Basic Principles, Concepts, and Definitions PDF

# 1. Basic Principles, Concepts, and Definitions ## 1.1 Introduction The study of thermodynamics is like a journey and must have a beginning. It is assumed that the reader has had first college courses in physics, calculus, chemistry, and mechanics. This chapter lays the foundation upon which subs...

# 1. Basic Principles, Concepts, and Definitions ## 1.1 Introduction The study of thermodynamics is like a journey and must have a beginning. It is assumed that the reader has had first college courses in physics, calculus, chemistry, and mechanics. This chapter lays the foundation upon which subsequent chapters will build. The student should have a crystal clear understanding of the basic principles, concepts, and definitions presented in this chapter before leaving it. After covering the basic laws and language and the methods of handling different substances thermodynamically, we will move into application chapters. These provide a greater insight into the subject but also give an introduction to specialized applications. Most students will find a personal interest in certain applications. Except for some local needs, the symbols and abbreviations are defined in the inside cover. ## 1.2 The Working Substance Engines produce work and refrigerators produce cooling because there are actions occurring on a working substance, usually a fluid, in which energy can be stored or from which it can be removed. A fluid exists, or is regarded as existing, as a continuum characterized by low resistance to flow and the tendency to assume the shape of its container. Examples of working substances include steam in a steam turbine, air in an air compressor, air and fuel mixture in an internal combustion engine, and water in hydraulic turbine. Hereafter, by the word substance, we will mean something that usually is made up of molecules; sometimes atoms may be involved (in reactive systems). Hence, it will not mean radiation or electrons or other subatomic particles unless they are specifically included. A pure substance is one that is homogeneous in composition and homogeneous and invariable in chemical aggregation. For example, if pure water exists as a solid, a liquid, a vapor, or any combination of these, it is a pure substance. On the other hand, if air exists as a combination of liquid and vapor, it is not a pure substance since the liquid will be richer in nitrogen than the vapor. A simple substance is one whose state is defined by two independently variable intensive thermodynamic properties; see § 1.5 for discussion of properties and state. The state postulate (or principle), § 3.2, will reveal that a simple substance will have only one relevant work mode. ## 1.3 The System A system is that portion of the universe - an atom, a galaxy, a certain quantity of matter, or a certain volume in space - that one wishes to study. It is a region enclosed by specified boundaries, which may be imaginary, either fixed or moving. We partition off a system or region on paper because we wish to study transformations of energy occurring within the boundaries, and with the passage of energy or matter or both across the systems' boundaries to or from the surroundings. The region all about the system is called the surroundings or environment. **The surroundings contain systems, some of which may affect the particular system under study, such as a source of heat.** The free body in analytical mechanics is a system in which the mode of analysis is based on Newton's laws of motion. In thermodynamics, the principal mode of analysis is based on the mass-energy balance of the system under study. Systems can be defined in several ways. Here, for present purposes, we wish to distinguish between three kinds: * **Closed system:** no exchange of matter with the surroundings - mass does not cross its boundaries - though energy may cross. * **Open system:** flow of mass across its boundaries. Energy may also cross. * **Isolated system:** completely impervious to its surroundings - neither mass nor energy cross. **Example of a Closed System:** An ideal gas system contained within a cylinder with a movable piston. If heat is applied externally to the cylinder, the gas will undergo an increase in temperature, raising the piston. With the rise of the piston, the boundary moves. Energy (heat and work) crosses the boundary during this process while the mass of gaseous matter remains constant. ## 1.4 Control