hi Jt The forge, F, of the wind blowing against building ia given by F = Cop¥? A/2, winere Vis the wind speed, p the density of the ai, ‘A the etoss-scctional area of the building, and Cy is a constant termed hedimg cosiicies. Determine the dimesions ofthe dng coelfckeaL Fet,eVA or Oy =2F/eV'A, where Fem e= mL? Va be Thus, AeL* Cy UML) / [ome ML TF (LY) = MPT? Hence, Cy is dimensionless.J2 LLB. Verify the dimensions, in both the FLT and MLT systesns, of the following quantities which appear in Table 1.1: (a) vol. ume, (b) acceleration, (¢) mass, (d) moment of inertia (aren), and (e) work. (2) amass = {4 er with F2m.r? anges BLn' 7? (A) moment sf inertia (area) = s€cead moment of Aree. & sar) = 2? (@) work = force x distance 3S FL erowith Fam.r72 works pad? 77?3 13 Determmne the dimensions, in both the FLT system and_the MLT system, for (a) the product of force times acceleration, (b) the prod ‘et of fores times velocity vided by area, and () momentum divided by volume. force x acceleratwon = (F)(LT™ 4) = FLT Since FSPLT, Force x acceleretion = (mer NLT) 3 MT (a) 4 toree x velocity . (FMLT") 2 pint pm eS ee Gt Net) sp? aS ee Me ce) Mementum _ mass £ velocity volume ve herne. CereMer . ker i. a “ ss UM ea = me*771. Verify the dimensions. in both the FLT system and the MLT system, of the following ‘quantities which. appear in Table 11: (ab fee= quency. (B) stress, (c) strain, (€) torque, end (e) work (8) tregeeney = SE 2 fore FS Fr h) stress = on Since Fe 17-4, stress MAT = me tr? (¢) strain = Change ta lengtn 2 heng7h © (dimensionless) nix (4) torque = Force distence = FL Gaur Dt) = Mr (6) werk = force distance = Fl. = ter Wh) = I-4#Ls 1.5 Ifwisa velocity, xa length, and ta time, ‘what are the dimcasions (in the MET system) of a) du/de, dbo) Pucfacde, and (ec) f (ausat) da? ge 2 LT" “2 ai =e 62° Sgt ae = es ad) 2% a LT aeae — (LjeT) f buy, 2 OT) 3 Ltr «) te dx — d[Tey 1.6 Ip is & pressure, Va velocity, and p a fluid density, ‘what are the dimensions (in the MILT system) of (a) ap, (>) P¥e, and (6) p/aV#? b . mer | pe pne eer = LF te) pp= Gac'r*)(.7~) (me?) * ir? ohnz see ee = M*LET? (damenuentess) Ceres AV? i) (iT-)* I-6“7 17 If Visa velocity, € x length, and v 0 faid property dhe ine: matic viscosity) having dimensions of 1°F"', which of the {ol lowing combieations ar lncnalonea: () VED, 0) VE). (e VP, (a ¥/ev2 car VAY S (LT yarr) SLT * (nt dinemsials) VE a tere) | ay) ME es LEF* Caimessionbas) (2 VAS Gr PT) S 27 Coad desensenlas) Moog Wer). @) sae eae, = 2 L* (not _ dimensionless)18 I Vis a velocity, determine the dimensions of Z.c, and C. Which appear in the dimensionally homogeneous equation V=Za- +e V= Zz @-1I) +6 Ler d* [ej[e-] + [2] Since each term ri The eguatiwn must have the same dimensions, 1 fallonls thad es ur od = FEL°T*® C dimensenless snie Combiicd with 4 number) 6 = 277!LG. 14 The volume rate of flow, Q, throush & pipe containing a slowly moving liquid is piven by the equasion One sehere Ri the pipe radius, Ap the pressure diop long the pipe, 4.0 fluld property called viscosity (FE °T). and @ the length of pipe. What are the dimensions of the constant 1/8? Wold you slassify this equation as a general homogencous equation? Exalain, er (s\led fuer) 2[2\ [et] The Constant Te 1s dimensionless, and the ehuction 13 a. general homogeneous Cqustiba that i valid im any Consistent Uni system. Yes. es.140 Ascotling to information found in an old hydraulics book, the energy lost per unit weight of fuid flowing throagh a nozzle connected ta ahose cam be estimated by the focus f= 01 O09, D/d7'V2/2¢ where h isthe energy Joss per unit weight, D the hose diameter, d the norale tip diameter, V the ftuid velocity in the hose, and ithe acceleration of gravity, Do you think this equation is vid in any system af unite? Explain. 