can someone help me please Steam is being condensed on the s
can someone help me please...
Steam is being condensed on the surface of a pipe in a power plant heat exchanger. The condensation process implies that the pipe wall temperature will be fixed at 135degreeC. The length of the circular pipe is L = 40 m, and the diameter is D = 4.5 cm. There are two possible fluids that can be pumped through the pipe: water, or air, but the flowrate is fixed at m = 0.025 kg/s, and the inlet temperature is T_m, j = 20degreeC. Assume that the fluids become fully developed in the pipe, Determine the average heat transfer coefficient for both fluids, Determine the outlet temperature for each of the fluids. (Partial answer: T_m(L) for water = 114degreeC). Determine the total heat transfer rate from each of the fluids, If the enthalpy of vaporization for the steam is 4.25 kJ/kg, which fluid will condense more steam over a given time?Solution
For water properties are:
Density rho_w = 1000 kg/m^3
Specific heat Cp = 4180 J/kg-K
Dynamic viscosity meu = 0.000547 Pa-s
Thermal conductivity k = 0.644 W/m-K
Area A = 3.14/4*4.5^2 = 15.9 cm^2
Perimeter P = 3.14*4.5 = 14.13 cm = 0.1413 m
Velocity V = m / (rho*A) = 0.025 / (1000*15.9*10^-4) = 0.0157 m/s
Reynolds number Re = rho*V*D/meu = 1000*0.0157*4.5*10^-2 / 0..000547 = 1291
Since Re < 2300, flow is laminar and hence Nusselt number Nu = 3.66 for uniform wall temperature.
Nu = hD/k
h = Nu*k/D
h = 3.66*0.644 / (4.5*10^-2)
h = 52.38 W/m^2-K
(Twall - Tout) / (Twall - Tin) = e^(-PLh / mCp)
(135-Tout) / (135 - 20) = e^(-0.1413*40*52.38 / (0.025*4180))
Tout = 128.2 deg C

