Chapter 3

General Differential Equations for Heat Conduction

3.1 Introduction

In heat transfer analysis, one of our objectives is to determine the temperature distribution within the body at any given instant, i.e. the temperature at every point in the body, taken as a continuum. Then, we can calculate the heat transfer rate at any point in a given direction by applying the Fourier’s law. Knowledge of temperature distribution is also required in other fields of engineering analysis; e.g. in calculation of thermally induced stresses, thermal expansion , optimum thickness of insulation, etc.

General technique to obtain the temperature distribution over the entire body is to consider a differential control volume within the body and apply the law of conservation of energy to this differential control volume. This results in a differential equation. Solution of this differential equation with appropriate initial and boundary conditions gives the temperature field, i.e. the temperature at any point within the body.

In this chapter, first, general differential equations for conduction is derived in Cartesian (i.e. rectangular or x-y-z) coordinates. This is useful to analyse problems of heat transfer in rectangular-shaped bodies such as squares, rectangles, parallellopiped, etc. Next, the general differential equation of conduction is stated in cylindrical and spherical coordinates; these are useful to solve heat transfer problems in cylinders and spheres. Simplifications to the general differential equations for different, possible practical conditions are presented. Typical boundary conditions encountered in practice and the methods to represent them mathematically are explained. A summary of the equations is given at the end of the chapter for ready reference.

3.2 General Differential Equation for Heat Conduction in Cartesian Coordinates

This is also known as heat diffusion equation or, simply heat equation. Consider a homogeneous body within which there is no bulk motion and heat transfer occurs in this body by conduction. Temperature distribution within the body at any given instant is given by: T (x, y, z, τ). The coordinate system used in this derivation is given in Fig. 3.1.

Consider a differential volume element dx.dy.dz from within the body as shown. It has six surfaces and each surface may be assumed to be isothermal since the differential element is very small. Further, the body is assumed to be rigid, i.e. negligible work is done on the body by external mechanical forces.

Let us make an energy balance on this differential element. Let us list out the various energy terms involved: first, there is energy conducted into the element; second, there is energy conducted out of the element; third, for generality, let there be energy generated within the element, say, due to joule heating, chemical reaction or nuclear fission, etc. Net heat conducted into the element in conjunction with the heat generated within the element, will obviously cause an increase in the energy content (or the internal energy) of the element. We can write it mathematically as

 

where,	Ein = energy entering the control volume per unit time
	Eout = energy leaving the control volume per unit time
	Egen = energy generated within the control volume per unit time
	Est = energy storage within the control volume per unit time.

Let us calculate these quantities, one by one.

To calculate E_in.   Energy enters the differential control volume from all the three sides by conduction only, since the control volume is embedded within the body considered.

Let the energy entering the control volume in the X-direction through face ABCD be Q_X. Similarly, Q_Y and Q_Z enter the control volume from the faces ABFE and DAEH as shown in the Fig. 3.1.

 

To calculate E_out. Energy entering the control volume in the X-direction at face ABCD leaves the control volume at the opposite face EFGH. This is designated as Q_x+dx. Similarly, Q_y+dy and Q_z+dz leave the control volume from the surfaces opposite to the ones at which they entered. Therefore, we write,

 

Now, from calculus, we know that Q_x+dx etc. can be expressed by a Taylor series expansion, where, neglecting the higher order terms, we can write,

To calculate E_gen. Let there be uniform heat generation within the volume at a rate of q_g (W/m³). Heat generation is a volume phenomenon, i.e. heat is generated throughout the bulk of the body—so, note its units (W/m³). As mentioned earlier, heat may be generated within the body due to passage of an electric current, a chemical reaction, nuclear fission, etc. Then, for the differential control volume dx.dy.dz, we can write,

 

To calculate E_st. As a result of the net energy flow into the control volume from all the three directions and the heat generated within the control volume itself, internal energy of the control volume increases. This will manifest itself as an increase in the temperature of the control volume. Let the temperature of the control volume increase by dT in time dτ Then, if ρ is the density and c_p, the specific heat of the material of the control volume, rate of increase of internal energy of control volume is given by,

Now, substituting for all terms in Eq. 3.1, we get,

 

i.e.

i.e.

Now, let us bring in Fourier’s law of heat conduction. If, for generality, we assume k_x, k_y, k_z to be the thermal conductivities of the material in the x, y and z-directions respectively, and A_x, A_y and A_z to be the areas normal to the respective heat flow directions, we can write for the heat flow rates,

Substituting Eq. (3.8) in (3.7), and dividing throughout by dx.dy.dz, we obtain,

This is the general form of heat diffusion equation in Cartesian coordinates, for time dependent (i.e. unsteady state) heat conduction, with variable thermal conductivity and uniform heat generation within the body. This is a very important basic equation for conduction analysis. It has to be solved with appropriate initial and boundary conditions to get the temperature distribution within the body as a function of spatial and time coordinates. Of course, the heat transfer rate is calculated applying the Fourier’s law, once the temperature distribution is known.

