2.44 × 970 = 2366.80 B.t.u.
The vapor is now raised in temperature, to that of the furnace, which we may assume is 1200°F. The furnace being at atmospheric pressure the vapor merely expands in volume as a gas. The specific heat of steam at atmospheric pressure is 0.464; that is, 1 pound of steam requires only 0.464 B.t.u. to raise it a degree, and 2.44 pounds of water will absorb
0.464 × 2.44 × 1200 = 1356.00 B.t.u.
Of this last amount of heat, approximately 50 per cent. is recovered as the gases pass through the furnace. The total loss of heat due to the evaporation of the water is
| Raising temperature from normal to 212° | 395 B.t.u. |
| Evaporation | 2,366 B.t.u. |
| Changing temperature of vapor, less 50 per cent. | 678 B.t.u. |
| ————— | |
| Total heat loss | 3,439 B.t.u. |
In the 100 pounds of coal under consideration, there is 100 pounds, less 2.44 pounds of water, or 97.56 of dry coal, each pound of which contains 13,732 B.t.u. as given by the table on [page 193]. This gives
97.56 × 12,682 = 1,339,753 = practically 1,340,000 B.t.u.
From this quantity is subtracted the loss of heat, 3439.
1,340,000 - 3439 = 1,336,561 B.t.u.