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.

why is accumulation taken as d(energy)/dt ???

Wouldn't you have to multiply the WCp terms by rho to get things in mass terms since we are expressing enthalpy on a per mass basis?