What is heat transferred through current

3.1.1 Conduction Heat Transfer

Conduction heat transfer is the transfer of heat by means of molecular excitement within a material without bulk motion of the matter. Conduction heat transfer in gases and liquids is due to the collisions and diffusion of the molecules during their random motion. On the other hand, heat transfer in solids is due to the combination of lattice vibrations of the molecules and the energy transport by free electrons.

To examine conduction heat transfer, it is necessary to relate the heat transfer to mechanical, thermal, or geometrical properties. Consider steady-state heat transfer through the wall of an aorta with thickness Δx where the wall inside the aorta is at higher temperature (Th) compared with the outside wall (Tc). Heat transfer, Q˙(W), is in the direction of x and perpendicular to the plane of temperature difference, as shown in Fig. 3.1.

Heat transfer is a function of the higher and lower temperatures of the aorta wall, and aorta geometry and properties and is given by [1]:

(3.1)Q˙∝(A)(ΔT)Δx

or

(3.2)Q˙=kA(Th−Tc)Δx=−kA(Tc−Th)Δx=−kAΔTΔx

In Eq. (3.2), thermal conductivity (k,WmK)is transport property. Parameter A is the cross-sectional area (m2) of the aorta and Δx is the aorta wall thickness (m). In the limiting case of Δx → 0 Eq. (3.2) reduces to Fourier's law of conduction:

(3.3)Q˙=−kAdTdx

where dTdxis the temperature gradient and must be negative based on the second law of thermodynamics. A more useful quantity to work with is heat flux, q″(Wm2), the heat transfer per unit area:

(3.4)q″=Q˙A

Separating the variables in Eq. (3.3), integrating from x = 0 and rearranging gives:

(3.5)Q˙=kAΔTL

where L is the aorta wall thickness. For steady flow, one-dimensional heat conduction with no shaft work and no mass flow, the first law reduces to:

(3.6)dQ˙=dQ˙(x)dx=0

Combining Eqs. (3.3) and (3.6) gives:

(3.7)ddx(kAdTdx)=0

If properties are assumed constant and by using the chain rule, the energy equation based on temperature is:

(3.8)d2Tdx2+(1AdAdx)dTdx=0

Solving Eq. (3.8) provides the temperature field in a plane wall. Steady flow and one-dimensional heat transfer rate in cylindrical coordinate is:

(3.9)Q˙=−kAdTdr

Separating the variables in Eq. (3.9), integrating from r = 0 and rearranging gives [1]:

(3.10)Q˙=(2πL)kΔTln(ro/rin)

where A = 2πrL, and ro and rin are the outside and inside wall radii.

Thermal Resistance Circuits: For steady one-dimensional flow with no generation of heat conduction equation, Eq. (3.5) can be rearranged as:

(3.11)Q˙=ΔTL/KA=ΔTRcond

Conduction thermal resistance (Rcond) is represented by:

(3.12)Rcond=LKA

It is obvious that the thermal resistance Rcond increases as wall thickness (L) increases, and area (A) and K decrease. The concept of a thermal resistance circuit can be used for problems such as composite wall thickness (see Fig. 3.2).

The heat transfer rate for composite wall is given by:

(3.13)Q˙=ΔT∑Rcond=ΔTR1+R2

5.1.1 Conduction heat transfer

Conduction heat transfer is the transfer of heat by means of molecular excitement within a material without bulk motion of the mater. Conduction heat transfer mainly occurs in solids or stationary mediums such as fluids at rest. For instance, transfer of heat in solids is due to the combination of lattice vibrations of the molecules and the energy transport by free electrons, while in gases and liquids it is due to the collisions and diffusion of the molecules.