4= food eo (B)" [e]= [owt 008] [ Fae] [E Ae} [1] = [o.04 40.09) [LL] Since each term th The eguation must have the Same dimensiens the constant Herm (0.04 b 0.08) must be di'rmensiinless. Thus, the egaatiin 4 & general hemegencons €, ef uaticn That is Walid pe any systent of anits, Yes. 1.11 The pressure difference, Ap, across 2 casity (FL-°7), p the blood density (ML), BY "partial blockage in an artery (called 1 stenosis) isthe artery diameter, Ay the area of the unob- approximated by the equation structed artery, and A, the area of the stenosi w 1 Determine the dimensions of the constants Ap = KE + (4: aw? and X,.. Would this equation be valid in any sys- tem of units? Ap= Kae ¢ bale Be ate [re Da [a] CaEKE)) Cols) ET) [ec] = a ec’] + [k.)[Fe*] Since each term must have the same dimensions, Ky and Ky, are dimensionless. This, the eguation is @ genera! homogenceus €44 ee “that woule be ' valid 7h aay Consistent SYS of units, Yes 1-10ee 112 Assume that tho speed of sound, ¢, ina fnid depends ‘90 an elastic modulus, £,, with dimensions FL ~*, and the fluid eonity. p. in the form c= (B,¥"a), AF this ig wo be-a dimen: sionally homogeneous equation, what ure the valves for ¢ und Ye Is Your result comsistent with the xtandard formula for the peed of sound’ (See Eq. 1.19.) ete te? Duce cant! £2 FE? p= Pit? (4): [Eee fi a {724 Ts For a dimensionally homogeneous 2fuation each term in The epfuatin must have the same domennons. Thas, the right hand side of 3.0) must have the dirhensious of LT! There fre, A+b20 (to elimnate F) 2Bbh=-l Che sansty conhtren onT) Zat¢b=-!) Ch athsty endition an L) Et fllows Met a24 ana 42-4 So Het ez The: resilh ye eenatatens with The staat Wowele Ke the Speed of sound. Yes.113 1.13 A formula to cstimate the volume rate of the dam (ealled the hea ‘This formula gives of flow, Q, lowing over a dam of length, 4, is Qin ffs when A and Hf in feet, Is the con- siven by the equation fant, 3.09, dimeasionless? Would this equation = 3.0080 ‘be yal femi oter tan feet and seconds were where H is the depth of the water above the top Q= 307 Bu : fer] * Bore] iI" (e¢7-] [x00] [0] Stata. Gath hem tn flat epaation mast tase the eme dimensions the conslant 2.04 rmust have dimensions of L*T~! and is therefore not dimenstoaless. Vo. Since the constant fas dimenssons its value will chenge. with wchane ta uaits, Me.1.15 Make use of Table 1.3 to express the following quantities in SI units: (a) 10.2 in./min, 4b) 4.81 slugs, (c} 3.02 tb, (@) 73.1 {els (e) 0.0234 To-s/ft?. @) jo2% = (a2 ™,) (2s40xi0 ; =) jeu) Gos 2 a2 xioe am = %32 23m (6) 4at slagse (421 shgs) (4¥or4/0 #£)- 70,2 4g (2) 30a = (202% )( noe X)= sana A) 13446 = (7242) (2, avec! Be toes | fbes @) o.o2as ES « (2. 0024 ES S fees1.16 Make use of Table 1.4 to express the following quantities in BG units: (a) 14.2 km, a) 8.14 Nim’, (¢) 1.61 kg/mm’, (8) 0.0320 N-mis, 4e) 5.67 mm/h. CA) 142 bm = (anim) (3201 t)- 406 x10" H i A za: tS = (bm By = (Bie BM) (lb se6<10 B= cc B, ame 4 4 <2 Slugs. = cep 4b) 48 = (067 &) (1 t90x10 2). 312 «10 sae ms (d) d.0320 2” (p, 0300 Me) asia! 4 ) ew = = 236x107 eee (2) 5.67 rite (sez x0? 2) (3201) Lhe \ shoo s a saa FL SstI7 1.17 Express the following quantities in SI units: (a) 160 acre, (b) £5 gallons (U.S.), (6) 240 miles, {d) 79.1 hp, (e) 60.3 °F. a a2 a? (a) Ibo acre = (lod tered 4are xo )(9.240 xl ‘s) = 647 x08 im? (L) 15 gallons = (15 getlons (3, 198 fie) (gin at = 5b.2 x10 rm? Gallon int = (2) ayo me = (240 mi) (e280 )(ame si 22)- 9b x105 rm hP and J = \ Wo Se That 74.1 bp= 5.90 x10" W ve) hes (oo.g'F -32) = 15.7"C a (A) 79.1 hp = (74) hp) (e50 te Naz) Sapane z SIG 7°C 4.273 = 284 Ic ints| LAS 1.48 For Table 1.3 verify the conversion re- lationships for: (a) area, (b) density, (¢) velocity, and (d) specific weight. Use the basic conversion, relationships: 1 ft = 0.3088 m; 1 Ib = 4.4482 Nj | and | slug = 14.594 kg, (4) f #t*= (140°[fa sone)'20) - 2.09290 am® Thus, multiply ft? by 9.440 E-2 te convert to m*, 3 shg 2 f) steg 4: | If ab) (/ sted ) (4 59% #4) i aa ae = S/S. H Thus, multiply slug ft? by SIS Et2 to convert to ke lm ce) | ft 2 () #) (030% 2)= 0.3049 & Thus, multiale fl/s by 2042 E-1 comerl to m/s. i #2? a) tf A) )Gssea 4 #)| ee | = ist & Thus, multiply lef £6? by LSI Et2 +> convert fo Mam,(oe) L197” For Table 1.4 verify the conversion re- lationships for: (a) acccleration, (b) density, 4c) pressure, and (d) volume flowrate. Use the ‘basic conversion relationships: 1m = 3.2808 fi IN = 0.22481 Ib: and 1 kg = 0.068521 slug, (a) Paap (debe He B20) & Thus, rrultiply m/s? by 3.48) toe cenvert to ft/s, tb As. Le slugs) [ lan | d | ak (; #8 , )(o.0 52) iP) (acre ra ~3 $i = [, F4o «10 aol Thus, multiply g/m? by [440 E-3 to convert to slugs fe}, N N ib te) ie os .) (aaya 4 hs = 2 Ib = Loaf £i0° pL Thus, multiply W/m? by 208 E-2 to converd fo Ib / fe, [ ) Pe - ca () | 2 ( a) [ 208 |. 