# Convection Hmt

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Q. 4.1. Define convection. Differentiate between free and forced convection.

Ans. Convection is the phenomena of heat transfer by actual bodily movement of particles of themedium. Convection is linked to heat transfer by transportation of medium itself and heat

transfer takes place due to intermixing of the particles. Free convection takes place due to

temperature changes which changes the density of the fluid. The hot fluid becomes lighter andmoves up whereas cold fluid moves down to take its place and so convectional currents are setup and energy transfer takes place by intermixing of the particles.

In forced convection, the flow of fluid (convectional currents) are caused by pump, fan, or

atmospheric winds. Heat transfer in forced convection is more than in free convection.

Q. 4.2. Write Newton - Rickman law/Or state Newtons law of cooling?

Ans. Newtons law of cooling states that

Q. convective heat flow rateh = co efficient of convective of heat transfer

A -, surface area exposed to heat transfer

= surface temperature of the solid

= undisturbed of the fluid.

Q. 4.3. Define thermal boundary lay.

Ans. When hot fluid flows over the place exists a region near the plate where temperature

gradient exists. The region r tie plate where temperature gradient exists is called thermal

boundary layer.

Thermal Boundary layer thickness (6k) is defined as the value of transverse

distance y from the plate surface at which

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Q. 4.4. What is a Nusselt number? Explain its significance.

Ans. Consider the flow of fluid past the hot plate

At plate surface0 the energy transport can occur only by conduction.

From energy balance, heat conducted through the plate must be equal to heat transfer by

convection into rest of the fluid.

The dimensionless number hl/A is called Nusselt number

Nusselt number is the ratio of temperature gradient at the surface to the overall

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Q. 4.5. Define:

(i) Reynold No. (ii) Grashof No. (iii) Prandtl No. (iv) Stanton number.

Ans. (i) Reynold No. : It is the ratio of inertia forces to viscous forces

ReNo.= Inertia force = = pVlViscous force

(ii) Grashof Number (Gr) : It is the relative strength of buoyant to viscous forces.

(iii) Prandtl Number (er) is the indicative of the idafive ability of the fluid to

diffuse momentum and internal energy by molecule

= Kinematic visocity

Thermal diffusity

(iv) Stanton number (St) is the ratio of 1 ..- ni to the flow of heat per unit temperature rise due tothe velocity of the fluid

Q. 4.6. Prove the free convection correlation

Ans. Buckingham 3t method

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The functional relationship between variables affecting free convection can be

premised as;

There are 8 physical quantities (Bg and AT are counted separately) and 5fundamental units hence there are (8-5) = 3ir terms

Equating the exponents of fundamental dimensions on both sides

M :0a+1L: 0-a-b+c+d-3

T: 0 - a - b - 2c

0: 0=-bH: 0 = b

Equating the exponents of fundamental dimensions on both sides;

M: a1 0L: a-b+c+d=0

T: -a-b-c=O (1: -b -10

H: b+1=0

Solution of these equations give;

b = -1

a = 1c = 0

d= 0

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M: a + b = 0

L: a + b + c + d + 2= 0

T: -a -3 h c - 2 = 00: b-1 = 0

Solving we get

b = -1

a = 1

c = 0d = 0

Solving we get

b= -1, a 0, c = 0, d = I

Thus the functional relation becomes:

Nu = (Re, Pr)Hence proved

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Q. 4.8. Define bulk temperature and mean film temperature.

Ans. Bulk temperature (tb) is defined as the arithmetic mean of the temperatures at inlet and exit

of heat exchanger tube.

Mathematically;

Mean film temperature (9 is the arithmetic mean of the surface temperature t5 of

a solid and the undisturbed temperature t of the fluid which flows past it.

Mathematically

Q. 4.9. Define local and average convective coefficient.

Ans. Consider the flow of fluid with velocity and temperature past astationary flat plate of length l and width

The local heat flux is given by;

Where is the local convective coefficient The total heat transfer rate Q is given by;

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Total heat transfer rate can also be expressed as;

where h = Average value of convective coefficient based on total surfa . Equating (i) and (ii)

Q. 4.10. Explain the formation of hydrodynamic boundary layer.

Ans. When a fluid flows around an object there exists a thin layer of fluid close to the solidsurface where velocity gradient exists. This thin layer of changing velocity has been called the

hydrodynamic boundary layer.

