Question

A hot brass plate is having its upper surface cooled by impinging jet of the air at temperature of 15 oC and convection coefficient of k=180, 190, 200, 210 220 and 230 W/m2 oK. The 20-cm wide and 10-cm length brass plate (ρ=8530 kg/m3, and Cp=380 J/kg K. k=220 w/mK had a uniform initial temperature of 650 oC, and lover surface of the plat is well insulated. Using uniform node for Δx = Δy = 2.5 cm and left side and right side of the plate cooled by natural convection of h=35 W/m2K and ambient temperature of 20 oC. Derive finite differential equation for all nodes at steady state.

We are suppose to write the equations for the 16 nodes that form in the grid, if someone could help me just showing the nodes, showing if I can apply similarity and writing a couple node equations so I can at least see how to start the problem, that would help. Thank you! I will like your work.

Air at 15°C h = 220 W/m2K L = 10-cm Insulation w = 20-cm

Answer 1

The detailed solution is given below.

Top 15°C hq=-220W/m%% 6 7 8 9 1 2 3 0 13 14 is 16 14 es D 0 22 21 23 24 25 24 27 1235W/mik To = 20°c 20 1 B D 19 h₂35w the To= 20°c 30 31 29 322 33 34 35 36 P 3 200 42 431 4 1745 38 41 39 40 37 Symmetly plane At steady state Sum of all heat your into a control Egin 20 Volume is equal to zelo

Terminology let An=oy = a *درة 14) - ذر - "Ag АкhЭ*1 An ya ", = = 9 Р Ay- Диx1 . 1 г Д д Thus, An= Ay= on=oy=a

Mode #1 и 0 h, Съю, - т )+ h,, Сте, -т) Е 2. + кө, C-) + кву Стрь-т)=0 оо Node #2 2. 3 ыс h, 4, СТ», -ъ) + KA, Cта --) + КА» СТ-1) + кв» ст. - 1)=0 Дж 1 are similah. Equations for node #384 Node #5 S 4 At a symmetry plane & Insulated can treat them as melhor boundary. Thus, boundary we Symmetry чо plore ічі КА» СТ-y+ K2 (СТ, ТО + клу CT, - со Дж

lo 11 Nocle #lo 1 p h₂ An (Tooz-Tio) + KAY (Ti-To) + Karya (Tig- Tia) + KANCTU-Zoe o sy Equations are for mode # 19 8 28 are similar Ah 19 - an 11 12 10 he can atta alo write, can be Mode #11 Kay (TA-Tv) + Kay(Teó-T2 + kan lio-T +KAn Tiz-Tai) -0 Since Ay= An (Alla, onasy) Į + Tie + Tio + Tio - 45120 Equations for all internal modes written as this one. (20, 11, 12, 13, 20, 21, 22, 29,30,31) Nele #14 AB can see this node can abo be treated as an interior node because of the miloh boundary condition. This, Tst Tist T23 t Tiz - 471420 we OS - 14 13 13 23 Ts + 27,3t T23 - 491470

are similar Equations for mode #23 6 32 Node #37 28 KAy CT2g-T37) + KAM CT3-737 AY than (Tø2-T37) =0 sn 37 38 2 Node #38 Insulated boundary can be treated as misoals 29 T2g + Tag + Tag + Toa - 4738 20 37 38 39 T2g + 2T3g+ T37 - 4738.50 Similare equations can be written for node # 39 & 40. 29

Node #41 4T41 zo 32 -040 bondaly Culing 2732 +2740 mirol for incubated & Symmetry boundary 40 3 2

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