To examine conduction heat transfer let us, for instance, look at the steady-state heat transfer rate, Q·(W), through the solid electrolyte layer thickness, Δx, which is a function of the hot fluid temperature, TH, and cold fluid temperature, TC, the geometry and property are given as

(5.1)Q·=f(TH,TC,geometryandproperty)

where the hot fluid, TH, and the cold fluid, TC, temperatures are in absolute Kelvin. It is also possible to express the heat transfer rate based on the hot and cold fluid temperature difference, TH − TC, as

(5.2)Q·=f[(TH–TC),geometryandproperty]

Fourier’s law of heat conduction relates the heat transfer to mechanical, thermal, and geometrical properties of the medium. Fourier has shown that heat transfer rate is proportional to the temperature difference across the solid layer and the heat transfer area and inversely proportional to the solid layer thickness. That is,

(5.3)Heattransferrate∝(Area)(Temperaturedifference)Thickness=(A)(ΔT)Δx

The cross-sectional area, A, is in square meter and the thickness of the slab, Δx, is in meter. The proportionality factor in Eq. (5.3) is replaced by transport property (k) called thermal conductivity (W/mK) which is a scalar property. Therefore, Eq. (5.3) becomes:

(5.4)Q·=kATH−TCΔx=−kATC−THΔx=−kAΔTΔx

Thermal conductivity is the measure of ability of a material to conduct heat. Thermal conductivity is a well-tabulated property for a large number of materials and can be found in the different heat transfer or thermodynamics references.

In the limit, the heat transfer rate equation, Eq. (5.4), for any temperature difference, ΔT, across a slab length, Δx, as both approach

(5.5)Q·cond,n=−kAdTdx

dTdx(Km)is the temperature gradient as shown in Fig. 5.2. The minus sign appearing in the above equation is due to heat transfer and temperature gradient directions are in opposite direction.

By rearranging Eq. (5.5) and comparing with electric current flow, the conduction thermal resistance in Cartesian coordinate, Rcond, is as follows:

(5.6)Rcond=ΔxkA

Conduction thermal resistance, Rcond, is the measure of wall resistance against heat flow. It is obvious that the thermal resistance, Rcond, increases as thickness increases and as surface area and thermal conductivity decreases. Conduction thermal resistance for cylindrical and spherical coordinate is determined from the one-dimensional energy equation in the relative coordinate and are as follows, respectively [1]:

(5.7)Rcond=ln(ro/rin)2πk

(5.8)Rcond=1rin−1ro4πk

where ro and rin are the outside and inside diameters of the cylinder as well as sphere.

The general steady one-dimensional conduction heat transfer equation with no generation is written as

(5.9)1RNddR(RNkdTdR)=0

The general unsteady one-dimensional conduction heat transfer equation with source term is written as

(5.10)1RNddR(RNkdTdR)+q‴=ρCp∂T∂t

where R and N in both Eqs. (5.9) and (5.10) are x and 0 for slab, r and 1 for cylinder, and r and 2 for sphere, respectively. q′′′ (W/m3) is the heat generation, ρ (kg/m3) is density, Cp (kJ/kg·K) is heat capacity, and t (s) is the time. For constant thermos-physical properties Eq. (5.10) becomes

(5.11)1RNddR(RNkdTdR)+q‴=ρCp∂T∂t

where α=kρCp(m2s)is thermal diffusivity.

The rate of conduction heat transfer for isotropic medium is a vector quantity. The general three-dimensional constant properties heat conduction equations for isotropic medium in rectangular (x, y, z), cylindrical (r, φ, z), and spherical (r, φ, θ) coordinates are as follows, respectively:

(5.12)∂2T∂x2+∂2T∂y2+∂2T∂z2+q‴k=1α∂T∂t

(5.13)1r∂∂r(r∂T∂r)+1r2∂∂ϕ(r∂T∂ϕ)+∂2T∂z2+q‴k=1α∂T∂t

(5.14)1r∂∂r(r2∂T∂r)+1r2sin2θ∂2T∂ϕ2+1r2sin2θ∂∂θ(sinθ∂T∂θ)+q‴k=1α∂T∂t

The rate of conduction heat flux for anisotropic medium, q·→(Wm2), is also a vector quantity and in Cartesian coordinate system it is as follows

(5.15)q·→cond=−(kxx∂T∂x+kxy∂T∂y+kxz∂T∂z)iˆ−(kyx∂T∂x+kyy∂T∂y+kyz∂T∂z)jˆ−(kzx∂T∂x+kzy∂T∂y+kzz∂T∂z)kˆ