Consider the flow of fluid past a thin flat plate with sharp leading edge (fig. 4.4)

(I)For Re 5 x 1O the boundary layer is

turbulent. For 3 x

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(iii) The turbulent boundary layer does not extend to the solid surface underlying it is an

extremely thin layer called laminar sub layer.

(iv) For zone within the boundary layer, du / dy 0 while the condition for flow beyond theboundary layer and its outer edge are;

(v) The boundary layer thickness () is defined as the distance measured normal to the plate

surface at which u 0.99 u

(vi) For laminar boundary layer the velocity gradient becomes smaller along the flow directionand so does the shear stresses whereas for turbulent boundary layer the shear stresses at plate

surface have high value with steeper velocity gradient

Q. 4.11. What is a heat exchanger? What is its purpose? Give examples.

Ans. Heat exchanger is process equipment designed for effective transfer of heat energy between

two fluids, a hot fluid and the coolant. Noticeable examples are:

(i) Boi1ers(ii) Super heaters

(iii) Condensers

(iv) Automobile radiators(v) Evaporator of an ice plant

(vi) Water and air heaters or coolers.

The purpose of heat exchanger is to either remove heat from the fluid or to addheat to the fluid. The heat transferred may be in form of latent heat (Boilers and condensers) or

sensible heat (Heaters and coolers)

Q. 4.12. Classify different types of heat exchangers.

Ans. Heat exchangers can be classified according to:

(i) Operating principle(ii) Arrangement of flow path

(iii) Design and constructional detail

Classification of Heat Exchangers: Many types of heat exchangers have been developed tomeet the widely varying applications. Based upon their

. Operating principle

Arrangement of flow path Design and certain constructional features,

The heat exchangers can be classified into the following categories:

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Nature of heat exchange process Based upon the nature of heat exchange process, the heat

exchangers are classified into direct contact, generator arid respirators.

In direct contact or open heat exchangers, the energy transfer between the hot and cold

fluids is brought about by their complete physical mixing; there is simultaneous transfer

of heat and mass. Use of such units is restricted to the situations where mixing between the fluids is either

harmless or is desirable.

Examples are water cooling towers and jet condensers in steam power plants. Figure 4.5

represents a direct contact heat exchanger.

Steam is being bubbled into water; steam gets condensed and releases heat that warms up

the water.

In a regenerator, the hot fluid is passed through a certain medium called matrix. The heat

is transferred to the solid matrix and accumulates there; the operation is called heating

period.

The heat thus stored in the matrix is subsequently transferred to the cold fluid by

allowing it to pass over the heated matrix. The regenerators are quite often used in

connection with engines and gas turbines

.Other applications are:

Regenerators of open hearth and glass melting furnaces and air heaters of blast furnaces.

The operation of a regenerator is intermittent; the matrix alternately stores heat extracted

from the hot fluid and then delivers it to the cold fluid.

However in some of the regenerators the matrix is made to rotate through the fluidpassages arranged side by side and that renders the heat exchange process continuous.

The effectiveness of a regenerator depends upon the heat capacity of the regenerating

material and the rate of absorption and release of heat.

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In a recuperator, the fluids flow simultaneously on either side of a separating wall; the heat

transfer occurs between the fluid streams without mixing or physical contact with each other.The wall provides an element of thermal resistance between the fluids.

and the heat transfer consists of:

Convection between the hot fluid and the wall

Conduction through the wall Convection between wall and the cold fluid

Such exchangers are used when the two fluids cannot be allowed to mix, i.e., when

the mixing is undesirable. Majority of the industrial applications have exchangers of therecuperator type. Notable examples are:

(i) Boilers, superheaters and condensers ; economisers and the air preheaters in steam power

(iii) Condensers and evaporators in refrigeration units.

(iv) Oil heaters for an airplane(v) Heat exchanger inside a gas furnace etc.

