4. Natural Convection - Heat an= d Mass Transfer, 2nd Edition

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4 Natural Convection

= 4.1 INTRODUCTION

In natural or free convection= heat transfer mechanism, the flow of fluid takes place when there is a dif= ference between the body and the fluid temperatures. The difference in dens= ity between the cold and hot fluids in the region causes the hot fluid to m= ove in the upward direction. The force causing this flow is known as buoyan= cy force. The natural convection heat transfer coefficients are usually ver= y low. Natural convection heat transfer mechanism is applied in many fields= of engineering for cooling purposes. The free convection mechanism involve= s transferring of heat, for example, from electric transformers, rectifiers= , pipes carrying refrigerants and steam, walls of the furnaces, and from th= e walls of air-conditioned houses.

KEY CONCEPTS

Cooking in a gas oven or boiling water are good illustrations = of heat being transferred by convection.
When heated, the particles of a liquid or gas become lighter and ri= se, while colder particles sink to the bottom and are then heated in turn.<= /p>
Convection involves the transfer of heat by the motion and mixing o= f =E2=80=98macroscopic=E2=80=99 portions of a fluid.
The term natural convection is used if this motion and mixing is ca= used by density variations resulting from temperature differences within th= e fluid
The heat transfer inside are frigerator is an example of natura= l convection.

= 4.2 NATURAL CONVECTION HEAT TRANSFER OVER A VERTICAL PLATE

Fluid velocity in free convec= tion is very low compared with that in forced convection. A boundary layer = forms near the surface and fluid velocity is zero at the solid surface.

The density of the fluid nea= r the hot surface is less than that of the cold fluid away from the heated = wall, and this creates a buoyant force which causes the heated fluid to mov= e in the upward direction, and a boundary layer is formed near the heated w= all.

The t= emperature and velocity variations at a specific cross section are shown in= Fig. 4.1. The fluid velocity incr= eases from zero at the wall to a maximum and then decreases to zero as the = ambient condition is reached because the effect of viscosity is less away f= rom the wall and the effect of buoyant force is predominantly away from the= surface.

Fi= gure 4.1 Natural Convection Heat Iranster on Hot vertical Plate

The heat flows from the soli= d surface to the fluid only inside the boundary layer and not to the ambien= t air in the room. Air inside the boundary layer becomes hotter in the = x direction and will entrain cold air at the outer edge of the boundar= y and widen the thickness of the boundary along the x direction. T= emperature and velocity gradients become zero at one point, and the hydrody= namic and thermal boundary layers will have equal thickness.

4.2.1 = Integral Method

The problem of natural convec= tion can be solved using the method proposed by Von-Karman. To find the hea= t transfer from a vertical plate to the surrounding fluid by natural convec= tion, continuity, momentum, and energy equations are to be solved.

The boundary formed on a hea= ted vertical flat plate by natural convection is shown in Fig. 4.1.

The continuity equation for = two-dimensional incompressible fluid flow is given as

With = the flow in vertical direction and body force acting in the downward direct= ion, the momentum equation for the boundary layer of natural convection is = given by

where =E2=80=93=CF=81g represents the body force. The pressure gradient in the x direction upto the height of the plate, can be written as

The suffix =E2=88=9E indicat= es the ambient condition of the fluid outside the boundary.

Substituting the value of = =E2=88=82p/=E2=88=82x from Equation (4.3) into Equation = (4.2), we get

Assuming that the density =CF=81 depends only on temperature and not on pressure, the buoyancy= force g(=CF=81_=E2=88=9E =E2= =88=92 =CF=81)and substituting in Equations can be written in the = following form:

where =CE=B2 is the = volume coefficient of thermal expansion whose value is given by

The volume coefficient = =CE=B2 may be determined from the property tables for specific fluid a= nd for ideal gases, =CE=B2 =3D 1/T, where T is the absolu= te temperature of the gas.

By substituting Equation (4.5), Equation (4.4) becomes

Considering that the flow is= incompressible, the energy equation for the boundary layer is given as

Integ= rating Equation (4.6) between the l= imits of zero and boundary layer thickness =CE=B4

Considering the flow as inco= mpressible, the left-hand side of Equat= ion (4.8) can be written as

Substituting from the continuity = Equation (4.1)

As both u and v= are zero at y =3D 0 and y =3D =CE=B4, it b= ecomes

In addition, considering the= last term in the right-hand side of Eq= uation (4.8)

Substituting the values from= Equations (4.9) and (4.10) into = Equation (4.8), we get

Equation (4.11) is the momentum integra= l equation for the boundary layer.

Similarly by integrating the= energy Equation (4.7) in the bound= ary layer, we can obtain the energy integral equation as

Defining =CE=B8 =3D= (T =E2=80=93 T_=E2=88=9E) and = substituting into Equation (4.12),= it becomes

Assuming the temperature dis= tribution as

to satisfy all the temperatur= e boundary conditions mentioned below:

T =3D T_w at y =3D 0,

T =3D T_=E2=
=88=9E at y =3D =CE=B4,

at y =3D =CE=B4.

Assuming a polynomial functi= on of y multiplied by some arbitrary function of x for th= e velocity in the boundary layer as

where u_x is any arbitrary function of x.

The constants a, b, c, and d can be calculated using the following = boundary conditions in Equation (4.15)= :

u =3D 0 at y =3D = =CE=B4,

u =3D 0 at y =3D = 0,

from <= a href=3D"https://learning.oreilly.com/library/view/heat-and-mass/978933251= 5680/xhtml/chapter004.xhtml#eq4.6">Equation (4.6) because u an= d =CF=85 are zero at the surface.

