This screencast will take a look at developing a mathematical model for where an energy balance
is required. The problem statement is provided on the screen, which discusses the stirred
tank heater which allows for the heating of a fluid. So the inlet stream comes in at a
mass flow rate W, and an inlet temperature Ti, and exits at that same flow rate W, and
at temperature T. The fluid is heated by the use of a heating coil, provided at the bottom
of the diagram, in which a constant rate of energy of rate Q is provided to the heater.
The volume of the liquid inside the stirred tank heater is volume V. So first before we
enter the energy balance lets show the fact that the mass balance for this does not really
provide much guidance. So if we look at our mass balance first, we recognize the fact
that there is no generation, no reaction going, so there is no chemical reaction occurring.
In the inlet flow rate is W, and the outlet flow rate is W, so what that means is the
fact that the accumulation in this mass balance is zero, so all this is really telling us
is that the volume inside the tank is a constant. So now that we've shows that our mass balance doesn't
provide any fruitful information, we can now go to our energy balance which is that
accumulation of energy equals in minus out, energy in minus energy out, plus energy from
a chemical reaction, plus energy from the surroundings. An important point to note here
is the fact that the signs for the chemical reaction surroundings have been provided here as
positive. You have to be very aware of the signs that you are using. So for example,
for an exothermic chemical reaction the value for this term should be a positive because
if the temperature of the process inside the system is going to increase. So to begin this
energy balance we'll skip the accumulation term for the time being and we'll focus on
the right hand side. The information we have for our in term is the mass flow rate and
the temperature. What we need is an energy, so recall from thermodynamics that we can
get some idea of an energy per mass space by the use of enthalpy, H. So if we multiply
enthalpy, which has units typically of energy either per mole or per mass, here we'll go
on a per mass basis. And if we multiply it by the mass flow rate W, which has units of
mass per time, we get our desired units of energy per time. Our in term will be W times
the inlet enthalpy of the fluid minus the outlet stream, which is W, now will be times
H out, the enthalpy of the exiting stream. There is no chemical reaction going on, so
that term goes to zero, and now we have to look at the surroundings term. So our system
here has a heating coil which is providing heat at a rate of Q, for our control volume
is the inside of the stirred tank heater. This is energy being added to the system,
therefore we're going to want to have this as a plus Q, and our accumulation term for
now we'll leave as d energy dt. So d energy dt inside this derivative, what we need to
find is the amount of energy that is inside the tank that the fluid has. So to determine
the energy inside the tank we actually take a very similar approach to how we handle the
inlet and the outlet streams, where we're going to be interested in multiplying the
mass of fluid in the tank, multiplied by the energy per mass. So first lets find the mass
of fluid in the tank. We know the fact that the volume inside the tank of fluid is constant
based on the mass balance at value of V. In order to convert from volume to mass we multiply
by rho. In order to get the energy per mass that again is just simply the enthalpy, but in
this case the enthalpy of the tank, and this will equal the energy term thats in our differential
equation. With this information we can restate our energy balance as d rho V H tank dt equals
the right hand side. From thermodynamics we can express what the value of the enthalpy
is . We can say that for systems with constant heat capacity that H can be expressed as the
heat capacity multiplied by the temperature of which the fluid is at minus some reference
temperature. And what we'll now do is substitute this definition of enthalpy into our energy
balance where enthalpy is located, which is at the tank, the inlet stream, and the outlet
stream. The derivative will now become rho V Cp times T, which is the temperature inside
the tank. Since the tank is well mixed that means the fact that the temperature of the
outlet stream of T is identical to the temperature inside the tank, so therefore this will just
be T minus T ref dt. Our in term becomes W times Cp. Here the inlet stream enters at
Ti, so it will Ti minus Tref. We can do a very similar evaluation for the outlet stream,
which is at temperature T, and then we add the Q term. We can now take advantage of some
simplifications. On the left hand side, the reference temperature is a constant value, therefore we take the
derivative with respect to time the value of the T ref part will go to zero. Although
dTref dt goes to zero, dT dt does not. On the right hand side we see the fact that we
have negative W Cp Tref, and then for our second term we have the same terms except
with an additional minus sign, so therefore the Tref terms are going to cancel out, an
occurrence that is relatively common when doing these energy balances. After these simplifications
we're left with the following. The last step here is to take out a few constants out of
the derivative. The problem statement stated that rho and heat capacity were constants,
and based on our mass balance we know that the volume is a constant, so therefore we
can take all those out of the derivative to get us to our final answer, which is that
rho V, Cp, dT dt equals W Cp multiplied by the difference between the inlet temperature
and the outlet temperature, which is also the tank temperature, plus Q.