8 Thermo-electricity 8.1. Seebeck Effect. When two dissimilar metal wires are joined together so as to form a closed circuit i junctions are maintained at different tet a- tures, an emf is developed in the circuit (Fig. 8.1). This causes a currént to flow ion in the galvanometer G. This phe- nomenon is called the Seebeck effec jis arrangement is called a thermo- couple. The emf developed is called thermo emf. The thermo emf so devel- oped depends on the temperature dif- 3 ference between the two junctions and Hot Cold the metals chosen for the couple. See- Fig. 81 beck arranged the metals in a series as follows : Bi, Ni, Pd, Pt, Cu, Mn, Hg, {Pb)Sn, Au, Ag, Zn, Cd, Fe, Sb. When a thermocouple is formed between any two of them, the thermoelectric current flows through the hot junction from the metal occurring earlier to the metal occuring later in th list. The more removed are the two metals in the list, the greater is the thermo emf developed. The metals to the left of Pb-afe called thermoelectrically negative and those to its right are thermoélectrically positive. @f thermo em.f. Law of Intermediate Metals, The introduction of any additional metal into any thermoelectric circuit does not alter the thermo emf provided the metal introduced is entirely at the same temperature as the point at which the metal is introduced, If ,E; is the emf for a couple made of metals A and B, and ,E, that for the couple of metals B and C, then the emf for couple of metals A and Cisgiven by aE, = oFy + oe 105106 Electricity and Magnetism 4 Nate of intermediate Temperatures. The thermo emf E of thermocouple whose junctions are maintained at temperatures T, and T, is ‘ ; ; . equal to the sum of the emfs E, and Ey when the junctions are maintained at temperatures T,, T, and Ty, T, respectively. Thus B= E+E 8.3. Measurement of Thermo EMF using Potentiometer Thermo emfs are very small, of the order of only a few millivolts. Such small emfs are measured using a potentiometer. A ten-wire potentiometer of resistance R is connected in series with an accumulator and resistance boxes P and Q (Fig. 8.2). A standard cell of emf E is connected in the secondary circuit. The positive terminal of the cell is connected to the positive of Q. The negative terminal of the cell is connected to a galvanometer and through a key to the negative of Q. A resistance of 100 ER ohms is taken in Q. The resistance in P is adjusted so that on closing the key, there is no deflection in the galvanometer. Now, the PD across 100 ER ohms is equal to E. ° PD across Rohms. | _ ER. 1 7 cates TOOER volt = che 709 Volt = 1Omillivon. Thus the fall of potential per metre of the potentiometer wire is | millivolt. So we can measure thermo emf up to 10 millivolt. Without altering the resistances in P and Q, the positive of the thermocouple is connected to the positive terminal of the potentiometer and the negative of the thermocouple to a galvanometer and jockey. One junction is kept in melting ice and the other junction in an oil bath or in a sand bath. The jockey is moved till a balance is obtained against the small emf e of the thermocouple. Let AJ = / cm be the balancing length. Then, thermo emf ¢ 7 I millivolt. Keeping the cold junction at 0°C, the hot junction is heated to different temperatures. The thermo emf generated is determined for different temperatures of the hot junction. A graph is drawn between thermo emf and the temperature of the hot junction (Fig. 8.3). The graph is a parabolic curve. The thermo emf E varies with temperature according toThermo-electricity 107 max — Copper - Iron E = at + br, where a and b emf thermocouple are constants, The thermo & emf increases as the tem-f perature of the hot junction increases, reaches a maxi- mum value 7,, then de- creases to zero at a parti-cular temperature T;. G 7 On further increasing the dif- 270°C —+ \540°C ference of temperature, emf Temperature is reversed in direction. Fig. 83 For a given temperature of the cold junction, the temperature of the hot junction for which the thermo emf becomes maximum is called the neutral temperature (T,) for the given thermocouple. For a given temperature of the cold junction, the temperature of the hot junction for which the thermo emf becomes zero and changes its direction is called the inversion temperature (T,) for the given thermo- couple. T, is a constant for the pair of metals. 