3. Transient Heat Conduction - = Heat and Mass Transfer, 2nd Edition

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3 Transient Heat Conducti= on

= 3.1 INTRODUCTION

In transient heat conduction,= the thermal boundary conditions are time dependent. The mathematical model= for a transient heat conduction problem includes the temperature distribut= ion within the body and the thermal boundary conditions at the surface of t= he body.

= 3.2 LUMPED PARAMETER ANALYSIS (INFINITE THERMAL CONDUCTIVITY)

If a body at uniform temperat= ure is suddenly exposed to a different temperature, then the temperature ch= ange in the interior of the body depends on the resistance to heat transfer= internally and at its surface. If the resistance at the surface is much gr= eater than at internal, then the internal body temperature will be almost u= niform. In transient heat conduction problems, this situation can be obtain= ed neglecting the internal thermal resistance of a body. The analysis of he= at transfer under such an assumption is called lumped system analysis.

The application of the first= law of thermodynamics to a body shown in Fig. 3.1, whose internal thermal resistance is neglected, gives

Figure 3.1 Nom= enclature for Single-Lump Heat-Capacity Analysis

where = =E2=80=98V=E2=80=99 is the volume (m³),= =E2=80=98=CF=81=E2=80=99 is the density (kg/m³), =E2=80=98C=E2=80=99 is the specific heat (J/kgK), =E2= =80=98T=E2=80=99 is the temperature of the body, =E2=80=98A=E2=80=99 is the surface area (m²), =E2=80=98<= em>t=E2=80=99 is the time (s), and =E2=80=98h=E2=80=99 is the= heat transfer coefficient (W/m²K) at the surfac= e of the body.

Rewriting Equation (3.1) and integrating, we get

where =E2=80=98D=E2= =80=99 is the constant and it can be evaluated using the initial condition = t =3D 0, T =3D T_i as

D =3D ln (T_i =E2=88=92 T_a).<= /div>

Equation (3.2) becomes

The parameters can be non-di= mensionalized as follows:

Temperature, =CE=B8= =3D (T =E2=88=92 T_a)= ; =CE=B8_i =3D (T_i =E2=88=92 T_a); = characteristic length, L_c =3D V/A;

Thermal diffusivity, =CE= =B1 =3D k/=CF=81C, where k is the thermal conductivi= ty of the solid body.

The equation can be written = as

where

3.2.1 = Significance of Biot and Fourier Numbers

The Biot number is the ratio = of internal conduction resistance to external convection resistance. The lu= mped parameter analysis is valid for Bi < 0.1 with an error of = less than 5%. This shows that the lumped parameter analysis should be used = when convective heat exchange controls the heat transfer with larger convec= tive resistance.

KEY = CONCEPTS

Physical properties of fluids generally vary with the temperat= ure.

But these variations are assumed negligible and the following prope= rties are considered to be constants

Thermal conductivity K (W/mK) of materials

Specific heat c (J/kgK)

Density =CF=81 (kg/m³)

iscosity =CE=BC (kg/ms).

At ambient temperature and pressur= e, values of these properties for air and water are

The Fourier number is also a= non-dimensional one. It indicates the degree of heating or cooling effect = that penetrates into the solid for a given time. Higher Fourier number indi= cates the more penetration of heat into a body. For small values of =CE= =B1/L²_c, more ti= me is required for temperature change in the solid.

The instantaneous heat trans= fer rate relationship in lumped parameter analysis can be derived from Equations (3.2) and (3.4) as

From Equation (3.3), we get

The total heat loss or gain = to the solid body up to a time =E2=80=98t=E2=80=99 is

= 3.3 SEMI-INFINITE SOLIDS

A solid is said to be semi-in= finite solid if it is extended to infinity in the y and z= directions as shown in Fig. 3.2. = In a semi-infinite solid, at any instant of time, there is always a point w= here the effect of heat transfer at one of its boundaries is not felt at al= l. The temperature at the point remains unchanged. Consider a semi-infinite= solid as shown in Fig. 3.2. The e= ntire body is maintained at uniform temperature (T_=
i) when t =3D 0. Then the surface temperature is sudden= ly changed and maintained at that temperature (T₀).

Figure 3.2 Nom= enclature for Transient Heat Flow in a Semi-Infinite Solid

The differential equation fo= r the temperature distribution T(x, t) which is = a function of distance and time is given by

The boundary and initial con= ditions are

T(x, 0) = =3D T_i and

T(0, t) =3D T₀for t > 0.

The solution of this may be = obtained using Laplace transform technique as

where erf is known as the Gaussian error function and is defined as

The value of erf is taken from the Tables for a given erf

The heat flow at any positio= n (x) may be obtained from

Also <= img id=3D"img0222" class=3D"Eq1" src=3D"https://learning.oreilly.com/librar= y/view/heat-and-mass/9789332515680/images/pg79_img01.png" alt=3D"image" wid= th=3D"108" height=3D"42" data-mfp-src=3D"/library/view/heat-and-mass/978933= 2515680/images/pg79_img01.png"> [=E2=88=B5 partial differentiation of Equation (3.8)]

Then Q at x =3D 0

3.3.1 = Constant Heat Flux on Semi-infinite Solid

The semi-infinite solid is ex= posed to sudden application of a specified heat flux q₀as, for example, exposing the surface to radiation.

The initial and boundary con= ditions are

=20

The temperature distribution= for this case is

3.3.2 = convection Boundary Conditions

In most of the practical situ= ations, the solid surface is exposed to a fluid at a different temperature = through a uniform and constant heat transfer coefficient (h).

The initial and boundary con= ditions are

=20

The temperature distribution= for this case is

Figures 3.3 and 3.4 s= hows the temperature distributions.

Figure 3.3 Tem= perature Distribution in the Semi-Infinite Solid

<= /p>

Figure 3.4 Tem= perature Distribution in the Semi-Infinite Solid with Convection Boundary C= ondition

3.3.3 = Use of Transient =E2=80=93 Temperature Charts

In lumped system analysis, th= ere is no temperature gradient within the solid, i.e., the temperature is c= onstant. But in many situations, temperature changes in the solid with resp= ect to the positions from the surface. In heat conduction problems, the tem= perature distribution is a function of time and position and it is a compli= cated one. Simple problems such as one-dimensional, time-dependent heat con= duction, as in the following cases with no internal energy generation, can = be solved by the method of separation of variables.

