1 The Nature of Fluids 1.1 OBJECTIVES OF THIS BOOK The termfiuid mechanbs refers to the study of the of fiui s, either at rest or in motion. Fluids can be either liquids (such as water, oil, gasoline, or glycerine) or gases (.uch as air, oxygen, nitrogen, or helium). The behav-ior of fluids affects YOl r daily life in many ways. When you turn OII a faucet, water is delivered to you by a distribution system composed of pt: mps, valves, and pipes. The source of the water may be a storage tank, rese-voir, river, lake, or well. The flow of water from its source to your faucet is controlled by the principles of fluid mechaniCs. These principles must be understood in order to -select properly the size and type of pumps and pipes, to design the storage tanks, to select flow control valves, and to monitor the performance of the system. The performance of an automated manufacturing machine, which is controlled by fI':Iid systems, is dependent on the flow of the hydraulic oil and the pressure at the actuators. A typical system is shown in Fig. 1.1. The greater the preSSLIe of the oil in a cylinder, the greater the force it can exert. The greater the flow rate of Qil into the cylinder, the faster it will move. You will learn how to analyz.e such a system using the material in this book. A buoy marking a boating cBannel seems like a very simple device, and it is. H;owever, the mLterial from which the buoy is made and its geometry must be specified acc( rding to the la s of buoyancy and stability of floating bodies, which you wil learn in eh.apter 5 of this book. In your auto motile , fuel is dlivered to the carburetor from the tank by a pump. How m\JICh poweT is being supplied by the engine to drive the pump? material h Chapter 7 will help you to caiculate this. A highway si directions to motorists may have to withstand higH winds. In or-der k:J determine the forces exerted on the sign due to the the pnncipl.e \I impulse-momentum as presented in Chapter 16 must be unckr UQ!!. . Automation ment for manufacturing systems often employs high-pressure compressed ;air to actuate pneumatic cylinders and air motors. The flow of air in piping srstems must be analyzed to ensure thatthe air pressure at the point of use is -sufficient. Heating, ventilL1ing, and air-conditioning systems deliver air at low pressure to living an. working spaces to improve the comfort of the occu-pants. The methods presented in Chapter 19 will help you analyze such 1 2 1.2 OBJECTIVES OF THIS CHAPTER 1.3 DIFFERENCE BETWEEN LIQUIDS AND GASES Chapter 1 The Nature of Fluids FIGURE 1.1 Typical piping system for fluid power. fi res Pump Fluid power cylinder actuator . Return line Conveyor These are just a few of the many practical problems you are likely to encounter that require an understanding of the principles of fluid mechanics for their solution. The objective of this book is to help you solve these kinds of problems. Included in each chapter are problems representing situations from many fields of technology. Your ability to solve these problems will be a measure of how well the objective of this book has been accomplished. In order to understand the behavior of fluids, it is necessary to understand the of the fluids themselves. This chapter defines the properties of fluids , introduces the symbols and units involved, and discusses the kinds of calculations required in the study of fluid mechanics. Mter completing this chapl er, you should be able to: 1. Differentiate between a gas and a liquid. 2. Identify the units for the quantities of time, length, force, and in the SI system (metric unit system). 3. Identify the units for the quantities oftime, length, force, and .. in the U .S. Customary 4. Properly set up equations to ensure consistency of units. 5. Define compressibitity and bulk modulus. 6. Define pressure. 7. Define the bet ween force and mass. 8. Define density. 9. Define specific weigltt. 10. Define specific gravity. 11. the relationskips between specific weighty speGifi'? gravity, and density, and solve problems using these relationship-so When a liquid is held in a container, it tends to take the sllape of the con-tainer, covering the bottom and sides. The top in with the atmosphere above it, maintains a uniform level. As the is tipped, the liquid tends to pour out; the rate of pouring is depengent on a property called viscosity, which will be defined later. 