The inter-cooler is the vital feature of the two-stage or multi-stage machine. In this construction the air is partially compressed in one cylinder; it is then passed through an inter-cooler where it is cooled and finally is compressed to the desired degree in the second or other additional cylinders.
An inter-cooler is shown in Fig. [373]. The cooling surface consists of a nest of small brass water tubes. These tubes break up the stream of air entering the cooler, while their thin walls insure rapid conduction. The receiver volume formed by the connecting pipes and inter-cooler body results in a nearly uniform discharge pressure in the low-pressure cylinder. The air being outside of the tubes encounters practically no frictional resistance, and its slow passage allows time for cooling. A pocket, with gauge glass attached, is so placed as to catch any precipitated moisture which might otherwise enter the high-pressure cylinder.
An after-cooler is shown in Fig. [374]. This serves to reduce the temperature of the air after the final compression.
The heat of compression, as may be judged from the foregoing, relating to inter and after-coolers is a feature of interest. The temperature to which it finally attains depends, 1, upon the initial temperature; 2, upon the degree of compression, or in other words, the amount of work expended upon the compression.
The extent of this heating is shown in the following table, for dry air when compression is performed with no cooling.
| Temperature of air before compression, | 60° | 90° |
| Temperature of air compressed to 15 lbs. | 177° | 212° |
| „„„„ 30 lbs. | 255° | 294° |
| „„„„ 45 lbs. | 317° | 362° |
| „„„„ 60 lbs. | 369° | 417° |
| „„„„ 75 lbs. | 416° | 465° |
| „„„„ 90 lbs. | 455° | 507° |
| „„„„ 105 lbs. | 490° | 545° |
| „„„„ 120 lbs. | 524° | 580° |
The Norwalk compound compressor is shown in outline by the cut 375. The large air cylinder on the left determines the capacity of the compressor; for illustration assume its piston at 100 square inches area; the small air cylinder can have an area of thirty-three and one-third square inches.
The small piston only encounters the heaviest pressure; at 100 pounds pressure the resistance to its advance is 3,333 pounds. The resistance against the large piston is its area multiplied by the pressure which is caused by forcing the air from the large cylinder into the smaller cylinder. In this case it is thirty pounds per square inch. But as this thirty pounds pressure acts on the back of the small piston, and hence assists the machine, the net resistance to forcing the air from the large into the small cylinder is equal to the difference of the area of the two pistons multiplied by the thirty pounds pressure. This is sixty-six and two-thirds by thirty, and equals 2,000 pounds.
Hence 2,000 pounds, the resistance to forcing the air from the larger into the smaller cylinder, plus 3,333 pounds, the resistance in the smaller cylinder to compressing it to 100 pounds, is the sum of all the resistances in the compound cylinders at the time of greatest effort. This is 5,333 pounds. By thus reducing the work to be done at the end of the stroke, more work is done in the first part, and the resistance is made nearly uniform for the whole stroke.