Now, if the material is isotropic, i.e. the thermal conductivity is the same in all the three directions, i.e. k_x = k_y = k_z = k say, then we can write,

Eq. 3.10 is the general form of heat diffusion equation in Cartesian coordinates, for time dependent (i.e. unsteady state) conduction, when thermal conductivity varies with temperature (i.e. with position) and uniform heat generation occurs within the body.

If k is constant and does not vary with temperature, i.e. k does not change with position, the Eq. 3.10 can be written as,

 

where,	α = K/(ρ cp) is thermal diffusivity, and
	∇ = Laplacian operator.

Solution of general form of heat diffusion equation as given in Eq. 3.10 or 3.12 is rather complicated. However, in many practical applications, we make simplifying assumptions and the resulting equations are easily solved. For example:

1. Steady state   This means that the temperature at any position does not change with time, i.e. . So, Eq. 3.12 becomes:
   

   This is known as Poisson equation and is for steady state, three-dimensional heat conduction with heat generation, with constant thermal conductivity, in Cartesian coordinates.

2. With no internal heat generation   This means that q_g term is zero. So, Eq. 3.12 becomes,
   

   This is known as Diffusion equation, and it represents time dependent, three-dimensional heat conduction, with no internal heat generation, and with constant thermal conductivity, in Cartesian coordinates.

3. Steady state, with no internal heat generation   This means that q_g and are zero. So, Eq. 3.12 becomes,

    

   ∇² T = 0              …(3.15)

   This is known as Laplace equation, and it represents steady state, three-dimensional heat conduction with no internal heat generation, with constant thermal conductivity, in Cartesian coordinates.

4. One-dimensional, steady state, with no internal heat generation   This means that,
   

   So, Eq. 3.12 becomes,

   

Note that now, partial derivative is written as full derivative since temperature is dependent on one coordinate only.

You may be wondering why we have to consider one-dimensional heat flow when we are dealing with three-dimensional bodies. You will be surprised to know that solution of this simplified version of heat conduction equation for cases of simple geometries gives results with acceptable accuracy for engineering applications. One-dimensional conduction implies that temperature gradient is considerable only in one direction and is relatively negligible in the other three directions; consequently, heat flow is also in only one direction. Examples of such practical cases are: large slab (or wall) where length in one dimension (say, its thickness) is small compared to the other two-dimensions—then, temperature varies only along its thickness; long cylinder, whose temperature may be assumed to vary only along its radius; sphere, whose temperature may be assumed to vary only along its radius.

3.3 General Differential Equation for Heat Conduction in Cylindrical Coordinates

Eq. 3.10 derived earlier is suitable to analyse heat transfer in regular bodies of rectangular, square or parallelopiped shapes. But, if we have to analyse heat transfer in cylindrical-shaped bodies (which are commonly used in practice), then, working with cylindrical coordinates is more convenient, since in that case, the coordinate axes match with the system boundaries.

Nomenclature for cylindrical coordinate system is shown in Fig. 3.2.

Differential equation for heat conduction in cylindrical coordinates may be derived by considering an elemental cylindrical control volume of thickness dr and making an energy balance over this control volume, as was done in the case of Cartesian coordinates, or, coordinates transformation can be adopted; for this purpose, transformation equations are,

    x = r cosϕ

    y = r sinϕ

    z = z

    ϕ = tan^–1(y/x)

The resulting general differential equation in cylindrical coordinates is,

Eq. 3.17 is the general differential equation in cylindrical coordinates, for time dependent, three-dimensional conduction, with constant thermal conductivity and with internal heat generation.

For one-dimensional conduction in r direction only, we get from Eq. 3.17,

i.e.

Eq. 3.19 represents one-dimensional, time dependent conduction in r direction only, with constant k and uniform internal heat generation, in cylindrical coordinates.

And, for steady state, one-dimensional heat conduction in r direction only, with constant k and uniform heat generation, Eq. 3.19 reduces to,

3.4 General Differential Equation for Heat Conduction in Spherical Coordinates

To analyse heat transfer in spherical systems, working with spherical coordinates is more convenient, since the coordinate axes match with system boundaries. Nomenclature for the spherical coordinates is shown in Fig. 3.3.

Differential equation for heat conduction in spherical coordinates may be derived by considering an elemental spherical control volume and making an energy balance over this control volume, as was done in the case of Cartesian and cylindrical coordinates, or, coordinate transformation can be adopted using the following transformation equations,

        x = r sinθ sin ϕ

        y = r sinθ sin ϕ

        z = r cosθ

The resulting general differential equation in spherical coordinates is,

Eq. 3.21 is the general differential equation in spherical coordinates, for time dependent, three-dimensional conduction, with constant thermal conductivity and with internal heat generation.

For one-dimensional conduction in r direction only, we get from Eq. 3.21,

i.e.

Eq. 3.23 represents one-dimensional, time dependent conduction in r direction only, with constant k and uniform internal heat generation, in spherical coordinates.

