The products of combustion from burner are routed to an industrial application through a thin-walled metallic duct of diameter Diand length L. The gas enters the duct at atmospheric pressure and a mean temperature and velocity of Tmiand um, respectively. It must exit the duct at a temperature that is Tmo. The thickness of the tube is twith conductivity of kw. The outer surface is exposed to ambient air at T[infinity]. Determine the required outer flow velocity to achieve the exit temperature Tmo. You may assume both the inner and outer flows are laminar. For the required fluid properties, please leave them as symbols. You can also assumePr for all gases are >0.7(25 pt.)

Start by calculating the heat lost falling from Tm,i to Tm,o

As a first approximation use an average Cp of (Tm,i + Tm,o)/2 and an average density at (Tm,i + Tm,o)/2

Cp air at (Tm,i + Tm,o)/2 (1)

Density of air at (Tm,i + Tm,o)/2 : PV=nRT, V=T1/T2, air density at 293K = 1.204kg/m^3

Air density at (Tm,i + Tm,o)/2 1.204×293/(Tm,i + Tm,o)/2

Energy lost per kg for (Tm,i - Tm,o)K drop: (Tm,i - Tm,o)K × (1) =

Time of travel: t

Energy lost per kg drop/t seconds = 24.3 kW

Volume occupied by 1kg air at at mean tempera ture = 1/Air density at mean temperature

Length of pipe ( Di m diameter) needed to hold Volume occupied by 1 kg of air at mean temperature :

Cross section = π/4 Di^2, Volume occupied by 1 Kg of air at mean temp. ÷ π/4Di^2

Surface area of pipe Di m diameter by L m long = Length of pipe to hold Volume of air in m × π*Di

Q/A=k (delta T)/ thickness,

Thickness of insulation = Area × k ×dT / Q

Explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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explanation: