SOLID MATTER BROUGHT INTO BOILER 319.-Necessity for occasionally "blowing out" the water from the boiler arises from the impurities contained in the water, as the process of evaporation leaves them in the boiler. Steam consists of pure water alone, hence its formation causes accumulation in the bottom of the boiler, which would soon prove its destruction; for such solid matter being a bad conductor of heat, the furnace would soon overheat the plates of the boiler, where covered internally with such deposits; and such becoming red-hot, the danger of explosion and rupture of the plates would be the consequence. So that the blowing out the boiler is the ejection of this accumulation of "brine" before it becomes heated into a solid form. The principle upon which this question depends, is that the brine blown off must evidently contain the same proportion of solid matter as the feed has brought into the boiler since last blown out. Now, as regards the saline matter held in solution in the "brine" of the boiler, it must be remembered that sea water in its ordinary state contains about of such impurities;* for one cubic foot of sea water weighs 64°3 lbs., and 3 = 1'95, and 1·95 + 62°425 (weight of cubic foot of fresh water) = 64°375 = (cubic foot of sea water). 64.3 = the And if B represents the quantity of water blown off in any given time, and b = the quantity of water lost from the boiler by conversion into steam in the same time; then B+ b B+b will the quantity of water used in the boiler in the same time, and 32 quantity of solid matter formed in the boiler in the same time; and if the increased strength or accumulation of saline matter blown out in such time, then quantity (at such strength) blown out at such time; B=. Therefore, the strength of the brine being, the quantity blown out would be equal to half the quantity evaporated. RULE CXXXIII. 32 3 B will the Bx b 3 B 32 32 = or B+b=3B, or Take the difference of the numerators of the given fractions; the remainder is the denominator of another fraction, having 1 for its numerator, which fraction expresses the ratio of the quantity blown off to the quantity evaporated. : EXAMPLES. Ex. I. Sea water contains of solid matter, and it is required that the water in the boiler may not contain more than what proportion must the quantity blown off bear to the quantity evaporated? As the saturation of the water in this case is doubled ( being the double of 4), onehalf of it must have been evaporated. So to carry the water at, as much must be blown off as is converted into steam. Water blown off : water evaporated :: I : I. * See Hydrometer and Salinometer. EEE Ex. 2. The required saturation of the water is; then, since the saturation of the water in the boiler is trebled, it must have received from the evaporated water; so that in this case we blow off of the water that enters the boiler, and the quantity blown off equals of the quantity evaporated; or, Water blown off : water evaporated :: I : 2 = }. Similarly, we may show that— = If required saturation, water blown off of water entering the boiler, and Water blown off : water evaporated :: I : 3=}. Ex. 3. and If required saturation=3, water blown off of water entering the boiler, Ex. 6. The water in a boiler connected with a jet condenser shows 1: what proportion of the feed is used as steam? If i be the quantity blown off, then 17 will be the full feed and 7. 32 Then the answer to the question will be the ratio to 17; clearing the terms of the ratio of fractions by multiplying by 8, we have 7 × 8 = 7, and 13 × 8=15, .. the ratio is 7 to 15, ¿e, 7 gallons out of 15 gallons of feed. The water in the boiler is 21 what proportion of the feed is used as steam? 32 Answer. 7 gallons out of 15 gallons. The water in the boiler is 2: what proportion of the feed is used as steam? 32 Answer. 3 gallons out of 5 gallons. Ex. 7. The boiler contains 50 tons of fresh water on starting on a voyage; the water in the hot-well, through some leak, is of the saltness of the sca: how many tons of water must have been evaporated during the voyage, if at the end of the voyage the boiler has 2} times the saltness of the sea, there having been no blow-off? The number of times the boiler was filled = 2}÷}= 20. The weight of water 50 tons X 20 1000 tons. Answer. QUESTIONS RELATING TO TEMPERATURE. EXAMPLES. Ex. 1. Fahrenheit's thermometer is marked 32° at freezing point, and 212° at boiling point, while Centigrade is marked o° at freezing and 100° at boiling: what mark on Fahrenheit should agree with 89° on Centigrade. (See article on "Thermometers, &c.") There are 180° between freezing and boiling point on Fahrenheit. Now this temperature, 89° C., is 89° above the temperature of the melting point of ice. Find how many of the F. degrees will correspond to this by the proportion 100° : 180° :: 89° : 160°2 F., the answer. To this 160°2 F. degrees above the freezing point add 32°, and we get 192°2 F. as the required reading. = Ex. 2. What mark on the Centigrade scale will indicate the temperature of 47° Fah. Here 47° F. means 15° of Fahrenheit's degrees (47° — 32° — 15°) above the melting point of ice. Find the number of degrees on the Centigrade scale which corresponds to 15 on the Fahrenheit scale by this proportion 180° : :: 15° 8°.3 the answer. 