(Solution Manual) Chemical Engineering Fluid Mechanics

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SOLUTIONS MANUAL FOR byCHEMICAL ENGINEERING FLUID MECHANICS Ron Darby Raj P. ChhabraSOLUTIONS MANUAL FOR by Boca Raton London New Y ork CRC Press is an imprint of the Taylor & Francis Group, an informa businessCHEMICAL ENGINEERING FLUID MECHANICS Ron Darby Raj P. ChhabraCRC Press Taylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742 © 2017 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20160713 International Standard Book Number-13: 978-1-4987-2447-0 (Ancillary)This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.comAcknowledgement We would like to thank Ms Swati Patel and Mr Anoop Gupta, graduate students in the Department of Chemical Engineering, Indian Institute of Technology Kanpur, for their help with the word processing of this solutions manual. Ron Darby Raj Chhabra July 2016. 1 Chapter 1 1-1 Write equations that define each of the followi ng laws: Fick’s, Fourier’s, Newton’s, and Ohm’s. What is the conserved quantity in each of th ese laws? Can you represent all of these laws by one general expression? If so, does this me an that all of the processes represented by these laws are always analogous? If they aren’t, wh y not? Solution: (a) Fick: A Ay AB A AB A dC 1 D: n =-D , 3 D: n =-D C dy − − ∇ /harpoonrightnosp/harpoonrightnosp Fourier: ()v ydρC T dT 1 D: q =- α =-k , 3-D: q=-k T dy dy − ∇ /harpoonrightnosp/harpoonrightnosp vkα= =Thermal Diffusivity ρC Newton: ( )()( )()x x yx m mdρvdv 1 D: τ=- ν=- µ, 3-D: τ=- µvdy dy v+− ∇ + ∇ /harpoonrightnosp /harpoonrightnosp/harpoonrightnosp/harpoonrightnosp /harpoonrightnosp /harpoonrightnosp νKinematicviscosity = (3–D form is not analogous to other laws) Ohm: y e e de 1 D: i =-k , 3-D: i =-k e dy − ∇ /harpoonrightnosp/harpoonrightnosp (b) What is the conserved quantity? Fick: Mass of species A Fourier: Heat Newton: x-momentum Ohm: Charge (c) General Expression: Q y T dC Q =-k applies to 1-D formsof alllaws dy T Q 3 D: Q=-k C applies to all except momentum − ∇ /harpoonrightnosp/harpoonrightnosp d) All but momentum equations are analogous. (in 3 –D) 2 1-2 The general conservation law for any conserved quantity Q can be written in the form of Eq. (1-12). We have said that this law can also be applied to ‘‘dollars’’ as the conserved quantity Q. If the ‘‘system’’ is your bank account, (a) Identify specific ‘‘rate in,’’ ‘‘rate out,’’ an d ‘‘rate of accumulation’’ terms in this equation relative to the system (i.e., each term co rresponds to the rate at which dollars are moving into or out of your account). (b) Identify one or more ‘‘driving force’’ effects that are responsible for the magnitude of each of these rate terms, i.e., things that influen ce how fast the dollars go in or out. Use this to define corresponding ‘‘transport constants’ ’ for each ‘‘in’’ and ‘‘out’’ term relative to the appropriate ‘‘driving force’’ for e ach term. Solution: Balance: Rateof Rateof Rateof Rateof $in $out Generation of$ Accumulation of $         − + =                 ∑ Rateof Paycheck Gifts InvestmentIncome .... $in   = + + +     ∑ Rateof School Rent Food Transport Clothing Entertain ment ... $out   = + + + + + +     ∑ ( )Rateof Interest Thiscan also beconsidered a "Rate in" term Genetationof$   =     ∑ ()x xRateof X k DF , DF drivingforce = ⋅ = () ( )1 Paycheck DF k Education Ingenuity Experence ... ≅ + + + ()( )2 Gifts DF k Generosityof Parents ≅ () ( )3 Investment DF k Amount of Savings/Portfolio ≅ ()( )4 School DF k Where no. hours ≅ + ()( )5 Rent DF k Size,Loationof Apt. ≅ ()( )6 Food DF k "Taste",e.g.Steakvs.Hamburger ≅ Σ($)out Bank Σ($)in 3 () ( )7 Transportation DF k Costofcar "image" ≅ → () ()8 Clothing DF k "Style" ≅ () ( )9 Entertainment DF k "Lifestyle",Costof Beer ≅ () ( )10 Interest DF k Sizeof Account,Interstrate ≅ 4 1-3 A dimensionless group called the Reynolds numbe r is defined for flow in a pipe or tube 2 Re DV V NVDρ ρ µµ= = where V is the average velocity in the pipe, ρ is the fluid density, µ is the fluid viscosity, D is the tube diameter. The second form of the group ind icates that it is a ratio of the convective (turbulent) momentum flux to the molecular (viscous ) momentum flux, or the ratio of inertial forces (which are destabilizing) to viscous forces (which are stabilizing). When viscous forces dominate over inertial forces, the flow is l aminar and fluid elements flow in smooth, straight streamlines, whereas when inertial forces dominate, the flow is unstable and the flow pattern break up into random fluctuating eddies. It is found that laminar flow in a pipe occurs as long as the value of the Reynolds number is less than about 2000. Calculate the maximum velocity and the correspondi ng flow rate (in cm 3/s) at which laminar flow of water is possible in tubes with the following diameters: D = 0.25, 0.5, 1.0, 2.0, 4.0, 6.0, 10.0 in. Solution: Reynolds No: 2 Re ρVD ρVN = = µ µ V D Water: ρ = 1 gm/cc, µ = 1 cP = 0.01 P =0.01 g/(cm s) Laminar: NRe ≤ 2000 = ρVD ,µ,2000 µV= Dρ2πDQ= V 4 For, D = 0.25 in : ( )g2000 0.01 cm s cm =31.5 s g25×2.54cm 1 cc Re Nµ DρV=       =      ( )( )22π0.25×2.54cm cm cc = 31.5 =9.97 s s 4 4 D V Qπ= Similarly, for air at standard temperature and pres sure (see Appx.A): ρ = 1.205 kg/m 3, µ = 1.983 ×10 -5Pa s (i.e.1 Pas = Ns/m 2 = kg/(m s) 5 D(in) Water Air V (cm/s) Q(cc/s) V (cm/s) Q(cc/s) 0.25 31.5 9.97 5.18 164 0.5 15.7 19.9 2.59 328 1 7.87 39.9 1.30 657 2 3.94 79.8 0.648 1310 4 1.97 160 0.324 2630 6 1.31 239 0.216 3940 10 0.787 399 0.130 6570 (Note: Although the fluid properties are given to 4 digits, the number of digits in NRe = 2000 is undefined, so that we give the solution to only 3 d igits. See Ch. 2 for details on Units and Dimensions.) 6 1-4 A layer of water is flowing down a flat plate that is inclined at an angle of 20° to the vertical. If the depth of the layer is 1/4 in., what is the shear stress exerted by the plate on the water? (Remember: Stress is a dyad , with two directions associated with each component.) Solution: ( )()atplate x x w yx yx w y y F Fτ= , τ=A A Fx = force in x-direction on water Ay = area of water contacting plate Force in x direction on plate = x-component of weight of water = ρg ∇co s 20 °, ∇ = δWL (W = width, L = length) Ay = WL Fx on water = – Fx on plate = – ρg WL cos 20 ° ( )o o yx w-ρgWLcos20 τ= =-- ρg cos20 WL This is (–), since wall exerts a force in – x direc tion on water. (Note: The "system" is the water. τyx is the force on the water in the x-direction actin g on the surface Ay. (the +A y surface outward normal is in + y direction, with +y direction from the surface inward). ∴ Ay surface of water at plate is (+); ( )x yx yx m yFτ=- τ= = A( )x yx yFτ= is - A (the stress exerted by the plate, on the water is ( –)) δ y x 20 ° 7 1-5 The relationship between the forces and fluid velo city can easily be illustrated by considering the simple one-dimensional steady flow between two parallel plates in Fig. 1.1. The area of each plate 21m yA=and the gap between the two plates is 0.1 mm (very thin). Fluids of different viscosities, as listed below, a re contained between the two plates. A tangential force of one Newton is applied tangentia l to the top plate in the x direction at a fixed velocity V0 , while the bottom plate is held stationary. Calcu late the value of V0 for each of the following fluids: Substance Viscosity (cP) Olive oil 