccoorrrreellaattiioonn ooff tthhee ffoorrmm,,

nDD cRak

DhNu ((11))

cc aanndd nn aarree ccooeeffffiicciieennttss tthhaatt ddeeppeenndd oonn tthhee RRaayylleeiigghh nnuummbbeerr

RRaayylleeiigghh nnuummbbeerr cc nn

1100--1100

1100--22

00..667755 00..005588

1100--22

110022

11..0022 00..114488

110022

110044

00..885500 00..118888

110044

110077

00..448800 00..225500

110077

11001122

00..112255 00..333333

TThhee RRaayylleeiigghh nnuummbbeerr iiss ccaallccuullaatteedd ffrroomm,,

PrD)TT(g

Ra2

3

as

wwhheerreefilmT

1 aanndd

2

TTT asfilm

CChhuurrcchhiillll aanndd CChhuu rreeccoommmmeenndd aa ssiinnggllee ccoorrrreellaattiioonn ffoorr aa wwiiddee rraannggee ooff RRaayylleeiigghh nnuummbbeerr,,

2

27/816/9

6/1

D

Pr)/559.0(1

Ra387.060.0Nu

1210Ra ((22))

FFrroomm ccoorrrreellaattiioonn ((11)) aanndd ((22)) wwee ccaann ddeetteerrmmiinnee hhCC11tthh aanndd hhCC22tthh aanndd ccoommppaarree wwiitthh hhcc

oobbttaaiinneedd ffrroomm tthhee eexxppeerriimmeenntt..


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QQiinnppuutt hhrr hhCC RRee NNuu11 NNuu22 hhCC11tthh hhCC22tthhSSeett

WW WW//mm22

KK WW//mm22

KK -- -- -- -- --

11

22

33

44

55

66

77

TThhee ttoottaall hheeaatt iinnppuutt iiss::

QQiinnppuutt == VVII

TThhee hheeaatt ttrraannssffeerr rraattee bbyy rraaddiiaattiioonn iiss::

QQrraadd == AA ((TTss44 TTaa

44)) == hhrr AA ((TTss TTaa))

SS

oo

,,

as

asr

TT

TTh

)( 44

TThhee hheeaatt ttrraannssffeerr rraattee bbyy ccoonnvveeccttiioonn iiss::

QQccoonnvv == QQiinnppuutt -- QQrraadd

FFrroomm NNeewwttoonnss llaaww ooff ccoooolliinngg

)( ascconv TTAhQ

aanndd

)TT(A

Qh

as

convc

FFoorr aann iissootthheerrmmaall lloonngg hhoorriizzoonnttaall ccyylliinnddeerr,, HHiillppeerr ssuuggggeessttss,,

3/1m

DD PrReC

k

DhNu ((33))


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wwhheerree CC aanndd mm aarree ccooeeffffiicciieenntt tthhaatt ddeeppeenndd oonn tthhee RReeyynnoollddss nnuummbbeerr::

RReeDD CC mm

00..44--44 00..998899 00..333300

44--4400 00..991111 00..3388554400--44000000 00..668833 00..446666

44000000--440000000000 00..119933 00..661188

4400000000--440000000000 00..002277 00..880055

AAllll pprrooppeerrttiieess aarree eevvaalluuaatteedd aatt tthhee ffiillmm tteemmppeerraattuurree

2

TTT asfilm

CChhuurrcchhiillll aanndd BBeerrnnsstteeiinn pprrooppoosseedd tthhee ffoolllloowwiinngg ccoorrrreellaattiioonn ffoorr RRee PPrr>>00..22

5/48/5

D

4/13/2

3/12/1

D282000

Re1

Pr

4.01

PrRe62.03.0Nu

((44))

wwhheerree aallll pprrooppeerrttiieess aarree eevvaalluuaatteedd aatt tthhee ffiillmm tteemmppeerraattuurree..

FFrroomm ccoorrrreellaattiioonn ((33)) aanndd ((44)) wwee ccaann ddeetteerrmmiinnee hhCC11tthh aanndd hhCC22tthh aanndd ccoommppaarree wwiitthh hhcc

oobbttaaiinneedd ffrroomm tthhee eexxppeerriimmeenntt..


