Ht 034 Solution

CHEMICAL ENGINEERING SERIES: HEAT TRANSFER SOLVED PROBLEMS

A horizontal rod 5 mm in diameter is immersed in water maintaind at 18 °C. If the rod surface temperature is 56 °C, estimate the free convection heat trasfer rate per unit length of the rod.

Source: Fundamentals of Heat and Mass Transfer 6 th edition, by Incropera, et al

SOLUTION: NATURAL CONVECTION For flow on isothermal horizontal cylinders, using equation 5-36 CHE HB 8th edition

{

´ L = 0.60+ Nu

0.387 RaD 1/ 6 9 /16 8 / 27

[ ( ) ] 0.559 1+ Pr

2

}

Water properties at film temperature

T f=

18+56 =37 ° C=310 K 2

ρ=993.342

C μ Pr= p = k

kg W kJ ; μ=693.54 μPa ∙ s ; k =0. 62605 ;C p=0.075294 3 m∙K mol ∙ K m

(

kg 1 kJ kmol 1,000mol 1 Pa m∙ s 0.075294 x x 693.54 μPa ∙ s x x mol ∙ K 18 kg 1 kmol Pa 1 x 106 μPa

)

(

kJ W s 0. 62605 x m ∙ K 1,000W 1

Pr=4.6339 D o3 ρ f 2 βg ∆T o Gr= μf 2

ENGR. RONNIE V. FLORES

(

Page 1

)

)

CHEMICAL ENGINEERING SERIES: HEAT TRANSFER SOLVED PROBLEMS

β=

( 997.695−984.3786 ) 3.536 x 10−4 = ( 991.0368 ) ( 56−18 ) K

3

(

( 0.005 m ) 993.342 Gr=

(

kg m3

2

)(

)(

3.536 x 10−4 m 9.80665 2 ( 56−18 ) K K s

kg 1 1 Pa m∙ s 693.54 μPa ∙ s x x 6 Pa∙ s 1 x 10 μPa

)

2

)

Ra=GrPr Ra=( 33,789.44903 ) ( 4.6339 )=156,576.9279

{

´ L = 0.60+ Nu

0.387 (156,576.9279 )

[ (

0.559 1+ 4.6339

)

1 /6 2

9/ 16 8/ 27

]

}

=10.4077

´ L = hD Nu k kJ W s ( 10.4077 ) 0.62605 x m∙ K 1,000 W kJ h= =1.30315 2 0.005 m m ∙K∙s

(

1

)

q=hπDL ∆T q kJ 1000W = 1.30315 2 x π ( 0.005 m )( 56−18 ) K L kJ m ∙ K ∙s 1 s

(

)

q W =777.85 L m

ENGR. RONNIE V. FLORES

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=33,789.44903