7; is variable. T, is as much above the neutral ee as the cold junction is below it. Thermo Ti 8.4. Peltier Effect Consider a copper-i 84) allowed to pass through the the: ii irect (from A to B), heat is absorbed at the e Pros junction B and generate function A. This absorption or evolution of heat at_a yunction when a current is sent through a thermocouple is ier—Heated effect. The Peltier effect is a Teversible phenomenon. If the direction of the current is reversed, then there will be cooling at the junction A and heating at the junction B. The amount of heat H absorbed evolved at a junction is proportional to the charge q passing through the junction. ie, fi<«qg of Halt or H=nlt108 Electricity and Magnetism where m is a constant called Peltier coefficient. When J = 1A and t= Is, then H = 1. The energy that is liberated or absorbed at a junction dissimilar metals due to the passage of unit quantity of electric Peltier coefficient, It is expressed in joule/coulomb i.e., volt. The Peltier coefficient is not constant but depends on the temperature of the junction. ‘The Peltier effect is different from the PR Joule heating effect. The main differences are given below. Peltier Effect Joule Effect 1. tis areversible effect. Itis an irreversible effect. 2. Ittakes place at the junctions only. | It is: observed throughout the conductor. 3. I may be a heating or a cooling| It is always a heating effect. effect. 4, Peltier effect is directly proportional] Amount of heat evolved is directly tol =+ nh) _ proportional to the square of the current. 5. It depends upon the direction of the | It is independent of the direction of current. the current... Demonstration of Peltier effect - S.G. Starling Method. Fig. 8.5’ Shows ’a, bismuth bar-between two bars of antimony. Two coils C, and C, of insulated copper wire are wound over the two junctions J, and Jy. These coils are connected across the two gaps of a metre-bridge and the balance-point is found on the bridge wire. Now a current is passed through the rods from Sb to Bi, The junction J, is heated and Jy is cooled. The resistance of cop- per varies rapidly with change of temperature. Hence the balance in the bridge is immediately upset. The galvanometer shows a deflec- tion, If the current is reversed, the deflection in the galvanometer, also gets reversed. This shows that the junction J, is now cooled and Jy is heated. Jy Heated J2 CooledThermo-electricity 109 sharon Effect Consider a copper bar AB heated in the middle at the point C (Fig. 8.6). A current is passed from A to B. It is observed that heat is absorbed 2 potent “en, “anay get ast fayy"t fa" i abiorbea~| Hear = = _| | Heat. a{ ster, Pp +1 absorbed cine e & }> sont fo» A ae B A B Flame = ror Flame x Fig86 Geo" Fig. 87 in the part AC and evolved in the part CB. This is known as Positive Thomson effect. Similar effect is observed in metals like Ag, Zn, Sb and Cd. In the case of an iron bar AB, heat is evolved in the part AC and absorbed in the part CB (Fig. 8.7). This is known as Negative Thomson effect. Similar effect is observed in metals like Pt, Ni, Co and For lead, the Thomson effect is zero. The Thomson effect is reversible. In the case of copper, the hotter parts are at a higher potential than the colder ones. It is opposite in the case of iron. Heat is either absorbed or evolved when current passes between two points having a difference of potential. Therefore, the passage of electric current through a metal having temperature gradie: ufts inn absorption or evolution of heat in the known as ‘Thomson effect’. Thomson Coefficient. The Thomson coefficient 6 of a metal is defined as the amount of heat energy absorbed or evolved when a charge of | coulomb flows in the metal between two points which differ in temperature by 1°C. Thus, if a charge of q coulomb flows in a metal between two points having a temperature difference of 1°C, then heat energy absorbed or evolved = 6 q joule. But if E volt be the Thomson emf developed between these points then this energy must be equal to Eq joule. a oq = Eq or o=E.