Introduction to Chemical Engineering Processes/How to use the mass balance

A Typical Type of Problem

Most problems you will face are significantly more complicated than the previous problem and the following one. In the engineering world, problems are presented as so-called "word problems", in which a system is described and the problem must be set up and solved (if possible) from the description. This section will attempt to illustrate through example, step by step, some common techniques and pitfalls in setting up mass balances. Some of the steps may seem somewhat excessive at this point, but if you follow them carefully on this relatively simple problem, you will certainly have an easier time following later steps.

Single Component in Multiple Processes: a Steam Process

Example:

A feed stream of pure liquid water enters an evaporator at a rate of 0.5 kg/s. Three streams come from the evaporator: a vapor stream and two liquid streams. The flowrate of the vapor stream was measured to be 4*10^6 L/min and its density was 4 g/m^3. The vapor stream enters a turbine, where it loses enough energy to condense fully and leave as a single stream. One of the liquid streams is discharged as waste, the other is fed into a heat exchanger, where it is cooled. This stream leaves the heat exchanger at a rate of 1500 pounds per hour. Calculate the flow rate of the discharge and the efficiency of the evaporator.

Note that one way to define efficiency is in terms of conversion, which is intended here:

$efficiency={\frac {{\dot {m}}_{vapor}}{{\dot {m}}_{feed}}}$

Step 1: Draw a Flowchart

The problem as it stands contains an awful lot of text, but it won't mean much until you draw what is given to you. First, ask yourself, what processes are in use in this problem? Make a list of the processes in the problem:

1. Evaporator (A)
2. Heat Exchanger (B)
3. Turbine (C)

Once you have a list of all the processes, you need to find out how they are connected (it'll tell you something like "the vapor stream enters a turbine"). Draw a basic sketch of the processes and their connections, and label the processes. It should look something like this:

Remember, we don't care what the actual processes look like, or how they're designed. At this point, we only really label what they are so that we can go back to the problem and know which process they're talking about.

Once all your processes are connected, find any streams that are not yet accounted for. In this case, we have not drawn the feed stream into the evaporator, the waste stream from the evaporator, or the exit streams from the turbine and heat exchanger.

The third step is to Label all your flows. Label them with any information you are given. Any information you are not given, and even information you are given should be given a different variable. It is usually easiest to give them the same variable as is found in the equation you will be using (for example, if you have an unknown flow rate, call it ${\dot {m}}$ so it remains clear what the unknown value is physically. Give each a different subscript corresponding to the number of the feed stream (such as ${\dot {m_{1}}}$ for the feed stream that you call "stream 1"). Make sure you include all units on the given values!

In the example problem, the flowchart I drew with all flows labeled looked like this:

Notice that for one of the streams, a volume flow rate is given rather than a mass flow rate, so it is labeled as such. This is very important, so that you avoid using a value in an equation that isn't valid (for example, there's no such thing as "conservation of volume" for most cases)!

The final step in drawing the flowchart is to write down any additional given information in terms of the variables you have defined. In this problem, the density of the water in the vapor stream is given, so write this on the side for future reference.

Carefully drawn flowcharts and diagrams are half of the key to solving any mass balance, or really a lot of other types of engineering problems. They are just as important as having the right units to getting the right answer.

Step 2: Make sure your units are consistent

The second step is to make sure all your units are consistent and, if not, to convert everything so that it is. In this case, since the principle that we'll need to use to solve for the flow rate of the waste stream (${\dot {m_{3}}}$) is conservation of mass, everything will need to be on a mass-flow basis, and also in the same mass-flow units.

In this problem, since two of our flow rates are given in metric units (though one is a volumetric flow rate rather than a mass flow rate, so we'll need to change that) and only one in English units, it would save time and minimize mistakes to convert ${\dot {V_{2}}}$ and ${\dot {m_{5}}}$ to kg/s.

