7 Electrical Properties This chapter covers the principles and applications of electrical behavior,empha- sizing those related to composite materials.This includes thermoelectric behavior and the effects of temperature and stress on electrical resistivity,as well as ap- plications such as thermoelectric energy generation,thermocouples,thermistors, heating (such as deicing),and electrical resistance-based sensing of strain,stress and damage.The materials used to make electrical connections are also covered. 7.1 Origin of Electrical Conduction Unless noted otherwise,an electrically conductive material utilizes electrons and/or holes as the charge carriers(mobile charged particles)for electrical con- duction.A hole is an electron-deficient site. In the presence ofan electric field(i.e.,a voltage gradient),a hole drifts toward the negative end of the voltage gradient due to electrostatic attraction,thus resulting in a current(defined as the charge per unit time)in the same direction.In the presence of an electric field,an electron drifts toward the positive end of the voltage gradient,thus resulting in a current(defined as the charge per unit time- not the magnitude of the charge per unit time)in the opposite direction.Hence,the currents resulting from electron drift and hole drift occur in the same direction, even though electrons and holes drift in opposite directions.This means that in a material with both holes and electrons,the total current is the sum of the current due to the holes and that due to the electrons.Drift is actually a term that refers to the movement of charge in response to an applied electric field. The drift velocity(v)is the velocity of the drift.It is actually the net velocity,since electrons or holes are quantum particles that obey the Heisenberg Uncertainty Principle,and so they constantly move,even in the absence of an applied electric field.The drift velocity is proportional to the applied electric field E,with the constant of proportionality being the mobility(),which describes the ease of movement of the carrier in the medium under consideration.Hence, v=uE. (7.1) Since the units of E are V/m and the units of v are m/s,the units of u are m2/(V s). For a given combination of carrier and medium,v depends on the temperature. 203
7 Electrical Properties This chapter covers the principles and applications of electrical behavior, emphasizing those related to composite materials. This includes thermoelectric behavior and the effects of temperature and stress on electrical resistivity, as well as applications such as thermoelectric energy generation, thermocouples, thermistors, heating (such as deicing), and electrical resistance-based sensing of strain, stress and damage. The materials used to make electrical connections are also covered. 7.1 Origin of Electrical Conduction Unless noted otherwise, an electrically conductive material utilizes electrons and/or holes as the charge carriers (mobile charged particles) for electrical conduction. A hole is an electron-deficient site. Inthepresenceofanelectricfield(i.e.,avoltagegradient),aholedriftstowardthe negative end of the voltage gradient due to electrostatic attraction, thus resulting in a current (defined as the charge per unit time) in the same direction. In the presence of an electric field, an electron drifts toward the positive end of the voltage gradient, thus resulting in a current (defined as the charge per unit time – not the magnitude of the charge per unit time) in the opposite direction. Hence, the currents resulting from electron drift and hole drift occur in the same direction, even though electrons and holes drift in opposite directions. This means that in a material with both holes and electrons, the total current is the sum of the current due to the holes and that due to the electrons. Drift is actually a term that refers to the movement of charge in response to an applied electric field. The drift velocity (v) is the velocity of the drift. It is actually the net velocity, since electrons or holes are quantum particles that obey the Heisenberg Uncertainty Principle, and so they constantly move, even in the absence of an applied electric field. The drift velocity is proportional to the applied electric field E, with the constant of proportionality being the mobility (μ), which describes the ease of movement of the carrier in the medium under consideration. Hence, v = μE . (7.1) Since the units of E are V/m and the units of v are m/s, the units of μ are m2/(Vs). For a given combination of carrier and medium, v depends on the temperature. 203