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10. DOUBLE PIPE HEAT EXCHANGER

OBJECTIVES

The objective of this experiment is to investigate the effect of Reynolds number on the

individual heat transfer coefficients and the performances of co-current and counter-current

flow heat transfer in a double pipe heat exchanger.

INTRODUCTION

Double pipe heat exchanger consists of outer tube and inner tube. Fluids flow through inner

tube and annulus.

The amount of heat transferred in the hot stream and in the cold stream are

Applying overall energy balance at steady state and rearranging gives

RATE OF HEAT TRANSFER

HEAT FLUX: The rate of heat transfer per unit area is called the heat flux. In a heat-transfer

equipment which the transfer surfaces are constructed from tubes or pipe, heat fluxes may

then be based on either the inside area or the outside area of the tubes. Although the choice is

arbitrary, it must be clearly stated, because the numerical magnitude of the heat fluxes will

not be the same for both.

OVERALL HEAT TRANSFER COEFFICIENT: The heat flux depends on the localtemperatures different between the hot side and the cold side and the local heat transfer

coefficient. In practise, it is easier to work with the average of all the local temperature

differents and the local heat transfer coefficient. The heat trasfer rate from the hot stream to

the cold stream can then be written as

where

ThcThh

Tcc

Tch


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The LMTD is not always the correct mean temperature difference to use. It should not be

used when U changes appreciably or when T is not a linear function of q. As an example,consider an exchanger used to cool and condense a superheated vapor.

INDIVIDUAL HEAT TRANSFER COEFFICIENT: Overall heat transfer coefficient consists

of many parts. Graphically it can be shown as

The rate of heat transfer based on the outside area and the individual heat transfer coefficient

can be calculated from the following realtion.

And for the inside area

PRELIMINARY WORK

Study the equipment and familiarize yourself with its operation. Draw a detailed diagram of

the equipment.

The data about the dimensions of the exchanger are as follows:

Number of double pipes: ?

Length of each pipe: ? m

Inner pipe dimensions: ? Sch. 40 steel pipe

Outer pipe dimensions: ? Sch. 40 steel pipe

Prepare a data sheet to record the data acquired during the experiment.

EXPERIMENTAL PROCEDURE

1. Perform at least four separate runs for each of the co-current and counter-current flow

configurations. During the runs vary the flow rate of the cold stream in the turbulentregion and keep the flow rates hot stream constant.

Thot

Tcold

hot fluidcold fluid

metal wall

TwcTwh


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Page 44 of 54

2. Take readings for both rotameter during each run at suitable time intervals. Record the

mass flowrates for the streams.

3. Wait for the attainment of steady state conditions during each runtime after changing the

flow rate of the cold stream. Check the steady-state conditions by recording the streams

temperatures at suitable time intervals.

CALCULATIONS

1. Plot the temperature of each stream vs. the length of the heat exchanger for each run.

Indicate the ranges and approaches.

2. Calculate the overall heat transfer coefficients based on the cold water stream. Plot the

overall heat transfer coefficients of cold stream vs. corresponding Reynolds numbers.

3. Calculate the overall heat transfer coefficients based on the hot water stream. Plot the

overall heat transfer coefficients of hot stream vs. corresponding Reynolds numbers.

4. Calculate the heat fluxes to cold stream once using the overall heat transfer coefficientsand log-mean temperature differences; and once using the flow rate and temperature data

of the cold stream.

5. Determine the heat fluxes to the surroundings by difference in heat fluxes obtained using

the data for cold and hot streams. Calculate the percentage of heat loss to surroundings.

6. Make an energy balance.

DISCUSSIONS

1. The effect of Reynolds number on the individual heat transfer coefficients.

2. Performance of the co-current flow heat exchanger vs. the counter-current flow heat

exchanger.

3. Possible causes of deviations, if any, between the results of heat fluxes obtained by the

two different calculation procedures, for the same run.

4. Make recommendations to increase the heat transfer rate to the cold stream, and for the

better control of variables to obtain more accurate experiment results.

5. Explain how you can calculate the heat transfer coefficient on the annular side of the

inner pipe in double pipe heat exchanger.

6. Under which conditions would you recommend to have hot fluid flowing in the inner pipe

of a double pipe heat exchanger.

7. Where and when do you use double pipe heat exchangers? What is the most reasonable

advantage of a double pipe exchanger when compared with shell-and-tube exchangers?

