Engineering Bulletin No 1: Boiler and Furnace Testing | Page 5

B. t. u. Every pound of coal fired,
then, carried into the furnace 13,500 heat units, and this value therefore
is the input to be used in calculating the boiler efficiency.
During the test 5,000 pounds of coal was fired and 35,000 pounds of
water was fed and evaporated. This means that 35,000 ÷ 5,000 = 7
pounds of water was evaporated per pound of coal burned. This is the
"actual evaporation," and the heat required to evaporate this 7 pounds
of water is the output to be used in calculating the efficiency.
Every fireman knows that it takes more coal, and therefore more heat,
to make steam with cold feed water than with hot feed water; also, that
it is somewhat easier to make steam at a low pressure than at a high
pressure. So it is plain that the heat required to evaporate 7 pounds of
water into steam depends on two things, namely, (1) the temperature of
the feed water and (2) the pressure of the steam in the boiler. From the
data of the test, both the average feed-water temperature and the
average steam pressure are known, and so it is a simple matter to find
out the amount of heat needed to evaporate 7 pounds of water from the
average temperature to steam at the average pressure.
A pound of water at 212° F. must have 970.4 B. t. u. added to it to
become a pound of steam at 212° F., or zero gage pressure. This value,
970.4 B. t. u., is called the latent heat of steam at atmospheric pressure,
or the heat "from and at 212° F." It is the heat required to change a
pound of water from 212° F. to steam at 212° F., and is used by

engineers as a standard by which to compare the evaporation of
different boilers.
In a boiler test the temperature of the feed water is usually something
less than 212° F., and the steam pressure is commonly higher than zero,
gage. In the test outlined previously, the feed-water temperature was
180° F. and the pressure was 100 pounds per square inch, gage. It must
be clear, then, that the amount of heat required to change a pound of
water at 180° to steam at 100 pounds gage pressure is not the same as
to make a pound of steam from and at 212° F.
To make allowance for the differences in temperature and pressure, the
actual evaporation must be multiplied by a number called the "factor of
evaporation." The factor of evaporation has a certain value
corresponding to every feed-water temperature and boiler pressure, and
the values of this factor are given in the accompanying table. Along the
top of the table are given the gage pressures of the steam. In the
columns at the sides of the table are given the feed-water temperatures.
To find the factor of evaporation for a given set of conditions, locate
the gage pressure at the top of the table and follow down that column to
the horizontal line on which the feed-water temperature is located. The
value in this column and on the horizontal line thus found is the factor
of evaporation required. If the feed water has a temperature greater than
212° F., obtain the proper factor of evaporation from the Marks and
Davis steam tables.
Take the data of the test, for example. The average steam pressure is
100 pounds, gage. The average feed-water temperature is 180° F. So, in
the table locate the column headed 100 and follow down this column to
the line having 180 at the ends, and the value where the column and the
line cross is 1.0727, which is the factor of evaporation for a feed-water
temperature of 180° F. and a steam pressure of 100 pounds, gage.
This factor, 1.0727, indicates that to change a pound of water at 180° F.
to steam at 100 pounds requires 1.0727 times as much heat as to change
a pound of water at 212° F. to steam at atmospheric pressure. In other
words, the heat used in producing an actual evaporation of 7 pounds
under the test conditions would have evaporated 7 × 1.0727 = 7.5

pounds from and at 212° F. Hence, 7.5 pounds is called the "equivalent
evaporation from and at 212° F." per pound of coal used.
As already stated, it takes 970.4 B. t. u. to make a pound of steam from
and at 212° F. Then to make 7.5 pounds there would be required 7.5 ×
970.4 = 7,278 B. t. u. This is the amount of heat required to change 7.5
pounds of water at 212° F. to steam at zero gage pressure, but it is also
the heat required to change 7 pounds of water at