The open-type (direct contact) heat exchangers and recuperators operate under steady state

conditions ; and the transfer of heat inside a regenerator takes place essentially under transient

conditions.Relative direction of motion fluids: According to the direction of flow of fluids,

the heat exchangers are classified into three categories : parallel flow, counter flow and the cross

flow.In the co-current or parallel flow arrangement, the fluid (hot and cold) enter the unit from the

same side, flow in the same direction and subsequently leave from the same side (Fig. 4.7).

Obviously the flow of fluids is unidirectiamal and parallel b eh other.

In the counter-current or counter-flow arrangement0 the fluids (hot and cold) enr the unit from

opposite ends, travel in opposite directions and subsequently leave from opposite ends.

Obviously the flow of fluids is opposite in direction to each othet. For a given surface area, thecounter-flow arrangement gives the maximum heat transfer rate and is naturally preferred for the

heating and cooling of fluids.

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In the cross-flow arrangement, the two fluids (hot and cold) are directed at right angles to each

other. Figure 4.9 shows two common arrangements of cross-flow heat exchangers. In figure 4.9

(a) the fluid A flows inside the separate tubes and its different streams do not mix. The fluid Bflows over the tube banks and gets perfectly mixed. In Figure 4.9(b), each of the fluid stays in

prescribed paths and are not allowed to mix as they flow through the heat exchanger. When

mixing occurs, the temperature variations are primarily in the flow direction. When unmixed,

there is temperature gradient along the steam as well as in the direction perpendicular to it.

Apparently, temperatures of the fluids leaving the unit are not uniform for the unmixed steams.The cross flow heat exchangers are commonly employed in air or gas heating and cooling

applications, e.g. the automobile radiator and the cooling unit of an air-conditioning duct.

Mechanical design of heat exchange surface:

(i) Concentric tubes : Two concentric pipes are used, each carrying one of the fluids. The

direction of flow may correspond to unidirectional or counter flow arrangement.

(ii) Shell and tube : One of the fluids is carried through a bundle of tubes enclosed by a shell.

The other fluid is forced through the shell and flows over the outside surfaceof tubes. The direction of flow for either or both fluids may change during its passage

through the heat exchanger.

(iii). Multiple shell and tube passes : The two fluids may flow through the exchanger only once(single pass), one or both fluids may traverse the exchanger more than once (multi-pass). By

suitable header design, the fluid within the tubes (tube side fluid) can be made to traverse back

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and forth from one end of the shell to the other. Quite often longitudinal baffles are provided

within the shell which cause the fluid surrounding the tubes (shell side fluid) to travel the length

of shell a number of times. An exchanger having n-shell passes and rn-tubes passes is designatedas n-rn exchanger.

Q. 4.13. Define overall heat transfer coefficient.

Ans. Rate of heat transfer between two fluids is given by:

WhereA T = Temperature difference between two bodies.

R = Thermal resistance

Also,

whereU = overall heat transfer coefficient

A = surface area

From (1) and (ii)

When two fluids of heat exchanger are separated by plane wall (fig. 4.11) the thermal resistance

comprises

1. Convection resistance due to fluid film at inner surface,

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1. Wail conduction resistance, /KA

1. Convection resistance due to fluid film at outside surface,

A.=A0= AFor plane wail

Q. 4.14. What is fouling ? What are its effects?

Ans. Consider heat flux through a cylindrical wall separating two fluids (Fig. 4.12)

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The thermal resistance for clean and uncorroded surface is given by;

(Based on inner surface)

For small wall thickness

However during normal operation the tube gets covered by deposits of ash, soot, dirt and

scale etc. This phenomena of rust formation and deposition of fluid impurities is calledfouling.

Due to fouling the surface deposits increase thermal resistance with corresponding drop in

performance of heat exchange equipment. If h51 and h50 denote heat transfer coefficientsfor scale formation on inside and outside surface then thermal resistance due to scale

formation on inner and outer surface is;

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(Based on inner surface)

Fouling factors are determined experimentally by testing the heat exchanger inboth the clean and dirty conditions.

Q. 4.15. Sketch the variation in temperature during

(a) Condensation (b) Boiling

Ans

Q. 4.16. Draw the variation temperature in case of parallel and counter flow heat

exchanger. Also define logarithmic mean temperature difference.