Surface and Volume Often the system under analysis is of the **open type**, as an automobile engine shown in Fig. 1/2. With open systems we normally specify the boundary as a **control surface** and the **volume encompassed** by that surface as the control volume. Thus, the control volume may be defined as a volume in space in which one has interest for a particular study or analysis. The mass of working substance within the volume may be constant (though not the same molecular matter at any given instant) as in the case of the automobile engine or that of a simple water nozzle, or it may be varying, as in the case of a tire undergoing inflation. ## 1.5 Properties and State To compute changes of energy that have occured in a system or working substance, we must be able to express the behavior of the system in terms of **descriptive characteristics** that are called properties. Macroscopic properties that are familiar to the reader from prior study include pressure (p), temperature (T), density (p), and specific volume (v), each of which is to be discussed shortly. Properties may be classified as **intensive** or **extensive**. **Intensive properties** are **independent of the mass**. Examples include temperature, pressure, density, and voltage. **Extensive properties** are **dependent upon the mass of the system** and are total values such as total volume and total internal energy. **Specific properties** are those for a unit mass, and are **intensive** by definition, such as specific volume. Thus, generally, we note, as examples, that total volume is an extensive property and that temperature and pressure are inherently intensive. When we speak of the **state of a pure substance, or system**, we are referring to its condition as identified through the properties of the substance; this state is defined generally by particular values of any two independent properties. All other thermodynamic properties of the substance have certain particular values whenever a certain mass of the substance is in this particular macroscopic state. Examples of thermodynamic properties, besides p, v, and T, are: internal energy, enthalpy, and entropy (all to be studied later). Other properties of systems in general include: velocity, acceleration, moment of inertia, electric charge, conductivity (thermal and electrical), electromotive force (emf), stress, viscosity, reflectivity, number of protons, and so on. No matter what happens to a particular mass of a pure substance, be it compressed, heated, expanded, or cooled, if it is returned to the stipulated defining properties, the other thermodynamic properties also return to values identical, respectively, with their original values. See Fig. 1/3. Consider for a moment the qualification, **independent properties.** As we know, the density is the __reciprocal__ of the specific volume; hence these properties are not independent of each other. During the boiling or freezing of a liquid, the pressure and temperature of the two-phase mixture are not independent; the boiling temperature is a certain value for a particular substance, depending upon the value of the pressure. ## 1.6 Systems of Units Isaac Newton made the momentous statement that the acceleration of a particular body is directly proportional to the resultant force acting on it and inversely proportional to its mass, a = kF/m where k is a proportionality constant. Rearranging the equation form of the statement: $F = ma/k$ or dimensionally F→ ML/T², enables us to define a force unit in terms of the mass, length, and time in any system of units. Those consistent systems of units most commonly used for which k is unity, but not without dimensions, include these definitions of force: * **cgs system:** 1 dyne force accelerates 1 gm mass at 1 cm/sec² * **mks system:** 1 newton force accelerates 1 kg mass at 1 m/sec² * **fps system:** 1 lb force accelerates 1 slug mass at 1 ft/sec² Unfortunately, where the same word is used for both mass and force in a given system, we find the value of k is neither unity nor dimensionless. Several definitions of force include: * 1 lb force accelerates a 1 lb mass at 32.174 ft/sec² * 1 gm force accelerates a 1 gm mass at 980.66 cm/sec² * 1 kg force accelerates a 1 kg mass at 9.8066 m/sec² Rearranging equation (1-1A) produces k = ma/F. Applying the preceding definitions of force shows * k = (1lbm)(32.174 fps²/lby) → 32.174 lb-ft/lby-sec² * k = (1 gmm)(980.66 cm/sec²)/gm, → 980.66 gm-cm/gmy-sec² * k = (1 kgm)(9.8066 m/sec²)/kg, → 9.8066 kg-m/kg-sec² At this point, the student must realize that the value of k may be other than unity and must have units consistent with the system of units being used. ## 1.7 SI Units In view of the relative newness, uniqueness, and universal acceptance of this particular set of metric units, it is believed at this point that a short discussion of the SI units is most appropriate. Definitions of the seven base units are given in order to impress upon the reader their physical concepts, realizing that the inadequacies of some definitions may produce a lack of full comprehension at this time. In 1872 an international meeting was held in France and attended by representatives of 26 countries including the United States. Subsequently, in 1875, 17 countries (including the United States) signed an international treaty, the Metric Convention, to formulate an international metric standard. In 1960 the standard was modernized and called the International System of Units (SI Units). SI units are divided into three classes: base units, derived units, and supplementary units. See Tables 1.1, 1.2, 1.3. **Table 1.1 SI Base Units** | Quantity | Unit Name | Unit Symbol | | -------- | -------- | -------- | | Length | meter | m | | Mass | kilogram | kg | | Time | second | s | | Electric current | ampere | A | | Amount of matter | mole | mol | | Thermodynamic temperature | Kelvin | K | | Luminous intensity | candela | cd | **Table 1.2a Examples of SI Derived Units Expressed in Terms of Base Units** | Quantity | Name | SI Unit | Symbol | | -------- | -------- | -------- | -------- | | Area | square meter | m² | | Volume | cubic meter | m³ | | Speed; velocity | meter per second | m/s | | Acceleration | meter per second squared | m/s² | | Density | kilogram per cubic meter | kg/m³ | | Specific volume | cubic meter per kilogram | m³/kg | | Current density | ampere per square meter | A/m² | **Table 1.2b Examples of SI Derived Units with Special Names** | Quantity | Name | Symbol | SI Unit | SI Base Units | | -------- | -------- | -------- | -------- | -------- | | Force | newton | N | m.kg/s² | | Pressure | pascal | Pa | kg/(m.s²) | | Frequency | hertz | Hz | 1/s | | Energy, work, heat | joule | J | m².kg/s² | | Power | watt | W | m².kg/s³ | | Quantity of electricity | coulomb | C | s.A | | Electric potential | volt | V | m².kg/(s³A) | | Capacitance | Farad | F | kg/(s²A²) | | Electric resistance | ohm | Ω | m².kg/(s²A²) | | Conductance | siemens | S | s²A² / (m²kg) | | Magnetic flux | weber | Wb | m².kg/(s²A) | | Magnetic flux density | tesla | T | kg/(s²A) | | Inductance | henry | H | m².kg/(s²A²) | | Luminous flux | lumen | lm | cd.sr | | Illuminance | lux | lx | cd.sr/m² | ## 1.8 Acceleration Derived units are expressed algebraically in terms of base units. A number of derived units have been given special names; subsequently, numerous derived units are expressed by means of these special names. See Tables 1.2a, 1.2b, 1.2c **Table 1.2c Examples of SI Derived Units Expressed by Means of Special Names** | Quantity | Name | SI Unit | SI Base Units | | -------- | -------- | -------- | -------- | | Heat capacity, entropy | joule per kelvin | J/K | m².kg/(s²K) | | Specific heat capacity | joule per kilogram kelvin | J/(kg.K) | m²/s²K) | | Thermal conductivity | watt per meter kelvin | W/(m.K) | m.kg/(s³K) | | Dynamic viscosity | pascal second | Pas | kg/(m.s) | | Moment of force | meter newton | N.m | m².kg/s² | | Surface tension | newton per meter | N/m | kg/s² | | Molar energy | joule per mole | J/mol | m².kg/(s²mol) | ## 1.9 Mass The mass of a body is the absolute quantity of matter in it, an unchanging quantity for a particular mass when the speed of the mass is small compared to the speed of light (no relativistic effects). Newton's universal gravitation law relates the force of attraction between two masses and, in equation form, is: $F = G \frac{m_1 m_2}{r^2}$ where F_g is the force of attraction (for the earth’s attraction on an earthly system, this is the force of gravity) between the masses m_1 and m_2 that are r distance apart, and G is the gravitational constant. For F_g lbm, r ft, and m slugs, a consistent system, (G = 3.44 x 10^-8 lbm-ft²/slug² (ft³/lbm-sec)). For F newtons, meters, m kg. G = 6.670 x 10^-11 N-m²/kg². On earth, the change in the force of gravity is seldom enough to affect significantly the usual engineering problem. But, for example, at a distance of 1600 mi from the