Since the physical properties of all materials used in different layers of SOFCs do not vary with the direction [2], thermal conductivities of these materials are also scalar quantity and therefore Eq. (5.15) is rewritten as

(5.16)q·→cond=−k∂T∂xiˆ−k∂T∂yjˆ−k∂T∂zkˆ=−k∇T

∇Tis the temperature gradient and is given as

(5.17)∇T=∂T∂xiˆ+∂T∂yjˆ+∂T∂zkˆ

11.1 Introduction

Conduction heat transfer and single-phase convective heat transfer were presented in the previous chapters. Two-phase heat transfer, which involves conversion of liquid into vapor, referred to as boiling, or conversion of vapor into liquid, referred to as condensation, takes place in equipment such as boilers and condensers in thermal power plants, evaporators and condensers in refrigeration systems, water-cooled nuclear reactors, major equipment in process industry, and modern heat sinks for thermal management of electronics. Of particular importance in the design of the equipment are the rates of heat transfer or heat transfer coefficients and the associated pressure losses. In boiling and condensation, the coupling between the fluid dynamic process and the heat transfer process is stronger than what exists in single-phase flows. In a two-phase flow with phase change, there is a continuous variation in the fraction and distribution of each phase and hence the flow pattern, which influences the local heat transfer processes. Therefore, the flow at any axial location in the tube can never be fully developed thermally or hydrodynamically, unlike a single-phase flow. The flow also involves transient properties and deviations from thermodynamic equilibrium. In this chapter, the commonly observed regimes of pool boiling and flow patterns in flow boiling are discussed. Prior to the presentation of the correlations used to predict the heat transfer coefficients and the critical heat flux, the chapter presents a brief discussion on the wall superheat required for nucleation from a heating surface. This is followed by a discussion on the film condensation that occurs on flat plates and horizontal tubes and a presentation of the respective heat transfer coefficient correlations. The chapter ends with an introduction to the prediction of pressure drop in two-phase flows with phase change.

1.3 Conduction

Transfer of heat by conduction occurs in solids and in essentially nonmoving liquids and gases. It has been observed (through experimentation) that the rate of heat transfer per unit area is proportional to the temperature gradient in the material. That is,

(1.1)qnAn∝∂T∂n

where n designates the direction of the heat flow; e.g., x, y, or z in Cartesian coordinates.

In this equation, qn is the rate of heat flow in direction n, An is the cross-sectional area through which the heat flows, and ∂T/∂n is the temperature gradient in direction n. (Note: The area A is the area perpendicular to the heat flow.)

If we introduce a proportionality parameter into Eq. (1.1) and introduce a minus sign so that heat flow in the positive n direction will be numerically positive, we get Fourier's law [1,2]:

(1.2)qnAn=−k∂T∂n

Eq. (1.2), named for French mathematician and physicist Joseph Fourier (1768–1830), gives the heat flow rate per unit area in the n direction at a point in the solid or fluid. Heat flow rate per unit area is called “heat flux.” Heat flow is a vector quantity which has direction and magnitude. If we are using Cartesian coordinates, there can be heat flow and heat flux components in all three coordinate directions. The heat flow components are qx, qy, and qz. The heat flux components are

(1.3)(qA)x=−k∂T∂x(qA)y=−k∂T∂y(qA)z=−k∂T∂z

where k is the thermal conductivity of the material. If we are using the SI unit system where q is watts, the area is m2, and the temperature gradient is °C/m, then k has the units of W/m °C. (Note: As the size of a Celsius degree is the same as the size of a Kelvin degree, the numerical value of k in W/m °C is the same as its value in W/m K. That is, 1 W/m °C = 1 W/m K.)

[Note: In this text, we will usually leave out the symbol used for a temperature degree. That is, we will often use W/m C rather than W/m °C for the units of thermal conductivity. And, when we are talking about a temperature of 20 degrees Celsius, we will use 20 C rather than 20°C.]