And, for steady state, one-dimensional heat conduction in r direction only, with constant k and uniform heat generation, Eq. 3.23 reduces to,

3.5 Boundary and Initial Conditions

Let us rewrite the heat diffusion eqn. i.e. Eq 3.11 in Cartesian coordinates,

Of course, temperature distribution in the body is obtained by solving this differential equation.

We observe that this is a second order differential equation in spatial coordinates; therefore, its solution will need two boundary conditions to eliminate the two constants of integration. Also, this equation is of first order in time coordinate; so, it will require one initial condition, i.e. the condition at τ = 0.

Commonly encountered Boundary Conditions (B.C.’s) are:

1. Prescribed temperature conditions at the boundaries—known as B.C. of the first kind or Dirichlet condition
2. Prescribed heat flux condition at the boundaries—known as B.C. of the second kind or Neumann condition
3. Convection boundary condition—known as B.C. of the third kind
4. Interface boundary condition—known as B.C. of the fourth kind.

   We will study below the method of representing the boundary conditions mathematically.

3.5.1 Prescribed Temperatures at the Boundaries (B.C. of the First Kind)

This situation is represented in Fig. 3.4. Here, it is granted that the temperatures at the boundaries are specified and are constant. For example, temperature at a surface is constant if that surface is in contact with a melting solid or boiling liquid. Or, in more general case, the variation of temperature at the surface may be specified as a function of position and time. Referring to Fig. 3.4, the surface at x = 0 is maintained at a constant temperature T₁ and the surface at x = L is maintained at constant temperature T₂.

Mathematically, these conditions are represented as:

 

 

3.5.2 Prescribed Heat Flux at the Boundaries (B.C. of the Second Kind)

Here, the heat flux at the boundaries is assumed to be known. For example, if a surface is heated by an electric heater, the heat flux entering the surface is known. This condition is depicted in Fig. 3.5 (a).

In Fig. 3.5 (a), a plate of thickness L is shown. At x = 0, i.e. at the left face, a heat flux q₀ is supplied; this is conducted into the material as shown. At x = L, i.e. at the right face, a heat flux q_L is supplied and this is also conducted into the material. This situation is mathematically represented as follows, remembering that the conduction flux at a surface is equal to the heat flux supplied. Also, note clearly that −k.(∂T/∂x) represents the heat flux in the positive X-direction, i.e. from left to right; if the direction of heat flux in a slab is from right to left, obviously, it is equal to +k.(∂T/∂x).

 

 

Note again that in Eq. 3.26 b, RHS is positive since the heat flux at x = L is in the negative X-direction, i.e. from right to left as shown in the Fig. 3.5 (a). Similar relations can be written if the geometry is cylindrical or spherical.

There are two special cases of this boundary condition,

1. Insulated boundary Many times, to reduce the heat loss (or gain), the boundary is insulated with an appropriate insulating material. Even though theoretically heat loss will be reduced to zero only with an infinitely thick insulation thickness, heat loss may be practically assumed to be zero with a sufficiently thick insulation; we call this as perfect insulation.
   

   FIGURE 3.5(a) Prescribed heat flux at the boundaries (B.C. of second kind)

   

   FIGURE 3.5(b) Insulated boundary at x = 0

   

   FIGURE 3.5(c) Thermal symmetry at x = L/2

   So, for a perfectly insulated boundary at x = 0, shown in Fig. 3.5 (b), the heat flux across the boundary is zero and we represent this condition mathematically as follows,

           k.(∂T(0, τ)/∂x) = 0

           or, (∂T(0, τ)/∂x) = 0           …(3.26c)

    

2. Thermal symmetry In many cases, there is thermal symmetry over a plane inside the system being analysed; for example, consider a copper plate, initially heated to a high temperature and then hung in air for cooling. It is intuitively clear that heat flow is from the centre of the plate to the two sides and the centre plane will be the plane of symmetry. In other words, no heat will cross this plane, i.e. this plane is equivalent to an insulated boundary.

So, for the centre plane, we can write,

 

3.5.3 Convection Boundary Condition (B.C. of the Third Kind)

This is a more common practical situation, where heat transfer occurs at the boundary surface to or from a fluid flowing on the surface at a known temperature and a known heat transfer coefficient, e.g. in heat exchangers, condensers, reboilers etc.

Consider, again, a slab of thickness L as shown in Fig. 3.6.

At the left surface (x = 0), a hot fluid of temperature T₁ is flowing with a heat transfer coefficient h₁, supplying heat into the body. At the right surface (x = L), a cold fluid at a temperature T₂ is flowing on the surface, removing heat from the body with a heat transfer coefficient h₂.

Equating the conduction heat flux to the convection heat flux on either surface and remembering to note the direction of heat flow (i.e. whether it is in the positive X-direction or negative X-direction), we can represent this boundary condition mathematically as follows,

 

 

In Eq. 3.27 b, we write for the conduction heat flux at x = L: = −k.(∂T/∂x)|_{X = L} since heat is flowing in positive X-direction.

Using the same principles, expressions can be written for convection boundary conditions at surfaces of cylindrical and spherical geometries.