35:3) mam ah (3 arag)* Fe® Thus, multiply am7/s by 3.531 Et] to convert to Ft%5,120, Water flows from 2 large deainage pipe al a rate of 1300 gal/min: What is this volume rate of flow in (a) my"/s, (bi) essen, and (€) 1/8? (ay im? flowrate = (/200 zat ) (6.308 e109 A a prin = 257 xijo a (b) Since | fier + OA m3, Humax (131 ast 222) (tte )( liters min = #540 C6) flowrate = (257x108 2) (3 sat x00 = 247 ale wld ola ) bes mia 3 ) )i nL. 0. = | 1.24 An important dimensionless parameter the Froude number using SI units for V, g, and in certain types of uid flow problemsis the Froude _‘f. Explain the significance of the results of these number defined as V/V gl, where Vis a velocity, calculations. 8 the acceleration of gravity, and fa length. De- termine the value of the Froude number for V = 10 ftls, g = 32.2 fi/e!, and ( = 2 fr, Recalculate Ln 86-«atls, ME tie j Wen aa Zn SI units: Velio )¢4 sre = 3.05 2 g= 431% hz (2st) (o30u8 a) O.b10 am Thus, 305 F an AS a Ife Var % ).c10m) The value of a dimensionless Parameter is independent of The unit system,41.23. tank contains 500 g of a tiguidt whose specific grav 2 Demise the vole of the lig in ie ae m=@V = 56 Quo ¥ isi = M/(S6 Ogg) =-500kg U2) 999 8 )) = 0.250m* hey 1.24 Clouds can weigh thousands of pounds due to their | liguid water content. Often this content is measured in grams | {| per cubic meter (pim?), Assume that a cumulus clond occupies | ‘a volume of one cubic kilometer, and its liquid water conte:tt is 0.2 ghm?. (a) What is the volume of this cloud in cubic | ~~ miles? (b) How much does the water in the cloud weigh in pounds? See S (he ) ees Since tm = 2 ees (a seeping Lge, base \Cotm?) = |. 24z x10°N a =() = Chez xp ta lle ad? B= ger x Ib i-20Oo 4a) Determine its weight in pounds and in now ‘tons at the earth's surface, (b) What would be its amass (in slugs) and its weight (in pounds} if lo- cated on the moon's surface where the gravita- tional attraction is approximately one-sixth that at the earth's surface? ca weight = mass n = (25 slugs ) (322 #) = (25 slugs) (4.59 Es (6) mass = 25 slugs (tmass does not depend on gravite tions) attraction >? weight = (a5 slugs) es #) = /34 Lb = af Ah 1.26 A-cerain object weighs 300 N at the eurth's surface, Determine the mass of the object (in kilograms) and its weight (im newton) when located on a planet with an acceleration of gravity equal to 4.0 fi/s*. Muse = gets ‘ # = 860 Moo. 326 49 Ret Ze ee fap 42 He Fé fe>, weight = (30.0 dg) (40 #) (0.3042 2) = 394 4 leat127 1.27 The density of a certain type of jet foe! is 715 kg/m’. Determine its specific gravity and specific weight. 4 e _ 775 38 5 S6= Ber Teg 7 SUE to an y=pg = (775 42) (4.0) m)s reo 241.28 A hydromerer ls used w measure the specific gravity of lige nuids, (See Video V2.8.) For s certain tiquid & hydrumeter read- ing indicates a specific gravity of 1.15. What a the Lguid's den- sity and specific welght? Express your answer In SI units, pa table 2) EP aes Harllee bey EN 4 ewe ags (1150 Ss) fen @)e 3hue L2H An open, rigit-walled, eylindcical tank comains $f ‘of water at 40 °F, Over 4 24-hour period of tinse the water temperature varies from 49 °F w 90°F, Make use of the data in Appendix B to determine how mich the volume of water-wil chimge. For a tank diameter of 3, would the corresponding citange in water depth be very noticeable? Explain tress of water = Wn P where & tne volume and 2 The density. Since the Mass must remain constant as The tempereture ehanges Ws? Gos an * he From Table 8.1 Momus L498 ae wy be Cae ® guep = 181 BB CF Th a nae ad CLAN. 90 ge) = Keith £2 n° 14a} 2s Thus, The socvease th volume ps 4 pel —Hooo= Oo188 ft? The change wi water depth, AL, tpul + 4y OSE = ea = DiC risen teadbTibin ™ @Ft)* z This Smell change on depth weuld not be very fofteable, Ne. Mote! A slightly dtlertat value fer AL witl be obtained if spectre werget ef water tr used reTher Thaw density This 13 dice te tae feck qhot there is seme ancertuity int tne Rurth Sipnttant Ligare oF these Hite values, and Tre Solution 1 Sensitive tp Tess dncerteinty Inn131 1.31 A mouniain climber’s axygen tank comains 1 tb of oxygen when he begins his trip a sea level where the don of grav ity is 32.174 fs? What isthe weight of the oxygen in the tank when he reaches to tap of Mt, Everest where the acceleration of pet fs 32.02 TUs"T Assume that no oxygen has been femoved the tani: it will be used on the desoemt partion of the climb, W= mg ue a C dey denote sea level and (ye denotethe typ of MH. Everest v5, Wiz = |b=m, ei and Wuie =, ME Gonie However Mey = Myye 50 thal since m= ¥, me Me Eres Wipe a nL “aa ty Sow 32,082 H/s* = Le 19a2 H/s® _ Wire Mr ga, = Nb secrete = 0.9471 I 1-25132 The information on a can of pop indicates thu the can + Contains 355 ml... The mast of a full can of pop is. 0.369 ky while an empty ean weighs 0.153 N. Determine the specific ‘weight, density, and specific gravity of the pop and compare your results with the comesponding Values for watcr at 20 °C Express your results in $1 units _ Weight of Lhavel tr) 1 Gelame oF Plant toh! weight = messx go = (a.3bt by (281% ) = rb2v weight of Cans O53 Volume of Plaid = Gs5x 7k) 07-2 )= 385x106 “om? Thus, from Eg. (1) £42 - 0153 fw t= ee BSS 10 aw Ww . & . Sess, _ Nes? | 4g PoE pga = Ov = HE ot a 44 SG sf = 74m? = 0 996 Tyo@ec — 1000 4£ For water of 20°C Csee Table 2 14 Append 3) 2 gs 4s. oe iF 7789” te PAE» SG 0782 A companson of These Yalees Be water with Those for The pop shes Thad the zpecihi Weignt, aeasity, Aad gpecifie gravity of The pep are all Slightly Jower Than the corresponding values For weder,"133 ‘The variation in the density of water, p, with fem: perature, , in the range 20°C: = T=: 50°C, is given in the following table, Denn he? | 982 | ert | oN | SAKE | Ee | R02 | RA Faroe se bas Te tas bw 1a | Use these data to determine an empirical equation of the form # = 6 + «sl + ca” which can be used to pred the density ‘over the range indicated. Compare the predicted values with the am given, What is the density of water at 42.1 °C? Frt the data do 9 second arder 2elynome/ asing A standard Curve fe Hig Pregrim such a5 tend in EXCEL. Thus, ff) = (80f = 2.083997 ~ 0.084) 7* en ee eee As shown in the table below, (predicbet) fron Eg 18 1 geod agreement with fp Cater). a ‘908.3 cS 997.4 sar 30 9057 B57 35 So47 aa 40 22 8 9923 4s 902 soa cy 998.1 988.1 At Ts F240 °C . pe (eel 00838 (waste) ~ ooov1 Haire) = Fais *E I-27ASH 1.8% IF cup of cream having a density of 1005 ky/m? is turned into 3 cups of whipped cream, detetmine ihe specific gravity and specific weight of the whipped cream. Mews of cream, am = (toor#t)« (Vb, ) where + ~ volume. Since ereamy = tipped OM whippe h [aintreed ~ “diegedl . (loos = J as even Fy cape Wp cies | eos 3S = 3a5 * Zs rm 3 hi bk SG: ty tt . 335 58 0, 335 Bs et 1000 ae Sedippes = Pea eg? (23s #4.) (ei 2) = 3240 %1.36 1.36 Determine the mass of airin a 2 m’ tank ifthe air is at room m=OV where V=2m' and C= P/RT with Te 20% @ (20+273)K= 293K and pr= 200kPa = 200x10° m Thus, 0 = (200% 0? Be )/| (2.809 x1 4% )299 KY] = 2,39 ro Hence, meV = 2.3044 (2m) « 476 kysee : 1.37 Nitrogen is compresed to a density of 4 kg/m? under an absolute pressure of 400 KPa Determine the temperature in degrees Celsius P foaxre” TeH- eR es Bat PES eM Yim) a < Te - 173 = 337K -273 = 6% 1.38 The temperature and pressure at the surface of Mars toring a Martian spring day were determined to'be ~S0 °C and 900 Pa, cespectively, (a) Determine the density of the Marian ‘atmosphere for these conditions if the gas constant for the Manian atmosphere is