FUNDAMENTAL CONCEPT OF MATERIAL AND ENERGY
BALANCES
Syllabus: Fundamental concept of material and energy
balances. Units and dimensional analysis.
Questions:
1.
What fundamental concepts are involved for material and
energy balances ? [1991] 12
2.
What is dimensional analysis? [1991] 4
3.
Derive the dimension of Reynold’s number. [1994] 4
PHYSICAL
QUANTITIES
Any physical
quantity has two parts:
a unit, which tells what the quantity is an gives the standard by
which it is measured, and
a
number, which tells how many units
are needed to make up the physical
quantity.
Explanation:
The statement that ‘the distance between two points is
3 m’ means: a definite length has been
measured; to measure it, a standard length, called the meter, has been chosen
as a unit; and three 1-m units; laid end-to-end, are needed to cover the
distance. No physical quantity is defined until both the number and the unit
are given.
SI Units
The SI system covers the entire field of science and
engineering, including electromagnetics and illumination. By international
agreement, standards are fixed arbitrarily for the quantities mass, length, time, temperature, and the
mole.
Base units of SI system
Mass: The standard of mass is the kilogram (kg), defined as the
mass of the international kilogram, a platinum cylinder preserved at Sèrves,
Frances.
Length: The standard of length is the meter (m), defined (since
1983) as the length of the path traveled by light in vacuum during a time
interval of 1/299,972,458 of a second.
Time: The standard of time is the second (s) defined as
9,192,631.770 frequency cycles of a certain quantum transition in an atom of 133Ce.
Temperature: The standard of temperature is the Kelvin (K), defined
by assigning the value 273.13 K to the temperature of pure water at its
triple-point (temperature at which liquid water, ice, and steam can co-exist at
equilibrium.
Mole: The (abbreviated mol) is defined as the amount of a substance
comprising as many elementary units as there are atoms in 12 g of 12C.
Derived units if SI system
Energy and work are measured in newton-meters, a unit called the joule (J), and so
1
J º
1 N-m = 1 kg-m2/s2
Power is measured in joules per
second, a unit called the watt (W).
Heat Like work heat is
measured in joules.
Pressure units: The natural unit of pressure in the SI system is
the newton per square meter.
This unit, is called the pascal (Pa), is inconveniently small,
and a multiple, called the bar, is
also used. It is defined by
1 bar º 1 x
105 Pa = 1.01325 bars
Value of acceleration
due to gravity : g = 9.80665 m/s2.
CGS Units
Mass: The standard for mass is the gram (g) defined
by
1
g º
1 x 10-3
kg
Length: The standard of length is the centimeter
(cm), defined by
1
cm º
1 x 10-2
m
Time: The standard for time is second.
Thermodynamic temperature: Kelvin
Temperature: The standard of temperature is Celsius.
T0C
º
T K -
273.13
Mole: The standard is same as SI i.e. mole (mol).
Energy & Work: The standard is erg.
1
erg º
1 dyn-cm = 1 x 10-7 J
Heat: Defined as calorie (cal)
1
cal º
4.1840 x 107 erg = 4.1840 J
Value of acceleration
due to gravity : g = 980.665 m/s2.
FPS Engineering Units
Mass: The standard of mass is the avoirdupois pound (lb), defined by
1
lb = 0.4536 kg
Length: The standard length is the inch (in), defined as 2.54 cm. This is
equivalent to defining the foot (ft)
as 1 ft º 2.54
x 12 x 10 -2
m = 0.3048 m
Time: Standard of time is second.
Thermodynamic temperature: The thermodynamic
temperature scale is called the Rankine
scale, in which temperatures are denoted by degrees Rankine and defined by
10R
º
1/ 1.8 K
The ice-point on Rankine scale is 273.15 x 1.8 = 491.67 R0.
