Engineering Bulletin No 1: Boiler and Furnace Testing | Page 6

180° F. to steam at 100
pounds gage pressure, because 7.5 pounds from and at 212° F. is
equivalent to 7 pounds from 180° F. to steam at 100 pounds. Therefore,
the 7,278 B. t. u. is the amount of heat usefully employed in making
steam per pound of coal fired, and so it is the output. Accordingly, the
efficiency of the boiler is--
Output 7,278 ~ Efficiency = ------ = ------ = 0.54, nearly. Input 13,500
In other words, the efficiency of the boiler is 0.54, or 54 per cent,
which means that only a little more than half of the heat in the coal is
usefully employed in making steam.
The chart shown in figure 3 is given to save the work of figuring the
efficiency. If the equivalent evaporation per pound of coal is calculated
and the heating value of the coal is known, the boiler efficiency may be
found directly from the chart. At the left-hand side locate the point
corresponding to the equivalent evaporation and at the bottom locate
the point corresponding to the heating value of the coal. Follow the
horizontal and vertical lines from these two points until they cross, and
note the diagonal line that is nearest to the crossing point. The figures
marked on the diagonal line indicate the boiler efficiency.
Take the case just worked out, for example. The equivalent evaporation
is 7.5 pounds and the heating value of the fuel is 13,500 B. t. u. At the
left of the chart locate the point 7.5 midway between 7 and 8 and at the
bottom locate the point 13,500 midway between 13,000 and 14,000.
Then follow the horizontal and vertical lines from these two points
until they cross, as indicated by the dotted lines. The crossing point lies
on the diagonal corresponding to 54, and so the efficiency is 54 per
cent.

BOILER HORSEPOWER OR CAPACITY.
The capacity of a boiler is usually stated in boiler horsepower. A boiler
horsepower means the evaporation of 34.5 pounds of water per hour
from and at 212° F. Therefore, to find the boiler horsepower developed
during a test, calculate the evaporation from and at 212° F. per hour
and divide it by 34.5.
Take the test previously mentioned, for example. The evaporation from
and at 212° F. or the equivalent evaporation, was 7.5 pounds of water
per pound of coal. The weight of coal burned per hour was 5,000 ÷ 10
= 500 pounds. Then the equivalent evaporation was 7.5 × 500 = 3,750
pounds per hour. According to the foregoing definition of a boiler
horsepower, then--
3,750 Boiler horsepower = ----- = 109. 34.5
The "rated horsepower" of a boiler, or the "builders' rating," is the
number of square feet of heating surface in the boiler divided by a
number. In the case of stationary boilers this number is 10 or 12, but 10
is very commonly taken as the amount of heating surface per
horsepower. Assuming this value and assuming further that the boiler
tested had 1,500 square feet of heating surface, its rated horsepower
would be 1,500 ÷ 10 = 150 boiler horsepower.
It is often desirable to know what per cent of the rated capacity is
developed in a test. This is found by dividing the horsepower
developed during the test by the builders' rating. In the case of the
boiler tested, 109 horsepower was developed. The percentage of rated
capacity developed, therefore, was 109 ÷ 150 = 0.73, or 73 per cent.
HEATING SURFACE.
The heating surface of a boiler is the surface of metal exposed to the
fire or hot gases on one side and to water on the other side. Thus, the
internal surface of the tubes of a fire-tube boiler is the heating surface
of the tubes, but the outside surface of the tubes of a water-tube boiler
is the heating surface of those tubes. In addition to the tubes, all other

surfaces which have hot gases on one side and water on the other must
be taken into account. For instance, in a fire-tube boiler from one-half
to two-thirds of the shell (depending on how the boiler is set) acts as
heating surface. In addition to this, the surface presented by both heads,
below the water level, has to be computed. The heating surface of each
head is equal to two-thirds its area minus the total area of the holes cut
away to receive the tubes.
COST OF EVAPORATION.
The cost of evaporation is usually stated as the cost of fuel required to
evaporate 1,000 pounds of water from and at 212° F. To find it,
multiply the price of coal per ton by 1,000 and divide the result by the
product of the equivalent evaporation per pound of coal and the number
of pounds in