204 7 Electrical Properties 7.2 Volume Electrical Resistivity The electric current(or simply "current,"I,with units of ampere,or A)is defined as the charge (in coulomb,or C)flowing through the cross-section per unit time. It is the charge,not the magnitude of charge.Thus,the direction of the current is the same as the direction of positive charge flow and is opposite to the direction of negative charge flow. The current density is the charge flow per unit cross-sectional area per unit time.The units of current density are A/m2. The presence of a current in a material requires the presence of charges that can move in response to the applied voltage.These mobile charges are called carriers (i.e.,carriers ofelectricity).The carrier concentration(with units ofm3)is defined as the number of carriers per unit volume. An electrical conductor may be an electronic conductor(electrons are the main carriers),an ionic conductor (ions are the main carriers),or a mixed conductor (both ions and electrons are the main carriers).In this context,electrons constitute a class ofcarriers that include both electrons and holes.Holes are electron-deficient sites that are present in semiconductors and semimetals,but not in conventional metals.Holes are positively charged,since the removal of a(negatively charged) electron leaves something that is positively charged. The current due to mobile charges with a charge g per carrier is given by I=nqvA, (7.2) where n is the carrier concentration.When the carriers are electrons,q=-1.6 x 10-1C.Equation 7.2 derives from the simple argument that each electron drifts by a distance v in one second,and that nvA electrons (the number of electrons in a volume of vA)move through a particular cross-section in one second,as illustrated in Fig.7.1.Dividing Eq.7.2 by A gives I/A nqv. (7.3) Area A Current Figure 7.1.An electric current in a wire of cross-sectional area A.The current results from the drift of a type of charge carrier with a charge g per carrier at a drift velocity of v
204 7 Electrical Properties 7.2 Volume Electrical Resistivity The electric current (or simply “current,” I, with units of ampere, or A) is defined as the charge (in coulomb, or C) flowing through the cross-section per unit time. It is the charge, not the magnitude of charge. Thus, the direction of the current is the same as the direction of positive charge flow and is opposite to the direction of negative charge flow. The current density is the charge flow per unit cross-sectional area per unit time. The units of current density are A/m2. The presence of a current in a material requires the presence of charges that can move in response to the applied voltage. These mobile charges are called carriers (i.e., carriers of electricity). The carrier concentration (with units of m−3) is defined as the number of carriers per unit volume. An electrical conductor may be an electronic conductor (electrons are the main carriers), an ionic conductor (ions are the main carriers), or a mixed conductor (both ions and electrons are the main carriers). In this context, electrons constitute a class of carriers that include both electrons and holes. Holes are electron-deficient sites that are present in semiconductors and semimetals, but not in conventional metals. Holes are positively charged, since the removal of a (negatively charged) electron leaves something that is positively charged. The current due to mobile charges with a charge q per carrier is given by I = nqvA , (7.2) where n is the carrier concentration. When the carriers are electrons, q = −1.6 × 10−19 C. Equation 7.2 derives from the simple argument that each electron drifts by a distance v in one second, and that nvA electrons (the number of electrons in a volume of vA) move through a particular cross-section in one second, as illustrated in Fig. 7.1. Dividing Eq. 7.2 by A gives I/A = nqv . (7.3) v Current Area A Figure 7.1. An electric current in a wire of cross-sectional area A. The current results from the drift of a type of charge carrier with a charge q per carrier at a drift velocity of v
7.2 Volume Electrical Resistivity 205 The current density (J,with units of A/m2)is defined as the current(I)per unit cross-sectional area (A): J=I/A. (7.4) Hence,Eq.7.3 can be written as J=nqv. (7.5) Dividing Eq.7.5 by the electric field E and using Eq.7.1,we get J/E=nqp· (7.6) The electrical conductivity (o)is defined as J/E.In other words,it is defined as the current density per unit electric field.Hence,Eq.7.6 becomes o=nqμ. (7.7) From Eq.7.7,the units of o are 1/(m),i.e.,m-,since the units of n arem, the units of g are C(coulomb),and the units of u are m2/Vs).Alternate units for o are Sm-,where S(short for siemens)equals-1.Note that,by definition, ampere coulomb/second (7.8) and,from Ohm's law, volt/ampere. (7.9) The electrical resistivity (o)is defined as 1/o.Hence,the units of e are m.The electrical resistivity is also known as the volume electrical resistivity in order to emphasize that it relates to the property of a volume of the material.It is also known as the specific resistance. The electrical resistivity ois related to the electricalresistance(R)by theequation R=ol/A, (7.10) where I is the length of the material in the direction of the current.This length is perpendicular to the cross-sectional area A.Equation 7.10 means that the re- sistance R depends on the geometry such that it is proportional to l and inversely proportional to A.On the other hand,the resistivity o is independent of the geome- try and is thus a material property.Similarly,o is a material property.Equation 7.7 allows the calculation of o from n,q and u. The rearrangement of Eq.7.10 gives o=RA/I. (7.11)
7.2 Volume Electrical Resistivity 205 The current density (J, with units of A/m2) is defined as the current (I) per unit cross-sectional area (A): J = I/A . (7.4) Hence, Eq. 7.3 can be written as J = nqv . (7.5) Dividing Eq. 7.5 by the electric field E and using Eq. 7.1, we get J/E = nqμ . (7.6) The electrical conductivity (σ) is defined as J/E. In other words, it is defined as the current density per unit electric field. Hence, Eq. 7.6 becomes σ = nqμ . (7.7) From Eq. 7.7, the units of σ are 1/(Ωm), i.e., Ω−1 m−1, since the units of n are m−3, the units of q are C (coulomb), and the units of μ are m2/(Vs). Alternate units for σ are Sm−1, where S (short for siemens) equals Ω−1. Note that, by definition, ampere = coulomb/second (7.8) and, from Ohm’s law, Ω = volt/ampere . (7.9) The electrical resistivity (ρ) is defined as 1/σ. Hence, the units of ρ are Ωm. The electrical resistivity is also known as the volume electrical resistivity in order to emphasize that it relates to the property of a volume of the material. It is also known as the specific resistance. Theelectricalresistivity ρ isrelatedtotheelectricalresistance(R)bytheequation R = ρl/A , (7.10) where l is the length of the material in the direction of the current. This length is perpendicular to the cross-sectional area A. Equation 7.10 means that the resistance R depends on the geometry such that it is proportional to l and inversely proportional toA. On the other hand, the resistivity ρ is independent of the geometry and is thus a material property. Similarly, σ is a material property. Equation 7.7 allows the calculation of σ from n, q and μ. The rearrangement of Eq. 7.10 gives ρ = RA/l . (7.11)
206 7 Electrical Properties Hence, O=V/RA. (7.12) Since,by definition,o =J/E, J/E V/RA. (7.13) Based on Eq.7.4,Eq.7.13 becomes I/E=VR. (7.14) Since,by definition,E V/I,Eq.7.14 becomes V=IR, (7.15) which is known as Ohm's law.Thus,the equation o =J/E is the same as Ohm's law. 7.3 Calculating the Volume Electrical Resistivity of a Composite Material The volume electrical resistivity of a composite material can be calculated from the resistivities and volume fractions of all of the components.Various mathematical models are available for this calculation.The simplest model is known as the rule of mixtures(ROM),as described below for two configurations:the parallel configuration(each component is continuous and oriented in the direction of the current)and the series configuration(each component is continuous and oriented in the direction perpendicular to the current). 7.3.1 Parallel Configuration Consider an electric current I flowing in a composite material under a voltage difference of V over a distance of l,such that the composite material consists of two components,labeled 1 and 2,as illustrated in Fig.5.16a.The cross-sectional areas are A and Az for Component 1 (all the strips of Component 1 together)and Component 2(all the strips of Component 2 together),respectively.The electrical resistivities are and for Components 1 and 2,respectively.The current(ie., the charge per unit time)I through the composite is given by I=1+12, (7.16) where li is the current in Component 1(all the strips of Component 1 together) and I2 is the current in Component 2(all the strips of Component 2 together). Using Ohm's law, I1=V/R (7.17)
206 7 Electrical Properties Hence, σ = l/RA . (7.12) Since, by definition, σ = J/E, J/E = l/RA . (7.13) Based on Eq. 7.4, Eq. 7.13 becomes I/E = l/R . (7.14) Since, by definition, E = V/l, Eq. 7.14 becomes V = IR , (7.15) which is known as Ohm’s law. Thus, the equation σ = J/E is the same as Ohm’s law. 7.3 Calculating the Volume Electrical Resistivity of a Composite Material The volume electrical resistivity of a composite material can be calculated from the resistivities and volume fractions of all of the components. Various mathematical models are available for this calculation. The simplest model is known as the rule of mixtures (ROM), as described below for two configurations: the parallel configuration (each component is continuous and oriented in the direction of the current) and the series configuration (each component is continuous and oriented in the direction perpendicular to the current). 