Heat current

Transcription

let’s anlyze, thermal conduction analysis in steady state. as we’ve already discussed that if we have a uniform media, and the ends are maintained at temperature t-1 and t-2, such that t-1 is more than t-2. then obviously heat will flow from high temperature end to low temperature end. in this situation, say if the length of this uniform media is l. and say its cross sectional area is given as, a. if in this situation, if t-1 and t-2 are maintained temperature, say these are maintained, temperatures. in this situation, the rate of flow of heat. through the medium, is also written as, rate of flow of heat is also called, heat current, through the medium is experimentally determined. and it is found that it depends on various factors, out of which most important are. the 1st 1 is, we can write, d-q by d t is directly proportional to temperature difference, t 1 minus t 2. if the difference of these temperature is more, it’ll flow relatively fast. The second one is this rate of flow of heat can be written as directly proportional to the cross sectional area, of the medium. and the 3rd 1 is, d q by d t, can be written as, inversely proportional to the length of the medium. now in this situation, if we combine these relations we can see we’ll get, d q by d t is equal to, a proportionality constant, that we substitute as, k. t-1 minus t-2, multiplied by area divided by, length. this is the relation we’re getting if we combine these. here, k we call a proportionality constant, which basically depends on the material of medium. and, this proportionality constant is called, thermal conductivity. of medium. if the value of k is high, the heat flow will be more, or the heat current through the medium will be more. and if this thermal conductivity is low, heat flow through the medium will be less. so just keep this relation in your mind because, this’ll be the basis of several different kinds of numerical problems related to thermal conduction in steady state. now this, same analysis can also be obtained, by analogy of conduction with ohm’s law. as, ohm’s law is defined for electrical conduction, which we’ve already studied. where we define that if there’s a resistance between 2 ends, or 2 terminals, ay and b. their potentials are, v-a and v-b respectively such that, v-a is more than v b. and we know that, electric current always flow, from high potential to low potential end. so as v-a is more than v-b, electric current will flow from, high potential end to low potential end. and, by ohm’s law we write, this electric current is, v-a minus v-b upon, r. where r we define as electrical resistance. and, this electrical resistance, r, can be written as, ro l upon, ay. where ro is, the electrical resistivity of the material. l is its length, and ay is its area. we can also define it in terms of conductivity. electrical conductivity sigma can be written as, 1 by ro. so this can be written as, 1 by sigma, l upon ay. in this situation, we can simply define, the electrical resistance in terms of conductivity, sigma. now, similar analysis we can execute, for thermal conduction. if we just analyze thermal conduction, in the similar way of ohm’s law, we’ve studied that if we’re given with a uniform media, with temperature t-1 and t 2, with cross sectional area a, and its length l. then in this situation again, the rate of flow of heat or, heat current, can be given in terms of, this ohm’s law. like here, 1st we define thermal resistance of medium. thermal resistance can be given as r t h, which is given by, here as this expression is given 1 upon conductivity into l by ay, so thermal resistance can be given as, 1 by thermal conductivity into, l by ay. this is defined as thermal resistance. and similarly we can define heat current, by using this expression of ohm’s law. here we can say it is, d q by d t. as electric current is, coulomb per second or rate of flow of charge, heat current is defined as, joule per second or rate of flow of heat. it can be written as, just similar to this, here it is potential difference upon r. and as current flows from high potential to low potential, similarly here heat current flows from, high temperature to low temperature and temperature can be regarded as, thermal potential for, heat conductions. so this can be written as, t-1 minus t-2 upon, thermal resistance. here we can see, instead of potential we’ve substituted temperature. so we substitute the value of thermal resistance we get, d q by d t, is equal to, if we substitute value of r it is, k, a, t-1 minus t-2, divided by l. you can say this is the same relation which we’ve obtained, by experimental analysis, with some proportional relationships. now here we’re getting it by using this, ohm’s law application on, a uniform medium in which, heat conduction is taking place in steady state.