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# Heat Transfer ME 3345

GPA 3.62

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This 0 page Class Notes was uploaded by Chloe Reilly on Monday November 2, 2015. The Class Notes belongs to ME 3345 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/234248/me-3345-georgia-institute-of-technology-main-campus in Mechanical Engineering at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

5 t quot ME 3345C Heat Transfer spring 2006 Objective Introduction To Steady State 2D Heat Transfer 2D and Transient Heat Conduction Equations 2 Partial differential equations Chapter 4 2 2 V2 T 9 aT 0 How can we solve this equation given 4 BCs 2 2 6x 53 my LaPiaee Equation speciai form of the Helmholtz and Poisson Equations Chapter 5 2 a T 1 aT How can we solve this equation given 6x2 a at 2 BCs and the initial condition Parabolic equation Tx t can only be obtained for limited cases With regular BC s complicated functions and series are involved 39 serve as a tool to Verify numerical solutions excellentsources of analytical solutions in many found in many texts e g 1 Carslaw and J aeger ConduCtion of Heat in Solids 1959 physcally intuitive and tedious limited use for simple geometries only Not good for high accuracy calculations powerful easy to get wrong results be careful 1 amp 2 are often used to ensure that the numerical solutions are correct FLUX PLOT iss herms Haat aw nes Fimm l 4 i Twadimemianal atmdxmtian Main points Isotherms and Heat Flow Lines y Long sql39i ire duet Main points Symmetry lines Lines of symmetry are adiabatic Symmetry lines are alsoyheat ow lines We need only to study 18 of the whole structure to save time Thermal resistance qt Applicable to 1Dor 2D irT1 and T2 R t are identi able Sometimes Rt is inversely proportional to the thermal c0nductivity ie 1 Rt 2 kS q kSAT i Iere S isa geometric factor that is Independent of the materlals We call S the shape factor or conduction shape factor Table 41 Conduction Shape Factors Case Circular cylinder of length L centered in a square solid of equal length as B ZWL L 45 W 275kT1 T2 1nl08wD q39kS39ltT1 T2 Effective conduction resistance Rm 2 k5 391 A 3cmdiameter disk is maintained at 50 C by electriCal heating The diskis placed on the groundk 12 Wm C Whose average temperature is about 5 Assume that the top and side of the disk is insulated What is the electrical power that must quotbe provided to maintain the disk temperature P e e q T2 122003505 2396 W Case 10 S 2D 2D steadyeSt ate 5x3 ayz T0yT1 TLyT1 Tx0T1 TxWT2 T Tl quot Tz T1 V 0 i i Let 011quot aze 620 6x 6y 60 y O 9L y 0 9x0 O 6x W 1 Separation of Variables Assume 9x y fx go Hope to obtain equations for f x and gy separately Find all solutions and the one that meets all the BC s The Fourier series is often involved Let 0x y Xx Y y The SOIUtiO t0 the differential equation for this problem is inthe form of v 39 i 0xiy mrx Sin y nl H i L 7 L You can verify that it satis es the partial differentequation asylwell as three of the BCquots Forit to be the solution we must have th 9x W 1 2 C sinEZ xsinh nl Solving for mee obtain quotIntroduction to Finite Di erenceMethod r Several different numerical methods have beendeveloped Finite difference method is more intuitive and easy to apply Other methods include Finite element method 0 Finite volume method Boundary element method etc Numerical method continued Discretization anodal network called mesh or grid We can re ne the mestht we cannot obtain continuous solutions A coarse mesh A ne meSh Nodal points or Simply nodes nodal property such as nodal temperature it is the temperature of the node but it also represents the temperature around the node average Basic principle Through approximation convert the differential equations to a set of algebraic equations that can be solved numerically 2 w m f 2 1 f f mg i3 wees 3 213 a Tm average temperature around node Pm n 7 Discretizati39on of a COntinuousFunction i 7 m 12 7 39 m12 dx m12 Ax dT T T dx m 12 Ax QT 6T 121 z 6quot m12 ax m 12 Tm1Tm1 2Tm dxz m Ax 2D Steady State 82T 82T 527quot Tm ll Tm 1n 2Tmn 0 ax2 Ax2 2 2 m 8x 8y 62T Tmn1 Tmn 1 2Tmn W3 f11 7 2 39 4 5y m M m i 1 E m E 1m If Ax Ay then we obtain 33 m 3 a Tmln Tm 1n Tmnl Tmn l 4Tmn 0 n i Fimife difference mummy foe mg m Use N linear algebraic equations to solve N unknowns 39 is Energi39 Balance Method w 512966 Tmn n 11 mm W m1n 9 gm Tm1 1 5 i 2D SteadyState with Heat Generation m72391 mn1 Ay V mn l Ax Ax kTmln Tmn 16 Tm Ln Tmn Ein Eg 0 kTmnl Tmnk Tmn 1Tmn q39AxAy 0 39 ng Ax Ay 4ltAxgt2 z T Tin 111 Tm39n1 Tmn 1 4Tmn 0 m1n Corner Nodes and others Read pp 200 205 11 12 1 R m n v 1 What do we endup with is a set of linear algebraic equations r 7 Assume there are N unknown nodal temperatures 01111a12T2 a1NTNC1 11 12111 022212 aZNTN gtC2 21 aN1 aN1T1aN2T2 aMVTNCN N 2 N 391 12 022 N2 1N 1 021v aNNJ T A 1C TNJ ANXNTNXI Cle Matrix inversion solution FCI I C2 CNJ Iteration Method Why Reorder matrix such that the largest absolute value coef cient in a row is ail ell1T1 611sz aiNTN Ci was ME 3345C Heat Transfer Spring 2006 Objectives 1 Electrical Resistance Analogy and Contact Resistance 2 1D Heat Flow in Cylindrical Coordinates 3 Electrical Resistance Analogy in Cylindrical Coords Thermal Contact Resistance Thermal contact resistance gives rise to a temperature discontinuity at the interface A surface energy balance would be 5 Very Small 0 qx is continuous Unit Area kg 9 kb 91 dx dx qquotx Temperature is discontinuous qu TA TB tc In order to develop the electrical resistance for each mode of heat transfer we must look at the linear heat transport coef cients kA V L 1 Conductlon q L AT gt Rm a e W EWE 1 2 Convectlon q hA AT 3 Rm hA 1 k h A AT gt Rm 2 vlng 4 Interface Contact Resistance Rt 0 9 Radial Heat Transfer F Or 1D steady state constant k at WO heat generation We have in the cylindrical coordinates dT dr dr dr Tr C1 lnrC2 dT dT kA k 2an qr c dr dr n dT qr ZqFACZ kcl 1quot For a cylindrical shell with known surface temperatures T C1 lnr C2 T1 C11n 1C2 1 T2 C11n 2C2 2 Solving for C1 and C2 yields the following temperature distribution T1 l x T2 11 H 1r lnlnkr2 12 2 What about temperature gradient q k constant A and dTdr not constant T T1 gt T2 T T1 lt T2 T1 T2 T2 T1 iki Aii n W 1 VT QW r1 r2 r1 r2 Note that dT drl is getting smaller as the radius is getting larger Heat transfer rate and thermal resistance To 553 1115 T2 lnW 7392 r2 2mL ii 1n39 239 1 i 4513 T1T2 A dr r lnr2r1 Thermal Resistance T T 2k L 1 2 qr 72 1110270 T2 T1 1 qr 1110413 AT min3 2k7rL 2k7rL 3amp2 Twz 7392 W3 1 mfgr3 1 klzxi gL h22m 2L The amsgmggcimmi i amp8 am i smm mum ma ag V89 mm mdiE 8 Composite Cylindrical Walls i E Tw v 24 TM v n t f l g V 5 q UAAT E l I I g l f I x I q MT T 73 T3 7 TM B bmm awmfi igmi i a 1 mm magr3 QM gmgtglgfw mef mmw 12er ZxkAL Zxkgl EMkCL h 22m L Spherical shell qrkA 4 29E dr 61397quot j qr 7sz 475k r2 T d dT gt qr4nkTi T 1 lrl lr TABLE 33 Onedimensional steadystate solutions to the heat equation with no generation 39 Plane Wall 39 Cylindrical Wall Spherical Wall 39 14 4 111 2 41 Heat equatlon dxz 0 r dr r dr 0 r2 dr 739 dr O In rr I rlr Temperature g T 2 T5 AT distribution T ATL quot2 AThu flr2 quot 1 Inr2 Heat u k9 k AT k AT x q L rln rzlrl r21r 115 A 27rLk AT 477k AT Heat rate 1 M L lnr2r1 1m 15 mannal A 11172r1 Ur quot llrz resistance Rgmnd kA 27139Lk 4 H39k lquotThe critical radius of insulation is rc fell for the cylinder and rcr 2kh for the sphere Example 1 The wall of an oven is made by sandwiching insulation