The values of the constants = obtained are

a =3D 0, = ; &n= bsp;

and

Using the values of these co= nstants in Equation (4.15), we can= obtain

Adding (T_w =E2=88=92 T_=E2=88=9E= ), =CE=B4² and =CE=B2 g/4 = v in u_x, the expression becomes

To find the maximum velocity=

= ; =CE=B4² =E2=88=92 4y += 3y² =3D 0,

= ; (=CE=B4 =E2=88=92 y)(=CE=B4 =E2=88=92 3y) =3D 0,

= ; y =3D =CE=B4= or =3D"image"

At y =3D =CE=B4, th= e velocity becomes zero and hence at the second value of=E2=80=99 y =3D= =CE=B4/3, the maximum velocity occurs and the magnitude of the maximu= m velocity is

The mean velocity.

Subst= ituting the values of u/u_x and= =CE=B8/=CE=B8_m in Equation (4.13), we get

Solving the equation, we get=

Substituting the values of <= em>u/u_x and =CE=B8/=CE=B8_m in Equati= on (4.11), we get

Solving the equation, we get=

Assuming that u_x and =CE=B4 vary as

u_x =3D C₁₌ x^m and =CE=B4 =3D C= ₂xⁿ

and substituting in Equations (4.19) and (4.20), we get

From Equations (4.21) and (4.22),

m + n =E2=88=92 1 =3D =E2= =88=92n

2m + n =E2=88=92 1 =3D n + m =E2=88=92 n =3D n.

Solving these equations, we = get

m =3D 1/2 and n = =3D 1/4.

Substituting m and = n in Equations (4.21) and= (4.22), we get

Solvi= ng Equations (4.23) and (4.24), we get

Substituting C_l and C₂ to find = u_x and =CE=B4 and also v/=CE=B1 =3D P_n, we get

and

Using this relationship, the= variation of boundary layer thickness along the height of the plate can be= found out.

4.2.2 = Heat Transfer Coefficient

For steady-state, the heat fl= ux is

Local heat transfer coeffici= ent,

Using the value of T from Equation (4.14) in Equation (4.30), we get

where Nu_x is the local Nusselt number.

Substituting the value of x/=CE=B4 from Equation= (4.29) to Equation (4.31), we= get

The average Nusselt number,<= /p>

Using the value of Nul from Eq= uation (4.32),

=3D 0.667 (P_n)^1/2=(P_n + 0.952)^{=
=E2=88=921/2}(Gr_l)^1/4. (4.34)

The above equation can be us= ed only for the laminar boundary layer, and the characteristic length used = in the above equation is the height of the plate. The critical Grashof numb= er for air is Gr_l =3D 4 =C3=97 10⁸ (Eckert and Sochngen).

If the plate is inclined wit= h an angle =CE=B8to the vertical, the average Nusselt number is

If the flow is turbulent, th= en the average heat transfer coefficient over the whole surface of the plat= e is given by

The physical properties of t= he fluid are to be taken at the average temperature value of wall surface t= emperature and ambient air temperature.

= 4.3 EMPIRICAL RELATIONS FOR FREE CONVECTION

The average free convection h= eat transfer coefficient for various circumstances is obtained using

where f represents t= hat the properties should be evaluated at the film temperature

The characteristic dimension= used in Gr and Nu are (i) height of the plate L= for vertical plates and (ii) diameter d for horizontal cylinder. = The approximate values of C and m for various geometries = used are given inTable 4.1.

Table = 4.1 Constants for isothermal surfaces

4.3.1 = Vertical Planes and Cylinders

4.3.1.1 Uniform wall temperature= or isothermal surface

McAdams (1954) developed the = mean heat transfer coefficient in the form

and

The properties of the fluid = are to be found at film temperature T_f = =3D ((T_w + T_=CE=B1) / 2). In Equ= ations (4.37) and (4.38), the = Prandtl number range is not mentioned.

Churchill and Chu (1975) pro= posed the following correlations with the limitations of Prandtl number:

for laminar flow only (0 <= Ra < 10⁹ and 0 < Pr < =E2=88=9E),

For liquid metals (Pr &l= t; 0.03), the equation is

4.3.1.2 Uniform wall heat flux
The following correlations we= re proposed by Vliet and CKL in (1969) for uniform heat flux conditions for= local Nu. In this, modified Grashof number (Gr^*_x) is used to find th= e type of flow.

where q_w is the constant wall heat flux,

Then mean Nu for Equations (4.42) and (4.43) are

and

4.3.2 = Horizontal Plates

The average Nu used = for free convection heat transfer on a horizontal plate based on whether th= e hot and cold surfaces are facing up or down as shown in Fig. 4.2.

Fi= guer 4.2 Natural Convection on Horizontal Plates

4.3.2.1 Uniform wall temperature=

Here, the mean Nusselt number=

and

Grasof number

where =E2=80=98L =E2= =80=99 is the characteristic length of the plate which is defined as

The following correlations w= ere given by McAdams (1954) for mean :

For hot surface facing upwar= d,

In case of hot surface facin= g down or cold surface facing up, the plate blocks the fluid flow. Since th= e velocity is low, only laminar flow is existing in this case. The mean Nu is given by

4.3.2.2 Uniform heat flu= x

Heated surface facing upward<= /p>

Hot surface facing downard

4.3.3 Inclined Plates

The inclination of a plate fr= om the vertical is shown in Fig. 4.3. Here the transition from laminar to turbulent flow depends on the angle= of inclination. The angle is positive if the hot surface faces upward and = negative if the hot surface faces downward.

Fi= gure 4.3 Co-ordinate System for Inclined Plates

4.3.3.1 Uniform wall temperature=

Fujii and Imura (1972) recomm= ended that the horizontal plate correlations can be used for inclination be= tween 60=C2=B0 and 90=C2=B0 and for less than 60=C2=B0, the vertical plate = corrections can be used.

4.3.3.2 Uniform heat flux

If the hot surface faces down= ward.

and

If the hot surface faces upw= ard, then the correlation for angles between =E2=80=9315=C2=B0 and =E2=80= =9375=C2=B0 is

where Gr_c is the transition Gr which depends on the angle o= f inclination as given Table 4.2. = The properties of a fluid are taken at mean temperature.

Table = 4.2 Angle of inclination

=CE=B8 Gr_C<= /th>

15 5 =C3=97 10⁹<= /td>

30 2 =C3=97 10⁹<= /td>

60 10⁸

and =CE=B2 is taken = at T_=CE=B1+ 0.5 (T_w=E2=80=93 T_=CE=B1).

4.3.4 Vertical Cylinders

In the case of isothermal cyl= inders with their areas vertical, the expressions presented for vertical pl= ate can be used provided the curvature effect is negligible. A vertical cyl= inder can be evaluated using correlations for vertical plates when

where D and L are diameter and height of the cylinders, respectively.

In case of a vertical cylind= er subjected to uniform all heat flux, the local Nu can be determi= ned from Equations (4.42) and (4.43).

4.3.5 = Horizontal Cylinders

Churchill and Chu (1975) have= proposed the following correlation for isothermal horizontal cylinder.

where a= nd Ra_D are determined using the cylinde= r diameter as characteristic length.

For laminar range alone the = following equation is used:

Properties for the dimension= al numbers in the above equations are to be determined based on film temper= ature.

For rotating horizontal cyli= nder, the correlation is given as

where Re_D =3D w=CF=80D² / v<= /em>, w is the rotational speed of the cylinder.

4.3.6 = Spheres

The following correlations ar= e used to calculate the average Nu for isothermal spheres:

Churchill suggested the foll= owing general formula for spheres:

= 4.4 FREE CONVECTION IN ENCLOSED SPACES

The configuration and nomencl= ature of a rectangular enclosure (solar collector) are shown in Fig. 4.4. Heat transfer rate depends on the a= ngle of tilt (=CE=B8), aspect ratio (H/L) and dimensionle= ss parameters (Pr and Ra).

Fi= gure 4.4 Natural Convection in an Enclosure

Fluid properties are evaluat= ed at mean temperature ((T₁+ T= ₂) / 2) and Rayleigh number

For vertical enclosure, = =CE=B8 =3D 90=C2=B0.

Catton (1978) proposed the f= ollowing corrections for vertical enclosure:

for 2 < H/L < = 10, Pr < 10⁵ and Ra < 10= ¹⁰,

for 10 < H/L < 40, 1 < Pr < 2 =C3=97 10⁴ and 1= 0⁴ < Ra < 10^=
7.

For horizontal enclosure: =CE=B8 =3D 0.

Globe and Dropkin (1959) pro= posed the following correlation:

where =CE=B8 =3D 180= =C2=B0.

When Ra < 1700, = the convection between the surfaces is negligible, and the heat transfer ta= kes place by pure conduction, and Nu tends to unity.

For inclined enclosure 0 <= ; =CE=B8 < 90:

Holland et al. (1976) have g= iven the following correlation for inclined enclosures:

for

The bracket [ ]* implies tha= t if the quantity is negative, it must be equal to zero.

Ayyaswamy and Catton (1973) = have recommended the following correlations for all aspect ratios (H/L<= /em>):

= 4.5 CONCENTRIC CYLINDERS

Heat transfer in the annular = space between long horizontal concentric cylinders by natural convection du= e to temperature difference between the walls has been studied by Raithby a= nd Hollands (1975). The correlations for heat transfer rate (q) pe= r unit length for configuration as in = Fig. 4.5 is expressed as

Fi= gure 4.5 Annual space Between Concentric Cylinders or Spheres

where k_eff is effective thermal conductivity of fluid. It is defined= as thermal conductivity of a stationary fluid should have to transfer the = same amount of heat is the moving fluid. It can be calculated as

where

Here

= 4.6 CONCENTRIC SPHERES

Raithby and Holland (1975) pr= oposed the following equations=E2=80=99 correlation for concentric spheres = for total heat transfer (Q):

The effective thermal conduc= tivity (k_eff) is

where

This correlation is valid fo= r the range 10² < Ra_=
c < 10⁷.

= 4.7 COMBINED FREE AND FORCED CONVECTION (MIXED C= ONVECTION)

In forced convection, the flo= w induced by density gradient is negligible compared with the fluid motion = imposed by external devices, but, in case of low flow velocity, the free co= nvection factor also plays an important role. A dimensionless parameter Gr/Re² predicts the relative importance of= free convection with forced convection. The various conditions of convecti= on conditions are:

for free convection.

for mixed convection (free += forced convection),

for forced convection,

Correlations for local N= u_x for mixed convection on vertical plates = are given as

and

where A =E2=89=83 0.= 6 for Pr =E2=89=A4 10, and A =E2=89=83 1.0 for Pr =3D= 100.

In case of horizontal plate = where (Gr_x/Re^{25_x) < 0.083, the following equatio=
n may be used:}

For mixed convection through= horizontal tubes with laminar flow (Re_D =E2=89=A4 2000), the recommended correlation is

where Gz (Graetz num= ber) =3D Re_D Pr (D/L) and = =CE=BC_b and =CE=BCs=E2=80=93 viscosities of the fluid at the bul= k mean temperature and surface temperature, respectively.

For mixed convection with tu= rbulent flow in horizontal tubes, Metais and Eckert (1966) recommend

= 4.8 COMBINED CONVECTION AND RADIATION HEAT TRANS= FER

In many practical problems, t= he heat lost by radiation from the surface is of considerable magnitude com= pared with the heat loss by natural convection. This is particularly true w= hen the temperature of the surface is considerably high.