assamed a0 be equivalent 1 thai of carbon dioxide, (b) Compare the answer froin part (a) with the easity of the earth's atmosphere during a spring day whn the temperature is 18°C and the presse 101.6 RPA (abs). Foo fs 4 = Pm ay 28 oie a (es) [sorc+a734 im? ) e. Po leluxwigs iaa Se RT (24h 2) [Cate +23)x) Thus, a aoe: Wears _ oh as = 0.0198 = 1.95% Penetn i2z te, —— 1-30Oo 134 A closed tank having a volume of 2 f? is fitted with 0.30 fb of a aos. A pressure gage attached to the tank neads 12 Psi when the gas temperawie is 80 *F, There is some question 5 to whether the gas i0 the tank is oxygen oF helium, Which do you think itis? Explain how you azrived at your answer. Fe vime Ca72 8 Vie we) = F slags Hebe 10 33 Density of Fes in bank f= weight _ 2308 nee Pr x em ope liar ter) poe Catmosphene pressure assumed te be & #7 pria ) and with T= (oF +hbo)*e 4 Ailes thet _ (207 tt ie) Cw de Riz shes tn) — = +2 5 (s#0°R) = fee From Table (7 Rsk 55x10? for oxygen ana R= favexit 46 fy belum. lag 8K Thus, From Eg) of the gas is oxygen = fife Shas 2 egy ip? slugs a L55¢ LDS fF 4 és ana te heluim P = heh ® aie? shegs £73 f- kavento* #2? A Compensoe af These values tite The actual density of fhe 948 t@ The tank imdreates That The Gas must be OX ygen. f-3t1a 140A compreised sir tank contains $y of fir ata temperature of 86°°C. A page on the tank reads 300 kPa. Determine the wolume of the nk. velume = T8*3 Po. (Be0+ ro1)x10 Pe = ay BE = = ens at C> RF (annr cg (sore + 273)4] - volume = sa = Laban? 5% St =| 7a = i i i pn 4.1 A rigid tank contains air at a pressure of $0 psia and at lemperature of 60 °F. By bow much will the pressure increase ‘as the tempensture is increased to 110 "FP pe PRT - C&q. 1.8) For \a rigra Closed tank The ath mass and are tonshant so f= constant. Thins, G48 (wits R eoastont) an) bk +460 = 520°R, = S10, Fram FRC) 7 76°R OE ) (ta pein) = $8.9 psa 1-32142 42 ‘The hetium-Mted blimp shown in Fig. PL.A2 is used at var- jous aihlesc events. Deiemmine the nomber of pounds of holiwm within it fs volume is 68,000 1 and the temperature and pres- sure are BOF and 14.2 pa, respectively, BFIGURE Piao W=SV where V=68 0008 and H<09-(p/RT)g Thus, B= Pogo (one Bien wt hs “g)(8o4 ¥s0)"R)]] (32.2 £2) = 982 xiO* SE ( | /Csloytt/s4)) = 9, 62x10 Fs Hence, We terre (sa000ff?) = 6¢8 1b f-33Muster Typing Sheet BL i ta °C Then "1.43 Develop 2 computer program for calculating the density ‘of an Ideal gas whea the ges pressuce in pascals (abs), tho tem- perature in degrees Celsius, and the gas constant in J/kg K ace | ‘specified Plot the density of helium a8 a function of temperature fom © °C to 200 °C and pressures of $0, 100, 180, and 200 kPa (abs). | Be dn teal gas | { \ pe par se That Spit [ES seep attes | Tine p & sbsolute pats R Thelgas autshaat, and To 1s abselite Hemperadiee, Tos, if The tempe rd sure IS. [This program calculates the density of an ideal gas | Ie na seo recur Pasa 18 tempe Pressure, ees aaa Pa kg k TOIESOS 3 2669 [Formula [=A10/(B10+273,15)°C10) te fp hr p= 200k Pa, temprrature = » aaa Ra 287 fae i i143] Ceen't) The density of helivm js ploited inthe graph belew. Density of Helium 04 0.35 —}———. 03 a kg/m? 525 ~~ — |as 1.45 For Nowing wate, what isthe magnitude ofthe velocity era lem needed to peoduce a shear somss of 1.0 Nim d aa T= Gy where wea haxi* HE ond 7 ato Thos, dale. wm ! dy ~ % * Tay wa 7 O93 = ee = — 146 Make use of the data in Appendix B to determine the dynamic viscosity of glycerin at 85 “P. Express your answer in ‘both ST and BG uniss, Tet (Fe -38)= F Cos -22) = anne From Fig, BL in A ppendic B: A Glytena of bs (Cayuse) = 06 YS (51 units) fe 6 EE) (d.009 610? RY 310? tes (26 wit) ne Fe ‘me 36LT One type of capillary-the viscometer is shown in ‘Video V1.5 and in Fig. PLAT. For this device the liquid to be tested is drawn into the tube to a level above the top ‘thed line, The tite is then obtained forthe liquid to dexin 0 the bottom etched line. The kinematic viscosity, w, in mls fs then obtained from the equation v — KR where A is a constant, F is the radius of the capillary tube in mm, and ¢ is the drain time in seconds. When glycerin at 20° C is used as a calibration fluid in a panieutar viscometer the drain time is 1430s, When a liquid having a density of 970 kg'm? is tested In the same viscometer the drain time is 900. What is the dynamic viscosity of this liquid? Z v= KR ¢ m FIGURE P1.41 Por glycerin @ 20°C DM= Liu Imi a ble onthe = eR, 430 5) KR"? 