Question

A heat recovery device involves transferring energy from the hot flue gases passing through an annular region to pressurized water flowing through the inner tube of the annulus. The inner tube has inner and outer diameters of 24 and 30 mm and is connected by 8 struts to an insulated outer tube of 65-mm diameter. Each strut is 3 mm thick and is integrally fabricated with the inner tube from carbon steel (k = 50 W/m-K). 71-1=3 mm D. D.2 Di h Water Gas Consider conditions for which water at 300 K flows through the inner tube at 0.161 kg/s while flue gases at 800 K flow through the annulus, maintaining a convection coefficient of 100 W/m2.K on both the struts and the outer surface of the inner tube. What is the rate of heat transfer per unit length of tube from gas to the water? Use the Dittus-Boelter equation to obtain the water-side convection coefficient Determine the rate of heat transfer per unit length of tube from gas to the water, in W/m. ! W/m q= i

Answer 1

Solution. Gas, Th=sook. 3mm D. – 30mm Di, 1=24mm 12 Do=65mm لی h=100k/nk. k=50 wlink. Water To=3ook. ni= 0.161 kgls. Dil Di,o=3omm. 2 1 Stout length , 2 Giren, = 24mm. Inneo tube innes diameter Inner tube outer diameter outer diameter of insulated tube, Do = 65mm. Stout thickness t=3mm. 65-30 Ls 17.5mm = 0.0175m. The smal conde K=50 wink. Convection coefficient, hi= 100 W/m²k. Mass flow rate of water, in = 0.161 kgls. Tom = 3ook. flue gas temperature, Thim = 8ook: Wateo temperature Heat transfers sate per unit length of tube, q = Uc Ac (Thim - Tam 0 Where 1 + + 2 U cac hac nohnah,

Resistance due to tube wall Conduction, ln (0.2/ Dil) R4 = 2TKL ln (30/24 = 7.11 x10 klw. 25x50x I Foo wated at Tam = 300k. The anal Candcanity, K = 0.613 w/mk K = 0.613 6u/mk , Peopodtl number, Pg = 5.83 . viscosity, M= = 855x10 Ns/m² Reynolds number Re = 4 mi * Di, il 4xo.161 Tx(24x102); X855x10 55x106 = 9994.90 Re > 2300, so flow is turbulent. using Dittus - Boelted equation , Nusselt number, Ny = ha Dil 45 0.4 = 0.023 Re x Po k k 0.4 he X0.023 Re Po Dil X0.023 x(9994.9) uk HIS 0.613 0.4 x 5.83 0.024 = 1883.94 W/m²k. Suxface area of heat exchanged Act = TX 24x10²x1 = 0.075 m² A Di,,L

hc Ac = 1883.94 %0.075 = 141.97 Wlk. Ein efficiency No = 1- Af (i-nes A sudface area of entire fin, Ag = N*2 (4gw) = 8x2 (0.0175x1) = 0.20 m 2 Total surface area , A = Ag+ (TDiiz - 8 ts) = 0.28+ GTX 0.03 - 8x0.003) = 0.35 m² Efficiency ng of single fin, tanh (my) mLs 12hni kts Where 2x 100 - 1 36.51m 50X0.003 tanh ( 36.51XO, 0175 24 = 0.88 36.51X0.0175 0.28 ni = 1- (- 0.88 = 0.90 0.35

nohn An= 0.90 x 100 x 0.35 = 31.5 Wlk. Equation 2 + Rt + Uc Ac hc Ac nahah -4 - + 7.11xlo + 141.97 31.5 0.0395 Uc Ac = 25.31 W/k Heat transfer sate per unit longth, q = Uc Ac (Thm - Tem - 25.31 ( 700 – 300) = 12657.98 W/m. () Tem) Result Heat tsansfer date per unit length, q = 12657.98 wlm. (Answed)