material with k 005 WmK between thin metal sheets the oven air is at 300C with h 30 WmzK the inner wall absorbs a radiant heat ux of q 100 Wm2 from hot elements within the oven The room air is 25C with h 10 WmzK What is the insulation thickness required to maintain the outer wall surface at a safetotouch temperature of 40C 2 A commercial grade cubical freezer 3 m on a side has a composite wall consisting of an exterior stainless steel face sheet 635 mm thick an intermediate layer of 100 mm cork insulation and an inner sheet of 2024 aluminum 635 mm thick Adhesive interfaces between the insulation and the metallic strips are characterized by a thermal contact resistance of R 25x10394 mZKW What is the steadystate cooling load of the refrigerator if the inner surface is 6C and the outer surface is 22C Example 3 The outer surfaces of a composite wall are exposed to uid at T25C and h IOOOWmZK The interior wall experiences uniform heat generation while there is no generation in the two outer walls The temperatures at the interfaces at T1 261C and T2 211C Assuming negligible contact resistance at the interfaCes determine the volumetric heat generation and the thermal conductivity of the interior wall T25C T25C h1000Wm2K h1000Wm2K K A 25WmK L A 30 mm KB WmK LB 30 mm KC SOWmK LC20mm mam am mg a 1m as K iii V m gag m m WW r 40mm kb 15 WmK r 20mm ka 015WmK h 50 WmZK Too 15 C lei An insulated Wire of diameter D 2mm has an electrical resistance of 001 Qm and a current of ZOA The insulation has an outer diameter of 3mm and thermal cOnductiVity of K001 WmK h5WM K Too 20 C eat 1s oss through convection What is the surface temperature of the rod and the insulation B What is the rate of heat transfer per unit length at r 0 05mm and 15 mm if the power density in the rod is 3X105 Wm3 l A long rod carrying electrical current has a volumetric energy generation of l x 105 Wm3 The rod is insulated by a cladding material and39cooled by forced convection with air at a temperature of 20C The thermal conductivity of the rod is 100 WmK and the insulation material is 10 WmK h T 200 q1x105Wm3 Draw the thermal resistance circuit for this system in symbolic notation What is the convective heat transfer coef cient necessary to ensure that the maximum w AerOspae Material Processing Three basic modes of heat transfer conduction convection and radiation Heat Conduction refers to thermal energy that is transferred as a result of the collision of energetic particles within a solid stationary uid or between them Dominant in solids and occurs in liquids 7 Thermal radiation tranSfers energy between objects by the emission and absorption of electromagnetic waves it can transfer energy without any intervening mediuml Convective heat transfer or simply convection involves the transfer of energy associated with conduction into a uid accompanied by uid motion advection Fourier39s law Empirical Observation lead to the concept that the rate of heat ow in solids was proportional to the temperature gradient and cross sectional area dT n K q quot dx Concept true regardless of heat carriers Thermal conductivity carries all material speci c information Heat Transport is a vector quantity q What is Fourier s Law for the following system Fourier39s law applies to solids as well as liquids and gases as long as the medium is macroscopically stationary Thermal conductivity carries all of the material speci c information concerning heat ow and is de ned as q dTdx Wm 39 K or Wm C Room temperature conductivity values see Appendix A Diamond IIa Copper Glass SiOz Water Air 2300 401 138 061 0026 Convection Newton39s Law of Cooling In 1701 Isaac Newton performed experiments by placing heated objects in uids and found that the rate of temperature change of the obj ect39is