An infinite plate of width =E2=80=982L=E2=80=99 =E2= =80=93 plate width is small compared to the other dimensions.

An infinite cyli= nder of radius =E2=80=98R=E2=80=99 =E2=80=93 diameter is small com= pared to length.

A sphere of radius =E2=80=98R<= /em>=E2=80=99.

The boundary and initial con= ditions for all three geometries are similar. Results of the analysis for t= hese geometries have been presented in graphical form by Heisler charts.
3.3.4 = Slab

Consider a slab wall of thick= ness =E2=80=982L=E2=80=99 as shown in Fig. 3.5. Initially, the slab is at a uniform temperature = T_i. Then both boundary surfaces of the slab= are subjected to convection with heat transfer coefficient h into= ambient temperature T_aand are maintai= ned when t > 0. Since the wall is symmetric about the center pl= ane, we consider the conduction for only half the region. The basic Equation (3.7) for transient conduction pr= oblem is given as

Figure 3.5 Inf= inite Plate of Thickness 2L

The boundary conditions are<= /p>

=20

The above equation can be wr= itten in non-dimensional from as

<= /p>

=CE=B8 =3D 1 in 0 =E2=89= =A4 x =E2=89=A4 1, for Fo > 1

where

The results for the dimensio= nless temperatures at the center plane for different values of Bi = and Fo are given in the chart as shown in Figs. 3.8=E2=88=923.= 19. The chart gives temperature distribution with respect to 1/Bi for different x/L values. The chart gives the ratio =CE=B8= /=CE=B8₀ against the parameter Bi²Fo for different values of Bi.= Q is the total heat loss by the body up to time =E2=80=98t=E2=80=99 and =CE=B8₀ is the initial = internal energy of the body relative to ambient temperature (T= _s), i.e., Q_{0<=
/sub> =3D =CF=81C_pV(T_i=E2=88=92 T_a).}

<= /p>

Figure 3.8 Mid= plane Temperature for an Infinite Plate of Thickness 2L: (a) Full = Scale

3.3.5 = Long Cylinder

Consider one-dimensional tran= sient heat conduction in a long cylinder of radius r₀as shown in Fig. 3.6. Similar to the slab, the basic equation in the dimensionless form is

Figure 3.6 Inf= inite Cylinder of Radius r₀

The boundary conditions are<= /p>

= =CE=B8 =3D1 in 0 < R < 1 for F₀ =3D 0;

where

The equation is solved, and = the results for the center temperature are given in Figs. 3.8=E2=88=923= .19 as a function of Fo and 1/Bi. The figures also sh= ows the ratio of temperature at any radius to center temperature (=CE= =B8/=CE=B8₀) as a function of r/r<= sub class=3D"EBsub">0 and 1/Bi and the dimensionless heat tr= ansfer Q/Q₀ as a function of Fo and Bi.

3.3.6 = Sphere

A sphere of radius r= ₀ is considered as shown in Fig. 3.7. Similar to the cylinder problem, the dimensi= onless parameters are defined in sphere problems. The basic equation in the= dimensionless form is

Figure 3.7 Sph= ere of Radius r₀

The boundary conditions are = the same as a cylinder problem. The temperature distribution and heat trans= fer are given the Heister charts, as shown in Figs. 3.8=E2=88=923.19.

<= /p>

Figure 3.9 Exp= anded Scale for 0 < Fo < 4, from Reference 2

<= /p>

Figure 3.10 Ax= is Temperature for an Infinite Cylinder of Radius r₀: (a) Full Scale

<= /p>

Figure 3.11 Ex= panded Scale for 0 < Fo < 4, from Reference 2

<= /p>

Figure 3.12 Ex= panded Scale for 0 < Fo < 3, from Reference 2

<= /p>

Figure 3.13 Te= mperature as a Function of Center Temperature in an Infinite Plate of Thick= ness 2L, from Reference 2

Figure 3.14 Te= mperature as a Function of Axis Temperature in an Infinite Cylinder of Radi= us r_0=E2=80=B2 from Reference 2

<= /p>

Figure 3.15 Te= mperature as a Function of Center Temperature for a Sphere of Radius r<= /em>_0=E2=80=B2 from Reference 2

Figure 3.16 Ce= nter Temperature for Plates, Cylinders and Spheres, for Small Values of h, According to Reference 2 (s =3D L for plate; s = =3D r₀ for Cylinder and Sphere)

<= /p>

Figure 3.17 Di= mensionless Heat Loss Q/Q₀ of an Infini= te Plane of Thickness 2L with Time, from Reference 6

Figure 3.18 Di= mensionless Heat Loss Q/Q₀ of = an Infinite Cylinder of Radius r₀with = Time, from Reference 6

<= /p>

Figure 3.19 Di= mensionless Heat Loss Q/Q₀of = a Sphere of Radius r₀with Time, from R= eference 6

= 3.4 MULTIDIMENSIONAL SYSTEMS

When the temperature gradient= s change in two different directions within a solid, then the problem is tw= o-dimensional transient heat conduction in a solid. If there is no heat gen= eration in the medium, the solution for a two-dimensional problem can be ob= tained by combining the one-dimensional solutions from the charts. The basi= s for obtaining two- and three-dimensional solutions from one-dimensional c= harts in the manner in which partial differential equations can be separate= d into the product of two or three ordinary differential equations. Such an= approach is known as product solution method.

The product solution method = can be explained by considering a short cylinder as shown in Fig. 3.20. The cylinder is initially at a unifo= rm temperature T_i. Then the cylinder su= rface is exposed to convection heat transfer. The length and diameter are c= omparable in finite cylinder. Suppose we wish to determine the transient te= mperature at a point P, the temperature depends on the radial and = the x direction heat transfers.

Figure 3.20 Mu= ltidimensinal Systems

The appropriate heat transfe= r equation is

The solution may be expresse= d in the following form:

Simila= rly the results for other multidimensional geometries shown in Fig. 3.20 can be written. The one-dimensional= solutions can be written as

for semi-infinite solid    =   (3.20)

for plane wall     &nb= sp;(3.21)

<= /p>
for infinite solid.    &nbs= p; (3.22)

Here x is measured = from the surface for semi-infinite solids, where as for the plane wall it i= s measured from the mid plane.