The total heat transfer from= the surface by combined natural convection and radiation is given by

Q_t =3D Q_c + Q_r

      = ;            &n= bsp;            = ;             &= nbsp;         =3D h_c A(T_w =E2=88=92 T_=E2=88=9E) + h_r A(T_u= =E2=88=92 T_=E2=88=9E),   &nb= sp;  (4.83)

where h_c and h_r are the average = heat transfer coefficients by natural convection and radiation.

where U is known as = overall heat transfer coefficient which takes the heat flow by both the mod= es into account.

The values of h_c is given by the equation

and we can calculate the heat= lost by convection.

The heat loss by radiation i= s given by

After calculating the heat l= oss by convection and radiation and adding the two, we can find out the com= bined average heat transfer coefficient by using Equation (4.79).

= 4.9 CONCLUSIONS

When fluid flows over a solid= surface or inside a body with different temperatures of the fluid and the = solid surfaces, heat transfer between the fluid and the solid takes place b= y physical contact of the particles. Moving fluid particles exchange heat w= ith the surface continuously and this mechanism of heat transfer is known a= s convection heat transfer.

In convection heat transfer,= if the fluid motion is set up by buoyancy force because of density differe= nce caused by temperature difference in the fluid layers, then the heat tra= nsfer mechanism is known as free convection or natural convection.

Natural convection heat tran= sfer mechanism is applied in many fields of engineering for cooling purpose= s.

Fluid velocity in free conve= ction is very low compared with that in forced convection. A boundary layer= forms near the surface and fluid velocity is zero at the solid surface.
To find the heat transfer fr= om a vertical plate to the surrounding fluid by natural convection, continu= ity, momentum, and energy equations are to be solved.

Nusse= lt number is defined as the ratio of convection heat transfer to fluid cond= uction heat transfer under the same conditions.

The Gr is the ratio= of the buoyancy to viscous force acting on a fluid.

The characteristic dimension= used in Gr and Nu are (i) height of the plate L= for vertical plates and (ii) diameter d for horizontal cylinder.<= /p>
EXAMPLE PROBLEMS

4.1. Find the rate= of convective heat transfer from a 0.4 m (L) vertical square plate maintai= ned at an uniform temperature (T_w) 130=C2=B0C in= ambient air at 25=C2=B0C (T_a).

Film temperature,

=CE=BD =3D 20.75 =C3=97 10^=E2=88=926 m²/s; Pr = =3D 0.697; K =3D 0.03 W/mK

Grashof number,

where

FromTable 4.1 or Equat= ion (4.3), we get the value of Nusselt number,

Heat transfer coefficien= t,

Heat transfer rate from both= sides,

U= sing Churchill's correlation,

Both correlations give c= lose values of .

4.2. Find the rate= of heat transfer from a vertical plate of height 4 m (L) and width 10 m (w= ) which is maintained at 333 K (T_w) and exposed = to atmospheric air at 283 K (T_w).

Film temperature,

v =3D 16.5 =C3=97 10^=E2=88=926 m²/s; Pr =3D= 0.7; K =3D 0.027 W/mK

Grashof number,

From Equation (4.38),

Nusselt number,

Heat transfer coefficien= t

R= ate of heat transfer from one side of the plate,

4.3. Find the surf= ace temperature of 4 m (L) square vertical metal plate insulated on one sid= e with the other side exposed to a solar radiation flux (q) of 800 W/m². The exposed surface has 100% absorptivity and the = incoming radiation is lost by free convection to the ambient air at 30=C2= =B0C (T_=CE=B1).

As the surface temperatu= re is not known, let h =3D 8 W/m²K

Film temperature, T<= sub class=3D"EBsub">f =3D (T_w + T= _a)/2 =3D (130 + 30)/2 =3D 80=C2=B0 C

v =3D 20.8 =C3=97 10^=E2=88=926 m²/s; Pr =3D= 0.697; K =3D 0.03 W/mK

Assuming uniform heat fl= ux, the modified Grashof number, Gr* =3D g=CE=B2qx⁴/kv² |_{x =3D L}

Then,

Nusselt number,

Heat transfer coefficien= t,

Then average heat transf= er coefficient,

=3D 1.136 =C3=97 h_x =3D 1.136 =C3=97 6.55 =3D 7.4 W/m²K

T= he wall temperature,

The film temperature, T_f =3D (138.1 + 30)/2 =3D 84=C2=B0 C

Since assumed film tempe= rature (80 =C2=B0C) is very close to 84=C2=B0C, the calculation need not be= repeated.

4.4. A plate of si= ze 2 m =C3=97 1 m (l =C3=97 b) is kept horizontally in water and m= aintained at a temperature of 20=C2=B0C (T). Find the heat transfer coeffic= ient and the rate of heat transfer if (i) the top surface of the plate is m= aintained at 60=C2=B0C (T_w) and (ii) the bottom = surface of plate is maintained at 60=C2=B0C (T_w)= .

Film temperature

The top surface is heated:

Characteristic length,

Then

The flow is turbulent an= d from Equation (4.51)

R= ate of heat transfer,

Q =3D A= (T_w =E2=80=92 T_=CE=B1)

      = ;            = =3D 1870 =C3=97 2 =C3=97 1(60 =E2=80=92 20)

=3D149.6 kW

The bottom surface is heated:

The correlation is

Rate of heat transfer

Under similar conditions, the= hot surface facing downwards transfers less heat due to the blocking of th= e upward flow of the fluid stream.

4.5. Find the rate= of heat transfer from a 2 m (L) square plate maintained at temperature of = 210=C2=B0C (T_w) in ambient air at 30=C2=B0C (T_=CE=B1) in (i) vertical and (ii) horizont= al positions.

Vertical position:

Grashof number,

From Equation (4.39),

Nusselt number,

Rate of heat transfer on= both sides,

Horizontal position:

Grashof number,

Then, GrPr =3D 8.6 1= 0⁸ =C3=97 0.686 =3D 5.9 =C3=97 10⁸. From Equation (4.51), when the top surface is heated the Nusselt number,

Heat transfer from top surfac= e,

From Equation (4.52), when the bottom surface is heated, the Nu= sselt number

The heat transferred from the= bottom surface,

Q_b =3D 1.84kW

Total heat transfer =3D Q= _t + Q_b = =3D 7.458 kW.

