Cube Equilibrium in Liquid: Heat Transfer & Gas Thermodynamics, Exams of Physics

Download Cube Equilibrium in Liquid: Heat Transfer & Gas Thermodynamics and more Exams Physics in PDF only on Docsity! \ ' l/' I | ,i,-- t " t. f i . ' lJ.y'" i"t t- l. t I t^ . Sf tD+; t' t\ fd PHY 206 EXAM I ID: 1- .l l. ',. .l /_. 3. 1, p + pgg * ;pu" : const. dQ : n'LCdT dQ -- ncdTL L,L-: aLgLT LV : pVa'LT ,# - *lO,.-, ' dt -' l' .' P :-oAeTa .py : nRT dQ : dU + dW" . dW : pd,V "pV1 :'cons!. ^y - cp/cv,) l. ^'\ T)- 17.D J (110) <td- 1. This problem consists of two i,ndependent problems. a) A hollow, massless cube with side tr is par- . tially fitled with a volumdV ota liquid with density p and the remainder with a gas of negligible density. The cube is in equilib- rium with its midpoint at the interface be- tween two liquids'of density p1 and pz t pr as shown. Find the length tr in terms of the . densities and V. b) An tank has a small hole of cross-sectional area o at the bottom. Liquid of density p enters the initially empty tank at a volume flow rate ft (i.e. in m3/s). Find the height' gr to which the water level will rise.. ?-, fl= Lt<B- L'fl- Solution:' a) In"equilibrium means'the weight of the liquid in the cube must be bal- anced by the sum of buoyant forces from each of the displaced liquids: pvg:)orr'o +)orL's + L: (-ryL)"'. \Pt + P2,1 b) The water level will risi: until the flow rate exiting the hole equals ft. The flow rate out the hole is ou, where u is the speed of effiux which we compute in the usual way using Bernouilli's'equation. Taking a point at the water level, height y .frorn the bottom, and the second poidt at the output of the hole, Bernouilli reads: PolPgA:Po*lO"4u-frW So now,we have: R : ou : or1Zfu. Squaring both sides and solving for y gives the desired result:. l, a: +(.u)' /i\ t-Q.v,'; ."."2 I f- { I I I I irt - I : t :-" I Mar.."5, 2008 PHY 206'EXAM II NAME: SIGNATURE: I. ) a a'J. ID: ! rl 1,. , . Some relations you may need: "*t '1 L -t P + PgA -t ;Pu' - corlSt.z Er' l - LL:."aL'oLT LV : pVoLT dQ : rncdT dQ : ncdT -' #::f,"' i ,-R Na W . lQoutl '7: e;,:'r - ad DV -D "'f (110) I t' 1. In the following three 'independent problems irll gases are ideal and vibrational degrees of freedorh are to be ignored. Yoir are to use: ly'a : 6 x'1023 rholecules/mol, R: 8 J/mol.K, | ' ''" and the table beiow as needbd n- ' a)''Suppose Cl2 and N2 gases are in equilibrium in* a contd,inef at temperature ?. What is the ratio of their rms speeds, uffi"lrY,A"? b),A d,i,atomi',c gas from the table below is measured to have a con- stant-volum"e specific heat per, mass, C : (5000/8)J/kg.K. Use' ' the equipartition of energy principle to find'the molar mass and identify the gas. c) The specifib heat of a mixture of gases is the weighted sum of the respective "specific heats of the gases: "_n(rLnzci t , I 1-* 71,2 . *hete fi,1 and. n2 are the number of moles of each gas. A student has a cohtairier with 57g.of dr,atom'ic Fl2 gas in it. She wants to add monatomic helium gas to create a mixture with constant" uolume molar specific'heat (2L110)8. How many mols of helium' must she add? R f, Molecule M (e/mol H2 2 N2 28 o2 32 FI, 38 CI, 70 mole: M- a, 5n r / 1 r.:6JlmOI .nzl (5000/8) l lkg-K (5000/8)J lke.K *y \PV= A simple dimensional analysis tells you that the specific heat per mol divided by the'spiecific heat per mass is the mass per =Fr A $r/ =1. /P ffi in 32 1 | r ,: ,O*Ug/mol :32glmol + 02 ,i I ,t l!1- {, I a lr' It- ,I LI c) First we determine that 57g of Fl2 corresponds to nz : 57 g/(3S S/mol) -'",/ * mol. Then the specific heht equation reads:..2 ']n:tiff i 2r., 3. 3 15 i_(n1 +_):_rt1 +_. 10.--. 2, 2.'4 1 ' 21(2nt + 3) : 30nr + 75 jr .'.(42 - 30)n1 :75 - 63 + nr - 1'. tt- ti i I.r r I '. j" I''t -l' t''".i 1',t'-'l Irf I PHY 206 EXAM III ; Apf. 9, 2008 C . NAME: . . SIGNATURE: I I ID: .t. I I 1. 