1.4 FORCE AND MASS C WEIGHT-MASS RELATIONSHIP 1.5 THE INTERNATIONAL SYSTEM OF UNITS (SI) 1.5 The International of Units (SI) 3 When a gas is he(j in a closed container , it tends to expand and com-pletely fill the container. If the container is opened, the gas tends to continue to expand and escape trom the container. In addition to familiar differences between gases and liquids, another difference is inportant in the study of fluid mechanics: . Liquids are only sliglltly compressible. Gases are readily co:npressible. Compressibility refers to the change in volume of a substance when the pressure on it changes These distinctions are suffioient for mosJ plli-poses. The continuing jiscussion of pressure, ompressit>ility; and other properties of fluids re:}uires an understanding of the units in: which quantities are measuroo as presented in the following An understanding of fuid properties requires a art;ful betWeen mass and weight. The fo definitions apply. Mass is the of a body of flu d that is a measure of its inertia or resistance to a chaLge in mokom. It is also a measure of the quantity of fluid. We will use the symb:Jl m for IJIlaSS. Weight is the amolLlt that a body weighs, that is, the force with which a body is attracted toward the earth by gravitation. We will use the symbol w for weight. Weight is relate:! to mass and the acceleration due to gravity, g, by Newton's law of gra';tation, w = mg (1-1) In this book, we will :Jse g = 9.81 m/s2 in the SI system and g = 32.2 ft/S2 in the U .S. Customary!)ystem. These are the standard values for g to three significant digits. To L greater degree of precision, the standard values of g = 9.806 65 m/s2 or g = 32.1740 ft/S2. For high precision work and at high elevations (such as aerospace operations) where the actual value of g is different from the stLndard, the local value .should be used. In any technical walk the units in which physical properties are measured must be stated. A syftem of units specifies the units of the basic quantities of length, time, force, cod mass. The units of other terms are then derived from these. Le Systeme ILternational d'Unites, or the International System of Units (abbreviated SI), is used in this book. The units for the basic quantities 4 Chapter 1 The Nature of Fluids are length = meter* (m) time = second (s) mass = kilogram (kg) force == newton (N) or kg m/s2 An equivalent unit for force is the kg'm/s2, as indicates above. This IS derived from the relationship between force and mass froD physics, F= ma where a is acceleration expressed in units of m/s2. the derived unit for force is F = ma = kg m/s2 = newton Thus a force of 1.0 N would give a mass of 1.0 kg an acceleration of 1.0 m/s2. This means that either newtons (N) or kg m/s2 may be useB as the units for force. In fact, some calculations in this book require that you be able to use ' both or to convert from one to the other. For example, we may say that a rock with a mass 01" 5.60 kg is sus-pended by a wire. Then, in order to determine what force i:s exerted on the wire, Newton's law of gravitation (w = mg) should be used: w = mg = mass x acceleration of gravity But, under standard conditions at sea level, g = 9.81 m/s2. Then we have w = 5.60 kg x 9.81 m/s2 = 54.9 kg m/s2 = 54. 9 N Thus, 5.60 kg of the rock weighs 54.9 N. Because the actual size of physical quantities in the study of fluid mechanics covers a wide range, prefixes are added to the basic quantities. Table 1.1 shows these prefixes. Standard usage in the SI 3ystem calls for only those prefixes varying in steps of 103 as shown. Results of calculations should normally be adjusted so that the number is betwee. 0.1 and 10 000 times some multiple of 103 t Then the proper unit with prefiX: can be specified. Some examples follow. * The American National Metric Council (ANMC) recommends tile spellings "me-ter" and "liter" instead of "metre" and "litre." We will use tlae preferred nota-tion of the ANMC throughout this book . . t Because commas are used as decimal markers in many countrie3, we will not use commas to separate groups of digits. We will separate the digi ts into groups Qf three, counting both to the left and to the right from the decimal point, and !! space to separate the groups of three digits. We will not use a space if there Me only four digits to the left or right of the decimal point unless required in ta!mlar matter. TABLE 1.1 SI unit prefixes 1.6 THE V.S. CUSTOMARY SYSTEM 1.6 The U.S. Custom3ry System Computed Result 0.0042: m 15700 kg 86330 N Reported Result 4.23 X 10-3 m, or 4.23 mm (millimeters) 15.7 x 103 kg, or 15.7 Mg (megagrams) 86.33 x 103 N, or 86.33 kN (kilonewtons) 5 Sometimes called Hoe English gravitational unit system or the pound-Joot-second system, the U .S. Customary System defines the basic quantities as follows: length = foot (ft) time = second (s) force = pound Ob) mass = slug Ob-s2/ft) Probably the most difficult of these units to understand is the slug, since we are familiar with measuring in terms of pounds, seconds, and feet. It may help to note the relationship between force and mass from physics, F=ma where a is aocelerCJliol1 expressed in units of ft/s2. Therefore, the derived unit for mass is F lb tb-s2. m = a = ftls = ft = slug This means that yOu may use either slugs or ib-s2/ft for the unit of mass. In fact, some calculations m this book require that you be able to use both or to convert from one b the other. Mass Expressed s Ibm (Pounds-Mass) Some , in the analysis of fluid flow systems, have used the unit Ibm (pounds-mass) for the unit of mass instead of -the unit of slugs. In this system, an object (I a quantity of fluid having a weight of 1.0 -lb would have a mass of 1.0 Ibm. Tie pound-force is then sometimes designated lbf. It must 1.7 CONSISTENT UNITS IN AN EQUATION Chapter 1 The Nature of Fluids be noted that the numerical equivalence of lbf and Ibm applies only when the value of g is equal to the standard value. This system is avoided in this book because it is not a COterent system. When one tries to "relate force and mass units using New_on's law, the following is obtained. F = m' a = Ibm(ft/s2) = Ibm-ft/s2. This is not the same as the Ibf. To overcome this difficulty, a conversion constant, cOIlmonly called gc, is defined having both a numerical value and units . That is, 32.2 Ibm 32.2 Ibm-ft/s2 gc = Ibf/(ft/s2) = Ibf Then to convert from Ibm to lbf, a modified form of Newtons law is used, F = m(a/gc) Letting the acceleration a = g, F = m(g/gc) For example, to determine the weight of material in lof tial has a mass of 100 Ibm,and assuming that the local value of g is equal El lhe slandaFd value of 32.2 ft/S2, g 32.2 ft/S2 w = F = m gc = 10Q fum 32.2 lOrrr-ft/S2 = 100 l::Jf lDf This shows that in lbf is numerically equal to mass i. Ibm provided g = 32.2 ft/51. But if analysis IQ he done for an objeGor fluid on the earth's moon where g is app o&imatcly 1/6 that o-n thee-arth. say, 5.=t ft/S2, " g 3,4 ft/52 w = F = m gc = 100 Ibm 32.2 Ibm-ft/s2 = 16.8:bf lbf "" This is dramatic difference. In summary, because of the cumbersome nature of 6e relationship between Ibm and Ibf, we avoid the use of Ibm in this Mass will be expressed in the unit of slugs when problems are in the U_S. Customary System of units. The analyses required in fluid mechanics involve the algebraic manipulation . ' . of several terms. The equations are often complex, and it is extremely im-portant tliat the results be dimensionally correct. That is, ther must have the proper units. Indeed, answers will have the wrong numerit=al value if the units in the equation are not consistent. 1.7 Consistent Units in :an Equation A simpl e straightj;orward procedure called unit cancellation will ensure proper units in any kinrder to cancel. Now that we have :he time unit of seconds we can proceed with Step 6. The correct answet is t = 67.5 s. Pressure is defined a1 the amount of force exerted Qn a unit area of a substance. This can b( stated by the equation F p==-A (1-2) Two important principes about pressure were described by Blaise Pascal, a seventeenth-century scientist. Pressure acts unifornly in all directions on a small volume of a fluid . In a fluid confined bl solid boundaries, acts perpendicular to the boundary. These principles, sometimes called Pascal's laws, are iHustrated in Figs. 1.2 and 1.3. . FIGURE 1.2 Pressureacting uniformly in all on a small volume Qf ftuig. FluLd surface LO Chapter 1 The Nature of Fluids l"IGURE 1.3 Direction of fluid Jressure on boundaries. (a) Furnace duct (b) Pipe or tube (c) Heat exchanger (a pipe inside another pipe) (d) Reservoir (e) Swimming pool (f) Dam (g) Fluid power cylinder Using Eq. (1-2) and the second of Pascal's laws, we can compute the of in a fluid if we know the amount of force exerted on a given area. o EXAMPLE PROBLEM 1.2 Figure 1.4 shows a container of liquid with a movable piston supporting a load. Compute . the magnitude of the pressure in the liquid . under the piston if the tolal weight of the piston and the load is 500 N and the area of the piston is 2500 mmi . Solution It is reasonable to assume that the entire surface of the fluid under the piston is sharing in the task of supporting the load. The second of Pascal's laws states that tlTe fluid pressure acts perpendicular to the piston. Then, using Eq. (1-2), _ F _ 500 N _ .. 2 P - A 2500 mm2 - 0.20 N/mm The standard unit of pressure in the SI system is the NI m2, called the pascal (Pa) in . . bonor of Blaise Pascal. The conversion can be made by using the factor 1()3 mm = 1 m. . 0.20 N (103 mm)2 .. . p = x = 0.20 X 106 N/m2 = 0.20 MPa . mm2 m2 . Note that pressure in N/mm2 is numerically equal to pressUre 10 MPa. It is not unusualto encounter pressure in the range of several megapascals (MPa) or several hundred kilopascals (kPa). FIGURE 1.4 Illustration of fluid Pressure in the V.S. Customary System is illustrated in the following example pressure supporting a load. problem. . 1.9 Compressibility 11 o EXAMPLE PROBLEM 1.3 A load of 200 pounds Ob) is exerted on a piston confining an oil in a circular cylinder with an inside diameter :Jf 2.50 inches (in). Compute the pressure in the oil at the piston. See Fig. lA. Solution Using Eq. (1-2) , we muft compute the area of the piston. 1.9 COMPRESSIBILITY 0 , BULK MODULUS A = 1TD2/4 = 1T(2.50 in)2/4 = 4.91 in2 Then, F 200lb p = A = 4.91 in2 = 40.7 Ib/in2 Although t he standard u.it for pressure in the U .S. Customary System is pounds per 09t (lb/ft2), it iSllot often used because it is inconvenient. Length measure-ments are more convenllCntly made in inches, and pounds per square inch (lb/in2), psi, is used 1I10st often for pressure in this system. The pressure in the oil ! 40.7 psi. This is a faiAy low pressure; it is not unusual to encounter pressures of I hundred or sevtr.tl thousand psi. The bar is anotler unit used by some people working in fluid mechan-ics and thermodynatrics. The bar is defined as 1()5 Pa or 105 N/m2 Another way of expressing th( bar is 100 x 103 N/m2, which is equivalent to 100 kPa. Since atmospheric pJlessure near sea level is very' nearly the same, the bar has a convenient" pont of physical reference. This, plus the fact that pres-sures expressed in bus yield smaller numbers, makes this unit attractive to some practitio'ners. -: ou must realize, however, that the bar is not a part of the coherent SI system and that you must carefully convert to N 1m2 (pas-cals) in problem soh:ing. Compressibility refe-s to the change in volume (V) of a substance that is subjected to a chang= in pressure on it. The usual quantity used to measure this phenomenon is 1l1e bl!lk modulus of elasticity or, simply, bulk modulus, E. -6.p , E= (6.V)IV (1-3) Because the quantibes 6. V and V would have the same units, the denomina-tor of Eq: (1-3) isdimensionless. Therefore, the units for E are the same as , those for the , . As stated befo::-e, liquids are very slightly compressible, indicating that it would take a ver; large change in pressure to produce a small change in volume. Thus, the -nagnitudes ofE for liquids, as shown in Table 1..2, are very high. For this ::-eason, liquids will be considered incompressible in this . book, unless stated otherwise. The term modulus is not usually applied to gases, and the princi-ples of must be 'applied to determine the change in volume of a gas with a cha:tge in pressure. u Chapter 1 The Nature of Fluids TABLE 1.2 Values for bulk modulus for selected liquids . - . . , , ':"" Bllk Modulus ' - -" ,- -, Liquid (psi) C (MPa) . , ' - - . - o EXAMPLE PROBLEM 1.4 Compute the change in pressure that must be applied to water to change its vglume by 1.0 percent. Solution The l.O-percent volume change indicates that AV!V = -0.01. Tnen, change in pressure is 1.10 DENSITY, SPECIFIC WEIGHT, AND SPECIFIC GRAVITY Ap = - E[(AV)IV] = [-316000psi][ - 0.Ol] = 3160 psi Because the study of fluid mechanics typically deals with a continuously flowing fluid or witha small amount of a fluid at rest, it is most convenient to relate the mass and weight of the fluid to a given volume of fluid. Thus, the properties of density and specific weight are defined as follows. Density is the amount of mass per unit volume of a substance. Therefore, using the greek letter p (rho) for density, , DENSITY p = m/V (1-4) c:" SPECIFIC WEIGHT I where V is the volume of the substance having a mass fr.. The for density are kiiograms per cubic meter in the SI system and slugs per cubic foot 'in the U.S. Customary System. The American Society for Testing and Materials (ASTM) has published several standard test methods for measuring density that describe vessels having precisely known volumes, called pycnometers. proper filling, handling, temperature control, and reading of these devices are prescribed. Two types are known as the Bingham pycnometer and th6 Lipkin bicapil/ary pycnometer. The standards also call for the precise deter:nination of the mass Of the fluids in the pycnometers to the nearest 0.1 mg using an analyti- balance. See References 2, 3, 5, and 6. Specific weight is the amount of weight per unit volumt of a substance. U sing the Greek letter 'Y (gamma) for specific weiglJit t, 'Y = w/V (1-5) o SPECIFIC GRAVITY 1.10 Density, Specific Weight, and Specific Gravity 13 where V is the volume of a substance having the weight w. The units for specific weight are newtons per cubic meter (N/m3) in the SI system and pounds per cubic foot (lb/ft3) in the U.S. Customary System. It is often convenient to indicate the specific weight or density of a fluid in terms of its relatior.ship to the specific weight or density of a common fluid. When the term specific gravity is used in this book, the reference fluid is pure water at 4C. At that temperature, water has its greatest density. Then, specific gravity can be defined in either of two ways: 3. Specific gravity is the ratio of the density of a substance to the density of water at 4C. b. Specific gravity is the ratio of the specific weight of a substance to the specific weight of water at 4C. These definitions for specific gravity can be shown mathematically as (1-6) where the subscript s refers to the substance whose specific gravity is being determined and the subscript w refers to water. The properties of water at 4C are constant, having the values shown below. 'Yw @ 4C = 9.81 kN/m3 pw @ 4C = 1000 kg/m3 'Yw @ 4C = 62.4 Ib/ft3 or Pw @ 4C = 1.94 slugs/ft3 Therefore, the mathematical definition of specific gravity can be written 'Ys Ps 'Ys . Ps sg = 9.81 kN/m3 = 1000 kg/m3 or sg = 62.4 Ib/ft3 = 1.94 shigs/ft3 This definition holds regardless of the temperature at which the specific gravity is being determined. The properties of fluids do, however, vary with temperatu e. In: n-eral, the density (and therefore, the specific weight and speeifie gravity) decreases with increasing temperature. The properties of water at various temperatures are listed in Appendix A. The properties of other liquids at a few selected temperatures are listed in Appendix B and Appendix C. You should seek other references for data on -sp.eclllc gravity at speci-fied temperatilres if they are not reported in the and if high preci-sion is desired. One estimate that gives reasonable accuracy for petroleum oils, as presented more fully in References 8 and 10, is that the specific gravity of oils decreases approximately 0.036 for a 100F (37.8C) rise in temperature. This applies for nominal values of s.g. from 0.80 to 1.00 and for temperatures in the range from approximately 32F to 4000P (OC to 204C). Some industry prefer modified demif ons for specific gravity. Instead of usinR the properties of water at 4C (39.2P) as the basis, the petroleum industry and others use water at 60F (JS.6C). This makes little 1.10.1 Relation Between Density and Specific Weight "> -y-p RELATION Chapter 1 The Nature of Fluids difference for typical design and analysis. Although the demity of water at 4e is 1000.00 kg/m3, at 600P it is 999.04 kg/m3 The differeoce is less than 0.1 percent. References 2, 3, 5, 6, 7, and 10 contain more extensive tables of the properties of water at temperatures from ooe to 1000e C3:2P to 212P). Specific gravity in the Baume and API scales is discu8Sed in Section 1.10.2. We will continue to use water at 4C as the basis for pecific gravity in this book. The ASTM also refers to the property of specific gra'-ity as relative density. See References 2-6. Quite often the specific weight of a substance must be found hen its density is known and vice versa. The conversion from one to the othc=r can be made by the following equation: "I = pg (1-8) where g is the acceleration due to gravity. This equation can be justified by referring to the definitions of density and specific gravity and by using the equation relating mass to weight, W = mg. The definition of specific weight is W "1=-V By multiplying both the numerator and the denominator ofthi. equation by g we obtain W "I .