earth, the force of gravity is only half as great as at the earth’s surface. As implied by the foregoing, the attractive force between masses may be used to establish units for measuring mass. A handy and preponderant reference body is the earth; hence a unit of mass is defined by the force interaction between a mass and the earth at a point on the earth’s surface where gravity is defined as standard (g = 32.174 fps) -45° north latitude, close to sea level. A certain platinum mass in France is accepted as an international standard kilogram, and the pound is defined in terms of this kilogram. Having such a standard, we may now compare masses on a balance scale (gravitational forces the same on each side and theoretically no friction or air buoyancy). Of course, other means are used to determine the masses of molecules, atoms, and planets. Since a pound of mass located at a point of standard gravity g_o is subjected to a force of gravity of 1 lb, it weighs 1 lb there. Then, for a mass of m lb, we write m/k = F_g/g_o = F/a, or the force in pounds is $F = \frac{m}{k}a$ where, of course, m/k is the mass in slugs (a fps²). In this text, we shall use m for the mass in pounds, kilograms, and so on. In many instances, for example, the many energy balances that we shall make, the use of a consistent system of units is unnecessary; mass units and energy units cancel. Hence, the mass may be in any unit (lb, gm, kg, etc.) and the energy may be in any unit (Btu, cal, J, ft-lb, etc.), but each term must involve the same unit of energy and the same unit of mass. Where there can be little doubt as to whether the pound abbreviation is for force lb, or for mass Ibm, we shall not use the subscripts. Remembering that the consistent energy unit comes from the work concept in mechanics (involving force) and keeping the concepts of force and mass distant, one should have no difficulty. Of course, the use of subscripts is optional with the reader. **Example** A car whose mass is 2 metric tons is accelerated uniformly from standstill to 100 kmph in 5 sec. Find the mass in pounds, the acceleration, the driving force in newtons, and the distance travelled. **Solution** $m = (2 m\ tons)(1000 kg/m\ ton)(2.205 lb/kg) =4410 lb_m$ $a = (v_2-v_1)/t$ $= (100-0 km/hr)(1000 m/km)/(5 sec)(3600 sec/hr)$ $= 5.56 m/sec²$ $F = ma/k = (2000 kg)(5.56 m/sec²)/(1 kg-m/N-sec²) = 11,120 N$ $d = v_{avg}t$ $= (1/2)(100 km/hr)(1000 m/km)(5 sec)/(3600 sec/hr)$ $ = 69.4m \rightarrow 227.6 ft$ ## 1.10 Weight The weight of a body means the force of gravity F_g on the body; it may be determined by a so-called spring scale. Gravity produces a force field and a body in this field is subjected to a **body force**. Since the force field on the moon is much less than that on the earth [see equation (1-2)], the weight of a body is less there. In accordance with Newton's law (1-1A) that the acceleration of a particular body is proportional to the resultant force on it, we write F_g/g = F/a, where g is the acceleration produced by F_g alone (in vacuum), and a is the acceleration produced by another force F. If the analogous symbols for the moon's gravitational field are F_m and g_m, then F_g/g = F_m/g_m. If we let the acceleration produced by a gravitational field, wherever it is, be represented by g, then Newton’s equation (1-1A) says that the force of gravity is $F_g = mg/k$ in which the units must be consistent. Thus, for the force to be in pounds, the mass must be in _slugs_ when g is in feet per second per second. **Example** Two _masses_, one of 10 kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec². The combined weight of these two masses is 174.06 N. Find the unknown mass in kg and lbm. **Solution.** Using equation (1-1B) $F_g = mg/k$ $m = F_gk/g$ = $ \frac{(174.06 N)(1 kg m/sec²-N)}{9.67 m/sec^2}$ = 18 kg (total mass) unknown mass = 18-10 = 8 kg (8 kg)(2.205 lbm/kg) = 17.64 lbm. ## 1.11 Specific Volume and Density The density p of any substance is its __mass__ (not weight) per unit volume. $Average\ density\ = \frac{mass}{volume} = (\frac{m}{V}) or p = lim_{\Delta V\to 0} \frac{\Delta m}{\Delta V}$ where, for the density at a point, the volume must contain enough molecules to classify as a continuum. If the mass m is measured in pounds and the volume V in cubic feet, then the average density of a particular volume V is p = m/V lb/ft³; other units include slugs/ft³, lb/in³., gm/cm³, kg/m³. The __specific volume v__ is the volume of a unit mass, say, cubic feet per