2. a J. I ' Som'e.relations vou mav"need: a2o-a_Lo-a, r 0r2 - u2 0t2 .t,' n.1^sin 0t : n2sin02 " +?Z - U4l,sstR (i+il: cos r cos yasiir r siny , sin2r+cos2h:I 1* H 5 (1?o) 1. The following expression is proposed as'a solution of the wave equation: t- a@, t) - A sinlk (L - r)] cos(c,.'l), ' " where A, k,, L, aind a are constants. _ . a) Show that it is a solution of the wave equation, or more precisely, find a condition involving some of the constants that ensures it is a solution. b) This expression"is to represent standing waves on a string (7 : 2rf w and ) - 2nlk'are the peiiod and wavelength). One end of, the string is at r - 0 and ihe other end is at tr : tr. Suppose the end at r - 0 is tied down so that A : 0 there forall time. What does this condition imply about the relationship between ) and Z , (this specifies the normal modes of vibration)? c) Using the axes below,'plot the two longest-wavelength ' modes of vibration at time t : T 12. H_tlt Solution: a). The wave equations is on the -A \N I l : - k A coslk(L-r)] cos(c..'t) ;' - -w A sinlk (L -r)l sin(alf) ; Now put these results in the wave equation: -k'a(*,t) : \w2y(r,t)u. which is satisfie d if u2 - (w lk)2 . >Y r ora ./ l'c:-- u2,0t2 A2o, # : -t*Asinfk(r-r)] cos(r,.'t) : -k2a@,t) 0'a ) t . 17 /r \ 1 / ;# - -r2 Asinfk(I-r)l cos(urt) :'-r2,y(r,t) o'a' A"' 0a 0r 0y a 1'' h b; 'l\' a\,t) - Asih(kp)cos(ut) :0+.sin(k,t) :0 +kL\-Lmtr, in-0,!,2,...U\ J / .i i'-- =* .\ : 2Lf m, . TrL : I:2,,'3, ... ' I 6) 'ih" t*o longest wavelength riormal modes "oir"rporrd tb rn : 1, and ni:'2. Note that y(Ltr) -- 0, so the right'hand'end rnust also be tied down (i:e. a node). To plot these two modes at t : T 12, check a point (that's not a node!) such as r-: Ll2 (for.m _- 1) and r = L/4 (for T:2):'. u ' : .nL - 1 + ,\ - 2L: y(Ll2,T12) - Asinlk(L - Ll2)]c.os@r12)n.t' f l* r.1 - .4 sin l;; I cos(zr) - .:A - L-" - ) ,. rrL:2 + ),: L; y(Ll4,f l4 - Asinlk(L -'Ll4)lcos(uT12) .,--', lzTgrl , \ , ' .r ='i, qrnl+;1cos(r) -1 ' * L- J ,I This is,enough to draw the plots shown. t. ,* I P I i I I I I I I I, j I 'i I I I I i I I I I t I "l I I l .t (1 Jt' 1 l r lI l 1 I .1 J I .J l I l' 1 l ,LI I I l , -) l PHY 206 FINAT EXAM NAME:M.y. 5, 2OO8 ' lrr -: , ( SIGNATURE: a ,ID: 1. '2. t.1 J. 4. 5. . $ome relations you may need: .1 2p + psa * ioy? - const. dQ = mCd'T dQ : :",1:T- . L,L-: aLsLT Lv -- pvoar # : "f+, P - oAeTa pV : ?RT -," dQ - d,(l+d,W .. dW - pdV p.V' : cbns,t. 'y : cp/cv f - 1 , 3- _ R ", : iR I KEporttcle ): ,*u;*r: ifu, , kB : lrla ,-!a--1- W4 ds-Q ",- Qun, t lQo"l @u T 0'a L*a : a.r,: ,"# 'Tr1sind1 - rl2Sindz "ry .-? : *#' o' "" 70) 1. A Vpnturi tube consists of a pipe having regions sectional areas, l.1 and 42, resp€ctively, through po is flowing. The air flow causes liquid (density p) to be displaced so that the level heights on either t. with diftering crosS- which air of. density in a separate u*tube side differ by h. terms cancel since we take our a) Find the aiq speed u2 (in the region witb cross-sectional area A2) in terms oj ut, Ar, and A2. b) Compute the pressure difference, pt-pz,-between the two sides of the u-tube in terms of po, ut,At, and A2 fif you couldn't solve a), use pa, 't)1, and u2] c) F ind an expression for the height h. Solutioni. , a) Continuity of the air volume'flow rate dictates, Apt - Azuz ) uz: *r, b) Appl;ling the Bernouilii equation to point, in th" airflow just above the ' sides of the u-tube yields: h*t^r^u?:Pz+!P"3zz (The gravitational potential energy points at the same height) 7 ,, ,' 1? Pr - p? - |0"(u3 - "?) - )o"u? c) The weight of the liquid of height h is balanced by forces associated.with the pressure difference. Let sectional area of the u-tube: . frr\'- l\tz / , the A 1l I difference in be the cross- PgAh:(P'r-Pr)A*h: p,b?[f*l'-1]/\ \Pr - Pz) 2pg '2. A liquid with volume exp-ansion coeffici ent p just 'fills a thin-walled, spherical container <jf volume V at temperature T. Th'e'"container has volume expansion coefficient go < 0. When the'terhperature in- cr€ases by A" the liquid is free to expand into an.open tube with c_ross-sectional area A proje,"cting from the top of .the sphere'. Assume the container and liquid are in thermal equilibrium throughout and the tube does not erpand. . nA a) Write down expressitins for the vol- .ume changes of th'e container and liquid. b) fo what height h does the liquid rise in the tube T!AT Solution: a) container : LV" - PyV LT, liquid : LU - PV LT .i b) The difference in the volume chdnges must equal the volume.of liquid that goes up the tube: '. -s tl \#v T "V' Ah + h: A(P - 0o)LT. \ p\I