4.6. A pipe with o= uter diameter 20 cm (D) and length 3 m (L) carries saturated steam at 100= =C2=B0C (T_w). The temperature of the surrounding= air is 30=C2=B0C (TJ. Find the rate of heat loss if the pipe is kept in (i= ) a vertical position and (ii) a horizontal position.

K =3D 0.03 W/m² K; Pr =3D 0.697;

=CE=BD =3D 20.76 =C3=97 10^=E2=80=92⁶ m²/s

Vertical position:

Grashof number,

Ra =3D GrPr

=3D 12.7 =C3=97 10¹⁰ =C3=97 0.697 =3D 8.85 =C3=97 10¹⁰

For this vertical cylinder,

The correlation for the Nusse= lt number, can be obtained from the expression for th= at of a vertical plate and from Equati= on (4.39), this is found to be

Heat loss,

Q =3D h(=CF=80DL)= (T_w=E2=80=92T_=CE=B1)

      = ;            &n= bsp;            = ;       =3D 5.75 =C3=97 (=CF=80 =C3=97 0= .2 =C3=97 3)(100 =E2=80=93 30) =3D 758.7 W

Horizontal position:
Grashof number,

Ra_D = =3D Gr_DPr =3D 3.77 =C3=97 10⁷ =C3=97 0.697 =3D 2.63 =C3=97 10⁷

F= rom Table 4.1, the value of may be obtained.

Heat loss,

4.7. A pipe of dia= meter 0.2 m (D) and length 1 m which is maintained at a temperature of 50= =C2=B0C (T_w) is kept under water at 30=C2=B0C (T= _a). Find the rate of heat loss.

For this case,

Grashof number,

For the above range of <= em>Ra_D, the correlation for Nusselt number,= fromTable 4.1, is

H= eat loss per metre length of the pipe,

4.8. A wire of dia= meter 0.2 mm (D) and length 1 m is placed horizontally and its surface is m= aintained at a constant temperature of 60=C2=B0C (T_w). The wire is exposed to air at 20=C2=B0C (T_a). Find the heat loss from the wire.

Film temperature,

v =3D 16.96 =C3=97 10^=E2=88=926 m²/s; K =3D= 26.56 =C3=97 10^=E2=88=923 W/mK; Pr =3D= 0.7

Grashof number,

For Ra_D in the above range, fromTable 4.1, the Nusselt number, is given by<= /p>

Heat loss per metre leng= th,

4.9. A 90-W bulb a= t a temperature of 120=C2=B0C (T_w) is exposed to= air at a temperature of 20=C2=B0C (T_a). If the = bulb diameter is 60 mm (D), find the heat loss; assuming that it i= s a sphere.

Film temperature,

K =3D 29.64 =C3=97 10^=E2=88=923 W/mK;

v = =3D 20.2 =C3=9710^=E2=88=926m²/s; Pr =3D 0.695

Grashof number,

From Equation (4.65),

Nusselt number,

Heat loss by convection,=

4.10. A hot metal = cylinder of diameter 0.5 m (D) rejects heat while machining at a speed of 3= 00 rpm (N) in air. The metal and air temperatures are 65=C2=B0C (T_w) and 15=C2=B0C (T_a), respectively. Find the rate of heat loss.

Film temperature,

G= rashof number,

Using Equation (4.63)

The Nusselt number,

Rate of heat transfer

4.11. Find the rat= e of heat transfer by free convection across the air space between two squa= re parallel vertical plates of area 1 m =C3=97 1 m separated by a distance of 5 cm (L) enclosing air at a pressure = of 1 atm. The temperatures of hot (T_o) and cold = (T_i.) plates are 100=C2=B0C (T_i.) and 20=C2=B0C (T₀), respec= tively.

Film temperature,

v =3D 18.97 =C3=97 10=E2=88=926 m²/s; K =3D 28.96 =C3=97 10^=E2=88=923 W/mK; Pr =3D 0.696

T= he Rayleigh number, Ra, is given by

Therefore,

The aspect ratio, H/L =3D 1/0.05 =3D 20

The suitable correlat= ion for this range of H/L and Ra is (4.69)

Heat transfer rate,

4.12. The surface = of a tube of outer diameter (D_i.) 0.1 m is maint= ained at a temperature of 400 K (T_i). An insulat= ion is provided concentric to the tube with an air gap of 15 mm and the ins= ulation is at 310 K (T₀). Find the heat lost per= metre length of tube.

Film temperature,

v =3D 21.1 =C3=97 10^=E2=88=926 m²/s; K =3D = 30.47 =C3=97 10^=E2=88=923 W/m K; Pr =3D= 0.692

For concentric cylinders= , the heat lost per unit length, Q, is given by (4.75)

F= rom Equation (4.76)

where

Therefore,

The effective thermal co= nductivity,

Heat loss,

4.13. Air at 5 bar= fills the space between two concentric spheres having diameters 10 cm (Di) and 15 cm (D₀),= respectively. The surface temperatures of the inner and outer spheres = are maintained at 50=C2=B0C (T_i) and 20=C2=B0C (= T₀), respectively. Find the heat transf= er rate and compare it with the heat transfer rate for air at atmospheric p= ressure.

Film temperature,

Density,

T= herefore,

Grashof number,

Therefore,

From Equation (4.78), the heat transfer rate,

Therefore,

Heat transfer rate,

For air at atmospheric p= ressure:

=CF=81 =3D 1.15kg/m³

v =3D 16.43 =C3=97 10^=E2=88=926 m²/s

Grashof number,

Therefore

Therefore

Heat transfer rate,

Higher the air pressure = between concentric spheres, higher is the heat transfer rate.