The heat transfer solution f= or a multidimensional body formed by the intersection of two bodies is

and for intersection of three= bodies

= 3.5 PERIODIC HEAT FLOW

Solids with high value of the= rmal conductivity when exposed to fluid with periodic temperature variation= have the following temperature distributions:

3.5.1 = Semi-infinite Solid-surface Temperature Varied Periodically

3.5.1.1 Surface temperature vari= ation

T_f =E2=80=93 fluid temperature at any time

T_a =E2=80=93 Amplitude of temperature variation

t =E2=80=93 Time

=CE=B4 =E2=80=93 Angle of lag in the temperature variatio= n of the solid with reference to fluid temperature variation

=CF=81 =E2=80=93 Solid density

C =E2=80=93 Solid specific heat

V =E2=80=93 Solid volume

A =E2=80=93 Solid surface area

T_s =E2=80=93 Temperature of the solid

3.5.1.2 Depth of penetration

<= /p>

=E2=88=92 Amplitude ratio

T =E2=88=92 Surface temperature

=CE=B1 =E2=88=92 Thermal diffusivity of solid

=CE=B4 =E2=88=92 Angle of lag in the temperature at depth= x, with reference to surface temperature.

x_o =E2=88=92 Depth upto which the temperature variati= on penetrates

=E2=88=92 Heat flow rate at time W/m² or kcal/m²hr

=CE=B4_q =E2=88=92 Time lag of heat flow with reference to = the surface temperature variation

=E2=88=92 Total heat flow during half period

T_s =E2=88=92 Temperature at depth x from sur= face

k =E2=88=92 Solid thermal conductivity

h =E2=88=92 Convective heat transfer coefficient.

3.5.1.3 Semi-infinite solid surf= ace exposed to fluid with periodic temperature variation

where A =3D (= =CF=80k² / =CE=B1=CF=84₀h²).

= 3.6 FREEZING/MELTING

3.6.1 = Flat Liquid/Solid Surface

Liquid or solid at melting te= mperature T_fr suddenly exposed to fluid= at T_=E2=88=9E:

x*[2=CF=84^* + 1]^0.5 =E2=88=92 1      (3.37)

=CF=84^* =3D x^* +0.5x^*²   &nb= sp;  (3.38)

where

(melting)

(freezing)

(melting)

<= /p>
(greezing)

x =E2=88=92 thickness melted/frozen from surface upt= o time t

x* =E2=88=92 dimensionless distance parameter

t* =E2=88=92 dimensionless time parameter

k_s, k₁ =E2=88=92 thermal conductivity of solid, liquid

=CF=81_s= =CF=81₁ =E2=88=92 density of solid, liquid

h_sf<= /td> =E2=88=92 enthalpy of melting/freezing

h_=E2=88=9E= =E2=88=92 convective heat transfer coefficient at = surface

T_=E2=88=9E= =E2=88=92 free stream temperature of fluid flowing= over surface

T_fr<= /td> =E2=88=92 melting/freezing temperature

t =E2=88=92 time

3.6.2= Freezing of Liquid at a Temperature Higher than Freezing Point

where

h₁ =E2=88=92 convection coefficient at the liquid sol= id interface

^T₁ =E2=88=92 Temperature of liquid

x =E2=88=92 thickness from surface frozen upto time = t.

r =E2=88=92 radius at the (freezing) solid / liquid = interface at time t

r ₀ =E2=88=92 tube outside radius

= 3.7 CONCLUSIONS

In solid to fluid heat transf= er process, if the internal resistance is assumed negligible in comparison = with its surface resistance, then it is called as Newtonian heating or cool= ing process. If the heat flow and temperature distribution at any location = of the system vary continuously with time, then it is known as transient st= ate. Examples are cooling of IC engines, heating and cooling of metal bille= ts, and brick burning.

When the surface temperature= of a solid body is suddenly changed, the temperature within the body begin= s to change with the position and time. If the variation of temperature wit= h position is negligible during the transients, the analysis of heat transf= er with time only is known as lumped system analysis.

The Biot number is defined a= s the ratio of the internal resistance of a body to the external resistance= for heat transfer. Lumped parameter analysis is applicable only when the B= iot number is less than 0.1. Time constant is the time required for a tempe= rature-measuring system to reach 36.8% of the initial temperature differenc= e. The Fourier number is a non-dimensional parameter used in transient heat= conduction analysis. It is defined as the ratio of the rate of heat conduc= ted to the rate of storage of thermal energy in a solid.

A solid which extends itself= infinitely in all directions of space is known as infinite solids. If an i= nfinite solid is split in to two parts, then each half part is known as sem= i-infinite solids. The Heisler chart is used to determine the transient tem= perature at different locations of large plane wall, long cylinder, and sph= ere.

The mathematical expressions= of the thermal condition at the boundaries of the system are called bounda= ry conditions. Two boundary conditions are required for one-dimensional pro= blem, four for two-dimensional problems, and six for three-dimensional prob= lems. Boundary conditions mostly used are the specified temperature, specif= ied heat flux, convection, and radiation boundary condition.

If the condition of paramete= r is specified at time equal to zero, then the condition is called initial = condition. In heat transfer, only one initial condition is needed irrespect= ive of the dimension since the conduction equation is first order in time.<= /p>
= EXAMPLE PROBLEMS

3.1. A slab made u= p of carbon steel is initially at a temperature of 500=C2=B0C (T_i) and it is exposed to air at 30=C2=B0C (T_a). The thickness of the slab is 1 cm (L). What will be= the temperature of the slab after one minute (t)? Assume that the convecti= ve heat transfer coefficient (h) on the surface of the slab is 40 W/m²K.

The properties of carbon= steel are given as

=CF=81 =3D 7833kg/m³, C =3D 0.465 kJ/kg K,

K =3D 38.5W/mK and =CE= =B1 =3D 1.474 =C3=97 10^=E2=88=92⁵ m²/s

This problem can be solv= ed by the lumped system analysis if Bi < 0.1.

Characteristic length,

=20

Biot number,

As the Biot number is le= ss than 0.1, lumped parameter analysis can be used.

Fourier number,

The temperature distribu= tion is given by

Temperature of the slab = after one minute,

=20

= 3.2. An aluminium rod which is initially at a temperature of 200=C2=B0C= (T_i ) is exposed to air at 30= =C2=B0C (T_a). The diameter of a pipe is 10 cm. F= ind the time required for the rod to reach a temperature of 50=C2=B0C (T). = Assume that the surface convective heat transfer coefficient, h is= 50 W/m²K.