3.5.4 Interface Boundary Condition (B.C. of the Fourth Kind)

When a system is made up of one or more layers of different materials, solution of the problem requires that the conditions at the interface between the layers A and B is specified. Perfect thermal contact at the interface presupposes the following requirements:

1. both the bodies must have the same temperature at the interface
2. heat flux on both the sides of the interface must be same.

Interface boundary condition is depicted in Fig. 3.7.

We write, at the interface,

 

 

Of course, the four boundary conditions explained above do not cover all the possible boundary conditions that may be encountered in practice. However, in any given situation, correct B.C. can be derived by applying the energy balance at the surface (i.e. to a control volume of zero volume—which means that no energy storage is possible at the control surface—and, heat entering IN = heat going OUT), as was done in deriving Eq. 3.27.

As a further example of this technique, consider a slab of thickness L; at its left surface, it receives heat by radiation and at its right face, loses heat by radiation. This situation is represented mathematically as shown in Fig. 3.8.

On the left hand side, energy impinging on the surface by radiation is equated to the energy conducted into the slab; since the heat is conducted in the positive X-direction the conduction term (flux) has a negative sign as per Fourier’s law. Similarly, on the right hand face, radiation impinging on the surface is conducted into the slab from right to left, i.e. in the negative X-direction; therefore, we put a positive sign in the conduction term, as shown in the Fig. 3.8.

Example 3.1.   Temperature variation in a slab is given by: T(x) = 100 + 200 x – 500 x², where x is in metres; x = 0 at the left face and x = 0.3 m at the right face. Thermal conductivity of the material k = 45 W/(mC). Also, c_p = 4 kJ/(kgK) and ρ = 1600 kg/m³. Determine:

1. Temperature at both surfaces
2. Heat transfer at left face and its direction
3. Heat transfer at right face and its direction
4. Is there any heat generation in the slab? If so, how much?
5. Maximum temperature in the slab and its location
6. Time rate of change of temperature at X = 0.1 m if the heat generation rate is suddenly doubled
7. Draw the temperature profile in the slab
8. Average temperature of the slab.

Solution.   Temperature profile is given; so, temperatures at the left and right faces are easily determined by substituting x = 0 and x = 0.3 m. Maximum temperature is determined by first differentiating T(x) w.r.t. x and equating to zero to get the position (x max) where the maximum occurs and then substituting this x max in T(x). Temperature profile is graphed using Mathcad. Time rate of change of temperature at X = 0.1 m is found by applying the time dependent, one dimensional, heat conduction equation in Cartesian coordinates. Procedure to determine the average temperature of the slab is explained at the end. We shall solve this problem in Mathcad, with suitable comments at each step.

Data:

i.e.

α = 7.031 × 10⁻⁶ m²/s

T(x): = 100 + 200x − 500x²

 

(Define T(x)… i.e. temperature as a function of x)

Temperature at left face, i.e. at x = 0     T(0) = 100°C

Temperature at right face, i.e. at x = 0.3 m: T(0.3) = 115°C

To find max. temperature

Define the first derivative of T(x):

Also, define the second derivative of T(x):

By hand calculation: we get: T′(x) = 200 − 1000.x

We set T′(x) equal to zero to get the position x max where temperature is maximum

i.e.              200 − 1000.x = 0.

This gives x = 0.2 m. Substitute this value of x max in T(x) to get the value of Tmax.

So, Tmax = T(0.2) = 100 + 200 × 0.2 – 500 × (0.2)² = 120°C.

However, in Mathcad, all this procedure is very simple. Read the comments in Mathcad solution below.

Set T′(x) = 0 and find out the value of x max. To do this, use the root function, which solves the root of T′(x) = 0. First, assume a trial value of x; then use the root function which gives the true value of x

 

	x : = 0.15	(Trial value of x)
	x max : = root(T′(x), x)	x max is obtained from the root function)
i.e.	x max = 0.2 m	(value of x where T is a max.)

To get T max: Substitute this value of x max in T(x):

To Sketch the temperature distribution in the slab:

 

(Define the range variable x, i.e. x to vary from 0 to 0.3 m in steps of 0.01 m)

To draw the graph:

Just select the x – y plot from pallete, plug in x and T(x) in the place holders:

x is in metres and T(x) in deg. C. Click anywhere outside the graph region; immediately the graph appears.

Note from the graph that the maximum temperature occurs at x = 0.2 m and its value is 120°C, as already calculated.

To calculate the heat fluxes at the left and right faces:

Apply the Fourier’s law at x = 0 and x = 0.3 m, remembering that temperature gradient is given in T′(x), already defined.

