air (2) Examine the effect of taking alternate strokes of an pump and of a condenser attached to a receiver. Prove that if the barrel of each pump is one twentieth of the receiver, and the condenser be worked for 20 strokes and then the air pump for 14 strokes, the density of the air will be practically unaltered. (3) Prove that if a bladder occupies one nth of the volume of the receiver of an air pump, and if it bursts when the pressure is reduced to one mth of an atmosphere, the mercurial gauge will fall h(m −1)/mn, when h is the height of the barometer. (4) Prove that if the temperature is constant the work required to increase q fold, or to diminish to oneqth, the density of atmospheric air of pressure P in a receiver of volume V is, respectively, PV(1– PV(log q1—q+1) and PV(1-1-log/g). q Calculate the work required and the change of temperature in these cases when the compression or rarefaction takes place adiabatically (§ 233). Work out V-1 m3, P=10 kg/m2, q=100. (5) Examine the change in the indications of the siphon barometer of § 179, placed in the receiver of an air pump or condenser, when the dimensions of the barometer are taken into account; and prove that, in one stroke of the air pump, the barometric column falls a distance, approximately, Bh 1 A aßh while, in n strokes of the condenser, it rises, where A denotes the original volume of atmospheric air in the receiver. Prove also that if v denotes the volume and a the height of the air pump gauge in § 269, the mercury will rise in one stroke (neglecting the square of v), CHAPTER IX. THE TENSION OF VESSELS. CAPILLARITY. 276. The vessels employed for containing a fluid under great pressure are generally made cylindrical or spherical for strength; and it is important to determine the stress in the material for given fluid pressure, or the maximum pressure allowable for given strength of material, for the purpose of calculating the requisite thickness. The simplest case of a vessel in tension is a circular pipe or cylindrical boiler, exposed to uniform internal pressure, so that there is no tendency to distortion from the circular cross section. With an internal pressure p, a circumferential pull will be set up in the material of magnitude T per unit of length suppose, acting across a longitudinal section or seam of the cylinder; and to determine T we suppose a length of the cylinder to be divided into two halves by a diametral plane, and consider the equilibrium of either half. Denoting by d or 2r the internal diameter of the tube, the resultant fluid thrust on the curved semicircular surface is equal to the thrust across the plane base, and is therefore pld; and this thrust being balanced by the pull Tl on each side of the diametral plane, therefore G.H. 2T-pld, or T=1pd=pr. 2B ..(1) 385 386 TENSION OF CYLINDRICAL, SPHERICAL, 277. When the cylindrical vessel is closed there is in addition a longitudinal tension in the material; denoting by T′ the longitudinal pull per unit length across a circumferential seam or section, the fluid thrust 1πď2р on the end of the cylinder must be balanced by the longitudinal pull dТ across a circumferential seam, and therefore TdT = d2p, or T=1pd=pr=&T........ (2) Thus the longitudinal tension is half the circumferential tension; or in the cylindrical shell of a boiler the circumferential joints and rows of rivets which resist the longitudinal pull need be only half the strength of the longitudinal joint, which resists the circumferential pull. 278. A spherical surface in tension is beautifully illustrated by a soap bubble as a complete sphere in air, or as a hemisphere on the surface; denoting the internal diameter of a spherical vessel by d or 2r, the tension per unit length across a diametral section, due to internal pressure p, is also T or pr; for considering the equilibrium of either hemisphere into which the sphere is divided, the fluid thrust on the hemisphere is equal to the thrust on the base, or d2p; and this is balanced by TdT, the resultant pull round the circumference; so that, as before, TdTdp, or T=1pd=pr. = Thus if a cylindrical boiler is made with hemispherical ends, these ends need have only half the thickness of the cylindrical shell; but they will weigh the same as flat ends of the same thickness as the shell. The same results are obtained by considering the equilibrium of the part cut off by any plane parallel to the AND CONICAL VESSELS. 387 axis of the cylinder; if this part subtends an angle 20 at the axis, the fluid thrust on it, 2lr sin 0. p, is balanced by the components of the pull on each side, perpendicular to this plane, 21T sin 0; and sin divides out; so also for the spherical surface. 279. If the cylinder has a conical end or shoulder, of vertical angle 2a, the stresses in the surface will be no longer uniform; taking a circular cross section PMQ, of centre M and diameter 2y, cutting off the conical end POQ, and denoting by T, and T2 the tensions per unit length across this section and across the straight section of the surface, then from the equilibrium of POQ, 2πуT cos α=у'p, or T1=py sec a. To determine T2, consider the equilibrium of either half of the surface cut off by two adjacent circular sections pmq, p'm'q', equidistant from PMQ; therefore 2T2.pp' p.PQ.mm'+iptana.pm.pq-iptana.p'm'. p'q', T2=py cos a+py tan a sin a=py sec a = 2T1. or = 2 Thus T1 and T2 become large when the conical end is nearly flat; so that the ends require strengthening with longitudinal stays. Suppose however that there is a conical shoulder, as in the Coney Island Stand Pipe (§ 42); then if 2a denotes the diameter of the upper small end, and p the average pressure at the shoulder, 2πуT cos a = upward thrust=(y2 — a2)p. 280. If the thickness of the material is e, then T/e and Te are the average circumferential and longitudinal tensions, per unit of area, in the cylinder; denoting them by t and ť, t=2t' =pr/e. |