PROCESS
and PROCESS VARIABLES
LESSON
OBJECTIVE: Your objectives in completing this lesson
are to:
1. understand the mole unit, definition of
density, and convert mole fractions to mass fraction and vice-versa.
2. choose an appropriate basis for solving
problems.
REQUIRED
INSTRUCTIONAL
NOTES:
In
this lesson we learn some definitions and methods of calculating variables that
characterize operation of processes.
mass
in g (or lb)
Moles: Number
of g (or lb) moles = ________________________
molecular
weight
Make
sure you write g moles (or lb moles or ton moles, etc.), not just moles. Note that gas analyses are always
assumed to be reported in mole percent (= vol. %) or volume
fraction, unless otherwise stated. Liquid
or solid analyses are
reported usually in mass percent or mass fraction.
Other commonly used definitions are:
molarity
of A = g moles
A / liter of solution
molality
of A = g moles A
/ kg solvent
ppm
of A = g A
/ 106 g solvent
Density,
r º
Mass/Volume: Density is the mass of a substance per unit
volume. Thus, it has units of lbm/ft3,
g/ml or kg/m3, etc. Although
the law of conservation of mass allows us to set up balances in terms of masses
of various species, volumes may not be conserved, especially for gases. For liquids and solids, volumes are usually
additive. However, for reacting
systems even for liquids and solids, conservation of volume is a poor
assumption. In many cases, therefore,
when data on volume and density are given, it is a good idea to convert these
to masses and then attempt to solve the problem.
It is
very useful to memorize the density of water in CGS (1 g/cc) and FPS (62.4 lbm/ft3)
units. Quite often the specific gravity (SG) of a substance is given and needs
to be converted to density of the substance using the following relationship
(for liquids and solids):
specific
gravity of A = Density of A / Density of water. Or SGA = rA/rwat
Thus, Kerosene with a SG of 0.81, has a density =
SGA g/cc = 0.81 g/cc, or (in FPS system) SGA*62.4 lbm/ft3
= 50.54 lbm/ft3.
For gases the reference substance is air at standard conditions, instead
of water as in the above case for liquids and solids.
Ex. Pure
solid NaCl (table salt) has a specific gravity of 1.6. A 75 mass% solution of NaCl in water has a
density of 1.5 g/cc. Calculate the
molarity, molality, grams NaCl per lit. sol., and pounds NaCl per ft3
sol. Also calculate the density of the
solution if volumes were additive.
Sol. First
note that we have to contend with three different densities, that of the pure
water (1 g/cc), of pure NaCl (1.6 g/cc) and of the mixture (1.5 g/cc). Often a source of error is applying the
density value of one to the other. Let's
start with a basis (because mass% are given) of,
Basis: 100 g solution.
Sol.
Vol. =
mass/r =
100g/(1.5g/cc) = 66.667 cc =
0.002354 ft3
g NaCl
= 75g gmol NaCl = 75/(23+35.5) = 1.282 g
mol
lb
NaCl = (75g)(1lb/453.6g) = 0.1653 lb
(a)
Molarity = g mol NaCl/lit. Sol. = 1.282/0.0667
= 19.23 Molar
(b)
Molality = g mol NaCl/Kg water =
1.282/(0.025) = 51.28 Molal
(c) g
NaCl/lit Sol. = 75/0.0667
= 1125 g/lit.
(d) lb
NaCl per ft3 Sol. = 0.1563/0.002354 = 70.24
lb/ft3
(e)
NaCl vol = (75g)(1cc/1.6g) = 46.875 cc Water
vol. = 25 cc Tot. Vol = 71.875 cc
Hence
rSol = mass/vol. =
100/71.875 = 1.39 g/cc.
Concentration: To set up
material balances for mixtures or solutions of two or more substances, we need
to know their concentrations. It is
often expressed in terms of mass (weight) fraction or mass percent and mole
fraction or mole percent. Note that
molarity, molality and ppm are also ways of expressing concentration or
composition. We often need to convert
composition given in terms of one unit (say mass %) into another (say
mole%). Examples 3.3-1 through 3.3-5 (pp
48-55 of text) are good examples. We
will rework ex. 3.3-3 in the following.
Ex.