proportional to the temperature difference between the object and the uid ie dT oc T T 00 dt The cooling mechanism between uids in solids in this case was due to convection which can be expressed as q Too This is called Newton s law of cooling and h WmZK is called convection heat transfer coe eient or lm coej cient v Types of Convection We Will concentrate on forced and natural convection circu t boards c d mem 15 Convection heat transfer processes a Forced convection 1 Natural convection c Boi ng d Condensation TABLE 1 1 Typical values if the Pmcess convectiuni heat transfer cmef ci em h Wlm2 K Free mnvectizon Gases Liquids Forced canvectien Gases Liquids Convectian with phase change Bailing or condensation 24 25 504000 255250 1 OOH 201000 2500400300 Examples of Natural Convection in Electronics Graphics Framer 925 mm thick 42mm Spamshieldm m P adhesi lely attached to lower case Sony Vaio Dampmsihle pad underneam Spreadershield m manta gaod mutant Ween Spmdexst ld andamphi mm Thermal Radiation Radiation is electromagnetic energy emitted by matter due to molecular and atomic Vibrations Thermal radiation is the portion of the electromagnetic speCtrum that 0 lies in the wavelength regime of 04 1000 um Majority of heat is transmitted in the IR regime THE NETIc SPECTRUM Wm 10 103 1101 1 w 1039 W3 wquot 59 WE w w w WWJUquot w l if F K I in mm 1 l l l u c a f 1 a I Show a 1 lt 39 I H Damn mmdm Thermal Radiation Mgth m l l l a c l l m l L i la 1 I ll second 10 to 20 109 W1quot 19 w 1013 w w w 10 mm 101339 miquot 39 Wei Vim 7 v my 7 I R I 1 E t 1 4i 3 l I l I i l i mm me W9 lo3 wquot w 195 NFquot W3 w toquot a m to 103 10quot mi 10 In 1879 Joseph Stefan 183 51893 published an empirical relation that the radiant heat ux from a surface is proportional to T4 Five years later LudWig Boltzmann 18441906 theoretically derived the equation known as the StefanBoltzmann law It expresses the maximum emissive power radiant heat ux of a surface as n 4 qrad Eb quotGTs Where 039 567 X 10 8 WmZK4 is called the StefanBoltzmann constant This equation is only for a perfect of ideal emitter known as a blackbody E b is the blackbody emissive power Temperature is always in K Radiation and Real Surfaces Mostsurfaees are not ideal thus their emissive power is lessgthan a blaekbody and do not strictly follow the Stefan Boltzmann Law The ratio of the emiSsive power of a real body to a blackbody is known as emissivity i4 which is between 0 and 1 E1 GT The emissive power emitted radiative heat ux is 8 I 4 qrad e 3quot E SGT Irradiation and Absorption EmissiVe Power speaks of thermal radiation which is leaving a surface Radiation can also be incident upon a surface and is called Irradiation G Has units of Wrnz For an ideal body black body all energy is absorbed For real surfaces the fraction that is absorbed is described by a material property known as absorptivity 0t between 0 and 1 I n qrada OLqmdj OLG Energy Balance Key to Solving Problems in Heat Transfer iV 39 Eg 39 t quotquot39 Eout E0ut dts Em Est Thermal Energy Storage Section 131 dEt dT d Est pch dt Thermal Energy Generation Section 131 Egq39V Surface Energy Balance Surroundings Steady state no mass in the control volume quot I Mgr TI 1 i I E02 39 12 l i It II n qcond qconv grad T T2 dT quot 2 4 ha Too was 19 LL WWW T T K 1L2 hTz Too 80734 11 Methodology for Applications of Conservation Laws f 1 The appropriate control volume must be de ned and 39 identi edquot i 39 2 The appropriate time basis must be identi ed for transient problem 3 The relevant energy transfer processes must be identi ed and shown on the control volume 4 The conservation equations must be written using rate form Only then can the numerical values be substituted into the equations and to solve for the unknowns Example