### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ProcessHeatTransfer CHE303

Drexel

GPA 3.96

### View Full Document

## 45

## 0

## Popular in Course

## Popular in Chemical Engineering

This 40 page Class Notes was uploaded by Ms. Adaline Stark on Wednesday September 23, 2015. The Class Notes belongs to CHE303 at Drexel University taught by VibhaKalra in Fall. Since its upload, it has received 45 views. For similar materials see /class/212543/che303-drexel-university in Chemical Engineering at Drexel University.

## Reviews for ProcessHeatTransfer

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/23/15

Heat TransferReview Lecture CHE 303Fa112011 Vibha Kalra Outline I Introduction to Heat Transfer l Conduction Heat Transfer IntroductionHeat Equation 1D Steady State Conduction with and Without generation 2D Steady State Conduction Transient Conduction l Convection Heat Transfer l Heat Exchangers Heat Transfer and Thermal Energy cont Heat Transfer Thermal energy transport due to temperature gradients Direction of Heat Transfer is parallel to the temperature gradient direction Quantity Meaning Symbol Units 1 lczll Munster Thermal energy transport due to temperature gradients l Icul Amount of thermal energy transferred Q J over a time interval A t gt 0 lcul Rule Thermal energy transfer per unit time C W chl 1m Thermal energy transfer per unit time and q VVl39n2 surface area Modes of Heat Transfer d f f Conduction through a solid Convection from a surface Net radiation heat exchange or a stationaw fluid to a moving fluid between two surfaces TgtTM x Moving fluid T Conduction Heat transfer in a solid or a stationary uid gas or liquid due to the random motion of its constituent atoms molecules and or electrons Convection Heat transfer due to the combined in uence of bulk m id random mouun for uid ow over a surface Energy that is L mlllL39d h muIlL39l due to changes in the electron configurations of its atoms or molecules and is transported as electromagnetic waves or photons Radiation Conduction and convection require the presence of temperature variations in a material medium Although radiation originates from matter its transport does not require a material medium and occurs most efficiently in a vacuum Heat Transfer Rates Conduction Heat Transfer Rates Conduction General vector form of Fourier s Law 397 q kVT Yl chl llux lhcrmul conducln m lcmpcramuc gradient Wm2 Wm K Cm or Km Onedimensional steadV conduction across a plane wall of constant thermal conductiVit with no heat eneration 12 Heat rate W 9x q 39A Heat Transfer Rates Convection Heat Transfer Rates Connection Velocity distribution distribution My TO E v HEated 39 surface Iewlon s law of cooling qrrh7 TOO h Convection heat transfer coef cient Wm2 K 13a Heat Transfer Rates Radiation Heat Transfer Rates Radiation Heat transfer at a gas surface interface involves radiation emission from the surface and may also involve the absorption of radiation incident from the surroundings irradiation G as well as convectionif TS 2 TOO Energy out ow due to emission E gEb 8074 15 E Emissive powcrWm2 5 Surface emissivity O S 8 31 Surface of emissivity Eb Emissive power of a blackbody the perfect emitter and a StefanBoltzmann constant 567X10 8Wrn2 K4 1 Energy absorption due to irradiation Gm 05G Ga A bsorbed incident radiation Wmz a Surface absorptivity 0 S a S 1 G Irradiation Wmz Heat Transfer Rates Radiation Cont Heat Transfer Rates Irradiation Special case of surface exposed to large surroundings of uniform temperature TSW 4 G GSMV 01110 Gas TN 1 Surroundings at T sur liiad ilqgonv Surface of emissivity e or area A and temperature T b If a 8 the net radiation heal ux from the surface due to exchange with the surroundings is grid 8E aG 80114 71 17 ConductionIntroduction Fourier s Law of Conduction Cartesian Coordinates Txyz 671 6 6 q k i k j k k 23 3x k 6y 32 615 61 q Cylindrical Coordinfiles Tr z gt T gt T gt 671 q ka z ka Jk k 23922 6r 75 62 J ell 61 612 Spherical Coordinates Tr 9 6T 7 3T 7 3T gt k k 225 6r r66 1 krsm66 j Heat Equation The Heat Equation 0 A differential equation whose solution provides the temperature distribution in a stationary medium Based on applying conservation of energy to a differential control volume through which energy transfer is exclusively by conduction Cartesian Coordinates i kalj3 kal 2 ljqucpal 217 0x 0x 3y 3y 