Ans. Temperature variation of fluids for : (a) Parallel flow and (b) Counter flow is

shown below

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Logarithmic Mean temperature difference (LMTD) is given by;

Where

Parallel flow heat exchanger

Counter flow heat exchanger

= Inlet temperature of hot fluid= Outlet temperature of hot fluid

= Inlet temperature of cold fluid.

= Outlet temperature of cold fluid.

Q. 4.17. Derive the relation for logarithmic mean temperature difference (LMTD) for

parallel flow and counter flow heat exchanger.

Ans.

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(a) Parallel flow heat exchanger. Consider an elementary strip at the distance end of heat

exchanger. Let dx be the length of strip. Let at this section. the temperature of hot fluid be

th and that of cold fluid be t which are assumed to be constant

dQ = Heat flow through the elementary length dx.

Heat exchange between the fluids can be written as;

Heat capacity or water equivalent of hot fluid

Heat capacity or water equivalent of cold fluid.

(b) Counter flow heat exchanger

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In general

+ve sign refers to parallel flow heat exchanger and -ye sign refers to counter flow heatexchanger.

Integrating (III) we get;

From equation (A) and (III)

Or

Integrating between inlet and outlet sections;

From (IV)

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Put this value in (VI) we get;

Q. 4.18. Define effectiveness and numb of transfer units and capacity ratio.

Ans. The effectiveness of heat exchanger is defined as ratio of energy actually

transferred to the maximum theoretical energy transfer.

Number of transfer units (NTU) : The number of transfer units (NTU) is a measure of thesize of heat exchanger, it provides some indication of the size of heat exchange_ It is

defined as;

Capacity ratio (C) : It is defined as ratio of minimum to maximum capacity rate.

Q. 4.19 Derive the relation for effectiveness of parallel flow heat exchanger.

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Ans. Consider the heat exchange through an incremental area dA as shown in fig.

From (I) and (II)

Integrating the above expression we get;

Solving

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Substituting this value in equation (III)

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Q. 4.20. Derive the relation for effectiveness of counter flow heat exchanger.

Ans. Consider the heat exchange through an incremental area dA of the exchanger

From (I)

Integrating the above equation between inset and outlet of heat exchanger

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From the definition of effectiveness;

Substituting these values in (II)

Let

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If c is assumed to be minimum i.e. c < Ch

Q. 4.21. Derive momentum equation for the hydrodynamic boundary layer.

Ans. For a two-dimensional infinitesimal control volume (dx x dy x unit depth)within the boundary layer region, the viscous forces acting along with the momentum of

fluid entering and leaving the elementary volume have been indicated in fig. 4.18.

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Momentum change : The momentum flux in x-direction is the product of mass flowing in x-direction and the i-component u of velocity. A fluid mass enters the left face at the rate p

udy producing an I-momentum influx

The momentum efflux through the right face is

Since we are concerned only with momentum in i-direction, the momentum of the fluid

moving in y-direction is obtained by multiplying the mass moving in y-direction also withthe i-component u of the velocity. Therefore the momentum influx from the bottom face is

mu = (p vdx) u

= p u v dx

and the momentum effux from the top face is

The resultant momentum change in x-direction is,

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= momentum efflux from the right and top faces - momentum influx from the left and

bottom faces

For the continuity equation (au/ax) + (av/ay) =0 and therefore the rt momentum

transfer in x-direction becomes:

Viscous forces : The shearing stress due to fluid viscosity is proportional to the

velocity gradient and is given by Newtons law c viscosity.

where u is the dynamic viscosity of the fluid

The shearing stress at the lower face of the control volume is r = u ( u/dy) aid the

corresponding shearing force for the area (dx x 1) is s (au/ay) dx. The shearing due to

viscosity at the upper face of the control volume is x + (r/.3ij) 4) and the correspondingshearing force for the area (dx x 1) is

Since the main stream flows in x-direction, the shearing force in y-direction can be

neglected. Therefore the net viscous force in the direction of motion is:

For a fixed control volume and steady flow, the Newtons second law of motion

stipulates that the resultant applied x-force equals the net rate of x-momentum transfer out of

the volume, i.e.,[(x-momentum efflux) - (x-momentum influx)]

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Neglecting the product of small quantities

Net energy convection

Likewise the net energy convection in the y-direction would be:

The conductive terms follow from the Fourier law, and the conduction heat rate in

the y-direction is

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Due to relative motion of fluid in the boundary layer (fluid on the top face of the

control volume moves faster than fluid on the boftom face), there will be viscous effectswhich will cause heat generation.

