δ/2 T base = 82.5 °C T s = 86.85 °C w · 7 Forced convection in a variety of configurations 8...
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w δ/2 T_base = 82.5 °C T_s = 86.85 °C
Transcript of δ/2 T base = 82.5 °C T s = 86.85 °C w · 7 Forced convection in a variety of configurations 8...
w
δ/2
T_base = 82.5 °C
T_s = 86.85 °C
Reynolds number, friction factor and heat transfer coefficient
f = 0.024936
Δp = f*(⍴/2*u2)*L/Dh = 0.81 kPa
Nu = 130.22
h = 1.4107e+04 W/m2K
Dh = w
Mechanical power = 10*m*Δp/rho = 0.59 W
Fin efficiency
T_base = 60 °C
δ/2
L
eta = 0.95 mL = 0.392
Energy balance
δQfin
δQbas
e
δQbase = h*2w*dx*(Tbase-T)
δQfin = h*2w* *dx*(Tbase-T)
δQbase + δQfin = m*cp*dT
m*cp*dT = 2*h*w*(1+ )*dx*(Tbase-T)
Tout = 32.715 °C
Q = 10*m*cp*ΔT = 1.739e+4 W
Validation
Q/2 = kcopper*Abase/(δ/2)*(Ts-Tbase) Ts = 86.847 °C = 360 K
T_base = 82.6 °CT_s = ? °C