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First Law of Thermodynamics
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.
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