From the properties tabl= e, =CF=81 =3D 2707 kg/m³, the average t= hermal conductivity K =3D 210W/mK, =E2=88=9E =3D 8.418 = =C3=97 10^=E2=88=92⁵m<= sup class=3D"EBsup">2/s and C =3D 0.896 kJ/kgK.

Check whether Bi < 0.1 to apply lumped system analysis.

The characteristic lengt= h,

As Bi < 0.1,= lumped parameter analysis can be used.

=20

Time taken for the rod t= o reach a temperature of 50=C2=B0C =3D 106, 863 s =3D 29.68hr.

3.3. A carbon stee= l ball, initially at a temperature of 900=C2=B0C (T_i) is suddenly dropped into water at a temperature of 30=C2=B0= C (T_a ) The radius of the ball is 5 cm (r_o). Assuming that the convection heat transfer coeffici= ent (h) is 210 W/m²K, find the centre p= oint temperature of the sphere after 30 min (t).

The properties of carbon= steel are K =3D 54 W/mK; =CE=B1 =3D 1.474 =C3=97 10 =E2=88=92⁵ m²/s

As Bi < 0.1,= lumped parameter analysis can be used.

<= /p>
=20

Therefore T =3D= 31.78=C2=B0C.

3.4. A sphere of 1= 0 mm diameter (d_o) made of Tungsten steel is ini= tially at a temperature (T_i) of 300=C2=B0C. It i= s suddenly exposed to a stream of air at a temperature of 30=C2=B0C (T_a). The surface convective heat transfer coefficient = (h) is 100 W/m²K Find the time required for the = sphere to reach a temperature of 100=C2=B0C (T). Also calculate the instant= aneous heat transfer rate after one minute from the start of the cooling pr= ocess and the total heat transferred from the sphere in one minute.

The properties of tungst= en steel are =CF=81 =3D 7897kg/m³, = C =3D 0.452kJ/kgK, =CE=B1 =3D 2.026 =C3=97 10^=E2=
=88=925m²/s and K =3D 73 W/mK.

Characteristic length,

Biot number,

As Bi < 0.1,= we can apply lumped system analysis in this problem.

Fourier number,

Temperature distribution

Time taken to reach 100=C2=B0= C, t =3D 81.2 s.

The instantaneous heat transfer rate a= fter 1 min is

=20

The total e= nergy transfer in one minute

3.5. A copper slab= of surface area 0.25 m² (A_=
s) and thickness 10 mm (L) has a uniform temperature of 300=C2=B0C. I= ts temperature is suddenly lowered by convection in ambient temperature of = 40=C2=B0C (T_a) with h =3D 90 W/m²=C2=B0C. Calculate the time required for the slab to reach the = temperature of 110=C2=B0C (T) Take =CF=81 =3D 9000 kg/m³; C =3D 0.38 kj/kg=C2=B0C and K =3D 370 W/m=C2=B0C.

Characteristic length,

As Bi < 0.1.= Therefore, lumped system analysis can be applied.

Therefore,

Therefore, t = =3D 249.2 s.

3.6. A stainless s= teel ball of diameter 2 cm (d_o) (=CF=81 =3D 7865= kg/m³, C =3D 0.46 kj/kg=C2=B0C and K = =3D 61 W/m=C2=B0C is uniformly heated to a temperature T_i =3D 800=C2=B0C. It is to be hardened by suddenly dropping it into= an oil bath at T_a =3D 50=C2=B0C. If the quenchi= ng occurs when the ball reaches 100=C2=B0C (T) and the heat transfer coeffi= cient (h) for the oil and sphere is 300 W/m²=C2= =B0C, how long should the ball be kept in the oil bath?

=

Therefore

As Bi < 0.1,= therefore we can use lumped system analysis.

But

Therefore, t = =3D 107.71s

Time taken for the ball = to reach 100=C2=B0C=3D 107.71 s.

Semi-infinite So= lids

3.7. An oil bath m= aintained at 30=C2=B0C (T_a) is used to cool a la= rge carbon steel flat slab at a uniform temperature of 800=C2=B0C (T_i). Find the time needed (t) to cool the slab to 500=C2= =B0C at a depth of 2 cm (x) from the surface.

The thermal diffusivity = of carbon steel, =CE=B1 =3D 1.474 =C3=97 10^{=E2=88=925<=
/sup> m²/s. As the thickness of the slab is not =
given, to find the temperature at a depth of 2 cm from the surface, the sem=
i-infinite solid solution should be used.}

From error function tabl= e, for the value of erf the value of

Therefore,

=

3.8. Soil at an un= iform temperature of 20=C2=B0C (T_i) has its surf= ace subjected to an uniform temperature of =E2=88=9210=C2=B0C (T_o) continually for two months. A water pipe is to be burie= d in the soil at a depth from the surface to prevent freezing of water in w= inter. What is the minimum depth needed to prevent freezing in the pipe? As= sume =3D 0.2 =C3=97 10^=E2=88=926 m²/s for the soil, as pipe temperature should be more = than 0=C2=B0C.

To find the depth from t= he surface of the soil, semi-infinite solid solution should be used.

Here (T =E2=88=92 T<= sub class=3D"EBsub">o)/(T_{i =E2=88=92 T_o}) =3D 0 + 10/20 + 10 =3D 0.3= 3

But (T =E2=88=92 To)/(T_i=E2=88=92 T_o) =3D erf

From the error function = table, for the value of erf , the value of= x/2 =3D 0.3, that is, 0.491 = =C3=97 x =3D 0.3.

Therefore, x = =3D 0.61m.

3.9. A thick coppe= r slab (=CE=B1=3D1.1 =C3=97 10^=E2=88=924m²/s and K=3D380 W/m=C2=B0C) is initially at an uniform = temperature of 10=C2=B0C (T_i). Suddenly the surf= ace is exposed to 100=C2=B0C (T₀). Calculate the= heat flux at the surface at 5 (t₁) and 10 min (= t₂) after the exposure of the surface to the tem= perature of 100=C2=B0C. How long will it take for the metal position at a d= epth of 5 cm (x) from the surface to reach 90=C2=B0C (T)?

