the weight of air in lb which flows up the chimney per second is Avpg=Ap√(2gh)√(0—02) = Ꮎ = Ap√(2gh) √ { 4 − ( 8 − 1 )"}, a maximum Ap/(2gh), when 0′ = 20. 2 Thus if the outside air is at 17 C, 0=290, and 0′=580, 7=307 C, nearly the temperature of melting lead; and so that a chimney 100 ft high produces a draught of inch of water, when the flow through it is a maximum. In Metric units, a litre of air at 0 C weighs 1.29 g; so that a chimney h m high will produce a draught of z cm of water, given by for instance, if 0'20, and T=0, z=0·0645 h. Variations of barometric height or of temperature will cause air to enter or leave a given space, such as a room or a mine. This is illustrated by a "whistling well," in which a whistle placed in the lid is blown by the current of air which enters or leaves the well; also by the liberation of gas in a coal mine when the barometer falls. Thus if the barometer falls from h to h', or the temperature rises from 0 to 0', the density of the air falls from p to p', where p_h'e ρ and the percentage of air which leaves a room is 100(1 – P.) = 100(1–10). A room 10 m by 6 m by 5 m will, with a barometric height 76 cm and a temperature 0 C, contain 10 × 6 × 5 × 1.2932-388 kg of air; and now, if the barometer falls to 75 cm and the temperature rises to 15° C, the room will lose about 6 per cent or 25 kg of air. Examples. (1) Prove that if volumes V1 and V2 of atmospheric air 1 2 are forced into vessels of volume U1 and U2, and if communication is established between them, a quantity of air of volume at atmospheric pressure will pass from one to the other. (2) Prove that, if a partially exhausted siphon with equal arms is dipped into mercury; and if the sum of the heights to which the mercury rises in an arm, when first this and then the other arm is unstopped, is equal to the height of the barometer; then the original pressure of the air in the siphon is due to a head of mercury whose height is equal to the difference of the lengths of the mercury in the siphon in the intermediate and final stages. (3) Prove that, if a piston of weight W lb is in equilibrium in a vertical cylinder with a ft of air beneath it, and if it is depressed a small distance x ft, the energy of the system when the temperature is unaltered is increased by about Wa2/a ft-lb. (4) The mouth of an inverted cup is submerged to a depth of 6 ins in warm mercury, and it is found that no air escapes. Prove that, if the barometer stands at 30 ins, the mercury cannot be more than 100° F. warmer than the atmosphere. (5) A gas saturated with vapour, originally at a pressure p, is compressed without change of temperature to one-nth of its volume, and the pressure is then found to be Pn the Prove that the pressure of the vapour and of gas in its original state is respectively (6) Prove that, if the height of the column of mercury in § 211 is reduced from y to y' by the introduction of a bubble of water into the vitiated Torricellian vacuum, just small enough to evaporate completely, the pressure of the vapour, in mercury head f, is given by f = (b+h—y—y') ?—Y b-y' reducing to y-y' when the vacuum is perfect, or y=h; and that now (b−y)(h—y—ƒ) is constant, when b, h, and y change, the temperature remaining constant. E.g., If h=2981 ins; and y = 29 when b=32, y' = 28.5 when b'=30}; then ƒ=0·47 in, and the dry air would occupy a length 004 in of the tube at atmospheric pressure. (7) A piston, of weight A, in a closed vertical cylinder of height a and section A is in equilibrium at a height a/n from the base, the pressure of the air underneath it being p. Prove that a small rise C of the temperature of the air underneath will raise the piston through a height approximately equal to (8) An air thermometer is made of a bulb and tube inverted vertically in a reservoir of mercury of depth b so that the tube rests on the bottom. Prove that if the volume of the bulb and tube is equal to a length c of the tube, and h is the height of the barometer, the graduation for the temperature is at a height above the bottom of the tube or where 0 is the absolute temperature at which the (9) Assuming that the relative distribution of oxygen and nitrogen at different heights in an atmosphere in equilibrium follows the law that one is not affected by the other, find at what height in an isothermal atmosphere the proportion of oxygen would be reduced to half what it is at sea level, where the proportions by weight may be taken to be 80 parts of nitrogen to 20 of oxygen, and where the densities are in the ratio of 14 to 16. CHAPTER VIII. PNEUMATIC MACHINES 235. The Montgolfier Hot-Air Balloon. This balloon, invented by the Montgolfiers in 1783, is historically interesting as the first employed by the aeronauts Pilâtre de Rozier and d'Arlandes to make an ascent in the atmosphere. The principle is the same as that of the ordinary hotair toy balloon; the air in the balloon is rarefied by heat to such an extent that the total weight of the balloon, of the hot air it contains, of the car and of the aeronauts is equal to or less than the weight of the external cold air displaced, when the balloon begins to rise. Denote by W lb the weight of the balloon, car, and aeronauts, as weighed in vacuo, or corrected for the buoyancy of the air; and denote by W'lb the weight of cold air they displace, so that W- W' lb is the apparent weight when weighed in air; denote also by V ft3 the capacity of the balloon, so that M= Vp lb denotes the weight of cold air which fills the balloon, p denoting the density, in lb/ft3, of the surrounding cold air. Then when the air inside is raised in temperature from ◊ to ' degrees absolute, part of the air will flow out, |