industrial chemical technology

LONG TERM TRAINING PROGRAMME

ON

“INDUSTRIAL CHEMICAL TECHNOLOGY”

MODULE – I

(03/10/2010 TO 20/02/2011)

COURSE MATERIAL

COURSE CO-ORDINATOR

Prof. Latesh B. Chaudhari Asst. Prof., F.E.T.R., Bardoli.

CONDUCTED BY

Nodal Institute Chemicals and Petrochemicals Sectors N.G. Patel Polytechnic, Isroli-Bardoli.

&

ANCHOR INSTITUTE

Chemicals and Petrochemicals Sectors Promoted by

Industries Commissionerate Government of Gujarat

In Partnership with Faculty of Technology, D.D.U., Nadiad

Long Term Training Programme on "Industrial Chemical Technology"

1.1 HEAT TRANSFER FUNDAMENTALS 1.1.1 Introduction

Practitioners of the thermal arts and sciences generally deal with four basic

thermal transport modes: conduction, convection, phase change, and radiation. The

process by which heat diffuses through a solid or a stationary fluid is termed heat

conduction. Situations in which heat transfer from a wetted surface is assisted by the

motion of the fluid give rise to heat convection, and when the fluid undergoes a liquid–

solid or liquid–vapor state transformation at or very near the wetted surface, attention is

focused on this phase-change heat transfer. The exchange of heat between surfaces, or

between a surface and a surrounding fluid, by long-wavelength electromagnetic radiation

is termed thermal heat radiation.

It is our intent in this section to describe briefly these modes of heat transfer, with

emphasis on an important parameter known as the thermal resistance to heat transfer.

Simple examples are given for illustration; detailed descriptions of the same topics are

presented in specialized chapters.

1.1.2 Conduction Heat Transfer

One-Dimensional Conduction Thermal diffusion through solids is governed by

Fourier’s law, which in one-dimensional form is expressible as

where q is the heat current, k the thermal conductivity of the medium, A the

crosssectional area for heat flow, and dT /dx the temperature gradient, which, because it

is negative, requires insertion of the minus sign in eq. (1.1) to assure a positive heat flow

q. The temperature difference resulting from the steady-state diffusion of heat is thus

related to the thermal conductivity of the material, the cross-sectional area A, and the path

length L (Fig. 1.1), according to

Prepared By: Prof. Latesh Chaudhari, Asst. Prof. ,F.E.T.R., Bardoli


The form of eq. (1.2), where k and A are presumed constant, suggests that in a way that is

analogous to Ohm’s law governing electrical current flow through a resistance, it is

possible to define a conduction thermal resistance as

One-Dimensional Conduction with Internal Heat Generation Situations in which a

solid experiences internal heat generation, such as that produced by the flow of an

electric current, give rise to more complex governing equations and require greater care

in obtaining the appropriate temperature differences.



Direct Contact HeatTransfer Direct contact heat transfer can occur whenever two substances at different temperatures

touch each other physically. The implication is that there is not an intervening wall

between the two substances. Heat transfer where there is a surface between the two

streams is sometimes called indirect, or the heat transfer device is one of the closed types.

The physical interaction of the two streams can accomplish heat transfer very efficiently.

Without an intervening wall, the energy transport between the two streams can take place

across small thermal resistances. In addition, the fact that a wall is not present can allow a

mass transfer process to take place. In some cases, this is a desirable phenomenon (open

cooling towers), but in other cases it may not be.

Costs are often more favorable for direct contact heat transfer devices than for

their closedcounterparts. The thermal resistances present in closedheat exchangers result

in less heat transfer than might be accomplishedin direct contact, andthis often translates

to lower operating costs for the latter. In addition, the equipment to accomplish the direct

contact processes is generally less expensive than the counterpart closed heat exchangers.

Both aspects can result in considerable life-cycle cost savings for the direct contact

approach over that of the closed type of heat exchanger. Some potential limitations are

inherent in direct contact processes. There is a requirement that the two streams be at the

same pressure. Although this requirement does not often cause significant problems, it

could be very important. Also, as noted above, the mass transfer possibility in direct

contact may not be desirable. Direct contact heat transfer is a fieldwith a wide range of

potential applications. The actual situation is that with some notable exceptions, such as

open feedwater heaters andwet cooling towers, few of these applications have been

usedto any great extent. Reasons for this are numerous, but one of the main reasons is

that engineers are not as knowledgeable as they might be about the design of these types

of systems. This chapter is an attempt to expose some of these possibilities so that the

design of more efficient industrial processes might result.

To present a description of direct contact processes within the limited space of

this chapter, some restriction in scope is necessary. Because the direct contact of any



number of generic streams is possible (and most have in fact been proposed to transfer

heat), only some of the more important applications will be notedhere.

Solid–solid transfer processes are not covered, nor are high-temperature situations

included where radiative heat transfer is important. Open cooling towers are not

discussed here to any great extent, even though they are the single most widely applied

type of direct contact heat exchangers. Although some information related to cooling

towers is provided in the Section 19.4, this will by no means cover a very significant

fraction of the total literature on the subject. The previous work on cooling towers is

voluminous and tends s to use quite specialized design approaches. Interested readers can

find current overviews by ASHRAE (2000) and Mills (1999). An earlier review of the

literature on the numerical modeling developed to predict cooling tower performance was

given by Johnson et al. (1987). The literature contains reports of studies of interaction

between different substances as well as the same substances. For example, a great deal of

interest has been directed to the use of water in phase-change situations, particularly the

condensation of steam on liquid water. On the other hand, interest has also been directed

to the use of fluids of different types. For example, a great deal of literature on heat

transfer in immiscible liquids was cited years ago by Sideman (1966), and many studies

on this topic have appeared since then.

Generally, the prediction of mass transfer has been given more attention over the

years than has the fieldof direct contact heat transfer. Because of mass transfer and heat

transfer analogies, some information from the mass transfer literature can be used in

direct contact heat transfer design. This analog is exposed minimally here, however.

Emphasis here is on surveying predictions and applications, the main thrusts of benefit to

designers.

19.2 SENSIBLE HEAT EXCHANGE

19.2.1 General Comments



Heat transfer from a continuous fluid to droplets or bubbles of another fluid is a

complicated situation involving not only the typical convection-related variables (e.g.,

geometry, velocity, and thermo physical properties) but also the proximity of the objects

to one another where more than one is present. The latter characteristic can be handled in

an overall way through the definition of the void fraction or holdup, the latter denoted

here by the symbol φ. Either of these relates to the volumetric ratio of the amount of the

dispersed phase (droplet or bubble) to the total volume. Holdup has a profound effect on

direct contact heat transfer, as noted several times in this chapter. Another aspect that

influences the heat transfer to droplets or bubbles is the shape of these objects. It has been

well documented that droplets and bubbles can experience a wide variety of shapes,

depending on the object size and the flow situation (Grace, 1983). Despite this, much

work has been done on a variety of systems assuming that the droplet is spherically-

shaped. A great deal of the early work was reviewed by Sideman (1966). EVAPORATION AND CONDENSATION

General Considerations

Lock (1994) has summarized the various modes in which condensation and evaporation

can take place in simple direct contact systems. These include the evaporation–

condensation interactions with droplets and jets in the presence of an incondensable gas.

Consider a representation of the regimes shown in Fig. 19.1. In this figure the italic

notations relate to the state of the vapor, with four regions defined by the intersection of

the interface isobar and the interface isotherm. The non shaded areas, hot evaporation on

the upper right and cold condensation on the lower left, are the normally anticipated

regions where those phenomena occur. Above the isotherm TI the region of vapor heat

gain is demarcated from the region of vapor heat loss. In the hot evaporation region, the

vapor–gas mixture is at a higher temperature than the liquid. This superheated mixture

drives the process by the transfer of heat to the cooler liquid. Below the isotherm, but still

on the same side of the isobar, the process can take place only if the liquid is superheated

and thus furnishes the heat necessary for the evaporation process. Cool evaporation is

limited by the liquid superheat, as would be represented by the liquid Jak ob number.

Condensation processes take place in the region shown to the left of the interface



isobar in Fig. 19.1. These are normally considered to occur when the vapor is cooler than

the liquid, and this region is denotedas cold condensation (lower left). Clearly, the liquid

source or sink available for evaporation or condensation is limited in real situations.

Hence the finite heat capacity of the liquid is a critical element in the determination of the

amount of heat and mass transfer that can occur. For the cool evaporation and warm

condensation regions, this can be assessed by examining the Jakob number for the liquid.

A practical issue that can arise is the presence of a non condensable gas, often air, in a

condensing or evaporating system. In evaporating systems, small amounts of a

component of this type do not cause much effect, and the phase-change processes are

relatively unimpeded. In condensing systems, the situation is quite different. Here the



phase-change process can be grossly hindered. Thus it is very important in condensing

systems to know the extent of the presence of non condensables and account for them

appropriately.

Condensation of a Vapor on or in a Liquid

Condensation of a vapor on or in a liquid, whether or not that liquid is the same substance

as the vapor, is commonly encounteredin engineering systems. Direct contact processes

differ from their indirect counterparts in many respects already noted. In indirect transfer,

the extent of the process is limited by the area of the surface andthe heat transfer rate

possible through the surface. In direct contact processes, the situation is limitedby the

interplay between the latent heat of condensation and the amount of sensible heat the

liquidcan absorb. The amount of the liquid used for condensing purposes and its

subcooling determines the extent of condensation that can be accomplished. The presence

of noncondensables during the condensation process affects performance in negative

ways. Details of the many studies of this will not, in general, be summarized here because

of the effects of the multitude of variables that influence individual situations.

Condensation of a vapor that contains noncondensable elements on a surface finds that a

noncondensable layer builds up near the surface. This causes both a heat transfer and a

mass transfer resistance that impedes the basic condensation process. In many respects

this is similar to what is foundin surface condensers when noncondensable gases are

present. In vapor droplet direct contact condensation in a liquid, this same phenomenon is

present at the inner interface of the bubble. Additionally, though, the presence of

noncondensables results in a decreased vapor pressure inside the bubble compared to that

of a pure substance in the same situation. This then lowers the condensation temperature,

decreasing the driving potential for heat transfer.

Many studies of direct contact condensation are reported in the literature. An

extensive review of this literature was given by Sideman and Moalem-Maron (1982).

Film Condenser We start with a description of a situation that will be of value

primarily for visualization purposes. This is the direct contact film condenser shown

in Fig. 19.2. In this situation, a bulk vapor is condensed on another liquid, the latter

serving as the sink. Patternedafter the model usedfor Nusselt’s solution for laminar



film condensation, and described in most elementary heat transfer texts, the concepts

form a basis for other systems that follow below.

Consider the heat transfer processes that occur in the direct contact film condenser.

The pure saturatedv apor condenses on the liquidof the same substance. Heat then

flows by conduction through the condensed liquid to the sink liquid. Since it is

assumedhere that there is no other place for the heat to be absorbedultimately than

the sink liquid, the heat capacity of that liquid is the determining factor for the duty

possible from devices of this type.

The heat transfer processes that must be analyzedfor this type of system are the

diffusion phenomena through the two liquidfilms. For short times, the sink liquidis

consideredas being semi-infinite at the interface between it andthe condensedliquid



INDUSTRIAL CHEMICAL TECHNOLOGY MODULE – I

HEAT TRANSFER TECHNOLOGY DATE: 21/11/2010 BY: L. B. Chaudhari Conduction Mode of Heat Transfer: Transfer of heat from one part of a body to another part of a same body or from on body to another body which is in physical contact with it, without appreciable displacement of particles of a body. Fourier’s Law: The physical law governing the transfer of heat through uniform material by conduction mode is known as Fourier’s Law. Q ∝ A x [ dT / dx] Q = k x A x [ dT / dx] Where, Q = Rate of heat transfer, W A = Area of heat transfer, m2 dT / dx = Rate of change of temperature with distance in the direction of heat flow, oC/m k = Thermal conductivity, W/mK Q/A = q = k x [ dT / dx] q = Heat Flux, W/m2 Fourier’s Law states that “the rate of heat flow by conduction through uniform material is directly proportional to the area normal to the direction of heat flow and temperature gradient in the direction of heat flow”. Thermal Conductivity: Q = k x A x [dT / dx] k = (‐ Q x dx) / (A x dT) k = ( W x m) / (m2 x K) k = W / mK (Unit of Thermal Conductivity) Thermal conductivity is defined as “the quantity of heat passing through a quantity of material of unit thickness with unit heat flow area in unit time when unit temperature difference is maintained across the opposite face of material”

OR

“The ability of material to conduct heat through itself is known as thermal conductivity”.



• Thermal conductivity depends on the nature of material and its temperature. • Thermal conductivity of solids is higher then that of liquids and liquids is higher then that of

gases.

State of Matter Solids Liquids Gases Range of Thermal Conductivity (W/mK) 2.3 to 420 0.09 to 0.7 0.006 to 0.6

The thermal conductivities of some materials at room temperature:

Material Thermal Conductivity W/m · °C Material Thermal Conductivity

W/m · °C Diamond 2300 Brick 0.72Silver 429 Water (l) 0.613Copper 401 Human skin 0.37Gold 317 Wood (oak) 0.17

Aluminum 237 Helium (g) 0.152Iron 80.2 Soft rubber 0.13

Mercury (l) 8.54 Glass fiber 0.043Glass 0.78 Air (g) 0.026

The range of thermal conductivity of various materials at room temperature



Classification of Materials according to thermal Conductivity: HEAT CONDUCTORS: The materials having higher value of thermal conductivity are referred to as heat conductors. Eg. Silver metal (k = 420 W/mK), Red Copper (k = 395 W/mK), Gold (k = 302 W/mK), Aluminums (k = 210 W/mK) HEAT INSULATORS: The materials having low value of thermal conductivity are referred to as heat insulators. Eg. Cork (k = 0.025 W/mK), Glass Wool (k = 0.10 W/mK), Asbestos (k = 0.024 W/mK), 85% magnesia (k = 0.04 W/mK) Thermal Insulation: • Process equipments such as reaction vessels, Reboilers, distillation column, evaporators etc or

steam pipe will lose heat to the atmosphere by conduction, convection and radiation. • Thus to overcome this heat loss, some form of lagging is applied to the hot surfaces. Eg. In a

furnace surface temperature is reduced by making use of series of insulating bricks that are poor conductor of heat (Heat Insulators).

