Question

2) (35 pt) Atmospheric air at 25 Centers a duct having a cross section of an equilateral triangle 2 cm on a side with a velocity of 1.5 m/s. The length of the duct is 10 m. A constant heat flux is imposed on the duct wall. The mean temperature of the air at the duct exit is 75 °C. a. Is the flow at the duct exit hydrodynamically and thermally fully developed? Show. T-75 °C b. Calculate the convection heat transfer coefficient in the duct. c. Evaluate the duct wall temperature at the exit. T-25°C 1.5 m/s Note: Area of equilateral triangle: A = 42 constant Ls 10 m 2 cm

Answer 1

Ti T. V-1.5 ms 25°C Te = 75°c evaluate the properties of air at bulk mean Temperature To Iit te 25+75 2 50°c 3 P= 1:0925 kg/m3 V = 3.794x10-5 m/s ep= 1807 lkyk pr=0.71 W mk K=0.0278 . enoss section area of the pipe (Ac) 3 2 a - Ą = 4 Vi fix (0.02)2 m² A = 1.732x10-4 ma Perimeter (P) = 3Q = 3X0.02 = 0.0CM hydralic Diameter (0) 4 de 4x 1.732 x10-4 3 Pn P 0.06 D 0.0115 m h

Reynold number (Re) v Dn 1.5 X 0.0115 Re- = 961:53 1.799 x 16 Re=961.53 <2000 flow is laminar entry lenth @ hydrodynamie a La Oru5 Rex De as = 0.05 x 961.53 X 0.0115 0.5 52 m thermal entry lenth 0,552 70.71 -0.392m 4 Lexpr are Both Lh 6. Both Lh and hz Less than the leath of the pipe (L=10m). .. flow is fully Developed at the Duet enit.

D Nusselt number is constant in ease of fully Developed laminar flow. For constant heat flun and triangular enoss section with Goo angle NU: 3.11 109 coo Goo or Nu= h Dn k 3:11 0.0278 hak W x 3.11 m²K x 3:11 = coa 0.0115 h=7.518 W/m2K | (Ans) for heat flun is mep (Te-T;) = Ås xas Now mess flow rate (m) in= PXA, XV =li0925 X 1.732 x 100 -4x1.5 Kg/

or ñ = 2-838 X10-4 Kyls 4 2.83% X10-4X 10607* (75-25)= 95 * PXL 2.838X10 -4 x 1007 X50 =) 다 2 0.06 x 10 => 1 4s= 23.815 W/m2 at enit g= h (Ts - Te = > 23.815 = 7.518 ( TS-75) -> Is =75 = 3.16 this is the TS = 78.16 Duet wall Temperature at enit