In this segment I'm going to continue our discussion of fluid properties. In particular properties related to density and also some discussion of normal stresses, which are pressure. In the hand book we have several definitions which are related to fluid density. The first one, density which we normally denote by the Greek symbol or the Greek letter rho. It's the mass per unit volume, defined as delta m over delta v, where delta m is the mass of the fluid contained in an infinitesimally small volume, delta v. A close related prop, property is the specific weight, which we normally denote by the Greek letter gamma. Where gamma is the weight per unit volume, delta w over delta v. And a final related property is the specific gravity which doesn't have a specific symbol associated with it. We usually express it as SG. And the specific gravity, is defined as the weight of the fluid, divided by the specific weight of water. Or the density of the fluid divided by the density of water. Where the properties of water, are taken under typical standard conditions of temperature and pressure. So these definitions rho the density of the fluid has dimensions of mass per unit volume. For example, kilograms per cubic meter. Gamma, the specific weight is a weight per unit volume. Weight is a force so it has predominant fundamental dimensions of force per unit volume. For example, newtons per cubic meter, or pound force per cubic foot. The specific gravity is a pure ratio. It's a pure number. It doesn't have any dimensions or any units. There is a simple relationship between the specific weight and the specific gravity and that relationship is gama, the specific weight is equal to the density rho multiplied by g, the acceleration due to gravity. And, one other piece of terminology is that if the density is constant, in other words, the density of the fluid doesn't change as we change the pressure, then we call that fluid an incompressible fluid, or constant density. And, generally for this course, we will assume that all of the fluids are incompressible. In other words the density is constant and this will even apply to air. Particular values that we'll be encountering very often is the density of water rho W, is approximately 1,000 kilograms per cubic meter, or 62.4 pounds mass per cubic foot. The specific weight of water under standard conditions is either 9,810 newtons per cubic meter or 62.4 pounds force per cubic foot. Now let's do some examples and the first one is that we're given that gasoline has a density of 680 kilograms per cubic meter. The specific weight is most nearly which of these alternatives? And usually, at this point, I will pause briefly so you can try these examples, and then check to see if you've got the right answer. So here's the solution, definition of specific weight is gama is equal to rho times g. In this stake, case, we given that rho is equal to 680 kilograms per cubic meter, g is 9.18 per second squared, therefore the specific weight is 6,671 newtons per cubic meter. And the closest answer is B. Similar problem. Gasoline has a density of 680 kilograms per cubic meter. The specific gravity is most nearly which of these? So again, the solution just follows from the density. The definition of specific gravity is equal to rho over rho w, or equivalently gamma over gamma w. In this case, we're given the density is 680 kilograms per cubic meter. So the specific gravity is 680 divided by 1,000, taking the density of water as 1,000 kilograms per cubic meter. And the answer is 0.68, so the answer is D, 0.68. Now stress is force per unit area. And it's defined in this way, the limit as the area tends to 0 of delta F over delta A, where delta F is the force acting. And in the FE manual, they decompose the stresses into the normal and tangential component where of the normal component. The magnitude of the normal component is the, is the pressure, or the normal stress. So, pressure is normal force per unit area, in other words, F over A, force per area. So therefore it will have fundamental dimensions of force over area, which would be either Newton's per square meter typically, or pounds force per square inch, in the system of units. We have some special units here. A newton per square meter is written as Pa, where Pa is a pascal. So pascal is the metric unit for pressure, newton per square meter. And pound force per square inch we usually abbreviate as psi, pounds per square inch. The pascal, however, is a very small unit. For example, the conversion. One pound per square inch is almost 7,000 pascals. So because the pascal is a small unit, more commonly we will, we will, we will be using kilopascals, which is 1,000 pascals. For example typical tire pressure say of 30 pounds per square inch is approximately 207 kilopascals. Now, another very important convention in expressing pressure is whether we express it as a Gage or an absolute pressure. The difference between these is that the Gage pressure is relative to atmospheric pressure, but the absolute pressure is relative to absolute zero. So, if we're expressing our pressure as a Gage pressure, we would, we denote that by psig for Gage pressure. However, almost always we're only interested in Gage pressures. In other words, pressures relative to atmospheric so we don't normally need to add on the g at the end there. So, if we simply write psi or kPsa. It's implicit that what we're referring to there is a Gage pressure, in other words a pressure relative to atmospheric. If it's an absolute pressure, we have to put an a on the end here or an absolute after kPa to denote that we're talking about an absolute pressure. And that is essential in that case to distinguish it from a Gage pressure. So, normally, if we just say psi, then, we were referring to a, a Gage pressure. Standard atmospheric pressure that we'll use fairly often is approximately 101.3 kilopascals absolute, or 29.921 inches of mercury, or 14.7 psia will be typical values that we'll use. And atmospheric pressure, by definition, is always, of course, zero psi Gage, or zero kilopascal gauge by definition. And the relationship between the two is that the absolute pressure is equal to the atmospheric pressure plus the Gage pressure. We also have another convention, the vacuum pressure, and vacuum pressures are useful if the pressure is below atmospheric. In other words, it's a negative Gage pressure. And the vacuum pressure is simply the negative, pressure expressed as a positive number. So, in that case, we would write it as psi vac, or kPa vac, to denote that we're referring to a vacuum pressure. So, let's do some examples on that. We're given that the liquid in a closed tank has a pressure of 50 kPa, kilopascals. And that implies that it's a gauge pressure because it's not specified. If atmospheric pressure is 101 kilopascals absolute, the pressure is most nearly which of these alternatives? So here's the solution, simple definition. Absolute pressure is equal to atmospheric pressure plus gauge pressure. In this case, the atmospheric pressure is 101, the Gage pressure is 50, so the answer is the pressure is 151 kilopascals absolute, and the answer is C. Similarly, liquid in a closed tank has a pressure of minus 20 kilopascals. Again, that's an atmosphere. A Gage pressure. In other words, it's below atmospheric pressure. If atmospheric pressure is 101 kilopascals absolute, the pressure is most nearly which of these alternatives? So, the first one is obviously not correct, because a vacuum pressure is always a positive number. It can never be a negative number. So, here is the solution. The vacuum pressure is equal to atmospheric pressure minus absolute pressure is equal to in this case 121 minus 101 minus 20. And the answer is 20 kilopascals vac, so the answer is B. It's just the Gage pressure expressed as a positive number.