7.3.1 Parallel Configuration Consider an electric current I flowing in a composite material under a voltage difference of V over a distance of l, such that the composite material consists of two components, labeled 1 and 2, as illustrated in Fig. 5.16a. The cross-sectional areas are A1 and A2 for Component 1 (all the strips of Component 1 together) and Component 2 (all the strips of Component 2 together), respectively. The electrical resistivities are ρ1 and ρ2 for Components 1 and 2, respectively. The current (i.e., the charge per unit time) I through the composite is given by I = I1 + I2 , (7.16) where I1 is the current in Component 1 (all the strips of Component 1 together) and I2 is the current in Component 2 (all the strips of Component 2 together). Using Ohm’s law, I1 = V/R1 (7.17)
7.3 Calculating the Volume Electrical Resistivity of a Composite Material 207 and 2=V/R2, (7.18) where R is the resistance of Component 1(all the strips of Component I together) and R2 is the resistance of Component 2(all the strips of Component 2 together). Hence,Eq.7.16 becomes I=V[(1/R)+(1/R2)月. (7.19) Based on Eq.7.10, R1=PlA1, (7.20) and R2=el/A2. (7.21) Based on Eqs.7.20 and 7.21,Eq.7.19 becomes I=(V/0[(A1/e)+(A2/e2)】. (7.22) On the other hand,from Ohm's law, I=V/R, (7.23) where R is the resistance of the composite.Based on Eq.7.10,R is given by R=ol/A, (7.24) where p is the resistivity of the composite and A is the total cross-sectional area of the composite.Combining Eqs.7.23 and 7.24 gives I=VA/el. (7.25) Combining Eqs.7.22 and 7.25 gives A/el=(V/0[(A/e)+(A2/e2】, (7.26) or A/e=[(A1/e1)+(A2/e2】. (7.27) Dividing by A gives 1/e=(A/A)1/e)+(A2/A)(1/e2)=v/e1+v2/P2, (7.28) where v and v2 are the volume fractions of Components 1 and 2,respectively. Equation 7.28 implies that,for the parallel configuration,the reciprocal of the resistivity of the composite is the weighted average of the reciprocal of the resistiv- ities of the two components,where the weighting factors are the volume fractions of the two components.In other words, 0=y101+V202, (7.29) where o is the electrical conductivity of the composite material,and o and o2 are the conductivities of Components 1 and 2,respectively.Equation 7.29 is a mani- festation of the rule of mixtures
7.3 Calculating the Volume Electrical Resistivity of a Composite Material 207 and I2 = V/R2 , (7.18) where R1 is the resistance of Component 1 (all the strips of Component 1 together) and R2 is the resistance of Component 2 (all the strips of Component 2 together). Hence, Eq. 7.16 becomes I = V [(1/R1) + (1/R2)] . (7.19) Based on Eq. 7.10, R1 = ρ1l/A1 , (7.20) and R2 = ρ2l/A2 . (7.21) Based on Eqs. 7.20 and 7.21, Eq. 7.19 becomes I = (V/l)[(A1/ρ1)+(A2/ρ2)] . (7.22) On the other hand, from Ohm’s law, I = V/R , (7.23) where R is the resistance of the composite. Based on Eq. 7.10, R is given by R = ρl/A , (7.24) where ρ is the resistivity of the composite and A is the total cross-sectional area of the composite. Combining Eqs. 7.23 and 7.24 gives I = VA/ρl . (7.25) Combining Eqs. 7.22 and 7.25 gives VA/ρl = (V/l)[(A1/ρ1)+(A2/ρ2)] , (7.26) or A/ρ = [(A1/ρ1)+(A2/ρ2)] . (7.27) Dividing by A gives 1/ρ = (A1/A)(1/ρ1)+(A2/A)(1/ρ2) = v1/ρ1 + v2/ρ2 , (7.28) where v1 and v2 are the volume fractions of Components 1 and 2, respectively. Equation 7.28 implies that, for the parallel configuration, the reciprocal of the resistivity of the composite is the weighted average of the reciprocal of the resistivities of the two components, where the weighting factors are the volume fractions of the two components. In other words, σ = v1σ1 + v2σ2 , (7.29) where σ is the electrical conductivity of the composite material, and σ1 and σ2 are the conductivities of Components 1 and 2, respectively. Equation 7.29 is a manifestation of the rule of mixtures
208 7 Electrical Properties Example Problem 1.Calculate the volume electrical resistivity of a composite material consisting of three components that are in parallel electrically.The resistivity required is the resistivity in the direction in which the components are parallel.Component 1 has a volume electrical resistivity of 1.4 x 10-3cm and a volume fraction of 0.35.Component 2 has a volume electrical resistivity of 4.4 x 10-4cm and a volume fraction of 0.25.Component 3 has a volume electrical resistivity of 9.5 x 10-4cm and a volume fraction of 0.40. Solution: From Eq.7.28, 1/resistivity T=0.35/(1.4×10-32cm)+0.25/(4.4×10-42cm) +0.40/(9.5×10-42cm) =1.24×103(2cm)-1, Resistivity =8.1 x 10cm. 7.3.2 Series Configuration Consider an electric current I flowing in a composite material under a voltage difference of V over a distance of l,such that the composite material consists of two components,labeled I and 2,as illustrated in Fig.7.2b.The lengths are h and k2 for Component 1(all the strips of Component 1 together)and Component 2(all 1 2 Current I ,, 2 2 Current I b Figure 7.2.An electrically conducting composite material.The composite consists of Component 1(dotted regions)and Component 2(whiteregions).a Components 1 and 2 of the composite in parallel,bcomponents 1 and 2 of the composite in series
