Consider a 5-m long horizontal pipe which has inner radius (r_i) of 0.005m and a

Question

Consider a 5-m long horizontal pipe which has inner radius (r_i) of 0.005m and an outer radius (r_o) of 0.0075m . Water at 0.01 kg/s enters the pipe at 10C and exits at 85C. The water is heated by the pipe walls, where heat is generated uniformly at 1.28x 10^7 W/m^3 (q(dot)). Cold Air (10C) flows over the pipe in cross flow at a velocity of 20m/s . the outer surface temperature of the pipe was determined to be relatively constant. Assume Steady state. Assume constant properties for the water and the air and use properties given below. Find a)what is the outer surface temperature of the exposed pipe? b)What is the maximum inner surface temperature of the pipe? Water Properties ( density =993 kg/m3 , heat capacity = 4178 j/kg-k , viscosity = 6.95 *10^-4 , kinematic viscosity = 7*10^-7 , thermal conductivity = 0.628 , Volumatric expansion coeffecient = 0.0003619 , themral diffusivity = 1.51*10^-7 , Pr = 4.62)

air Properties ( density =1.1614 kg/m3 , heat capacity = 1007 j/kg-k , viscosity = 1.846 *10^-5 , kinematic viscosity = 1.59*10^-5 , thermal conductivity = 0.0263 , Volumatric expansion coeffecient = 0.0033, themral diffusivity = 2.25*10^-5 , Pr = 0.707)

Explanation / Answer

First off, continuity equation: the amount of water pass flowing through the pipe is always the same, that is mass flow = constant the mass flow can be calculated as mass flow = density * velocity * Area = density * V * A since the density is the same throughout the pipe, at any point V A = constant In your problem, you have V1 A1 = V2 A2 thus V2 = V1 A1 / A2 = 0.55*1.7E-2/6.5E-4 = 14.38 m/s To calculate the pressure, you need to use the Bernoulli Principle, this is V^2/2 + gz + P/density = constant Thus for your problem V1^2/2 + P1/density = V2^2/2 + P2/density P2 = P1 + density(V1^2 - V2^2)/2 = 5E5 + 1*(0.55^2-14.38^2)/2 = 4.9E5 Pa

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