Answered: Estimate the natural convection heat…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Question

Estimate the natural convection heat transfer coefficient of the horizontal pipe. The outer surface temperature of the pipe is 120 °C. The ambient air temperature is 40 °C. The outer diameter of the pipe is 18 cm. Convection coefficient = W / m2 C.

Expert Solution

Step 1

Outer surface temperature of the horizontal pipe, T_S = 120°C =393 K

Ambient air temperature, T_A = 40°C = 313 K

Outer pipe diameter, D_o = 18 cm = 0.18 m

Step 2

The average temperature is calculated as:

$\begin{matrix} T_{a v g} = \frac{T_{S} + T_{A}}{2} \\ = \frac{120 + 40}{2} \\ = 80^{\circ} C \\ = 353 K \end{matrix}$

Take the properties of air at this average temperature from the standard table as:

$\begin{matrix} ρ = 1.00 \frac{kg}{m^{3}} \\ μ = 20.88 \times 10^{- 6} \frac{kg}{m \cdot s} \\ k = 0.03023 W/m \cdot K \\ \Pr = 0.708 \\ β = \frac{1}{T_{a v g}} \end{matrix}$

Step 3

Calculate the Rayleigh number as:

$\begin{matrix} R_{L} = \frac{g β (T_{S} - T_{A}) D_{o}^{3} \Pr}{{(μ / ρ)}^{2}} \\ = \frac{g (\frac{1}{T_{a v g}}) (T_{S} - T_{A}) D_{o}^{3} \Pr}{{(μ / ρ)}^{2}} \\ = \frac{(9.81 \frac{m}{s^{2}}) (\frac{1}{353 K}) (393 K - 313 K) {(0.18 m)}^{3} (0.708)}{{(\frac{20.88 \times 10^{- 6} \frac{kg}{m \cdot s}}{1.00 \frac{kg}{m^{3}}})}^{2}} \\ = \frac{(9.81) (\frac{1}{353}) (393 - 313) {(0.18)}^{3} (0.708)}{{(\frac{20.88 \times 10^{- 6}}{1.00})}^{2}} \\ = 2.11 \times 10^{7} \end{matrix}$

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