832x)0 7 ant/s® For anknoun ligaid with £> Fods v= (9.22210 mh) Go0s) = R49 x jo * m*/s Since fer pe G10 taps) (149.210 m4 ) = 0.727 22 = 0.727 MS rs mt set 1-37Master Typiny Sheet (0% Reduction x 1.48 The viscosity of « soft drink was determined by wstag = veupillary tube viscometer similar to that shown in Fig. Pl “7 and Video V1.5; For this device the Rinematic viscos tireey proportional te te time, t, tal I kes for # ven amount of Iiquid to flow throug! all capillary tube, That is, » = Ki. The following data were obtained from regular pop and diet pop. The corresponding measured specific gravities are also given. Based on these data, by what percent is the absolute viscosity, 1, of regular pop greater than that of diet op? Regular pop Diet pop. a 3718 3003 | so 10a | Mace Adit | Since vihh, ae meee aed £2 Cog, ure [is ae Ratha of peed (ae 7 toe 1 (be 56) rea {| leo Che shlaai =] (377.85)(Law) _ |) x 10 | (Cee. BADE! 003) gaan pode — 4-38ae 1.44 Determine the ratio of the dynamic vis- cosity of water to air at a temperature of 6) °C. ‘Compare this value with the corresponding ratio of kinemtie viscosities, Assume the niris at stan- dard atmospheric pressure Frem Table 8,2 tm A ppendis B= From Table BY im Appendia 8: CBr am ab wore) = tara 2S 5s vs hebnie & a Thus, “+ - hobs x10 = 237 497 K107F 7 AUSSI oe derajo? Lebxige | nt 1-39Master 10% K 480 (sr ] Alp t nga Sheet 7 T 1.50 The viscosity of a certain fluid is 5 x 10 poise. Determine its viscosity in both ST and BG units. Frem Appendrn E, i0' 3 = | poise, Thes, fe (520 poise ).( to" BA )= Sa1e® wee Peive ——"" ana Frey Table |.# = 5 As pt Les — ST does = (S210 ae ) (2.084 x10 )= axis Ee ws Se LSI Thekinematic viccosity of oxygen st 20°C and a pressure of 150 kPa (abs) is 0.104 stokes. Determine the dynamic viscosity of oxygen at this temperature and pressure, os rene 3 Si (259, re zee c va3)i] a DV O./ox oo = bson ae 4 ies (0.10% 20) (ro™* a% (1.97 =) Ae gos ee MS -- = fos 2 /b SA = Pones amMaster Typing Sheet vi 52 ~ "L52 Fluids evils eceaareaic 4s noe Hinsearly, —————— —— related to the rte of shearing strsin, 7, of¢ designated 23 non= Bi ‘Newtonian fluids. Such Aids are commonplace and can exhibit | | ‘unusual behavior as shown in Videv V1.6. Some experimental I data aba for a pardcular non-Newtonian Nida 30 ae sec oO 2 732 Ws 317 | Fy tye ee eae cerms ce 8 suitable graphing program, What is the apparent viscasity of this fluid when the race of shearing strain is 70 4-!? ts this ‘apparent viscosity larger or smaller than that for water at the | samme temperature? Rate of Shearing shearing stress, ssrain, ‘Is. Ibis ft © 0 ° & so 2th ] 1007.82 i 1500 18S. 200 a7 z 3 a 0 50 100 180 200 250 Rate of shoaring stain, us Pe eee le ; | ‘om the graph “p= 2. 0008/5" + 0.0097 % where | Shbaving strain a arf Fastlane SE = cle 2009}}-+-t-d035 tae He opis _ Pappanat = aye. A008 pee i tee De 810045 parc Ibs Waid hes) @ meh larger-value, tts fhe sheanns Stress i (6/ Pt" and ¥ 43 the rate | Bl 1a 4 mB, My eter = 7H ei os Ba Sace ave an Chala This Value 1s db = of ¥ + TAWS, The wnknewn Hon-New tonesTyping Sheet 53 Wier hows near a Mat surfiee and some measure: - — ~~ Mentsof the water velocity «, parallel the surfoce. wt different heights, y, above the surface are obiained. At the surface 9 = 0. —| Afier an analysis of the data, the lab technieiun reports that the £ ~ velocity distribution in the range 0 < y < Qi ft is given by ——-— the equation | um OBL + 92y 441 x 10? Hes | with w in fs when y isin ft, (a) Do-you think that this equation | would be valid in any system of units? Explain. (b) Do you ‘think this equation is comreet? Explain. You may want to look