3.3-3. A mixture of gases contains 16% O2,
4.0% CO, 17% CO2, and 63% N2 by weight. The mol. wt. of the gases are: O2
= 32, CO = 28, CO2 = 44, and N2 = 28 g/g mol.
What
is the molar composition and the average molecular weight of this gas?
Sol. We need a basis for calculation. We use 100 g of the gas mixture as the basis
because the composition is given on mass
basis, allowing us to calculate the mass of each component gas. Thus,
BASIS: 100 g gas mixture.
Compo. Mass, g Mol.
Wt. g moles mole%
O2 16 32 0.50 15.25
CO 4 28 0.1429 4.36
CO2 17 44 0.3864 11.78
N2 63 28 2.25 68.61
TOTAL 100 3.2793 100.00
The
mol. wt. for the gas mixture
=
mass of the mixture / moles of mixture
= 100/3.2793 = 30.49 g/g mol
Pressure Measurement
and Units
LESSON
OBJECTIVE: Your objectives in completing this lesson
are to:
1. understand the difference between relative and
absolute pressures, learn about pressure measurement, and be able to manipulate
various units of pressure.
2. calculate the pressure exerted by a column of
fluid from its density and height and calculate pressure measured by
manometers.
3. be conversant with various temperature scales.
REQUIRED
INSTRUCTIONAL
NOTES:
Pressure
is defined as the force exerted per unit area.
Thus, P º F/A,
P =
Force/Area, hence the units are lbf/ft2 (or lbf/in2),
N/m2 (=Pa), etc.
Example: A block weighing 20 lbm is sitting
on a two in2 area. Hence, at
the bottom of the mass,
Force =
(mass) (acceleration)
= 20 lbf
Hence, P =
20 lbf/(2 in2) =
10 psi.
In the
example shown above, we assumed that there was no pressure at the top of the
block at b. If somebody is pressing down on the block at the top, obviously
pressure at the bottom (a) would be that much greater. So, we should actually write Pa =
10 psi + Pb. In the normal
earth environment, air is pressing down on everything at a nearly constant pressure
of about 14.7 psi. This pressure, known as barometric pressure,
is so pervasive that we have a special way of reporting it. Instead of adding the barometric pressure, we
report the pressure as Gauge
Pressure. Thus, in the example above, if Pb is
barometric pressure, then we could either report that Pa = 10 psia +
Pb or that Pa = 10
psig, where “a” and “g” after psi stands for “absolute” and “gauge”
respectively. Note that barometric
pressure varies with time and place.
Pressure
due to a liquid column:
Pa = Pb +
(Force due to mass of liquid) / Area
= Pb + (mass of liquid) (acc. due to gravity) / Area
Therefore:
Pa ‑ Pb = (Volume)
(density) (g) / (Area)
= (Area)(h)(r) (g) / (Area)
= h r g
where r is the fluid density. Note that the units of pressure, and hence of
(Pa ‑ Pb), in the
Pa
‑ Pb = h r (g/gc)
Numerically g/gc = 1. Hence if you are using height in ft. and r in lbm/ft3, (Pa
‑ Pb) is numerically = hr psf.
Also, when Pb = 0 (as in the
case of a mercury barometer) P = hr(g/gc). Note that gc = 1 in SI or CGS system.
If the
same pressure is being read using two manometers, each containing
different fluids, the fluid heights in
the manometers will be different. Then
we may write,
P
= h1r1
= h2r2
or r1/r2 = s1/s2
= h2/h1
where s is specific gravity.
This
leads to another set of pressure units, liquid columns heights. For example:
29.92
in Hg = (29.92 / 12
= 2116 lbf/ft2 =
14.69 lbf/in2
= 1 atm.
Two
liquid manometer:
As
shown in the adjacent diagram, two different fluids are often used in the two
legs of a manometer to measure pressure difference between two points, a and
b. In this type of manometer problems,
we start at a two fluid interface point (here point c) and choose another point
such that it is on the other leg of the manometer and at the same height as the
first point. So we start at point c and
d. Since both these points are inside
the same fluid (liquid 2) and at the same height, the pressure at these two
points must be equal. Therefore Pc
= Pd. We now need to find Pc
in terms of Pa and Pd in terms of Pb so that
we have an expression for Pa ‑ Pb. We know how to calculate pressure due to a
liquid column. Hence:
Pc = (h +
z) r1(g/gc) + Pa
and Pd = hr2
(g/gc) + zr1(g/gc) + Pb
Because Pc = Pd,
we have,
hr1(g/gc)
+ zr1(g/gc) + Pa = hr2(g/gc)
+ zr1(g/gc) + Pb
Therefore:
Pa
‑ Pb = h(g/gc)(r2
‑ r1)
What
if liquid 1 is actually a gas (air)?
