EQUATIONS (EFFECTIVE DRAFT)
Area of the plate
2 ft 2 ft 4 sqft ( 13)
Areaof theplate, sqin.
4 sq ft 144 sq in/sqft 576 sqin. ( 14)
Force of atmospheric pressure over plate surface
576 sq in. 2 in. 1,152 cu in. ( 15)
Pulling force onplate
1,153cuin. 0.58 664oz ( 16)
Pulling force onplate, lb
644 oz 16 oz 42lb ( 17)
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EQUATIONS (STACK DRAFT)
Draft
0.023lb/cu ft 100 ft 2.3lb/sq ft ( 18)
Draft,lb/sqin.
2.3lb/sq ft 144 ft 0.016lb/sqin. ( 19)
Draft,in.H 0
0.016 28in.H 0/psi 0.045inH 0 ( 20)
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FIG. 3 TYPICAL NATURAL-DRAFT FIRED HEATER
Hot fue gas
Damper
Stack draft
measurement
Fire box draft
measurement
Radiant
section
Burner
Air resistor
gauge will read 1.0 psi (Fig. 2). This means that to convert
from psi to in. H2O, multiply the psi value by 28. The draft
in our 100-ft high stack full of 300° F. stationary air, then, is
0.45 in. H2O (Equation 20).
Don’t we need fire to create draft?
What we require to create draft in a stack is heat to warm
the air from 70° F. to 300° F. The pressure at the top of the
stack is ambient.
If we heat air inside the stack with an electric heating coil
and cap the stack’s top to prevent flow, we would still measure the inside pressure at the bottom of the stack to be 0.45
in. H2O lower than ambient air pressure outside of the stack.
As the gas inside the stack gets hotter and less dense,
draft increases. As the air outside the stack cools and becomes denser, draft also increases.
The author has omitted three elements from this analysis
because of their minimal impact on draft. These include:
• The frictional losses of flue gases in the stack, as stacks
are made fat enough for friction to not play a role (e.g., ve-
locities less than 50 fps).
• The molecular weight difference between air and flue
gas.
• Wind blowing across the top of the stack.
• Ambient air temperature.
Let’s say a stack is 100 ft high. The ambient air temperature is 70° F., and the temperature of gas inside the stack
is 300° F. The molecular weight of the gas inside the stack
is pretty close to air. If you refer back to the example of the
giant hot-air balloon calculation, the density difference between the cool ( 70° F.) ambient air and the heated (300° F.)
air inside the balloon was 0.023 lb/cu ft. This density difference applies to the entire 100-ft height of the stack so that
the effect is similar to that inside the balloon, where cold,
denser air pushes up the lighter, hotter air. In other words,
the 100-ft column of colder air forces up the 100-ft column
of hotter air inside the stack.
This pressure difference is called draft. It is a result of the
density difference between hot flue gas in the heater stack
and the ambient air outside multiplied by the 100-ft height
of the stack, which in this instance, is calculated as 2. 3 lb/
sq ft (Equation 18).
Anyone who has worked on a process heater knows draft
is never expressed in lb/sq ft, but in inches of H2O. Converting lb/sq ft to in. H2O requires a two-step calculation that
involves first converting draft in lb/sq ft into psi by dividing
stack draft by 144 sq in./sq ft, which works out to be 0.016
psi. (Equation 19).
But how do you convert from psi to in. H2O? If you fill a
glass tube with cool 60° F. water to a height of 28 in., and
then place a pressure gauge at the bottom of the tube, your