Viscous force = shear stress x area upon which it acts

This force will act through a distance s which can be determined by the relative

velocity of fluid flow at the upper and lower faces of element; s = (au/ ay) dy

With stipulations of steady state condition, the algebraic sum total of heat due to

convection, conduction and viscous effect equals zero. Thus

For a two-dimensional boundary layer flow, (au/ax) + (av/oy) = 0, and therefore the aboveequation can be recast as

Which represents the differential energy equation for flow past a flat plate. If the heat

generation due to viscous effects is neglected, the energy equation takes the form

Q. 4.23. Draw the profile of thermal boundary layer during flow of cold fluid over a

warm plate.

Ans.

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If the approaching free steam temperature S above the plate surface temperature the

thermal boundary layer will have the shape depicted in fig. 4.20.

The temperature a the fluid changes from a minimum at the plate surface to the

temperature of the main stream at a certain distance from the surface.

At point A, the temperature of the fluid is the same as the surface temperature t. Tl

fluid temperature increases gradually until it acquires the free stream temperature t.

The distance AB, measured perpendicularly to the plate surface, denotes the

thickness of thermal boundary at a distance x from the leading edge of the plate.

The concept of thermal boundary layer is analogous to that of hydrodynamic boundary layer; the

parameters affecting their growth are however different. The velocity profile of the

hydrodynamic boundary layer is dependent primarily upon the viscosity of the fluid. In a thermalboundary layer the temperature profile depends upon the flow velocity, specific heat, viscosity

and thermal conductivity of the fluid The thermo-physical properties of the fluid affect the

relative magnitude of 6 and

and the non-dimensional Pradtl number (Pr = M ce/k) constitutes the governing parameter:

\

Q. 4.24 What is meant by dimensional homogeneity ? What is dimensionless number ?

How and where they are used in heat transfer?

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Ans. In heat transfer there are 5 fundamental quantities units i.e. mass (M), length (L), Time (T),

Heat (H) and temperature (0). For any equation to be dimensionally homogeneous the

dimensions of each term must be same. Hence principle of dimensional homogeneity states thatfor an equation to be dimensionally homogeneous the fundamental units of each term must be

same.

A dimensionless number is the ratio of different forces. For example:(1) Reynold Number = Inertia force

Viscous force

(ii) crash of Number = buoyant force x Inertia force(Viscous force)2

(iii) Nusselt Number = Temperature gradient at surface

(iv) Prandtl Number = Kinematic viscosityThermal diffusivity

free and forced convection correlations and hence heat transfer can be determined as

follows

Example 4.1. A nuclear reactor with its core constructed of parallel vertical plates 225 m

high and 15 wide has been designed on free convection heating of liquid bismuth.

Metallurgical considerations limit the maximum surface tempratur of the plate to 975C

and the lowest allowable temperature of bismuth is 325C Estimate the maximum possibleheat dissipation from both sides of each plate.

The appropriate correlation for the convection coefficient is

where the different parameters are evaluated as the mean film temperature.

Solution : At the mean film temperature t = (975 + 325)/2 = 650C, the thermo Length physical

properties of bismuth are:

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Using the given correlation

This gives a heat transfer of: Q = 2 hA At

The factor 2 accounts for two sides of the plate.

= 153 MW.

Example 4.1 A. A Hot plate 1x1.5m is maintained at 300C. Air at 25C blow

overs the plate. If the surface heat transfer co-efficient is 20 W/m22C, Calculate the rate of

heat transfer.

Solution: We know that

Q = 20 x 1)C1.5(300 - 25)

= 20 x x 1.5 (275) 8250 W.

Example 4.2. Estimate the heat transfer from a 40 W incandescent bulb at 125C to

25C in quiescent air. Approximate the bulb as a 50 mm diameter sphere. What

percent of the power is lost by free convection?