 

	qleft : = −k·T′(0)	(applying Fourier’s law at left face, i.e. at x = 0)
i.e.	qleft = −9 × 103	(Heat flux at the left face (W/m2); note that −ve sign indicates heat flowing from right to left)
	qright := −k·T′(0.3)	(applying Fourier’s law at right face, i.e. at x = 0)
i.e.	qright = 4.5 × 103	(Heat flux at the right face (W/m2); note that +ve sign indicates heat flowing from left to right)
	qtotal : = |qleft| + |qright|	(Total heat generated per m2 of surface)
i.e.	qtotal = 1.35 × 104	W/m2 (Total heat generated/m2)

Therefore, q_g, the volumetric heat generated rate is given by total heat generated per unit volume:

(volumetric heat generation rate in the slab)

To calculate the time rate of change of temperature at x = 0.1 m when q_g is suddenly doubled:

We have the time dependent differential equation for heat conduction in Cartesian coordinates

Therefore,

From the given equation for temperature distribution, it is clear that does not depend on x, i.e. depends only on q_g:

(define dT/dτ as a function x. Now, we can get dT/dt at any x by simply substituting that value of x in the function defined)

i.e.

Note that this is true for all x since T″(x) does not depend on x for the temperature distribution given

To determine the average temperature of the slab:

For a differential element of thickness dx, amount of heat energy contained in the element is equal to A.dx.ρ. c_p. T(x). Total amount of energy in the slab is obtained by integrating this from x = 0 to x = 0.3. Now, if the average temperature of slab is T_av, amount of energy in the slab can also be written as: ρ.A.L.c_p.T_av. Equating these two expressions, we get

In Mathcad, evaluating the integral within given limits is very easy. First, define T_avg and then just plug in the limits; Mathcad automatically evaluates the integral and gives the value.

(Mathcad easily does the integration of T(x) within the limits specified)

i.e.     T_avg = 115°C

(Average temperature of the slab.)

Note that Mathcad directly gives the value of the integral within the limits specified; there is no need to expand the integral and write down as you do in hand calculations.

Example 3.2.   Uniform internal heat generation at q_g = 5 × 10⁷ W/m³ occurs in a cylindrical nuclear reactor fuel rod of 50 mm diameter, and under steady state conditions the temperature distribution is of the form:

T(r) = 800 – 4.167 × 10⁵r², where T is in deg. Celsius and r is in metres. The fuel rod properties are: k = 30 W/(mK), ρ = 1100 kg/m³ and c_p = 800 J/(kgK)

1. What is the rate of heat transfer per unit length of the rod at r = 0 (i.e. at the centre line) and at r = 25 mm (i.e. at the surface)?
2. Sketch the temperature distribution along the radius.
3. If the reactor power is suddenly increased to 10⁸ W/m³, what is the initial time rate of temperature change at r = 0 and r = 25 mm?
4. Find the average temperature of the rod in the first case.

Solution.   Here, temperature distribution is given; so, heat flux can be calculated at any radius r from Fourier’s law: q = – k (dT/dr). Temperature distribution along the radius is easily graphed with Mathcad. Time rate of change of temperature when q_g changes is found out by applying the time dependent, one-dimensional heat conduction equation in cylindrical coordinates. Average temperature of the cylinder is obtained from first principles as done in the case of slab in Example 3.1.

Data:

        R : = 0.025 m

        k : = 30 W/mC

        c_p : = 800 J/kgK

        ρ : = 1100 kg/m³

        L : = 1 m

T(r) := 800 − 4.167 × 10⁵ · r²

 

(Define T(r)… i.e. temperature as a function of r)

(Define first derivative of T(r))

(Define second derivative of T(r))

Q_centre = 0

 

(heat transfer rate at the centre is zero since temperature at centre is maximum and dT/dr = 0 at r = 0.)

To find the heat transfer rate at the surface (i.e. at r = 0.025 m):

Apply Fourier’s law:

(heat transfer rate at the surface is obtained by applying Fourier’s law at the surface, i.e. at r = R; T′(R) is the temperature gradient at r = R)

i.e.     Q_surface = 9.818 × 10⁴ W/metre length

 

(heat transfer at the surface)

Then, select the x-y graph from pallete and fill in the place holders in both the axes. On x-axis, fill in r and on y-axis, fill in T(r). Click anywhere outside the graph region and immediately, the graph appears.

To calculate the time rate of change of temperature at x = 0.1 m:

We have the one-dimensional, time dependent differential equation, with constant k, for heat conduction in cylindrical coordinates:

Therefore, time rate of change of temperature is given by:

Our aim is to find out dT/dτ when q_g changes suddenly to 10⁸ W/m³

(thermal diffusivity.)

…Define second derivative of T(r) w.r.t. r = d²T/dr²

(define the desired time rate of change of temperatire as a function of r)

At the surface i.e. at r = R:

(define the desired time rate of change)

i.e.     dT by dt (0.025) = 56.814 C/s

At the centre, i.e. at r = 0:

(since at r = 0, dt/dr = 0)

i.e.     dT by dt (0) = 85.225 C/s.

To determine the average temperature of the cylinder:

For a differential element of thickness dr, amount of heat energy contained in the element is equal to 2π.r.dr.L.ρ.c_p.T(r). Total amount of energy in the cylinder is obtained by integrating this from r = 0 to r = R =0.025 m. Now, if the average temperature of cylinder is T_av, amount of energy in the cylinder can also be written as: ρ.π.R².L.c_p.T_av. Equating these two expressions, we get,

All the above calculations are done just in one step easily in Mathcad:

(define T_avg Mathcad easily does the integration of T(r) within the limits specified)

i.e.     T_avg = 669.781°C

 

(Average temperature of the cylinder)

Note that Mathcad directly gives the value of the integral within the limits specified; there is no need to expand the integral and write down as you do in hand calculations.