at Video 1 +10 help you arrive at your answer | { ¥ fap |e OG 92 ge hI elo gy? | fails fe ae] + fer] | Bath! term oa Tae agi tien rout have the same ol mensions, Ot rust Have demnswns of kT Ty add hy X17 drmensiens of ET! ants fn The tpuation Have dimenswns Their | Gases dante soa Riubascdee a cannot be torveck piace at gre az asitth, value which would violate tne no=slip Condition. dine rrack. | |Master OM R 154 Calculate the Reynolds numbers for the fow of water ‘and for air through » 4-mm-diameter tube, ifthe mean velocity is 3 m/s and the temperature is 30 °Cin both cases (see Example 1). Assume the sit is at standard atmospheric pressure. (Fer wader ad 30°C (From Table B2 1 Appendie B)+ -4 Mes a 4g See ; P= F957 wv L475 x10 ae fe = OD . (7987 &) (3 ¥) (2.00% 07) = ($000 z ERIS ig Nes ome Ee acr at Sore ( from Table B.4 in Appeniy 8) : i = Ae = -F Abs | i Ales =, x Ade 440 = | - 4% ca Ht f= Vea (ted ren) (3 FP) ( e.cotan) x h8b aio’? M8 a ees Se L1.85 Forair at standard atmospheric pressure the values of the constants that appear in the ‘Sutherland equation (Eg. 1.10) are C = 1.458 x 10°* ke/(ms-K!*} and § = 110.4 K. Use these values to predict the viscosity of air at 10°C and -90 °C and compare with values piven in Table B.4 in Appendix B. ere sp xia 4 ) 3 Tr ho4ak For T= /0°C = fe°C + ATzis = 28RI5K, ae 34, tase nie) (agsisk) a Less ® N ARLISK + 110m From Table G4 jf? Le eso Hes For Ts Fo°l = GC + aI /8 = 3b3.15K, (hase x to7") 304 ie i ae i = 23K S435 Ke sO. 4 From Table oe, f = dex sar Ms ane Oo1,56" Use the values of viscosity of air given in Table B.4 at temperatures of 0, 20, 40, 40, 81, and 100°C te determine the constants C and § which appear in the Sutherland equation (Eq 1.10). Compare your results with the values given in Problem 1.55, (Hint: Rewrite the equation in the form = i s fe (2)r+ € and plot 7p versus T. From the stope and in- tercept of this curve € and S can be obtained.) Equation 410 Cam be whiten iy The form a ee INA) r+ § w and with The data from Table BH: Tee) THe) lists) — Ta [yh] e ATs Liew 2.640.208 ae Aaal6 heaxn® 275P x it co 813.15 LITA a ene bo B88 IS saree Borneo’ go 35815 2.07 Ko 2206 x10 {00 37315 apres 3, R2xI0 A plat of Tu ve. T 1a shawn below 35K" 280 ‘Bao aso “Feo ~ Te)[46*] (eon't) Since the data plot as an approsmede straight line, F911 aim be reprewated by gn eguetey of the Pr, ” yaaxra where ya THe peed, b~ Vo, ana ar S/e Fit the dade +o a linear eguahon using a Standard surve~f Hing program Such as found In Excel. Thus, Y= 699x108 + T. gu) 10" and gabe baegewe =f y so tat C= 13 xIo 4a fons ™) —— EE and Ssa= 74d x10" (ss and Therefore S= /07 K These values tor Cand S are in good agreemmt w/ty values given in Problem 55 1-46Masicr Typing Sheit lay 1S? _ The viscosity of 4 Quid plays a very important role in determining how a Muid flows. (See Video V1.3) The value of the viscosity depends not only on the specific fluld but also on the flsld temperature, Some experiments show that when a liquid, under the action of a constant driving pressure, is forced | with a low velocity, ¥, through a small horizontal tube, the =P velocity is given by the equation V = X/u, In this equation. 4g constant fora given tube and pressure, and isthe dynamic vssosity. Fora particular ligidof interes, the viscsiy in given by Andrade’s equation (Eq 1.11) with D = 5 10" Ib «s/f? and.B = 4000 °R, By what percentage will the velocity increase 85 the liquid temperature ia increased from 40 'F co 100 °F? ‘Assume all other factors remain constant, Vb? Vin i —-, My = i | pbs | aa ae lk oh increase inV= [tee et = [ and from Pq) @@) “ef anevease tn Y= ieee - ewe = [Jae ein Andvades @eua equation Yoo Par = xe Ve (pers Abe) Hove ra aap waite é [oar Pe ‘Freel By.) rs a 2 = 1/% tee Fer o Tecvease v= | an C23 @) I-47“LSB Use the value of the viscosity of water given in Table B.2 at temperatures of 0, 20, 40, ‘80, 80, and 100°C to determine the constants D and B which appear in Andrade’s equation (Eq. 1.11), Calculate the valuc of the viscosity at $0 °C ‘and compare with the