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Chapter 14: Problem 199

Air flows in a 4-cm-diameter wet pipe at \(20^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}\( with an average velocity of \)4 \mathrm{~m} / \mathrm{s}$ in order to dry the surface. The Nusselt number in this case can be determined from \(\mathrm{Nu}=0.023 \operatorname{Re}^{0.8} \mathrm{Pr}^{0.4}\) where \(\operatorname{Re}=10,550\) and \(\mathrm{Pr}=\) \(0.731\). Also, the diffusion coefficient of water vapor in air is $2.42 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$. Using the analogy between heat and mass transfer, the mass transfer coefficient inside the pipe for fully developed flow becomes (a) \(0.0918 \mathrm{~m} / \mathrm{s}\) (b) \(0.0408 \mathrm{~m} / \mathrm{s}\) (c) \(0.0366 \mathrm{~m} / \mathrm{s}\) (d) \(0.0203 \mathrm{~m} / \mathrm{s}\) (e) \(0.0022 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
Question: Determine the mass transfer coefficient inside the pipe for fully developed flow, given the Reynolds number (Re) as 10,550, the Prandtl number (Pr) as 0.731, the Nusselt number formula as Nu = 0.023 * Re^0.8 * Pr^0.4, the diffusion coefficient of water vapor in air (Dm) as 2.42×10^-5 m²/s, and the pipe diameter (Dp) as 0.04 m. Answer: The mass transfer coefficient inside the pipe for fully developed flow is 0.0366 m/s.

Step by step solution

01

Find Reynolds number (Re)

Given in the problem, Reynolds number (Re) = 10,550.
02

Find Prandtl number (Pr)

Given in the problem, Prandtl number (Pr) = 0.731.
03

Calculate Nusselt number (Nu)

Using the given formula: Nu = 0.023 * Re^0.8 * Pr^0.4 Nu = 0.023 * (10,550)^0.8 * (0.731)^0.4 Nu ≈ 171.48
04

Determine the mass transfer coefficient using analogy between heat and mass transfer

The mass transfer coefficient (k_m) can be determined using the analogy between heat and mass transfer, which states that Nu/Re*Pr = k_m*Dm / Dp, where Dm is the diffusion coefficient of water vapor in air, and Dp is the pipe diameter. We need to rearrange this formula to isolate k_m: k_m = Nu * Dm * Pr / (Re * Dp) Substitute the known values: k_m = 171.48 * (2.42×10^-5 m²/s) * 0.731 / (10,550 * 0.04 m) Solve for k_m: k_m ≈ 0.0366 m/s So the mass transfer coefficient inside the pipe for fully developed flow is 0.0366 m/s, which makes the correct answer (c).

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Most popular questions from this chapter

What is diffusion velocity? How does it affect the mass-average velocity? Can the velocity of a species in a moving medium relative to a fixed reference point be zero in a moving medium? Explain.

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One way of increasing heat transfer from the head on a hot summer day is to wet it. This is especially effective in windy weather, as you may have noticed. Approximating the head as a 30 -cm-diameter sphere at $30^{\circ} \mathrm{C}\( with an emissivity of \)0.95$, determine the total rate of heat loss from the head at ambient air conditions of $1 \mathrm{~atm}, 25^{\circ} \mathrm{C}, 30\( percent relative humidity, and \)25 \mathrm{~km} / \mathrm{h}$ winds if the head is \((a)\) dry and (b) wet. Take the surrounding temperature to be \(25^{\circ} \mathrm{C}\). Answers: (a) \(40.5 \mathrm{~W}\), (b) $385 \mathrm{~W}$
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