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

    reference temperature gradient.

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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.


    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.


    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


    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

    plants.(ii) Automobile radiators

    (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


    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


    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


    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;


    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.


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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)


    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;


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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)


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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.


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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


    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

    Overall reference temperature gradient

    (iv) Prandtl Number = Kinematic viscosityThermal diffusivity

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


    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


    Density (kg/rn3)



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    Sp. heat (J/kg )



    Kinematic viscosity (m2/s)

    Fouling factor (m2 1(/W)



    1.0 x 10-6




    6.8 x 10.6


    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


    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


    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


    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:


    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.


    (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.