pound, and is the reciprocal of the density, v = V/m = 1/p. For homogeneous substances a __continuum__ is a quantity of matter involving a very large number of molecules. The density of a cubic centimeter with 2 or 3 molecules in it is not practically useful (except in the sense of "population density"-1 person per square mile). Moreover, if the substance is not homogeneous, the density can only be an average value; in this case, either the whole volume is used for the computation or the sample is large enough to be representative of the whole. ## 1.12 Specific Weight The specific weight γ of any substance is the force of gravity on unit volume. $Average\ specific\ weight = \frac{force\ of\ gravity}{volume} = \gamma = \frac{dF_g}{dV} = lim {_{\Delta V\to 0}\frac{\Delta F_g}{\Delta V}}$ usually lbf/ft³ or lby/in³ in English units. Since the specific weight is to the local acceleration of gravity as the density is to the standard acceleration, γ/g = ρ/k, conversion is easily made; $\rho = \frac{k}{g}\gamma or \gamma = \rho \frac{k}{g}$ If the mass is on or near the surface of the earth, then numerically g = k, and the two quantities are nearly the same. ## 1.13 Pressure-Kinetic Theory The pressure of a gas, if gravitation and other body forces are negligible (as they generally are for a gas), is caused by the pounding of a large number of gas molecules on the surface. The elementary kinetic theory presumes that the volume of the molecule itself is negligible, that the molecules are so far apart they exert negligible forces on one another, and that the molecules are rigid spheres that do have elastic collisions, with walls and with each other. Elastic collision means, for example, that when molecule A strikes the plane surface MN at an angle of incidence of α with the normal PN, it rebounds symmetrically on the other side of PN with angle α, and with no loss of kinetic energy or momentum. The _pressure_ is a consequence of the rate of change of momentum of the molecules striking the surface. **Consider a cubical container, L on a side. This assumption simplifies the physical concepts, but the result is just as general.** Since the momentum is a vector quantity and since pressure is defined as the normal force per unit area, we note that the change of momentum of molecule A in the y direction is m(v_2y - v_1y) = ma[(VA1y + (VA2y)]; or for each molecule, it is 2mv_y. Assume that this molecule moves to the opposite wall and bounces back to strike the wall MN again; the time taken for the round trip is τ= 2L/v_y. Dividing the change of momentum by the time, we get the force produced by one molecule; 2mv_y/τ = mv^2/L. If all the molecules are alike (same mass m), the total force is $F= \frac{2}{L}mN\sum v_y^2 = \frac{Nm}{L} \sum v_y^2 $ where we have multiplied and divided by N, the number of molecules in the container. The sum of the squares of the y-component velocities of all the molecules divided by the total number of molecules N gives the average of the square of v_y, symbolized by a bar thus, <v_y^2>. Then, since the average pressure is the total force divided by the area, p = F/L², we get $p = \frac{F}{L^2} = \frac{Nm<v_y^2>}{L^3} = \frac{Nm}{V}<v_y^2>$ where V = L³ = the volume of the container. Inasmuch as the molecules have random motion, as many will strike each of the other surfaces of the container as struck the one studied - evidently true because there must be some pressure on each wall for static equilibrium. This logic is verified by experimental observation that the pressure of the gas "at a point" is the same in all directions. Thus, the component velocities in the x, y, and z directions are such that <v_x^2> = <v_y^2>= <v_z^2>. Since these components are orthogonal components, the velocity vector that has these components is such that $v^2 = v_x^2 + v_y^2 + v_z^2 = 3v_y^2$ and equation (b) becomes (v^2 = 3v_y^2) $p = \frac{Nm<v^2>}{3V}$ $p = \frac{Nm<v^2>}{3V}$ where v^2 is the mean-square molecular speed; the square root of this mean square is called the root mean square; say <v^2>1/2 = v_rms. If a particular molecule strikes another molecule en route to a wall and thus fails to make the round trip, another identical molecule takes its place, producing the same effect, since all collisions are elastic and the average molecular velocity and kinetic energy (relative to the container) do not change. We speak of the pressure at a point, but actual pressure-measuring equipment (Fig. 1/5) shows one