Question

Forced air at T = 25°C and V = 13 m/s is used to cool electronic elements on a circuit board. One such element is a chip, 4 mm x4 mm, located 120 mm from the leading edge of the board. Experiments have revealed that flow over the board is disturbed by the elements and that convection heat transfer is correlated by an expression of the form Nuz = 0.04 Re0.85 Pp1/3 V.T. 3 V,T 1 = 4 mm Chip Board L= 120 mm Estimate the surface temperature of the chip if it is dissipating 30 mW. T, = i

Answer 1

The =25€ V=13m/s - L=4mm - L=4nom L=4 oom - B 1 = 0.004 m yx L=120mm Nur=0.04 Rege Pion Pos og eo. San 1- 120 m 1000 L= 0.12om

Assumptions :- #steady - state conditions * poures dissipated within chip i8 lost by convection across the uppes surface only. * chip susface is isothermal from ais properties table assume surface tems fesature To = 45c, * = 25°c (Giver) Means tenspersature (TE) = 45+25 = 35€ = 35+273K (Tp)= 308 k, latus y = 16. 69 X106 milo K=26998 103 wlmak Pos = 0.703

cony 30mW From an energy balance on the chip - Vony = Ég = Bon Ég = 30x 10"W Newton's law of cooling for the uppes chip Surface can be writters as To = The + Neonr ① b x Acer We know that - he e = 0.04 Rexe Pos substituting x=L on 0.85 1/3 Nuse k We know 0.85 % 0.85 1/3 Nese = bx L =0.04 (VL 2009 Shugo he = 0.04 K (ULP hx = 0.04 269X10(13 X0.120) (0.703) hoe = 133.87 Wilmek o 16-698106 Assume me that the average heat transfer Coefficent (h) over the chip susface is equivalent to the local coefficient equated at x=L. that is hchip = h (L)

from o a conv otot e Tg = The t r x Achip Tg = The + von ъх. Tg = 25 + 30x10° 133.87 x (0.004) To = 39.006 °C