Air (inner) hi = 5 W/m2K Ti = 4 C. Air (outer) ho = 5 W/m2K To = 20 C
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1 hi, Ti ho, To Steel (2 layers) Ksteel = 60 W/mK, Lsteel = 3 mm. Fiberglass Kf = W/mK, Lf = 50 mm. Air (inner) hi = 5 W/m2K Ti = 4 C Air (outer) ho = 5 W/m2K To = 20 C Steel Fiberglass
2 1. Identify the isothermal surfaces The isothermal surfaces are identified with the help of the boundary conditions. The convection on the two sides of the sandwich defines two isothermal surfaces. Then, between each layer of the sandwich, there are the other four. hi, Ti ho, To When between two isothermal surfaces there is only one contribution to the thermal resistance, then the scheme is in series. In a scheme in series, the heat rate (or the flux) is the same for each resistance. Steel Fiberglass
3 2. Calculate each thermal resistance In the case of the sandwich, it is convenient to calculate directly the specific thermal resistance. Ri Rsteel Rf Ri = 1/hi = 0.2 m2k/w Ro Rsteel Ro = 1/hi = 0.2 m2k/w Rsteel = L/ksteel = 5e-5 m2k/w Ti 3. T1 T3 T4 To Rf = L/kf = 1.09 m2k/w Calculate the overall thermal resistance (or overall heat transfer coefficient) Rtot = R = 1.49 K/m2K U = 0.67 W/m2K Q = UΔTtot = 10.7 W/m2
4 Air 27 C ho = 6 W/m2K r2 r1 Calcium silicate k = 0.10 W/mK Steel k = 35 W/mK r3 Steam 575 C The thin sheet of aluminum can be neglected for the calculation of the conduction thermal resistance, but it can t be neglected for the radiation resistance.
5 1. Identify the isothermal surfaces Air 27 C ho = 6 W/m2K r2 r1 Steam However, outside the tube there are two different mechanisms of heat transfer (convection and radiation). This scheme is called parallel. Steel k = 35 W/mK r3 For the conduction, there is the usual scheme in series. Calcium silicate k = 0.10 W/mK Tstea m T1
![2. Conduction thermal resistances and heat rate The general definition of conduction thermal resistance [K/W] in a tube is given in 2.22.](/docs-images/96/128741021/images/6-0.jpg "It is possible to define a specific conduction thermal resistance [m2k/w] by multiplying 2.22 by the inner or outer surface.")
6 2. Conduction thermal resistances and heat rate The general definition of conduction thermal resistance [K/W] in a tube is given in It is possible to define a specific conduction thermal resistance [m2k/w] by multiplying 2.22 by the inner or outer surface. Another choice would be defining a thermal resistance per unit tube length [mk/w] by multiplying 2.22 by the tube length. Rsteel = ln(r2/r1)/(2πksteel) = 8.29e-04 [mk/w] Rcs = ln(r3/r2)/(2πkcs) =? [mk/w] Qcond = (Tsteam-)/(Rsteel+ Rcs)
7 3. Convective and radiative thermal resistances and heat rates To define a convective thermal resistance [K/W], start from the heat rate by convection. Q = h A ΔT = h 2πr3 l ΔT R = ΔT/Q = 1/h 2πr3 l It is possible to define a convective thermal resistance per unit tube length [mk/w]: = ΔT/Q = 1/h 2πr3 Same procedure for the radiation thermal resistance. = ΔT/Q = 1/h 2πr3 =? [mk/w] Q = εσa(4-t4wall) = εσ 2πr3 l (4-T4wall) = 1/(εσ 2πr3 (2+wall)(+Twall)) =? [mk/w] = 1/(εσ 2πr3 (2+wall)(+Twall))
8 4. Energy balance The energy balance applied to the tube states that: Qcond = Qair + Qwall Qcond = (Tsteam-)/(8.29e-04+ln(r3/r2)/(2πkcs)) = = 525/(8.2907e-04+ln(r3/r2)/0.6283) Qair = h A ΔT = h 2πr3 (-Twall) = r3 Qwall = εσa(4-t4wall) = εσ 2πr3 (4-T4wall) = = r3 Qcond = W/m Qair + Qwall = W/m r3 = m; thickness = m
9 5. Total thermal resistance Rsteel = ln(r2/r1)/(2πksteel) = 8.29e-04 [mk/w] Rcs = ln(r3/r2)/(2πkcs) = [mk/w] = ΔT/Q = 1/h 2πr3 = [mk/w] = 1/(εσ 2πr3 (2+wall)(+Twall)) = [mk/w] Rtot = Rsteel + Rcs + (R-1air + R-1wall )-1 = [mk/w] Note: While in a series scheme the highest resistance is the dominant, in a parallel scheme the lowest resistance is the dominant.
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