100° That is, 47° F. is the same as 8°3 of Centigrade degrees above the freezing point, or 8°.3 0. For further exercises on the changing of the readings of the different thermometer scales see article "Thermometers." Ex. 3. The sea water is at 54°, and the water in the hot-well is at 115°; if the sea water gets to 74°, and if you cannot put on more injection, what will now be the temperature of the hot-well? 74° 54° 20°. Then 115° 20° = 135°. Answer. Ex. 4. What weight of water at 60° F. must be mixed with 20 lbs. of steam at atmospheric pressure to produce water at 120° F. ? The formula for working the above is obtained as follows: Let z lbs. of steam at a temperature t be injected into y lbs. of water at a temperature ť, and let the resulting temperature of the condensed steam and water be T. Also let H be the number of units of heat in steam at a temperature t, reckoning from zero of Fahrenheit's scale. The steam loses x (H — T) units of heat, and the water gains y (T—") units of heat. .. x (HT) = y (T− t). Ex. 5. If 37 lbs. of water at 32° Fah. be added to 43 lbs. at 212° Fah.: what will be the resulting temperature? Thermal units in the 1st quantity 37 lbs. X 32° 1184 2nd = 43 X 212° 9116 = 10300 Weight of the mixture = 37 lbs. + 43 lbs. = 80 lbs. .. Temperature of mixture = 1030080 = 128.75 lbs. Ex. 6. If 1 lb. of steam raises the temperature of 1000 lbs. of water 1°, the sea water is 50°, and the discharge water is 110°: how many pounds of circulating water passes the condenser for each pound of steam? Ex. 7. What is the pressure per square inch when the temperature is 75° Fahrenheit, and the barometer shows 29'4 inches of mercury? Formula 4907 × h 4907 X 29'4 144 == 15.798 lbs. Ans. Ex. 8. The temperature in the funnel is 593°, with a consumption of 25 tons per day; after a few days it increases to 29 tons per day to raise the same amount of steam: what is now the temperature of the funnel, nothing else having altered; and what is the cause of this increased consumption? NOTE.-If the cost of evaporation be increased by (n) per cent., the alteration in the temperature of the escaping gases will be increased by 22 (n) degrees. Ex. 9. Suppose the temperature of the injection water to be 60°, steam as it enters the condenser 212°, and water in the condenser at 110°: required the proportion of injection water to the water evaporated in the boiler. Now, then, we have 1068° 6 of heat to be destroyed, and only 50° to do it with; therefore, we must make up this difference in quantity; hence 1068°6 ÷ 50° = 21°·372 times the evaporated water to be admitted into the condenser to condense the steam and keep the condenser at the temperature of 110°. Ex. 10. If the sea water be at 60° Fah., and the water in the hot-well at 100° Fah.: how many lbs. of sea water will be required to condense 1 lb. of steam? where t the temperature of sea water, Rule. 114906 " t'-t 1149°.6 ť = and t' = 1149° 6- -100 hot-well. 100-60 Ex. 11. The total heat of steam may be found from the following formula, 1115° + 3 To, where T is the temperature or sensible heat of the steam. (See chapter on "Properties of Steam)." Example.-What is the total units of heat in steam of 212°? Ex. 12. 1115° + *3 × 212° — 1115° + 63o6 = 1178-6, the total heat. And what will be the latent heat in this case? Ex. 13. Find the total and latent heat of steam that is 60 lbs. by the gauge. 60 lbs. + 15 lbs. 75 lbs. gross. And the temperature of steam at 75 lbs. pressure is 3090, .. 1115° + 3 × 309 = 1115° + 92'7 = 1207°•7 total heat. 120707-3090898°.7 latent heat. Then if we know the temperature of the feed water, and subtract this temperature from the total heat of the steam, the remainder will be the units of heat to each pound of water turned into steam. Or, 1115° +3 T° t = units of heat required to turn 1 lb. of water into steam. Ex. 14. If the steam in the boiler be at 309°, and the feed water be at 110°: how many units of heat will it be necessary to add to this water to convert a pound of it into steam? 1115 +3 X 309 — 110 1097°7. Ans. Ex. 15. The temperature of the hot-well was 1120, and the evaporation was at the rate of 8 lbs. of water per 1 lb. of coal: what would be the evaporation if the temperature of the hot-well was increased to 150° ? .', 8+ ·276 = 8.276 the evaporation. Answer. Ex. 16. The I.H.P. of an engine is 682, and the weight of steam per hour per I.H.P. is 21 lbs.; if 1 lb. of steam condensed gives out sufficient heat to raise 1000 lbs. of water 1° Fah., the injection water having a temperature of 61°, and is discharged at 104°: how many tons of injection water are used per day of 10 hours? Temperature of discharge 104° injection 61 43 Then 198 23:256 lbs. of water to condense 682 × 21 = 14322 lbs. of steam condensed per hour. 14322 X 23°256 = 333072°432 lbs. of cold water per hour. 333072 432 X 10 = 1486.93 tons of injection water used per day. Ans. 2240 Ex. 17. If 1 lb. of coal evaporates 9 lbs. of water from and at 212° Fah.: how much will 1 lb. evaporate if the steam was at 320° Fah., and the feed water at 120° Fah. ? First, 1115° +3 X 212° — 212° — 966-6 units of heat to 1 lb. of steam, |