Requirement of lagging materials: 1. It should have low thermal conductivity and 2. It should suppress convection currents. Optimum insulation thickness: Increasing the thickness of insulation will decrease the heat loss but at the same time it will increase the fixed cost. Thus, optimum thickness which will help in minimum heat loss and minimum cost is calculated.

Optimum Insulation Thickness



Effect of Temperature on Thermal Conductivity:

Variation of the thermal conductivity with temperature

Effect on Thermal Conductivity of Cu & Al with temperature

Temperature, K Copper Aluminum

100 482 302200 413 237300 401 237400 393 240600 379 231800 366 218



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Prepared By: Prof. Kartik Desai, Sr. Lecturer,N.G. Patel Polytechnic, Bardoli


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BASIC LAW’S OF RADIATION 1) Stefan‐Boltzmann law: The total radiation energy emitted by a black body per unit

time and per unit surface area was given as:

Total blackbody emissive power, Eb = σ *T4 , where σ =5.67 x10‐8 W/(m2 ∙ K4)

2) Plank’s law: Spectral(or Monochromatic) blackbody emissive power (Ebλ) :

The amount of radiation energy emitted by a blackbody at a thermodynamic temperature (T) per unit time, per unit surface area, and per unit wavelength(λ).

Ebλ(λ,T) = C1/[λ5*(exp(C2/λ*T)‐1)

where, C1 = 2Π*h*c02, C2 = h*c0/k, h : Plank’s constant, k : Boltzmann’s constant 3) Wien’s displacement law: The wavelength at which the peak spectral blackbody

emissive power occurs for a specified temperature is given by Wien’s displacement law.

(λ*T)max power =2897 μm*K 4) Kirchhoff’s law: The total hemispherical emissivity of a surface at temperature T is

equal to its total hemispherical absorptivity for radiation coming from a blackbody at same temperature.

ε(T) = α(T) QUESTIONS: 1) Why the negative sign appears in Fourier's law of heat conduction? 2) A sealed cube containing argon floating in deep space is heated on one side by a nearby star. Will there be any heat transfer due to natural convection? 3) What is thermal radiation? 4) Define: Emmisivity, Absorptivity, Reflectivity, Transmitivity, Black body. 5) Explain Stefan‐Boltzman’s law. Kartik R. Desai Sr. Lecturer, N.G.Patel Polytechnic, Isroli‐Afwa.




HEAT TRANSFER TECHNOLOGY DATE: 24/10/2010 BY: H. S. DESAI What Is Heat Transfer? HEAT TRANSFER is the science which deals with the rates of exchange of heat between hot and cold bodies called the SOURCE and the RECIEVER. What are Heat Transfer Equipments? A heat transfer equipments are the devices built for efficient heat transfer from one medium to another. Classification of Heat Transfer Equipments:

1. According to Mechanical Design: ‐ Double Pipe Heat Exchanger ‐ Shell & Tube type Heat Exchanger ‐ Plat Type Heat Exchanger ‐ Extended Surface (Fin) type Heat Exchanger ‐ Mechanically Agitated Reactor (With Jacket and / or Coil)

2. According to Nature of Flow arrangement: ‐ Parallel (Co‐current Flow) ‐ Opposite (Counter Current) ‐ Perpendicular (Cross Flow)

3. According to Number of Passes: ‐ Single Pass ‐ Multiple Pass

4. According to Physical State of Heat Exchanging Fluids: ‐ Gas to Gas ‐ Liquid to Liquid (Coolers) ‐ Gas to Liquid (Economizer) ‐ Liquid or gas to condensing vapors (Condensers) ‐ Gas to Boiling liquid (Evaporators)

5. According to Operating Pressure Range: ‐ Low Pressure (0 to 8 atm) ‐ Medium Pressure (8 to 35 atm) ‐ High Pressure (35 atm and above)

Prepared By: Prof. Himanshu Desai, Lecturer,N.G. Patel Polytechnic, Bardoli


Types of Heat Transfer Equipments:

• Shell and tube • Double pipe • Plate • Finned tubes/gas heaters • spiral • Vessel jackets • Reboilers and vaporisers / evaporators Etc • Agitated Vessels or Agitated Reactors.

Important Terminologies: Range, Approach, Newton’s Law of Cooling, Heat Transfer Coefficient Temperature Profile & Logarithmic Mean Temperature Difference (LMTD):



What is Shell & Tube Heat Exchanger? A shell and tube heat exchanger is a class of heat exchanger designs. It is the most common type of heat exchanger in oil refineries and other large chemical processes, and is suited for higher‐pressure applications. As its name implies, this type of heat exchanger consists of a shell (a large pressure vessel) with a bundle of tubes inside it. One fluid runs through the tubes, and another fluid flows over the tubes (through the shell) to transfer heat between the two fluids. The set of tubes is called a tube bundle, and may be composed by several types of tubes. Types of Shell & Tube Heat Exchangers:



Construction of Shell & Tube Heat Exchangers: Shell & Tube heat exchangers consist of many components such as: Shell, Tube Bundle, Baffles, Tie Rod, Tube sheet, Pass, Inlet & Outlet Nozzles.

Baffles: Baffles are metal discs cut to certain area. They are used to increase the

residence time of shell side fluid, to increase the turbulence and to support the tubes.

Tube Sheet: Tube bundle is welded to the tube sheet. Pass: It is a metal plate used to create multiple passes in the flow. Tie Rod: They are used to keep the baffles at the prescribed positions. Nozzles: They are standard pipes used for entry and exit of shell and tube side fluids. Working of Shell & Tube Heat Exchanger:

• In the shell and tube heat exchanger shown above, the shell side fluid flows once through the shell and the tube side fluid flows twice through the tube bundle.

• In this type of heat exchangers the tube side fluid flows in co‐current and counter current fashion with respect to shell side fluid.




HEAT TRANSFER TECHNOLOGY Date: 05/12/2010 By: H. S. Desai

CONDENSERS & STEAM TRAP

CONDENSER:

• A condenser is a device or unit used to condense a substance from its gaseous to its liquid state, typically by cooling it.

• In doing so, the latent heat is given up by the substance, and will transfer to the condenser coolant. Condensers are typically heat exchangers which have various designs and come in many sizes ranging from rather small to very large industrial‐scale units used in plant processes.

• Condensers are used in air conditioning, industrial chemical processes such as distillation, steam power plants and other heat‐exchange systems. Use of cooling water or surrounding air as the coolant is common in many condensers.

Types of Condensers: Surface Condensers: Surface condenser is the commonly used term for a water‐cooled shell and tube heat exchanger installed on the exhaust steam from a steam turbine in thermal power stations. These condensers are heat exchangers which convert steam from its gaseous to its liquid state at a pressure below atmospheric pressure. Where cooling water is in short supply, an air‐cooled condenser is often used. An air‐cooled condenser is however significantly more expensive and cannot achieve as low a steam turbine exhaust pressure as water cooled surface condenser. Contact condenser: Contact Condenser is a device in which a vapor is brought into direct contact with a cooling liquid and condensed by giving up its latent heat to the liquid. In almost all cases the cooling liquid is water, and the condensing vapor is steam. In this type of condenser the steam and cooling water are mixed in a condensing chamber and withdrawn together. STEAM TRAP: A steam trap is a device used to discharge condensate and non condensable gases with a negligible consumption or loss of live steam. Most steam traps are nothing more than automatic valves. They open, close or modulate automatically. The three important functions of steam traps are:

1. Discharge condensate as soon as it is formed. 2. Have negligible steam consumption. 3. Have the capability of discharging air and other non‐condensable gases.



Types of Steam Trap:

1. Mechanical traps. They have a float that rises and falls in relation to condensate level and this usually has a mechanical linkage attached that opens and closes the valve. Mechanical traps operate in direct relationship to condensate levels present in the body of the steam trap. Inverted bucket and float traps are examples of mechanical traps.

2. Temperature traps. They have a valve that is driven on / off the seat by either

expansion / contraction caused by temperature change. They differ from mechanical traps in that their design requires them to hold back some condensate waiting for it to cool sufficiently to allow the valve to open. In most circumstances this is not desirable as condensate needs to be removed as soon as it is formed. Thermostatic traps, BiThermostatic traps and bimetallic traps are examples of temperature operated traps.

3. Thermodynamic (TD) traps. Thermodynamic traps work on the difference in

dynamic response to velocity change in flow of compressible and incompressible fluids. As steam enters, static pressure above the disk forces the disk against the valve seat. The static pressure over a large area overcomes the high inlet pressure of the steam. As the steam starts to condense, the pressure against the disk lessens and the trap cycles. This essentially makes a TD trap a "time cycle" device: it will open even if there is only steam present, this can cause premature wear. If non condensable gas is trapped on top of the disc, it can cause the trap to be locked shut.



‐ 0 ‐

Performance Assess.& Energy conservation approach for industrial boiler‐H.R.Jivanramjwala‐N.G.Patel Polytechnic

PERFORMANCE ASSESSMENT AND

ENERGY CONSERVATION APPROACH FOR

INDUSTRIAL BOILER

Prepared By: Prof. H.R. Jivanramjiwala, Lecturer,N.G. Patel Polytechnic, Bardoli


‐ 1 ‐


1. INTRODUCTION...........................................................................................1

2. TYPE OF BOILERS.......................................................................................2

3. ASSESSMENT OF A BOILER .....................................................................4

4. ENERGY EFFICIENCY OPPORTUNITIES .......................................... 14

1. INTRODUCTION

This section briefly describes the Boiler and various auxiliaries in the Boiler Room.

A boiler is an enclosed vessel that provides a means for combustion heat to be transferred to water until it becomes heated water or steam. The hot water or steam under pressure is then usable for transferring the heat to a process. Water is a useful and inexpensive medium for transferring heat to a process. When water at atmospheric pressure is boiled into steam its volume increases about 1,600 times, producing a force that is almost as explosive as gunpowder. This causes the boiler to be an equipment that must be treated with utmost care.

The boiler system comprises of: a feed water system, steam system and fuel system. The feed water system provides water to the boiler and regulates it automatically to meet the steam demand. Various valves provide access for maintenance and repair. The steam system collects and controls the steam produced in the boiler. Steam is directed through a piping system to the point of use. Throughout the system, steam pressure is regulated using valves and checked with steam pressure gauges. The fuel system includes all equipment used to provide fuel to generate the necessary heat. The equipment required in the fuel system depends on the type of fuel used in the system.

The water supplied to the boiler that is converted into steam is called feed water. The two sources of feed water are: (1) Condensate or condensed steam returned from the processes and (2) Makeup water (treated raw water) which must come from outside the boiler room and plant processes. For higher boiler efficiencies, an economizer preheats the feed water using the waste heat in the flue gas.



2


BOILER BURNER

Econo mizer

Water Source

STEAM TO PROCESS

EXHAUST GAS VENT

STACK

Econo mizer

DEAERATOR PUMPS

VENT BOILER

BURNER

BLOW DOWN SEPARATOR

FUEL

Water Source

CHEMICAL FEED

BRINE

SOFTENERS

Figure 1. Schematic diagram of a Boiler Room

2. TYPE OF BOILERS

This section describes the various types of boilers: Fire tube boiler, Water tube boiler, Packaged boiler, Fluidized bed combustion boiler, Stoker fired boiler, Pulverized fuel boiler, Waste heat boiler and Thermic fluid heater. 2.1 Fire Tube Boiler

In a fire tube boiler, hot gases pass through the tubes and boiler feed water in the shell side is converted into steam. Fire tube boilers are generally used for relatively small steam capacities and low to medium steam pressures. As a guideline, fire tube boilers are competitive for steam rates up to 12,000 kg/hour and pressures up to 18 kg/cm2. Fire tube boilers are available for operation with oil, gas or solid fuels. For economic reasons, most fire tube boilers are of “packaged” construction (i.e. manufacturer erected) for all fuels.

Figure 2. Sectional view of a Fire Tube Boiler (Light Rail Transit Association)



3


2.2 Water Tube Boiler

In a water tube boiler, boiler feed water flows through the tubes and enters the boiler drum. The circulated water is heated by the combustion gases and converted into steam at the vapour space in the drum. These boilers are selected when the steam demand as well as steam pressure requirements are high as in the case of process cum power boiler / power boilers.

Most modern water boiler tube designs are within the capacity range 4,500 – 120,000 kg/hour of steam, at very high pressures. Many water tube boilers are of “packaged” construction if oil and /or gas are to

be used as fuel. Solid fuel fired water tube designs are available but packaged designs are less common.

