Análisis de las alternativas de intervención Tabla Nº 8 b

Many units are combined units (i.e., are derived as combinations of the base units). For example, units of velocity are derived from length and time, hence mi/hr. Examples of a number of combined units are listed in Table 4.2 for the three measurement systems discussed above. A number of these combined units will be discussed later in this chapter. In addition, some units are defined in terms of combinations of other units (e.g., Pa and lbf). Some common defined units are summarized at the beginning of this textbook.

Table 4.2 Examples of Combined Units for Three Measurement Systems

System cgs System SI Systems American System density g/cm3 _kg/m3 _lb

m/ f t3

velocity cm/s m/s f t/s acceleration cm/s2 m/s2 f t/s2 volumetric flow rate cm3/s m3/s f t3/s mass flow rate g/s kg/s lbm/s

concentration gmol/L∗ kgmol/m3 lbmol/ f t3 *often abbreviated M (i.e., molarity)

To produce conversion factors for combined units, conversion factors of base units can be raised to any power. The following are a few examples:

area:  100 cm 1 m 2 =10 4_cm2 1 m2 volume:  12 in 1 f t 3 =1728 in 3 1 f t3

Notice that raising a conversion factor to a certain power raises its units to that same power.

4.1.6

Force

An important parameter that has combined units is force, and one kind of force is weight. The force F to accelerate a mass m at an acceleration rate a is defined by Newton’s second law as

F= ma (4.2)

One form of acceleration is that associated with Earth’s gravity. (This is an acceleration because holding all matter on the earth’s surface requires constantly changing the direction of travel to produce a circular arc.) In this case, the rate of acceleration is described by the gravitational acceleration, g, and the force that an object exerts on the earth’s surface (its “weight”), is

Fweight= mg (4.3)

In the American engineering system, g is expressed in ft/s2, so the weight of a person whose mass was measured in lbmwould be in units of lbmft/s2. These units are not very

convenient, so an equivalent unit of pound-force (lbf) has been defined as 1 lbf ≡ 32.174

lbm ft/s2(the triple equals sign,≡, denotes a definition). In this way, weight can be cal-

culated and expressed in the convenient unit of lbf (often simply called “pounds”). Table

4.3 summarizes the values of g (at sea level) and the equivalent units of force for the three units systems discussed previously.

Table 4.3 Gravitational Acceleration (at Sea Level) and Defined Units of Force System g Defined Unit of Force cgs 980.66 cm/s2 _{1 dyne≡ 1 g cm/s}2

SI 9.8066 m/s2 _{1 Newton (N)≡ 1 kg m/s}2

American 32.174 ft/s2 _{1 pound-force (lb}

f)≡ 32.174 lbmft/s2

Each of the definitions of force units in Table 4.3 (dyne, Newton, and pound-force) can also be rearranged into the ratio form of a conversion factor as follows:

IMPORTANT! 1=_s1 g cm2_dyne= 1 kg m s2_N = 32.174 lbmf t s2_lb f

This conversion factor is used so frequently that it is sometimes referred to with the symbol gc. Like all conversion factors, the inverse of each expression is also a conversion factor:

4.1 Units 49 1=1 s 2_dyne g cm = 1 s2N kg m = 1 s2lbf 32.174 lbm f t

The use of such conversion factors is illustrated in Examples 4.2 and 4.3.

EXAMPLE 4.2 An object has a mass equal to 1 lbm. What is its weight in pounds-force (lbf)?

SOLUTION From Table 4.3, we note that g = 32.174 ft/s2. From Equation 4.3,

Fweight= mg = (1 lbm) 32.174 f t s2  _{1 lb} fs2 32.174 lbmf t ! = 1 lbf IMPORTANT!

This result explains why a pound-force was defined as it was - so that one pound-mass would weigh one pound-force (on Earth). However, you should understand that it is never correct to simply write that one pound-force equals one pound-mass!

EXAMPLE 4.3 An object has a mass equal to 8.41 kg. What is its weight (a) in Newtons (N) and (b) in pounds-force (lbf)?

SOLUTION a. From Table 4.3, we note that g = 9.8066 m/s2_{. Thus, from Equation 4.3,}

Fweight= mg = (8.41 kg)  9.8066 m s2   1 s2_N kg m  = 82.5 N b. From the answer to (a) and a conversion factor from the front of the book,

Fweight= (82.5 N) _{0.22481 lb} f 1 N  = 18.5 lbf

4.1.7

Pressure

Another important parameter that has combined units is pressure, which is force ex- erted per area. As with force, certain units of pressure are commonly used in each of the different units systems (i.e., cgs, SI, and American). Converting between these units re- quires using the conversion factors presented in the discussion of force. The units of pres- sure most commonly used are presented in Table 4.4.