8. Explain, with respect to the fluids involved, the following terms used in the heat transfer

terminology: Exchanger, heater (or steam heater); cooler; condenser; reboiler; vaporizer;

waste-heat Boiler.

9. What is the main difference between a 1-2 shell-and-tube exchanger and a U-tube

exchanger?


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Suggested Reading

1. Mc Cabe, W.L., J.C. Smith, and P. Harriott, Unit Operations of Chemical Engineering,

6th ed., Mc Graw Hill, N.Y. (2001) Chapter 11, Principles of Heat Flow in Fluids.

2. Holman, J.P., Heat Transfer, 7th ed., Mc Graw Hill, N.Y. (1990).

3. Perry, R.H. and D. Green, Perrys Chemical Engineers Handbook, 7th ed., Mc Graw

Hill, N.Y. (1997) Section 10, Heat Transmission.

4. Foust, A.S., et al, Principles of Unit Operations, 3rd ed., John Wiley and Sons, N.Y.


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11. SHELL AND TUBE HEAT EXCHANGER

OBJECTIVE of this experiment is to measure the two inlet temperatures and the mass flows

through the shell and tubes, in order to predict the two outlet temperatures using the NTU method

andcomparethesepredictedvalueswithactualmeasuredoutlettemperatures.

INTRODUCTIONMany engineering applications involve a process of heat exchange between

two fluids. Heat exchangers are devices used to promote the heat transferred between two fluids;

e.g., a car radiator and the condenser units on air conditioning systems. Space heating, air

conditioning, power production, and chemical processing are typical areas of application.

There are many heat exchanger designs. The laboratory setup for this experiment contains three

heat exchanger types: a shell-and-tube exchanger, a concentric tube exchanger, and a tube bank

exchanger in cross flow. This particular experiment employs the shell-and-tube type heat exchanger

(see Figure 1). A shell-and-tube heat exchanger is constructed of tubes that are attached on each

end by a plate, called the tube sheet, through which the tubes pass. One fluid streams into the inlet

of the heat exchanger, flows through the tubes, and exits through the tube sheet at the opposite end

of the heat exchanger.

Figure 1Schematic of shell-and-tube exchanger

A shell encloses the internal volume where the tubes are housed. Another, fluid flows through

the shell and heat is exchanged between the tube-side fluid and the shell-side fluid. In a powerplant, most heat exchangers are of the shell-and-tube design. The number of passes commonly


47/54Page 47 of 54

presents a further description of a shell-and-tube heat exchanger. A single pass means the fluid

flows straight through the entire heat exchanger without changing direction and so, in this design,

the fluid moves past the length of the heat exchanger only a single time. In a two-pass heat

exchanger the fluid in the tubes goes in one end, flows to the other end, reverses direction then

flows back to the same end that the fluid entered through a second set of tubes. Thus, the fluid

travels the full length of the heat exchanger twice. Similarly, multiple pass heat exchangers are so

named because they make many passes. This experiment employs a shell-and-tube heat exchangerconsisting of two tube passes and one shell pass.

THEORY: HEAT EXCHANGER ANALYSIS

Thermodynamics and the First Law dictate the overall energy transfer in a heat exchanger. There

are two widely used methods of heat exchanger analysis, the NTU-Effectiveness method and the

Log-Mean-Temperature-Difference (LMTD) method. These are briefly discussed below.

Log-Mean-Temperature-Difference (LMTD) Method

For a heat exchanger between two fluids with given inlet and outlet temperatures, there are three

equations for the rate of heat transfer, Q,

(11)

(12)

(13)

Where

and . These temperature differences are called

temperature approaches, .

U= overall heat transfer coefficient, W/(m2

K)A = area of surface across heat transfer occurs, m2

= mass flow rate of fluid j, kg/s

= specific heat of fluid j, J/(kgK)

For known specific heats, U, A, and entering temperatures, the three equations above can be solved

for three unknownsT1,out, T2,out, and Qby successive substitution of one of the equations forQ

onto another. It is a simple matter to use the log-mean-temperature-difference method of heat

exchanger analysis when the fluid inlet temperatures are known and the outlet temperatures are

specified or readily determined from the energy balance expressions. The value ofTlmtdfor theexchanger may then be determined. However, ifonly the inlet temperatures are known, use of the

LMTD method requires an iterative procedure.

