Question

Air at a pressure of 1 atm and a temperature of 60 °C is in parallel flow over the top surface of a flat plate that is heated to a uniform temperature of 194 °C. The plate has a length of 0.30 m (in the flow direction) and a width of 0.20 m. The Reynolds number based on the plate length is 60,000. 1) Draw an annotated schematic of the problem. 2) What is the rate of heat transfer from the plate to the air? 3) If the free stream velocity of the air is doubled and the pressure is increased to 10 atm, what is the heat transfer?

Answer 1

Date: Page no: Soln Given :- Ambient ait tembezature, Too = 60°C uniform temperature of Plate Plate, Is=194 °C. Reynold's Numbes , Rei = 60,000 = 6 X 104 length of the plate , L = 0.30 m width of the Plate W = 0.20m CAIR) (a) Schematic : -Ts = 194°C loo, ReL = 6X 104 7 WE 0-20 m Too = 60°C P=Latm L=0.30m (6) Assumptions:- (1) steady-state Conditions 2) uniform surface tempezatuel (3) Negligible Radiation. (4) Rexe = 5x10 Properties :- Te - Too + Ts - 60+194 – 400K 2 2 Exom table A-4, Ait (T9= 400K, Latm); k = 0·0338 W/mok Pr=0.690 Analysis :- The heat rate is, q=hi CWXL) (Is-Too)

Dute: Page no Since the floue islaminać over the entire plate for ReL = 6 x 104 it follows that, ILL 1/2 4/3 0 664 ReL PE = 0.664C60,000342606905 NUL = K NU = 143.7 Hence, L hi = 143- = 143.7k 143.7 0.0338 W/m.k 0.30m = 16.19 W/m²K The = and q = 16 •19 W (0.2m x 0.3m) (194-600°C m²k q = 130.2 W (c) with P2 = 10 P1, it follows that P2 = 10 PL and V2 = 1 10 Hence, Re4,2= ( W4) = 2x20 (ML) Reaa = 20 Rew = 12X105 And mind boundary layet conditions exist on the plate Hence, NUL=hl fuL - (0.037 R25-871) P& NUL = [0.037 x (12x105,475 -871] co.690) All = 1617 Hence, hi = 1617 0·0338 W/mok 182.2 W/m²k. 0.30m q= 182.2 W (0.2m x 0.3m) (194-60) °C = 1464.9 W m²k ORlq=1465 W