02 02 at 4 cl lrunsl39u of lhemml energy into the Thermal energy Change in thermal control volume in owout ow generation energy Image Heat Equation Special Case OneDimensional Conduction in a Planar Medium with Constant Properties and No Generation aZT 1 6T 5x2 a at k E gt thennal dlffllSlVlt of the medlum pcp Boundary Conditions Boundary and Initial Conditions For transient conduction heat equation is rst order in time requiring speci cation of an initial temperature distribution Txtt0 Tx0 Since heat equation is second order in space two boundary conditions must be speci ed Some common cases 39onstanl Surface Temperature T OJ TS Constant Ilczu FIUV Applied Flux Insulated Surface 6T k a lx0 qs 6T k 2 Too T 0 ax ix O ConductionOneDimensional SteadyState Without Thermal Energy Generation Methodology ot a Conduction Analysis Governing Equationgt Specify appropriate form of the heat equation Specify Boundary Conditions Initial Conditions Solve for the temperature distribution 0 Apply Foun er s law to determine the heat ux Simplest Case OneDimensional SteadyState Conduction with No Thermal Energy Generation PlaneWall Consider a plane wall between two uids of different temperature Heat Equation d dT k 0 dxi dx 3391 Hot fluid Tm 1 hl Cold fluid Tw z I12 Implications Heat ux is independent of x Heat rate qx is independent of x Boundary Conditions TOTs 1 TLTs 2 0 Temperature Distribution for Constant k x TXTsi Ts2 Ts1z 33 16 Plane Wall cont Heat Flux and Heat Rate dT k qu kngasJ ng 35 dT kA Thermal Resistances R and Thermal Circuits q L Conductlon 1n a plane wall Rtgcond E 36 l Convectlon Rheum g 39 Thermal circuit for plane wall with adjoining uids Too 1 dr 393 w1339V 39 1 m m a h A 1 2 312 Too 1 Too 2 CI x Rmt l Composite Wall with Negligible Contact Resistance Plane Wall cont Hi at ut uid 39 39 I Too1 Too4 mm 1 qx ZRI Guld uid In ll 1 hl kA kB kC h4 ZRZZRIOIZZ A Overall Heat Transfer oe iciem U A modi ed form of Newton s Law of Cooling to encompass multiple resistances to heat transfer qx overall l7 1 RV a 319 18 Heat Transfer from Extended Surfaces I To improve heat transfer from a solid to an adjacent uid by increasing the heat transfer surface area l 4 special conditions at the fin tip see Table 34 for temperature distribution and heat transfer rates convection heat transfer Adiabatic tip ie heat transfer0 Fixed temperature at the tip In nite Length Fin ie Fin tip tempTOO l Fin performance de ned as the ratio of heat transfer rate with n to the heat transfer rate that would exist Without the n 20 ConductionOneDimensional SteadyState With Thermal Energy Generation The Plane Wall The Plane Wall Consider onedimensional steadystate conduction in a plane wall of constant k uniform generation and asymmetric surface C inditions 0 Heat Equation 2 I ilr qO gtOIITZO 339 dx dx a x2 k Is the heat ux q independent of X General Solution Tx asz x2 Clx C2 What is the form of the temperature distribution for q 0 q gt 0 q lt 0 How does the temperature distribution change with increasing 22 Conduction Unsteady State or Transient conductionrefer to supplement notes 23 Convection Heat Transfer stationary solid and moving uid Forced convection when the uid motion is externally created Natural or free convection when uid motion is solely due to temperature gradient that causes density differences Boiling and Condensation associated with phase change could be with or without forced convection 24 Boundary Layer Features Boundary Layers Physical Features Velocin Boundary Layer when a moving uid hits a stationary solid A consequence of viscous effects associated with relative motion between a uid and a surface Vdocny boundaw layer A region of the ow characterized by velocity gradients with no slip at the boundary between solid and uid A region between the surface and the free stream whose thickness 0 increases in the ow direction Why does 6 increase in the ow direction 25 Boundary Layer Features com flicnnal Boundary Layer when a moving uid at a temperature TOO hits a solid at temperature Ts such that A consequence of heat transfer between the surface and uid A region of the ow characterized by temperature gradients and heat uxes A region between the surface and the free stream whose thicknesst TS TOO 03999 increases in the ow direction Why does 6t increase in the q 2 k ow direction S f 5y F0 If TS Tw is constant how do q and