Heat flux (when t₁ =3D 5 min),

Similarly heat flux (whe= n t₂ =3D 10min),

Here

That is,

F= rom the error function table, for the value of erf , the value of

Therefore,

=20

3.10. A thick wood= piece is initially at an uniform temperature of T_{i =3D 20=C2=B0C. Suddenly the surface is raised to 120=C2=B0C (T₀). Calculate the temperature}, 2 cm from the sur= face (x) at 2 and 20 min after the exposure. Assume =CE=B1 =3D 0.8 =C3=97 1= 0^=E2=88=927 m²/s and = K =3D 0.15 W/m=C2=B0C.

This problem is one of a= semi-infinite solid, because the thickness of the wood piece is not given.=

t =3D 2 min,

Therefore, erf (from the error function table)

But

Therefore

=20

t =3D 20 min,

Therefore

=20

3.11. A large slab= made up of steel is initially at a temperature of 500=C2=B0C (T_i). It is suddenly exposed to the atmospheric air at 30=C2= =B0C (T_a). Assuming that the convective heat tra= nsfer coefficien(h) is 300 W/m²⁼ K find the centre-line temperature (T_m) o= f the slab after 2 minutes (t). If the slab thickness is 40 mm (L), what will be the temperature at a distance 10 mm (x) from the centre-line= ?

The properties of steel = are K =3D 42.5 W/mK; =CE=B1 =3D 0.043 m²/hr =3D1.2 =C3=97 10^=E2=88=925 m²/s

Slab thickness, L =3D 40 mm

Characteristic length, <= em>L_c =3D L/2 =3D 20 mm

Biot number,

Fourier number,

As the Biot number is gr= eater than 0.1, lumped parameter analysis is not appropriate for this probl= em. Heisler charts for the infinite slab can be used, as 0 < Bi= < 100.

Centre-line temperature

For 1/Bi =3D 7.1 and= Fo =3D 3.6, from Heisler's chart, we get

Therefore, and T_m =3D 349.6=C2=B0C

From position correction factor ch= art, for 1/Bi =3D 7.1 and x/L =3D 10/20 =3D 0.5

Temperature at 10 mm from the= centre-line, T =3D 333.62=C2=B0C.

3.12. A long cylin= drical carbon steel rod is heated to 700=C2=B0C (T_{i) in a furnace and then suddenly quenched in a coolant at 30=C2=B0C (Ta}). The diameter of the rod is 200 mm (r_o). Assuming that the convective heat transfer coefficient= (h) is 200 W/m²K, find the time taken = for the centre-line temperature of the rod to reach 100=C2=B0C (T_m).

The properties of carbon= steel are K =3D 54 W/mK; =CE=B1 =3D 1.474=C3=9710^=E2=88=92⁵ m²/s

A= s the Bi > 0.1, lumped parameter analysis is not valid. But 0 &= lt; Bi < 100, and hence Heisler's chart for an infinite cylinde= r can be used.

Here

From Heisler's chart for= a cylinder, Fo =3D 3.6 and =CE=B8_o= /=CE=B8_i =3D 0.104, we get

Infinite Solid P= roblems=E2=80=94Slab

3.13. An iron plat= e of thickness 5 cm (L) is initially at 250=C2=B0C (T_i=). Suddenly both surfaces are exposed to a fluid at T_a=3D50=C2=B0C with h=3D500 W/m²=C2=B0= C. Calculate the (i) centre temperature of the plate after (t) 120 seconds = of cooling; (ii) temperature at a depth (x) of 1.0 cm from the surface duri= ng this time; (iii) heat transferred to the fluid from one square metre sur= face area of the plate during this time. (Assume that K =3D 60 W/m =C2=B0 C= , C_p =3D 460 J/kg=C2=B0C; =3D 7850 kg/m= ³, =3D 1.6 =C3=97 10^=E2=88=925 m²/s.

Check for Bi &l= t; 0.1 to apply the lumped system analysis.

Since Bi > 0= .1, we have to use Heisler's charts to solve the problem.

From the chart, for 1/Bi =3D<= /em> 4.8 and Fo =3D 3.07, we get

Therefore,

=20

Here 1/Bi =3D 4.8, 2x/L =3D (2.5 =E2=88=92 1)=C3=972/5 =3D 0.6

From the position correction = chart for 1/Bi =3D 4.8 and x/L =3D 0.6, we have<= /p>

Therefore,

=20

Here Bi²Fo =3D 0.137

From the heat transfer ratio = chart for the value of 1/Bi =3D 4.8 and Bi²Fo =3D 0.137, we have Q/Q_o=3D 0.46

But

=20

Heat transfer to the fluid af= ter 120 seconds, Q =3D 0.46Q_o = =3D 0.46=C3=9735.3=C3=9710³ =3D 16.24 kJ/m².

3.14. A solid iron= rod of diameter 10 cm (d_o ), initially= at a temperature of 1000=C2=B0C (T_i ) is sudden= ly exposed to an oil bath at 50=C2=B0C (T_a ) wit= h a surface heat transfer coefficient (h) of 500 W/m^{2<=
/sup> C. Determine the centre-line temperature (t) of the rod 10 min after =
exposure to oil and time taken for the centre-line temperature to reach 200=
=C2=B0C and heat loss per one metre length of rod during this time. (Assume=
K =3D 60 W/m=C2=B0C}, =3D 2.0=C3=9710^=E2=88=925, C=3D460 J/kg K and =3D 7850 kg/m³.)

Since Bi > 0= .1, we have to use Heisler's chart to solve the problem.

In this case,

Centre-lin= e temperature after 10 min (t)

From the Heisler's chart for = the value 1/Bi =3D 2.4 and Fo =3D 4.8

We get

Therefore,

=20

Time taken for the centre-line tempera= ture to reach 200=C2=B0C

From the same chart, for the = value of 1/Bi =3D 2.4 and T_o =E2=88= =92 T_a/T_i = =E2=88=92 T_s =3D 0.16 we have =3D 2.5.

Therefore

Heat loss from the rod

Bi^{2Fo =3D 0.43}

From the dimensionless heat t= ransfer chart, for the value of Bi =3D 0.42 and Bi²Fo =3D 0.43, we get

Here

Therefore, Q =3D 24.= 79=C3=9710⁶ J/m.

3.15. An apple of = radius 5 cm (r_o) is initially at an uniform temp= erature of 30=C2=B0C (T_i). It is placed in a fri= dge in which the air temperature is 5=C2=B0C (T_a= ). Determine the time required for the centre of the apple to reach 10=C2=B0C (T_o). Also determine t= he amount of heat lost from the apple. (Assume that =3D 1.4 =C3=97 10<= sup class=3D"EBsup">=E2=88=92⁷ m²/s, K =3D 0.59 W/m=C2=B0C, h =3D 50= W/m²=C2=B0C, =3D 1000 kg/m³ and C =3D 4.2 kJ/kg K.)

