Air flows in a 4-cm-diameter wet pipe at 20°C and 1 atm with an average velocity of 4 m/s in order to dry the surface. The Nusselt number in this case can be determined from Nu = 0.023 Re0.8 Pr0.4 where Re = 10,550 and Pr ?= 0.731. Also, the diffusion coefficient of water vapor in air is 2.42 × 10?5 m2/s. Using the analogy between heat and mass transfer, the mass transfer coefficient inside the pipe for fully developed flow becomes
(a) 0.0918 m/s
(b) 0.0408 m/s
(c) 0.0366 m/s
(d) 0.0203 m/s
(e) 0.0022 m/s
D=0.04 [m]
T=20[C]+273 [K]
P=1 [atm]
V=4 [m/s]
Re=10550
Pr=0.731
D_AB=2.42E-5 [m^2/s]
Nus=0.023*Re^0.8*Pr^0.4 "Table 14-13"
Sh=Nus
h_mass=(Sh*D_AB)/D
"Some Wrong Solutions with Common Mistakes"
W_Sh=3.66 "Considering laminar flow"
W_h_mass=(W_Sh*D_AB)/D
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