100 Castor oil 600 Pure Glycerine (293K) 1500 Honey 10 4 Corn Syrup 10 5 Bitumen 10 11 Molten glass 10 15 In view of these results, what is the meaning of th e phrase from the Bible: “the mountains flowed before the Lord” and the significance of the phrase: “ Everything flows ”! ( )( ) ( )( )() ( )2 2 -3 2m-0 N Ns s N= =1Pa = m m 0.1×10 m mox yx yV Fτ µ A      ( )() ( ) ( )( )-4 -4 mPa s 10 Pa m s1Pa = or = =m/s Pa s 10 m o oµV Vµ Substance µµ µµ (Pa. s) Vo (m/s) Olive oil 100 ×10 -3 1×10 -3 or 1 mm/s Castor oil 600 ×10 -3 1.67 ×10 -4 or 0.167 mm/s Pure Glycerine (293 K) 1500 ×10 -3 6.67 ×10 -5 or 66.7 µm/s Honey 10 4×10 -3 1×10 -5 or 10 µm/s Corn Syrup 10 5×10 -3 1×10 -6 or 1 µm/s Bitumen 10 11 ×10 -3 1×10 -12 i.e ., 86.4 nm/day Molten Glass 10 15 ×10 -3 1×10 -16 i.e., 8.64×10 -3 nm/day 8 In contrast to the velocity of 1 mm/s attained by t he top plate with olive oil as the working fluid, the corresponding value is only 3.15 nm/year for the molten glass! Thus, everything flows but the relevant question to ask is at what t ime scale. Thus, “the Mountains flowed before the Lord” implies that even mountains flow o ver timescales much longer than we ordinary human beings can observe since Gods are im mortal and only they can see the mountains flow! In other words the distinction betw een solids and fluids is not as sharp as we think! 9 1-6 A slider bearing consists of a sleeve surroundi ng a cylindrical shaft that is free to move axially within the sleeve. A lubricant (e.g., greas e) is in the gap between the sleeve and the shaft to isolate the metal surfaces and support the stress resulting from the shaft motion. The diameter of the shaft is 1 in., and the sleeve has an inside diameter of 1.02 in. and a length of 2 in. (a) If you want to limit the total force on the sle eve to less than 0.5 lb f when the shaft is moving at a velocity of 20 ft/s, what should the vi scosity of the grease be? What is the magnitude of the flux of momentum in the gap, and w hich direction is the momentum being transported? (b) If the lubricant is grease with a viscosity of 400 cP (centipoise), what is the force exerted on the sleeve when the shaft is moving at 20 ft/s? (c) The sleeve is cooled to a temperature of 150°F, and it is desired to keep the shaft temperature below 200°F. What is the cooling rate ( i.e., the rate at which heat must be removed by the coolant), in Btu/hr, to achieve this ? Properties of the grease may be assumed to be: specific heat = 0.5 Btu/(lb m°F); SG (specific gravity) = 0.85; thermal conductivity = 0.06 Btu/(hr ft °F). (d) If the grease becomes contaminated, it could be corrosive to the shaft metal. Assume that this occurs and the surface of the shaft starts to corrode at a rate of 0.1 µm/yr. If this corrosion rate is constant, determine the maximum c oncentration of metal ions in the grease when the ions from the shaft just reach the sleeve. Properties of the shaft metal may be assumed to be MW = 65; SG = 8.5; diffusivity of metal ions in grease = 8.5 x 10 -5 cm 2/s. Solution : Slider Bearing (a) Fmax on sleeve = 0.5 lb f when V x = 20 ft/s , µ = ? ()()x x shaft sleeve F F = − assume grease is Newtonian: x x yx ydv F τ=µ=dy A ()() y y shaft sleeve A -A = A y is +ve (surface of grease in +y direction) F x = F on grease in x-direction Ay = πDL x x x dV ∆V V = = dy ∆y h so x x yF V =µA h Sleeve Shaft L = 2 in d = 1 in D = 1.02 in Vx, F x shaft Ay h sleeve x Grease y 10 ( ) ( )( ) ( )( ) ( )505 001 445 10 102 2 20 3048 254 f fx y x dyn . lb . in. . lb F h µ=ft AV cm cm . in. in. . . s ft in. π  ×    = = 0.224 dyn s/cm 2 =0.224 Poise = 22.4 cP Momentum flux = ( ) ( )( )( )5 3 205 4 45 10 538 10 102 2 254 f fx 2 yx yx m2ydyn . lb . lb F dyn