The coating on a plate is cured by exposure to an infrared lamp providing a uniform irradiation of 2000 Wmz It absorbs 80 of the irradiation and has an emissivity of 05 It is also exposed to the surroundings at 30 C and toran air ow at 20 C with a convection coef cient of 15 WmZK Determine the cure temperature of the plate Known a 08 a 05 Schematic G 2000 Wm2 c067 T h 1W 27330303 K qirr or G qconv 00 30 on Find TS Note We must use K for radiation Assumptions SChema Ci R Steady state 39 l I Neglect heat transfer to the back 39 39 V qry T h gm 0139 x goonv 00 9 Analysis Energy balance per unit area ESt Ein E0uz Eg a0 qzad qzonv 800 11 M 400 08x 2000 O5gtlt 567gtlt1084 3034 15TS 293 Solve by trialand error I 377 K 104 C Discussion What can we do to decrease T S Home ME 3345C Heat Transfer Spring 2006 12606 Objective 1 Introduction to Heat Transfer from Extended Surfaces hTS 4 TOO 7 qconv Asurface There are two Ways to enhance heat transfer Without increasing the temperature difference 1 Increase h by strong forced convection use fan use water instead of air spray or inject water etc 2 Increase the surface area A Radiator household heating Mobile Pentium Processors Simple Structures c Pin fir We will perform the analysis for simple cases and discuss engineering methods to deal withvcomplicated geometry a Rectangular n 13 Pin n a if v ins of Uniform Cross Section Analysis of Heat Transfer Enhancement The application of extended surfaces for heattransfer enhancement must be carefully considered proce sSes induces additional manufacturing 39costs and COmplexity Thus we must nd a way to quantify the added bene ts of using extended surfaces to justify their application A Determine the rate of heat transfer from an extended surface Involves nding the temperature distribution in the n structure B De ne some measure of ef ciency for extended surfaces Use this as a basis for determining when to use them 1D Temperature Distribution Heat Diffusion Equation P seudo 1D steadystate V Qx dx wi qx 13 1s because we have i included the convection qcom boundary in the control Volume i dT quotkA qx C d P Perimeter Ac Crosssectionaliarea kA 7 hPT T 3900 Constant k uniform crosS section 2 Solution of jzmze 0 dx 2 Linear homogeneous second order differential equationWith constant coef cients gt Tx Clem Cze39mx T00 g Need Boundary Conditions to s01ve for temperature distribution 39 H I 9x Clem Czefm c 1 Atvbase of n T b i N r 90 9t C1 2 At n tip A Convection at the tip surface h 6L kd 9dx at X L B Adiabatic tip dHdx 0 at X L 39 ti tem erature 6 6 atxL D In nite n L gt oo 6L 0 at x L X ms Tatum imsikxxtign an hea atssrfor af Warmamw a siiim 39 f isinh 4 v iiiz G 7 m 39w39hmag 93w 269 4 175 I M 39V PW 235 13 380 Rate of Heat Transfer from Fin qf Tb T00 Example A rod of diameter D and thermal conductivity k protrudes from aifurnace Wall that is at temperature TW The initial length of the rod Limp is insulatedwhile the remainder is exposed to convective heat transfen 39 AssumeaCOnvective heat transfer coef cient of h and temperature ToO Find an a 39 expression for the temperature of the rod at the insulation surface and the rate of heattransfer fOr the n h T O I A n 9 9 II ins Adiabatic Use and electrical resistance analogy Rins Lins k Ac 1 R n eb qf tanhmLothAc From conservation of energy W R n Rins R n ms 00 R n And the rate of heat transfer for entire n structure is given by Too h qx qxdx qconv d kAcflI hPT Too dx 39 39 dx What is n heat transfer rate 39 e What is the excess temperature P Perimeter 1906 TU quot Too Ac Crosssectional area M For constant k and uniform crosssection 0x Clemx Cze mx At the base of n 90 0 Tb T00 gt 0b C139 C2 At n tip A Convection at the tip surface B Adiabatic tip 7 C Prescribed tip temperature T L DIn nite