The appropriate correlation for the convection coefficient is

when the different parameters are evaluated at the mean film

temperature, and the chrematistic length is the diameter of the sphere.

Solution: At the mean film temperature,

properties for air are:

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Using the given correlation,

This gives a heat transfer of

Therefore the percent of heat loss by free convection is

Example 4.3. A hot square plate 40 cm x 40 cm at 100C is exposed to atmospheric air

at 20C. Make calculations for the heat loss from both surfaces of the plate, if:(a) the plate is kept vertical (b) plate is kept horizontal. The following empirical

correlations have been suggested:

Nu = 0.125 (Gr Pr)33 for vertical position of plate, and

Nu = 0.72 (Gr Pr) for upper surface

0.35 = (Gr Pr) for lower surface

where the air properties are evaluated at the mean temperature.

Solution: At the mean temperature, t (100 + 20)/2 = 60C

the thermo physical properties of air are

= 0.724

1. When the plate is oriented vertically,

This gives a heat transfer of : Q = 2h AAtThe factor 2 accounts for two sides of the plate

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Q = 2 x 5.508 x (0.4 x 0.4) x (10020) = 141 W

(b) When the plate is positioned horizontally

(1) For upper surface:

(ii) For lower surface

Comments : The above calculations show that the plate loses more heat when it is

oriented vertically. Obviously natural cooling can be achieved more effectively by keeping the

plate in vertical position

Example 4.4. The lubricating oil for a large industrial gasturbine engine is cooled in a

counter flow, concentric tube heat exchanger. The cooling water flows through the inner

tube (d = 25 mm) with inlet temperature 25C and mass flow rate 0.2 kg/s. The oil flows

through the annulus (d0 =50 mm) with mass flow rate 0.125 kg! s and its temperature at

entry and exit are 90C and 60C respectively. Neglecting tube wall thermal resistance,

fouling factors and heat loss to surroundings, make calculations for outlet temperature of

cooling water, overall heat transfer coefficient and length of the tube.

The relevant thermo-physical properties of engine oil and water are as given below:

Solution : For the flow arrangement and temperature distribution of hot fluid

(engine oil) and cold fluid (water) in a counter flow, concentric tube heat exchanger.1. From energy balance,

The subscripts Ii and c refer tu hot (oil) and cold (water) fluids respectively.

Output temperature of cooling water,

(b) Reynolds number for flow of water through the tube is

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Since Re> 2300, the flow is turbulent and accordingly the following correlationapplies for calculating the heat transfer coefficient on the inside (water side),

The oil flows through the annulus with hydraulic diameter,

Since Re

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Neglecting fouling effects and thermal resistance of the tube material, the overallheat transfer coefficient is

1. The logarithmic mean temperature difference for counter flow arrangement

The heating surface area also equal (2r 41) where d and 1 represent the pipe diamer

and length respectively.

Example 4.5. A chemical industry operates continuously and produces 2 kg of sulphuric

acid per day which needs to be cooled from 60C to 40C in a counter flow double pipe heat

exchanger. The acid flows through the inner pipe while water employed as cooling medium

flows through the annulus with temperature 15C at entry and 20C at exit. The inner

diameters for the inner and outer pipe are 70 mm and 120 mm respectively, and each pipe

is 5 mm thick. Presuming that thermal conductivity of inner pipe material is 48 W/m K,

make calculations for the mass flow rate of water and length of the heat exchanger.

The thermo-physical properties of water and sulphuric acid at the mean bulk temperature

are given below:

Property Water

Acid

Density (kg/rn3)

998

1800

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Sp. heat (J/kg )

Thermalconductivity

(W/mK)

Kinematic viscosity (m2/s)

Fouling factor (m2 1(/W)

4187

0.60

1.0 x 10-6

-

1465

0.30

6.8 x 10.6

0.0002

Solution : (a) The mass flow rate of water can be determined from an energy balance on the twofluids. That is heat gained by water = heat lost by acid

(h) For counter flow arrangement

For the given configuration, the ove7all heat transfer coefficient is prescribed bythe relation