4.14. Air at 1 bar= and 30=C2=B0C (T_a) flows with a velocity (U) of= 0.25 m/s through a horizontal tube of diameter 2 cm (D). If the tube surfa= ce is maintained at 130=C2=B0C (T_w), fi= nd the rate of heat transfer per unit length of the tube.

Film temperature,

v =3D 21.09 =C3=97 10^=E2=88=926 m²/s; Pr =3D 0.692; K =3D 0.03047 W/mK

Reynolds number,

Grashof number,

Therefore,

Hence,

Therefore the required c= orrelation is given by Equation (4.83)=

Here,

The dynamic viscosity at temperatur= e T_=CE=B1,

=CE=BC_b =3D 18.63 =C3=97 10^=E2=88=926 = Ns/m²

The dynamic viscosity at tube wall = temperature T_w,

=CE=BC_s =3D 23.29 =C3=97 10^=E2=88=926 Ns/m²

The Graetz number,

Therefore,

Therefore,

Heat transfer rate,

4.15. Calculate th= e rate of convection heat loss per unit length from a horizontal electrical= wire of diameter .002 mm and maintained at 124=C2=B0C. The temperature of = the room air is 30=C2=B0C. If the wire is kept in a carbon dioxide atmosphe= re at 30=C2=B0C, find the convection heat transfer.

F= or a film temperature of 77=C2=B0C, the Rayleigh number is

For a film temperature o= f 77=C2=B0C, the properties of carbon dioxide are such that

Ra_D<= /em> =3D 168.15

Nu_D<= /em> =3D 1.995

4.16. Find the ele= ctrical power required to maintain the vertical heater surface at 120=C2=B0= C (T_s in air at 30=C2=B0C (T_a). The plate is 20 cm high (L) and 15 cm wide (W). If the plate is = 500 cm high (L¹) then find the power required. A= ssume that the heat transfer coefficient for radiation, <= em>_r is 9 W/m²K.

20-cm-high plate
For film temperature of 75=C2= =B0C,

Gr_L = =3D 4.68 =C3=9710⁷; Pr =3D 0.71; <= em>GrPr =3D 3.32 =C3=97 10¹¹

As the Grashof number is less= than 10⁹, the flow is laminar. The average Nuss= elt number is 35.9.

500-cm-high plate
For film temperature 75=C2=B0= C, Ra =3D 3.78 =C3=97 10¹¹. The flow is= turbulent. The average Nusselt number is 950 and _c =3D 6.1 W/m^2<=
/sup>K

Q =3D (2 =C3=97 4.5 =C3=97= 0.15)(6.1 + 9.0) =C3=97 90 =3D 1834 W.

4.17. Find the rat= e of convection heat loss from the top and bottom sides of a flat horizonta= l square plate of area 2 m², heated to = 224=C2=B0C (T_w) and surrounded by air at 30=C2= =B0C (T_a).

C= haracteristic length (L_c) of the square= plate is L²/4L =3D 0.25 m. Fo= r properties of air at film temperature of 127 =C2=B0 C,

Using Equation (4.52), Nusselt number for heat transfer fr= om the bottom of the plate (hot surface facing downwards) is obtained.

The Nusselt number for h= eat transfer from the top surface (hot surface facing upwards) is determine= d using Equation (4.50) as follows= :

The total convection hea= t loss,

      = ;      Q =3D 2 =C3=97(3.18 + 7.53) = =C3=97 194

=3D 4155 W.

4.18. A closed pan= with water to a depth of 4 cm (L) is placed on a stove and the bottom of t= he pan is maintained at 105=C2=B0C (T_w). If the = water is initially at room temperature, 15=C2=B0C, (Ta) what will be the rate of heat transfer from the b= urner to the water? Assume the pan to be circular and its diameter to be 20= cm (d).

Properties of water at 6= 0=C2=B0C gives =CE=B3 =3D 0.478 =C3=97 10^=E2=80=926; Pr =3D 3.02

4.19. A steel shaft of diameter 15 cm is heated to 400=C2=B0C and then= allowed to cool in air at 25=C2=B0C while rotating about its own (horizont= al) axis at 5 revolutions per minute. Find the rate of convection heat tran= sfer from one metre length of the shaft when it has cooled to 95=C2=B0C.

The rotation speed of th= e shaft is

For properties of air at= 60=C2=B0C,

For a rotating horizonta= l cylinder,

4.20. Water enters= a tube at 20=C2=B0C (T_i.) and is heated to 30= =C2=B0C (T_o). The inner diameter of the tube is = 0.03 m (D) and the mass flow rate of water (m) is 0.01 kg/s. The tube is wr= apped with an insulated electric heating element which produces a uniform f= lux of 10, 000 W/m² (q) on the surface.= Find (i) the Reynolds number (ii) the heat transfer coefficient and (iii) = length of the pipe needed.

Properties of water at 2= 5=C2=B0C are

=CF=81 =3D 997 kg/m³

c_p =3D 4180 J/kgK

K =3D 0.608 W/mK

=CE=BC =3D 910 =C3=97 10=E2=88=926 N s/m²

For uniform heat flux,

The length = of pipe needed for a temperature rise of 30=C2=B0C is obtained from the hea= t balance

4.21. Engine oil f= lows through a copper tube of inner diameter 2 cm and thickness 0.03 cm at = the rate of 0.06 kg/s. The oil enters at 30=C2=B0C and is heated to 50=C2= =B0C by steam condensing on the outside. Find the heat transfer coefficient= assuming that the pressure of the steam is 1 bar.