τ=- τ=- . cm A cm . in. . in. π  ×    =− =− × i.e. Momentum flux is (–) w.r.t coordinates specifi ed. (+ve x-momentum transferred in –y direction), i.e. momentum goes from faster to slowe r fluid - from shaft to sleeve. (b) Grease: µ = 400 cP What is F on sleeve if Vshaft = 200 ft /s ? ( )( ) ()()()() ( )( )2 102 2 4 20 3054 254 001 254 2 y x xdyn ft cm cm . in. in. . . AµV s ft in. cm F= cm h . in. . in. π = 6 5397 10 892 445 10 f f. dyn . lb dyn .lb ×= = × (c) Sleeve T = 150 °F want shaft T ≤ 200 °F Cooling rate = ? Cv = 0.5 BTU/lb m°F, SG = 0.85, k = 0.06 BTU/ hr ft °F Fourier's Law: ()1 0 yT T dT q k k dy h −=− − ≃ ( )200 150 0.06 25 .0.01in 12 y o 2 F Btu Btu q in hr ft F hrin ft       −  =− =−                 ∘ (i.e., heat is transfer in - y direction to cool shaft) ( )( )( ) 25 1.02 2 . 160 0.0445 y y y 2Btu Btu Btu Q q A in. in hr s hrin. π = = − = − = − Q = ? T0 = 150 °F T1 = 200 °F x y 11 (d) Shaft corrodes at rate of 0.1 µm/year . What is maximum concentration of ions in grease when diffusing ions first reach the shaft? MW = 65, SG metal = 8.5, DmG = 8.5 ×10 -5 cm 2/s Fick's Law: 1 my mG mG C dC n -D -D dy h =≃ ( )() ( )-4 -6 -0.1×10 8.5 rateof corrosion =fluxof ions= =-1.308×10 65 32 my yg cm yr cm n mol cm yr g A mol = (dissolved ions diffuse in -y direction) Maximum C1 occurs when C0 = 0 (before concentration builds up in grease) 1 my mG Cn -D h≃ ( )( ) ( ) ( )( )( )6 -5 1308×10 0.01in 2.54 = 8.5×10 3600 24 365 2 my 12 mG mol cm in. -n h cm yr C = day D cm s hr s hr day yr       -11 -8 =1.24×10 =1.24×10 1 3 mol mol Cl cm (Note: Corrosion rate of " x cm " is equivalent to " x cm 3/ cm 2") C0 C1 x y nmy 12 1-7 By making use of the analogies between the molecul ar transport of the various conserved quantities, describe how you would set up an experi ment to solve each of the following problems by making electrical measurements (e.g., d escribe the design of the experiment, how and where you would measure voltage and current , and how the measured quantities are related to the desired quantities). (a) Determine the rate of heat transfer from a long cylinder to a fluid flowing normal to the cylinder axis if the surface of the cylinder is at temperature T0 and the fluid far away from the cylinder is at temperature T1. Also determine the temperature distribution within the fluid and the cylinder. (b) Determine the rate at which a (spherical) mothb all evaporates when it is immersed in stagnant air, and also the concentration distributi on of the evaporating compound in the air. (c) Determine the local stress as a function of pos ition on the surface of a wedge-shaped body immersed in a fluid stream that is flowing slo wly parallel to the surface. Also, determine the local velocity distribution in the fl uid as a function of position in the fluid. Solution: (a) Heat transfer from cylinder to fluid flowing ⊥ to axis. Cylinder at To, fluid at T1. Also want T = f(r, θ). Fourier’s Law: q=-k T ∇/combarrowextender/arrowrightnosp/harpoonrightnosp Ohm’ s Law: ei=-k e ∇/harpoonrightnosp /combarrowextender/harpoonrightnosp Measure: ( ) ( )()( )o e 1 o 1 o e r -e e r, θT r, θ, i=k e -e e -e   → ∇     /combarrowextender/arrowrightnosp/harpoonrightnosp ( ) ( ) ( )o 1 o 1 o T-T i r, θ δ r, θ, q=k T -T T -T   → ∇     /harpoonrightnosp /harpoonrightnosp /combarrowextender/harpoonrightnosp Experiment: Put a highly conducting disk in a mediu m of lower material (e.g., a fluid). Apply a voltage ( eo-e1) between the cylinder and the fluid far from the c ylinder, using electrodes. T(r, θ) T1 r θ To δr e(r, θ) e1 ir θ eo 13 Measure: Voltage at various points in the medium i. e., e(r, θ). Can also measure (or calculate) voltage differences (e.g., gradients) at