Question

Case Study 2 Convection) – 30 Marks Consider an air duct heater consists of electrical heating elements with an aligned arrangement and the longitudinal (5) and transverse pitches (S.) of 5 = 5, = 22 mm. There are 3 rows of elements in the flow direction (N = 3) and 3 elements in each row (N = 3). The atmospheric air at a velocity of V = 10 m/s, and a temperature of. = 15°C enter the duct and flow over the heating elements. Each heating elements has a diameter of D = 10 mm, a length of 200 mm, and is kept at a surface temperature of T - 277°C (Figure 2). Heating elements: 1,- 277°C D = 10 mm N = 3 N, 3 200 mm Air V = 10 m/s -15°C Figure 2. Schematic of air duct heater. a) Determine the total heat transfer to the air, and the temperature of the air leaving the duct. (12 Marks) b) Compare the average convection heat transfer coefficient obtained in your analysis with the value for a single heating element and discuss the reasons for the difference between the results. (8 Marks) c) of the longitudinal and transverse pitches increase to 5 = 5 = 28 mm, what will be the effect on the air temperature leaving the duct and total heat transfer rate? Discuss your observation. (10 Marks)

Answer 1

Solution:- Let us con Sidey mean temperature of aire 138.5°C properties of air at this temperature P = 0.8488 Kg/m3 (p= (ol 5.6 5legk v= 27.79x1ő mets K= 34.6x10 3 wink PY-0.689 Prst=277°C =0.683 Vmax *10 = 18.33 m/sec RCD ST 22 ST-D 22-10 Sumax D Vmax D 18.33x16x1ó? = 6595.90 시 7 27.79x156 0.63 0.36 0.25 0.27 Rco Py Pr pes 0.25 0-63 0:36 0.689 0.27 (6595.90) (0.689) 0-683 Mup llup - Muo = 60.285 = 60.29 but u This Nussalt lo is applicable for tube sank with Mc>16 Our ease no. of rows NL = 36 corresponding Correction factor F-0.86 for where arrangement = F Nup = 0-86% 60.29 = 51.85 51-85x 34.61x16? .=179.4 W/m2K 10x103 NUD NL-3 HUDNL- No DL = (3X3) fixloxió'x0.2 =0.0565 m2 -3 ń = five CMT STL) =0.8488X10X3 X 22x10°x0.2 = 0.112 kg/sec -LAS Te=Ts-(Ts-Ti) e me -179.4X0.0565 0.112 x 1015.6 Te= 277 (277-15) e = 37.38°C

Те- TC 37.38-15 ATm = =-250.640 lo Ts-Te TS-Tc en 277 37.38 277-15 Q= LASAT = 179.4*0.0565X250.64 = 2540.550 (6). 0.466 For Single heating element is case of cross flow over heating element PVD VD lox 10x10? Re= - 3598.42 >2300 д 27.798106 0.466 113 Ny = 0.683 Re Pr3 - 0.683 3598.02) 6.689) = 27.39 LD - 27.39 - La 27.39834.618163 = 94.81 w/mik 10X163 So the heat transfer coefficient value. obtained for one element is less than for that combination of tubes due to fact that as Huidlaik enters the tube sank How crea increases the reynold no. of thuid & then one nusselt no. which further increases heat transfer co-efficient. (c). On increasing S = S2 = 28mm flow area increases which reduces the velocity of ais & due to which Nusselt No decycases and thien heat transfer coefficient also decrease LAS Exit temperature of fluid is Te=Ts (TS-Tc) e mep Decrease as from formula its relation with Lis expenential So as L decreases exit fluid temperature also decreases and heat transfer also decreases as Q=LAS ATM @ Po dvectly propastion to L and exit temperature of fluid