{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

assignchap4-46

# assignchap4-46 - 3 l 5 PROBLEM 4.46 KNOWN Nodal...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3: _ l 5 PROBLEM 4.46 KNOWN: Nodal temperatures from a steady—state, ﬁnite-difference analysis for a one—eighth symmetrical section of a square channel. FIND: (a) Beginning with properly deﬁned control volumes, derive the ﬁnite-difference equations for nodes 2, 4 and 7, and determine T3, T4 and T7. and (b) Heat transfer loss per unit length from the channel. I q . SCHEMATIC: run: 300 K, h = 50 Wlmz- K _d_11q_'1_ _2T9'g _ 3Tq_'3_ _4lq’4 Node WC) 1 430 i Symmetry adiabat 3 394 5 t, 7 / 3‘ 6 492 l C><olr=1 Wim-K VI 8’9 (’00 r I Ax = a .01 m I 6. qu \é 6| y: X _?_ ASSUMPTIONS: (l) Steady~state conditions, (2) TWO-dimensional conduction, (3) No internal volumetric generation, (4) Constant properties. ANALYSIS: (a]Def'1nc control volumes about the nodes 2, 4, and 7, taking advantage of synunetry where appropriate and performing energy balances, Em e Bout = 0 , with Ax = Ay, Node 2: q; +q’rJ +012; WEI =0 T3 — To hAx(T,,°—T2)+k(Ay/2) A “+kAxT6_T2 x T3 {0511+ 0.5T3 + T6 + (naming/[2+ (hAx/k)] T3 =l:0.5x430+0.5><394+492+ (SOW/mz-KXODIm/lW/m-K)300]K/[2+0.50] < Node 4: q; + qij +qé = 0 T3 —T4 Ax T4 = [T3 + (hAx/k)r,,, ]/[1+ (hAxfkﬂ T4 =[394+0.5x3001K/[1+0.5]:363K < =0 h(Ax/2)(Tw —T4)+O+k(Ay/2) PROBLEM \$4.46 (Cont) K Symmetry plane Node 7: From the ﬁrst schematic, recognizing that the diagonal is a symmetry adiabat, we can treat node 7 as an interior node. hence T7 = 0.25(T3 +T3 + T6 +T6 ) :0.2s(394+394+ 492 + 492)K = 443K < (b) The heat transfer loss from the upper surface can be expressed as the sum of the convection rates from each node as illustrated in the ﬁrst schematic. CIEV =cti+ci'2+q:2+q'4 qgv = h(Ax/2)(T1-Tm)+ 1mm}; —rm)+ 1mm “Too )+ h (Ax/2)(T4 4m) qgv = sow/m2 -Kx0.1m[(430~300)/2+(422— 300)+ (394—300)+ (363—300)/2] K qgv =156W/m < COMMENTS: (1) Always look for symmetry conditions which can greatly simplify the writing of the nodal equation as was the case for Node 7. (2) Consider using the IHT Tool, F inite-Diﬂcrence Equations, for Steady—State, Two-Dimensional heat transfer to determine the nodal temperatures T. - T; when only the boundary conditions T3. T9 and (Tmh) are speciﬁed. ...
View Full Document

{[ snackBarMessage ]}