The features of water tube boilers are:

Figure 3. Simple Diagram of Water Tube Boiler (YourDictionary.com)

� Forced, induced and balanced draft provisions help to improve combustion efficiency. � Less tolerance for water quality calls for water treatment plant. � Higher thermal efficiency levels are possible

2.3 Packaged Boiler

To Chimney

The packaged boiler is so called because it comes as a complete package. Once delivered to a site, it requires only the steam, water pipe work, fuel supply and electrical connections to be made to become operational. Package boilers are generally of a shell type with a fire tube design so as to achieve high heat transfer rates by both radiation and convection.

Oil Burner

Figure 4. A typical 3 Pass, Oil fired packaged boiler

(BIB Cochran, 2003)



4


Stack Stochiometric Gas Excess Air

Un burnt

Ash

and Un-

burnt parts ofFuel in A

sh

Blow

Dow

n

Convection

& R

adiation

The features of packaged boilers are: � Small combustion space and high heat release rate resulting in faster evaporation. � Large number of small diameter tubes leading to good convective heat transfer. � Forced or induced draft systems resulting in good combustion efficiency. � Number of passes resulting in better overall heat transfer. � Higher thermal efficiency levels compared with other boilers.

These boilers are classified based on the number of passes - the number of times the hot combustion gases pass through the boiler. The combustion chamber is taken, as the first pass after which there may be one, two or three sets of fire-tubes. The most common boiler of this class is a three-pass unit with two sets of fire-tubes and with the exhaust gases exiting through the rear of the boiler.

3. ASSESSMENT OF A BOILER

This section describes the Performance evaluation of boilers (through the direct and indirect method including examples for efficiency calculations), boiler blow down, and boiler water treatment.

3.1 Performance Evaluation of a Boiler

The performance parameters of a boiler, like efficiency and evaporation ratio, reduces with time due to poor combustion, heat transfer surface fouling and poor operation and maintenance. Even for a new boiler, reasons such as deteriorating fuel quality and water quality can result in poor boiler performance. A heat balance helps us to identify avoidable and unavoidable heat losses. Boiler efficiency tests help us to find out the deviation of boiler efficiency from the best efficiency and target problem area for corrective action.

3.1.1 Heat balance

The combustion process in a boiler can be described in the form of an energy flow diagram. This shows graphically how the input energy from the fuel is transformed into the various useful energy flows and into heat and energy loss flows. The thickness of the arrows indicates the amount of energy contained in the respective flows.

Stack Gas

Stochiometric Excess Air Un burnt

FUEL INPUT STEAM OUTPUT

Figure 11. Energy balance diagram of a boiler



5


Dry Flue Gas Loss

BOILER

Heat in Steam

Heat loss due to dry flue gas

Heat loss due to steamDry Flue Gas Loss in flue gas

Heat loss due to moisture in fuel

Heat loss due to unburnts in residue

Heat loss due to moisture in air

Heat loss due to radiation & other unaccounted loss

BOILER

8.1 %

1.7 %

0.3 %

2.4 %

1.0 %

73.8 %

A heat balance is an attempt to balance the total energy entering a boiler against that leaving the boiler in different forms. The following figure illustrates the different losses occurring for generating steam.

100.0 %

Fuel

12.7 %

Heat loss due to dry flue gas Heat loss due to steam in flue gas

Heat loss due to moisture in fuel

Heat loss due to moisture in air

Heat loss due to unburnts in residue Heat loss due to radiation & other unaccounted loss Heat in Steam

Figure 12. Typical Losses from Coal Fired Boiler

The energy losses can be divided in unavoidable and avoidable losses. The goal of a Cleaner Production and/or energy assessment must be to reduce the avoidable losses, i.e. to improve energy efficiency. The following losses can be avoided or reduced: � Stack gas losses:

- Excess air (reduce to the necessary minimum which depends from burner technology, operation, operation (i.e. control) and maintenance).

- Stack gas temperature (reduce by optimizing maintenance (cleaning), load; better burner and boiler technology).

� Losses by unburnt fuel in stack and ash (optimize operation and maintenance; better technology of burner).

� Blow down losses (treat fresh feed water, recycle condensate) � Condensate losses (recover the largest possible amount of condensate) � Convection and radiation losses (reduced by better insulation of the boiler).

3.1.2 Boiler efficiency

Thermal efficiency of a boiler is defined as “the percentage of (heat) energy input that is effectively useful in the generated steam.”

There are two methods of assessing boiler efficiency: � The Direct Method: the energy gain of the working fluid (water and steam) is compared

with the energy content of the boiler fuel � The Indirect Method: the efficiency is the difference between the losses and the energy

input



6


3.1.3 Direct method of determining boiler efficiency

Methodology

This is also known as ‘input-output method’ due to the fact that it needs only the useful output (steam) and the heat input (i.e. fuel) for evaluating the efficiency. This efficiency can be evaluated using the formula:

Heat Output

Boiler Efficiency (η) = x 100 Heat Input

Boiler Efficiency (η) = Q x (hg – hf)

q x GCV

x 100

Parameters to be monitored for the calculation of boiler efficiency by direct method are:

� Quantity of steam generated per hour (Q) in kg/hr. � Quantity of fuel used per hour (q) in kg/hr. � The working pressure (in kg/cm2(g)) and superheat temperature (oC), if any � The temperature of feed water (oC) � Type of fuel and gross calorific value of the fuel (GCV) in kcal/kg of fuel

And where

� hg – Enthalpy of saturated steam in kcal/kg of steam � hf – Enthalpy of feed water in kcal/kg of water

Example

Find out the efficiency of the boiler by direct method with the data given below: � Type of boiler: Coal fired � Quantity of steam (dry) generated: 10 TPH

2 0 � Steam pressure (gauge) / temp: 10 kg/cm (g)/ 180 C � Quantity of coal consumed: 2.25 TPH

0 � Feed water temperature: 85 C � GCV of coal: 3200 kcal/kg

2 � Enthalpy of steam at 10 kg/cm pressure: 665 kcal/kg (saturated) � Enthalpy of feed water: 85 kcal/kg

10 x (665 – 85) x 1000

Boiler Efficiency (η) = x 100 = 80.56 percent 2.25 x 3200 x 1000

Advantages of direct method

� Plant workers can evaluate quickly the efficiency of boilers � Requires few parameters for computation



7


� Needs few instruments for monitoring � Easy to compare evaporation ratios with benchmark figures

Disadvantages of direct method

� Does not give clues to the operator as to why efficiency of the system is lower � Does not calculate various losses accountable for various efficiency levels

3.1.4 Indirect method of determining boiler efficiency

Methodology

The reference standards for Boiler Testing at Site using the indirect method are the British Standard, BS 845:1987 and the USA Standard ASME PTC-4-1 Power Test Code Steam Generating Units.

The indirect method is also called the heat loss method. The efficiency can be calculated by subtracting the heat loss fractions from 100 as follows:

Efficiency of boiler (n) = 100 - (i + ii + iii + iv + v + vi + vii)

Whereby the principle losses that occur in a boiler are loss of heat due to:

i. Dry flue gas ii. Evaporation of water formed due to H2 in fuel iii. Evaporation of moisture in fuel iv. Moisture present in combustion air v. Unburnt fuel in fly ash vi. Unburnt fuel in bottom ash vii. Radiation and other unaccounted losses

Losses due to moisture in fuel and due to combustion of hydrogen are dependent on the fuel, and cannot be controlled by design.

The data required for calculation of boiler efficiency using the indirect method are: � Ultimate analysis of fuel (H2, O2, S, C, moisture content, ash content) � Percentage of oxygen or CO2 in the flue gas � Flue gas temperature in oC (Tf) � Ambient temperature in oC (Ta) and humidity of air in kg/kg of dry air � GCV of fuel in kcal/kg � Percentage combustible in ash (in case of solid fuels) � GCV of ash in kcal/kg (in case of solid fuels)

A detailed procedure for calculating boiler efficiency using the indirect method is given below. However, practicing energy managers in industry usually prefer simpler calculation procedures.



8


Step 1: Calculate the theoretical air requirement

= [(11.43 x C) + {34.5 x (H2 – O2/8)} + (4.32 x S)]/100 kg/kg of fuel

Step 2: Calculate the percent excess air supplied (EA)

= O2 percent x 100 -------------

(21 - O2 percent)

Step 3: Calculate actual mass of air supplied/ kg of fuel (AAS)

= {1 + EA/100} x theoretical air

Step 4: Estimate all heat losses

i. Percentage heat loss due to dry flue gas

= m x Cp x (Tf-Ta) x 100 ----------------------------

GCV of fuel

Where, m = mass of dry flue gas in kg/kg of fuel m = (mass of dry products of combustion / kg of fuel) + (mass of N2 in fuel on 1 kg basis ) + (mass of N2 in actual mass of air we are supplying). Cp = Specific heat of flue gas (0.23 kcal/kg )

ii. Percentage heat loss due to evaporation of water formed due to H2 in fuel

= 9 x H2 {584+Cp (Tf-Ta)} x 100

-------------------------------------- GCV of fuel

Where, H2 = percentage of H2 in 1 kg of fuel

Cp = specific heat of superheated steam (0.45 kcal/kg)

iii. Percentage heat loss due to evaporation of moisture present in fuel

= M{584+ Cp (Tf-Ta)} x 100 ---------------------------------

GCV of fuel

Where, M – percent moisture in 1kg of fuel Cp – Specific heat of superheated steam (0.45 kcal/kg)

iv. Percentage heat loss due to moisture present in air

= AAS x humidity factor x Cp (Tf-Ta)} x 100

--------------------------------------------------- GCV of fuel



9


C: H2:

84 percent 12.0 percent

S: 3.0 percentO2: 1 percent

�

GCV of Oil: 10200 kcal/kg

�

Percentage of Oxygen: 7 percent

Where, Cp – Specific heat of superheated steam (0.45 kcal/kg)

v. Percentage heat loss due to unburnt fuel in fly ash

= Total ash collected/kg of fuel burnt x GCV of fly ash x 100 -----------------------------------------------------------------------

GCV of fuel

vi. Percentage heat loss due to unburnt fuel in bottom ash

= Total ash collected per Kg of fuel burnt x G.C.V of bottom ash x 100 -----------------------------------------------------------------------------------

GCV of fuel

vii. Percentage heat loss due to radiation and other unaccounted loss

The actual radiation and convection losses are difficult to assess because of particular emissivity of various surfaces, its inclination, airflow patterns etc. In a relatively small boiler, with a capacity of 10 MW, the radiation and unaccounted losses could amount to between 1 percent and 2 percent of the gross calorific value of the fuel, while in a 500 MW boiler, values between 0.2 percent to 1 percent are typical. The loss may be assumed appropriately depending on the surface condition.

Step 5: Calculate boiler efficiency and boiler evaporation ratio


Evaporation Ratio = Heat utilized for steam generation/Heat addition to the steam

Evaporation ratio means kilogram of steam generated per kilogram of fuel consumed. Typical Examples are:

� Coal fired boiler: 6 (i.e. 1 kg of coal can generate 6 kg of steam) � Oil fired boiler: 13 (i.e. 1 kg of oil can generate 13 kg of steam)

However, the evaporation ratio will depend upon type of boiler, calorific value of the fuel and associated efficiencies.

Example

� Type of boiler: Oil fired � Ultimate analysis of Oil



10


� Percentage of CO2: 11 percent� Flue gas temperature (Tf): 220 0C � Ambient temperature (Ta): 27 0C � Humidity of air : 0.018 kg/kg of dry air

Step-1: Calculate the theoretical air requirement

= [(11.43 x C) + [{34.5 x (H2 – O2/8)} + (4.32 x S)]/100 kg/kg of oil = [(11.43 x 84) + [{34.5 x (12 – 1/8)} + (4.32 x 3)]/100 kg/kg of oil = 13.82 kg of air/kg of oil

Step-2: Calculate the percent excess air supplied (EA)

Excess air supplied (EA) = (O2 x 100)/(21-O2) = (7 x 100)/(21-7) = 50 percent

Step 3: Calculate actual mass of air supplied/ kg of fuel (AAS)

AAS/kg fuel = [1 + EA/100] x Theo. Air (AAS)

= [1 + 50/100] x 13.82 = 1.5 x 13.82 = 20.74 kg of air/kg of oil

Step 4: Estimate all heat losses

i. Percentage heat loss due to dry flue gas

m x Cp x (Tf – Ta ) x 100 = -----------------------------

GCV of fuel

m = mass of CO2 + mass of SO2 + mass of N2 + mass of O2

0.84 x 44 0.03 x 64 20.74 x 77 m = ----------- + ---------- + ----------- (0.07 x 32)

12 32 100

m = 21.35 kg / kg of oil

21.35 x 0.23 x (220 – 27) = ------------------------------- x 100

10200

= 9.29 percent



11


A simpler method can also be used: Percentage heat loss due to dry flue gas

m x Cp x (Tf – Ta ) x 100

= ----------------------------- GCV of fuel

m (total mass of flue gas)

= mass of actual air supplied + mass of fuel supplied

= 20.19 + 1 = 21.19

= 21.19 x 0.23 x (220-27)

------------------------------- x 100 10200

= 9.22 percent

ii. Heat loss due to evaporation of water formed due to H2 in fuel

9 x H2 {584+0.45 (Tf – Ta )}

= --------------------------------- GCV of fuel where H2 = percentage of H2 in fuel

9 x 12 {584+0.45(220-27)}

= -------------------------------- 10200

= 7.10 percent

iii. Heat loss due to moisture present in air

AAS x humidity x 0.45 x ((Tf – Ta ) x 100

= ------------------------------------------------- GCV of fuel

= [20.74 x 0.018 x 0.45 x (220-27) x 100]/10200

= 0.317 percent

iv. Heat loss due to radiation and other unaccounted losses

For a small boiler it is estimated to be 2 percent

Step 5: Calculate boiler efficiency and boiler evaporation ratio


i. Heat loss due to dry flue gas : 9.29 percent



12


ii. Heat loss due to evaporation of water formed due to H2 in fuel : 7.10 percent iii. Heat loss due to moisture present in air : 0.317 percentiv. Heat loss due to radiation and other unaccounted losses : 2 percent

= 100- [9.29+7.10+0.317+2]

= 100 – 17.024 = 83 percent (approximate)

Evaporation Ratio = Heat utilized for steam generation/Heat addition to the steam

= 10200 x 0.83 / (660-60) = 14.11 (compared to 13 for a typical oil fired boiler)

Advantages of indirect method � A complete mass and energy balance can be obtained for each individual stream, making

it easier to identify options to improve boiler efficiency

Disadvantages of indirect method � Time consuming � Requires lab facilities for analysis

4. ENERGY EFFICIENCY OPPORTUNITIES

This section includes energy efficiency opportunities related to combustion, heat transfer, avoidable losses, auxiliary power consumption, water quality and blow down.