Since Bi > 0= .1, we have to use Heisler's chart to solve the problem

But in chart

From Heisler's chart, fo= r the value of 1/Bi =3D 0.24 and (T_{o =E2=88=92 T_a})/(T_=
i =E2=88=92 T_a) =3D 0.2 we get Fo= =3D =3D 0.5.

Therefore,

Rate of heat transfer

Bi^{2Fo =3D 8.7}

From the dimensionless h= eat transfer chart for the value of Bi =3D 4.2 and Bi2Fo =3D 8.7, we have Q/Q_o =3D 0.8

But

Therefore,

Q =3D 0.8 Q_o =3D 44kJ.

3.16. A potato of = 5 cm radius (r_o), initially at an unifo= rm temperature of 30=C2=B0C (T_i) is suddenly dro= pped into boiling water at 100=C2=B0C. Determine the time required for the = centre temperature of the potato to reach 95=C2=B0C (T_=
o) and the energy transferred to the potato during this time. (Take h= =3D 6000 W/m²=C2=B0C; =3D 1.6 =C3=97 10^=E2=88=927 m²/s; K =3D 0.68 W= /m=C2=B0C; C =3D 4.2 kJ/kg=C2=B0C; =3D 1000 kg/m^3.)

Here also we have to use= Heisler's chart to solve the problem because Bi > 0.1

In charts,

F= rom the Heisler chart, for the value of 1/Bi =3D 0.002 and (T<= sub class=3D"EBsub">o =E2=88=92 T_a)/(= T_i =E2=88=92 T_a) =3D 0.071.

We have

Therefore,

Energy transferred to po= tato, Q =3D Bi²Fo =3D 68,068.<= /p>

Since Bi²Fo is higher than the maximum value in the ch= art,

Therefore,

Multidimensional= Problems

3.17. A short cyli= ndrical bar of diameter d_o =3D 6 cm and height L= =3D 3 cm is initially at an uniform temperature of T_i= =3D 175=C2=B0C. The bar is suddenly subjected to convective cooling = with h =3D 250 W/m²=C2=B0C into ambient air at T= _a =3D 25=C2=B0C. Calculate the centre temperatur= e T_o of the cylinder at t =3D 1 min after the st= art of the cooling. (Assume K =3D 200 W/m=C2=B0C and =3D 8.4 =C3=97 10^=E2=88=925 m²/s.)=

The solution of this pro= blem is a product of the solution for an infinite cylinder of diameter = d_o =3D 6 cm and the solution for an infinit= e slab of thickness L =3D 3 cm.

Temperatur= e calculation for a cylinder

From the Heisler chart for in= finite cylinders, for the value of and = Fo =3D 5.6, we get

Temperature calculation for an inf= inite slab

From the Heisler chart for a = slab, for the values 1/Bi =3D 53.3 and Fo =3D 22.4, we ge= t

Then the dimensionless centre= temperature for a short cylinder is given by the relation

=20

Therefore,

=20

3.18. In the above= problem, calculate the temperature at a radial position of 1.5 cm= and at a distance of 0.5 cm from one end of the cylinder and calculate the= heat loss from the cylinder. (Take =3D 2707 kg/m^{3 and C =3D 896 J/kg K.)}

Temperature calculation

For an infinite cylinder, for= the same values of above problem,

=

From the position correction = chart, for the value of r/r_o =3D 5 and 1/Bi =3D 26.7 we have

(T =E2=88=92 T_a)/(T_o =E2=88=92 Ta) =3D 0.95

=20

For an infinite plate,

From the position correction = chart, for the value of x/L =3D 0.67 and 1/Bi =3D 53.3, we get (T =E2=88=92 T_a)/(T<= sub class=3D"EBsub">o =E2=88=92 T_a) = =3D 0.98.

=20

Then we have

Therefore,

=20

Heat loss calculation

For an infinite cylinder, Bi =3D 0.04 and Bi²Fo =3D (0.04)²=C3=97(5.6) =3D 0.09

We have (Q/Qo)_i_c =3D 0.3.

For an infinite slab, from th= e dimensionless heat transfer chart, for the values of Bi =3D 0.02= and Bi²Fo =3D 0.08, we have (= Q/Q_o)_=
ip =3D 0.4.

Then

=20

But

=20

Theref= ore the actual heat loss in 1 min,

=20

3.19. A brick colu= mn with cross-sectional area 20 cm =C3=97 10 cm is initially at an uniform = temperature of T_i =3D 225=C2=B0C. Suddenly the s= urfaces are subjected to convective cooling with h =3D 60 W/m²C into an ambient at T_a =3D 25=C2= =B0C. Calculate the centre temperature T_o at t = =3D 1 hr after the start of cooling. (Take K =3D 0.7 W/m=C2=B0C and =3D 0.5= =C3=97 10^=E2=88=926 m²/s.)

The brick column is an i= nfinite bar, the solution is obtained as a product of the solutions of two = infinite slab problems.

For the slab of thicknes= s, L₁ =3D 20 cm

From the Heisler chart f= or an infinite slab, for the value of 1/Bi =3D 0.12 and Fo =3D 0.18, we get (T_o =E2=88=92 T_a)/(T_i =E2=88= =92 T_a) =3D 0.8.

For the slab of thicknes= s, L₂ =3D 10 cm

F= rom the same chart, for the value 1/Bi =3D 0.28 and Fo =3D 0.72, we have (T_o =E2=88=92 T_a)/(T_i =E2=88=92 T_a) =3D 0.3.

Therefore

Temperature of the centr= e of the column,

=20

3.20. A semi-infin= ite cylindrical iron bar of diameter d_o =3D 5 cm= is initially at an uniform temperature of T_i = =3D 300=C2=B0C. The surfaces are suddenly subjected to convection with h = =3D 200 W/m²=C2=B0C into ambient air at T_a =3D 30=C2=B0C. Determine the temperature at the axis = of the bar 3 cm from the flat surface 2 min (t) after the start of the cool= ing. (Take K =3D 60 W/m=C2=B0C and =3D 1.6 =C3=97 10^=
=E2=88=925 m²/s.)