type) typically registers not dozens of molecular strikes, but ordinarily millions, in a small fraction of a second. Exceptions to this generalization include extreme vacuums and the outskirts of the earth's atmosphere. At an altitude of 30 mi, the mean free path (MFP) of a molecule is about 1 in., relatively quite far; at 400 mi, the MFP is about 40 mi. This decreasing density means fewer strikes, and if the pressure probe is _struck by a molecule only_ now and then, there is no meaning to the "pressure at a point." A cubic inch of atmosphere (a handful) contains some 4 × 10^19 molecules. In dealing with so many particles making up a system, it may occur (microscopically), and probably does, that the pressure on an extremely small area is momentarily quite high (or low) because by chance a number of high velocity (or low velocity) molecules have just happened to strike that area. Such an event involves such a small area that no pressure-measuring instrument could identify it-not to mention the virtually infinitesimal duration of the event. In short, nearly all our gas systems will involve enough molecules that the systems easily qualify as continuums, and the pressure (and other) instruments record a statistical average number that applies to the system (at rest), the macroscopic pressure. **Barometers are used to measure atmospheric pressure.* It is convenient to have a standard reference atmospheric pressure, which is 760 mm Hg or 29.92 in. Hg at 32°F, or 14.696 psia (14.7 for slide rule), or 1 atm. Pressure is one of the most useful thermodynamic properties because it is easily measured directly. (Precision measurements of anything are difficult.) Pressure-measuring instruments read a difference of pressures, called gage pressure; in pounds per square inch, we shall use the abbreviation psig, the g standing for gage. The absolute pressure, say psia, can be determined from the gage pressure as follows: $Absolute\ pressure = atmospheric\ pressure \pm gage\ pressure$ where the positive sign applies when absolute pressure is greater than atmospheric, and the negative sign for absolute pressure less than atmospheric. The negative sign is for a gage reading called vacuum pressure of pressure. Each term in (1-8) should, of course, be in the same pressure unit. Equation (1-8), written to apply when the gage proper is located in the atmosphere, may be generalized by this statement: the gage pressure is the difference in pressures of the region to which it is attached (via the "threaded connection," Fig. 1/5) and the region in which the gage is located. **Example** A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. **Solution.** Using equation (1-8) $P =P_{atm}+P_g = 14.25 + 50 = 64.25 psia$ also $p = (64.25 psia)(6894.8 Pa/psi) = 442,991 Pa = 443 kPa$ ## 1.14 Fluid Pressure The foregoing discussion applies particularly to homogeneous systems in equilibrium, imperceptibly affected by body forces (gravity, magnetism). In a liquid system, while the movement of the molecules is considerably more restricted than in a gas, the molecular bombardment does produce pressure but not the total pressure since the body force of the gravitational field is more likely to produce a significant effect. In a boiler the pressure of the vapor above the water and the pressure at the bottom of the boiler drum (Foreword) are nearly enough the same that the difference is usually ignored. The decision as to whether or not to account for such differences is an engineering decision, to be made in the context of the actual situation, depending on its relative magnitude and on precision requirements. The following discussion applies to any fluid at rest, but is particularly intended to apply to manometers, a common pressure measuring instrument. There is no pressure gradient in any horizontal direction, but there is in a vertical direction because of gravity; and the pressure p on a cross-sectional area A is uniform. Thus, we may use the differential volume dV = A dz, shown in Fig. 1/6, as the free body. The force of gravity on this free body is dF_g = -γA dz acting through the c.g., where the negative sign is used because z is positive as measured upward, while the force vector dF_g points down. The resultant force of the pressure on the upper side of the element dV is pA; on the lower side, (p + dp)A. Summing forces on the element, we get $(p + dp)A - pA - dF_g = A dp + yA dz = 0$ $dp = -y dz$ the basic relation; the units must be