Thermal conductivity is highest for pure metals and a bit lower for alloys. The conductivity for liquids is generally lower (except for liquid metals) and for gases even lower. Building materials such as wood, plaster, and insulation have low conductivity. Multilayer evacuated insulation used for insulating cryogenic tanks has very low conductivity.

For most materials, thermal conductivity is isotropic; that is, it is the same in all directions. However, for some materials, it varies with direction. For example, although a single average k value is often given for wood, the conductivity of wood is actually different in the across-the-grain and with-the-grain directions.

Table 1.1 gives typical thermal conductivity values for some materials at about 20 C. Appendices A through E give detailed information on the properties of common solids, liquids, and gases.

Table 1.1. Typical thermal conductivity at 20 C.

Materialk (W/m C)Aluminum240Copper400Gold315Silver430Carbon steel40Stainless steel15Plasterboard0.8Brick0.7Cement (hardened)1.0Hardwoods0.16Softwoods0.12Styrofoam0.03Water0.60Engine oil0.14Air (1 atm pressure)0.025He (1 atm pressure)0.15

In general, thermal conductivity decreases with temperature for metals, decreases slightly with temperature for many liquids, and increases with temperature for gases.

1.3.1 Conduction through a plane wall

In this section we develop the equation for the rate of heat flow through a plane wall. The equation is very useful for estimating heat flow through large walls of finite thickness, such as the exterior walls of a building. It is also useful in other situations where the heat flow can be considered to be one-dimensional; that is, the heat flow is in a single direction.

Consider the plane wall shown in Fig. 1.1. The wall has a thickness L and is very large in the y and z directions. The left face of the wall is at temperature T1 and is at location x = 0. The right face of the wall is at temperature T2 and is at location x = L.

Figure 1.1. Heat conduction through a plane wall.

Let us assume that T1 > T2. The heat flow will be one-dimensional and will be in the positive x direction.

As all the heat flow is in the x direction, Eq. (1.2) can be modified by changing “n” to “x” (or just simply use “q”) and changing the partial derivative to a total derivative. That is,

(1.4)qA=−kdTdx

Moving dx to the left-hand side of the equation, we can then integrate both sides of the equation:

(1.5)qA∫0Ldx=−∫T1T2kdT

(1.6)qAL=∫T2T1kdT

In general, the conductivity k is a function of temperature. However, if k is constant or if we use an average k value for the problem, then k may be taken outside of the integral sign in Eq. (1.6), and we have

(1.7)qAL=k(T1−T2)

Rearranging this equation, we reach its final form:

(1.8)q=kAL(T1−T2)

Details of conduction heat transfer are given in Chapters 2–4.

Example 1.1

Heat flow through a plane wall

Problem

A large concrete wall is 300 mm thick and has a thermal conductivity of 1.2 W/m C. The heat flux through the wall is 100 W/m2. The higher-temperature surface of the wall is at 42 C. What is the temperature of the other surface of the wall?

Solution

Heat flux is heat flow per unit area, or q/A. From Eq. (1.8), we have

q=kAL(T1−T2)

We want to obtain T2. Rearranging the equation, we get

T2=T1−(qA)(Lk)=42−(100)(0.31.2)=17C

The temperature of the cooler surface of the wall is 17 C.

4.13.1 Conduction

The basic equation of conduction heat transfer is Fourier's law:

(4.73)Q˙cond=−ktA(dTdx)

where Q˙condis the conduction heat transfer rate, kt is the thermal conductivity of the material, A is the cross-sectional area normal to the heat transfer direction, and dT/dx is the temperature gradient in the direction of heat transfer. The algebraic sign of this equation is such that a positive Q˙condalways corresponds to heat transfer in the positive x direction, and a negative Q˙condalways corresponds to heat transfer in a negative x direction. Since this is not the same sign convention adopted earlier in this text, the sign of the values calculated from Fourier's law may have to be altered to produce a positive when it enters a system and a negative when it leaves a system.