208 7 Electrical Properties Example Problem 1. Calculate the volume electrical resistivity of a composite material consisting of three components that are in parallel electrically. The resistivity required is the resistivity in the direction in which the components are parallel. Component 1 has a volume electrical resistivity of 1.4 × 10−3 Ωcm and a volume fraction of 0.35. Component 2 has a volume electrical resistivity of 4.4×10−4 Ωcm and a volume fraction of 0.25. Component 3 has a volume electrical resistivity of 9.5 × 10−4 Ωcm and a volume fraction of 0.40. Solution: From Eq. 7.28, 1/resistivity T =0.35/(1.4 × 10−3 Ωcm) + 0.25/(4.4 × 10−4Ωcm) + 0.40/(9.5 × 10−4 Ωcm) =1.24 × 103 (Ωcm)−1 , Resistivity =8.1 × 10−4 Ωcm . 7.3.2 Series Configuration Consider an electric current I flowing in a composite material under a voltage difference of V over a distance of l, such that the composite material consists of two components, labeled 1 and 2, as illustrated in Fig. 7.2b. The lengths are l1 and l2 for Component 1 (all the strips of Component 1 together) and Component 2 (all Figure 7.2. An electrically conducting composite material. The composite consists of Component 1 (dotted regions) and Component 2 (white regions). a Components 1 and 2 of the composite in parallel, b components 1 and 2 of the composite in series
7.4 Contact Electrical Resistivity 209 the strips of Component 2 together),respectively.The electrical resistivities are o and ez for Components 1 and 2,respectively. The voltage drop V is given by V=V1+V2, (7.30) where Vi is the voltage drop in Component 1 (all the strips of Component 1 together)and V2 is the voltage drop in Component 2(all the strips of Component 2 together).From Ohm's law,and using the fact that the two components have the same cross-sectional area A and the same current I, V=IR Iol/A, (7.31) V1 IR1 Ieh/A, (7.32) and V2 IR2 Ieb/A. (7.33) Combining Eqs.7.30,7.31,7.32 and 7.33 gives Iel/A Ioh/A+Iob/A, (7.34) or ol oih +ok. (7.35) Dividing by I gives P=P1(l/+e2(l/)=Q1M+P22, (7.36) where vi and v2 are the volume fractions of Components 1 and 2,respectively. Equation 7.36 implies that,for the series configuration,the resistivity of the com- posite material is the weighted average of the resistivities of the two components, where the weighting factors are the volume fractions of the two components. Equation 7.36 is a manifestation of the rule of mixtures. 7.4 Contact Electrical Resistivity An interface is associated with an electrical resistance in the direction perpendic- ular to the interface.The contact electrical resistance refers to the resistance of an interface between two objects(or between two components in a composite mate- rial)when a current is passed across the interface in the direction perpendicular to the interface.When the interface has air voids,for example,the contact resistance will be relatively high.This is because air is an electrical insulator and the objects themselves are more conductive than air.When the interface has an impurity that is less conductive than the objects themselves,the contact resistance will also be
7.4 Contact Electrical Resistivity 209 the strips of Component 2 together), respectively. The electrical resistivities are ρ1 and ρ2 for Components 1 and 2, respectively. The voltage drop V is given by V = V1 + V2 , (7.30) where V1 is the voltage drop in Component 1 (all the strips of Component 1 together) and V2 is the voltage drop in Component 2 (all the strips of Component 2 together). From Ohm’s law, and using the fact that the two components have the same cross-sectional area A and the same current I, V = IR = Iρl/A , (7.31) V1 = IR1 = Iρ1l1/A , (7.32) and V2 = IR2 = Iρ2l2/A . (7.33) Combining Eqs. 7.30, 7.31, 7.32 and 7.33 gives Iρl/A = Iρ1l1/A + Iρ2l2/A , (7.34) or ρl = ρ1l1 + ρ2l2 . (7.35) Dividing by l gives ρ = ρ1(l1/l) + ρ2(l2/l) = ρ1v1 + ρ2v2 , (7.36) where v1 and v2 are the volume fractions of Components 1 and 2, respectively. Equation 7.36 implies that, for the series configuration, the resistivity of the composite material is the weighted average of the resistivities of the two components, where the weighting factors are the volume fractions of the two components. Equation 7.36 is a manifestation of the rule of mixtures. 7.4 Contact Electrical Resistivity An interface is associated with an electrical resistance in the direction perpendicular to the interface. The contact electrical resistance refers to the resistance of an interface between two objects (or between two components in a composite material) when a current is passed across the interface in the direction perpendicular to the interface. When the interface has air voids, for example, the contact resistance will be relatively high. This is because air is an electrical insulator and the objects themselves are more conductive than air. When the interface has an impurity that is less conductive than the objects themselves, the contact resistance will also be