Energy losses and therefore energy efficiency opportunities in boilers can be related to combustion, heat transfer, avoidable losses, high auxiliary power consumption, water quality and blow down. The various energy efficiency opportunities in a boiler system can be related to: 1. Stack temperature control 2. Feed water preheating using economizers 3. Combustion air pre-heating 4. Incomplete combustion minimization 5. Excess air control 6. Radiation and convection heat loss avoidance 7. Automatic blow down control 8. Reduction of scaling and soot losses 9. Reduction of boiler steam pressure 10. Variable speed control for fans, blowers and pumps 11. Controlling boiler loading 12. Proper boiler scheduling 13. Boiler replacement



Prof. Latesh B. Chaudhari,

Asst. Professor,

Faculty of Engineering Technology & Research, Isroli- Bardoli.

Prepared By: Prof.Latesh Chaudhari, Asst. Prof.,F.E.T.R., Bardoli


CONVECTIONCONVECTION

C ti i thConvection is the transfer of heat by the actual movement of the warmed matter.

•Convection is the transfer of heat energytransfer of heat energy in a gas or liquid by

fmovement of currents.




H t d iHeated air rises, cools,

then falls

Hot water rises, cools,

and falls

Air near heater isreplaced by cooler

air andthen fallsand falls air, andthe cycle repeats



SEA BREEZE AND SEA BREEZE AND LAND BREEZELAND BREEZE

SEA BREEZE LAND BREEZE



CONDUCTIONCONDUCTIONCONDUCTIONCONDUCTION

• Conduction is the • Conduction is the transfer of energy through matter from through matter from particle to particle.

• It is the transfer and distribution of heat energy from atom to atom within a atom within a substance.



HOW DOES CONDUCTION HOW DOES CONDUCTION HAPPEN?HAPPEN?HAPPEN?HAPPEN?



RADIATIONRADIATIONRADIATIONRADIATIONSunlight is a form of

radiation that is radiated through space to our planet

ith t th id f fl idwithout the aid of fluids or solids.

Since there are no fluids inSince there are no fluids in space, convection is not responsible for transferring theresponsible for transferring the heat. Thus, radiation brings heat to our planet.heat to our planet.



RADIATIONRADIATIONRADIATIONRADIATIONPrepared By: Prof.Latesh Chaudhari, Asst. Prof.,F.E.T.R., Bardoli


RADIATION AMONG RADIATION AMONG RADIATION AMONG RADIATION AMONG OBJECTSOBJECTS



HEAT TRANSFER

Prof Latesh B ChaudhariProf. Latesh B. ChaudhariAsst. Professor

l f h & hFaculty of Engg. Tech. & ResearchIsroli-Bardoli



What we are going to learnWhat we are going to learn

How heat spreads from one region to another.

Examples:

How the whole copper rod get hot when we heat one pp d g aendHow does the heat spread throughout the whole pot of

t h l h t th b ttwater when we are only heating the bottom

How does the heat from the Sun reaches us



Heat TransferHeat Transfer

Heat always moves from a warmer place to a cooler place.pHot objects in a cooler room will cool to room temperatureroom temperature.Cold objects in a warmer room will heat up to room temperatureto room temperature.



QuestionQuestion

If a cup of coffee and an ice lolly were left on the table in this room what would happen to them? Why?

The cup of coffee will cool until it reaches room temperature The lolly will melt androom temperature. The lolly will melt and then the liquid will cool to room temperaturetemperature.



What is happening?What is happening?

All things are made up of moleculesWhen things get heated, they absorb heat energy.This means that the molecules are absorbing the heat energyWith th l l bl tWith more energy, the molecules are able to move fasterWhen the molecules move faster the temperatureWhen the molecules move faster, the temperature of the object increases.Temperature increase means the object gets p j ghotter.



3 Processes of Heat Transfer3 Processes of Heat Transfer

Conduction

ConvectionConvection

Radiation



CONDUCTIONCONDUCTION

Occurs mainly in solids

Two types of conductionMolecular vibrationFree electron diffusion

Note: Conduction is not the main form of heat transfer in liquids and gases because their s e qu ds d g ses bec use emolecules are spaced further apart.



STEADY STATESTEADY STATESTEADY STATESTEADY STATE��THERMAL CODUCTIVITY�THERMAL CODUCTIVITY�

STEADY STATE IS THE STATE WHEN THERE IS NO MORE ABSORPTION OF HEAT AND THE TEMPERATURE BECOMES CONSTANTSTEADY STATE IS THE STATE WHEN THERE IS NO MORE ABSORPTION OF HEAT AND THE TEMPERATURE BECOMES CONSTANT��

DURING STEADY STATE� DIFFERENT PARTICLES MAY HAVE DIFFERENT TEMPERATURES BUT DIFFERENT PARTICLES HAVE CONSTANT TEMPERATURE� DURING STEADY STATE� DIFFERENT PARTICLES MAY HAVE DIFFERENT TEMPERATURES BUT DIFFERENT PARTICLES HAVE CONSTANT TEMPERATURE�



Molecular VibrationMolecular Vibration

When heat is supplied to one end, the molecules at the hot end start to vibrate more vigorously.

In the process, they ‘bump’ into their neighboring molecules. In doing so, some energy is transferred to the neighbour.

The neighbour molecule gains energy and starts to ib t i l Th l tivibrate more vigorously. The cycle continues.



Conduction AnimationConduction Animation

The conduction animation is on the left.Notice the movement of the moleculesNotice the movement of the molecules along the rod.

Cli k h f i tiClick here for animation



Free electron DiffusionFree electron Diffusion

This form of conduction takes place only in metals. As only metals have free electronsonly metals have free electrons.

The electrons are freed from the molecule when heated and they travel towards the cold end.and they travel towards the cold end.

At the cold end they collide into a molecule therefore passing all their energy to the molecule.p g gy



Comparing the 2 mechanismsComparing the 2 mechanisms

Molecular vibration Free electron diffusion

Occurs in all solids Occurs in metals onlyOccurs in all solids

Slow process

Occurs in metals only

Fast processSlow process Fast process

This explains why metals heat up faster:

1. Metals have 2 mechanisms of conduction occuring at the same time.

2. In metals, free electron diffusion is the main mechanism, which is faster.



Conductors and InsulatorsConductors and Insulators

Materials that can conduct heat easily and readily (eg. Metals) are known as y ( g )conductors.Materials that do not conduct heat easilyMaterials that do not conduct heat easily (eg. Water, air, plastic) are known as insulatorsinsulators.



ConductionConduction

CONDUCTION is when heat moves through a ________. Metals are better conductors of heat than plastic or wood.than plastic or wood.

This is whyThis is why…These parts are made of plastic ormade of plastic or wood:

These parts areThese parts are made of metal:




In this experiment we saw that liquid water is a poor conductor of heat. The ice cube did not melt even though the water above it wasnot melt even though the water above it was boiling.

A piece of ice fixed (with gauze) at the bottom of the testA piece of ice fixed (with gauze) at the bottom of the test--tube will not tube will not melt even when the water is boiling at the top. Melting has to await melt even when the water is boiling at the top. Melting has to await conduction of energy through the water. conduction of energy through the water.

The hot water carries its own thermal energy with it when it moves. The hot water carries its own thermal energy with it when it moves. That is the process of convection. It happens in a fluid whether it is a That is the process of convection. It happens in a fluid whether it is a liquid or a gas On a large scale it happens in the oceans and in theliquid or a gas On a large scale it happens in the oceans and in theliquid or a gas. On a large scale it happens in the oceans and in the liquid or a gas. On a large scale it happens in the oceans and in the atmosphere, in hot water heating systems and in ventilation. Winds are atmosphere, in hot water heating systems and in ventilation. Winds are just convection currents.just convection currents.




Air is also a poor conductor of heat. This is how many types of house insulation and clothing keep us warm.

•Conduction and convection both need co vec o bo eedparticles to carry energy when heat is transferred



Applications of conductionApplications of conduction

Trapping air as insulationDifferent sensations from conductors andDifferent sensations from conductors and insulatorsUses of good conductors: cooking utensilsUses of good conductors: cooking utensilsUses of good insulators: table mats, h dlhandles



ConvectionConvection

Occurs in liquids and gasesDoes not occur in solids because theDoes not occur in solids because the molecules are not free to move around



What happens during convectionWhat happens during convection

Taking the example of heating waterWater at the bottom is heated firstHeated water expandsWhen water expands density decreasesHeated water of lower density starts to riseCooler water of higher density rushes in from sides to take its placesides to take its placeThe cooler water gets heated and the cycle repeatsrepeats.Convection currents are set up.



Convection AnimationConvection Animation

Refer to the animation on the right.The animation shows the process ofThe animation shows the process of convection taking place.The blue circles represent unheatedThe blue circles represent unheated molecules.Th dd h l l h hThe redder the molecules get the more heat they are absorbing.



Applications of convectionApplications of convection

Air con is usually placed at the top of a room.Heating coil of a kettle is usually at the bottombottomFormation of land and sea breezes



Convection

Words to use: expands, radiators, dense, heated, current, floatfloat



Heating iceTemp/Temp/OO

CC Heating ice150

CC

This flat line shows where energy

100is being used to push the particles further apart for evaporation

50

00

50

Time/sThis flat line shows where energy is being used to break bonds –-50 is being used to break bonds –this has to be done during melting



Particles gain energy and become less dense and rise. They pass theirThey pass their energy on and become more d d i kdense and sink. Heat up, less dense rise……

Current



A hot air balloon usesA hot air balloon uses…..

Explain.pWhen the air particles in the balloon are heated they move apart. The air y pexpands and becomes less dense. This causes the hot air to rise and the balloon does toothe balloon does too.



Why does the balloon float back down?

As it rises the warm air meets cooler air and passes the energy on. Having passed p gy g pon the energy it cools down and becomes denser which sinks.

This sets up a convection currentThis sets up a convection current.



RadiationRadiation

Radiation does not require a medium to transfer heat. (can occur in a vacuum)( )Sun releases electromagnetic waves (heat is contained in the waves as infra-red)is contained in the waves as infra red)Hotter objects radiates more heat.



Emitters and absorbersEmitters and absorbers

The Sun gives out the heat.It is known as an emitter / radiatorIt is known as an emitter / radiator

Th E h k i h hThe Earth takes in the heat.It is known as an absorber.



RadiationBlack ________ infrared radiation (i.e. heat) from the sun better than white does.

I’m very hot!I’m cool!

RadiationRadiation cancan happenhappen throughthroughpppp ggabsolutelyabsolutely ________________ becausebecause thethe heatheatisis movingmoving byby ____________.. ThisThis isis howhow heatheatreachesreaches usus fromfrom thethe ________..

Words – sun, absorbs, nothing, warmer, waves



Good and Bad Emitters/Absorbers

A good emitter would also be a good absorber.A poor emitter would be a poor absorber.

Good emitter/absorber Poor emitter/absorberGood emitter/absorber Poor emitter/absorber

Dull, black surface Shiny, silver surfaceRough surface Smooth surface



Factors affecting radiationFactors affecting radiation

Colour and texture of the surface (refer to previous slide)p )Surface temperatureHigher surface temperature higher rate ofHigher surface temperature, higher rate of transfer. S fSurface areaLarger surface area, higher rate of transfer



Applications of radiationApplications of radiation

TeapotsThe greenhouseThe greenhouseColour and texture of clothingsSkiSkin cancer



The vacuum flaskThe vacuum flask

The vacuum prevents conduction and convectionThe silvered surfaces

d di ireduces radiationCap and base are

d f dmade of good insulators to reduce conductionconduction



HEAT TRANSFER BY

RADIATIONRADIATION

INDUSTRIAL CHEMICAL TECHNOLOGY -2010MODULE-1

Heat Transfer Operationp

KARTIK R. DESAISr Lecturer Chemical Engineering DepartmentSr. Lecturer, Chemical Engineering Department,

N.G. Patel Polytechnic, Isroli-Afwa



OUTLINE

Introduction

Electromagnetic Waves

Thermal Radiation

Terms involved

Basic Law’s of Radiation



INTRODUCTION

HEAT is a form of energy which flow as a result of temperature gradient.



INTRODUCTION

HEAT TRANSFER is the science which deals with the rates of exchangeof heat between hot and cold bodies called the SOURCE and theRECIEVERRECIEVER.

