Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 15

10.5 Fluid Dynamics 337 Example 10.18 Flow Speed from a Reservoir (Torricelli’s Law) A tank is filled with water to a height y1 = 3.5 m The tank has a small hole in one of its walls at a height y2 = 1.5 m, see Fig 10.27 (a) What is the speed v2 of the water emerging from the hole? (b) What is the horizontal distance x from the base of the tank to the point at which the water stream strikes the floor? Fig 10.27 y y1 Pa h y1 - y2 Pa y2 x Solution: (a) The pressure at the top of reservoir and at the hole is the atmospheric pressure Pa , because both of them are exposed to the atmosphere If we assume the tank has a large cross-sectional area A1 compared to that of the hole’s, i.e A1 A2 , then water at the top of the reservoir will be almost stationary, i.e v1 In addition to this, h = y1 − y2 Using Bernoulli’s equation, see Eq 10.38, we get: Pa + 21 ρ × (0) + ρgy1 = Pa + 21 ρv22 + ρgy2 v2 = That is: Thus: v2 = 2gh = ⇒ v22 = 2gh 2gh 2(9.8 m/s2 )(2 m) = 6.26 m/s The relation v2 = 2gh is the same as that for an object falling freely from rest through a height h This is known as Torricelli’s law (b) As in Sect 4.3, the initial water velocity v→o ≡ v→2 is horizontal (i.e θ◦ = 0) and the initial position is y◦ ≡ y2 = 1.5 m Since water strikes the floor at y = 0, then we use y − y◦ = (v◦ sin θ◦ ) t − 21 g t to first find the duration of descent for the water as follows: 338 10 Mechanical Properties of Matter − 1.5 m = − (9.8 m/s2 ) t ⇒ t = 0.553 s The horizontal distance x covered by the water in that time is: x = (v◦ cos θ◦ ) t = (6.26 m/s)(cos 0◦ )(0.553 s) = 3.46 m Example 10.19 In the Venturi meter of Fig 10.28, air of density ρair = 1.3 kg/m3 flows from left to right through a horizontal pipe of radius r1 = 1.25 cm that necks down to r2 = 0.5 cm The U-shaped tube of the meter contains mercury of density ρmer = 13.6 × 103 kg/m3 If the speed of the air entering the meter is v1 = 10 m/s, then find the mercury-level difference h between the two arms A1 Air flow A2 r1 r2 P2 P1 h Mercury Fig 10.28 Solution: Since the air moves horizontally, then Bernoulli’s equation at the two openings of the U-shaped tube becomes: P1 − P2 = 21 ρair (v22 − v12 ) Also, from the continuity Eq 10.32 we have: A1 v1 = A2 v2 ⇒ v2 = v1 π r2 A1 = v1 12 = v1 A2 π r2 r1 r2 Substituting with the form of v2 in the pressure difference, we get: P1 − P2 = 21 ρair v12 r1 r2 −1 10.5 Fluid Dynamics 339 The pressure difference between the two openings of the U-shaped tube produces a mercury-level difference h given by P1 − P2 = ρmer gh Thus, by combining the last result with this equation we get: h= ρair v12 2ρmer g r1 (1.3 kg/m3 )(10 m/s)2 −1 = r2 2(13.6 × 103 kg/m3 )(9.8 m/s2 ) 1.25 cm −1 0.5 cm = 0.019 m = 1.9 cm ∗ Example 10.8 The sketch in Fig 10.29 shows a perfume atomizer before and after compressing its bulb When the bulb is compressed gently, air with density ρair flows steadily through a narrow tube, reducing the pressure at the position of the vertical tube Liquid of density ρL can rise in this vertical tube and enter the horizontal tube and be sprayed out If the pressure in the bulb is Pa + P, where Pa is the atmospheric pressure, and v2 is the speed of air in the horizontal tube, then find the pressure formula in the horizontal tube What must the value of v2 be in order to raise the liquid to the horizontal tube? Air Pa Air flow P2 P1 = Pa + P Pa Bulb Air Pa Liquid h Compressed bulb Before Pa h Liquid After Fig 10.29 Solution: By applying Bernoulli’s equation on the bulb and the narrow horizontal tube we get the following: (Pa + P) + 21 ρ × (0) = P2 + 21 ρair v22 That is: P2 = Pa + P − 21 ρair v22 340 10 Mechanical Properties of Matter So, the decrease in P2 depends on the square of the speed v2 When the liquid rises a distance h to the horizontal tube, we have: Pa = P2 + ρL gh By equating the expression of P2 from the last two results we get: P2 = Pa + P − 21 ρair v22 = Pa − ρL gh Thus: v2 = 2(P + ρL gh) ρair Viscosity Fluids cannot withstand a shearing stress However, fluids show some degree of resistance to shearing motion, and this resistance is called viscosity The degree of viscosity can be understood by considering a fluid between two sheets of glass where the lower one is kept fixed; see the sketch in Fig 10.30 It is easier to slide the upper glass if the fluid is oil as opposed to tar because tar has a higher viscosity than oil F d c y f F e Δx d At time t+ Δ t At time t a b a b Fig 10.30 A fluid between two sheets of glass where the lower one is kept fixed while the upper one moves to the right with a speed v under the action of an external force of magnitude F We can think of fluids as a set of adjacent layers Thus, a reasonable shearing stress produces smooth relative displacement of adjacent layers in fluids, called laminar flow When we apply a force of magnitude F to the upper glass of area A, it will move to the right with a speed v As a result of this motion, a portion of the fluid with shape abcd will take a new shape abef after a short time interval t 10.5 Fluid Dynamics 341 According to Sect 10.2, we can define the shearing stress and the shearing strain on the fluid of Fig 10.30 as follows: Shearing stress = F , A Shearing strain = x d (10.41) Since the upper sheet is moving with speed v,the fluid just beneath it will move with the same speed Thus, in time t, the fluid just beneath the upper sheet moves a distance x = v t Accordingly, we define the rate of shear strain as: Rate of shear strain = shear strain = t x/d v = t d (10.42) By analogy to the shear modulus in solids, we define in fluids, at a given temperature, the ratio of the shear stress to the rate of shear strain This ratio, η, is known as the coefficient of viscosity or simply the viscosity: η= F/A Fd = v/d Av (10.43) The SI unit of viscosity is N.s/m2 = Pa.s which is called the poiseuille (abbreviated by Pl) while in cgs it is dyne.s/cm2 which is called the poise (abbreviated by P) Thus, N.s/m2 = Pa.s = Pl = 10 P = 103 cP Table 10.6 depicts some viscosity values Table 10.6 The viscosity of some fluids at specific temperatures Fluid Temperature T (◦ C) Viscosity η (N.s/m2 = Pl) Benzene 20 0.07 × 10−3 Water 100 0.3 × 10−3 Water 20 × 10−3 Whole blood 37 2.7 × 10−3 10-wt Motor oil 30 250 × 10−3 Glycerin 20 830 × 10−3 ∗ Equation 10.43 is valid only when the velocity of the fluid varies linearly with the perpendicular distance to the fluid velocity In this case, it is common to say that the velocity gradient is uniform In case of non-uniform velocity gradient, the viscosity has the general form: η= F/A dv/dy (10.44) 342 10 Mechanical Properties of Matter Stokes’ law The drag force of a small object that moves with a low speed v through a viscous medium was given in Sect 5.2 of Chap by the relation FD = bv, where b is a proportionality constant When the small object is a sphere of radius r and it moves with a terminal speed vt through a viscous medium with viscosity η, it experiences a drag force Fvis which by Stokes’ law has a magnitude: Fvis = 6π ηrvt (10.45) As an application to Stokes’ formula, Fig 10.31 displays the fall of a small metallic spherical ball of volume Vs = 43 π r , density ρs , and mass ms = Vs ρs in a viscous liquid of density ρ The forces that act on the sphere when it reaches its terminal (constant) speed vt will be: The sphere’s weight W = ms g = ρs Vs g (downwards) The buoyant force FB = ρ V g (upwards) The viscous force Fvis = 6π ηrvt (upwards) Fig 10.31 A small sphere falling with terminal speed vt in a liquid of density ρ and viscosity η Fvis FB t ms g y We must equate the volume of the liquid V that was displaced by the sphere with the volume of the falling sphere Vs Thus: ms g = FB + Fvis ⇒ ρs Vs g = ρ Vs g + 6π ηrvt From this equation we get the following relation for the viscosity η: η= (ρs − ρ)r g vt (10.46) 10.5 Fluid Dynamics 343 Example 10.21 A steel plate of area A = 0.2 m2 is placed over a thin film of lubricant of thickness d = 0.4 mm sprayed over the flat horizontal surface of a table, see Fig 10.32 When connected via a cord that passes over a massless and frictionless pulley to a mass m = 10 g, the steel plate is found to move with a constant speed v = 0.05 m/s Find the viscosity of the lubricant oil Fig 10.32 A Fv is = constant T d T Lubricant mg Solution: Since the steel plate moves with a constant speed, its resultant force must be zero Thus, the magnitude of the tension force must equal the magnitude of viscous force exerted by the lubricant on the plate, i.e Fvis = T Also, the magnitude of the tension in the cord is equal to the magnitude of the suspended weight, i.e T = m g Thus: Fvis = T = mg = (10 × 10−3 kg)(9.8 m/s2 ) = 9.8 × 10−2 N The layer of lubricant in contact with the horizontal surface of the table is at rest The lubricant speed increases across the film, reaching a maximum, v, at the layer in contact with the steel plate which moves with speed v If we assume that the rate of shear strain is constant, i.e the velocity gradient is uniform, then we can use Eq 10.43 to evaluate the viscosity as follows: η= Fvis d (9.8 × 10−2 N)(0.4 × 10−3 m) Fvis /A = = v/d Av (0.2 m2 )(0.05 m/s) = 3.92 × 10−3 N.s/m2 = 3.92 cP 344 10 Mechanical Properties of Matter Example 10.22 A tiny glass sphere of density ρs = 2.6 × 103 kg/m3 falls with a terminal velocity vt through oil which has a density ρ = 950 kg/m3 and a viscosity coefficient η = 0.2 N.s/m2 It is experimentally observed that the sphere drops a distance d = 20 cm between the two points A and B in time t = 50 s, see Fig 10.33 Find the radius r of the glass sphere Fig 10.33 A t=0 t d B t t Solution: From experimental observations, the terminal speed of the sphere will be given by: vt = d 20 × 10−2 m = = × 10−3 m/s t 50 s The forces that act on the sphere when it reaches its terminal (constant) speed vt will be: the sphere’s weight W = ms g = ρs Vs g (downwards), the buoyant force FB = ρ V g (upwards), and the viscous force Fvis = 6π ηrvt (upwards), see Fig 10.34 The volume of the liquid V that was displaced by the sphere equals the volume of the falling sphere, i.e V = Vs = 43 π r Since the sphere moves with constant speed vt , its resultant force must be zero Thus: ρs Vs g = ρ Vs g + 6π ηrvt Solving for the sphere’s radius, we obtain: r= 9ηvt = 2(ρs − ρ)g 9(0.2 N.s/m2 )(4 × 10−3 m/s) 2(2.6 × 103 kg/m3 − 950 kg/m3 )(9.8 m/s2 ) = 4.7 × 10−4 m = 0.47 mm 10.6 Exercises 345 Fig 10.34 Fvis FB t 10.6 ms g Exercises Section 10.1 Density and Relative Density (1) A solid cube of mass of 0.04 kg has a side of length cm What is the density and the specific gravity of the cube? (2) A solid sphere has a radius of cm and a mass of 0.04 kg What is the density and the specific gravity of the sphere? (3) Lead bricks like the one in Fig 10.35 are used to shield people from the hazards of radioactive materials If the lead density is ρ = 11.36 × 103 kg/m3 , then find the mass and weight of such a brick Fig 10.35 See Exercise (3) cm 20 cm 10 cm Section 10.2 Elastic Properties of Solids (4) A mass of kg is suspended from the end of a copper wire that has a diameter of mm Find the tensile stress on the wire? (5) A m long structural steel rod with a cross-sectional area of 0.5 cm2 stretches mm when a mass of 250 kg is from its lower end Find the value of Young’s modulus for this steel (6) An iron rod 10 m long and 0.5 cm2 in cross section, stretches 2.5 mm when a mass of 300 kg is from its lower end Find Young’s modulus for the iron rod (7) A wire has a length L = m and a radius r = 0.75 cm, see Fig 10.36 A force acting normally on each of its ends has a magnitude F⊥ = × 104 N 346 10 Mechanical Properties of Matter Find the change in the wire’s length and radius, when its Young’s modulus Y is 190 × 109 N/m2 and its Poisson’s ratio μ is 0.25 Fig 10.36 See Exercise (7) A r F⊥ Initial shape r _ | Δr | Final shape L F⊥ L+Δ L (8) A uniform platform is suspended by four wires, one on each corner The wires are 2-m long and have a radius of mm, and their material has a Young’s modulus Y = 190 × 109 N/m2 How far will the platform drop if an 80 kg load is placed at its center? (9) A block of gelatin resting on a rough dish has a length L = 60 cm, width d = 40 cm, and height h = 20 cm, see the vertical cross section abcd in Fig 10.37 A force F = 0.6 N is applied tangentially to the upper surface, leading to a new shape abef and hence a displacement x = mm for the upper surface relative to the lower one Find: (a) the shearing stress, (b) the shearing strain, and (c) the shear modulus f d Vertical cross section e c x F h a b L Fig 10.37 See Exercise (9) (10) Two parallel but opposite forces, each having a magnitude F = × 103 N, are applied tangentially to the upper and lower faces of a cubical metal block of side a = 25 cm and shear modulus S = 80 × 109 N/m2 , see Fig 10.38 Find the displacement x of the upper surface relative to the lower one, and hence find the angle of shear θ 352 10 Mechanical Properties of Matter of water emerging from a small pipe at the bottom of the tank (b) If y2 = m, then find the horizontal distance x (from the base of the tank) that the water stream travels before striking the floor Fig 10.50 See Exercise (27) P2 h y2 P1 y1 M Fig 10.51 See Exercise (28) y A1 y1 Pa h= y1 - y2 y2 A2 Pa x (29) Assume the faucet in the previous exercise is closed In terms of the crosssectional areas A2 and A1 of the small pipe and the tank, respectively, show that v2 and x depend on the variable height h as follows: v2 = 2gh y2 h , x=2 [1 − (A2 /A1 )2 ] [1 − (A2 /A1 )2 ] (30) Figure 10.52 shows a tube of uniform cross section that is filled with water to siphon water at a steady rate from a large vessel (a) Use Torricelli’s approach to find an expression for the speed at level C (b) Use Bernoulli’s equation to find an expression for the pressure at level B in terms of Pa , H, and h (c) Find the maximum value Hmax of H for which the siphon will work (d) Calculate the answers to the previous parts when ρw = 103 kg/m3 , Pa = 100 kPa, H = m, and h = m 10.6 Exercises 353 Fig 10.52 See Exercise (30) B H Pa A h Pa C (31) A child tows a thin piece of wood of surface area A = 200 cm2 through a water puddle at a constant speed v = 15 cm/s, see Fig 10.53 The depth of the water puddle is d = mm and its viscosity is η = cP Assume that the velocity gradient is constant from the bottom to the surface of the water Find the horizontal force component F exerted by the tow cord Fig 10.53 See Exercise (31) = constant A Water F Fv i s Ground d (32) A steel plate of weight W = 0.5 N and area A = 0.2 m2 is placed over a thin film of lubricant (oil) of thickness d = 0.2 mm sprayed over a flat surface inclined at an angle θ = 30◦ , as shown in Fig 10.54 What is the value of the constant speed of the plate v assuming that the rate of shear strain, v/d, is constant across the thickness of the film and the viscosity of the oil is η = 0.05 Pa.s Fig 10.54 See Exercise (32) d fv is A =c ons tan t Lubricant mg (33) How fast will an aluminum sphere of radius mm fall through water at 20◦ C once its terminal speed has been reached, see Fig 10.55? Assume that the 354 10 Mechanical Properties of Matter viscosity of water is η = 8.5 Pl and that the aluminum sphere’s density is ρs = 2.7 × 103 kg/m3 Fig 10.55 See Exercise (33) Fvis FB t msg (34) The viscous force on a liquid flowing steadily through a cylindrical pipe of length L is given by: Fvis = 4π ηLvm where η is the viscosity of the liquid and vm is the maximum speed of the liquid which occurs along the central axis of the pipe, see Fig 10.56 If the pressures in the rear and front horizontal segments of the pipe are respectively P1 and P2 , where P1 > P2 , then show that vm will be given by: vm = (P1 − P2 ) r /4ηL where r is the radius of the cylinder P1 Flow P2 P1 Flow P2 r v L Cross sectional Streamlines r L Cross sectional velocity distribution Fig 10.56 See Exercise (34) (35) Blood of viscosity η = × 10−3 PI is passing through a capillary of length L = mm and radius r = μm If the speed of this blood as it travels through the center of this capillary is found to be vm = 0.66 mm/s, calculate the blood pressure (in pascal and mm Hg) using the result from the previous exercise Part III Introductory Thermodynamics Thermal Properties of Matter 11 In this chapter, we introduce a physical quantity known as temperature, which is one of the seven SI base quantities Temperature is associated with our sense of hot and cold Physicists and engineers measure temperature more objectively using the Kelvin scale, which is independent of the properties of any substance We will study the effect of temperature on matter; solid, liquid, and gas 11.1 Temperature The Kelvin Scale The limiting temperature of a body is taken as the zero of the Kelvin scale, and called the absolute zero To set up the Kelvin temperature scale, we select a standard fixed point and give it a standard fixed point temperature We select the triple point of water, where liquid water, solid ice, and water vapor can coexist in thermal equilibrium at only specific values of pressure and temperature By international agreement, at water vapor pressure of 4.58 mm Hg, the temperature of this mixture has been assigned a value 273.16 Kelvins, written as 273.16 K That is: T3 = 273.16 K (Triple-point temperature) (11.1) where the subscript denotes the triple point This agreement sets the size of the kelvin as 1/273.16 of the difference between absolute zero and the triple-point temperature of water This scale is used mostly in basic scientific calculations and studies H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_11, © Springer-Verlag Berlin Heidelberg 2013 357 358 11 Thermal Properties of Matter On the Kelvin scale, measurements show that the lowest reached temperature is ∼10−10 K, and the freezing and boiling (at atm Pressure) temperature points are: Tice = 273.15 K and Tsteam = 373.15 K The Celsius Scale The symbol ◦ C stands for degrees Celsius The size of ◦ C on the Celsius scale is the same as the size of K on the Kelvin scale However, the zero of the Celsius scale is shifted by 273.15 ◦ with respect to the absolute zero of the Kelvin scale On the Celsius scale, any temperature value TC is related to its Kelvin equivalent T by: TC = T − 273.15, (◦ C) or T = TC + 273.15, (K) (11.2) For example, the Celsius temperature of the triple point of water is 0.01 ◦ C, because T3 = 273.16 K and the ice point (273.15 K) corresponds to 0.00 ◦ C, and the steam point (373.15 K) corresponds to 100.00 ◦ C Note that we not use a degree mark in reporting Kelvin temperatures The Fahrenheit Scale The symbol ◦ F stands for degrees Fahrenheit The Fahrenheit scale has a smaller degree size than the Celsius and has a different zero (the freezing point of a certain concentration of salt water) The relation between the Celsius and Fahrenheit scales is: TF = 95 TC + 32 (◦ F) (11.3) Accordingly, degrees on the Fahrenheit scale equals degrees on the Celsius scale Moreover, ◦ C = 32 ◦ F and 100 ◦ C = 212 ◦ F Table 11.1 shows some corresponding temperatures and Fig 11.1 compares graphically the Kelvin, Celsius, and Fahrenheit scales 11.1 Temperature 359 Table 11.1 Some corresponding temperatures ◦C ◦F Temperature K Boiling point of water (at atm.) 373.15 100 212 Normal body temperature (average) 310.15 37 98.6 Triple point of water 273.16 0.01 32.02 Freezing point of water 273.15 32 Triple point of hydrogen 13.81 −259.34 −434.82 Absolute zero −273.15 −459.67 Fig 11.1 The Kelvin, temperature scales Steam Ice Absolute zero Temperature indicator Celsius, and Fahrenheit Fahrenheit o F Kelvin K Celsius o C 373.15 100 212 273.15 32 -273.15 -459.67 Example 11.1 The normal boiling point of nitrogen is −195.75 ◦ C (a) What is this temperature in Kelvin and in Fahrenheit? (b) If the temperature changes from −195.75 ◦ C to −100 ◦ C, find the change in the temperature on the Fahrenheit scale Solution: (a) Substituting TC = −195.75 ◦ C into Eq 11.2, we get: T = TC + 273.15 = −195.75 + 273.15 = 77.4 K Also, from Eq 11.3, we get: TF = 95 TC + 32 = × (−195.75) + 32 = −320.35 ◦ F Thus, −195.75 ◦ C, 77.4 K, and −320.35 ◦ F are equivalent temperatures on different scales (b) For a change TC = [−100 ◦ C – (−195.75 ◦ C)] = 95.75 C◦ , we use Eq 11.3 to find the change in temperature on the Fahrenheit scale as: 360 11 Thermal Properties of Matter TF = TC = 95 [−100 − (−195.75)] = 172.35 F◦ Thus, a change 95.75 C◦ = 172.35 F◦ , where the notations C◦ and F◦ refer to temperature difference, not to be confused with actual temperatures, which are written in terms of symbols ◦ C and ◦ F 11.2 Thermal Expansion of Solids and Liquids Most bodies expand as their temperatures increase This phenomenon plays an important role in numerous engineering applications, such as the joints in buildings, highways, railroad tracks, bridges etc Such thermal expansion is not always desirable Microscopically, thermal expansion arises from the change in the separation between the constituent atoms or molecules of the solid To understand this, we consider a crystalline solid of a regular array of atoms or molecules held together by electrical forces A mechanical model can be used to imagine the electrical interaction between the atoms or molecules, as shown in Fig 