value given in Table B.2, (Hine, Rewrite the equation in the form Ing = (@) 24 WD and plot In yr versus 1/7. From the slope and intercept of this curve B and D can be obtained, Ifa nonlinear curve fitting program is available the constants can be obtained directly from Eq. 1.11 without rewriting the equation.) Egaation 11 Cen be written sin the form Inu BIE ¢ Ind end with The dite trom Table BZ * Tc) Ttk) Ifrte) eo apis 3.kbt £10 do IBS 24y 10" Ho BIBS S13 e107 bo 3338 Booz 407 Po 3535 aész 41077 feo 37345 goer? Pe (i ston) L770 heoLLio” 6 52F Ka & esta TKI LOY 2.818 £10" A plot of Inn vs, WT 1S shown beloed: ahe * tn * 6.327 — 6.906 — 733K -7. Io 7. 98H ~ BlTt5a Cent) Since the dete plot as an approximate straigyt hie, B90) Can be wsed to represent fnese data, Te ebtnin Bana D, t+ the deta to an exponential Pp aatiin ef the form Y=he Such ns fund if XCEL, Thus, “ D=a= 177 xI0~ Nslamt and 5 Bibs LPIOKD” K so thet “70 ~é fe f7e7 Elbo @ aa AL S0'C (323.5%), 1870 — “| fr Lpe7 xi @ RE 3s 5.76 x10 WS fn® TS From Table 8.2, Bag x10 N.S fow®6SF distance between plates is 2mm, a shearing stress of 150 Pa develops at the upper plaie when it is Pulled at a velocity of 1 m/s, Determine the vin- ‘cosity of the Muid between the plates. Express your answer in ST units. Tah = de. U, fe SB = foe S Aes rae 7) @,300 }-S0160 1,60) Two fat plates ie ortebted parallel above a fixed lower plate as shown in Fig. P1.60. The top plate, located a distance & nbove the fixed plate, is pulled along with speed V. The other thin plate {a located a distance ch, where 0 <<< 1, above the fixed plate ‘This plate moves with speed ¥, which is deiermalned by the wis- cous shear forees impased an it by the fluids on its top und bot- tom, ‘Tae flac. an the top #3 twice as viscous as that on the bot- tom, Plot the ratio W/W a8 a function of ¢ for < © |, mb —»—_+ FIGURE P19 For constant speed, V,, of the middle plate, the net force on the plate is © Hence, Fup * Fiettom, where F2 7A. Thus, the shear stress on the top and bottom of the plate must be equal, du / Cap = Coettom where F=f dy ‘ For the baton flid $8 = ML while fr the top Fid = CO) Hence, from Eqn. (i), (ay) 8 ewe , Which can be written as: acV-2eli =\Y-ch or Ving 26 Vv Sai I-5161 4.61. There are many fvids that exhibit son-Newionian behavior (see, for example, Video V1.6), For a given fluid the distinction between Newionian and non-Newtonian Behavior is usually based on measurements of shear sires und rate of shearing rain, As- ‘sume that the Viscosity Of Blood is to be determined by mensare~ tents of shear stress, 1, and fate Of sheating simin, dutty, obs tgined from a smal! blood sample tesed im 3 suitable viscometee, Based on the data given below determine if the blood is a New: tonfan or non-Newtonian fluid, Explain how you arsived at your answer, sin’) [004 | 0.06 | O12 Jou | 0.30 j 0.82 | 1612 | 210 auédy 6") 12.25 | 4.501 112512251 45.0 [on.0 [aes | as0 For a Newhanan Fluid the rabic ef tt dufdy ts 4 Constant, Far the data. given! ft —— CW -s/mt dfs ) The ratio #3 not a Constent but decreases 23 the rate of shecring Strain mereases. Thus This Fluid (pled) 4 pon-Mewtoaon Fluid. & plet ef the data & showy below, For a Hewlonren vind the Curve would be a strenght line wrtn & slope ef / to / 8. O73 | 0.073 0.0067 | a0sp \o.ensa |2.0087 p07 | 2080 10 1 7 Mt oe Mewlonian fluid 0.01 : Ht q 10 100 41000 mw 1 ay, 3 Nate: 7 ley , where a=l for a Metonian fvid. 1-2162 1.62 The sled shown in Fig. P1.62 slides along on x thin horizontal layer of water between the ice and the runners. The hhorizontal force that the water pots On the sunners is equal to 1-2 Ih when the aled’s speed is 40 fus. The ual arex af both Funners in contact with the water is 0.08 4, und the viscasity of the water is 3.5 % 10" tb 6/ft Determine the thickness of the water layer under the rumaen. Assume a tineut ¥elocity distriboston in the vier layer MFIGUARE Pez F re) = tA Ts ph 3 ¥ where d= thideness of water lenny Thus, pee ha ana ma , Axi? Les\/a0 #)/o, 09.44) de AA Cra on Bons) u Jie * Le }-$3