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What is the temperature loss along a duct with hot air entering?
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[QUOTE="Chestermiller, post: 6819720, member: 345636"] The game plan in this calculation is to first determine 3 things: 1. The forced convection heat transfer coefficient between the air and the inside surface of the tube 2. The equivalent heat transfer coefficient across the wall of the tube 3. The natural convection heat transfer coefficient between the outside wall of the tube and the outside air at 30 C. Item 1 can be determined from the heat transfer correlation presented in Fig. 14.3-2 of Transport Phenomena. This expresses the Nusselt number (dimensionless heat transfer coefficient) as a function of the Reynolds number for the flow and the Prantdl number of the air. The correlation is supposed to use the air physical properties averaged over the bulk flow and the wall temperature (arithmetic averages at inlet and outlet). However, since these are initially unknown, they would have to be determined iteratively. So, as a first approximation, we would be using the properties evaluated at the inlet air temperature (200 C). From the ideal gas law, the density of the inlet air is $$\rho=\frac{PM}{RT}=\frac{(101325)(0.029)}{(8.314)(473)}$$$$=0.747\ \frac{kg}{m^3}$$and the mss flow rate is $$\dot{m}=\rho Q=(0.747)(0.8)=0.598\ kg/min=0.00996\ kg/sec$$The viscosity of air qt 200 C is $$\mu=2.58\times 10^{-5}\ \frac{kg}{m.sec}$$So, the Reynolds number at the inlet temperature is $$Re=\frac{\rho v D}{\mu}=\frac{4\dot(m)}{\pi D\mu}=\frac{(4)(0.00996)}{(3.14159)(2.58\times 10^{-5})(0.1)}=4920$$This is beyond the transition between laminar and turbulent flow, and the flow is definitely turbulent. From the heat transfer = correlation in Fig. 14.3-2 of Transport Phenomena, at this Reynolds number, we have that the Nusselt number is given by $$Nu=\frac{hD}{k}=0.004(Re)(Pr)^{1/3}=(0.004)(4920)(0.7)^{1/3}=17.5$$The thermal conductivity of air at 200 C is ##0.0382\ \frac{W}{m.C}## so the initial estimate of the inside heat transfer coefficient is given by $$h_{inside}=\frac{k}{D}Nu=\frac{0.0382}{0.1}17.5=6.69\ \frac{W}{m^2.C}$$ CONTINUED IN NEXT REPLY [/QUOTE]
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What is the temperature loss along a duct with hot air entering?
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