Example 3.3.   Consider an orange, assumed to be a sphere of 8 cm diameter, producing an average internal heat generation of 2.25 × 10⁴ W/m³ during its ripening. Thermal conductivity of the material is 0.15 W/(mK) and its centre temperature is observed to be 50°C. Assuming one-dimensional, steady state conduction, find out:

1. temperature distribution along the radius,
2. surface temperature,
3. heat transferred at the surface of the sphere,
4. draw the temperature profile along the radius, and
5. average temperature of the sphere.

Solution.

Data:

(i) Temperature distribution   For steady state, one-dimensional conduction, for a sphere, we have the controlling differential equation:

To get the temperature distribution, we have to solve Eq. (a) with the following Boundary Conditions (B.C.’s):

(i) at r = 0, (dT/dr) = 0, since the temperature has to be maximum at the centre because the heat flows from centre to periphery and symmetry considerations.

(ii) Given: T_c = 50°C at the centre, i.e. at r = 0

Multiplying Eq. (a) by r²:

i.e.

Integrating, we get,

Integrating again,

Applying B.C. (i) to Eq. b: C₁ = 0

Then,

Applying B.C. (ii) to Eq. c: C₂ = 50

Substituting values of C₁ and C₂ in Eq. c, we get the temperature distribution in the sphere:

(ii) Surface temperature   Now, temperature at the surface is obtained simply by putting r = R = 0.04 m in Eq. d. It is easier to work in Mathcad; first, define the function T(r):

(Define T(r) … i.e. temperature as a function of r)

Surface temperature:

T(R) = 10°C

 

(surface temperature, i.e. at r = 0.04 m)

(iii) Next, heat transfer at the surface   This is determined by Fourier’s law since we already have the relation for the temperature distribution:

i.e. Q(r) = –k.(4πR²).(dT/dr)/r=R. Gradients such as dT/dr and d²T/dr² etc. are easily found in Mathcad, once the function T(r) is defined:

Heat transfer at the surface:

 

		(Define derivative of T(r))
	Q(r) := k ·(4 · π · R2) · T′(r)	(Q(r) is the heat transfer rate at radius r, by Fourier’s law)
i.e.	Q(R) = 6.032 W	(heat transfer rate at the surface, i.e. at r = R = 0.04 m)

Check:   this must be equal to heat generated inside the orange in steady state Q_gen.

Sketch the temperature profile along the radius   This is done very easily and conveniently in Mathcad. First, define a range variable r from 0 to R = 0.04 m, in steps of say, 0.001 m; then, select the x – y graph from the pallete and just fill in the place holders, i.e. fill in r in the place holder on the x-axis and T(r) in the place holder on the y-axis. Click anywhere outside the graph region and immediately, the graph appears:

Temperature profile along the radius:

r := 0, 0.001.. 0.04

 

(define the range variable, i.e. r to vary from 0 to 0.04 m, in steps of 0.001 m)

Note from the graph that maximum temperature occurs at the centre (r = 0); slope of the temperature curve, (dT/dr) tends to zero (i.e. the curve becomes almost horizontal) as it aproaches the y-axis at r = 0.

(v) Average temperature of the sphere   For a differential element of thickness dr, amount of heat energy contained in the element is equal to 4πr².dr.ρ.c_p.T(r). Total amount of energy in the sphere is obtained by integrating this from r = 0 to r = R. Now, if the average temperature of sphere is T_av, amount of energy in the sphere can also be written as: ρ.(4/3)π.R³.c_p.T_av. Equating these two expressions we get,

All the above calculations are done just in one step easily in Mathcad:

(define T_avg Mathcad easily does the integration of T(r) within the limits specified)

i.e.     T_avg = 26°C

 

(Average temperature of the sphere)

Note that Mathcad directly gives the value of the integral within the limits specified; there is no need to expand the integral and write down as you do in hand calculations.

3.6 Summary of Basic Equations

 

