Frst CIRCLE YOUR DIVISION: Dv. 1 (9:30 am) Dv. (11:30 am) Dv. 3 (:30 m) Prf. Ruan Prf. Na Mr. Sngh Schl f Mechancal Engneerng Purdue Unversty ME315 Heat and Mass ransfer Eam #3 Wednesday Nvember 17 010 Instructns: Wrte yur name n each age Clsed-b eam a lst f equatns s gven Please wrte legbly and shw all wr fr yur wn beneft. Wrte n ne sde f the age nly Kee all ages n rder Yu are ased t wrte yur answers t sub-rblems n desgnated areas. Only the wr n ts desgnated area wll be graded. Perfrmance 1 30 35 3 35 tal 100
Prblem 1 [30 ts] Frst Cnsder a well-nsulated cncentrc tube cunter-flw heat echanger used t heat cld water. Saturated steam at a temerature f sat cndenses n the uter tube at mass rate f m. Cld water enters the nner tube f dameter D and length L at a temerature f wth a mass flw rate m and an average heat transfer ceffcent h. he latent heat f varzatn f water s h fg and the average secfc heat f water s c. Assume that the nner tube has neglgble thcness. Neglect any surface fulng effects. (a) Fnd an eressn fr the utlet temerature f water n the nner tube. (b) Setch the temerature varatn f the tw fluds alng the length f the heat echanger and clearly label the temeratures. (c) Fnd an eressn fr the effectveness ε f the heat echanger nly n terms f sat and. Wrte an eressn fr NU n terms f ε. (d) Eress the average heat transfer ceffcent fr steam cndensatn h nly n terms f nwn quanttes and/r thse btaned n arts (a) and (c). Start yur answer t art (a) here [6 ts]: Assumng heat lst by cndensng steam t be equal t the heat ganed by cld water we can mh mc wrte: fg Outlet temerature f water: mh mc Start yur answer t art (b) here [6 ts]: Steam fg sat Water mh mc fg
Prblem 1 cnt. Frst Start yur answer t art (c) here [1 ts]: Fr cndensng steam n the uter tube temerature des nt change Ch m C steam Cmn herefre n ths heat echanger: Ch Cma andc c C mn mc Cr 0 Cma q q mc Effectveness: ; nwn frm (a) q C mc Fr 0 r ma mn h c sat C : 1eNU NU ln1 Start yur answer t art (d) here [6 ts]: r ln 1 1 " r " 1 Overall heat transfer ceffcent: Rf Rf ; r r ; A A A UA ha L ha 1 1 1 UA U DL ln 1 mc ; NU ln 1 U U h h C mc DL mn Heat transfer ceffcent fr steam cndensatn: h sat 1 ; U and h are nwn 1 1 U h
Prblem [35 ts] Frst Cnsder a lng n fn attached t a vertcal square ch (0 mm 0 mm). he ch has a unfrm surface temerature f 67C whle the surrundng stagnant ar s at 7C. (a) Calculate the average cnvectve heat transfer ceffcent (W/m -K) n the fn surface. Snce the surface temerature vares yu can use a reresentatve fn surface temerature equal t the average f the base and t temeratures n ths calculatn. (b) Calculate the average cnvectve heat transfer ceffcent (W/m -K) n the unfnned base. Cmment n whether t s clse t the value btaned n art (a) whch s an assumtn ften made n fn analyss. (c) Determne the ttal heat dssatn rate (W) frm the fn and the unfnned ch. At = 300 K ar = 15.89 10-6 m /s ar =.5 10-6 m /s ar = 0.063 W/m-K At = 310 K ar = 16.90 10-6 m /s ar = 4.0 10-6 m /s ar = 0.070 W/m-K At = 30 K ar = 17.90 10-6 m /s ar = 5.5 10-6 m /s ar = 0.078 W/m-K Start yur answer t art (a) here [15 ts]: b L b Average fn surface temerature: s fn 47 C s fn Fr the fn surface flm temerature: flm 37 C 310 K ar = 16.90 10-6 m /s ar = 4.0 10-6 m 1 1 /s ar = 0.070 W/m-K K flm 310 3 g s fn D Fr the cylndrcal fn surface the Raylegh number: RaD 1.48 hfnd 0.148 Usng able 9.1 NuD 1.0RaD 1.48 W Average cnvectve heat transfer ceffcent n the fn surface: hfn 0 m-k 1
