Determining Steam Consumption

When it is desired to determine accurately the water rate of an engine, a boiler test should be carried on simultaneously with the test upon the engine, from which the pounds of dry steam supplied may be determined as described in Machinery’s Reference Series No. 67, “Boilers.” Knowing the average weight of steam supplied per hour for the run, and the average indicated horsepower developed during the same period, the water rate of the engine is easily computed. Sometimes the average cylinder condensation for a given type and make is known for certain standard conditions. In this case an approximation may be made from an indicator diagram which represents the average operation of the engine during the test.

Fig. 53. Diagram for Calculating Steam Consumption

A diagram shows by direct measurement the pressure and volume at any point of the stroke, and the weight of steam per cubic foot for any given pressure may be taken directly from a steam table. The method, then, of finding the weight of steam at any point in the stroke is to find the volume in cubic feet, including the clearance and piston displacement to the given point, which must be taken at cut-off or later, and to multiply this by the weight per cubic foot corresponding to the pressure at the given point measured on the diagram. As this includes the steam used for compression, it must be corrected, as follows, to obtain the actual weight used per stroke. Take some convenient point on the compression curve, as Q, in [Fig. 53]; measure its absolute pressure from the vacuum line OX and compute the weight of steam to this point. Subtract this weight from that computed above for the given point on the expansion line, and the result will be the weight of steam used per stroke. The best point on the expansion line to use for this purpose is just before release, both because the maximum amount of leakage has taken place, and also because of the re-evaporation of a portion of the steam condensed during admission. The actual computation of the steam consumption from an indicator diagram is best shown by a practical illustration.

Example—Let [Fig. 53] represent a diagram taken from the head end of a 16 × 30-inch non-condensing engine, running at a speed of 150 revolutions per minute; the card is taken with a 60-pound spring; the clearance of the engine is 6 per cent; the average cylinder condensation is 20 per cent of the total steam consumption; the diameter of the piston rod is 3 inches.

Measuring the card with a planimeter shows the mean effective pressure to be 48.2 pounds. The area of the piston is 201 square inches; the area of the piston rod is 7 square inches; hence, the average piston area = (2 × 201) - 7⁄2 = 198 square inches, approximately. Then

198 × 48.2 × 2.5 × 300
I. H. P. = —————————— = 217.
33,000

In [Fig. 53], GH is the atmospheric line; OX is the line of vacuum or zero pressure, drawn so that GO = 14.7 pounds on the scale; and OY is the clearance line, so drawn that ON = 0.06 NX. The line PQ is drawn from OX to some point on the compression line, as at Q. From C, a point on the expansion line, just before release, the line CF is drawn perpendicular to OX. The following dimensions are now carefully measured from the actual diagram (not the one shown in the illustration), with the results given:

OX = 3.71OP = 0.42
NX = 3.50CF = 0.81
OF = 3.20QP = 0.81

On the indicator diagram, being taken with a 60-pound spring, all vertical distances represent pounds per square inch, in the ratio of 60 pounds per inch of height. The stroke of the engine is 30 inches or 2.5 feet. The length of the diagram NX is 3.5 inches; hence, each inch in length represents 2.5⁄3.5 = 0.71 feet. From the above it is evident that vertical distances in [Fig. 53] must be multiplied by 60 to reduce them to pounds pressure per square inch, and that horizontal distances must be multiplied by 0.71 to reduce them to feet. Making these reductions gives:

OX = 2.63 feet.OP = 0.30 foot.
NX = 2.49 feet.CF = 48.6 pounds.
OF = 2.27 feet.QP = 48.6 pounds.

As a card from the head end of the cylinder is taken to avoid corrections for the piston rod, the area is 201 square inches or 1.4 square foot. With the above data the volume and weight of the steam in the cylinder can be computed at any point in the stroke. When the piston is at C, the volume is 1.4 × 2.27 = 3.18 cubic feet. When the piston is at Q, the volume is 1.4 × 0.30 = 0.42 cubic foot. From a steam table the weight of a cubic foot of steam at 48.6 pounds absolute pressure is found to be 0.116 pounds. Therefore, the weight of steam present when the piston is at C is 3.18 × 0.116 = 0.369 pounds. The weight of steam present when the piston is at Q is 0.42 × 0.116 = 0.049 pound. That is the weight of steam in the cylinder at release is 0.369 pound, and the weight kept at exhaust closure for compression is 0.049 pound.

The weight exhausted per stroke is therefore 0.369 - 0.049 = 0.32 pound. The number of strokes per hour is 150 × 2 × 60 = 18,000, from which the steam accounted for by the diagram is found to be 18,000 × 0.32 = 5760 pounds, or 5760 ÷ 217 = 26.5 pounds per indicated horsepower per hour. If the cylinder condensation for this type of engine is 20 per cent of the total steam consumption, the water rate will be 26.5 ÷ 0.8 = 33.1 pounds per indicated horsepower per hour.

In the present case it has been assumed, for simplicity, that the head- and crank-end diagrams were exactly alike, except for the piston rod. Ordinarily, the above process should be carried out for both head and crank ends, and the results averaged.

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