© Copy Rights, www.cctech.co.in



INTRODUCTION

Heat is transferred by three primary modes:

1) CONDUCTION: It is the primary mode of heat transfer through solid. Conduction occurs by two h imechanisms:

i) Molecular Motion - Molecules of higher energy (motion) impart that energy to adjacent molecules of lesser energy.ii) Migration of free electrons - This is primarily associated with pure metals.) g p y p

Mathematically conduction is describe by Fourier's Law:Mathematically, conduction is describe by Fourier s Law:

Where A is the cross-sectional area that the heat is flowing through the solid, ∆T is the temperature gradient in the solid. (k) is the thermal conductivity of the material.

The thermal conductivity is a physical property of the solid. It the a measure of the materials ability to conduct heat.

Q: Why the negative sign appears in Fourier's law of heat conduction?



INTRODUCTION

2) CONVECTION: It occurs when a fluid exchanges energy with an adjacent solid. The fluid motion adjacent to the solid surface assists in the transfer of energy.

There are two types of convection heat transfer:There are two types of convection heat transfer: i) Forced convection - Fluid motion is induced by an external source such as a fan or pump.ii) Natural convection - Heating a fluid results in natural convective heating, Air willcirculate due to natural convective heating.

The temperature gradient in the fluid creates variations in density within the fluid. The colder fluid(heavier) will sink, and the hotter fluid (lighter) will rise.

Newton's law of cooling describes the rate of heat transfer from a solid surface into a fluid medium:Newton s law of cooling describes the rate of heat transfer from a solid surface into a fluid medium:This relationship is valid for both forced and natural convection.

(Tamb) is the ambient fluid temperature, (h) is defined as the convective heat transfercoefficient. This proportionality constant contains all the nonlinearities associated with convection.Q: A sealed cube containing argon floating in deep space is heated on one side by a nearby

star. Will there be any heat transfer due to natural convection?



INTRODUCTION

3) RADIATION: It is the energy transferred in the form of electromagnetic waves as a result of thechanges in the “electronic configuration” of the atoms or molecules of matter. Whenever a chargedparticle undergoes acceleration , energy possessed by the particle is converted into a form of energyknown as electromagnetic radiation .THERMAL RADIATION: It is the form of radiation emitted by bodies because of theirtemperature. i) Converting internal energy into out-flowing electromagnetic waves, ii) Absorbingincoming electromagnetic waves which are converted to internal energy.g g gy

The rate of radiant thermal energy transfer between two bodies is described by the Stefan-Boltzmann Law: “The emission of thermal radiative energy is proportional to the fourth power ofthe absolute temperature (Kelvin or Rankine)”.

Where E is the emissive power flux of the surface (W/m2) (σ) is the Stefan BoltzmannWhere E, is the emissive power flux of the surface (W/m2), (σ) is the Stefan-Boltzmann constant, and (ε) is the emissivity of the surface.

The rate of radiative heat transfer between two surfaces, a and b is:

Stefan-Boltzmann constant, (σ) = 5.6703 x10-8 (W/m2K4)



INTRODUCTION

Three modes of heat transfer:



ELECTROMAGNETIC WAVES

The theoretical foundation was established by physicist James Clerk Maxwell in1864. El t ti di ti i th ti fElectromagnetic radiation is the propagation of a collection of discrete packets of energy called “photons” or “quanta”.

where, h : Plank’s constant (=6.625*10-34 J.s)All Electromagnetic waves travel at the speed

e = h ν

of light (=2.99*108m/s) in a vacuum.Electromagnetic waves are characterized by their frequency (ν) or wavelength (λ).

where C is the speed of propagation of a wave in that media.

λ = C/ ν

The Electromagnetic wave spectrum




Earth’s energy budget:




Electromagnetic wave spectrum:




Electromagnetic spectrum:



RADIATION

Types of radiation:



THERMAL RADIATION

Every object emits thermal radiation from its surface due to its temperature.Thermal radiation is continuously emitted by all matter whose temperature isabove absolute zero (0 K).above absolute zero (0 K).Thermal radiation is electromagnetic radiation that is transmitted by theelectrons in the atoms of the radiating object. Everyday objects emit thermalradiation that is in the infrared region of the electromagnetic spectrum.radiation that is in the infrared region of the electromagnetic spectrum.A hot object, like the filament of a light bulb, emits infrared and visible thermalradiation.Thermal radiation is transmitted through air or through the vacuum of spaceThermal radiation is transmitted through air or through the vacuum of space.Thermal radiation transports heat from a hotter surface to a cooler surface, e.g.light from the surface of the Sun to the Earth.Th th l di ti i id l d i t ilib i b ll thThe thermal radiation inside an enclosed oven is at equilibrium, because all thewalls are at the same temperature.In heat transfer studies, we are interested in the energy emitted by bodiesbeca se of their temperat re onl

© Copy Rights, www.cctech.co.in14

because of their temperature only.



THERMAL RADIATION

How radiation differs from other phenomenon?

It does not require the presence of a material media (solid/liquid/gas) to take place.

Energy transfer by radiation is fastest (at the speed of light) and it suffers nott ti iattenuation in a vacuum.

Heat transfer through an evacuated space can occur only by radiation.E S l di i (0 3 3 )Ex: Solar radiation (0.3-3μm).

Radiation heat transfer can occur between two bodies which are separated by amedium colder than both the bodiesmedium colder than both the bodies.

Radiation is a volumetric phenomenon, except for opaque bodies (τ=0).

© Copy Rights, www.cctech.co.in15



TERMS INVOLVED

Radiation Intensity (Ie): The rate at which radiation energy is emitted in the (θ,Φ ) di ti it l t thi di tidirection per unit area normal to this direction and per unit solid angle about this direction.

I (θ,Φ) = dQ /(dA*cosθ*dω)Ie (θ,Φ) dQe/(dA cosθ dω)

Solid Angle (ω): It is the two- dimensional angle in three-dimensional space thatthree dimensional space that an object subtends at a point.

dω =ds/r2



Terms Involved

Black Body Radiation:• Thermal radiation from a black body is

termed as black body radiation.• A black body is an idealized object that

absorbs all e.m.radiation that falls on it.

Irradiation (G): • The radiation flux incident on a surface

from all directions.

Radiosity (J):• The rate at which radiation energyThe rate at which radiation energy

leaves a unit area of a surface in all directions.



Radiative properties

Emissivity (ε): The ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a black body at the same temperature.by a black body at the same temperature.

Monochromatic emissivity (ελ): The ratio of the monochromatic emissive power of the surface to the monochromatic emissive power of a black surface atmonochromatic emissive power of a black surface at the same temperature and wavelength.

ελ = Eλ/Ebλ

Absorptivity (α): The fraction of irradiation absorbedAbsorptivity (α): The fraction of irradiation absorbed by the surface.

• It depends strongly on the temperature of the source at which the incident radiation is originatingat which the incident radiation is originating.

Reflectivity (ρ): The fraction of irradiation reflected by the surface

Transmissi it ( ): The fraction of irradiationTransmissivity (τ): The fraction of irradiation transmitted by the surface.



Terms Involved

Properties of Opaque Surfaces :Ii = Ir+Iaρ+α =1

Semi-transparent surfaces:• The region into which the radiation is transmitted may orρ α 1

τ=0radiation is transmitted may or may not be part of the computational domain.

Ii = Ir+Ia+Iti r a tρ+α+τ=1

Types of diffusion :

The Gray body : The gray body is defined such that the monochromatic emissivity(ε ) of the body is independent of wavelength(ελ) of the body is independent of wavelength.



Terms Involved

View factor : The view factor from a surface i to j is (Fij) defined as the fraction of radiation energy leaving surface i that strikes surface j directly.j yIt is a purely geometric quantity and is independent of surface properties and temperature.Radiation analysis on an enclosure consisting of NRadiation analysis on an enclosure consisting of N surfaces requires the evaluation of N2 view factors.

Reciprocity relation : Ai*Fij = Aj*Fji

Summation rule : ΣF =1 for j=1 to NSummation rule : ΣFij =1 , for j=1 to NOptical Thickness : An important dimensionless number

in radiation.O i l hi k i di h l di i iOptical thickness indicates how strongly radiation is absorbed (and scattered).

Optical thickness = (α + σs) LL = Mean beam length(typically distanceL = Mean beam length(typically distance

between two opposing walls)



BASIC LAW’S OF RADIATION

Stefan-Boltzmann law: The total radiation energy emitted by a black body per unit time and per

it f i bunit surface area was given by ,Total blackbody emissive power, Eb = σ *T4

where, σ =5.67 x10-8 W/(m2 · K4)

Spectral(or Monochromatic) blackbody emissive power (Ebλ) :The amount of radiation energy emitted by a blackbody at a

th d i t t (T) it ti it f dthermodynamic temperature (T) per unit time, per unit surface area, and per unit wavelength(λ).

Ebλ(λ,T) = C1/[λ5*(exp(C2/λ*T)-1)] ……..Plank’s lawwhere C = 2Π*h*c 2where, C1 = 2Π h c0

C2 = h*c0/kh : Plank’s constant k : Boltzmann’s constant



Variation of Ebλ with wavelength

•The emitted radiation is a continuous function of wavelength.g

•At any wavelength , the amount of emitted radiation increases with increasing temperaturetemperature.

•As temperature increases , the curves shift to the left to the shorter wavelength region.

•The radiation emitted by sun, at (5800K) reaches its peak in the visible region of the spectrumspectrum.



Continued…

Wien’s displacement law: The wavelength at which the peak spectral blackbody emissive powerThe wavelength at which the peak spectral blackbody emissive power

occurs for a specified temperature is given by Wien’s displacement law.(λ*T)max power =2897 μm*K

Kirchhoff’s law:The total hemispherical emissivity of a surface at temperature T is equal toThe total hemispherical emissivity of a surface at temperature T is equal to

its total hemispherical absorptivity for radiation coming from a blackbody atsame temperature.

ε(T) = α(T)ε(T) = α(T)



Radiative transfer equation (RTE) :

Radiation intensity along an incremental step in any direction is decreased by absorption and out-scattering, and increased by emission and in-scattering.



QUESTIONSPrepared By: Prof.Latesh Chaudhari, Asst. Prof.,F.E.T.R., Bardoli


I d i l Ch i l T h lIndustrial Chemical TechnologyModule – I

Heat Transfer TechnologyHeat Transfer Technology05/12/2010

By:yHimanshu S. Desai

Lecturer, Chemical Engineering DepartmentN G P t l P l t h iN. G. Patel PolytechnicIsroli – Afwa, Bardoli



CondensersA condenser is a device or unit used to condense asubstance from its gaseous to its liquid state, typically bycooling itcooling it.Condensers are typically heat exchangers which havevarious designs and come in many sizes.g yCondensers are used in air conditioning,industrial chemical processes such as distillation,

l d h h hsteam power plants and other heat‐exchange systems.Use of cooling water or surrounding air as the coolant iscommon in many condenserscommon in many condensers.



Types of CondensersSurface condenser is the commonly used term for awater‐cooled shell and tube heat exchanger.Th d h h hi hThese condensers are heat exchangers which convertsteam from its gaseous to its liquid state.When cooling water is in short supply an air‐cooledWhen cooling water is in short supply, an air cooledcondenser is often used.



Types of CondensersContact condenser is a device in which a vapor isbrought into direct contact with a cooling liquid andcondensed by giving up its latent heat to the liquidcondensed by giving up its latent heat to the liquid.In almost all cases the cooling liquid is water, and thecondensing vapor is steamcondensing vapor is steam.In this type of condenser the steam and cooling waterare mixed in a condensing chamber and withdrawngtogether.



PurposeSteam traps exist to discharge air and condensate whilenot permitting the escape of live steamTheir goal is to:Improve efficiency (Excess water or air in the system preventsit f hi ti t t i kl d i t t )it from reaching operating temperature quickly during start‐up)

Protect system (Inadequate steam trapping can lead to waterhammer, corrosion, leakage, and high maintenance costs)

Provide maximum heat transfer (‘dry’ steam has best heattransfer properties in equipment like a heat exchanger)



TypesTypes1. Mechanical steam traps ‐ Have a float that rises and falls in

relation to condensate level and this usually has a mechanical linkageattached that opens and closes the valve. Operate in directrelationship to condensate levels present in the body of the steamtraptrap.



TypesTypes2. Temperature steam traps ‐ Have a valve that moves in/out of

position by either expansion/contraction caused by temperaturechange. Some condensate builds up as it cools sufficiently to allow thevalve to open. In most circumstances this is not desirable ascondensate needs to be removed as soon as it is formedcondensate needs to be removed as soon as it is formed.



TypesTypes3. Thermodynamic steam traps ‐ Work on the difference

in response to velocity change in flow of compressible andincompressible fluids. As steam enters, static pressure abovethe disk forces the disk against the valve seat. The staticpressure over a large area overcomes the high inlet pressure ofthe steam. As the steam starts to condense, the pressureagainst the disk lessens and the trap opens to allow condensateg p pout.



Thank You



Industrial Chemical TechnologyModule IModule ‐ I

Heat Transfer Technology

Thermal Conductivity & Its

Heat Transfer Technology

yMeasurement

By

21st November, 2010By

Prof. Latesh B. ChaudhariAsst. Prof. & I/C Head

Chemical Engineering Department,Faculty of Engineering, Technology & Research

Isroli – Afwa, Bardoli



Contents:Contents:

• Conduction mode of heat transfer

• Fourier’s Law

• Thermal Conductivity

Cl ifi i f i l b d h l C d i i• Classification of Materials based on Thermal Conductivity

• Effect of Temperature on Thermal Conductivity.



Conduction:Conduction:

• Transfer of heat from one part of a body to anotherp ypart of a same body or from on body to another bodywhich is in physical contact with it, without

i bl di l t f ti l f b dappreciable displacement of particles of a body.



Fourier’s Law:Fourier s Law:Q ∝ A x [ dT / dx]

Q = k x A x [ dT / dx]Q = k x A x [ dT / dx]Where,Q = Rate of heat transfer (Watts)

2A = Area of heat Transfer (m2)k = Thermal Conductivity (W/moC)dT/dx = Rate of change of temperature with distance in the

direction of heat flow, oC/m

“The rate of heat flow by conduction through uniformThe rate of heat flow by conduction through uniformmaterial is directly proportional to the area normal tothe direction of heat flow and temperature gradient inthe direction of heat flow”



Conduction Of Heat Through AConduction Of Heat Through A Material



Thermal Conductivity:Thermal Conductivity:

Q = k x A x [dT / dx]k = ( Q x dx) / (A x dT)

“The quantity of heat passing through a quantity of material of unith k h h fl hthickness with unit heat flow area in unit time when unit temperaturedifference is maintained across the opposite face of material”

OR“The ability of material to conduct heat through itself is known asthermal conductivity”.

• Thermal conductivity depends on the nature of material and itstemperature.

• Thermal conductivity of solids is higher then that of liquids andy g qliquids is higher then that of gases.



Thermal Conductivities of different materials:

State of Matter Solids Liquids GasesState of Matter Solids Liquids Gases

Thermal Conductivity (W/mK) 2.3 to 420 0.09 to 0.7 0.006 to 0.6

M i lThermal Conductivity

M i lThermal Conductivity

Thermal Conductivities of different materials at Room Temperature

Materialy

W/m°CMaterial

yW/m°C

Diamond 2300 Brick 0.72Silver 429 Water (l) 0 613Silver 429 Water (l) 0.613Copper 401 Human skin 0.37Gold 317 Wood (oak) 0.17

Al i 237 H li ( ) 0 152Aluminum 237 Helium (g) 0.152Iron 80.2 Soft rubber 0.13

Mercury (l) 8.54 Glass fiber 0.043Glass 0.78 Air (g) 0.026



The range of thermal conductivity of various materials at room temperature



Classification of Materials based on Thermal C d i iConductivity:

HEAT CONDUCTORS:The materials having higher value of thermalconductivity are referred to as heat conductors. Eg. Silver

t l (k 420W/ K) R d C (k 395W/ K) G ldmetal (k = 420W/mK), Red Copper (k = 395W/mK), Gold(k = 302W/mK), Aluminums (k = 210W/mK)

HEAT INSULATORS:The materials having low value of thermal conductivityare referred to as heat insulators. Eg. Cork (k = 0.025W/mK), Glass Wool (k = 0.10 W/mK), Asbestos (k = 0.024W/mK), 85%magnesia (k = 0.04W/mK)W/mK), 85%magnesia (k 0.04 W/mK)



Thermal Insulation:Thermal Insulation:

• Process equipments such as reaction vessels, Reboilers,distillation column, evaporators etc or steam pipe will lose heat tothe atmosphere by conduction, convection and radiation.

• Thus to overcome this heat loss, some form of lagging is applied tothe hot surfaces. Eg. In a furnace surface temperature is reducedby making use of series of insulating bricks that are poorconductor of heat (Heat Insulators).

Requirement of lagging materials:

• It should have low thermal conductivity and• It should suppress convection currents.



Optimum Thickness of Insulation:Optimum Thickness of Insulation:

Increasing the thickness ofginsulation will decreasethe heat loss but at the

ti it ill isame time it will increasethe fixed cost. Thus,optimum thickness whichpwill help in minimum heatloss and minimum cost iscalculatedcalculated.



Variation of the thermal conductivity with temperature




T K C Al i


Temperature, K Copper Aluminum100 482 302200 413 237200 413 237300 401 237400 393 240600 379 231800 366 218



THERMAL CONDUCTIVITY OF INSULATING POWDER

• Objective: To determine thermal conductivity of insulating powder.

• Theory: Consider the transfer of heat by conduction through the wall of hollow sphere form by the insulating powder layer packed between two thin copper spheres. Let ri = radius of inner sphere (m), ro = radius of outer sphere (m), Ti = avg. temperature of the inner surface (C) where,

• Ti = (T1+T2+T3+T4) / 4 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (1)

• Similarly,

• To = (T5+T6+T7+T8+T9+T10) / 6 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (2)

• From the experimental value of q, Ti the unknown thermal conductivity K can be determine as,

• K = [Q x (ro ‐ ri)] / [4 xΠ x ri x ro x (Ti – To)] ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (3)K = [Q x (ro ri)] / [4 x Π x ri x ro x (Ti To)] (3)



• Procedure:• Connect the equipment with electrical supply.q p pp y• Adjust input equal to any value between 20 to 60 watt maximum by volt &

ammeter.• Take thermocouple readings at frequent intervals till consecutive readings are

same indicating that steady state has been reached.g y

• OBSERVATION:– Radius of inner sphere, ri = ___________ mm– Radius of outer sphere ro = mmRadius of outer sphere, ro = ___________ mm– Voltmeter reading, V = ___________ volts– Ammeter reading, I = ___________ ampere– Heater input (Q) = V x I = ___________ watts

••

• CALCULATION:• K = [Q x (ro ‐ ri)] / [4 x π x ri x ro x (Ti – To)]• =• =

•RESULT: Thermal Conductivity of the insulating powder is_____________W / moC.



Thank You




HEAT TRANSFER TECHNOLOGY Date: 21/11/2010 By: L. B. Chaudhari

THERMAL CONDUCTIVITY OF INSULATING POWDER

Objective: To determine thermal conductivity of insulating powder.

Theory: Consider the transfer of heat by conduction through the wall of hollow sphere

form by the insulating powder layer packed between two thin copper spheres. Let ri =

radius of inner sphere (m), ro = radius of outer sphere (m), Ti = avg. temperature of the

inner surface (C) where,

Ti = (T1+T2+T3+T4) / 4 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (1)

Similarly,

To = (T5+T6+T7+T8+T9+T10) / 6 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (2)

From the experimental value of q, Ti the unknown thermal conductivity K can be determine

as,

K = [Q x (ro ‐ ri)] / [4 x Π x ri x ro x (Ti – To)] ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (3)

Procedure: 1. Connect the equipment with electrical supply.

2. Adjust input equal to any value between 20 to 60 watt maximum by volt & ammeter.

3. Take thermocouple readings at frequent intervals till consecutive readings are same

indicating that steady state has been reached.

OBSERVATION: 1) Radius of inner sphere, ri = ___________ mm

2) Radius of outer sphere, ro = ___________ mm

3) Voltmeter reading, V = ___________ volts

4) Ammeter reading, I = ___________ ampere

5) Heater input (Q) = V x I = ___________ watts



OBSERVATION TABLE: For Inner Sphere:

Thermocouple No.

T1 T2 T3 T4 Mean temperature Ti = (T1+T2+T3+T4) / 4

Temperature (oC)

For Outer Sphere:

Thermocouple No.

T5 T6 T7 T8 T9 T10 Mean temperature

To = (T5+T6+T7+T8 + T9 + T10) / 6 Temperature

(oC)

CALCULATION: K = [Q x (ro ri)] / [4 x π x ri x ro x (Ti – To)] = = RESULT: Thermal Conductivity of the insulating powder is __________________W / moC. CONCLUSION: ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________



EXPERIMENT NO: DATE:

STEFAN BOLTZMAN APPARATUS

AIM: To determine the Stefan Boltzman constant for a given apparatus.

THEORY:

The most commonly used law of thermal radiation is the Stefan Boltzman law which state

that the thermal radiation heat flux for emissive power of a black surface is promotional to the

fourth power of absolute temp. of the surface & is given by,

(Q / A) = e x b = σ x T4 (1)

Where constant of proportionality σ is called Stefan Boltzman constant & has the value of 4.88

x 10‐8 Kcal / hr x m2 x K4.

PROCEDURE: 1. Heat the water in the tank by the immersion heater up to a temperature of above 90oC.

2. The disc D is removed before pouring the hot water in the jacket.

3. The hot water is poured in the water jacket.

4. The hemispherical enclosure B & A will come to some uniform temperature in short time

after filling the hot water in the jacket.

5. The enclosure will soon come to thermal equilibrium conditions.

6. The disc D is now inserted in A at a time when its temperature is TD.

7. The radiation energy falling on the disc D from the enclosure is given by:

E = σ x AD x (T4) (2)

Where,

AD = area of disc D.

T = temperature of enclosure.

8. The emmisivity of the D is assume to be unity the radiant energy disc D is emitting into

enclosure will be:

E1 = σ x AD x (TD4) (3)

Where,

TD = Temperature of disc D.



9. Net heat input to disc D per unit time is given by equation (2) – (3),

E – E1 = σ x AD x (T4 – TD4) (4)

10. If the disc D has a mass m & Specific Heat S then a short time after Disc D is inserted in A,

m x s (dT / dt) t = 0 = σ x AD x (T4 – TD4) (5)

σ = (m x s (dT / dt) t = 0) / AD x (T4 – TD4) (6)

In this equation (dT / dt) t = 0 denotes the rate of rise of temperature of the disc D. It is clear

measure at time t=0 before heat conducted from A to D begins to have any significant effects.

This is obtaining from plot of temperature rise of disc D with respect to time and obtaining it

slop at t=0 when temperature is TD. The temperature sensor mounted on disc is to be use for

this purpose. Note that the disc D with its sleeve is placed quickly in a position & start

recording the Temperature at fixed time interval. The whole process must be completed in 30

seconds of time.

OBSERVATION:

1. Mass of disc D, (m) = __________ Kg.

2. Specific Heat of disc materials (S) = __________ Kcal/Kg x oC

3. Diameter of disc D = __________ mm = __________ m.

4. Area of disc D, AD = (Π / 4) x D2

=

= __________ m2

5. Temperature of enclosure, (T) = __________ oC = ____________ K

6. Temperature of disc D at instant when it is inserted:

TD = __________ oC = ____________ K



7. Temperature of disc D at time interval of 5 sec.

Time (t)

(sec)

Temperature of disc, (TD)

(oC) 5

10

15

20

25

30

35

40

8. Obtain slop from graph of temperature (TD) versus time (t)

dT/dt = __________ oC/sec

dT/dt = __________ oC/hr

9. T4 = ( __________ )4 = __________ K4

10. TD4 = ( __________ )4 = __________ K4

11. T4 – TD4 = ( __________ )4 – ( __________ )4

=

12. Value of Stephen Boltzman Constant (σ) can be obtain by using

σ = (m x s (dT / dt) t = 0) / AD x (T4 – TD4)

=

=



RESULT:

The value of Stefan Boltzman constant σ is ________________ Kcal/m2K4.

CONCLUSION: ____________________________________________________________________________

_____________________________________________________________________________________________

_____________________________________________________________________________________________

_____________________________________________________________________________________________

_____________________________________________________________________________________________

_____________________________________________________________________________________________




PROCESS HEAT TRANSFER Date: 05/12/2010 By: H. S. Desai

DOUBLE PIPE HEAT EXCHANGER AIM: To determine log mean temperature difference (LMTD) for Horizontal double pipe

heat exchanger, vertical double pipe heat exchanger using co‐current and counter current

flow.

APPARATUS: Horizontal double pipe heat exchanger, Vertical double pipe heat exchanger,

Bare & lagged pipe apparatus, stop watch, measuring cylinder, etc.

PROCEDURE:

1) Allow the cold water to flow through pipe of exchanger and maintain the steady flow of

water.

2) Allow the steam to flow through Heat Exchanger at constant pressure.

3) Note down the inlet & outlet temperature & flow rate of water at the steady state

condition.

4) Also note down the flow rate & temperature of condensate at steady state condition.

5) Perform the experiment in co‐current and counter current flow.

OBSERVATION:

1. Inner diameter of annulus (Di) = __________ inch = __________ meter

2. Outer diameter of annulus (Do) = __________ inch = __________ meter

3. Length of pipe = __________ inch = __________ meter

Prepared By: Prof.Himanshu Desai, Lecturer,N.G. Patel Polytechnic, Bardoli


OBSERVATION TABLE:

Sr. No.

Type of Flow

STEAM COLD WATER

Pressure (Kg/cm2)

Inlet Temp. (oC)

Outlet Temp. (oC)

Flow Rate (Kg/S)

Inlet Temp. (oC)

Outlet Temp. (oC)

Flow Rate (Kg/S)

FOR HORIZONTAL DOUBLE PIPE HEAT EXCHANGER: 1 Co current Flow

2 Counter Current

FOR VERTICA DOUBLE PIPE HEAT EXCHANGER:

1 Co current Flow

2 Counter Current

CALCULATION:

LOGERATHMIC MEAN TEMPERATURE DIFFERENCE:

FOR HORIZONTAL DOUBLE PIPE HEAT EXCHANGER:

• FOR COCURRENT FLOW:

LMTD = (∆T2 ∆T1) / ln (∆T2/∆T1)

=

=

• FOR COUNTER CURRENT FLOW:

LMTD = (∆T2 ∆T1) / ln (∆T2/∆T1)

=

=



FOR VERTICAL DOUBLE PIPE HEAT EXCHANGER:

• FOR COCURRENT FLOW:

LMTD = (∆T2 ∆T1) / ln (∆T2/∆T1)

=

=

• FOR COUNTER CURRENT FLOW:

LMTD = (∆T2 ∆T1) / ln (∆T2/∆T1)

=

=

RESULT:

Type of Heat Exchanger Type of Flow LMTD

Horizontal Double Pipe H.E. Co‐current Flow

Counter Current Flow

Vertical Double Pipe H.E. Co‐current Flow

Counter Current Flow

CONCLUSION: _______________________________________________________________________________________

_________________________________________________________________________________________________________

_________________________________________________________________________________________________________

_________________________________________________________________________________________________________

_________________________________________________________________________________________________________




PROCESS HEAT TRANSFER Date: 05/12/2010 By: H. S. Desai

BARE & LAGGED PIPE

AIM: To determine heat losses through the bare and lagged pipe APPARATUS: Measuring cylinder, stop watch, Bare & lagged pipe, thermometer, etc. PROCEDURE: 1. Allow the steam to flow through bare pipe at constant pressure & measure the outlet

temperature of condensate & flow rate of condensate by using the measuring cylinder & stop watch.

2. Allow the steam to flow through lagged pipe by operating the valve & after sometimes measure the temp. & flow rate of condensate.

3. Measure the outside diameter of pipe by vernier caliper. 4. Calculate heat losses by radiation through bare & lagged pipe & compare the result.

OBSERVATION: 1. Outer diameter of bare pipe = __________ cm 2. Thickness of bare pipe = __________ cm 3. Inner diameter of bare pipe = __________ cm 4. Inlet Temperature of steam in bare & lagged pipe = __________ oC 5. Specific Heat at steam at pressure _________ Kg/cm2 = __________ J/kg. k OBSERVATION TABLE:

BARE PIPE LAGGED PIPE

Pressure (Kg / cm2)

Inlet Temp. (oC)

Outlet Temp. (oC)

Flow Rate (LPM)

Pressure (Kg / cm2)

Inlet Temp. (oC)

Outlet Temp. (oC)

Flow Rate (LPM)



CALCULATION: [A] FOR BARE PIPE:

Heat loss from bare pipe :

QB = mB x Cp x ∆T

Where, mB = Mass flow rate of condensate for bare pipe Cp = Specific heat of steam for bare pipe QB = =

[B] FOR LAGGED PIPE:

Heat loss from lagged pipe:

QL = mL x CP x ∆T Where, mL = Mass flow rate of condensate for lagged pipe Cp = Specific heat of steam for lagged pipe QL = =

RESULT:

• Heat loss through Bare pipe = ____________________ J/sec. • Heat loss through Lagged pipe = ____________________ J/sec.

CONCLUSION: _______________________________________________________________________________________ ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________



Long Term Training Programme on “Industrial Chemical Technology’ Chemical Engineering Department

N.G. Patel Polytechnic, Isroli Experiment: Mechanically Agitated Jacketed Vessel with Coil Staff: Mr. T.C. Padhiyar, I/c Head, CHED Objective: To study the effect of agitation, inlet temp., Flow rate of heating medium, flow rate of cooling medium on the heat transfer characteristics when using either jacket or coil. Theory: Mechanically agitated vessels, equipped with cooling coils or heating jackets are in common use throughout the chemical process industry. Heat transfer rates agitated vessel is vertical cylinder vessel incorporation suitable type of agitator that brings even distribution of the liquid in the vessel. Agitated vessels are commonly employed in chemical industries for different proposes such as mixing, dissolution, dilution, neutralization, adsorption, crystallization. To done by providing a heat transfer surface, which may be in the form of a jacket fitted outside the vessel or a coil fitted inside the vessel depending upon the situation. Procedure: 1. Fill the agitated vessel with predetermined amount of liquid. 2. Note down the inlet temperature of liquid. 3. Start the supply of hot water either in a coil or jacketed by switching on the pump& by setting a

derived flow rate using the given valve. 4. Note down the change in temp. 5. Change the inlet temperature of liquid with flow rate and note down the similar sets of result. 6. Perform the scheme experiment with agitation and note down the reading.

Prepared By: Prof.Tarak Padhiyar, Lecturer,N.G. Patel Polytechnic, Bardoli


Observation: 1) For vessel: 1. Volume of liquid in vessel: 5lit. 2. Material of construction: SS‐316 3. Inside dia. of vessel: 213 mm 4. Height of vessel: 250 mm 5. Thickness of wall: 1.5 mm 2) For Jacket: 1. Material of construction: SS‐316. 2. Inner dia. of jacket: 260mm 3. Outer dia. of jacket: 310mm 3) For Coil: 1. Material of construction: copper 2. Coil length: 915 mm 3. Inner diameter of coil: 7 mm 4. Outer diameter of coil: 9 mm 5. Outer diameter of Spiral: 130 mm Observation Table: (A) Without agitation: (1) Using Jacket:

Sr. No.

Inlet temp. of liquid in vessel, Tc1

Outlet temp. of liquid in vessel, Tc2

Inlet temp. of liquid in jacket,

Th1

Outlet temp. of liquid in jacket, Th2

Flow rate

1 2

(2) Using Coil:

Sr. No.




Th1


Flow rate

1 2



(B) With agitation: (1) Using Jacket:

Sr. No.




Th1


Flow rate

RPM

1 2

(2) Using Coil:

Sr. No.




Th1


Flow rate

RPM

1 2

Calculation: (A) Without agitation: 1) Using Jacket: ∆T1 = Th1 – Tc1 ∆T2 = Th2 – Tc2 = = = ___________ =___________ LMTD ∆Tlm = (∆T2 – ∆T1) ln ∆T2 ∆T1



2) Using Coil: ∆T1 = Th1 – Tc1 ∆T2 = Th2 – Tc2 = = = ___________ =___________ LMTD ∆Tlm = (∆T2 – ∆T1) ln ∆T2 ∆T1 (A) With agitation: 1) Using Jacket: ∆T1 = Th1 – Tc1 ∆T2 = Th2 – Tc2 = = = ___________ =___________ LMTD ∆Tlm = (∆T2 – ∆T1) ln ∆T2 ∆T1 2) Using Coil: ∆T1 = Th1 – Tc1 ∆T2 = Th2 – Tc2 = = = ___________ =___________



LMTD ∆Tlm = (∆T2 – ∆T1) ln ∆T2 ∆T1 Result:

• LMTD using Jacket without agitation = ______________ • LMTD using Coil without agitation = ______________ • LMTD using Jacket with agitation = ______________ • LMTD using Coil with agitation = ______________

Conclusion: ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________



Compiled by: JJPandya

Module 1

Chapter 1 : Introduction to fluid and Fundamental Concepts

The chapter contains

Lecture 1

Definition of Stress Definition of Fluid Concept of Continum Fluid Properties

Lecture 2

Distinction between Newtonian and Non-Newtonian Fluid Compressibility Surface Tension of Liquids

•

Definition of Stress

Consider a small area δA on the surface of a body (Fig. 1.1). The force acting on this area is δF This force can be resolved into two perpendicular components

The component of force acting normal to the area called normal force and is denoted by δFn The component of force acting along the plane of area is called tangential force and is denoted

by δFt

Fig 1.1 Normal and Tangential Forces on a surface


When they are expressed as force per unit area they are called as normal stress and tangential stress respectively. The tangential stress is also called shear stress

The normal stress

(1.1)

And shear stress

(1.2)

Definition of Fluid

A fluid is a substance that deforms continuously in the face of tangential or shear stress, irrespective of the magnitude of shear stress .This continuous deformation under the application of shear stress constitutes a flow.

In this connection fluid can also be defined as the state of matter that cannot sustain any shear stress.

Example : Consider Fig 1.2

Fig 1.2 Shear stress on a fluid body

If a shear stress τ is applied at any location in a fluid, the element 011' which is initially at rest, will move to 022', then to 033'. Further, it moves to 044' and continues to move in a similar fashion.


In other words, the tangential stress in a fluid body depends on velocity of deformation and vanishes as this velocity approaches zero. A good example is Newton's parallel plate experiment where dependence of shear force on the velocity of deformation was established.

Distinction Between Solid and Fluid

Solid Fluid

More Compact Structure

Attractive Forces between the moleculesare larger therefore more closely packed

Solids can resist tangential stresses instatic condition

Whenever a solid is subjected to shearstress

a. It undergoes a definitedeformation α or breaks

b. α is proportional to shear stressupto some limiting condition

Solid may regain partly or fully itsoriginal shape when the tangentialstress is removed

Less Compact Structure

Attractive Forces between the molecules are smaller therefore more loosely packed

Fluids cannot resist tangential stresses in static condition.

Whenever a fluid is subjected to shear stress

a. No fixed deformation b. Continious deformation takes

place until the shear stress is applied

A fluid can never regain its original shape, once it has been distorded by the shear stress

Fig 1.3 Deformation of a Solid Body

Concept of Continuum


The concept of continuum is a kind of idealization of the continuous description of matter where the properties of the matter are considered as continuous functions of space variables. Although any matter is composed of several molecules, the concept of continuum assumes a continuous distribution of mass within the matter or system with no empty space, instead of the actual conglomeration of separate molecules.

Describing a fluid flow quantitatively makes it necessary to assume that flow variables (pressure , velocity etc.) and fluid properties vary continuously from one point to another. Mathematical description of flow on this basis have proved to be reliable and treatment of fluid medium as a continuum has firmly become established. For example density at a point is normally defined as

In continuum approach, fluid properties such as density, viscosity, thermal conductivity, temperature, etc. can be expressed as continuous functions of space and time.

Fluid Properties :

Characteristics of a continuous fluid which are independent of the motion of the fluid are called basic properties of the fluid. Some of the basic properties are as discussed below.

Property Symbol Definition Unit

Density ρ

The density p of a fluid is its mass per unit volume . If a fluid element enclosing apoint P has a volume Δ and mass Δm (Fig. 1.4), then density (ρ)at point P is written as

However, in a medium where continuum model is valid one can write -

(1.3)

kg/m3


Fig 1.4 A fluid element enclosing point P

Specific Weight γ

The specific weight is the weight of fluid per unit volume. The specific weight is given

by γ= ρg (1.4)

Where g is the gravitational acceleration. Just as weight must be clearly distinguished from mass, so must the specific weight be distinguished from density.

N/m3

Specific Volume v

The specific volume of a fluid is the volume occupied by unit mass of fluid.

Thus

(1.5)

m3 /kg

Specific Gravity s

For liquids, it is the ratio of density of a liquid at actual conditions to the density ofpure water at 101 kN/m2 , and at 4°C.

The specific gravity of a gas is the ratio of its density to that of either hydrogen orair at some specified temperature or pressure.

However, there is no general standard; so the conditions must be statedwhile referring to the specific gravity of a gas.

-

Viscosity ( μ ) :

Viscosity is a fluid property whose effect is understood when the fluid is in motion. In a flow of fluid, when the fluid elements move with different velocities, each element will feel

some resistance due to fluid friction within the elements. Therefore, shear stresses can be identified between the fluid elements with different velocities.


The relationship between the shear stress and the velocity field was given by Sir Isaac Newton.

Consider a flow (Fig. 1.5) in which all fluid particles are moving in the same direction in such a way that the fluid layers move parallel with different velocities.

Fig 1.5 Parallel flow of a fluid Fig 1.6 Two adjacent layers of a moving fluid.

The upper layer, which is moving faster, tries to draw the lower slowly moving layer along with it by means of a force F along the direction of flow on this layer. Similarly, the lower layer tries to retard the upper one, according to Newton's third law, with an equal and opposite force F on it (Figure 1.6).

Such a fluid flow where x-direction velocities, for example, change with y-coordinate is called shear flow of the fluid.

Thus, the dragging effect of one layer on the other is experienced by a tangential force F on the respective layers. If F acts over an area of contact A, then the shear stress τ is defined as

τ = F/A

Viscosity ( μ ) :

Newton postulated that τ is proportional to the quantity Δu/ Δy where Δy is the distance of separation of the two layers and Δu is the difference in their velocities.

In the limiting case of , Δu / Δy equals du/dy, the velocity gradient at a point in a direction perpendicular to the direction of the motion of the layer.

According to Newton τ and du/dy bears the relation

(1.7)


where, the constant of proportionality μ is known as the coefficient of viscosity or simply viscosity which is a property of the fluid and depends on its state. Sign of τ depends upon the sign of du/dy. For the profile shown in Fig. 1.5, du/dy is positive everywhere and hence, τ is positive. Both the velocity and stress are considered positive in the positive direction of the coordinate parallel to them. Equation

defining the viscosity of a fluid, is known as Newton's law of viscosity. Common fluids, viz. water, air, mercury obey Newton's law of viscosity and are known as Newtonian fluids.

Other classes of fluids, viz. paints, different polymer solution, blood do not obey the typical linear relationship, of τ and du/dy and are known as non-Newtonian fluids. In non-newtonian fluids viscosity itself may be a function of deformation rate as you will study in the next lecture.

Causes of Viscosity

The causes of viscosity in a fluid are possibly attributed to two factors:

(i) intermolecular force of cohesion (ii) molecular momentum exchange

Due to strong cohesive forces between the molecules, any layer in a moving fluid tries to drag the adjacent layer to move with an equal speed and thus produces the effect of viscosity as discussed earlier. Since cohesion decreases with temperature, the liquid viscosity does likewise.

Fig 1.7 Movement of fluid molecules between two adjacent

moving layers

Molecules from layer aa in course of continous thermal agitation migrate into layer bb

Momentum from the migrant molecules from layer aa is stored by molecules of layer bb by way of collision

Thus layer bb as a whole is speeded up


Molecules from the lower layer bb arrive at aa and tend to retard the layer aa

Every such migration of molecules causes forces of acceleration or deceleration to drag the layers so as to oppose the differences in velocity between the layers and produce the effect of viscosity.

As the random molecular motion increases with a rise in temperature, the viscosity also increases accordingly. Except for very special cases (e.g., at very high pressure) the viscosity of both liquids and gases ceases to be a function of pressure.