consistent. In short columns of liquids or gases, the specific weight is virtually constant. If γ varies and its manner of variation as a function of z is known, this equation can be integrated as shown. Integrating (1-9) with constant γ from a liquid surface where the uniform pressure is any value p_a gives $p-p_a = y(z_o - z)$ or $P = p_a + y(z_o - z) = p_a + yd$ where p is the pressure at any level marked by the coordinate z, and p_a is the pressure at z_o (perhaps the interface of a liquid and a gas or of the liquid and its own vapor); the final form p_a + yd is handy where d represents the vertical dimension (depth) of the fluid, usually a liquid in this application. It often happens that p_a is the ambient atmospheric pressure. The units in (1-9) or (1-10) must be consistent: for example, γ lb/ft³, z ft, p lb/ft²; or γ lb/in³., z in., p lb/in². An infinitesimal area would result in the same basic equation (1-9). Therefore, we can say that the fluid pressure at a point is the same in all directions. ## 1.15 Manometers Manometers give a reading as the length of some liquid column: mercury, water, alcohol, etc. If the length of column of liquid, of cross-sectional area A, is d, then the volume is V = Ad and the force of gravity on the column is F_g = yAd, where γ is the specific weight of the fluid; γ = (g/k)ρ, equation (1-6). The corresponding pressure is p = F/A = yd. That part of the fluid in the loop HJ, Fig. 1/7, is evidently, from symmetry, in equilibrium and can be ignored. The pressure at B is equal to the pressure p_a at G plus γd_nd = y(GH) - γg(KJ), where γg is the specific weight of the fluid in the part KJ. If this fluid in KJ is gaseous of ordinary densities (say the fluid in the vessel is a gas), the value of γg(KJ) may be small enough to be neglected. If this is true, the pressure in the vessel is taken as $p = p_a + γd = p_a + \frac{gρd}{k} = p_a + gd$ where p_a is the pressure of the surroundings, v is the specific volume, and the units are consistent. See Item B 38, appendix for conversion constants; as (in. Hg) (0.49 psi/in. Hg) → lb/in. The pressure p as it appears in most equations must be absolute pressure. Even when the conversion unit cancels, it is advisable to write it down (and then cancel it) because the habit of conversion is important. Also remember to convert gage pressure to absolute pressure. **Example** The vessel in Fig. 1/7 contains a gas; the manometer contains a liquid whose density is 300 lb/ft³ with d = 50 in. Find the gas pressure in psi and Pa; local g = 32.11 fps² and barometer = 14.55 psia. **Solution.** Using equation (1-11) $p = p_a + gpd/k$ $= 14.55 psi + \frac{(32.11 ft/sec^2)(300\ lb/ft^3)(50\ in.)(ft/12\ in.)}{(32.174 lb-ft/lby-sec²)(144\ in²/ft^2)}$ $= 14.55 + 8.66 = 23.21 psia $ also $P = (23.21 psia)(6894.8 Pa/psi) = 160,028 Paa$ ## 1.16 Archimedes Principle A body immersed in a fluid is subjected to a **buoyant** (upward) force equal to the weight of the fluid displaced. Since there is no _horizontal pressure gradient_, this effect is due to the vertical fluid force on surfaces that face downward being greater than the vertical force on the surfaces that face upward. The net force on the body is the force of gravity on the displaced fluid minus the force of gravity on the body. For a floating body at rest, the net force is zero. Observe that, strictly, when a body is weighed in the atmosphere, a correction should be made for the weight of the displaced air, a negligible correction when the density of the body is much greater than the density of the air. ## 1.17 Temperature - With a Microscopic View As Maxwell said, the _temperature_ of a body is its thermal state considered with reference to its ability to communicate heat to other bodies. [1.21] (See Zeroth Law, §1.20.) It is an intensive property that, as we shall learn, is a measure of the intensity of the stored molecular energy in a system. Consider for a moment the microscopic view, which necessitates recalling certain things. Let M equal the molecular mass (also called molecular weight, formula mass); let n equal the number of moles, where, for example, a so-called pound-mole is M lb of a substanice. The quantity that is a mole is different in different systems of units: * 1 pmolet (pound-mole) = M lb * 1 gmole (gram-mole) = M gm * 1 kgmole (kilogram-mole) = M kg For instance, for O₂, M = 32; therefore 1 pmole of O₂ is 32 lb, 1 gmole is 32 gm, and so on. ## 1.18 Scales of Temperature Avogadro’s number