For steady conduction heat transfer through a plane wall (Figure 4.20), Fourier's law can be integrated to give

(4.74)(Q˙cond)plane=−ktA(T2−T1x2−x1)

and for steady conduction heat transfer through a hollow cylinder of length L, Fourier's law can be integrated to give

(4.75)(Q˙cond)cylinder=−2πLkt[Tinside−Toutside ln (rinside/routside]

Table 4.7 gives thermal conductivity values for various materials.

Table 4.7. Thermal Conductivity of Various Materials

MaterialThermal Conductivity ktTemperature (°C/°F)Btu/(h · ft · R)W/(m · K)Air (14.7 psia)27/810.0150.026Hydrogen (14.7 psia)27/810.1050.182Saturated water vapor (14.7 psia)100/2120.0140.024Saturated liquid water (14.7 psia)0/320.3430.594Engine oil20/680.0840.145Mercury20/685.028.69Window glass20/680.450.78Glass wool20/680.0220.038Aluminum (pure)20/68118.0204.0Copper (pure)20/68223.0386.0Carbon steel (1% carbon)20/6825.043.0

12.4.6 Influence of pipe embedment

Reducing the pipe embedment facilitates the conduction heat transfer between the heat carrier fluid and the tunnel air, thus minimising the conduction heat transfer resistance of the lining (Cousin et al., 2019). This influence increases with a successive decrease of the pipe embedment and an increase of the Reynolds number, because the latter reduces the convection thermal resistance in the pipes.

In the following, the influence of the pipe embedment on the thermohydraulic behaviour of energy tunnels is expanded with reference to the results of Cousin et al. (2019). Values of pipe embedment from the tunnel intrados of si/tl=0.5, 0.625 and 0.75, where siis the distance from the tunnel intrados and tlis the lining thickness, are considered. A graphical representation of the considered pipe embedment is provided in Fig. 12.30.

Fig. 12.31 presents an example of thermal power harvested per unit surface of tunnel lining, considering three different values of pipe embedment and three different flow rates. A tunnel lining characterised by a pipe configuration with 20 mm diameter pipes installed perpendicular to the tunnel axis after 16 days of geothermal operation is considered. Examples of percentages of harvested thermal power variation for a change in the pipe embedment are provided in Table 12.7 for varying values of Reynolds numbers. Decreasing the pipe embedment markedly improves the harvested thermal power.

Table 12.7. Influence of the pipe embedment on the extracted thermal power through an energy tunnel equipped with a pipe configuration involving 20 mm diameter pipes installed perpendicular to the tunnel axis, for three Reynolds number, after 16 days of geothermal operation.

Reynolds number, Re[–]Pipe embedment, si/tl[–]0.75→0.6250.75→0.56000+8.45%+17.48%9000+9.04%+18.85%12,000+9.30%+19.46%

Source: Data from Cousin, B., Rotta Loria, A.F., Bourget, A., Rognon, F., Laloui, L., 2019. Energy performance and economic feasibility of energy segmental linings for subway tunnels. Tunn. Undergr. Space Technol. 91, 102997.

INTRODUCTION

In this chapter, we discuss conduction heat transfer in a porous medium. For steady heat conduction under local thermal equilibrium conditions it will be shown that the governing equation for heat conduction in a porous medium can be rewritten in a form similar to that of the classical heat conduction equation with an effective stagnant thermal conductivity consisting of two components: the first component represents the volumetric averaging of the thermal conductivities of the solid and the fluid phases while the second component represents the tortuosity effect due to the undulating thermal path across the fluid-solid interface. Exact numerical evaluations of the thermal conductivities of some idealized two-dimensional, spatially periodic media will be obtained based on closure modeling, or solving the two-dimensional conjugate heat conduction problem in the fluid and solid phase of a unit cell. Various analytical models for the calculation of the stagnant thermal conductivity will be reviewed.