Standard Temperature: Fahrenheit
T0F
º
T0R -
(491.67 -
32) = T0R - 459.67
The relation between the Celsius and the Fahrenheit scales
is given by the exact equation
T0F
= 32 + 9/5 0C = 32 + 1.8 0C
Heat: The unit of heat is the British thermal unit(Btu), defined by the relation
1
Btu / lb /0F º 1 cal / g/ 0C
Standard acceleration of free fall: g = 32.174 ft/s2.
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SI
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CGS
Absolute
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CGS-
Gravitational
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FPS
Absolute
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FPS-
Gravitational
|
|
Force
Work, Energy
Power
|
newton (N)
joule (J)
watt (W)
|
dyne
erg
erg/s
|
gram weight (gm-wt)
|
poundal
foot-poundal
ft-poundal/s
|
pound force (lbf)
foot-pound force (ft-lbf)
ft-lbf / s
|
Conversion of Units
Example-1: Convert newtons to pounds force.
1
N = 1 kg-m/s2 = (1
kg x 1 m) / s2.
=
(1/0.4536)lb x (1/0.3048)ft / s2.
=
1/ (0.4536 x 0.3048) poundal
=
(7.2329 / 32.167) lbf.
=
0.2248 lbf.
Example-2: Convert Btu to Cal.
1 Btu / lb-0F
º
1 Cal / g-0C
1 Btu = (1 Cal x lb x 0F) / (g x 0C)
=
{453.6 g x (5/9)0C} / (g x 0C)
=
252 Cal
Example-3: Convert atmospheric pressure to pound force
per square inch.
1 atm = 76 cm of Mercury column
=
76 cm x 13.6 g/cm3 x 980.6 cm/s2.
=
1033.6 g/cm2. [in gravitational unit]
=
1033.6 x (1/453.6) lb / (1/2.54)2in2.
=
14.701 lbf / in2.
Example-4: Convert horse power to kilowatt.
1 hp = 550 ft-lbf /s
=
550 x (1 ft) x (1 lbf) / s
=
550 x (0.3048 m) x (0.4536 kgf) / s
=
76.0415 m kgf / s
=
76.0415 x 9.806 J/s
=
745.66 W
=
0.74566 kW
Material Balances
The law of
material balance states that matter cannot be either created or destroyed.
·
In other words the materials entering any
process must either accumulate or leave the process.
·
There can be no loss or gain during the process.
·
Material balances must hold over the entire
process or apparatus or over any part of it.
·
They must apply to all the material that enters
and leaves the process or to any one material that passes through the process
unchanged.
Energy Balances
The law of
energy balance states that energy cannot be either created or destroyed.
·
In other words the energy input and output of a
process must be equal.
·
To be valid, an energy balance must include al
types of energies: heat, mechanical energy, electric energy, radiant energy,
chemical energy, or other forms of energy.
Dimensionless
Equations
An equation
in which all terms have the same dimensions is a dimensionally homogeneous equation.
All equations that have been derived mathematically from
basic physical laws consists of terms that have the same dimensions.
For
example, the equation for the vertical distance Z traversed by a freely falling
body during time t, if the initial velocity is u0 :
Z = u0
t + ½
g t2. (1)
The dimension of [Z = L [i.e.
length]
[u0
t] = L
[½
g t2] = L
i.e. the dimension at both left and right hand sides are
equal hence this equation can be called dimensionally
homogeneous equation.
Equation (1) can be expressed in another way by dividing
both sides by the term Z:
All the terms in both sides of the equation have no
dimension hence the equation is called dimensionless
equation.
Advantages of
dimensionless equation:
If consistent units are used for
all the terms in equation (2) then either cgs or fps units can be used without
any conversion factors. e.g (u0 t / Z) can be expressed in cgs and
(gt2 / Z) can be expressed in fps unit system.
Dimensional equations
Equations
derived by empirical methods, in which experimental results are correlated by
empirical equations without regard to dimensional consistency, usually are not
dimensionally homogeneous ad contain terms in several different units.
Equations of this type are called dimensional
equations, or dimensionally
nonhomogeneous equations.
In
these equations two or more length units e.g., inches and feet, or two or more
time units e.g., seconds and minutes, may appear in the same equation.