Prblem - cnt. Frst Start yur answer t art (b) here [10 ts]: b Fr the unfnned base flm temerature: flm 47 C 30 K ar = 17.90 10-6 m /s ar = 5.5 10-6 m 1 1 1 /s ar = 0.078 W/m-K K flm 30 g 3 b L Fr the unfnned base the Raylegh number: RaL 149 hbasel 1/4 NuL 0.59RaL 7.15 W Average cnvectve heat transfer ceffcent n the unfnned base: hbase 9.95 m-k he tw cnvectve heat transfer ceffcents are dfferent snce the develment f the free cnvectn bundary layer s nt the same n cylndrcal and vertcal surface. Start yur answer t art (c) here [10 ts]: Heat transfer rate thrugh the fn: qfn hfnpfn A c b P 3 D6.8 10 m q fn 0.504 W Ac 4 6 D 3.14 10 m b b 40 K qbase hbase Abase Ac b ; A base 400 10 m qbase 0.1580 W tal heat transfer rate thrugh the ch: qch qbase qfn.e. q 0.66 W Heat transfer rate thrugh the unfnned base: ch 6
Prblem 3 [35 ts] Frst A flud s ntrduced at flw rate m and nlet mean temerature m thrugh a crcular tube f dameter D and length L. he tube s dvded nt tw equal sectns. he flud s subjected t a cnstant heat flu qs '' afr the frst half whle the tube wall f the secnd half s mantaned at a cnstant surface temerature. It s fund that the utlet mean temerature m s the same as the nlet mean temerature m. Assume that the flw s turbulent and fully develed everywhere nsde the tube. c and f the flud are nwn. Yu may treat the quanttes btaned n earler arts as nwn arameters n the later arts. (a) Recmmend crrelatns fr the cnvectve heat transfer ceffcent h nsde the tube fr the tw halves f the tube. (b) Derve an eressn fr the mean flud temerature m () alng the frst half f the tube. (c) Derve an eressn fr the tube surface temerature s () alng the frst half f the tube. (d) Obtan an eressn fr cnstant wall temerature sc fr the secnd half f the tube. (e) Obtan an eressn fr the mean flud temerature m () fr the secnd half f the tube. (f) Qualtatvely setch the varatns f m () and s () fr 0 < < L. Start yur answer t art (a) here [8 ts]: Fr turbulent fully-develed flw (ether cnstant flu r m cnstant temerature): m hd 4/5 n NuD 0.03ReD Pr Fr the frst half flud s beng hd 1 4/5 0.4 heated: 0.03ReD Pr Fr the secnd half flud s beng hd 4/5 0.3 cled: 0.03ReD Pr umd m D 4m where ReD ; Pr D D 4 Start yur answer t art (b) here [6 ts]: Cnsderng energy balance fr the frst half: Integratng: m d ap m mc m 0 qs " a C C " apd mc dm qspd dm mc ap d m m ; P D mc Cnstant wall temerature L/ L/ m = m
Prblem 3 - cnt. Frst Start yur answer t art (c) here [4 ts]: " Cnsderng heat flu at any sectn: qs h1 s m s m h 1 frm (b) and (a) resectvely Start yur answer t art (d) here [6 ts]: Fr the secnd half: sc m Ph L e L mc sc m L ap L where m m P D and h frm (a) mc Start yur answer t art (e) here [5 ts]: sc m a ; m () and h L Ph L m e mc Ph L 1e mc sc m Ph L Fr the secnd half: e L mc sc m L Ph L ms c s c m e ; where P D mc sc and h frm (d) and (a) resectvely Start yur answer t art (f) here [6 ts]: s 1 m m m m m sc