Thermal Conduction

The basic equation for the rate of steady conduction heat transfer is called Fourier's Law of Conduction. For steady conduction heat transfer through a flat plate or a plane wall (Figure 12.3), Fourier's Law is

(12.1)Q˙condplane=ktAThot−TcoldΔx

where Q˙condis the conduction heat transfer rate, kt is the thermal conductivity of the material (Table 12.1), A is the cross-sectional area perpendicular to the heat transfer direction, and Δx is the thickness of the plate or wall. The algebraic sign of this equation is such that a positive Q˙condalways corresponds to heat transfer from the hot to the cold side of the plane or wall.

Table 12.1. Thermal Conductivity of Various Materials

MaterialThermal Conductivity ktBtu⁎/(h·ft·R)W/(m·K)Air at atmospheric pressure0.0150.026Hydrogen gas at atmospheric pressure0.1050.182Liquid water0.3430.594Engine oil0.0840.145Mercury5.028.69Window glass0.450.78Glass wool0.0220.038Aluminum118.0204.0Copper223.0386.0Carbon steel (1% carbon)25.043.0

Example 12.1

The glass in the window of your dorm room is 2.00 ft wide, 3.00 ft high, and ⅛ inch thick. If the temperature of the outside surface of the glass is 21.0 °F and the temperature of the inside surface of the glass is 33.0 °F, what is the rate of heat loss through the window?

Need:Q˙cond= ?

Know:Thot = Tinside = 33.0 °F, Tcold = Toutside = 21.0 °F, A = 2.00 ft × 3.00 ft = 6.00 ft2, and Δx = 1/8 in × [1 ft/12 in] = 0.0104 ft.

Q˙condloss=ktAThot−TcoldΔx=0.45Btuhr⋅ft⋅R×6.00ft233.0−21.00.0104Rft=3120Btu/hr

Note that the temperature difference, Thot –Tcold, has the same numerical value regardless of whether Fahrenheit or Rankine temperatures are used, because

Thot−Tcold=460+33.0R−460+21.0R=33.0−21.0=12.0R or°F.

Edge Cooling

If a fuel cell flow field is made narrow enough, the heat generated may be removed on the sides of the cells instead of the more conventional way between the cells. In this case, heat is conducted in the plane of the bipolar plate rather than through it. Active cooling may still be needed at the edge of the bipolar plates. To enhance heat transfer, fins may be added and/or high thermal conductivity material may be used.

Obviously, in this case the maximum temperature will be achieved in the center of the flow field. In a narrow and long flow field, the heat transfer may be approximated as one-dimensional (i.e., heat removed through the long sides is significantly larger than the heat removed at the narrow sides).

The equation that describes one-dimensional heat transfer (conduction) in a flat plane with internal heat generation is:

(6-70)d2Tdx2+QkAdBPeff=0

where:

Q = heat generated in the cell (either given by Equation 6-65 or by a detailed energy balance analysis), W

k = bipolar plate in-plane thermal conductivity (in some cases, in-plane conductivity may be significantly different from through-plane thermal conductivity), Wm−1 K−1

A = cell active area, m2

deffBP = effective (or average) thickness of the bipolar plate in the active area, m

Although the heat is not actually generated inside the plate but rather in a thin catalyst layer above the plate, Equation (6-70) may be used with sufficient accuracy because the thickness of the plates is very small compared with the width (i.e., a few millimeters vs. a few centimeters). In that case, the solution of Equation (6-70) for symmetrical cooling on both sides with T(0) = T(L) = T0 is:

(6-71)T−T0=QkAdBPeffL22[xL−(xL)2]

where:

T0 = the temperature at the edge of the active area

L = width of the active area

The maximum temperature difference is between the edge (x = 0 or x = L) and the center (x = L/2):

(6-72)ΔTmax=QkAdBPeffL28

In addition, the heat must be conducted from the edge of the flow field to the edge of the plate or to the fin over the flow field border with width b. The thickness of the plate at the border is dBP. For this reason, the temperature will further decrease, according to Fourier's law:

(6-73)T0−Tb=Q2kALdbb

where Tb is the temperature at the edge of the bipolar plate.

Therefore, the total temperature difference between the center of the plate and the edge of the plate, or the base of a fin, is:

When heat is transferred through empty space it is called?

What is heat transfer called?

How do you find heat from current?