Known: long, aluminum cylinder acts as an extended surface.

Transcription

1 Prolem : A long, irular aluminium rod attahed at one end to the heated all and transers heat through onvetion to a old luid. (a) I the diameter o the rod is triples, y ho muh ould the rate o heat removal hange? () I a opper rod o the same diameter is used in plae o aluminium, y ho muh ould the rate o heat removal hange? Knon: long, aluminum ylinder ats as an extended surae. Find: (a) inrease in heat transer i diameter is tripled and () inrease in heat transer i opper is used in plae o aluminum. Shemati: T, h Aluminum Or opper D Assumptions: () steady-state onditions, () one-dimensional ondution, (3) onstant properties, (4) uniorm onvetion oeiient, (5)rod is ininitely long. Properties: aluminum (pure): k=40w/m. K; opper (pure): k=400w/m. K Analysis: (a) or an ininitely long in, the in rate is r M ( hpka ) ( hdkd / 4) ( hk) D 3 Where P=D and A =D /4 or the irular ross-setion. Note that D 3 /. Hene, i the diameter is tripled, (3D) ( D) And there is a 50 % inrease in heat transer.

2 () in hanging rom aluminum to opper, sine k /, it ollos that ( ) Cu k 400 Cu ( ) Al k Al 40.9 And there is a 9 % inrease in the heat transer rate. Comments: () eause in eetiveness is enhaned y maximum P/A = 4/D. the use o a larger numer o small diameter ins is preerred to a single large diameter in. () From the standpoint o ost and eight, aluminum is preerred over opper.

3 Prolem : To long opper rods o diameter D= m are soldered together end to end, ith solder having melting point o 650 C. The rods are exposed to air at 50 C ith a onvetion heat transer oeiient o 0W/m. K. Estimate the minimum eletrial poer input needed or the soldering to our? Knon: Melting point o solder used to join to long opper rods. Find: Minimum poer needed to solder the rods. Shemati: Copper D=m Juntion T =650 0 C Air T = 5 0 C H=0W/m. K Assumptions: () steady-state onditions, () one-dimensional ondution along the rods, (3) onstant properties, (4) no internal heat generation, (5) negligile radiation exhange ith surroundings, (6) uniorm, h and (7) ininitely long rods. Properties: opper T = (650+5) 0 C 600K: k=379 W/m.K Analysis: the juntion must e maintained at C hile energy is transerred y ondution rom the juntion (along oth rods). The minimum poer is tie the in heat rate or an ininitely long in,

4 (hpka ) (T T ) min sustituting numerial values, W 0 (π 0.0m) 379 min m.k thereore, min 0.9W W m.k π (0.0m) 4 (650 5) 0 C. Comments: radiation losses rom the rod are signiiant, partiularly near the juntion, therey reuiring a larger poer input to maintain the juntion at C.

5 Prolem 3: Heat transer enhanement is reuired rom a heated plane all. It is proposed that aluminium ins o retangular proile are attahed to the all. The ins are 50mm long, 0.5mm thik, and are eually spaed at a distane o 4mm. The in spaing is 50ins per metre. The heat transer oeiient assoiated ith the uninned plane all is 40W/m. K, hile that assoiated ith the in surae is 30W/m. K. Find the perentage inrease in heat transer resulting rom use o ins. Knon: Dimensions and numer o retangular aluminum ins. Convetion oeiient ith and ithout ins. Find: perentage inrease in heat transer resulting rom use o ins. Shemati: L=50mm N=50m - W=idth h = 30W/m.K(ith ins) h o =40W/m.K(ithout ins) Aluminium Assumptions: () steady-state onditions, () one-dimensional ondution, (3) onstant properties, (4) negligile in ontat resistane, (6) uniorm onvetion oeiient. Properties: Aluminum, pure: k40w/m. K Analysis: evaluate the in parameters

6 L L A L L 3/ 3/ p L t m 0.50 (h (h / KA / KA t m max / P ) / P ) it ollos that, 0.7h (0.0505m) hene L 3/ m W / m.k 0.05m ( 3 30W / m.k 40W / m.k 5.50 ) 6 m.6w / m / 6 m K( With the ins, the heat transer rom the alls is N ( Nt)h W ( ) (m m) 30W / m.k( m.k W ( ) ( ) 566 m.k Without the ins, h m 40. o Hene the perentage inreases in heat transer is o / ) ) o % Comments: I the inite in approximation is made, it ollos that = / (hpka ) = [h kt] / = (30**40*5*0-4 ) / =.68. Hene is overestimated.