== Vg But m = wIg. Therefore, we have Since p = m/V, we get _ mg "1 =. V l' = pg The following Problems il Clstrate the defi.nitions of the t.tsic fluid prop-erties presented above amd the relationships between the varbus properties. o EXAMPLE PROBLEM 1.5 Calculate the weight of a reservoir of oil if it has a mass of 825 kg. Solution Since w = mg, . w = 825 kg x 9.81 m/s2 = 8093 kg m/s2 . oC!wton for the unit kg ml S2, we have . w = 8093 N = 8.093 X 103 N = 8.093 kN 1.10 Density, Sp-ecifil> We'g1t, and Specific Gravity 15 o EXAMPLE PROBLEM 1.6 If the reservoir from Example f roblem 1.5 has a volume of 0.917 m3, compute the density, specific weight , and rhe specific gravity of the oil. SolutiQn Density: _ m _ 825 kg _ 3 P - V - 0.917 m3 - 900 kg/m Specific _ w _ 8.093 kN _ 3 'Y - V - 0.917 m3 - 8.83 kN/m Specific gravity: Po 900 kg/m3 . sg = Pw @ 4C = 1000 kg/m3 = 0.90 o EXAMPLE PROBLEM 1.7 Glycerine at 20C has a sp=cific gravity of 1.263. Compute its density and specific weight. Solution ' Density: Pg = (sgMl00c kg/m3) = (1.263)(1000 kg/m3) = 1263 kg/m3 Specific weight: 'Yg = (sgM9.81 =::N/m3) = (1.263)(9.81 kN/m3) = 12.39 kN/m3 o EXAMPLE PROBLEM 1.8 A pint of water weighs 1. ... 1 lb. Find its mass. Solution Since w= mg, the mass o ExAMPLE PROBLEM 1.9 Solution w 1.041 lb 1.041 Ib-s2 .,. = g = 32.2 ft/S2 = 32.2 ft = 0.0323 Ib-s2/ft = 0.0323 slugs . Remember that the units..of slugs and Ib-s2/ft are the same. One gallon.of mercury has a of 3.51 slugs. Find its weight. W = ni! = 3.51 slugs x 32.2 ft/S2 = 113 slug-ft/s2 . This is correct, but the u,its may seem confusing since weight is normally expressed in pounds. The units of:nass may be rewritten as Ib-s2/ft. ' lb-s2 . 32.2 ft . . w= mg = 3.51 Tt x --sr- = t131b 16 1.10.2 Specific Gravity in Degrees Baume or Degrees API Chapter 1 The Nature of Fluids The reference temperature for measurements in the Baume or API scale is 60F rather than 4C as defined before. To emphasize this difference, the API or Baume specific gravity is often reported as S'fi . 60 F peCl c gravity 600 This notation indicates that both the reference fluid (water! and the oil are at 60F. Specific gravities of crude oils vary widely depending on where they are found. Those from the western U.S. range from approximately 0.87 to 0.92. Eastern U.S. oil fields produce oil at about 0.82 specific gravity. Mexi-can crude oil is among the highest at 0.97. A few heavy asphaltic oils have sg > 1.0. (See Reference 7.) Most oils are distilled before use to enhance their quality of burning. The resulting gasolines, kerosenes, and fuel oils have specific gravities rang-ing from about 0.67 to 0.98. The equation used to compute specific gravity when the degrees Baume are known is fOf flUids lighter than and fluids beavier than water. For liquids he vier than wat r, 145 sg == 145 = aauiTI Or,to compute the degrees Raume fof a glYtm specinc gravity, deg Baume = 145 --sg For liquids lighter than water, 140 sg = 130 + deg Baume d B ' 140 eg aume = - - 130 sg (1-9) (1-10) (1-11) (1-12) The American Petroleum Institute (API) has developed the API scale, slightly different from the Baume scale, for liquids lighter than water. The formulas are 141.5 sg = 131.5 + deg API deg API = 141.5 - 131.5 sg (1-13) (1-14) Degrees API for oils may range from 10 to 80. Most fuel grades will fall in the range of API 20 to 70, corresponding to specific gravities from 0.93 to 0.70. Note that the heavier oils have the lower values of degrees API. ASTM Standards D287 and D1298 (References 1 and 4) describe stan-dard test methods for determining API gravity using a h),drometer. Figure . FIGURE 1.5 Hydrometer with built-in thermometer (thermohy-drometer). REFERENCES 1.10 Density, Specific Weight, and Specific Gravity 17 1.5 shows a sketch of 1 typical hydrometer incorporating a weighted glass bulb with a smaller dianeter stem at the top that is designed to float upright in the test liquid. Based on the principles of buoyancy (see Chapter 5), the l1yctrometer rests at a p)sition that is dependent on the density of the liquid. T h ~ ~ t e r n is marked Wi lh a calibrated scale from which the direct reading of density, specific gravity, or API gravity can be made. Because of the imp or-tan