fluid at the end of the time t from the beginning of the motion. Then, (Art. 58.) if the depth of the orifice below the surface of the fluid was a when t = 0. ..√(.2g)k. t = 2A(√α-√x); and the whole time of emptying A == K 136. To find the time of emptying a hollow sphere through a small orifice in its vertex. K Let a be the radius of the sphere, κ the area of the orifice, the depth of the orifice below the surface of the fluid at the end of the time t from the beginning of the motion. The area of the surface of the fluid at the end of the time t 137. To determine the motion of a fluid oscillating in an inverted syphon PDB (fig. 76.) of uniform bore. Let P, Q be the extremities of the column of fluid at the end of the time t from the beginning of the motion; A, B, the extremities of the column of fluid when at rest; a, ẞ the angles between AP, BQ and the vertical MN; AP = 8, к the area of a section of the tube. The moving force on the fluid = gpk.MN = gpê(AP cos a + BQ cos B) = gp (cos a + cos ẞ)AP. ρκ. The mass of the fluid = pк. ADB, and, since the bore of the tube is uniform, every part of the fluid moves with the same velocity, therefore the effective accelerating force at any point tending to make the fluid return to its position of equilibrium 138. A weight is raised by a rope wound round the axle of an undershot wheel; to find the velocity of the wheel. Let K be the area of each float-board; u the velocity of the wheel; v the velocity of the stream; a the radius of the wheel; b the radius of the axle; W the weight. The relative velocity of the stream is vu, and, therefore, the force with which it impels the wheel = pK (v – u). And when the velocity of the wheel is uniform, this force is in equilibrium with the weight W, .. bW = a д pK (v – u)2. The work performed by a water wheel is measured by the product of the weight lifted multiplied by the velocity of the α The wheel would be kept at rest by a weight pKv2, therefore ་ ་ ་ b the work performed by the wheel is a maximum when the weight lifted is of the weight that would keep the wheel at rest. 139. To find the position of the rudder of a ship, when the effect of the rudder in turning the ship is a maximum. Let AP (fig. 33.) be the keel of the ship, PE perpendicular to the rudder. The resolved part of the resistance on the rudder estimated in a direction perpendicular to AP, oc (cos APE). sin APE. (Art. 69. Cor. 2.); And this is a maximum when 0 = (cos APE)3 - 2 (sin APE)2. cos APE, or sin APE = √/3. The theory of resistances (Art. 68.) is the same as that given in the Note, Page 188, of Mr. Moseley's Hydrostatics. The correctness of the application of the theorem (Art. 53.) in this case, appears doubtful. Example of the comparison of the specific gravities 140. of two fluids. A glass flask being filled with mercury at 20,6, the mercury appeared to weigh 1340,893 grammes; when filled with water at 20,5, the water appeared to weigh 98,7185 grammes; the weight of the air contained in the flask = 0.1186 grammes; therefore the true weight of the water = 98,8371; and the true weight of the mercury 1341,0116. The apparent expansion of mercury in glass, between 0° and 100°, = 0.0154, therefore, the true weight of the mercury contained in the flask at 20°,5 1341.0116+ (.0000154) (1341) = 1341,0323; .. (S.G.mercury) (S. G. water), at 20°,5 =. (1341,0323): (98,8371) = 13,5681. Between 0 and 20°,5 the expansion of mercury = 0.00369, and the expansion of water = 0.0017; therefore (S. G. mercury) :(S. G. water), at 0° = 13,5681 (1 + .00369 − .0017) = 13,5952. 141. Daniell's barometric formula. Let u, v be the altitudes of the columns of mercury supported by the pressure of the vapour contained in the air at the lower and upper stations respectively: then, retaining the notation of (Arts. 35, 36.) the mean ratio of the pressure to the density at 0° will be x= μ {1 + & § (+)}, (Art. 66.) 3 h } 1 Therefore - log, 101++) (+) {1 + ↓ E(8 + T)}, .. x = 16 h (10467) (29,9218) inches = 26099 feet (Art. 31.) {60095 + (112,7) (S + T) + [1127 + 21 (S + T)] (2 +155.cos 2x + (.0029)x} + h k × {logioh - log10k - (0.000078) (s − t) + (0.0000000416)x}. When a is not very large, x = {60282 + 113 (S + T) + [1130 + 21 (S + T)] ( 142. If T be the temperature of steam, u its pressure expressed in inches of mercury at 0°, it is found that as long as T is not much greater than 100, log10=1,4759877-(0,01537278) (100-T)-(0,00006732) (100 — T2) + (0,00000003374) (100 — T)3. And for temperatures between 100o and 230o, log10 = 1,4759877 +5.log10 {1 + (0.007153) (T-100)}. 143. According to Dr. Young, the expansion of water is expressed very nearly by the formula E = (0.0000063475) (T - 3,9)2 - (0.000000013865) (T-3,9)3. 144. Ratios of the specific gravities of different substances to that of water at 60° F or 15,5°C. |