The heat transfer coefficients h1 for inner side (for acid side) and h0 for outer sideof pipe (for water side) are worked out as follows

For inner side of pipe:

Since Re > 2300, the flow is turbulent and the following correlation applies forCooling

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For outer side of pipe:Hydraulic diameter of annulus,

Since Re> 2300, the flow is turbulent and accordingly Dittus Boeltier correlation

for heating applies

Substituting the relevant values in expression (i), we get

The length of the heat exchanger can then be calculated as

67824 = 130.04 x (j x 0.08 x 1) x 31.91 ; 1 = 65.07 m

Comments : The heat exchanger has a long length. It would be appropriate

replace the double pipe heat exchanger by a shell and tube type heat exchanger.

Example 4.6. A counter flow heat exchanger is used to cool 2000 kg/hr of oil (c = 2.5 kJfkg

K) from 105C to 30C by the use of water entering at 15C. If the overall heat transfer

coefficient is expected to be 1.5 kW/m2 K, make calculations for the water flow rate, the

surface area required and the effectiveness of heat exchanger. Presume that the exit

temperature of the water is not to exceed 80C. Use NTU-effectiveness approach.

Solution (a) The mass flow rate of water can be determined from an energy balance on the two

fluids, i.e., heat lost by oil (hot fluid) = heat gained by water (coolant)

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1. Thermal capacity of the water stream (coolant)

Thermal capacity of the oil steam (hot fluid)

When hot fluid has the minimum heat capacity, then

The number of transfer units (NTU) can be computed from the following expression for

effectiveness of a counter flow exchanger

Rearranging

Therefore the heat transfer area is

Example 4.7. A counter-flow exchanger of surface area 8 m2 is to be used to heat a process

liquid by using a high temperature water available from another part of the plant. If the

overall coefficient of heat transfer is 450 W/m2 K, find the exit temperatures of the process

liquid and water stream from the data given below:

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Solution : Thermal capacity rate of the lt (water stream) and cooling (process liquid) fluid

are:

The effectiveness for a counter flow heat exchanger is,

Alternatively using the parameters C = 0.667 and NTU = 0.857, we get E = 0.495.The hot fluid has the minimum thermal capacity and therefore in terms of

temperature differences

Outlet temperature of the hot fluid (water),th2 = 3650.495 (365-300) = 332.82 K

From energy balance on the two fluids,

Outlet temperature of the cold fluid (process liquid),

Example 4.8. The engine oil at 150C is cooled at 80C in a parallel flow heat exchanger by

water entering at 25C and leaving at 60C. Estimate the exchanger effectiveness and the

number of transfer units. If the fluid flow rates and the inlet conditions remain unchanged,

work out the lowest temperature to which the oil may be cooled by increasing length of the

exchanger.

Solution: (a) From energy balance on the hot (oil) and cold (water) fluids,

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Apparently, the hot fluid has the minimum thermal capacity and

In terms of capacity ratio and number of transfer units,

exp [-1.5 NTUI = 1-1.5 0.56 = 0.16

Number of transfer units, NTU 1.221.

1. The exit temperature of oil would be minimum corresponding to the situation when the

exchanger is increased infinitely or NTU -

Therefore, the minimum possible exit temperature of the oil is

= 150 - 0.667 (150 - 25) = 66.62C

Example 4.9. A parallel flow heat exchanger uses 1500 kg/hr of cold water entering at 25C

to cool 600 kg/hr of hot water entering at 70C. The exit temperature on the hot side is

required to be 50C. Neglecting the effects of fouling, make calculations for the area of heat

exchanger. It may be presumed that the individual heat transfer coefficients on both sides

are 1600 W/ni2 K. Use the mean temperature difference approach and the effectiveness-

NT1J approach.

Proceed to calculate the exit temperature of the cold and hot streams if the flow of hot

water is doubled, i.e., it becomes 1200 kg/hr. It has been stated than the individual heat

transfer coefficients are proportional to 0.8th power of flow rate. For water C 4180 J/kg K.

