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# Cal State Fullerton - EGME 407 - Class Notes - Week 1

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Cal State Fullerton - EGME 407 - Class Notes - Week 1

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##### Description: these notes covered the introduction to conduction, convection and radiation
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Unformatted text preview: introduction (week 1) Mon-day, January 23. 201? 11:53 .-'-‘.r-.-1 application of heat transfer: I double doors I double pain glass I built in coasters I all ofthese add air between for better installation Chapter Dne : Introduction I. Categories 1. conduction 2. connection 3. radiation |l. physicaloriginand rateequation Reason: temperature difference 1. Conduction "l— p. "— .2. heat transfers from hot to cold energy transfers from high to low l //l Rate Equation: 1—D Conduction: q“ : heat flux :1 : heat rate {Wat-t] W ' hermal conductivity, Material Proper-tar W- ' K) “Hi—(x) T05) :TCc) THE; ‘.I 11.1: gall—T" TJHT' a = w --*~ =- L : thickness {m} Example: wall of 0.15m thickness, k: 1.? wfmfk, steady state T,= moo k, 1,: 1150 k the wall is [3.5 H 1.2 rn on one side Find: q , T. “T: if?“ 1“: L Iii-no. — sca- __._— l.I_Il-.|i|'.gh?q( Isl-[Fa 2. Connection Mecha nism Diffu sio n M otion of fluid Rate Equation: _'!bl——~ h : convection heattransfer coefficient ( Mu ' #1) depends on the geometry and nature of the fluid T3 :surface Temp Tau:free stream temp 3. Radiation Energy emitted by matterthat is at a nonzero temperature {k} Ezcggi (“w—a E: heatflux emitted by a surface m E' : emissivity. radiative property ofthe surface (3 s E. s | T; :surface temp [K] S'?tefan-?p?man constant o' FEE—l “to”5 w/mT/it tantrum.- - a “35' F ifl E“ SItroumtl; " ' 7“? .3. — at 3absorbativity D 1-. 0‘» El. I! ~E—mQ 4L Net rate of radiation of the surface s d 5 (“if +T‘f1r>(T\$ 4 Thu} (TS- {gm} W r3: : ME to) hr: radiatio n heat t ra nsfe r coefficient Example: Steam pipe in a room T;.1,,=25'°C D=??rnm L=1m T: = zoo “c cf = o_a Find 1. E and G 2. if convective heat transfer coef?cient is h = 15 wy'm 11”: find total heat loss “Eelmt me 5 «a “f (f) Ewes—T3” = oi (so? 1155(300 + 97’5le E 2 1370 \$4.1 (3: g—TJH = 5.61xin’f(os+3a3‘)u 6: our] N/ma @1=%"a“= . w 4 m ‘l E 3 TM) = E -o4(':'. t h (T5 F = 63317:? so Mam) + I5 [mwm- (15+ mail = 14%? Li W/m1 3‘— gy?l 351=ITDL = ass“ from} x I) = “’I‘i‘E N 4. 1. Thermal Resistance [ UL o for conduction: T: H L Relation to therm odynamiu -u- ‘31— -:-__—— Hosp-o :253 '2 1;.- "thermal 1 —‘~. res.” 5.1%; E; a H H tit—I- f-U'FL lmih— lst Law lClosed System : Q - W “'- ?ll; that“: at olt: . I a lI13".ontro|".l'o|ume: 3— MN -I- “isms-Ev? +jE£>—m£(|1€ + [it‘ll 4.3%! _d E ,_.— alE Example: fuel cell generates = 11.25 W temp of the cell is 56.4%: E: must be remoyed by convection and radiation TM: as'o , h = ms 1: v9" wrmffk ‘v' is air velocity in mfs A = [3.005 m1 Find: ‘v' required to remove E5 steady state, and E. = 0.33 254% " 3hr: un?t M‘h‘l“ :C' ‘5 a. l—Et?m _ Eur-ml. 4' E3 ‘0 E3 :Emv * M1346,” End : 1 (fa—qu " ??tdl-rb?l :lE’TE 1' MT “(3'3 a: (Sin H +173? 11“ Ijm??rx , {451F113}? I .095 (bra-1f =0 "Fl yd ff.“ : E3 _ 6”“ “Ml (T: ‘Ts-m) lo: 1?; “3w: .3 H.115 -.:1.=ri fl (T: ’71.") was (Esq ‘35) l" 1 £15. “’I “Ni/ma?a en‘s 1.1110.qu ‘xi 26:95.“ ‘7“ gm wt. |ID."I 2. 2nd Law or thermodynamics T 7:“? 4i Chapter two: Conduction I. Rate Equation gi'f EK -— ‘li?? ? Eu : _k"A'.p[ti Fourier'sLaw kEJT '13? = F H

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