(i) Electricity and Magnetism Thus the Thomson coefficient of a metal, expressed in joule per coulomb per °C, is numerically equal to the emf in volt, developed between two points differing in temperature by I°C. Hence it may also be expressed in volt per °C. 6 is not a constant for a given metal. It is a function of temperature Demonstration of Thomson effect. Fig. 8.8. shows Starling’s method of demonstrating the Thomson effect. An iron rod ABC is Yent into U shape. Its ends A and C are dipped in mercury baths. C, and C, are two insulated copper wires of equal resistance wound round the {wo arms of the bent rod. C, and C, are. - connected in the opposite gaps of a it metre bridge. The bridge is balanced. —— Then the mid-point B.of the rod is Cy 4 strongly heated. A heavy current is 4 passed through the rod. Then this current will be flowing up the temperature gradient in one arm and down the temperature gradient in the AC other arm. As a result, one of the coils will be cooled and the other will be warmed. The balance in the bridge will be upset and the galvanometer in its circuit will show a deflection. If the -—(} direction of the current is reversed, the deflection i galvanometer will be reversed. B strongly heated Fig. 88 €cmodynamics of Thermocouple . ‘pressions for Peltier and Thomson coefficients.) Consider a thermocouple consisting of two metals A and B. Let Tand T + dT be the temperatures of the cold and hot junctions respectively (Fig. 8.9]. Let nandx + dr be the Peltier coefficients for the pair at the cold and hot junctions. Let a, and 6, be the Thomson coefficients for the metals A and B respectively, both taken as positive. When a charge flows through the thermocouple, heat will be absorbed and evolved at the junctions due to Peltier effect and all along the metal due to Thomson effect. Fig. 89Thermo-electricity Ww Let 1 coulomb of charge flow through the thermocouple in the direction from A to B at the hot junction. Heat energy absorbed due to Peltier effect at the hot junction = (x + dx) joules ~— Heat energy evolved due to Peltier effecr-at.the cold junction = x joules Heat energy absorbed in the metal A due to Thomson effect = 6, dT joules Heat energy evolved in the metal B due to Thomson effect = G,dT joules . Net heat energy absorbed in the thermocouple = (n + dn - n) + (0,dT - 0,47) = dn + (6, - 6,)dT This energy is used in establishing a P.D. dE in the thermocouple ” dE = dn + (0, - 6,)dT (I) Since the Peltier and Thomson effects are reversible, the thermocouple acts as a reversible heat engine. Here, (i) the heat energy (x + dr) joules is absorbed from the source at (T + dT)K anda, dT joule is absorbed ‘in’ metal A at mean temperature TK. (ii) Also x joule is rejected to sink at T K and 0, dT joule is given out in metal B at the mean temperature T K. Applying Carnot's theorem, we have n+dn 947 _n, dT T+ dT T T T n+dn_ nm _ (9, -9,MT or Eta _ Flos T+dT T T nT + dnT ~ nT - ndT _ (Oy - 9,MT i ee or dnT ~ ndT = (0, - o,)UT(T + dt) or dn.T ~ ndT = (, - 6,)TdT + (0, - 9,)dT* or (dn.T - Nd) = (0, - 6,) TAT { Neglecting(o, - 6,) dT"] or Tldx + (6, - 6)47) = RdT But dn + (6, ~ 0)dT = dE from Eq. (1) TdE = nT ym2 Electricity and Magnetism or (2) Saale ae” The quantity (dE/dT7) is called the thermoelectric era Thermoelectric power (P) is defined as-the thermo emf per unit difference of temperature between the junctions. +. Peltier coetficient = Absolnib temperature x thermoelectric power Differentiating Eq. (2), oe oe oe + & Substituting the value of (dE/dT) via Eq. (1), dx_ @E | dn Tit apt 4 2 @eE or (6, - 6) = - T.—> . ar’ or (-o)=7. £8 3) aT° If the first metal in the thermocouple is lead, then 6, = 0 2 ave oe 6, =T.—> we (4) » ar (4) Thomson coefficient = absolute temperatureof the cold junction x first derivative of thermoelectric power QE _%-o) | a) (©, - 6.) ar From Eq. (3), —> = Oh Tr T aT T Putting dE/dT from Eq. (2), we have A(z), =o) aT|T |” A(x) _{% n or ( T =0 + 5) at| T This gives the relation between Peltier and Thomson's coefficients. 