Question

Hot water with the mass flow rate of 3 kg/s and temperature of 90 C enters a 2 inch copper pipe with the length of 3 m. The pipe is exposed to cold air at the constant temperature of 1C and wind velocity is 5 m/s. Find the outlet temperature of water and the rate of heat loss from the pipe

Answer 1

Given, mass flow rate of hot water, in= 3kg/ water Inlet temp To To = 90°c diameter of pipe, d= 2 inch -0.0508 m v=sings L = 3m length of pipe, 79233 1d Ost. Ambient (cold air) temp, Too = °c 3 Air, Too L wind velocity, v=5m/s at Titti properties of water = 98.0°C are 2 Dynamic viscosity, M= 3.0705x10 x54 kg/ms specific heat, cp = 4. 0747 KJ/kg k Thermal Conductivity , k = 0.67606 prandtl number, Pr= 1.8506 Reynolds number, Re= 4m Tidu 403) T(0.0508)(3.0705x10 Re= 244883 4000) flow is Turbulent . since, -Re> 4000, flow, → for Flow through pipes and turbulent Dittus Boelter Equation is given by & 0.3 Nusselt number, Nu = 0.023 (Re) (PU) 0.8 0.3 hd 0.023 ( 24488338 (1-8506)

h (0.0508) ** 566.35 0.67606 Average heat transfer h= = 7537 W/m²K Coefficient i > Heat loss from the pipe, Q = hati-To) =hArde) (si-TA) Q = 7537 x 16.0508) 3 ( 90- Q= 321.17 kw Rate of Heat loss from pipe, Also, Q = mcp (Ti-To) 321.17 = 3 (4.0747) (90- To) 26.27 90-To I To = 63.73 °C outlet temp. of water