Properties of oil at 40= =C2=B0C, are as follows:

The flow is laminar and =
=20

4.22. Find the ave= rage heat transfer coefficient for flow of n-butyl alcohol at a bulk temper= ature of 293 K through a 0.2 m =C3=97 0.= 2 m square duct. The length of the duct is 6 m, with a wall temper= ature of 305 K. The average velocity of the fluid is 0.05 m/s.

Properties of n= -butyl alcohol at 293 K are as follows:

The flow is laminar

The average heat transfe= r coefficient,

4.23. A device is = cooled by water flowing through capillary holes drilled in its casing. The = temperature of the device casing is 355 K. The capillary holes are 0.5 m lo= ng and 3 =C3=97 10^=
=E2=88=923 m in diameter. If water enters at a temperature of 333 K a= nd flows at a velocity of 0.3 m/s, find the outlet temperature of = the water.

Properties of water at 3= 33 K are as follows:

The flow is laminar

At 355 K, =CE=BCs =3D 3.56 =C3=9710^=E2=
=88=924 Ns/m²

T= he mass flow rate is

Therefore,

T_{2 =3D 345 K.}

4.24. A printed ci= rcuit board is to be cooled by forced convection cooling with air at 21=C2= =B0C flowing at a velocity of 2 m/s. The board is 30 mm square and 15 mm hi= gh and spacing between the boards is 20 mm. Find the heat transfer coeffici= ents for the second integrated circuit along the flow path.

At 21=C2=B0C, the proper= ties of air are v =3D 15.7 =C3=97 10^=E2=88=926m²/s and K =3D = 0.0251W/mK.

4.25. Find the rat= e of convective heat loss from a horizontal wire with a surface area of 0.0= 02 m², held at 402 K in quiescent air a= t 302 K.

FromTable 4.1, Nu =3D 1.02(Ra)^0.151

The rate of heat loss pe= r unit length,

      = ;          q =3D= 35.56 =C3=97 100 =C3=97 0.002

=3D 7 W.

4.26. A pipe carrying steam has 0.2 m of outer diameter (D) with surfa= ce temperature (T_s) of 210=C2=B0C passes through= a room. The room walls and air are kept at 30=C2=B0C (T_a). Determine the heat loss from the pipe if emissivity of the pipe= (=CE=B5_s) is 0.75.

Air properties at (T= _f) 120=C2=B0C are

k =3D 33.38 =C3= =97 10^=E2=88=923 W/mK

Pr =3D 0.68

=CE=B3 =3D 25.4= 5 =C3=97 10^=E2=88=926 m^2/s

Grashof number

Heat loss by convection = per unit length,

Heat loss by radiation p= er unit length,

Total heat loss per unit= length (Q) =3D Q_c + Q_r=3D 2026.12 W/m.

REVIEW QUESTIONS

Explain the con= cept of natural convection heat transfer mechanism.

Sketch the variation of local velocity= and temperature variations in natural convection heat transfer boundary.
Derive an expression for the natural c= onvection heat transfer coefficient based on the integral method.

State the expression for average Nusse= lt's number for free convection laminar flow in a vertical plate.

How will you define the characteristic= length of a horizontal plate?

Sketch the coordinate system for free = convection in inclined plates.

Explain the method of calculation of n= atural convection heat transfer coefficient for a rotating shaft.

Discuss the free convection heat trans= fer in an enclosed space.

Discuss the concept of mixed convectio= n.

Define the dimensionless numbers u= sed in free convection heat transfer.

OBJECTIVE QUESTIONS

If there is a g= ravitational field acting on a fluid, then at any local region of the fluid= the gravitational force per unit volume is the product of acceleration due= to gravity and

density

viscosity

velocity

specific volume

The Reynolds number is the ratio of in= ertial forces to

gravitational forces

viscous forces

kinetic forces

potential forces

The Grashof number is the ratio of buo= yancy forces to

gravitational forces

potential forces

viscous forces

kinetic forces

In free convection, the product of Gra= shof and Prandtl numbers appears often and it is called the

Nusselt number

Reynolds number

Kirchoff number

Rayleigh number

In a free convection heat transfer in = a horizontal pipe of diameter 5 cm, the Nusselt number is 10 and thermal co= nductivity is 0.029 W/mK. The heat transfer coefficient in W/m² K is equal to.

5.8

8.5

5

8

If the Grashof and Prandtl numbers are= 2 =C3=97 10¹¹ and 0.7, respectively, then the = Rayleigh number is

2.7 =C3=97 10¹¹

1.4 =C3=97 10^11<=
/sup>

2.86 =C3=97 10^11<=
/sup>

0.35 =C3=97 10¹¹

In a vertical metal surface of height = 3.5 cm, the free convection based Nusselt number is 600 and thermal conduct= ivity is 0.03 W/m K, then the heat transfer coefficient in W/m²K is

5.1

1.5

5

1

The film temperature used to evaluate = the physical properties in free convection heat transfer analysis is the va= lue of surface and ambient temperatures

total

product

sum

average

The Nusselt number in free convection = heat transfer is determined from

Grashof and Prandtl numbReynolds = and Grashof numbers

Reynolds and Prandtl number

all of the above

Free convection arises from the differ= ences between the fluid in contact with a surface and the surrounding fluid= .

viscosity

mass

density

volume

Answers:

(a)

(b)

(c)

(d)

(a)

(b)

(a)

(d)

(a)

(c)

EXERCISE PROBLEMS

Find the free c= onvection heat transfer coefficient and heat loss per metre length of a hor= izontal pipe of diameter 30.5 cm which is maintained at a temperature of 25= 0=C2=B0C in a room where the ambient air is at 20=C2=B0C. (Ans. 7 W/m² K, 1.48 kW)

Find the heat gain per hour by a 2-m l= ong copper tube that is 1 cm in diameter carrying F₁₂ at =E2=88=9230=C2=B0C when exposed to still air at 40=C2=B0C= . (Ans. 53 W)

