Make a table as required in the illustrated example with commentary on the amount of heat transfer loss value Example 3: Repeat example 2 for finite length of 2, 4, 8,at the end (case 3)..128cm, a surrounding heat lossSolution.2,4,8,16,32,64,128 cm LFor case 3sinh mL+ (h/mk) cosh m Lq=vh PkA 60cosh mL+ (h/mk) sinh mLch/mk= 11/(3.416*377) = 8.54*10^-3!!h(rD)4h3.416[1/m]m =%3DkAV KD4At L=2cm, mL=0.06832, sinhmL=0,06837, coshmL=1.00233> q =0.993W

Given: m=3.416 h=11W/m2K (Taking SI units as proper units are not given in question) k=377 W/mK…

Answered: xample 3: Repeat example 2 for finite…

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

Make a table as required in the illustrated example with commentary on the amount of heat transfer loss value

Example 3: Repeat example 2 for finite length of 2, 4, 8,
at the end (case 3).
.128cm, a surrounding heat loss
Solution.
2,4,8,16,32,64,128 cm L
For case 3
sinh mL+ (h/mk) cosh m L
q=vh PkA 60
cosh mL+ (h/mk) sinh mL
ch/mk
= 11/(3.416*377) = 8.54*10^-3
!!
h(rD)
4h
3.416
[1/m]
m =
%3D
kA
V KD
4
At L=2cm, mL=0.06832, sinhmL=0,06837, coshmL=1.00233
> q =0.993W

Expert Solution

Step 1

Given:

m=3.416

h=11W/m²K (Taking SI units as proper units are not given in question)

k=377 W/mK (Taking SI units as proper units are not given in question)

$\begin{array}{rcl} m & = & \sqrt{\frac{h P}{k A}} \\ 3.416 & = & \sqrt{\frac{4 h}{k D}} \\ 3.416 & = & \sqrt{\frac{4 \times 11}{377 \times D}} \\ D & = & 0.01 m \end{array}$

$\begin{array}{rcl} \sqrt{h P k A} & = & \sqrt{h \times πD \times k \times \frac{{πD}^{2}}{4}} \\ = & \sqrt{\frac{h \times π^{2} D^{3} \times k}{4}} \\ = & \sqrt{\frac{11 \times 3 . 14^{2} \times 0 . 01^{3} \times 377}{4}} \\ = & 1.011 W / K \end{array}$

$\begin{array}{rcl} q & = & \sqrt{h P k A} θ_{o} \times \frac{\sin h m L + (\frac{h}{m k}) \cos h m L}{\cos h m L + (\frac{h}{m k}) \sin h m L} \\ 0.993 & = & 1.011 \times θ_{o} \times \frac{0.0687 + 8.54 \times 10^{- 3} \times 1.00233}{1.00233 + 8.54 \times 10^{- 3} \times 0.0687} \\ θ_{o} & = & 12.7499 \end{array}$

for L=2cm

q=0.993W

Step 2

i) for L=4cm =.04m

$\begin{array}{rcl} m L & = & 3.416 m^{- 1} \times . 04 m \\ = & 0.13664 \end{array}$

$\begin{array}{rcl} \sin h m L & = & \sin h (0.13664) \\ = & . 13707 \end{array}$

$\begin{array}{rcl} \cos h m L & = & \cos h (0.13664) \\ = & 1.009349 \end{array}$

$\begin{array}{rcl} q & = & \sqrt{h P k A} θ_{o} \times \frac{\sin h m L + (\frac{h}{m k}) \cos h m L}{\cos h m L + (\frac{h}{m k}) \sin h m L} \\ q & = & 1.011 \times 12.7499 \times \frac{0.13707 + 8.54 \times 10^{- 3} \times 1.009349}{1.009349 + 8.54 \times 10^{- 3} \times 0.13707} \\ q & = & 1.8584 W \end{array}$

Step 3

ii) for L=8cm =.08m

$\begin{array}{rcl} m L & = & 3.416 m^{- 1} \times . 08 m \\ = & 0.27328 \end{array}$

$\begin{array}{rcl} \sin h m L & = & \sin h (0.27328) \\ = & 0.27669 \end{array}$

$\begin{array}{rcl} \cos h m L & = & \cos h (0.27328) \\ = & 1.03757 \end{array}$

$\begin{array}{rcl} q & = & \sqrt{h P k A} θ_{o} \times \frac{\sin h m L + (\frac{h}{m k}) \cos h m L}{\cos h m L + (\frac{h}{m k}) \sin h m L} \\ q & = & 1.011 \times 12.7499 \times \frac{0.27669 + 8.54 \times 10^{- 3} \times 1.03757}{1.03757 + 8.54 \times 10^{- 3} \times 0.27669} \\ q & = & 3.5395 W \end{array}$

Step 4

iii) for L=16cm =.16m

$\begin{array}{rcl} m L & = & 3.416 m^{- 1} \times . 16 m \\ = & 0.54656 \end{array}$

$\begin{array}{rcl} \sin h m L & = & \sin h (0.54656) \\ = & 0.57418 \end{array}$

$\begin{array}{rcl} \cos h m L & = & \cos h (0.54656) \\ = & 1.15312 \end{array}$

$\begin{array}{rcl} q & = & \sqrt{h P k A} θ_{o} \times \frac{\sin h m L + (\frac{h}{m k}) \cos h m L}{\cos h m L + (\frac{h}{m k}) \sin h m L} \\ q & = & 1.011 \times 12.7499 \times \frac{0.57418 + 8.54 \times 10^{- 3} \times 1.15312}{1.15312 + 8.54 \times 10^{- 3} \times 0.57418} \\ q & = & 6.5009 W \end{array}$

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