Latent heat is that condition of caloric in which it exists in a state of chemical combination with bodies, without giving any indications of its presence. Sensible heat is that which affects the senses, or more accurately, the thermometer, with an impression of heat or cold. Sensible heat and temperature are convertible terms. The thermometer, like most of the instruments designed to measure an increase or diminution in the heat of bodies, depends for its utility on the general tendency of heat to expand bodies. Certain fluids, as mercury, spirits of wine, &c., expand in almost the same ratio with the increase of their temperatures; so that when a thermometer containing such fluid is immersed in any other fluid, or placed in contact with bodies in general, the properties of radiation and conduction of heat possessed by bodies cause an equalization in the temperatures of the fluid in the thermometer and the external substance. The particular temperature is registered by the rise of the fluid up a graduated scale to which the tube of the thermometer is attached. The use of this instrument applies only to the manifestations of sensible heat. Latent heat has been estimated, though not with great exactness, by an instrument called the Calorimeter. It depends on the principle that ice cannot subsist at a higher temperature than 32°; and the amount of ice which can be converted into water of the same temperature with ice by abstraction of heat from equal weights of various substances, is taken as the measure of the latent heat of those substances. See Turner's Chemistry. Thermometer. SECTION II. pressure which may be 1. "Draw a diagram of Bramah's hydrostatic press, and show how to determine the produced by means of it. The Hydrostatic Press is a simple machine, the power of which may be increased to an almost unlimited extent. Its action depends upon that property H of fluid bodies, by which, in consequence of the extreme ease with which the particles move among themselves, pressure applied to any point is transmitted equally in all directions. In the annexed diagram, A B is a pipe connecting the cylinders N A, and C B, and passing to a well, or cistern, which supplies the water necessary to work the machine; V and V', are two valves opening towards the left; P is a solid piston working in the cylinder C B, and to which motion is communicated by means of the handle or lever F H; P' is a piston moving in N A, and to which is attached the press-board K. Let us now trace the motion of the different parts of the engine. Suppose an upward stroke to be made by P; then the pressure being removed from A B C, the valve V' will be shut, and the water from the reservoir will force open the valve V and fill up the void created by the ascent of P. On the contrary, when a downward stroke is made, the valve V will close; and the pressure in C B A exceeding that in N A, the valve V' will be opened and water forced into the larger cylinder. This will necessarily raise P K, and it may be readily conceived, that by a fixed board above K, any materials placed between them may be compressed to any extent, limited only by the power of the engine and the strength of its parts. To calculate the power gained by the hydrostatic press. Let D and d be put respectively for the diameters of P' and P. From the nature of a fluid any pressure on P will be transmitted to P'; and as such pressure is propagated equally in all directions, the pressure upwards on any portion of P' is equal to the force applied to an equal portion of P, in the opposite direction. .. Power gained = Area P' Area P = D2 The same result may be obtained on the principle of virtual velocities. Suppose one stroke to be made by the piston, and let the length be 1 or unity; also let x = the height to which the larger piston is raised by the stroke. Water forced from C B = d2 x 7854 = And water forced into N A D2 X 7854. x .. d2 X 7854 = D2 x 7854. x = space passed over by P' while P passes 2. “There is a barge whose section is an equilateral triangle, each side being a feet in length, and whose length is b feet; how deep will it sink in the water, its weight and that of its lading being together w lbs?" Let D be the water mark, and let x = K C .. DE 2 DK 2 KC x tan. 30° = 2 x tan. 30° area A D CEDEX K C tan. 30° x2. 2 .. solidity of water displaced But the weight of water displaced and lading; and a solid foot of water .. 1000 tan. 30° bx2 = 16 W 2 x tan. 30° x tan. 30°. b x2 = 10 tan. 30° b weight of barge = 1000 oz. 3. "Define the centre of pressure of a fluid, and show that the centre of pressure of a rectangular floodgate is situated at two-thirds its depth." Suppose a rectangular vessel filled with water; then the horizontal pressure upon any point of its side will be equal to the perpendicular pressure at that point.. Suppose this point to move with equal additions from the surface of the vessel to its base; then the pressures on the side will increase by equal differences from nothing to a pressure equal to the depth of the vessel. Therefore the sum of the forces acting on the side will be correctly represented by an isosceles right-angled triangle. There is obviously some point in the side of the vessel upon which the opposing forces would balance themselves; and if a pressure equal to that of the whole fluid on the side were applied to this point, the place would remain at rest. The point so determined is called the centre of pressure. This point is evidently coincident with the centre of gravity; since they are both determined by the same circumstance, viz. equilibrium of pressure. To find the centre of gravity of a triangle, A, B, C. Bisect the sides A B and B C in the points D and E, and join the points of bisection with their opposite B angles. Then the centre of gravity will be in the line C D, because all lines drawn parallel to A B, and intercepted by the other sides of the triangle will be bisected by C D. For the same reason it is in A E; and since the centre of gravity is in each of these lines, it must be in O, their point of intersection. = Join the points O and B. Since the side B C is bisected in E, we have And A CAE =ABAE A COE A BOE Taking equals from equals we have equal differences .... A CA 0 Ξ Δ ΒΑΟ Similarly the CAO = A CBO And .. A CAO =AAB C. Hence, any side of A B C being assumed as a common base, the height of the ▲, whose vertex is O, will be one-third of the height of A B C, for it is on the same base and of one-third the area of the largest A. Therefore the position of the centre of gravity will be one-third of the height of the triangle from the base, or two-thirds from the vertex. |