3. • The greatest opportunity for savings in the
installation and operation of ventilation
system lies in the design of airway.
• The shape, size and surface conditions must
be exercise in design of airways.
3
4. Shape And Surface Characteristics:
• McElroy has made cost calculations for various
types of airways and represented in Table-8.1.
• In general the results of indicate that the
unlined cost about half as much lined and
circular airways are slightly cheaper than
circular.
4
5. • Economic Velocities:
• The most economical velocity to employ in
designing airways is a function of air flow and
size of airway selected.
• The following approximate ranges are there:
Unlined 600-10000fpm
Smooth-Lined 2000-2500fpm
Timbered 1000-1500fpm
5
6. • Economic Size of Airway
• It is the most important in economical
distribution of air to determine optimum size
of airway.
• Two methods are there for determine size:
a) Graphical method
b) Numerical method
6
7. Graphical Method:
Several sizes of airways are assumed and
estimated annual capital, operating and
overall cost for each assumption is calculated,
then these costs are plotted against size ,
then that size is selected that results in lower
over all cost.
7
8. Numerical Method:
• In this method we derive an equation for
annual over all cost in terms of airway size and
then take its first derivative of expression and
equate to zero to get minimum cost.
• This method involves less computation and
no plotting.
8
9. • To demonstrate the procedure employed in
second method, an equation for optimum
diameter of circular air shaft will be developed.
• Consider first annual operating cost.
• The equation of head loss in terms of D is:
HL=1.25K(L+Le)Q2/D5
• The input brake horsepower is:
Pm=Pa/Ƞ= HLQ/6350Ƞ
9
10. • Annual operating cost is:
Co = (co)(Pm)
co is cost per hp-yr
• For capital cost fist determine volume of rock:
Y=π/4D2L
• Annual capital cost is:
Cc=ccceY
10
11. • Overall cost per year is sum of both operating
and capital cost. So
C = Co + Cc
C = 1.97x10-4coK(L+Le)Q3D-5/Ƞ + 0.785ccceLD2
• Taking first derivative of C with respect to D
and then equate it to zero.
ccceLD7-6.27x10-4coK(L+Le)Q3/Ƞ=0
11
13. • If rectangular opening is involved then
calculate the dimensions of rectangle whose
area is equal to that of a circle of diameter D.
• If ratio of sides of rectangle is N then lesser
dimension Da can be found as:
A(rect) = A(circle)
NDa 2 = π/4D2
Da=D/2√π/4
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