Solution : The unknown exit temperature of the cooling water can be found from an energy

balance on the two fluids. That is

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From energy balance on the hot fluid, the heat transfer rate is

Mean temperature difference approach:

Effectiveness - NTU approach:Thermal capacity rates of the hot and cold fluids are:

Obviously

Capacity ratio, C

Effectiveness, E

The effectiveness for a parallel flow heat exchanger is given by

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Upon solution we obtain: NTU = 0.695

Heating surface area, A = 0.695 x 696.67800

= 0.605 m2

The second part of the problem can be worked out more easily by using thefectiveness - NTU approach.

Since the flow rate is doubled on the hot side,

Capacity ratio C = 1395.33/1741.67 = 0.801Transfer units NTU

Invoking the effectiveness relation for parallel flow heat exchanger,

Upon solution we get;

= 70 0.305 x 45 = 56.27C

From energy balance,

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Example 4.10. Oil with a mean specific heat of 2.5 kJ/kg. K is to be cooled from 110C to

30C in a single pass counter flow heat exchanger. The coolant is water which enters at

20C and leaves at 80 C and the overall heat ttansfer coefficient for this type of heat

exchanger is 1.5 kW/m2 K. If the water flow rate is 1500 kg/hr, determine the quantity of

oil that can be cooled per hour and the heat exchanger area. (b) What would be the fluid

exit temperatures when the water flow rate is decreased to 1000 kg/hr for the same oil flow

rate ? Comment upon the results.

The effectiveness of a counter flow heat exchanger is given by the following

expression where NTU is the number of transfer units and C is the capacity rate ratio

Solution: (a) The mass flow rate of oil can be determined from an energy balance on the two

fluids, i.e.,

Heat lost by oil (hot fluid) = heat gained by water (coolant)

mh x 2.5 (110 - 30) = 1500 x 4.186 (80 - 20) = 376740 kJ/hr

Therefore the oil flow rate is,

= 3767402.5x80

= 1884 kg/hr

Thermal capacity rate of the hot (oil) and cold (water) fluid are

Capacity ratio, C

When the hot fluid (oil) has the minimum thermal capacity, thenEffectiveness,

Rearrangement of the given expression gives:

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:. Heat transfer area, A

Note : The heat exchange area can be found more conveniently by the LMTD

approach.For the counter flow arrangement, the end temperature difference are:

Now, the heat exchange, Q = UAO,,,

(h) The changed thermal capacity rates of the hot (oil) and cold (water) fluid are:Ci =1884 x2.5=4710kJ/hrK

Cc =1000x4.186=4186kJ/hrK

Inserting the appropriate values in expression (i),

When the coolant (water) has the minimum thermal capacity, then

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Outlet temperature of the coolant (water)

t2 = 0.868 (110 20) + s 20

= 98.12C.

Example 4.11. In a center flow heat exchanger, oil (c=3kJ/kg k) at rate of 1400 kg/ hr is

cooled from 100C to 30C by water that enters the exchanger at 20C at rate

1300 kg/hr. Determine heat exchanger area from an overall heat transfer co-efficient 3975

kJ/m2 K. Also derive the relationship between oil and water temperature at any section of

heat exchanger.

Solution.

(ii) At any section of heat exchanger

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Q. 4.12. A heat exchanger is to be designed to condense an organic vapour at a rate of 500

kg/mm which is available at its saturation temperature 355 K. Cooling water at 286 K is

available at a flow rate of 60 kg/s. The overall heat transfer coefficient is 475 W/m2C.

Latent heat of condensation of the organic vapour is 600 kJ/kg. Calculate:

(i) The number of tubes required, if 25 mm o diameter, 2 mm thick and 4.87 m long tubes

are available, and

(ii) The number of tube passes, if the cooling water velocity (tube side) shouldnot exceed 2 iVs. (M.U.)

Ans. Given

d0 = 25 mm = 0.025 m; d. = 25 2 x 2 = 21 mm = 0.021 m ; L = 4.87 m;

(I) The number of tubes required, N:

Heat lost by vapour = heat gained by water

Logarithmic mean temperature difference (LMTD) is given by

Heat transfer rate is given by,

N = 470 tubes Ans.

(ii) The number of tube passes, p:

The cold water flow mass passing through each pass (assume p are number of

passes) is given by,

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= 4.91 = 5 Ans.