210 7 Electrical Properties relatively high.Thus,the contact resistance is highly sensitive to the condition of the interface. The contact resistance also depends on the area of the interface.The larger the area,the lower the contact resistance.Hence,the contact resistance Rc is inversely proportional to the interface area A,and the relationship between these two quantities can be written as Rc Pc/A, (7.37) where the proportionality constant Pc is known as the contact electrical resistivity. This quantity does not depend on the area of the interface;it only reflects the condition of the interface.The units of c are cm2,which are different from those of the volume resistivity e,cm. 7.5 Electric Power and Resistance Heating 7.5.1 Scientific Basis The electric power p associated with the flow of a current I under a voltage difference V is given by P=VI. (7.38) This power is dissipated as heat,thus allowing a form ofheating known as resistance heating.Applications of this include the deicing of aircraft and bridges.This effect, in which electrical energy is converted to thermal energy,is known as the Joule effect.The units of P are watts(W),with watt volt x ampere.Based on Eq.7.15, Eq.7.38 can be expressed as P=PR (7.39) and as P=V2/R. (7.40) Therefore,in order to obtain a high value for the power,both V and I should not be too small(Eq.7.38).The values of V and I depend on the resistance R.When R is small,V is small since V IR.When R is large,I is small since I V/R. Thus,an intermediate value of R is optimal for obtaining a high power.It is R rather o that governs P.Therefore,for a given material(i.e.,a given value of e),the dimensions can be chosen to obtain a particular value of R(Eq.7.10).Since the current direction does not affect the resistance heating,the heating can be carried out using DC or AC electricity. Consider an example in which R 1 MS (i.e.,10)and V 120V(which is a common voltage from an electrical outlet).Hence, I=V/R=(120V)/1×102)=1.2×10-4A=0.12mA, (7.41)
210 7 Electrical Properties relatively high. Thus, the contact resistance is highly sensitive to the condition of the interface. The contact resistance also depends on the area of the interface. The larger the area, the lower the contact resistance. Hence, the contact resistance Rc is inversely proportional to the interface area A, and the relationship between these two quantities can be written as Rc = ρc/A , (7.37) where the proportionality constant ρc is known as the contact electrical resistivity. This quantity does not depend on the area of the interface; it only reflects the condition of the interface. The units of ρc are Ωcm2, which are different from those of the volume resistivity ρ, Ωcm. 7.5 Electric Power and Resistance Heating 7.5.1 Scientific Basis The electric power P associated with the flow of a current I under a voltage difference V is given by P = VI . (7.38) Thispowerisdissipatedasheat,thusallowingaformofheatingknownasresistance heating. Applications of this include the deicing of aircraft and bridges. This effect, in which electrical energy is converted to thermal energy, is known as the Joule effect. The units of P are watts (W), with watt = volt × ampere. Based on Eq. 7.15, Eq. 7.38 can be expressed as P = I 2 R (7.39) and as P = V2 /R . (7.40) Therefore, in order to obtain a high value for the power, both V and I should not be too small (Eq. 7.38). The values of V and I depend on the resistance R. When R is small, V is small since V = IR. When R is large, I is small since I = V/R. Thus, an intermediate value of R is optimal for obtaining a high power. It is R rather ρ that governs P. Therefore, for a given material (i.e., a given value of ρ), the dimensions can be chosen to obtain a particular value of R (Eq. 7.10). Since the current direction does not affect the resistance heating, the heating can be carried out using DC or AC electricity. Consider an example in which R = 1MΩ (i.e., 106 Ω) and V = 120V (which is a common voltage from an electrical outlet). Hence, I = V/R = (120V)/(1 × 106 Ω) = 1.2 × 10−4 A = 0.12mA , (7.41)
7.5 Electric Power and Resistance Heating 211 and P=V7=(120V)(1.2×10-4A)=1.4×10-2W=14mW. (7.42) Consider another example in which R 1k (i.e.,I x 103D)and V =120V. Hence, I=V/R=(120V)/(1×1032)=0.12A, (7.43) and P=V1=(120V)(0.12A)=14W. (7.44) Now consider another example in which R =1 and V =120V.Hence, I=V/R=(120V)/(12)=120A, (7.45) which is a current that is too high to be provided by conventional power sources, which often limit the current to,say,1A.Thus,in spite of the high V that the power source can provide,the actual V across the heating element is just V=IR=(1A)12)=1V, (7.46) and the power is thus merely P=VI=(1V)1A)=1W. (7.47) Therefore,among the three examples mentioned above,the intermediate R of 1kn gives the highest P. Nichrome(Ni-20Cr alloy),with e 1x 10-40cm,is a material that is commonly used for resistance heating because it has a relatively high resistivity for a metal. We can calculate the length of the wire required for a nichrome wire of diameter I mm to provide a resistance of 1k using Eq.7.10: 1=RA/p=(1k)[π(0.5mm)2]/(1×10-42cm)=7.9×104cm=790m. (7.48) This length of 790m is quite long,so the wire cannot be packaged linearly;it needs to be coiled.This is why heating elements commonly take the form of coils. In order to pass current to a heating element,two electrical contacts are required, as illustrated in Fig.7.3.An electrical contact may,for example,be a soldered joint Current Current Electrical contact Heating element Figure 7.3.Electrical contacts that are used to pass current to a heating element contribute to the electrical resistance
7.5 Electric Power and Resistance Heating 211 and P = VI = (120V)(1.2 × 10−4 A) = 1.4 × 10−2 W = 14mW . (7.42) Consider another example in which R = 1kΩ (i.e., 1 × 103 Ω) and V = 120V. Hence, I = V/R = (120V)/(1 × 103 Ω) = 0.12A , (7.43) and P = VI = (120V)(0.12A) = 14W . (7.44) Now consider another example in which R = 1Ω and V = 120V. Hence, I = V/R = (120V)/(1Ω) = 120A , (7.45) which is a current that is too high to be provided by conventional power sources, which often limit the current to, say, 1A. Thus, in spite of the high V that the power source can provide, the actual V across the heating element is just V = IR = (1A)(1Ω) = 1V , (7.46) and the power is thus merely P = VI = (1V)(1A) = 1W . (7.47) Therefore, among the three examples mentioned above, the intermediate R of 1kΩ gives the highest P. Nichrome (Ni-20Cr alloy), with ρ = 1×10−4 Ωcm, is a material that is commonly used for resistance heating because it has a relatively high resistivity for a metal. We can calculate the length of the wire required for a nichrome wire of diameter 1 mm to provide a resistance of 1kΩ using Eq. 7.10: l = RA/ρ = (1kΩ) π(0.5mm)2 /(1 × 10−4 Ωcm) = 7.9 × 104 cm = 790m . (7.48) This length of 790m is quite long, so the wire cannot be packaged linearly; it needs to be coiled. This is why heating elements commonly take the form of coils. In order to pass current to a heating element, two electrical contacts are required, as illustrated in Fig. 7.3. An electrical contact may, for example, be a soldered joint Heating element Current Current Electrical contact Figure 7.3. Electrical contacts that are used to pass current to a heating element contribute to the electrical resistance
212 7 Electrical Properties between the heating element and a wire.Each electrical contact is associated with a resistance,which is the sum of (i)the resistance of the joining medium (e.g.,solder),(ii)the resistance of the interface between the joining medium and the heating element,and(iii)the resistance of the interface between the joining medium and the wire.Since the joining medium is typically a highly conductive material,the resistance of the joining medium is usually negligible compared to the two interfacial resistances.The interfacial resistance can be substantial due to air voids,impurities,reaction products or other species at the interface.The resistances of both of the contacts contribute to the resistance R encountered by the current I.Therefore,the design of a heating element must consider the contribution to R from the electrical contacts.In particular,the resistance from the two contacts must not be so large that it overshadows the resistance within the heating element material.If it is too large,the electrical contacts become the main heating elements,while the actual heating element contributes little to heat generation,resulting in nonuniform heating as it is concentrated at the electrical contacts. 