8.7. Thermo-Electric Diagrams A thermocouple is formed from two metals A and 8. The difference of temperature of the junctions is TK. The thermo emf E is given by the equation E=aT+0r A graph between E and T is a parabola.Thenno-electricity 13 # = a+ bT dEIAT is called thermoelectric power. A graph between thermoelectric power (dE/dT) and difference of temperature T is a straight line. This graph is called the thermo-electric power line or the thermo-electric diagram. Thomson coefficient of lead is E/dT zero. So generally thermo- electric lines are drawn with | 0 lead as one metal of the thermo- couple. The thermoelectric line of a Cu-Pb couple has a positive slope while that of Fe-Pb couple has a negative slope. Fig. 8.10 a shows the power lines for a Temperature ——> number of metals. Fig. 8.10 8.8. Uses of Thermoelectric Diagrams (i) Determination of Total emf. MN represents the thermo-electric power line of a metal like copper coupled with lead (Fig. 8.11). MN has a positive slope. Let A and B be two points corresponding to temperatures T, K and T, K respectively along the temperature-axis. Consider a small strip abde of thickness dT with junctions maintained at temperatures T and (T + dT). O A 458 The emf developed when Temperature —> the two junctions of the Fig. 8.11 thermocouple differ by dT is : = at{ fe). dE = ar a } = area abdc Total emf developed when the junctions of the couple are at temperatures T, and T, is r E,= de (Fle Area ABDC (ii) Determination of Peltier emf. Let , and m be the Peltieris Electricity and Magnetism coefficients for the junctions of the couple at temperatures T, and 7, respectively. The Peltier coefficient at the hot junction (T,) is m= 1] = OBxBD = area OBDF a ld Similarly, Peltier coefficient at the cold junction (7,) is dE m= 1%]. = OAXAC = area OACE i m™, and nm, give the Peltier emfs at 7, and 7, respectively. Peltier emf between temperatures T, and T, is E, = % — m, = area OBDF - area OACE = area ABDFECA (iti) Determination of Thomson emf. Total emf developed in a thermocouple between temperatures 7, and T; is a E,= (tm ~ 1) + J'(e, ~ 6) ar Here @, and Gy represent the Thomson coefficients of two metals constituting the thermocouple. If the metal A is copper and B is lead, then 6, = 0. T, E,=(m-%) +f ",an or [Poa = - [2 -%)- &] Thus, the magnitude of Thomson emf is given by a Ey, = (ly - %,) - E = Area ABDFECA - Area ABDC = Area CDFE (iv) Thermo emf in a general couple, neutral temperature and temperature of inversion. In practice, a thermocouple may consist of any two metals. One of them need not be always lead. Let us consider a thermocouple consisting of any two metals, say Cu and Fe. AB and CD are the thermo-electric power lines for ang Cu and Fe with respect to lead 172 Th Temperature (Fig, 8.12). Let 7, and T, be Fig. 8.12Thermo-electricity Ws the temperatures of the cold and hot junctions corresponding to points P and Q. Emf of Cu = Pb thermocouple = Area PQR,A, Emf of Fe ~ Pb thermocouple = Area POD,C, v- the emf of Cu — Fe thermocouple is EX = Area PQD,C, - Area PQB\A, = Area A,B,D,C, The emf Ef, increases as the temperature of the hot junction is raised and becomes maximum at the temperature 7,, where the two thermoelectric power lines intersect each other. The temperature T,, is called the neutral temperature. As the thermo emf becomes maximum at the neutral temperature, at T = T,, (dE/d7) = 0. Suppose temperatures of the junctions, 7, and 7, for a Cu — Fe ther- mocouple are such that the neutral tempera- ture T, lies between-7; ‘and T, (Fig. 8.13). Then the thermo emf will be represented by the difference between the areasdE/dT ‘ANC, and B,D,N because these areas repre. sent opposing emf’s. In the particular case when T, = (T, + T;)/2, these areas are equal and the resultant emf is zero. In this case, Tp is the ‘temperature of inversion’ for the Cu — Fe thermocouple. [ved Examples The emf of a thermocouple, one junction of which is kept at am. de en OC, is_give y E-= at + bt’. Determine the neutral temperature, femperature of inversion\and the Peltier and Thomson Coefficients. ‘ ad So. E=ar+ br awd Now, °C = (T — 273)°C, where Tis in absolute degrees. E = a(T ~ 273) + b(T - 273). Fig. 8.13, Differentiating, = a + 26(T - 273) and ae = 2b. a Atthe neutral temperature, T = T, and « =0. 