Example: A formula for the rate of heat loss from a horizontal pipe
to the atmosphere by conduction and convection is
where q = rate of heat loss, Btu / h
A
= area of pipe surface, ft2.
DT =
excess of temperature of pipe wall over that of surrounding atmosphere, 0F
D0
= outside diameter of pipe, in.
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Left hand side of the equation
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Right hand side of the equation
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Dimensional formula
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[q/A] = Qq-1L-2.
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 |
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Unit
|
Btu h -1
ft -2.
|
(0F) 1.25
in -
0.25.
|
Obviously the dimension or unit
of the left hand side of the equation is not equal to the right hand side,
hence the equation is a dimensional equation.
* The
quantities substituted must be expressed in the units given, or the equation
will give the wrong answer. If other
units are to be used then the coefficient (here 0.50) must be changed.
DIMENSIONAL ANALYSIS
A physical problem may be
expressed in the form of a mathematical equation. This mathematical equation
can be derived by two methods:
1)By empirical experimentation
In
this case a dependent (y) variable may
be expressed in the following form:
y
= f (a, b, c, d, ...) where y depends on some independent variables like
a, b, c, d, ... etc.
For
example the pressure loss (DP) from friction in a long, round, straight, smooth
pipe depends on all these variables: the length (l) and diameter (d) of the
pipe, the flow rate of the liquid ( v ) , and the density (r) and viscosity (h) of the liquid.
i.e.
DP = f (l, d, v, r, h)
Now
to relate left and right hand sides the variable in the left hand side is
varied individually keeping the other variables constant. Similarly all other
variable in the right hand side of the equation is varied and thus the over all
relationship is determined. But this procedure is laborious.
2)By mathematical derivation
For
example v = u + f t , this is an
equation derived mathematically for an object moving at an initial velocity, u
and acceleration f. After time t the
velocity of that object is v which is related to the right hand side of the
equation mathematically without any experimentation.
There
exists a method intermediate between formal mathematical development and a
completely empirical study. It is based on the fact that if a theoretical
equation does exist among the variables affecting a physical process, that
equation must be dimensionally homogeneous. Because of this requirement, it is
possible to group many factors into smaller number of dimensionless groups of
variables. The groups themselves rather than the separate factors appear in the
final equation.
This
method is called dimensional analysis,
which is an algebraic treatment of the symbols for units considered
independently of magnitude.
Significance of dimensional analysis:
1.
It drastically simplifies the task of fitting
experimental data to design equations,
2.
It is useful in checking the consistency of the units
in equations
3.
Useful in converting the units and
4.
In the scale up of data obtained in model test units to
predict the performance of full-scale equipment.
Dimensional analysis
of the motion of a perfect pendulum
Example: What
information you can be obtained from a dimensional analysis of the motion of a
perfect pendulum?
Solution:
It may be expected that the period P of a pendulum would
depend on
1.
the mass m
of the pendulum,
2.
the length l
of the pendulum,
3.
the angle f swept by the pendulum and
4.
the local value of the acceleration due to gravity g.
This assumptions can be expressed by the following equation:
P
= f (m, l, f, g) eqn. 1
If equation 1 is a relationship derivable from basic laws of
physical laws, all terms in the function
f(m,l, f, g) must have the same dimensions of P. So
this function may be assumed to have the following form:
P
= k ma, lb, gc, fd. eqn.
2
Where, the k is a
constant, a,b,c and d are integers or integral fractions
Since f is an angle it has no unit i.e. it has no dimension
hence d = 0 and the f
term will not appear in the further equations.
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Dimension of
left-hand side of eqn. 2
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Dimension of
right-hand side of eqn. 2
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[P] = q
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[k] = 1 (i.e.
no dimension)
[m] = M
[l ] = L
[g] = L q -2.
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Substituting the dimensions in equation 2
q = Ma,
Lb, (L q
-2)c. eqn.