11.2 At an ordinary temperature, the average spacing between the atoms is of the order of 10−10 m, and they vibrate about their equilibrium positions with an amplitude of about 10−11 m and a frequency of about 1013 Hz As the temperature increases, the atoms vibrate with larger amplitudes and the average separation between them increases Atom or a molecule Average separation Fig 11.2 A mechanical model representing the average spacing in a unit cell of crystalline solid at a given instant Neighboring atoms or molecules (red spheres) are imagined to be attached to each other by elastic stiff springs, which represent the inter-atomic electric forces If the thermal expansion of an object is sufficiently small compared to its initial dimensions, then the change in any dimension (length, width, or thickness) is, to a good approximation, a linear function of the temperature 11.2 Thermal Expansion of Solids and Liquids 361 11.2.1 Linear Expansion If a rod of length L and temperature T experiences a small change in temperature T , its length changes by an amount L, see Fig 11.3 For a sufficiently small change T , experiments show that L is proportional to both L and T We introduce a proportionally coefficient α for the solid and write: L = αL T (11.4) where the proportionality constant α is called the coefficient of linear expansion for a given material Note that Eq 11.4 describes an expansion when T is positive and a contraction when T is negative If we set1 T = 1C◦ , we see from Eq 11.4 that α represents the fractional change in length ( L/L) per one degree change in temperature Thus the unit of α is (degree−1 ) An α value of 24 ×10−6 (C◦ )−1 means that the length L of an object changes by 24 parts per million for every Celsius degree change (C◦ ) in temperature Fig 11.3 The length L of the rod will increase by T its temperature changes from T to T + L+ Δ L L L when T+ Δ T T The expansion is exaggerated in the figure Generally, the coefficient of linear expansion α varies with temperature, but this variation is negligible over the temperature range of most everyday measurements Table 11.2 depicts some values of α Table 11.2 Coefficients of linear expansion α for some materials at room temperature (approximate) Material Name α (C◦ )−1 Material Name α (C◦ )−1 Fused quartz 0.5 × 10−6 Concrete 12 × 10−6 Diamond 1.2 × 10−6 Copper 17 × 10−6 Glass (Pyrex) 3.2 × 10−6 Brass & bronze 19 × 10−6 Glass (ordinary) × 10−6 Aluminum 24 × 10−6 Steel 11 × 10−6 Lead 29 × 10−6 As in Example 11.1, temperature changes are expressed in units of Celsius degrees, abbreviated C◦ which should not to be confused with actual temperatures, written with the symbol ◦ C and read degree Celsius 362 11 Thermal Properties of Matter Example 11.2 A steel rod has a length L = m and radius r = 1.5 cm when the temperature is 20 ◦ C Take α = 11 × 10−6 (C◦ )−1 and Young’s modulus of the rod to be Y = 200 × 109 N/m2 (a) What is its length on a hot day when the temperature is 50 ◦ C? (b) If the rod’s ends were originally fixed, then find the compression force on the rod? Solution: (a) From Eq 11.4, we can find the increase L when the change in temperature is TC = 50 ◦ C − 20 ◦ C = 30 C◦ as follows: L = αL T = [11×10−6 (C◦ )−1 ](8 m)(30 C◦ ) = 2.64×10−3 m = 2.64 mm Therefore, the rod’s new length at 50 ◦ C is 8.00264 m (b) If the rod is not allowed to expand, we then calculate what force would be required to compress the rod by the amount 2.64 × 10−3 m From the definition of Young’s modulus Y = (F⊥ /A)/( L/L): π r2 Y L AY L = L L (3.14)(1.5 × 10−2 m)2 (200 × 109 N/m2 )(2.64 × 10−3 m) = (8 m) F⊥ = = 4.7 × 104 N Note that this answer is independent of the length L 11.2.2 Volume Expansion Not only does the length of an object increase with temperature, but its area and volume change as well The change in volume V at a constant pressure is proportional to the original volume V and to the change in temperature the following relation: V =βV T T according to (11.5) where the proportionality constant β is called the coefficient of volume expansion for a given solid or liquid Setting T = C◦ in Eq 11.5, we see that β is numerically equal to the fractional change in volume ( V /V ) per one degree change in temperature Thus, like α, the unit of β is (deg−1 ) For example, a β value of 11.2 Thermal Expansion of Solids and Liquids 363 24 × 10−3 (C◦ )−1 means that the volume V of a solid or liquid changes by 24 parts in 103 for every Celsius degree change (C◦ ) in temperature An isotropic solid is a solid that has a coefficient of linear expansion that is equal in all directions Accordingly, for an isotropic solid, the coefficient of volume expansion is approximately three times the linear expansion coefficient, i.e β = 3α Table 11.3 depicts some values of β Table 11.3 Coefficients of volume expansion β for some materials at room temperature (approximate) β (C◦ )−1 Material Name × 10−4 Material Name β (C◦ )−1 Water 6.3 × 10−4 Alcohol, ethyl 1.12 Benzene 1.24 × 10−4 Turpentine × 10−4 Acetone 1.5 × 10−4 Gasoline 9.6 × 10−4 Air 3.67 × 10−3 Helium 3.665 × 10−3 × 10−4 Mercury 1.82 Glycerin 4.85 × 10−4 Example 11.3 On a hot day, when the temperature was Ti = 45 ◦ C, an oil trucker fully loaded his truck from an oil station with 10,000 gal of gasoline (1 gallon 3.8 liter) On his way to a delivery city, he encountered cold weather, where the temperature went down to Tf = 20 ◦ C, see Fig 11.4 The coefficients of volume expansion of gasoline and steel are β = 9.6 × 10−4 (C◦ )−1 and β = 11 × 10−6 (C◦ )−1 , respectively How many gallons did the trucker deliver? Fig 11.4 Ho t pic kup day Vi Co ld de live ry day Movement Ti Vf Tf Solution: The change in temperature from the production city to the delivery city is TC = Tf − Ti = 20 ◦ C − 45 ◦ C = −25 C◦ From Eq 11.5, we can find the change in the gasoline volume V as follows: V = βV T = [9.6 × 10−4 (C◦ )−1 ](10,000 gal)(−25 C◦ ) = −240 gal 364 11 Thermal Properties of Matter Thus, the amount of gasoline delivered was: Vf = Vi + V = 10,000 gal − 240 gal = 9,760 gal The thermal expansion of the volume of the steel tank can also be calculated as follows: V = βV T = [11 × 10−6 (C◦ )−1 ](10,000 gal)(−25 C◦ ) = −2.75 gal This change is very small and has nothing to with the problem, since the decrease in the gasoline volume is much bigger than that of the steel Question: Who paid for the missing gasoline? The most common liquid, water, does not behave like other liquids, see Fig 11.5a Above ◦ C, water expands as its temperature rises, and thus its density decreases as shown in the Fig 11.5b Between ◦ C and ◦ C, however, water contracts as its temperature increases, and thus its density increases Hence, the density of water reaches a maximum value of 1,000 kg/m3 at ◦ C r (kg/m3) r (kg/m3) 1000.0 999.9 999.8 999.7 999.6 999.5 1000 990 980 970 960 950 20 40 60 80 Temperature ( oC) 100 (a) Maximum density oC Ice oC oC oC 10 Temperature ( oC) oC (b) Fig 11.5 (a) The density of water versus temperature at atmospheric pressure The maximum density of 103 kg/m3 occurs at ◦ C (b) The warmer water from to degrees stays below the ice because it is more dense than the ice This unusual thermal expansion behavior of water explains why a pond or lake frezes only at its surface As the water on the surface is cooled towards the freezing point, it becomes denser (heavier) than the water below it and sinks to the bottom Warmer, less dense (lighter) water rises upwards to take its place and this in turn is also cooled down The water only stops circulating this way when it has all cooled 11.2 Thermal Expansion of Solids and Liquids 365 to ◦ C (the maximum density) Further cooling below ◦ C makes the water on the surface less dense than the water below it, so it stays on the surface until it freezes In time, ice continues to build up at the surface, and the denser warmer water at the bottom is unlikely to cool any further because it does not circulate, and water near the bottom remains at ◦ C The water temperature stabilizes as shown in Fig 11.5b Fish can survive by staying in the warmer deeper water 11.3 The Ideal Gas Let us examine the basic thermal properties of gases from an elementary point of view To that, we will consider the properties of a gas of mass m confined within a container of volume V at absolute pressure P and temperature T The relation that interrelates these quantities, the equation of state, is complicated However, if the gas is maintained at a very low pressure (or density), this equation is found experimentally