communication from the air-pump was then formed by turning the stop-plug, the pencil then instantly fell to 8, in consequence of the piston of the pump forming a suction directly it moved, and thus causing the atmosphere to force the indicator piston down to the point of resistance. The pencil at this point becomes stationary, and was kept in that position by the atmospheric pressure during the stroke of the pump piston, marked SP in diagram, or the length from C to T. This piston then returned and forced the water directly in front of it in the barrel for the length of N, which shows that during that portion of the stroke there was but little, if any, pressure in the pump. The pencil having traced the (nearly) parallel line for the length of N, it rose and formed the remainder of the line R, while the piston of the pump moved until it reached the atmospheric line.

It may perhaps be necessary, in further explanation, to state that while the pencil formed the line R, the water in the pump and in front of the piston constantly maintained its natural level; but as there was some air amongst the water, that vapour rose as the liquid was disturbed, and thus caused the pressure to be below the atmospheric line.

The compression and discharge of the water commenced, and thus the load on the piston caused a further pressure under the indicator piston which forced the pencil to the height of D. Here it stopped, and the pressure maintained its level for a moment, which allowed the pencil to form D1. The main load on the piston was here overcome, because the bulk of the water left was a less quantity than what had been discharged; then, as the pressure decreased, the pencil descended and formed the line D, and it then resumed its original position at S. The arrows marked in the diagram indicate the motion of the pencil.

Fig. 105 is a diagram taken from a steamer's starboard air-pump, which is of the lifting kind, and the piston fitted with a large brass conical valve. The following is an explanation of the diagram:-At a, the piston being at the bottom of the stroke, starts to rise, compressing the air and vapour above it, until it arrives at b, at which place a sudden discharge of air and vapour seems to have taken place, and the pressure fell to c, from which point the pressure again gradually rose until it arrived at d, where the water began to be delivered, and continued to the end of the stroke.

Attached to the top of the air-pump is a pipe, running down into the bilge, for the purpose of pumping off the bilge water. Where this pipe is attached to the pump is fitted a valve, operating like an ordinary check-valve, a handle being made to screw down on the top of it to keep it firmly in its seat when there is no water in the bilge.

Steam 15 lbs.; revolutions 10; hot-well 1062; vacuum gauge out of order.

Resistance of vapour and water in air-pump = (6·6 + 312 × ·0969 =) ·6697 lb. per square inch of steam piston.

Calculated for full of water.

There being no water in the bilge at the time Fig. 105 was taken, Fig. 106 was taken five minutes after, for the purpose of ascertaining what effect the opening of this valve and admitting air would produce. It shows that no extra power, from the admission of this air, was required to work the pump, the average pressure being about the same as in Fig. 105, and that the vacuum in the pumps at no time was more than 4 lbs. There was no alteration in the vacuum, as shown by the gauge, however, attached to the condenser, and the engines continued to work in the same manner as before the bilge-valve attached to the air-pump was opened.

In these diagrams, the pressure at the termination of the up stroke, it will be seen, is about 24 lbs. per square inch above the atmosphere, which is due

to the height of the level of the water surrounding the ship above the top of the air-pump. The pressure increased to between 7 and 8 lbs. per square inch, as shown in other parts of the diagram, is occasioned by the friction of water and vapour through the delivery-pipes and valves. The slanting off in the diagram, Fig. 105, from x to y, we think partly owing to two causes : First, the decreased velocity of the piston as it approaches the end of the stroke does not expel the water with such force, and hence there is not so much friction; but this would not occasion the slanting off from h to z on the return stroke; and secondly, therefore, we are inclined to think that the string slipped or stretched a little from s to y, and recoiled again to its original place from h to z.

Fig. 107 is an illustration of diagrams from vertical pumps, and shows the difference in the discharge pressure. It will be seen that the injection-pump diagram shows a rapid discharge up to a pressure of 23 lbs. from the vacuum line.

The diagram from the circulating-pump indicates a continuous discharge, and that there was little clearance for the water in the barrel before pressure occurred, as the pressure or angular line commences at the beginning of the diagram, thus showing that the water began to leave the barrel at about one-third of the stroke of the piston instead of two-thirds, as in the injectionpump. The reason of the discharge limit line being horizontal for such a length is, that the piston had to force the water through the tubes of the condenser, and thus the resistance was continuous during that process.

Double-acting pumps. Vertical motion. Screw engines.

The surface-condenser air-pump diagram indicates below the atmospheric line, and shows that the discharge was never above the atmosphere, and that at the return stroke there was a little pressure in the barrel, as indicated by the angular line.

December 25th, 1874. S.S. Sibylla. Full gear; Throttle-valve full open; Kingston open.

December 25th, 1874. S.S. Sibylla. Full gear; Throttle-valve full open; Kingston 1 open.

We now proceed to ascertain the

Power required to work the Air-pump.

318. Ascertain the capacities of the steam cylinder and air-pump, by multiplying the areas of their cross-sections by the lengths of their stroke, and divide the latter by the former, which will give the ratio of the cylinder capacity to that of the air-pump. But the air-pump makes but one delivery stroke to every two strokes of the steam piston, consequently, divide this ratio by two, which will give the co-efficient for our present calculation, and this co-efficient, multiplied by the mean pressure per square inch of air-pump

Fig. 109.

piston-which can be ascertained from an indicator diagram-will give the mean pressure per square inch required to expel the air and vapour.

This of course must be augmented by the weight of the water raised.

The indicator diagram will show very nearly at what part of the stroke the pump begins to deliver the water, and, therefore, what fraction of the pump is filled, from which can be easily ascertained the number of cubic feet of water lifted; and this number, multiplied by 64°3 or 62.5, as the vessel may be running in salt or fresh water, will give the number of pounds.

And the number of pounds lifted, divided by the area of the air-pump piston, and multiplied by the co-efficient before obtained, will give the pressure per square inch of steam piston required to expel the water from the pump.

These results give the pressure per square inch of steam piston required to work the air-pump independent of friction, an amount that is usually estimated.

Ex. The capacity of cylinder, i.e., the space displaced by the steam piston per stroke is 267.25 cubic feet; ditto in air-pump 51.8 cubic feet; proportion of steam piston displacement to that of half of air-pump piston displacement 1000 to 0969; area of air-pump piston 2134 square inches. The pump was filled full of water, as shown by diagram, and the mean pressure throughout the stroke was 6.5 lbs. per square inch; hence 6.5 X 0969='6298 lbs. per square inch of steam piston resistance from vapour in air-pump, and 312 X 0969 = 0302 lb. per square inch of steam piston resistance from the weight of water lifted; total = (62980302) 66 lb. per square inch of steam piston, required to work the air-pump independent of friction.

Now, supposing the mean unbalanced pressure on the steam piston per inch to have been 20 lbs., we have 20: 66 :: 100: 33 per cent. of the total power of the engine required to work the air-pump.

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