Well, then r1
<< r2 and (Pa ‑
Pb) = hr2(g/gc).
More
than two fluids:
How
would you calculate Pa ‑ Pb for the
manometer shown in the adjacent diagram?
Again, we start at the fluid interface point c and choose the point d on
the other leg which is at the same height as c.
Thus, Pc = Pd (because of the same liquid / same
height principle). Hence, we write,
Pc = Pa + (h1 + h2)r1(g/gc)
Pd = Pb + (h2 + h3)r3(g/gc)
+ h1r2(g/gc)
Therefore:
Pa
‑ Pb = [(h2
+ h3)r3
+ h1r2
‑ (h1 + h2)r1][g/gc]
Example: Fluid 1 is water (s1 = specific
gravity = 1), Fluid 2 is Hg (s2
= 13.56) and fluid 3 is an oil (s3 = 1.5). h1 = 2 in., h2 = 12 in. and h3 = 16 in. (r2
= s2rwater = 13.56 x 62.4 lbm/ft3, r3
= s3rwater = 1.5 x 62.4 lbm/ft3
)
Therefore:
=
286 lbf/ft2 = 1.99 psi.
Calculations
in Terms of Liquid Height. As
pointed out earlier, liquid column heights can be used as pressure units. Some students find that manometer problems
are solved with greater ease in terms of liquid heights. As an example, let us solve the last problem
using inches of water as the primary unit of pressure. Of course, in order to carry out the
computations, all pressures must be converted into the same units. Thus,
Pc = Pa +(h1 + h2)liq.
1 = Pa + 14 in water
Pd = Pb + (h3 + h2)oil
+ (h1)liq. 3
= Pb + 28 in oil + 2
in. Hg
= Pb + [28"
oil x 1.5" water/1" oil] + [2" Hg x 13.56" water/1"Hg]
=
Pb + 42" water + 27.12" water = Pb
+ 69.12" water
Since: Pc
= Pd
Pa
+ 14" water = Pb + 69.12" water
Pa ‑
Pb = 55.12" water (which is = 286 psf).
The problem could have been solved with any other
liquid (e.g. mm Hg, or inches of oil, etc.) as the primary liquid. One advantage of using water as the primary
liquid is that the specific gravities of liquids are with reference to water,
making computations less prone to error.
We recommend using this approach.
Manometer
open to atmosphere:
Referring
to the adjacent diagram:
Pa = h r (g/gc) + Pb
= h r (g/gc) + baro. pressure
where barometric pressure refers to the outside
pressure which is also often (but not always) the atmospheric pressure. Please note, however, that atmospheric (or
barometric) pressure is not always equal to 1 atmosphere (1 atm), which is
always the same and is equal to 14.7 psi.
Because of this possible confusion, it is better to refer to outside
pressure as barometric.
Therefore:
(Pa
‑ baro. pressure) = h r (g/gc)
As
mentioned earlier, (Pa ‑ barometric pressure) is defined as Gauge
Pressure. Hence, gauge
pressure shows how much the system pressure is above the outside pressure. Thus,
(Pa
‑ baro. pressure)
= (Pa)gauge = h r (g/gc)
For example, in the above diagram if h = 15"
Hg, and baro. pressure = 30" Hg., then,
Pa =
15" Hg gauge ( =
45" Hg absolute.) = 7.36 psig
Vacuum Pressure.
Sometimes pipes and containers are maintained at pressures lower than
the barometric pressure. In such cases,
instead of reporting the absolute pressure, the system pressure is often
reported in terms of how much lower it is compared to the outside
pressure. This is called the vacuum
pressure. Hence, by definition,
Vacuum
Pressure ≡ barometric pressure ‑
absolute pressure
Note the similarities in the definitions of gauge and vacuum pressures. Indeed, the numerical values of the gauge and vacuum pressures are t