Question

Water is to be heated from 10 °C to 83 °C as it flows through a 0.015 m-internal-diameter, 13-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. The system provides hot water at a rate of 5 L/min Assumelp = 990.1 kg/m .k=0.637 W/m-K, U = 0.000596 N-s/m. Cp = 4180 J/kg-K Pr = 3.91 - kg/m3 Find Flow velocity -m's Mass flow rate Reynolds number-- Nusselt number. Nu = -- Heat Transfer coefficient, h=- Heat Rransfer rate, Q = ---- Surface temperature of the pipe at the exit-- W/m2K OC

Answer 1

Te = 83° c, d = 0.01sm, L= 13 Ti = 10°C flow rate = Shwin = SX10 3 = 8.3333x10's mo sec ww 8.3333x10-5 flow rare @ ve flow reward cross sectional Area Axo.oIS 2 flow velocity = 0.4715 7 m/sec ION velou 36 Mass flow rate = flow rate x 8 = 8.3333x10sx ggo.1 | Mass dersity flow rate 0.082508 = I kg - - m3 kg Unit sel Reynolds No = PVD & 990.1 x 0.47157 x 0.015 0.000596 Reynolds No = 11750.8756 - Reynolds no > 2000 turbulent Flow Nu - 0.023 (Re) 0:3 (0) 04 = 0.0 23 x (11950.8456) 313.9, 104 Nussell do, Nu = 71.557 - 16 NU= nd » 710557 = hx 0.015 0. 637 h = 3038.7 87 watt m2K

(6) Heat Tramfer rate is = mcp [Te - Ti} = 0.082508 x 4180 x [83 - 103 c = 25176.49) watt (9) let surface Temperature - T e = wx ax dx L x [T- Te] => 25146.49 = 3038.789 x A x 0.015* 13 * (T-83] J osefone » Te 96.524 °C