3
The exponents in the right hand side and the left hand side
of eqn. 3 are as follows:
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Exponents of left
hand side of equation 3
|
Exponents of left
hand side of equation 3
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Exponent of q : 1
Exponent of M: 0
Exponent of L: 0
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Exponent of q : -2c
Exponent of M: b
+ c
Exponent of L: a
|
Equating both sides:
1 = -2c
0 = b +
c
0 = a
Solving this equation a
= 0
b
= -
c = 1/2
c
= -
1/2
and thus equation 3 becomes
q = M 0
L 1/2 ( L q -2 ) -1/2.
Hence, P
= l 1/2 g -1/2.
= (l
/ g) 1/2.
The quantity m drops out, i.e. the period in independent of
the mass of the pendulum.
Therefore,
P
= k (l / g) 1/2.f1(f)
where f1(f) is a function of the
angle f
. The dimensional analysis yield no information about the numerical value of
the constant k or of the function of the angle. The value of k and f1(f)
should be determined experimentally.
Now the by experimentation only the value of (l
/ g) is varied and thus it was found that k = 2p and when value of f is
very small f1(f) » 1.0
Hence the final equation describing the time period of a
pendulum is:
Thus the dimesional anlysis drastically simplifies the task of fitting experimental
data to design equations.
Testing the dimensional consistency of an equation
Important principles in testing
the dimensional consistency of an equation are as follows:
1.
The sum of the exponent relating to any given dimension
(e.g. length) must be the same on both side of the equation.
2.
An exponent must itself be dimensionless - a pure
number.
3.
All the factors in the equation must be collectible
into set of dimensionless groups. The groups themselves may carry exponents of
any magnitude, not necessarily whole number.
Example:
Check the dimensional consistency of the following empirical equation
for heat transfer coefficient
h = 0.023 G 0.8 k 0.67 cp 0.33
D -0.2 m - 0.47.
where h = heat transfer coefficient
G = mass velocity
k = thermal conductivity
cp= specific heat
D = diameter
m
= absolute viscosity
Solution
The dimension of length, mass, time, temperature and heat
(energy or work) are expressed in L, M, q, T and H respectively.
Dimension of left hand side of the equation is:
[h] = H 1L
-2
q- 1
T -
1.
Dimension of right hand side of the equation is:
[G] = M q -1
L -2.
[k] = Hq-1
L-1
T-1.
[cp]
= HM-1T-1.
[D] = L
[m] = ML-1 q-1.
Substituting this in the right hand side of the equation :
= (M q -1
L -2
)0.8 (Hq-1 L-1 T-1
)0.67 (HM-1T-1 )0.33 L -0.2 (ML-1
q-1
) -0.47.
= M0.8
-
0.33 -
0.47 L -
1.6 - 0.67 - 0.2
+ 0.47 q -0.8 - 0.67 + 0.47 H 0.67
-
0.33 -
0.47 T -
0.67 - 0.33.
= M0 L- 2 q- 1 H1 T- 1.
Now equating both sides of the equation:
M 0 L
-2
q- 1 H
1 T -
1. = M0 L- 2 q- 1 H1 T- 1.
Since the exponents on the left hand side is equal to the
corresponding exponents of the right hand side of the equation hence the
equation is dimensionally homogeneous.
Dimensionless groups
A number of
important dimensionless groups have been found by dimensional analysis or by other
means. Many are used so frequently that they are given special names and
symbols.
where D = internal diameter of a pipe (ft)
u
= velocity of the fluid within the pipe
(ft/s)
r = density of the fluid (lb/ft3)
m = viscosity of the fluid (lb/ft-s)
where cp = specific heat of gas at
constant temperature (Btu/lb-0F)
m = viscosity of the gas or vapor (lb/ft-h)
k = thermal
conductivity of gas (Btu/ft-h-0F)
The
numerical value of a dimensionless group for a given case is independent of the
units chosen for the primary quantities, provided consistent units are used
within that group. The units chosen for one group need not be the same with that
of another dimensionless group. For example, second may be chosen in Reynolds
number while hour may be used in Prandtls number.