Ex. 18. The temperature of the water in the hot-well is 120°, and the vacuum is 11 lbs., but by some accident the temperature of the water in the hot-well is increased to 150°: what should now be the vacuum ? Ex. 19. On starting on a voyage the consumption was 83 tons per day, and the temperature of the funnel was 560°, but at the end of the voyage the temperature was found to be 690°: what was the increased consumption? F 2200 increase, where F = diff. of temperature X by the original consumption. ... present consumption 8.5 +5029'002 tons. Answer. Ex. 20. If the loss by blowing off is '11 of the fuel used, the temperature of the steam being 250°, and that of the feed being 110°: what should the boiler water be at above the feed density? = n= T-t 320. In condensing engines, the steam is condensed (when it exhausts), either by being brought into contact with a jet of cold water, as in the case of a common jet condenser, or by passing through or around a series of tubes made cool by water being passed around or through them, which is the action when a surface condenser is employed. When a jet condenser is used, salt water is pumped into the boilers, but a surface condenser, if tight, saves the fresh water derived by condensation, and it can be returned to the boilers again and again. By this means nearly all the feed water is fresh, and but little blowing off is required to keep the water at a certain point of saturation. A calculation of the loss by blowing off will show us the gain derived from fresh water condensers, supposing them to be tight, and to condense all the steam used by the engines. We will show the method of calculating the loss by blowing off. The number of degrees of heat that must be imparted to the water converted into steam will be the number in the total heat of the steam minus the number of degrees in the feed water. The heat lost by blowing off will be the difference between the temperature of the feed water and the sensible heat of the steam. A comparison of this loss with the entire number of degrees of heat required, being the sum of the heat imparted to the steam, and that to the water blown off, will give the per cent. of loss. EXAMPLES. Ex. 1. What is the per cent. of loss by blowing off, supposing the water entering the boiler to be, and that the water is carried at a saturation of; supposing also the temperature of the water entering the boiler to be 100° Fah., and the temperature of the water in the boiler to be 248°. (The total heat in steam, having 248° for the sensible heat, is 1189°58). Therefore, since one part (requiring 1089°58) is converted into steam, and the other part (requiring 148°) is blown off, the total heat imparted to the water is (1089°58 +148° =) 1237°′58, and as 148° of this is blown off, we have 1237°.58 : 148° :: 100 : 11.95 &c., per cent. of heat lost by blowing off. Ex. 2. What is the per cent. of heat lost carrying the water at a density of 1, the other data being the same as in the preceding example? 32 Here only as much water is converted into steam as is blown off, so that 1189°•58 — 100° = (1089°•58) × ‡ = 817°185 = heat required from the fuel for the water to be evaporated. 248° 8170185148° 100° 148° heat lost by blowing off. Therefore, 956°185: 148° :: 100°: 15°33, &c., per cent. of heat lost by blowing off. Ex. 3. What is the per cent. of loss by blowing off, where the water which enters the boiler at 3, having a temperature of 110°, and is carried at a density of 3, and having a temperature of 260°. (The total heat of steam having 260° for the sensible heat is 1193°45.) 1193°45 110° 1083°45 = heat imparted to the water converted into steam. But twice as much water is made into steam as is blown off. - 1083°45 X 2166°9 = heat producing effect for every 260°. blowing off. = 2166°9 + 150° 2316°9 2316°9: 150° :: 100 110° 150° heat lost by total quantity of heat imparted to the water. These per centages are the losses in fuel, combustible, minus that lost by radiation and heated gases passing up the chimney. The above calculations apply only to cases when the water enters the boiler at a density of; should it enter at a lower density, the loss will be less, or a greater density more, because to retain the water at the density assumed in the above examples, there would either have to be a less or greater quantity blown off than we have considered to be the case. 321. Gain by pumping water into the boiler at an increased temperature. Ex. 1. Steam 248°, feed water 100°, and density: now suppose the feed water, instead of entering the boiler at 100°, is made to enter at 150°: what will be the saving of fuel by so doing? 1189°58 (total heat of steam) 100° (temperature of feed water) = 1089°58 heat required from the fuel to evaporate one part of water. 248° (temperature of water blown off) 100° (temperature of feed water) 148° (heat lost by blowing off), and 1089°58 +148° = 1237°58 = total heat required from the fuel when the water enters the boiler at 100°. Let us now see what the total heat will be when the water is pumped in at 150°, and the difference of these results will be, of course, the saving. 1189°·58 — 150° = 1039°58 = heat required from the fuel to evaporate one part of water; 248° 150° 98° heat lost by blowing off, = and 1039°58 +98° = 1137°58 = total heat required from the fuel (when the water enters the boiler at 150°). 1237°58 Therefore 100° saving in degrees; whence 1237°58: 100° :: 100: 8·08 per cent.; that is to say, if with the feed water at 100° the boilers consumed 100 tons of coal per day; with the increased temperature of 150° they would produce the same quantity of steam with (1008.08) 91′92 tons. 12 32 and all the Ex. 2. Suppose that the density of the water in the above example was other conditions remain the same: what would be the saving in that case? 1189°′58 — 100 = 1089°58 × 3 =817°185 = heat required from the fuel for the water that is evaporated. 248° 817°185 +148° 965°185 1189°58 — 150° 1039°58 100 148° heat lost by blowing off. total heat required from the fuel when the water is pumped into the boiler at 100°. heat required from the fuel to evaporate one part of water. 1039°58 × = 779°685 = heat required from the fuel for the water that is evaporated. 248° — 150° 98° heat lost by blowing off. 779°•685 + 98° = 877°685 = total heat required from the fuel when the water is pumped into the boiler at 150°. Therefore, 965°185 : 87°5 :: 100: 9'06 per cent. Ex. 3. Water in the boilers, carried at a density of 1 per saline hydrometer; temperature of the condenser and water entering the boilers 105° Fah.; vacuum in condenser 27.82 inches. Compare the economic performance of the engine under these circumstances with the same engine carrying the water in the boiless at the same density, but the water in the condenser at 120° Fah.; the mean pressure of steam in both cases 20 lbs. per sq. inch. Neglecting the difference of power in the two cases required to work the air-pump, taking the boiler pressure at 20 lbs., and 2 inches of mercury to be equal to 1 lb. pressure, we proceed thus 105 +228·5 · 105 : 228°.5 - 105° :: 100 : 1323 per cent. loss — 120 × & + 228′5 · 20 X 2 by blowing off in the first case. - 120 : 228°5 120° :: 100 : 11.96 per cent. loss by blowing off in the second case. : 2.18 (back pressure) :: 100 : 5'45 per cent. of the effect of the engine lost by back pressure in the first case. : 3'33 (back pressure) :: 100 : 8.325 per cent. of the effect of the engine lost by back pressure in the second case. Now, then, letting the fuel represent the power, we observe, in the first case, that only (100 1323) 86.77 per cent. reaches the engine, of which 5'45 per cent. of 86.77 per cent. = 4°73 per cent. of the total effect lost by back pressure, leaving (86·77 — 4°73 =) 82'04 per cent. to be applied to working the engine. In the second case (100 1196) 88.04 per cent. of the power reaches the engine, of which 8.325 per cent. is lost in back pressure, and 8-325 per cent. of 88.04 per cent. = 7.33 per cent. of the total effect lost by back pressure; leaving (88.04 -7°33) 80.71 per cent. to be applied to working the engine. Therefore, under the conditions of the question, the engine in the first case performs the same amount of work with (82'04 ·80'71) 133 per cent. less fuel. This calculation can be made accurate by taking diagrams from the cylinder and air-pump, under the conditions of the example, and estimating the power in each case; then, the power to work the air-pump is considered. EXAMPLES FOR PRACTICE. I. How much water should you allow for the condensation of each cubic foot of steam at 212° during the working of an engine? 2. What quantity of water is required to obtain one cubic foot of steam at 212°? 3. A pound of steam at 212° Fah. is passed into 20 lbs. of water at 70° Fah.: what is the temperature of the water at the close of the operation? Answer 122°·79 Fah. 4. If a pound of water at 212° be mixed with x pounds of water at 60°: what is the value of a when the resulting temperature is 100°? Again, if a pound of steam at 212° be mixed with y pounds of water at 60°: find y when the resulting temperature is 100°. Account for the difference between x and y. Ans., 2·8 and 26·965. 5. Zinc boils at 1204° Fah., mercury at 608° Fab.; change the readings into Centigrade scale. Answer. 650° C. and 320° C. 6. Change the following readings from Fah. scale into C. scale:-Polished steel is of a deep blue colour at 580° Fah. Polished steel is of a pale straw colour at 460° Fah. Answer. 304°5 C., 237°75 C., and 22 C. Sea water freezes at 28° Fah. 7. The following table of remarkable temperatures is given according to the C. scale: what are the corresponding readings on the Fah. scale? Answers. 220° Fah.; 28°90; 32°; 39°2; 97°.88; 212°; 9688; 2786°. 8. A unit of heat is the heat given to lb. of water to raise its temperature 1° C.; how many units of heat are involved in raising 3 lbs. of water through 30 of the Fah. degrees? Answer. 50. I 9. The latent heats of 1 lb. of water and 1 lb. of steam are respectively 79 and 573 units, according to the C. scale: what are they according to the Fah. scale ? 142 and 966-6 units. IO. Answer. A mixture is made of 4 lbs. of water at 7° C. with 6 lbs. of water at 12° C.: find the temperature. Answer. 10° C. FFF QUESTIONS RELATING TO STRAINS, &c. 322. Given the strain by which a bar one inch square is torn asunder; it is required to find the force that will break another bar a given number of inches square. It has been found by experiment that a strain of 4500 lbs. can tear asunder an iron bar of 1 inch square, so that we have I RULE CXXXIV. As the square of 1 inch is to the square of the number of inches in the bar, so is 4500 pounds to the force required to break the bar. EXAMPLES. Ex. 1. If a bar of iron one inch square is torn asunder by a strain of 4500 lbs.: what force will be required to break a bar 3 inches square? Ex. 2. If a bar of iron one inch square is torn asunder by a strain of 4500 lbs.: what force will be required to break a bar 4 inches square? In each of the following examples it is required to find the force that will break a bar of a given number of inches square. Ex. 3. A bar of iron 24 inches by inch: what is the strain at 5000 lbs. per sq. inch? Bar of iron 2.5 *75 125 175 1.875 Strain 9375 000 in lbs. |