7.5.2 Self-Heating Structural Materials 7.5.2.1 Introduction Resistance heating (or Joule heating)involves the conversion of electricity into heat.It requires a heating element through which electric current is passed.Heating elements are commonly used in toasters and hair dryers.Other applications include deicing,home improvement,ovens,space heating,warming vests,heating pads, electric blankets,and seat heating. The materials most commonly used for heating elements are metal alloys,such as nichrome(an alloy with a common composition of80%nickel and 20%chromium), which is attractive due to its low cost,high resistivity (compared to many other metals),and its resistance to oxidation or degradation in air at elevated temper- atures.In spite of its high resistivity compared to many metals,its resistivity is low compared to carbon or ceramics.Because of this low resistivity (typically 10-5cm),sufficient resistance in the metallic heating element is commonly at- tained by using a long length of the metal,which may be coiled or take the form of an etched foil or a patterned thick film.The thick film is made from a paste that contains the metal in powder form. Compared to metals,carbon is attractive due to its higher resistivity(typically 10-3cm),high temperature resistance,ability to radiate heat,and its availability in the form of fibers.Carbon heating elements take the form of continuous carbon fiber,unwoven carbon fiber mats/felts,woven carbon fiber fabrics,and flexible graphite.Flexible graphite(also known as Grafoil)is a graphite sheet that is formed by compressing exfoliated graphite in the absence of a binder. Heating elements that are also structural components are attractive,particularly when the heating is needed for purposes such as deicing in aeronautical and au- tomotive systems,transportation infrastructures(e.g.,aircraft,driveways,airport runways,highways and bridges),and industrial systems.This can be achieved by
212 7 Electrical Properties between the heating element and a wire. Each electrical contact is associated with a resistance, which is the sum of (i) the resistance of the joining medium (e.g., solder), (ii) the resistance of the interface between the joining medium and the heating element, and (iii) the resistance of the interface between the joining medium and the wire. Since the joining medium is typically a highly conductive material, the resistance of the joining medium is usually negligible compared to the two interfacial resistances. The interfacial resistance can be substantial due to air voids, impurities, reaction products or other species at the interface. The resistances of both of the contacts contribute to the resistance R encountered by the current I. Therefore, the design of a heating element must consider the contribution to R from the electrical contacts. In particular, the resistance from the two contacts must not be so large that it overshadows the resistance within the heating element material. If it is too large, the electrical contacts become the main heating elements, while the actual heating element contributes little to heat generation, resulting in nonuniform heating as it is concentrated at the electrical contacts. 7.5.2 Self-Heating Structural Materials 7.5.2.1 Introduction Resistance heating (or Joule heating) involves the conversion of electricity into heat.It requiresaheatingelement through which electric current ispassed.Heating elementsarecommonlyusedintoastersandhairdryers.Otherapplicationsinclude deicing, home improvement, ovens, space heating, warming vests, heating pads, electric blankets, and seat heating. Thematerialsmostcommonlyusedforheatingelementsaremetalalloys,suchas nichrome(analloywithacommoncompositionof80% nickeland20%chromium), which is attractive due to its low cost, high resistivity (compared to many other metals), and its resistance to oxidation or degradation in air at elevated temperatures. In spite of its high resistivity compared to many metals, its resistivity is low compared to carbon or ceramics. Because of this low resistivity (typically 10−5 Ωcm), sufficient resistance in the metallic heating element is commonly attained by using a long length of the metal, which may be coiled or take the form of an etched foil or a patterned thick film. The thick film is made from a paste that contains the metal in powder form. Compared to metals, carbon is attractive due to its higher resistivity (typically 10−3 Ωcm), high temperature resistance, ability to radiate heat, and its availability in the form of fibers. Carbon heating elements take the form of continuous carbon fiber, unwoven carbon fiber mats/felts, woven carbon fiber fabrics, and flexible graphite. Flexible graphite (also known as Grafoil) is a graphite sheet that is formed by compressing exfoliated graphite in the absence of a binder. Heating elements that are also structural components are attractive, particularly when the heating is needed for purposes such as deicing in aeronautical and automotive systems, transportation infrastructures (e.g., aircraft, driveways, airport runways, highways and bridges), and industrial systems. This can be achieved by