TABLE 3.1

Sl. No.	Equation	Remarks
1		Three-dimensional, time dependent heat conduction equation with heat generation and temperature dependent k, in Cartesian coordinates.
2		Three-dimensional, time dependent heat conduction equation with heat generation and constant k, in Cartesian coordinates.
3		Poisson equation, i.e. three-dimensional, steady state heat conduction equation with heat generation and constant k, in Cartesian coordinates.
4		Diffusion equation, i.e. three-dimensional, time dependent heat conduction equation with no heat generation and constant k, in Cartesian coordinates.
5		Laplace equation, i.e. three-dimensional, steady state heat conduction equation. with no heat generation and with constant k, in Cartesian coordinates.
6		Three-dimensional, time dependent heat conduction equation with heat generation and constant. k, in cylindrical coordinates.
7		Three-dimensional, time dependent heat conduction equation with heat generation and constant k, in spherical coordinates.
8		One-dimensional, time dependent heat conduction equation with heat generation and temperature dependent k, in Cartesian coordinates.
9		One-dimensional, time dependent heat conduction equation with heat generation and temperature dependent k, in cylindrical coordinates.
10		One-dimensional, time dependent heat conduction equation with heat generation and temperature dependent k, in spherical coordinates.
11	Equations 8, 9, 10 are compactly written as, where, n = 0 for Cartesian coordinates, use x as variable instead of r n = 1 for cylindrical coordinates. n = 2 for spherical coordinates.	Compact form of one-dimensional, time dependent, heat conduction equation with heat generation and temperature dependent k
12	where, n = 0 for Cartesian coordinates, use x as variable instead of r n = 1 for cylindrical coordinates. n = 2 for spherical coordinates.	Compact form of one-dimensional, time dependent, heat conduction equation with heat generation and constant k
13	where, n = 0 for Cartesian coordinates, use x as variable instead of r n = 1 for cylindrical coordinates. n = 2 for spherical coordinates.	Compact form of one-dimensional, steady state, heat conduction equation with heat generation and constant k
14		Alternate form of one-dimensional, steady state, heat conduction equation with heat generation and constant k, in Cartesian coordinates.
15		Alternate form of one-dimensional, steady state, heat conduction equation with heat generation and constant k, in cylindrical coordinates.
16		Alternate form of one-dimensional, steady state, heat conduction equation with heat generation and constant k, in spherical coordinates.
17		Alternate form of one-dimensional, time dependent, heat conduction equation with heat generation and constant k, in Cartesian coordinates.
18		Alternate form of one-dimensional, time dependent, heat conduction equation with heat generation and constant k, in cylindrical coordinates.
19		Alternate form of one-dimensional, time dependent, heat conduction equation with heat generation and constant k, in spherical coordinates.

3.7 Summary

This chapter lays the foundation for the study of heat transfer by conduction. First, general differential equation for conduction was derived in Cartesian (or, rectangular) coordinates. This equation has to be solved for a given system applying the appropriate boundary and initial conditions to get the temperature field. To do this, mathematical representation of more common types of boundary and initial conditions are explained. Once the temperature distribution within the body is known, rate of heat transfer (or heat flux) at any point is calculated easily by applying Fourier’s law. Cartesian coordinates are used while dealing with rectangular geometries such as squares, rectangles, walls, parallelopipes, etc; these geometries find applications in furnaces, boiler walls, walls of buildings, air conditioning ducts, etc. Next, general differential equations for conduction in cylindrical and spherical systems are stated. They are useful in solving heat transfer problems involving cylindrical tanks, pipes, spherical storage vessels, reactors, etc. Summary of the basic relations is given in Tabular form for ready reference.

In engineering practice, we ordinarily deal with three-dimensional objects; however, solution of three-dimensional general differential equation is rather complicated. So, a simplifying assumption is made sometimes, of one-dimensional conduction, i.e. temperature variation is substantial only in one-dimension and the temperature variation is considered to be negligible in the other two-dimensions. Many practical problems fit into this category: e.g. walls whose thicknesses are small compared to other dimensions, long cylinders, spheres, etc. In such cases, analytical solutions for one-dimensional heat transfer problems are very much simplified.

In the next chapter, we shall study one-dimensional, steady state conduction as applied to a few regular geometries such as slabs, cylinders and spheres.

Questions

1. Derive the general differential equation in rectangular coordinates (i.e. Cartesian coordinates). Therefrom, write down the governing differential equations for the following cases:
   1. 3-dimensional, constant k, unsteady state conduction with heat generation
   2. 3-dimensional, constant k, steady state conduction without heat generation
   3. 3-dimensional, temperature dependent k, steady state conduction with heat generation
   4. One-dimensional, constant k, unsteady state conduction with heat generation
   5. One-dimensional, temperature dependent k, unsteady state conduction with heat generation
   6. One-dimensional, constant k, steady state conduction without heat generation
   7. One-dimensional, constant k, steady state conduction with heat generation.
2. Derive the general equation for the 3-dimensional unsteady state heat conduction with uniform rate of heat generation in an isotropic solid. Hence, deduce Laplace’s equation.

    

   [V.T.U., Aug. 2001]

3. Write down the two-dimensional, steady state heat conduction equation in x and y variables in rectangular coordinate system, for the case of temperature dependent k and with uniform heat generation in the body.
4. Write down the one-dimensional, time dependent heat conduction equation in spherical and cylindrical coordinate systems, in the r variable, with temperature dependent k and with uniform heat generation in the body.
5. In a medium, heat conduction equation is given in the following form:
   
   1. Is the heat transfer steady or transient?
   2. Is heat transfer one, two or three-dimensional?
   3. Is there heat generation in the medium?
   4. Is the thermal conductivity of the medium constant or variable with temperature?
6. In a medium, heat conduction equation is given in the following form:
   