From the previous section, the equation relating volumetric flowrate to mass flow rate is:

${\dot {V}}_{i}*{\rho }_{i}={\dot {m}}_{i}$

Therefore, we need the density of water vapor in order to calculate the mass flow rate from the volumetric flow rate. Since the density is provided in the problem statement (if it wasn't, we'd need to calculate it with methods described later), the mass flow rate can be calculated:

${\dot {V_{2}}}={\frac {4*10^{6}{\mbox{ L}}}{1{\mbox{ min}}}}*{\frac {1{\mbox{ m}}^{3}}{1000{\mbox{ L}}}}*{\frac {1{\mbox{ min}}}{60{\mbox{ s}}}}=66.67{\frac {m^{3}}{s}}$

$\rho _{2}=4{\frac {g}{m^{3}}}*{\frac {1{\mbox{ kg}}}{1000{\mbox{ g}}}}=0.004{\frac {kg}{m^{3}}}$

${\dot {m}}_{2}=66.67{\frac {m^{3}}{s}}*0.004{\frac {kg}{m^{3}}}=0.2666{\frac {kg}{s}}$

Note that since the density of a gas is so small, a huge volumetric flow rate is necessary to achieve any significant mass flow rate. This is fairly typical and is a practical problem when dealing with gas-phase processes.

The mass flow rate ${\dot {m_{5}}}$ can be changed in a similar manner, but since it is already in terms of mass (or weight technically), we don't need to apply a density:

${\dot {m_{5}}}=1500{\frac {lb}{hr}}*{\frac {1kg}{2.2lb}}*{\frac {1hr}{3600s}}=0.1893{\frac {kg}{s}}$

Now that everything is in the same system of units, we can proceed to the next step.

Step 3: Relate your variables

Since we have the mass flow rate of the vapor stream we can calculate the efficiency of the evaporator directly:

$efficiency={\frac {{\dot {m}}_{2}}{{\dot {m}}_{1}}}={\frac {0.2666{\frac {kg}{s}}}{0.5{\frac {kg}{s}}}}=53.3\%$

Finding ${\dot {m_{4}}}$, as asked for in the problem, will be somewhat more difficult. One place to start is to write the mass balance on the evaporator, since that will certainly contain the unknown we seek. Assuming that the process is steady state we can write:

$In-Out=0$

${\dot {m}}_{1}-{\dot {m}}_{2}-{\dot {m}}_{4}-{\dot {m}}_{6}=0$

Problem: we don't know ${\dot {m}}_{6}$ so with only this equation we cannot solve for ${\dot {m}}_{4}$. Have no fear, however, because there is another way to figure out what ${\dot {m}}_{6}$ is... can you figure it out? Try to do so before you move on.

So you want to check your guess? Alright then read on.

The way to find ${\dot {m}}_{6}$ is to do a mass balance on the heat exchanger, because the mass balance for the heat exchanger is simply:

${\dot {m}}_{6}-{\dot {m}}_{5}=0$

Since we know ${\dot {m}}_{5}$ we can calculate ${\dot {m}}_{6}$ and thus the waste stream flowrate ${\dot {m}}_{4}$.

Step 4: Calculate your unknowns.

Carrying out the plan on this problem:

${\dot {m}}_{6}-0.1893{\frac {kg}{s}}=0$

${\dot {m}}_{6}=0.1893{\frac {kg}{s}}$

Hence, from the mass balances on the evaporator:

${\dot {m}}_{4}={\dot {m}}_{1}-{\dot {m}}_{2}-{\dot {m}}_{6}=(0.5-0.2666-0.1893){\frac {kg}{s}}=0.0441{\frac {kg}{s}}$

So the final answers are:

${\mbox{Evaporator Efficiency}}=53.3\%$

${\mbox{Waste stream rate}}=0.0441{\frac {kg}{s}}$

Step 5: Check your work.

Ask: Do these answers make sense? Check for things like negative flow rates, efficiencies higher than 100%, or other physically impossible happenings; if something like this happens (and it will), you did something wrong. Is your exit rate higher than the total inlet rate (since no water is created in the processes, it is impossible for this to occur)?

In this case, the values make physical sense, so they may be right. It's always good to go back and check the math and the setup to make sure you didn't forget to convert any units or anything like that.