n L 00 I theat transfer rate 139 def i1 qf qbc0nd kAcb q f geonv hT ToodA hedA I f I f qconv Af Af Raf 9b qf qf Rtf Tb M T00 Fin thermal resistance 0 0 Thermal Conductivitv Affects Heat Transfer Rate Assume in nite cyIindrical n with h 100 wmZK D 5mm 9b 75 C Copper Aluminum 39 SS 39 k WmjK 400 180 14 qf W N 83 39 56x 16 106 Tm Too Tb Tmgte quot 88 2 QC 339 mm Performance of a Fin qCOI lV a a With n b Without n qf Fin effectiveness 2 11bi Fin ffectjiveness if We may also de neeffectiveness in terms of thermal resiStance quot Rtaf and Rtab V qf hAcb Rtb it moot be greater then i for the fin to oe 8 f Raf For in nite n with uniform cross section The larger the thermal conductivity the better the nperformance i We may de ne theymaximum rate of heat transfer from a n an in nite thermal Conductivity by 39 39 quot 39 qmax This lead to a de nition for Fin Ef ciency 1n e 1c1ency n 0 lt n lt 1 f gmax Rtf f The de nitions of n effectiveness and ef ciency apply to ns of nonuniform crosssectional areas as Well In practice we generally use an arrayef ns A b allarea NOT alcovered by the n or area of un nned base 39 SQVe rall HeatTransferRate a quot39 qt hAbeb h NAfn fab Af n surface area A 7 Overall Surface Ef ciency Nnumber of n s qt F MAI nfNAfeb Tl0 qmax hAbNAfeb Moe ME 3345C Heat Transfer Spring 2006 11706 Objectives 1 1D Heat Conduction in a Plane Wall 2 Electrical Resistance Analogy 3 Interface Resistance Steady State Temperature Distribution in a Plane Wall System 1 With Uniform Volumetric Energy Generation 6 02 631 q 1 57 1 2 2 2 gt 39 0 6x 6y 2 k k at 5x2 Solution q 2 Tx x CxC 2k 1 2 2 What if there is no heat generation Tx Clx C2 Note The form of the equation giving the steady state temperature distribution is independent of boundary conditions However the Temperature magnitude does depend on BC s 1 Prescribed Temperature at Boundary e g T 4 To 2 Uniform Heat Flux at boundary q qo 3 Adiabatic at Boundary dTdX O 4 Convection and Radiation at Boundary k hT Too 80T4 711r We may also use an electrical resistance analogy to help solve heat conduction problems Electrical Analog of Heat Conduction V A U n e gt I kA AT IVR qx AT L Rm Electrical resistance Thermal resistance L L R Rt CA CA In order to develop the electrical resistance for each mode of heat transfer we must look at the linear heat transport coef cients M V 1 L 39 AT gt R 1 Conduction q L m M no energy generation 2 Convection q hA AT 3 Rm 2 i 11A 1 3 Radiation 6 hrA AT gt Rm Heat Transfer Through a Plane Wall Cold Fluid Thermal Circuits qx E where Rm 1L is the total thermal resist Rm hlA kA A 39 qx UAAT Where U RWAY1 is the overall heat transfer coe Thermal w Contact Resistance 5 Velry Small TA TB I x RM RZCA Ric m2 40W qquotx Contact resistance values range from Umt Area 10 6 10 3 mZKW depending on 1 materials 2 surface roughness and 3 contact pressure Effective thermal conductivity kc 8 c Example A composite wall withcontact resistance R What is the heat ux if the temperatures T1 and T2 are known 7 W qxquot Answer 61521114 tot A B 2 Ti Tz LA LB LAkARgLBk3 kA 1 3 Which material has the larger thermal conductivity A or B Some Comments Regarding Contact Resistance Tables 31 and 32 givcfvalues forrsome solidsolid contact resistance Types of contact metalmetal metal crystal etc Interfacial materials vacuum air soft metal solder grease plastic etc Contact resistance is often important in practice but only empirical theories exist w V1 fow 7 quotM a wf 30247 M magi 39 M5quot 3quot x 7 IW i WM 0 we 41 klt a l i k 7 5300 j5 BOMAfZK i c 90 m ack 15 C DZ OWW K ru 0 0 jglt n tiaz K o 1 Z A 00 C a 1239 7 GS ghgbgg T 739 7 7 Kg 9 gr a I if kve i T 27730 a 4 L r MM 2 27 3396 L U V l Vat Jat 7 f 7gw I M6 3 AT g E W

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