This section is from the book "Wrinkles And Recipes, Compiled From The Scientific American", by Park Benjamin. Also available from Amazon: Wrinkles and Recipes, Compiled From The Scientific American.
Find the product of the diameter of the boiler, the length, and 1.5708, and add this to the product of the length of the boiler, the number of tubes, and the beating surface of a tube per foot of length.
The preceding table gives the beating surface per foot of length for the standard sizes of tubes.
Locomotive Boii.ER
Example.- A cylindrical tubular boiler has a diameter of 42 inches, is 16 feet long, and contains 40 tubes, each 4 inches out Bide diameter.
The product of 3-1/2, 16, and 1.5708 is nearly 88.
The product of 16, 40, and 0.977 (the internal surface of the tube, per running foot) is about 615, so that the whole heating surface is 703 square feet.
(d) Locomotive boilers.-I. Add together the following quantities: (1) The product of the Length of the line hounding the cross-section of the furnace, and the length of the furnace. (2) Twice the area of the cross- section of the furnace. (8) The pro duct of the length of the tubes, the number of tubes, and the heating surface of a tube per footof length.
II. Subtract from this sum the sum of the following quantities: (4) The area of the furnace-door. (5) The produt of the number of tubes, the square of the internal diameter of a tube and 0.7854
As an example of the use of this rule, suppose it is required to determine the heating surface of a boiler.having the dimensions noted in the engravings, Fig. 1 being a cross-section of the boiler at the furnace, showing also the furnace door in dotted outline, and Fig. 2 a longitudinal section. (1) The length of the line bounding the cross-section of the furnace is the sum of twice 3.5, 1.5708, and 2.5, or 11.07, and the product of 11.07 and 4 is 44.28. (2) The area of the cross-section of the furnace is the sum of 3.5 squared, 2-1/2 times 1/2 , and 0.7854 divided by 2, or about 13.89. Twice 13.89 is 27.78." (3) The product of 8, 20, and 0.977 is 157.32. The sum of 44.28, 27.78, and 157.32 is 229.38. (4) The area of the furnace-door is the sum of 1.5 times 1.25, and 0.3927 times 1.5 squared, or about 2.76. (5) The product of 20, 0.311 squared, and 0.7854, is about 1.52.
Locomotive Boiler
The sum of 2.76 and 1.52 is 4.28. The difference between 229.38 and 4.28 is about 225 square feet, the heating surface required.
(e) Vertical boilers.-I. Take the sum of the following quantities: (1) The product of the diameter of furnace, height of same, and 3.1416. (2) The product of the diameter of the furnace squared and 0.7854. (3) The product of the number of tubes, length of same, and heating surface per foot of length.
II. Subtract from this sum the product of the number of tubes, the internal diameter of a tube squared, and 0.7854.
Example.-Required, the heating surface of a vertical boiler with the following dimensions: Diameter of furnace, 24 inches; height of furnace, 18 inches; 40 tubes, each 2 inches outside diameter, 6 feet long. (1) The product of 2, 1.5, and 3.1416 is 9.42. (2) The product of 4 and 0.7854 is 3.14. (3) The product of 40, 6, and 0.4739 is 113.74. The sum of 9.42, 3.14, and 113.74 is 126.3. The product of 40, 0.02274 (the square of 0.1508), and 0.7854 is about 0.72. The heating surface is the difference between 126.3 and 0.72, which is about 125.6 square feet. B.
 
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