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

Rufus T. Strohm
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 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
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