Table 4.4 Commonly Used Units of Pressure

System Units of Pressure Abbreviation Defined and Equivalent Units

cgs Pascals Pa 1 Pa≡ 1N/m2_{= 1 g cm/s}2

SI kiloPascals kPa 1 kPa≡ 1000N/m2_{= 1000 kg m/s}2

A more extensive discussion of pressure with example problems is presented in Chapter 7. Now we are ready to discuss some of the variables that will be important to the prob- lem we are trying to solve.

4.2

SOME IMPORTANT PROCESS VARIABLES

As we design the process for neutralizing the HCl, we will need to describe the flow rate of material into and out of that process. For that description, we need to understand how to express fluid density, flow rate, and chemical composition.

4.2.1

Density

The density of a material is the mass of a unit volume of that material. A common sym- bol to represent density is the Greek letterρ(pronounced “rho”). Expressing the definition of density in equation form,

ρ = m

V (4.4a)

or, rearranging,

m=ρV (4.4b)

For most liquids and solids, the densities are listed in standard reference books, such as the CRC Handbook of Chemistry and Physics1or Perry’s Chemical Engineering Handbook.2 For liquids, the density is relatively independent of such variables as pressure and temper- ature. A handy density to remember is the approximate density of water at room tempera- ture, which is

ρwater,25◦C≈ 1.0g /cm3= 1000 kg /m3

As we would expect, the densities of gases are much less than those of liquids. For com- parison, the density of air at room temperature is

ρair,25◦_C≈ 0.0012g /cm3= 1.2 kg /m3

For gases, the density varies significantly with the temperature and pressure of the gas, as you may have learned in your chemistry classes. Typical units for density are listed in Table 4.2.

The density of a material can be used to relate the volume (V) of a material to its mass. For example, the density of liquid propane is 36.53 lbm/ft3. Therefore, from Equation 4.4b,

the mass of 500 gal would be m=ρV = 36.53 lbm f t3  (500 gal)  1 f t3 7.4805 gal  = 2442 lbm

4.2.2

Flow Rate

We often describe the rate at which the stream flows using the flow rate (the amount of material that passes a reference point within a unit time). Three common types of flow

4.2 Some Important Process Variables 51 rates used are

mass flow rate(symbol ˙m): the mass of a material that passes a reference plane within a unit time interval

molar flow rate(symbol ˙n): the number of moles of a material that passes a reference plane within a unit time interval

volumetric flow rate(symbol ˙V): the volume of a material that passes a reference plane within a unit time interval

With the symbols assigned, the “dot” over the letter indicates that the variable is a “rate” (i.e., a measure of amount per time). Typical units for volumetric and mass flow rate are listed in Table 4.2. To better understand these flow rates, it might be helpful to visualize water coming out of a kitchen faucet, where the mouth of the faucet could represent the reference plane in the above definitions. For example, if the volumetric flow rate of water from the faucet is 2.5 gal/min, we could collect the water coming from the faucet for one minute, and we would find that we had collected 2.5 gallons. If we collected for two minutes, we would find that we had collected 5 gallons, and so forth.

Conversion between mass flow rate (mass/time), volumetric flow rate (volume/time), and density (mass/volume) is

NOTICE

IMPORTANT EQUATION

˙

m= ρ ˙V (4.5)

You should understand the concepts and units in this equation and be able to reproduce this equation without looking at the book. Now let’s apply these variables to the acid- neutralization problem.

EXAMPLE 4.4 The average flow rate of HCl produced by our company is 11,600 L/hr (see Chapter 3). Its density is approximately the same as that of water: 1000 kg/m3, or 1 kg/L.

a. What is the equivalent of this flow rate in units of cm3_/s?

b. What is the mass flow rate for this stream in kg/hr?

SOLUTION a. From the units of the given flow rate (L/hr), we recognize that we have been given a volumetric flow rate ( ˙V). Applying the appropriate conversion factors,

˙ V=  11, 600L hr   1000 cm3 1 L   _{1 hr} 60 min   1 min 60 s  = 3, 222cm 3 s

Again, notice that the units canceled out to leave the units we were seeking, indicating that we used the conversion factors correctly.

b. From Equation 4.5, ˙ m= ρ ˙V= 1000 kg 1 m3   11, 600 L hr   1 m3 1000 L  = 11, 600kg hr