0 =a + 2b(7, - 273)16 Electricity and Magnetism Let 7, be the temperature o} ~ T, = Q73 + Tyla a IF, - 213 = 2{2 - 5) fo-}G The Peltier coefficient, x = ré inver: or T, The Thomson coefficient, ¢ = =T.2b 2. A thermo-couple is made of iron a le Find the emf developed per °C difference of temperatures between the junctions, given that thermo-emfs of iron and constantan against platinum are +1600 and -3400 pV per 100°C difference of temperatures. Sol. According to the law of intermediate metals, = EE BN = te Now, Ep, = 1600p V/100°C = 16p V°C and Ep, = — 3400p W100°C = - 34 VC. ES = 16 - (- 34) = SOnW/C. con 3. The thermo-electric power of iron is 17.3 1 VPC at 0°C and 12.5 BYPC at 100°C and that for silver 2.9 tVPC and 4.1 2 VPC at the same two temperatures. Calculate (1) Peltier co-efficient between iron and silver at 100°C. (2) Thomson co-efficient of iron at 50°C. (3) thermo emf. when the two junctions are at 0°C and 100°C, (4) Neutral temperature of the couple formed by the 2 metals. (5) Temperature of inversion when the cold junction is at 50°C,Thermo-electricity 7 Sol. (1) Thermoelectric power between the metals at 100°C = 123 - 4.1 = 84pVPC 373 x84 = 3133 pV dT Peltier emf at Lo0'C Peltier co-efficient = 3.133x 10°* JIC (2) Slope of the thermoelectric diagram _ a (dE) 25-173 48 | -#(#)- = = - 0.048 pv? Thomson coefficient o at 50°C = TLE 2 — 323x0,048% 10° = - 15.5x 10K"! ar (3) Thermo emf = area of trapezium ABCD = Fx (44 + 84) 100 D 173 = 1140 pv GE Cc dT 12.5 Fe 2.9) 4.4 Ag {B= -] OC 100°C Th —T Fig. 8.14 (5) If 7, is the temperature of inversion when the cold junction is at 50°C, “T, + 50 +y = T= 240 T, + 50 = 480 T, = 430°C. EXERCISE VIII 1, Explain Seebeck effect. Describe the laws of thermoelectric effects. 2. Deseribe a method of measuring the thermo emf. 3. What is meant by Peltier effect ? How would you demonstrate it experimentally ? How does it differ from Joule heating effect ? Define Peltier coefficient. 4, Explain Thomson effect. Define Thomson coefficient. Describe an experiment to demonstrate Thomson effect.Us . The thermo-electric power of iron is 17.5 microvolt per degree at Electricity and Magnetism Prove that the Peltier coefficient of a pair of metals is the product of the absolute temperature and thermoelectric power. Applying thermo-dynamic considerations to the working of a thermo-couple show that dE ae ROT Ged Go - a Tos where the symbols have their usual meaning. What is a thermo-clectric diagram ? Show how Peltier and Thomson c.m.f’s, neutral temperature and the temperature of inversion can all be represented in this diagram. |. The emf (micro volts) in a lead-iron thermocouple, one junction of which is at 0°C is given by E = 17841 - 2.47 where ¢ is the tempera- ture in degree centigrade. Find the neutral temperature and Peltier and Thomson coefficients. [Ans. 371.7°C, T(1784 - 4.81). - 4.8 7) If for a certain thermo-cbuple E = at + b? where (°C is the tempera- ture of the hot junction, the cold junction being at 0°C a = 10microvolv“C,b = - 1/40 microvolv’C’, find the neutral temperature and the temperature of inversion. (Ans. 200°C, 400°C] ‘Thermo emf ina circuit in which the cold junction is at 0°C and the hot junction at :°C is found to be 3.5 Vat 100°C and 9 pV at 200°C. Assuming the thermo emf to be governed by the equation E = at + bf, find the constants a and b. Hint. 3.5x10°6 = ax 100 + b 1007; 9x 10° = ax 200 + bx 200, Solving, a= 3.51x10°° pvc; = 10° pvec?. c and zero at 360°C; that of copper is 5 microvolts per degree at 450°C and zero at - 50°C. Find the value of the e.m.f. of a copper-iron couple when the cold junction is at 0°C and the hot junction at the neutral temperature (Assume the thermo-electric lines to be straight). (Ans, 2465 micto-vols} Find the thermo e.m.f generated in a thermocouple, when its junctions are kept at 160°C and SOC respectively, if e.m.fs generated in the thermo-couple, when its hot junctions are at 160°C and SU*C are 25 microvolts and 5.6 microvolts tespectively. (The temperature of the cold junction being 0°C). (Ans. 19.4 1 V1