kfaTayy0 h vary in the ow direction T T 26 Local and Average Coefficients Distinction between Local and Average Heat Transfer Coef cients local lleat Flux and Coef cient q39h7 Too 0 Axerage lleal Flux and Coef cient for a Uniform Surface Temperature q 51301 Too q Law15 7 Tm AS hdAS 1 h XjAshdAs S o For a at plate in parallel ow Ezljjhdx L Laminar to Turbulent transition for ow on a at plate Transition criterion for a at plate in parallel flow Re critical5X105 x i Rex C E amp gt critical Reynolds number Note that the critical Reynolds number or ow inside a cylindrical pipe is 2300 and Re number de nition is based on the diameter of the pipe 28 Refer to supplement notes for more on convection and boiling Heat Exchangers Types Heat Exchanger l ypes They involve heat exchange between two uids separated by a solid and encompass a wide range of ow con gurations ConcentricTube Heat Exchangers l39urullcl l on fnunlcrllou gt Simplest con guration gt Superior performance associated with counter ow Types com Tube outlet Baffles Shequot Tube outlet inlet gt The number of tube and shell passes may be varied eg I lSheH inlet iSheH Inlet Tube outle Tube inlet LMTD Method A Methodology for Heat Exchanger Dcsign Calculations The Log Mean 39l39emperature Difference LMT D Method Aform ofNewton s Law of Cooling may be applied to heat exchangers by using a logmean value of the temperature difference between the two uids q 2 U A A T1 Equation 8 in notes ATM ATI AT2 1nAT1AT2 Evaluation of A I and A T 2 depends on the heat exchanger type 0 CounterFlow Heat Exchanger C a Th dq 33 a N Heal transfer 39 3quot surface area A T1 E Tm Tm Jim 110 A T2 E Tm Tc2 lm T ci LMTD Method com ParallelFlow Heat changer A T E T T masses 1 m a1 111 c z A T2 E Th2 Tc2 110 00 gt Note that Tao can not exceed T ho for a PF HX but can do so for a CF HX gt For equivalent values of UA and inlet temperatures AJlmCF gt A YlmPF Overall Energy Balance Assuming constant speci c heats q mh CM TM T1170 Ch Thai T1130 Equation 1 in notes q ma cm Tm Tm CC T630 Tm Equation 2 in notes C Cc gt Heat capacity mles minions l 2 and 8 can help conduct heat exchanger analysis using log mean temperature method Special Conditions Special Operating Conditions Ch gtgt 7 Ch ltlt Cror or a condensing an evaporating vapor C oo liquid Cs oo gt Case a ChgtgtCc or h is a condensing vaporCh gt oo Negligible or no change in Th TM gt Case b CcgtgtCh or c is an evaporating liquid Cc gt oo Negligible or no change in T0 Tm 2T gt Case c Cch A11AT2ATM General Considerations General Considerations 0 Computational F eaIuresLimitations oftlic LMTD Method gt The LMTD method may be applied to design problems for which the uid ow rates and inlet temperatures as well as a desired outlet temperature are prescribed For a speci ed HX type the required size surface area as well as the other outlet temperature are readily determined gt If the LMTD method is used in perlbmiance calculations for which both outlet temperatures must be determined om knowledge of the inlet temperatures the solution procedure is iterative gt For both design and performance calculations the effectivenessNTU method may be used De nitions Et t ectivenessNTU Method De nitions 0 Heat exchanger effectiveness 8 8 761 qmax Oltg 1 0 Maximum possible heat rate qmax ijn ChifChltCc c 2 111111 or C 1ch ltCh C gt Why is CW and not C used in the de nition of qmax m CDC De nitions com Number ot Transt er Units Nl39l NTU E UA C min gt A dimensionless parameter whose magnitude in uences HX performance q T with T NTU Two types of problems Type 1 Design Problem Where inlet temperatures are known and q and outlet temperatures ate prescribed Here we need to nd the appropriate length of the heat exchanger to enable the prescribed heat exchange 1 First nd a L and CminC max from given information max 2 Find NTU from correlations in Table 114 3 FindA and L using NTU E g A min Type 2 Performance Calculation of an existing heat exchanger Here the inlet temperatures are known You already know the length and area of the heat exchanger You need to evaluate the heat transfer and the outlet temperatures 1 First nd NTU Eg A and C min C from given information 2 Find from correlations in Table 113 q 3 Then f1nd q using 8 q Then from q f1nd outlet temperatures max Extra Problemssee supplement notes

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.