AGRORURAL ANA PSI

MESSAGE LEARN THESE RELATION- SHIPS!

Practice them until you can use them without looking at the book

˙

mA= xAm˙= MWA˙nA= MWAyA˙n= MWAcAV˙ (4.6a)

where Equation 4.6a presents various ways to express the mass flow rate of species A. Simple algebraic rearrangement of Equation 4.6a produces expressions for the molar flow rate of species A: ˙nA= ˙ mA MWA = xAm˙ MWA = yA˙n= cAV˙ (4.6b)

4.2 Some Important Process Variables 53 As with Equation 4.5, it is important that you fully understand and can reconstruct these relationships. For example, suppose that you know the value of the mass flow rate of a chemical compound (with known molecular weight) within a mixture, and that you also know the volumetric flow rate of that mixture. You should be sufficiently familiar with the relations in Equation 4.6 (both 4.6a and 4.6b) to be able quickly to compute the composition (in any of its various forms) of the compound in the mixture. Furthermore, you should also be very familiar with the relations and terms in Equation 4.5 so that you can use Equations 4.5 and 4.6 together. Make sure you understand the concepts and units of all variables in both equations and practice writing them until you can reproduce these equations without looking at the book.

EXAMPLE 4.5 For the acid-neutralization problem, the volumetric flow rate of the HCl solution coming from our manufacturing process is 11,600 L/hr, and the average concentration of HCl in that stream is 0.014

M, or 0.014 gmol/L. Based on this information and the answers in Example 4.4, a. How many gmol of HCl are in 88 m3of the solution?

b. How many gmol of HCl are flowing from the process per minute (i.e., what is the molar flow rate) when the volumetric flow rate of the solution is 11,600 L/hr?

c. What is the mass fraction of HCl in the solution?

SOLUTION a. Manipulating the definition of concentration,

nHCl= cHClV=  0.014 gmol L   88 m3 1000 L m3  = 1, 232 gmol HCl b. From Equation 4.6b, ˙nHCl= cHClV˙=  0.014 gmol L   11, 600 L hr   1 hr 60 min  = 2.71 gmol HCl /min c. From the definition of mass fraction,

xHCl= ˙ mHCl ˙ m = MWHCl˙nHCl ˙ m

Atomic weights are listed in the front of the book, giving MWHCl= 1.0 + 35.5 = 36.5 g HCl/gmol,

and ˙mis available from Example 4.4, so

xHCl=

 (36.5gHCl /gmol) (2.71gmol HCl /min) 11, 600 kg /hr   1 kg 1000 g   60 min 1 hr  = 0.00051

4.2.4

Conversion between Mole Fraction and Mass Fraction

It is common for a chemical engineer to know mass fractions (or mass percentages) of components in a mixture but to need the mole fractions – or to know the mole fractions but to need the mass fractions. In such cases, it is necessary to convert between the various types of fractions or percentages. When performing such conversions, the first step is to as-

sume an amount of material for the calculation (called the basis of calculation). The steps outlined in Figures 4.1 and 4.2 are suggested. You will note that when the given composi- tions are expressed as fractions or percentages, the most convenient basis of calculation is 100 units of mass or moles (e.g., 100 g, 100 kgmol) so that the amount of each component is easily calculated. Example 4.6 illustrates the application of the procedure outlined in Figure 4.1.

Assume a basis of 100 moles

From the known mole fractions,

calculate the number of moles

of each species

Using the molecular weights, convert the moles into mass for each species

Compute the desired mass fractions or percentages

Figure 4.1. Strategy for Converting Mole Fractions (or Percentages) to Mass Fractions (or Percentages)

Assume a basis of 100 mass units

From the known mass fractions,

calculate the mass of each

species

Using the molecular weights, convert the mass into moles for each species

Compute the desired mole fractions or percentages

Figure 4.2. Strategy for Converting Mass Fractions (or Percentages) to Mole Fractions (or Percentages)

EXAMPLE 4.6 One analysis of air produced the following approximate mole percentages: N2: 78.03 mole%

O2: 20.99 mole%

Ar: 0.94 mole%

What are the mass percentages of these components?

SOLUTION As outlined in Figure 4.1, we’ll select a basis of 100 gmol of air. As explained in the text, we select that number because it is convenient, and because it makes it easy to calculate how many gmol of each substance are present in those 100 gmol.

N2:

nN2= yN2n=

 78.03gmol N2

100 gmol total 

(100 gmol total) = 78.03 gmol N2

O2:

nO2= yO2n=

 20.99gmol O2

100 gmol total 

(100 gmol total) = 20.99 gmol O2

Ar:

nAr= yArn=

 _{0.94 gmol Ar} 100 gmol total



(100 gmol total) = 0.94 gmol Ar (continued on the next page)

4.2 Some Important Process Variables 55 EXAMPLE 4.6

(continued)

Continuing to follow the strategy in Figure 4.1, we’ll now calculate the number of grams of each substance:

N2: mN2= MWN2nN2= (28 g N2/gmol) (78.03 gmol N2) = 2185 g

O2: mO2= MWO2nO2= (32 g O2/gmol) (20.99 gmol O2) = 672 g

Ar: mAr= MWArnAr= (39.95 g Ar/gmol) (0.94 gmol Ar) = 38 g

Finally, the total mass of the mixture is the sum of the masses of the components, i.e., 2185 + 672 + 38 = 2895 g, so xN2= mN2 m = 2185 g N2 2895 g = 0.755 xO2= mO2 m = 672 g O2 2895 g = 0.232 xAr= mAr m = 38 g Ar 2895 g = 0.013

By the way, one should always check the calculation to see if the mass fractions add up to 1.000. Check: 0.755 + 0.232 + 0.013 = 1.000!

4.2.5

Dimensional Consistency

Equations that correctly describe physical phenomena must obey the rules of dimen- sional consistency:

1. Terms that are added together (or subtracted) must have the same units. For ex- ample, in the equation Q= ab + c2_{, the units of ab must be the same as those of c}2_.

2. Exponents must be unitless. Thus, if an exponent consists of several terms, the units of all those terms must cancel. For example, in the equation y= xab/c, the units in the term ab/c must all cancel out to leave no units.

CONCLUDING COMMENTS

Keeping track of the units in a calculation provides a number of benefits. One that has already been mentioned is that it provides a way of safeguarding against the incorrect application of conversion factors. It also provides a check on the equation being used for the calculation, because an error in that equation might produce erroneous units for the answer (e.g., kg m/s2 _{for a velocity). An error in the equation might also produce a violation of}

the rules of dimensional consistency, which would be discovered as units are carefully monitored. For these reasons, the habit of giving careful attention to units is vital to good engineering practice, and the reader should work hard to establish that habit in all technical calculations.