For Newtonian fluids, the coefficient of viscosity depends strongly on temperature but varies very little with pressure.

For liquids, molecular motion is less significant than the forces of cohesion, thus viscosity of liquids decrease with increase in temperature.

For gases,molecular motion is more significant than the cohesive forces, thus viscosity of gases increase with increase in temperature.

Fig 1.8: Change of Viscosity of Water and Air under 1 atm

No-slip Condition of Viscous Fluids

It has been established through experimental observations that the relative velocity between the solid surface and the adjacent fluid particles is zero whenever a viscous fluid flows over a solid surface. This is known as no-slip condition.


This behavior of no-slip at the solid surface is not same as the wetting of surfaces by the fluids. For example, mercury flowing in a stationary glass tube will not wet the surface, but will have zero velocity at the wall of the tube.

The wetting property results from surface tension, whereas the no-slip condition is a consequence of fluid viscosity.

Ideal Fluid

Consider a hypothetical fluid having a zero viscosity ( μ = 0). Such a fluid is called an ideal fluid and the resulting motion is called as ideal or inviscid flow. In an ideal flow, there is no existence of shear force because of vanishing viscosity.

All the fluids in reality have viscosity (μ > 0) and hence they are termed as real fluid and their motion is known as viscous flow.

Under certain situations of very high velocity flow of viscous fluids, an accurate analysis of flow field away from a solid surface can be made from the ideal flow theory.

Non-Newtonian Fluids

There are certain fluids where the linear relationship between the shear stress and the

deformation rate (velocity gradient in parallel flow) as expressed by the is not valid. For these fluids the viscosity varies with rate of deformation.

Due to the deviation from Newton's law of viscosity they are commonly termed as non-Newtonian fluids. Figure 2.1 shows the class of fluid for which this relationship is nonlinear.

Figure 2.1 Shear stress and deformation rate relationship of different fluids

The abscissa in Fig. 2.1 represents the behaviour of ideal fluids since for the ideal fluids the resistance to shearing deformation rate is always zero, and hence they exhibit zero shear stress under any condition of flow.

The ordinate represents the ideal solid for there is no deformation rate under any loading condition.

The Newtonian fluids behave according to the law that shear stress is linearly proportional to

velocity gradient or rate of shear strain . Thus for these fluids, the plot of shear stress against velocity gradient is a straight line through the origin. The slope of the line determines the viscosity.


The non-Newtonian fluids are further classified as pseudo-plastic, dilatant and Bingham plastic.

Classification of non-Newtonian fluids

Many mathematical models are available to describe the nonlinear "shear-stress vs deformation-rate" relationship of non Newtonian fluids. But no general model can describe the constitutive equation ("shear stress vs rate of deformation" relationship) of all kinds of non-Newtonian fluids. However, the mathematical model for describing the mechanistic behaviour of a variety of commonly used non-Newtonian fluids is the Power-Law model which is also known as Ostwald-de Waele model (the name is after the scientist who proposed it). According to Ostwald-de Waele model

(2.1)

where m is known as the flow consistency index and n is the flow behavior index.

When n = 1, m equals , the model identically satisfies Newtonian model as a special case. When n < 1, the model is valid for pseudoplastic fluids, such as gelatine, blood, milk etc. When n > 1, the model is valid for dilatant fluids, such as sugar in water, aqueous suspension of rice starch etc.

There are some substances which require a yield stress for the deformation rate (i.e. the flow) to be established, and hence their constitutive equations do not pass through the origin thus violating the basic definition of a fluid. They are termed as Bingham plastic. For an ideal Bingham plastic, the shear stress- deformation rate relationship is linear.

Compressibility

Compressibility of any substance is the measure of its change in volume under the action of external forces.

The normal compressive stress on any fluid element at rest is known as hydrostatic pressure p and arises as a result of innumerable molecular collisions in the entire fluid.

The degree of compressibility of a substance is characterized by the bulk modulus of elasticity E defined as

(2.3)

Where Δ and Δp are the changes in the volume and pressure respectively, and is the initial volume. The negative sign (-sign) is included to make E positive, since increase in pressure would decrease the volume i.e for Δp>0 , Δ <0) in volume.

For a given mass of a substance, the change in its volume and density satisfies the relation


Δm = 0, Δ( ρ ) = 0

(2.4)

using

we get

(2.5)

Values of E for liquids are very high as compared with those of gases (except at very high pressures). Therefore, liquids are usually termed as incompressible fluids though, in fact, no substance is theoretically incompressible with a value of E as .

For example, the bulk modulus of elasticity for water and air at atmospheric pressure are approximately 2 x 106 kN/m 2 and 101 kN/m 2 respectively. It indicates that air is about 20,000 times more compressible than water. Hence water can be treated as incompressible.

For gases another characteristic parameter, known as compressibility K, is usually defined , it is the reciprocal of E

(2.6)

K is often expressed in terms of specific volume . For any gaseous substance, a change in pressure is generally associated with a change in

volume and a change in temperature simultaneously. A functional relationship between the pressure, volume and temperature at any equilibrium state is known as thermodynamic equation of state for the gas. For an ideal gas, the thermodynamic equation of state is given by

p = ρRT (2.7)

where T is the temperature in absolute thermodynamic or gas temperature scale (which are, in fact, identical), and R is known as the characteristic gas constant, the value of which depends


upon a particular gas. However, this equation is also valid for the real gases which are thermodynamically far from their liquid phase. For air, the value of R is 287 J/kg K.

K and E generally depend on the nature of process

Distinction between an Incompressible and a Compressible Flow

In order to know, if it is necessary to take into account the compressibility of gases in fluid flow problems, we need to consider whether the change in pressure brought about by the fluid motion causes large change in volume or density. Using Bernoulli's equation p + (1/2)ρV2= constant (V being the velocity of flow), change in pressure, Δp, in a flow field, is of the order of (1/2)ρV2 (dynamic head). Invoking this relationship into

we get ,

(2.12)

So if Δρ/ρ is very small, the flow of gases can be treated as incompressible with a good degree of approximation.

For compressible fluid, if the flow velocity is small as compared to the local acoustic velocity, compressibility of gases can be neglected. Considering a maximum relative change in density of 5 per cent as the criterion of an incompressible flow, the upper limit of Mach number becomes approximately 0.33. In the case of air at standard pressure and temperature, the acoustic velocity is about 335.28 m/s. Hence a Mach number of 0.33 corresponds to a velocity of about 110 m/s. Therefore flow of air up to a velocity of 110 m/s under standard condition can be considered as incompressible flow.

Surface Tension of Liquids

The phenomenon of surface tension arises due to the two kinds of intermolecular forces (i) Cohesion : The force of attraction between the molecules of a liquid by virtue of which they are bound to each other to remain as one assemblage of particles is known as the force of cohesion. This property enables the liquid to resist tensile stress.

(ii) Adhesion : The force of attraction between unlike molecules, i.e. between the molecules of different liquids or between the molecules of a liquid and those of a solid body when they are in contact with each other, is known as the force of adhesion. This force enables two different liquids to adhere to each other or a liquid to adhere to a solid body or surface.


Work is done on each molecule arriving at surface against the action of an inward force. Thus mechanical work is performed in creating a free surface or in increasing the area of the surface. Therefore, a surface requires mechanical energy for its formation and the existence of a free surface implies the presence of stored mechanical energy known as free surface energy. Any system tries to attain the condition of stable equilibrium with its potential energy as minimum. Thus a quantity of liquid will adjust its shape until its surface area and consequently its free surface energy is a minimum.

The magnitude of surface tension is defined as the tensile force acting across imaginary short and straight elemental line divided by the length of the line.

The dimensional formula is F/L or MT-2 . It is usually expressed in N/m in SI units.

Surface tension is a binary property of the liquid and gas or two liquids which are in contact with each other and defines the interface. It decreases slightly with increasing temperature. The surface tension of water in contact with air at 20°C is about 0.073 N/m.

It is due to surface tension that a curved liquid interface in equilibrium results in a greater pressure at the concave side of the surface than that at its convex side.

Capillarity

The interplay of the forces of cohesion and adhesion explains the phenomenon of capillarity. When a liquid is in contact with a solid, if the forces of adhesion between the molecules of the liquid and the solid are greater than the forces of cohesion among the liquid molecules themselves, the liquid molecules crowd towards the solid surface. The area of contact between the liquid and solid increases and the liquid thus wets the solid surface.

The reverse phenomenon takes place when the force of cohesion is greater than the force of adhesion. These adhesion and cohesion properties result in the phenomenon of capillarity by which a liquid either rises or falls in a tube dipped into the liquid depending upon whether the force of adhesion is more than that of cohesion or not (Fig.2.4).

The angle θ as shown in Fig. 2.4, is the area wetting contact angle made by the interface with the solid surface.


Fig 2.4 Phenomenon of Capillarity

For pure water in contact with air in a clean glass tube, the capillary rise takes place with θ = 0 . Mercury causes capillary depression with an angle of contact of about 1300 in a clean glass in

contact with air. Since h varies inversely with D as found from Eq. ( ), an appreciable capillary rise or depression is observed in tubes of small diameter only.

Vapour pressure

All liquids have a tendency to evaporate when exposed to a gaseous atmosphere. The rate of

evaporation depends upon the molecular energy of the liquid which in turn depends upon the type of liquid and its temperature. The vapour molecules exert a partial pressure in the space above the liquid, known as vapour pressure. If the space above the liquid is confined (Fig. 2.5) and the liquid is maintained at constant temperature, after sufficient time, the confined space above the liquid will contain vapour molecules to the extent that some of them will be forced to enter the liquid. Eventually an equilibrium condition will evolve when the rate at which the number of vapour molecules striking back the liquid surface and condensing is just equal to the rate at which they leave from the surface. The space above the liquid then becomes saturated with vapour. The vapour pressure of a given liquid is a function of temperature only and is equal to the saturation pressure for boiling corresponding to that temperature. Hence, the vapour pressure increases with the increase in temperature. Therefore the phenomenon of boiling of a liquid is closely related to the vapour pressure. In fact, when the vapour pressure of a liquid becomes equal to the total pressure impressed on its surface, the liquid starts boiling. This concludes that boiling can be achieved either by raising the temperature of the liquid, so that its vapour pressure is elevated to the ambient pressure, or by lowering the pressure of the ambience (surrounding gas) to the liquid's vapour pressure at the existing temperature.


Figure 2.5 To and fro movement of liquid molecules from an interface in a

confined space as a closed surrounding Exercise Problems - Chapter 1 1. A thin film of liquid flows down an inclined channel. The velocity distribution in the flow is given

by

where, h = depth of flow, α = angle of inclination of the channel to the horizontal, u = velocity at a

depth h below the free surface, ρ = density of liquid, μ = dynamic viscosity of the fluid. Calculate the shear stress: (a) at the bottom of the channel (b) at mid-depth and (c) at the free surface. The coordinate y is measured from the free surface along its normal

[(a) α, (b) α , (c) 0]

2. Two discs of 250 mm diameter are placed 1.5 mm apart and the gap is filled with an oil. A power of 500 W is required to rotate the upper disc at 500 rpm while keeping the lower one stationary. Determine the viscosity of the oil.

[ 0. 71 kg/ms] [ 5.95]

Recap In this chapter you have learnt the following

A fluid is a substance that deforms continuously when subjected to even an infinitesimalshear stress. Solids can resist tangential stress at static conditions undergoing a definitedeformation while a fluid can do it only at dynamic conditions undergoing a continuousdeformation as long as the shear stress is applied.

The concept of continuum assumes a continuous distribution of mass within the matter


or system with no empty space. In the continuum approach, properties of a system canbe expressed as continuous functions of space and time. A dimensionless parameter

known as Knudsen number where λ is the mean free path and L is the characteristic length, aptly describes the degree of departure from continuum. Theconcept of continuum usually holds good when Kn< 0.01.

Viscosity is a property of a fluid by virtue of which it offers resistance to flow. The shear

stress at a point in a moving fluid is directly proportional to the rate of shear strain. For a

one dimensional flow, . The constant of proportionality μ is known as coefficient of viscosity or simply the viscosity. The relationship is known as the Newton's law of viscosity and the fluids which obey this law are known as Newtonian fluids.

The relationship between the shear stress and the rate of shear strain is known as the

constitutive equation. The fluids whose constitutive equations are not linear through origin (do not obey the Newton's law of viscosity) are known as non-Newtonian fluids. For a Newtonian fluid, viscosity is a function of temperature only. With an increase in temperature, the viscosity of a liquid decreases, while that of a gas increases. For non-Newtonian fluid, the viscosity depends not only on temperature but also on the

deformation rate of the fluid. Kinematic viscosity v is defined as .

Compressibility of a substance is the measure of its change in volume or density underthe action of external forces. It is usually characterized by the bulk modulus of elasticity

A flow is said to be incompressible when the change in its density due to the change in

pressure brought about by the fluid motion is negligibly small. When the flow velocity isequal to or less than 0.33 times of the local acoustic speed, the relative change indensity of the fluid, due to flow, becomes equal to or less than 5 per cent respectively, and hence the flow is considered to be incompressible

The force of attraction between the molecules of a fluid is known as cohesion, while that

between the molecules of a fluid and of a solid is known as adhesion. The interplay of these two intermolecular forces explains the phenomena of surface tension and capillaryrise or depression. A free surface of the liquid is always under stretched conditionimplying the existence of a tensile force on the surface. The magnitude of this force per unit length of an imaginary line drawn along the liquid surface is known as the surfacetension coefficient or simply the surface tension.