A plate of dimensions 20 cm =C3=97 20= cm is kept vertical and maintained at 100=C2=B0C. Find the average heat tr= ansfer coefficient and the heat lost by the plate per hour if the plate is = exposed to atmospheric air at 20=C2=B0C. (Ans. 6 W/m² K, 20 W)

A pipe of diameter 10 cm is maintained= at 170=C2=B0C and exposed to air at 30=C2=B0C. The length of the pipe is 2= m and it is kept horizontal. Find out the heat lost by the pipe per hour. = (Ans. 705 W)

A plate of size 0.6 m =C3=97 0.2 m is= exposed to atmospheric air at 305 K. If the plate is maintained at 385 K, = estimate the heat loss from both sides of the plate assuming equivalent len= gth as 0.2 m. (Ans. 120 W)

A pipe with radius 25 mm is used to ca= rry hot water. It is placed vertical in ambient air at 290 K. If the outer = surface of the pipe is 340 K, find the heat loss from the pipe per hour. Th= e length of the pipe is 1 m. (Ans. 21 W)

A duct of cross-section 30 cm =C3=97 = 20 cm carries cold air. If the temperature of the outer surface of the duct= is 5=C2=B0C and surrounding air temperature is 25=C2=B0C, calculate the he= at gain by the duct. The duct is in vertical position and its length is 1 m= . (Ans. 187 W)

A duct of 30 cm =C3=97 30 cm carries = air and its outside surface temperature is maintained at 15=C2=B0C. It is e= xposed to room air at 25=C2=B0C. Find the heat transfer coefficient and hea= t gained by the duct per metre length assuming the duct runs horizontal. (<= strong>Ans. 4 W/m² K, 48 W)

A plate 50 cm =C3=97 50 cm at 100=C2= =B0C is exposed to air at 20=C2=B0C. Calculate the heat loss from both surf= aces of the plate if (i) the plate is kept in vertical plane and (ii) if th= e plate is kept in horizontal plane. Find out the percentage change of heat= loss if the plate were kept vertical instead of horizontal. (Ans.<= /strong> 249 W, 207 W, 17%)

Find the free convection heat transfer= coefficient and heat loss from 30 cm diameter sphere maintained at 100=C2= =B0C and exposed to atmospheric air at 20=C2=B0C. (Ans. 5.= 4 W/m² K, 156 W

A plate 25 cm in height and 100 cm wid= e, whose temperature is 100=C2=B0C is exposed to the air at 20=C2=B0C. Dete= rmine

Boundary layer thickness at 10 cm= from leading edge.

Local heat transfer coefficient at 10 = cm from leading edge.

Average heat transfer coefficient for = the whole surface of the plate

Heat flow from both sides of the p= late. (Ans. 0.011 m, 5.5 W/m² K= , 5.75 W/m² K, 230 W)

A horizontal pipe of 5 cm in diameter = carries saturated steam at 10 bar. It is lagged with 85% magnesia. The thic= kness of the insulation is 2 cm. Calculate the heat transfer coefficient an= d heat lost from a length of 30 m of the pipe. Temperature of the insulated= surface is 80=C2=B0C and that of ambient air is 20=C2=B0C. (Ans. 6.8 W/m² K, 3.5 kW)

A hot plate of 20 cm =C3=97 30 cm is = dipped in water. The temperature of the plate is maintained at 120=C2=B0C a= nd temperature of water is 20=C2=B0C. Determine the heat transfer coefficie= nt and heat loss from the plate per hour if the 20 cm side of the plate is = in vertical plane. (Ans. 1.672 kW, 20 kW)

Water at 20=C2=B0C is heated by a stea= m pipe of length 50 cm and diameter 5 cm. If the surface temperature of the= pipe is maintained at 80=C2=B0C, determine the heat lost from the pipe per= hour and the heat transfer coefficient. (Ans. 4.6 kW, 0.9= 8 kW/m² K)

A 4-cm outer diameter pipe carrying st= eam at 1.5 bar is located where it is covered by water after a heavy rain b= ut is exposed to air under normal conditions. Find the rate of heat transfe= r to air and water assuming both fluids are at 10=C2=B0C. (Ans. 120 W, 16 kW)

A pipe of diameter 5 cm carrying steam= is covered with insulating material of thickness 2.5 cm. The surface tempe= rature is 80=C2=B0C and emissivity of the insulating material surface is 0.= 93. Calculate the total heat loss from 10 m length of the pipe by natural c= onvection and radiation. The temperature of the air surrounding the pipe is= 20=C2=B0C. Determine the overall heat transfer coefficient and heat transfer coefficient of radiation. (Ans. 2.6 kW, 14 W/m2 K, 7.3 W/m² K)

A plate of area 20 cm² and thickness 3 mm is maintained at a uniform temperature of 50= 0=C2=B0C and exposed to atmospheric air at 30=C2=B0C. If the plate is in a = vertical position, find out the approximate time required for the plate to = cool to 200=C2=B0C. (Ans. 690 s)

What is the minimum spacing which will= prevent interference of the free convection boundary layers of two vertica= l flat plates at 60=C2=B0C that are placed in a tank of water at 20=C2=B0C,= if the plates are 10 cm high. (Ans. 0.41 cm)

A plate of size 20 cm =C3=97 20 cm ma= intained at 65=C2=B0C is kept vertically in air at 15=C2=B0C. Find the free= convection heat transfer from this plate and that which would result from = forcing air over the plate at a velocity equal to the maximum velocity whic= h occurs in free convection boundary layer. (Ans. 14.3 W, = 23.8 W)

A shaft is heated to 695 K and the= n allowed to cool in air at 305 K while rotating at 5 rpm. Find the rate of= convective heat transfer from the shaft when its temperature comes to 385 = K. Assume that the shaft diameter is 0.3 m and its length is 1 m. (= Ans. 0.6 kW).

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