Air at atmospheric pressure and 200 C flows with 6 m/s velocity through main trunk duct of air conditioning system. The duct is rectangular in cross-section and measures 40 cm × 80 cm. Determine heat loss per meter length of duct corresponding to unit temperature difference. The relevant thermo physical properties of air are: $ν = 15 × 10{-6} m^2/s , \alpha = 7.7 × 10{-2} m^2/hr , k = 0.026 W/m-deg K. Use Nu = 0.023 (Re)^{0.8}× (Pr)^{0.4}$.

Given,

$t=20℃=293K$ (Temperature of air)

$v=6m/s$ (Velocity of air)

$Ac=40cm×80cm=3200cm^2=0.32m^2$ (Cross sectional area of duct)

$L=1m$ (Length of duct)

$Δt=1℃$ (Temperature difference)

$ν=15×10{-6}m^2/s$ (Kinematic viscosity)

$α=7.7×10{-2}m^2/h=2.138×10{-5}m^2/s$ (Thermal diffusivity)

$K=0.026W/mK$ (Thermal conductivity)

$Nu=0.023(Re)^{0.8}×(Pr)^{0.4}$

Assuming Cp=1.005kJkgK=1005J/kgK (Specific heat of air)

Find: (1) Q(Heat loss from duct)

Solution:

Characteristic length

l=4×c/s area wetted perimeter=4×0.322(0.4+0.8)

l=0.533 m

Thermal Diffusivity

α=KρCp

∴2.138×10-5=0.026ρ×1005

∴ρ=1.21kg/m3

Reynold’s number

Re=ρvlμ=ρvlρ.ν=v.lν=6×0.53315×10-6

Re=213200

Prandtl’s number

Pr=μCpK=ρ.ν.CpK=1.21×15×10-6×10050.026

Pr=0.701

Nusselt’s number

Nu=0.023(Re)0.8×(Pr)0.4=0.023(213200)0.8×(0.701)0.4

Nu=365.632

Also,

Nu=hlK

∴365.632=h×0.5330.026

∴h=17.835 W/m2K

Heat loss from duct per unit length per unit temperature

Q=h.As.(∆t)= 17.835×[2×(0.4+0.8)×1]×(1)

∴Q=42.804 W