various points:e e ,rθ∂ ∂ ∂ ∂ . Knowing electrical conductivity of medium (or measu re it if current flowing between 2 points is measured), and the thermal conductivity o f the fluid: apply ()() o 1 o 1 e -e T -T ⇒ , measure iw, calculate w wall ekq = i k: () ( )1 o w w e 1 o T -T kq =i k e -e (b) Evaporating mothball: Fick's Law: A A A n =-D C ∇/combarrowextender/combarrowextender/harpoonrightnosp /combarrowextender/harpoonrightnosp Ohm's Law: ei=-k e ∇/harpoonrightnosp /combarrowextender/harpoonrightnosp Use highly conducting ball in a lower conductivity medium. Apply voltage ()o 1 e -e between ball and medium far from ball. Measure iw∼ nAw . eo⇒ e1 (concentration at surface) (Must know solubility of compound in air, Co) (C o - 0) ∝ (e o - e1) Conductivity of medium is ke (or measure it) ,A r A o oCdn =-D C dr C       ( )1 r e o 1 o 1 e-e di =-k e -e dr e -e       Aw c A o nG , D C ≡ ( )r e e o 1 iGk e -e ≡ 1. ()o 1 A c e oe -e dC de G G , = dr C dr ⇒ 2. ( )oA Ar r e o 1 D C n =i =Rateof evaporation area k e -e 1. Relates concentration distribution (C A(r,θ)) to the voltage distribution (e(r,θ)). e1 nr Air C(r, θ) θ eo 14 2. Relates the diffusion flux, nAr (rate of evaporation/area) to the current density , ir (current through wall/area) (c) Wedge moving at velocity Vo in still fluid is equivalent to fluid moving at Vo past still wedge. Is 1-D in polar coordinates ( r, θ): Flux of r-momentum in θ - direction: r θrdV µτ=- r d θ Current in θ - direction: e ekde i =- r d θ ()()()() θrθ r τr i r , V θeθ ⇒ ⇒ Apply eo voltage to conducting wedge, e1 at boundary far from wedge. ()() 1 o o 1 V =0 V -0 e -e ⇒ At surface: r w wdV τ=- µrd θ      ()1 w e wd e-e i =-k rd θ      ()()() r 1 Vθ= e θ-e velocitydistribution ⇒ o w e o 1 Vµτ= wall stress related tocurrent through wall k e -e   ⇒    1-8 What is the “system” in Example 1-3? Answer: The System is the grease between the plate s. 1-9 Explain the "fruit salad law", and how does it appl y to the formulation of equations describing engineering systems? Answer: The “fruit salad law” states that “You can’t add apples and oranges unless you want fruit salad.” This means that every term in any equation must h ave the same dimensions (and units) – either “apples” or “orange s” - , if the equation is to be valid. θ r Vo τθr eo θ r e1 far from wedge 1 Chapter 2 2-1 Determine the weight of 1 g mass at sea level in u nits of: (a) dynes; (b) lb f; ; (c) g f ; (d) poundals. Solution: Define weight of 1 gm at sea level (L) in units of: (a) dynes: W = mg = 1 g (981 cm/s 2) = 981 dynes (b) lb f: 2 f m m 2 fft 32.2 1g sW = =0.00220 lb lb ft 453.6g lb 32.2 lb s            (c) g f: 2 f m c 2 fcm 1gm 981 mg sW = = =1g gcm g981 gs      (d) poundals: 2 m1gm ft W = mg = 32.17 =0.0709poundals g s453.6 lb                   2 2-2 One cubic foot of water weighs 62.4 lb f under conditions of standard gravity. (a) What is its weight in dynes, poundals, and g f? (b) What is its density in lb m/ft 3 and slugs/ft 3? (c) What is its weight on the moon (g = 5.4 ft/s 2) in lb f? (d) What is its density on the moon? Solution: 1 ft 3 H2O weighs 62.4 lb f at standard gravity (a) Find weighs in dynes, Poundals, g f 5 7 f f4.448×10 dyne W =62.4lb = 2.78×10 dyne lb       m m c 2 2 f f lb lb ft ft poundal g =32.17 =1 =32.17 lb poundal lb s s f fpoundal W =62.4lb 32.17 =2010 poundals lb       4 f f f fgW=62.4lb 453.6 =2.82× 10 g lb       (b) Find density in lb m/ft 3 and slugs/ft 3: m f 2 f c m m 3 2lb ft 62.4lb 32.2 lb s Fg lb M = = =62.4lb ρ=62.4 ft g ft 32.2 s      ∴ m 3 3 mlb slug slug ρ=62.4 =1.94 32.17lb ft ft (c) Weight on moon, in lb f ? (g = 6 ft/s 2) cmg W = g ; 2 2 1 1 W g =W g Since m, g c = constant 2moon moon earth f f earth 