to be quite simple, and this keeps the mathematics relatively simple This model is known as the ideal gas model and the low-pressure gas is commonly referred to as an ideal gas Most gases at room temperature and atmospheric pressure behave as ideal gases Equation of State of an Ideal Gas One mole (1 mol) of a substance is the amount of substance that contains as many particles as in exactly 12 g of the isotope carbon-12 Although the mole is one of the seven SI base units, it is convenient to introduce the One kilomole (1 kmol) of a substance as the amount of substance that contains as many particles as in exactly 12 kilograms of the isotope carbon-12 Thus: One kilomole (1 kmol): One kmol is the number of atoms in a 12 kg sample of pure carbon-12 This number is called Avogadro’s number, NA , after A Avogadro, who suggested that all gases contain the same number of particles (atoms or molecules) when they occupy the same volume under the same conditions of pressure and temperature Avogadro’s number is determined experimentally to be: NA = 6.022 × 1026 particles/kmol (11.6) Depending on the kind of study, atoms or molecules will replace the term “particles” 366 11 Thermal Properties of Matter In addition, the molar mass of each chemical element is defined as: Molar Mass (M): The molar mass M of a chemical element is its atomic mass expressed in g/mol or equivalently in kg/kmol For example, the mass of one 12 C atom is 12 u (12 atomic mass units, see Chap 1); then the molar mass M of 12 C is 12 kg/kmol and contains NA atoms For a molecular substance or chemical compound, we add up the molar mass from its molecular formula Thus, the molar mass M of nitrogen gas (N2 ) is 28 kg/kmol and it consists of NA molecules; this is because the mass of one nitrogen atom is 14 u Now, for an ideal gas of mass m (in kg) confined to a container of volume V (in m3 ) at a pressure P (in Pa) and temperature T (in K), it is convenient to express the amount of the gas in terms of the number of kilomoles n, see Fig 11.6 This number is related to the gas mass m and its molar mass M through the expression: n= m M (m in kg and M in kg/kmol) (11.7) Fig 11.6 An ideal gas defined by P, V, T, and n is contained in a cylinder with a Ideal gas movable piston to allow the P, V, T, n volume to be varied Moreover, we consider that the volume of the container can be varied, and hence the gas volume, by means of the movable piston of Fig 11.6 For this system, experiments show that at low densities, all gases tend to obey the following equation of state (which is known as the ideal gas law): PV = nRT (The ideal gas law) (11.8) where n is the number of kilomoles of gas present and R is the gas constant which is determined from experiments to have the same value for all gases, namely: R= 8.314 × 103 J/kmol.K if n is the number of kmol 8.314 J/mol.K if n is the number of mol (11.9) 11.3 The Ideal Gas 367 Using this value of R and the equation of state, Eq 11.8, we can find the volume occupied by kmol (kilomolar volume) of any ideal gas at the standard temperature and pressure (STP), which means temperature of ◦ C (273.15 K) and atmospheric pressure (1 atm), as follows: V= (1 kmol)(8.314 × 103 J/kmol.K)(273.15 K) nRT = 22.42 m3 = 22,420 L = P (1.013 × 105 Pa) where L ≡ liter = 103 cm3 = 10−3 m3 Thus: Volume of kmol: One kmol of any ideal gas at atmospheric pressure and at ◦ C occupies a space of 22.42 m3 = 22,420 L We can express the ideal gas law in terms of the total number of molecules N by using the fact that N equals the product of the number of kmol n and Avogadro’s number NA , i.e N = n NA Thus: PV = nRT = N R RT = N T NA NA or, PV = NkB T (11.10) where kB is called Boltzmann’s constant, which has the value: kB = R 8,314 J/kmol.K = = 1.38 × 10−23 J/K NA 6.022 × 1026 molecules/kmol (11.11) Equation 11.10 indicates that the pressure of a fixed volume of gas depends only on the temperature and the number of molecules in that volume Example 11.4 According to the periodic table of elements, see Appendix C, the molar mass of copper is M(Cu) = 63.546 kg/kmol Use this information to find the mass of one atom Solution: The molar mass of NA 63.5 Cu = 6.022 × 1026 atoms/kmol is M(Cu) = 63.546 kg/kmol and contains The mass of atom is then: