Question

A water heater consists of a copper tube that carries water through hot gas in a furnace, as shown in Fig. 1. hot gas at T, = 500° th=0.033 inch h, = 25 W/m2K Poytube = 0.25 inch water at Tv = 30° h = 500 W/m2K kube= 300 W/m-K Figure 1: Copper tube in a water heater. The copper tube has an outer radius, ro,tube = 0.25 inch and a tube wall thickness of th=0.033 inch. The conductivity of the copper is ktube=300 W/m-K. Water flows through the pipe at a temperature of Tw = 30°C. The heat transfer coefficient between the water and the internal surface of the pipe is hw = 500 W/m²-K. The external surface of the tube is exposed to hot gas at Tg = 500°C and the surroundings are also at 500 °C. The heat transfer coefficient between the gas and the outer surface of the pipe is hg = 25 W/m²-K. Neglect radiation from the tube surface. a.) How much heat is added to the water for a unit length of tube, L = 1 m, for this configuration (W/m)? b.) What is the dominant resistance to heat transfer in your water heater?

Answer 1

solution -* - Criven - lylindrical heated 20.033 inch TX HiFIIIIIIII 16sume a Osteody- Starde contu Heat franefer bei Radial-Direction heat flow No = 0.25 inch YiYo - thickenges Sincer I inch = 0.0254 moter = 0.25 -0.033 = 0.217 inch Y; = 0.017 inen 30 Y = 025X0.0154 meter = 6.358103 m Pic0:217X0.0254 meter - 5.5118x183 meter three Resistance - this is combination of Tue hue Krube hy . Ty Ruinten Reylinder Ruouer A = internal furluce area 1 = 2ITY L Hi 2x1T XL ScannedAdith outernal surface area CamScanner

Ao=257 rol Sincer No = 6.35x10m Lalm = 2 x TTX 6.35X10x1m = 0.03989 m 2 Ai = 2,7; L = 2XTX 5.5118x102x) Since- = 0.03463 m2 Pi = 5:5188 xiom Laim Stepe - Rnine = huar Soo X 0.03463 ml 12.k Since hue = 500 W m2.k - 0.05775 K 71FIETLLL91111991 o for Cylindrical Rosistance can be calculated as 8; Reylinder = Inte QTKL Yo wider Yudine Vi = internal radios ka thormal conduction La benth. = Sinco a in 6.38x103 155118 x 103 2X57 800 ly x Int mik = 7.510x1ős lleve 193 so miks r o = 6.31x10 m 1.Pi = 5.5118xiom Scanned with CamScanner

for cuter - Surface - Rhinsor & I Since- AO No = 0.03989 m2 Ky = 25 W m2k 25 w x 0.03989 m² hazka 1.002757 Klu Resistance can be calculated as a so - equivalent Rea = Rnince + Reylinder + Rn, miser = 0.05775+7.500xöre + 1.012757 like Hey = 1.0708821 Kler -6 Heat transfer from outer ceeater - surface to calct internal niven - Change in Kelvine Py = (500+273K The = 150+ 273) k O = Tg - Tue Rey = (500+273)– (507273) 1,0705821 E 5004 27/3 - 50-273 1.070 5821 - 450 1:0705821 Scanned with CamScanner E 42033 ut

so it is taken for por unite lenth ở : 44:33 |- | 50 © c = 490.33 Wilm 0 Resz 1.070589) Keel Scanned with CamScanner