2ft 6g sW = W =62.4lb =11.6 lb ft g32.17 s            (d) Density on moon? 3 mρ=VV VV m, VV VV independent of g m 3lb ρ=62.4 ft ∴ 4 2-3 The acceleration due to gravity on the moon is abou t 5.4 ft/s 2. If your weight is 150 lb f on the earth: (a) What is your mass on the moon in slugs,? (b) What is your weight on the moon, in SI units? (c) What is your weight on earth, in poundals? Solution: Moon: g = 5.4 ft/s 2. If you weigh 150 lb f on earth: (a) Mass on moon in slugs = ? Mass is independent o f g, so it is the same as on earth: W=mg ( )f 2 ft 150lb = m slugs 32.2 s M=4.66 slugs (b) Weight on moon, in SI units? (Scientific) W = m g can convert from slugs to kg using: m c 2 2 2 f f lb ft kgm slugft g =1 =1 =32.2 Ns lb s lb s or moon moon earth earth W g =W g f moon f f f5.4 lb W =150lb =25.155lb =25.155 =112N lb 32.2 0.2248 N      (c) Weight on earth in poundals? (Scientific) Convert from lb f to poundals: m m c 2 2 f f lb lb ft ft poundals g = =32.2 or 32.2 poundal lb lb s s earth f fpoundals W =150lb 32.2 =4830poundals lb       5 2-4 You weigh a body with a mass m on an electronic scale, which is calibrated with a known mass. (a) What does the scale actually measure, and what are its dimensions? (b) If the scale is calibrated in the appropriate s ystem of units, what would the scale reading be if the mass of m is (1) 1 slug; (2) 1 lb m (in scientific units); (3) 1 lb m (in engineering units); (3) 1 g m (in scientific units); (4) 1 g m (in engineering units). Solution: Weigh body with mass m on electronic scale Calculated with known mass (a) What does scale measure Force due to gravity = Wt 2ML [Wt]=F= t (b) What is scale reading if m is : (1) 1 slug : (Scientific, g c = 1) f 2ft W=mg=1slug 32.2 =32.2 lb s (2) 1 lb m: (Scientific, g c = 1) m 2ft W=mg=1lb 32.2 =32.2 Poundals s (3) 1 lb m(Engineering, g c = m 2 flb ft 32.2 lb s) 2 m f m c 2 fft 32.2 mg sW = =1 lb =1lb lb ft g32.2 lb s (4) 1 gm (Engineering, g c = m 2 fgcm 980 gs) 2 f m c 2 fcm 980 mg sW = =1 gm =1g gcm g980 gs m 6 2-5 Explain why the gravitational “constant’ ( g) is different at Reykjavik, Iceland than it is at La Paz, Bolivia. At which location is it gre atest, and why? If you could measure the value of g at these two locations, what would this tell you a bout the earth? Solution: Why is g different at Reykjavic than at La Paz ? Because: 1 2 m1 2mm W =G r where m 1 = "any mass", Weight W m1 m2 = earth, r = radius of earth r is smaller at Reykejavik than at La Paz because e arth isn't round but is slightly flat at the poles. ∴ gRey > g LaP since r Rey < rLaP : Measure of g gives a measure of r (radius of earth) Rey LaP 12 LaP Rey rg=r g         rRey rLaP 7 2-6 You have purchased a 5 oz bar of gold (100% pure), at a cost of $400/oz. Because the bar was weighed in air, you conclude that you got a bargain, because its true mass is greater than 5 oz due to the buoyancy of air. If t he true density of the gold is 1.9000 g/cm 3, what is the actual value of the bar based upon it s true mass? Solution: 5 oz gold bar cost $ 400/oz (weighted in air) ρgold = 1.900 g/cc. What is actual value based on true measure? f g a f f flb NetWt.=W -W =5oz =0.3125 lb 16oz       g a g N g a c c M g ρgW = W -W = - g g VV VV g g gM=ρVV VV a a N g g g c g c g gρ gρW = M - M = M 1- gρ gρ                cN g a ggWgM = ρ1- ρ            Assume, P = 1 atm, T = 70 °F m m air 3 3 std lb 29 lb M 492 lbmole ρ= = =0.0750 530 V ft std ft 359 lbmole       ( )m f 2 2 f g m lb ft ft 0.3125lb 32.2 32.2 lb s s M = =0.3127lb 0.0750 1- 1.90 62.4                     (True Mass) m true inair m air 0.3127lb 400 TrueValue= =$400.25 oz 0.3125 lb       (Made 25¢!) Gold Buoyancy = ρair VV VVg = W a Wt. = M gg = ρgVV VVg = W g