The solution for this ba= r is a product of the solutions for an infinite cylinder and a semi-infinit= e slab.

For an infinite cylinder=

From the Heisler chart f= or infinite cylinders, for the value of

we have

For a semi-infinite soli= d

=

Therefore,

=20

3.21. Material (d = =3D 0.5 mm) at temperature (Tsi) of 1200=C2=B0C moves into a fluidized bed = exchanger and gets cooled by air at room temperature (Ta) of 20=C2=B0C. Bot= h air and solid leave at a temperature (Tso) of 510=C2=B0C. If the air velo= city (Uo) is 0.5 m/s at 1 atm pressure (P), find how long it will = take to cool the solid to a temperature (T) of550=C2=B0C for 90% heat loss.=

The properties of solid = are found to be

=CF=81_s =3D 2600 kg/m³; k_s =3D 0.33 W/mK, C_ps =3D 800 J/kgK; =CE=B1 =3D 1.59 =C3=97 10^=E2=88=927 m²/s

Air properties at 510=C2= =B0C are:

=CE=BC_a =3D 36.19 =C3=97 10^{=E2=88=926 kg/ms=E2=80=99}

K_a =3D 0.0575 W/WmK

C_pa =3D 1093 J/kgK

=CF=81_a =3D 0.456 kg/m³

The correlation for heat= transfer coefficient (h) between the material and air is

Assuming particle conduction cont= rols the heat transfer:
For sphere, and ,

Fourier number,

Assuming film r= esistance controls the heat transfer:

Assuming for both resistances:

From chart, for Bi = =3D 0.084 and . The value of Fo = =3D 23

3.22. A multistage= fluidized bed is used to cool a stream of hot solids from (T_si) 840=C2=B0C to 240=C2=B0C (T_so) = with gas available at 40=C2=B0C (T_a). If the flo= w rates of gas and solid are adjusted to obtain the same thermal utilizatio= n for two streams, find the number of stages needed and the temperature of = the flowing streams in the exchanger. Assume counter flow in fluidized bed.=

For solids,

Equal thermal utilizatio= n gives =CE=B7_s =3D =CE=B7= _g

Number of stages, N = =3D 3

Since

=E2=88=B4 Exit temperatu= re of the gas (T₁) =3D 640=C2= =B0C.

But

T_{1 =E2=88=92 T_si =3D =CF=86(<=
em>T}₂ =E2=88=92 T₁)

where =CF=86 = =3D heat flow ratios of the two flowing streams =3D 1

=20

= REVIEW QUESTIONS

Discuss heat transfer using lumpe= d parameter analysis.

Explain the significance of Biot and F= ourier numbers.

Derive an expression for heat transfer= in semiinfinite solids.

Derive an expression for heat transfer= with constant heat flux on a semi-infinite solid.

Derive an expression for heat transfer= with convection boundary conditions.

Explain the use of transient-temperatu= re charts.

Discuss the use of multidimensional sy= stems for heat transfer analysis.

What is periodic heat flow?

Discuss semi-infinite solid, whose sur= face temperature is varied periodically.

Explain surface temperature variation.=

Explain depth of penetration.

Discuss heat transfer in a semi-infini= te solid surface exposed to a fluid with a periodic temperature variation.<= /li>
Discuss heat transfer in freezing or m= elting.

Discuss heat transfer in flat liquid o= r solid surface.

Discuss heat transfer in the proce= ss of freezing of a liquid at a temperature higher than its freezing point.=

OBJECTIVE QUESTIONS

If a body at an uniform temperatu= re is suddenly exposed to a different temperature, then the temperature in = the interior of the body depends on

resistance to heat transfer inter= nally

resistance to heat transfer at its sur= face

resistance to heat transfer at the cen= tre of the body

both (a) and (b)

The time-dependent, dimensionless temp= erature (=CE=B8) of the body can be written in terms of Biot (Bi) and Fouri= er (Fo) numbers as =E2=88=92(BiFo) (BiFo)

=CE=B8 =3D e^{=E2=88=92(BiFo)}

=CE=B8 =3D e^(BiFo)

<= /li>

Lumped parameter analysis is valid if = the value of Biot number is less than

0.01

0.1

10

100

The error expected if lumped parameter= analysis is used with Biot number less than 0.1, is less than

1%

10%

5%

0.5%

For small values of Fourier number tim= e is required for temperature change in the solid

less

more

very small

no

An aluminium rod of diameter 10 cm is = subjected to lumped parameter analysis. The characteristic length is

0.1 m

0.25 m

0.025 m

0.05 m

Heisler charts are preferred in transi= ent heat conduction if Biot number falls in the range of

0.1=E2=88=92100

0.1=E2=88=9210

0=E2=88=920.1

0=E2=88=92100

A solid is said to be an infinite soli= d if it extends itself infinitely in direction.

one

two

all

radial

The one-dimensional (x-direct= ion) steady-state transient-temperature distribution for the infinite slab = is calculated using

=

<= /li>

A carbon (K =3D 54 W/m K) s= teel ball of diameter 5 cm is suddenly exposed to a fluid with convection h= eat transfer coefficient of 210 W/m² K. The Biot= number is

0.0194

0.194

1.94

19.4

The Biot number of a sphere having rad= ius ro in the Heisler chart is

=

<= /li>

Fourier number of a cylinder having di= ameter do, is calculated from the equation =CE=B1t =CE=B1t

=

<= /li>

In lumped parameter analysis, the = temperature in a solid with respect to position.

varies

is a constant

zero

is very small

Answers:

(d)

(a)

(b)

(c)

(b)

(c)

(d)

(c)

(a)

(b)

(b)

(a)

(b)

EXERCISE PROBLEMS

A steel slab of thickness 20 mm a= t 798 K is suddenly exposed to an air stream at 298 K. Find the time for co= oling the slab to 298.5 K. Assume that the properties of steel are =CF= =81 =3D 7950 kg/m³; c =3D 0.455 kJ= /kg K; K =3D 46 W/m K and h =3D 36 W/m² K. (Ans. 116 min)

An aluminium pipe of diameter 1 m whic= h is initially at a temperature of 773 K is suddenly exposed to an air stre= am at 373 K. Find the time required for the pipe to reach 473 K. Assume the= convective heat transfer coefficient as 80 W/m²= K. (Ans. 10,510 s)

A mild steel ball of diameter 1.2 cm i= s initially at a temperature of 873 K and is exposed to air at 300 K. Find = the time needed to cool the steel ball to 368 K. Assume that the convection= heat transfer coefficient is 0.114 kW/m² K. Als= o find the heat transfer rate after 120s. (Ans. 2.1 min, 3= .9 W)

The initial temperature of a large ste= el plate is 698 K. Find the centre-line temperature of the plate after 180 = s if it is suddenly exposed to air at 338 K. Assume that the convective hea= t transfer coefficient is 0.285 kW/m² K, the the= rmal conductivity is 0.0425 kW/m K and the thermal diffusivity is 1.194 =C3= =97 10^=E2=88=925 m²/s= . (Ans. 554 K)

A cylindrical metal rod of diameter 15= cm is heated to 1088 K in a furnace and then suddenly quenched in a liquid= maintained at 311 K. Find the centre-line temperature of the rod after 35 = min. Assume that the convective heat transfer coefficient is 0.175 kW/m2 K and the thermal diffusivity is 5.14 =C3=97 10=E2=88=92⁶ m²/s. The thermal conductivity of the metal is 0.0175 kW/mK= . (Ans. 389 K)

A metal cylinder of diameter 200 mm an= d length 200 mm is heated to 673 K and then suddenly dropped in water maint= ained at 323 K with a surface convective heat transfer coefficient of 0.1kW= /m² K. Find the surface temperature, centre temp= erature and total heat transferred to water after 1200 s. Assume that the t= hermal conductivity and thermal diffusivity of the metal are 0.04 kW/mK and= 1 =C3=97 10^=E2=88=925 m^{2/s, respectively. (Ans. 458 K, 491 K, 53.3 MJ)}

A metal rod of radius 25 mm and length= 100 cm is heated to 473 K and then suddenly exposed to a fluid at 343 K. C= alculate the temperature at a radius of 10 mm. Also find the heat loss from= the rod after 60 s. The density of the metal is 2700 kg/m³, specific heat is 0.90 kJ/kg K, thermal conductivity is 0.25 kW= /mK and thermal diffusivity is 0.306 m²/hr. Assu= me that the convective heat transfer coefficient is 0.5kW/m² K. (Ans. 388 K, 434 kJ/kg K)

A metal ball of diameter 75 mm is = heated to 1173 K and then suddenly quenched in water at 308 K. Find the tim= e required for the surface temperature of the steel ball to reach 473 K. Th= e density, specific heat and thermal conductivity of the metal are 8100 kg/= m³, 0.465 kJ/kg K and 0.044 kW/m K, respectively= . Assume the convective heat transfer coefficient at the surface as 0.06 kW= /m²K. (Ans. 1368 s)

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=20 =20 =20 =20 Previous Chapter
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T_f	=E2=80=93	fluid temperature at any time
T_a	=E2=80=93	Amplitude of temperature variation
t	=E2=80=93	Time
=CE=B4	=E2=80=93	Angle of lag in the temperature variatio= n of the solid with reference to fluid temperature variation
=CF=81	=E2=80=93	Solid density
C	=E2=80=93	Solid specific heat
V	=E2=80=93	Solid volume
A	=E2=80=93	Solid surface area
T_s	=E2=80=93	Temperature of the solid

	=E2=88=92	Amplitude ratio
T	=E2=88=92	Surface temperature
=CE=B1	=E2=88=92	Thermal diffusivity of solid
=CE=B4	=E2=88=92	Angle of lag in the temperature at depth= x, with reference to surface temperature.
x_o	=E2=88=92	Depth upto which the temperature variati= on penetrates
	=E2=88=92	Heat flow rate at time W/m² or kcal/m²hr
=CE=B4_q	=E2=88=92	Time lag of heat flow with reference to = the surface temperature variation
	=E2=88=92	Total heat flow during half period
T_s	=E2=88=92	Temperature at depth x from sur= face
k	=E2=88=92	Solid thermal conductivity
h	=E2=88=92	Convective heat transfer coefficient.

x	=E2=88=92	thickness melted/frozen from surface upt= o time t
x*	=E2=88=92	dimensionless distance parameter
t*	=E2=88=92	dimensionless time parameter
k_s, k₁	=E2=88=92	thermal conductivity of solid, liquid
=CF=81_s= =CF=81₁	=E2=88=92	density of solid, liquid
h_sf<= /td>	=E2=88=92	enthalpy of melting/freezing
h_=E2=88=9E=	=E2=88=92	convective heat transfer coefficient at = surface
T_=E2=88=9E=	=E2=88=92	free stream temperature of fluid flowing= over surface
T_fr<= /td>	=E2=88=92	melting/freezing temperature
t	=E2=88=92	time

h₁	=E2=88=92	convection coefficient at the liquid sol= id interface
^T₁	=E2=88=92	Temperature of liquid
x	=E2=88=92	thickness from surface frozen upto time = t.
r	=E2=88=92	radius at the (freezing) solid / liquid = interface at time t
r ₀	=E2=88=92	tube outside radius

Table= of Contents for =20 Heat and Mass Transfer, 2nd Edition =20

3

Transient Heat Conducti= on

= 3.1 INTRODUCTION

= 3.2 LUMPED PARAMETER ANALYSIS (INFINITE THERMAL CONDUCTIVITY)

3.2.1 = Significance of Biot and Fourier Numbers

KEY = CONCEPTS

= 3.3 SEMI-INFINITE SOLIDS

3.3.1 = Constant Heat Flux on Semi-infinite Solid

3.3.2 = convection Boundary Conditions

3.3.3 = Use of Transient =E2=80=93 Temperature Charts

3.3.4 = Slab

3.3.5 = Long Cylinder

3.3.6 = Sphere

= 3.4 MULTIDIMENSIONAL SYSTEMS

= 3.5 PERIODIC HEAT FLOW

3.5.1 = Semi-infinite Solid-surface Temperature Varied Periodically

3.5.1.1 Surface temperature vari= ation

3.5.1.3 Semi-infinite solid surf= ace exposed to fluid with periodic temperature variation

= 3.6 FREEZING/MELTING

3.6.1 = Flat Liquid/Solid Surface

3.6.2= Freezing of Liquid at a Temperature Higher than Freezing Point

= 3.7 CONCLUSIONS

= EXAMPLE PROBLEMS

= REVIEW QUESTIONS

OBJECTIVE QUESTIONS

Answers:

EXERCISE PROBLEMS

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Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20