>> In this problem we're given, a mixture of benzene and toluene,
a re-feeding to a high-temperature heat exchanger,
and we're told how much heat's being added, 22 megajoules per hour,
and we want to know the outlet temperature,
and we're going to use heat capacity [inaudible] function of temperature.
So the form for the heat capacity is given in this equation, and the constants are indicated
in this table, where these constants meaning that the value
of D is 77.753 times 10 to the minus 9.
And the temperatures are in Kelvin, so we'll work everything in Kelvin.
So first thing we want to do is energy balance in the system, which would be the flow rate
of benzene times the enthalpy change of benzene plus the flow rate
of toluene, enthalpy change to toluene.
That's going to be equal to the rate that heat is being added,
which I'm going to write is 22 times 10 to the 1/3 kilojoules per hour.
And the flow rates -- so flow rate of benzene is the mole fraction
of benzene times the total flow rate.
And the mole fraction, 0.7; total flow rate, 540 moles per hour, and so 378 moles [inaudible]
of benzene, and we can do the same calculation for toluene.
Okay. So let's look at what this term looks like for benzene.
So it's going to be [inaudible] - 150C is 423 Kelvin.
The final attempt, we don't know; that's what we're trying to determine.
Then we have constants A, [inaudible] - these values out of the table.
And we have an equivalent for toluene.
So if we were to integrate this, and we have T evaluated
between these two limits, and then T squared over 2.
So T squared, again minus, the limits square.
And likewise T cubed over 3 minus that inlet temperature cubed.
And finally T ¼ over 4.
We did integration so T to 1/4 minus 423 [inaudible].
What we're going to have is the exact same form but for toluene.
So if we could multiply - oh let's remember this is delta H for benzene --
we can multiply them by the flow rates and factor out terms.
So let me write that down.
So I would multiply the constants by the flow rate and factored
out the common term T minus 423, and then I can go through and create the other terms.
And let me pause and writ eh next one.
So you can see, all I've done is factor out, in this case, the T squared minus 423,
and I have similar terms for C and D. And so I don't think we need to write those down.
And remember all of this is equal to Q dot.
So let me write one more just substitution.
Make sure we're all on the same page here.
So this number 378 moles per hour, and then A is minus 36.1 [inaudible] 3 plus low rate
of [inaudible] times minus 34.364 multiplied by T minus 423.
And then equivalent value substituted in.
But we're going to end up with a polynomial fourth order and we want to solve
for temperature, and so the easy way to do this is now put these numbers into spreadsheets.
So I'll show you the spreadsheet, and let me just first point
out that I made the substitution.
The right side is in joules.
So I either divide the right side by 1000 or I write the left side in joules.
It's 10 to minus 6.
Right. So this is joules per hour.
Make sure units are consistent because heat capacities are in joules, whereas the Q value
that I wrote down here was in kilojoule.
And the spreadsheet, I just worked in kilojoules; that seemed easier.
I end up with an outlet temperature of 625 Kelvin.
Let's look at that spreadsheet.
So in a spreadsheet I put in the flow rates here; put in the constants, inlet temperature.
Solve for this outlet temperature.
And so I added up like I showed the terms here, and likewise for the others, did the integration
for each term; summed them together.
Then divided by 1000; put everything in kilojoule per hour.
And so what the spreadsheet, what we want to do is to set this value equal to Q value,
and I did just -- I used just a slider to do this,
but we could certainly use solver and get more accurately.
This is close to 0 compared to 22,000.
So we can get it, you know, slightly more accurately.
But the reality is, this is probably not accurate to 5 degrees anyway,
so it doesn't really make sense to spend time trying to get it more accurate than that.
So the spreadsheets available on the LearnChemE site, so you can look at it in more detail.