   1. Is the heat transfer steady or transient?
   2. Is heat transfer one, two or three-dimensional?
   3. Is there heat generation in the medium?
   4. Is the thermal conductivity of the medium constant or variable with temperature?
7. Explain what do you understand by ‘one-dimensional heat conduction’.
8. State the general differential equation for steady state heat conduction in cylindrical and spherical coordinates.
9. What is the need to have the general differential equation for heat conduction in three separate coordinate systems? Give their applications.
10. What is meant by ‘Initial condition’ and ‘Boundary Condition’?
11. Explain the B.C.’s. of first, second and third kinds. Represent them mathematically.
12. Write down the mathematical formulation of the B.C.’s for heat conduction in a rectangular region 0 ≤ x≤ a, 0 ≤ y ≤ b, for:
    1. Boundary at x = 0: heat removed at constant rate of q₀ (W/m²)
    2. Boundary at x = a: heat dissipation by convection with heat transfer coefficient h_a into the ambient air at constant temperature T_a
    3. Boundary at y = 0: maintained at a constant temperature T₀
    4. Boundary at y = b: heat supplied into the medium at a rate of q_b (W/m²)
13. Write down the B.C. for the case of a cylindrical wall with inside radius r₁ and outside radius r₂, when the inside surface is heated uniformly at a rate of q (W/m²) and the outside surface dissipates heat by convection with a heat transfer coefficient h₂ (W/(m²C) into the ambient air at zero deg.C.
14. A spherical shell, inside radius r₁, outside radius r₂, is heated at the inner surface electrically at a rate of q₁ (W/m²); outside surface dissipates heat by convection with a convection heat transfer coefficient h₂, into ambient at temperature T_a Write down the B.C.’s.

Problems

1. A wall, 1.5 m thick has the following temperature distribution:

   T(x) = 60 + 18x – 6x³ where x is in metres and T(x) is in deg. C. Determine the location of maximum temperature and the heat flow per m² area at both the faces. Take k = 25 W/(mC). Also, find out the average temperature of the wall.

2. Consider a plane wall 2 cm thick, with uniformly distributed heat sources (q_g, W/m³) inside its volume; its left and right faces are maintained at temperatures T₁ and T₂, respectively. Steady state temperature distribution in this wall is given by:

   T(x) = 160 – 1000x – 10⁵x² . If q_g = 40 MW/m³, determine:

   1. Temperatures T₁ and T₂
   2. Heat flux at the left face
   3. Heat flux at the right face
   4. Heat flux at the centre of the plate
   5. Average temperature of the plate.
3. Temperature distribution in a slab of 1 m thickness is given by:

   T(x) = 900 – 300x – 50x². Heat transfer occurs across an area of 10 m² and there is uniform heat generation at a rate of q_g = 1000 W/m³. Assume density ρ = 1600 kg/m³, thermal conductivity k = 45 W/(mK) and specific heat c_P = 4 kJ/⁽kgK). Calculate:

   1. Maximum temperature in the slab
   2. Energy entering the left face (i.e. at x = 0)
   3. Energy leaving the wall at right face (i.e at x = 1 m)
   4. Rate of change of energy storage in the slab , and
   5. Time rate of temperature change at x = 0.5 m in the slab.
4. The temperature distribution across a large concrete slab 50 cm thick, heated from one side, as measured by thermocouples approximates to the relation:

   T(x) = 60 – 50x + 12x² + 20x³ – 15x⁴ , where T is in deg. C and x is in metres. Considering an area of 5 m², compute:

   1. heat entering and leaving the slab in unit time
   2. heat energy stored in unit time

      For concrete, take k = 1.2 W/(mK).          [V.T. U., Jan./Feb. 2003]

5. A hollow cylinder of inner radius r₁ and outer radius r₂ has temperature variation along the radius given by:

   T(r) = 400 – 400. ln(r/r₁). Thermal conductivity of the material, k = 45 W/(mC).

   If r₁ = 5 cm and r₂ = 10 cm, determine the direction and rate of flow of heat at the two surfaces for 1 m length of pipe.

6. A hollow sphere of inner radius r₁ and outer radius r₂ has the temperature along the radius varying as:

   T(r) = 400 + 400. ln (r/r₂). Assume k = 45 W/(mK). If r₁ = 5 cm and r₂ = 10 cm, determine the direction and rate of flow of heat at the two surfaces. Also, find out the average temperature of the sphere.

7. A 5 cm diameter cylindrical rod (k = 15 W/(mC)), with a uniform heat generation rate of q_g (W/m³) inside it, has a radial temperature distribution given by:

   T(r) = 315 – 2.1 × 10⁴ r² where T is in deg. C, r in metres. Determine:

   1. Maximum temperature in the rod
   2. Volumetric rate of heat generation
   3. Average temperature of the cylinder.
8. The steady state radial temperature profile in a 10 cm diameter solid sphere is given by:

   T(r) = 101.4 – 1390 r², where T is in deg. C and r, in metres. Its k = 10 W/(mC). The sphere is placed in an ambient of 30°C.

   1. What is the maximum temperature in the sphere?
   2. Is there heat generation in the sphere? If yes, at what rate?
   3. Calculate the convection coefficient at the outer surface.