Question

A device that recovers heat from high-temperature combustion products involves passing the combustion gas between parallel plates, each of which is maintained at 350 K by water flow on the opposite surface. The plate separation is 80 mm, and the gas flow is fully developed. The gas may be assumed to have the properties of atmospheric air, and its mean temperature and velocity are 1000 K and 60 m/s, respectively.

If a third plate, 20 mm thick, is suspended midway between the original plates, what is the surface heat flux for the original plates? Assume the temperature and flow rate of the gas to be unchanged and radiation effects to be negligible.

Co Ta 100 (mºis) 2.54 5.84 10.3 0.786 0.758 0.737 0.720 0.707 0.700 15.9 22.5 29.9 TABLE A.4 Thermophysical Properties of Gases at Atmospheric Pressure u. 107 v. 106 k. 103 (K) (kg/m) (kJ/kg K) ( Ns/m?) (m/s) (W/mK) Air, M= 28.97 kg/kmol 100 3.5562 1.032 71.1 2.00 9.34 150 2.3364 1.012 103.4 4.426 13.8 200 1.7458 1.007 132.5 7.590 18.1 250 1.3947 1.006 159.6 11.44 22.3 300 1.1614 1.007 184.6 15.89 26.3 350 0.9950 1.009 208.2 20.92 30.0 400 0.8711 1.014 230.1 26.41 33.8 450 0.7740 1.021 250.7 32.39 37.3 500 0.6964 1.030 270.1 38.79 40.7 550 0.6329 1.040 288.4 45.57 600 0.5804 1.051 305.8 52.69 650 0.5356 1.063 322.5 60.21 700 0.4975 1.075 338.8 68.10 750 0.4643 1.087 354.6 76.37 800 0.4354 1.099 369.8 84.93 57.3 850 0.4097 1.110 384.3 93.80 59.6 900 0.3868 1.121 398.1 102.9 62.0 950 0.3666 1.131 411.3 112.2 64.3 1000 0.3482 1.141 424.4 121.9 66.7 1100 0.3166 | 1.159 449.0 141.8 71.5 1200 0.2902 1.175 473.0 162.9 76.3 38.3 47.2 56.7 66.7 43.9 0.690 0.686 0.684 0.683 0.685 0.690 0.695 76.9 87.3 98.0 549 109 0.702 120 131 143 155 168 195 0.709 0.716 0.720 0.723 0.726 0.728 0.728

Answer 1

H = 80mm TE - 1000K UF 60m/sec H₂= 20mm from given table P = 0.348 kg/m3 Ka 0.0667 W/mK Pr= 0.726 Ma 424.4x107 was com we know that Continuity equation - Pu, Ai= f Uz Az Ust everything U2= wi WXHI WxHq Uz = uix ti U2= 60x80 U2 = 240m/sec . for each of the channel: Dn=aH Halomm Dn= 2x0.0) Dne 0.02 mls

Reda u Dn = 0.348 x 240x 0.02 424.4 x 107 = 39,859.095 so Red > 2300 so it is flow, Turbulent Convective heat transfer coefficient ha h * 0.023 * Read detais ha 0.0667 * (0.023) * (29,359.0999 * 60.726953 X 0.02 he 330.476 Heat Flum q=hC Tf - Ps) qe 330.476 (1000 – 3 50) la q- 214809.805 Wlas