T1,in T1,out

T2,in

T2,out


48/54Page 48 of 54

NTU-Effectiveness MethodThis method was developed to avoid the iterative calculation of LMTD method when only

the temperatures of inlet streams are known. Number of (heat) transfer unit is defined as

(14)

where

(15)

(16)

(17)

(18)

NTU is a function of geometric and material properties, and the mass flow rates. It does not include

any fluid temperatures.

Heat exchanger effectiveness is defined as the ratio of the actual heat that can be transferred by the

equipment and the maximum heat that possibly can be given or be received by the least stream

capacity (Cmin) at infinite area.

(19)

Effectiveness of heat exchanger equipment depends on many factors such as type, geometric

arrangement, etc. The following table presents the effectiveness of various cases.

The outlet temperatures can be easily calculated from Qactualand equation (1) and (2).


49/54Page 49 of 54

LABORATORY PROCEDURE1. Verify the dimensions and features of Figure 2.

2. Generally, small flow rates will generate better results but may take longer to reach steady

state. Also, do not let the air that comes out of entrainment accumulate in shell. Use bleed

taps as needed.3. For a hot water flow of about 15% of the maximum reading and a cold water flow of about

30%, take inlet and outlet temperatures of both flows until no further changes in

temperature are noted. This is the steady-state conditionuse only the associated flow

rates and temperatures for calculations.

Figure 2Experimental apparatus with dimensional data

DETAILED COMPUTATIONAL PROCEDURE

The NTU method will be described for a shell-and-tube heat exchange using only one tube; but

it could represent an entire tube bundle.

1. a. Determine cold and hot water flow rates, _mH_mH and _mC_mC , and their specific

heats,Hp

c andCp

c (look for average cp values). [ALWAYS CHECK UNITS!!]

b. Calculate a temperature specific energy flow known as the heat capacity rate, C, for both thecold and hot flows

Distance between Tube Sheets, 16-1/8

(inside face to inside face)

30 Tubes, each 0.25 diameterneglect wall thickness

Cold waterinlet

thermometer

Hot water inlet thermometer

Cold wateroutlet

thermometer

Hot water

outletthermometer

Shell: 5 OD4.5 ID

5 Baffles, 1.2 thick. Equally spaced to form 6

chamber. 23 tube penetrations per baffle.

The width of theflow coursevaries & thus theaverage velocity

SL = 0.475

ST= 0.548

SD = 0.548SL

ST

SD


50/54Page 50 of 54

smallertheand

istheseoflargerThe

min

max

C

C

cmC

cmC

hot

cold

phothot

pcoldcold

.

c. Calculate the heat capacity rate ratio, c = Cmin/Cmax.

2. Calculate the heat transfer coefficients at the inside and outside surfaces of the tubes, hinside and

houtside. These are used to compute the overall heat transfer coefficient, U. (See Figure 3)

Figure3Heattransfercoefficientsatinsideandoutsidetubesurfaces

a. Flow Inside Tubes: Even though there are many tubes in the bundle and there are parallel

and counter flows in this two-pass exchanger, the calculation may be performed by

considering the flow in just one of the tubes, but one must account for the direction of the

flow. That is, half of the tubes are associated with parallel flow and half the tubes areassociated with counterflow. Thus, the mass flow in the equivalent tubes is

tubeinside

flowsidetubetotalm

N

m1

-

2

where, N= total number of tubes.

From simple flow relations, it is known that the velocity inside a single tube is

A

mV insideinside

where, A = cross sectional area of one tube.

Given this velocity, a Reynolds number ( ) can be computed to indicatewhether the inside flow is laminar or turbulent. This will most likely be fully-developed,

laminar flow. For such with constant surface temperature, Ts, and :

where fluid properties are based on the mean (or bulk) temperature, .

If the flow is fully developed turbulent (Re 10,000),

.

Tube-side fluid properties should be evaluated at the average of the mean temperatures, .

b. Shell Flow (Outside of Tubes): For the staggered tube arrangement of the experiment

shown in Figure 4, use the following expression for the average Nusselt number

outsideh

insideh


51/54Page 51 of 54

Use Table 1 to determine m and C1. Note in the report which values ofm and C1 were

used. This relation applies when there are more than 10 tubes in a bundle (NL 10), 2000

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Chemical Engineering Laboratory i(1)