the first law of thermodynamics is basically the principle of conservation of energy. it says that energy will not vanish into nothing or arise out of nothing. this principle is as important to thermodynamic analysis as does newton's second law of motion in mechanics. what we want to learn in this tutorial is how to apply this simple law to problems in heat transfer.

applying the first law requires that we keep track of all forms of energy and make sure that every bit of energy is accounted for. in many cases, though, we ignore some of the forms that we deem small in comparison to other.

let's look at a simple example:

here a certain amount of air with a known temperature and flow rate enters a duct (or a pipe), gets heated up as it travels through the duct and exits at the other end. conservation of energy tells us that the energy of the air at the exit is equal to its energy at the entrance plus the amount of energy added to it during its travel through the pipe, or

in heat transfer we are mainly dealing with a form of energy called heat. heat is the form of energy that is transferred from one place to another as a result of temperature difference. there is an important distinction between what a thermodynamic analysis gives us versus what we get from a heat transfer analysis. in thermodynamics, we are looking at a system at two different states of equilibrium and can use the first law to determine the energy exchange in going from one state to another. heat transfer, on the other hand, deals mainly with the rate of heat transfer or the time it takes for the heat transfer to occur.

let's write the first law in terms of an equation starting from:

this simple equation says that the only way to change the total energy of a system is to have an imbalance between what we give to the system and what we take out of it (kind of like your bank account). writing this equation in a rate form and using e to represent energy, we have:

in heat transfer, we are mainly interested in the transfer of heat caused by a temperature difference. we separate the energy terms into the heat energy and other energies and lump all of this "other" forms of energy into a heat generation term. the above rate equations then become:

so, we have reduced the first law of thermodynamic to a simple energy balance. applying the first law to heat transfer problems is nothing but balancing the thermal energy items.

closed systems and open systems

it is a common practice to conside two different cases of closed systems (where mass does not cross the boundary of the system) and open or flowing system (where mass does cross the system boundary). for most applications the total energy is equal to the internal energy. the internal energy is the sum of all microspcopic forms of energy. for a constant volume system we define the specific heat at constant volume as the energy required to raise the temperature of a unit mass of the substance by one degree as the volume is held constant. mathematically:

which is basically the definition of the specific heat. we just took the energy required per unit mass and per degree and multiplied it by the mass and by the temperature difference to get the internal energy. if m and dt are unity, the internal energy is equal to the specific heat. with this, the energy balance for a closed system becomes:

where q is the net amount of heat transfer assuming that no work is crossing the boundary of the system. we can write the net rate of heat transfer for a steady flow (open) system as:

the mass flow rate through a given cross section a is calculated as:

let's now consider a simple example of air flow through a duct with a constant cross section. we know the inlet and outlet temperatures and we know the flow rate. we want to calculate the heat loss (or gain) through the duct

we will use the magic equation to calculate the heat transfer through the duct walls:

we get the value of cp and density from air property tables using the average temperature between inlet and out let. the inlet velocity is given and the cross-sectional area is equal to a x b. substituting in the heat balance equation gives us the net heat transfer.