pricipals of enhanced heat transfer by Ralph_L._Webb,_Nae-Hyun_Kim section 3 Q.1

Question

pricipals of enhanced heat transfer by Ralph_L._Webb,_Nae-Hyun_Kim section 3 Q.1

0.25 3.1 Simple Single-Phase PEC. A smooth tube hasf, = 0.079 Re and Js = //2. An internally roughened tube providesjs 0.079 Re-0.25 and has.f= 0.11 Re-0.2 Assuming all of the thermal resistance is on the tube side, calculate G/G, A/A, and h/hs using Case VG-1 for Re, = 20,000. What is AIA,2 [Ans. G/G, = 0.937, AA, = 0.525, h/hs = 1.90] Now consider Case FN-1. Outline the procedure you would use to obtain GIG and AlA,. Will A/A, be greater or smaller than for Case VG-1? Why?

Explanation / Answer

Let the length of tube be L, dia D and the water outlet tempearture be T

The steam outlet temperature is

The LMTD is

Heat transferred with plain tube is

Hence

Hence

With same operating conditions without considering increase in steam pressure drop the outlet temperature remains same as T and LMTD remains same.

With enhanced steam side area, considering water side area as reference area, the overall heat transfer coefficient is given as:

Or

U'= 5555.55 W/m2-K

With Q/Q'=1 and neglecting the increase in steam pressure drop

Hence

b) With steam pressure increasing 2.5 times

Hence

Let us consider L=1 m and D=0.05 m

For the original geometry for water side

Hence

In the above we have considered cp of water as 4187 J/kg-K

Solving the above

T=303.3913 K

Hence

Q=4187x(303.3913-300)=14199.4 W

Similarly for the revised configuration

Solving

T'=303.3224 K

Hence

Q'=4187x(303.3224-300)=13910.9 W

Hence

Q'/Q= 13910.9/14199.4=0.98

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