H6 - INTRODUCTION: Heat-exchangers are widely used in the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
INTRODUCTION: Heat-exchangers are widely used in the process industries so they have been highly developed. Most exchangers are liquid to liquid and some are gas and non condensing vapour. The double pipe heat exchanger and the shell and tube HE are some examples of H.E. . In a H.E. the shell side and the tube side and H.T.C. are of comparable importance as well as the velocity and the turbulence. There must be a minimum distance between the tubes to prevent weakening of tubes sheets. If the two streams are of comparable magnitude , the velocity on the shell side is low compared to the tube side. So that baffles are installed in the shell to decrease the cross sectional area of the shell side liquid and to force the liquid to flow across the tubes. In this exp.the study was done for co current and counter -current flow , and the flow rate of the hot water was kept constant, while that of the cold water was changed. THEORETICAL BACKGROUND: The heat transfer coefficient h i can be calculated by: (h i /C p G)(C p μ /k) 2/3 =(.023(1+(D i /L) .7 ) ) /(D i G / μ ) .2 -------------------------1) In this formula the properties are evaluated at the bulk temperature. The coefficient of the shell side is h o is calculated from the Donohue eq. h o D o /k =.2(D o G e / μ ) .6 (C p μ /k) .33 ----------------------------------------------2 the area for flow in baffle window Sb is calculated by Sb= f b *( π D 2 s /4)- N b *( π D 2 o )/4--------------------------3 and the area of cross flow in exchanger shell S c = PDs(1-Do/P)------------------------------4 the velocity term are Gb= mc/Sb ------------------5 Gc=mc/Sc----------------6 then , the effective mass velocity Ge is Ge=(Gb*Gc) .5 --------------------------7 the Re is RE= DH*Ge/ μ --------------------8 The correction of LMTD for cross flow If a fluid flows perpendicularly to a heated or a cooled tube bank, the LMTD is as given by / \T LM = (/\ T 2 -/\ T 1 )/ ln(/\ T 2 //\ T 1 )-------------------------------9
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The true mean temperature drop will be used in the following eq. To obtain the overall heat transfer coefficient U. q =UA/\T lm U = 1/((A o /A i h i )+A o (lnD o /D o )/2 π k gl L +1/h o +R o +R i )----------------------10 where q could be calculated from q =m*Cp*(^ T) ---------------------------------11 PROCEDURE : The unit consists of a graphite H.E. and a shell and tube H.E. .The hot water produced by the graphite H.E. is in the tube side wise the cold water is in the shell side. The cold water can be adjusted to cocurrent or counter current flow to the hot water.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/27/2011 for the course CHEMICAL E CHE 309 taught by Professor M.elgaily during the Fall '09 term at King Fahd University of Petroleum & Minerals.

Page1 / 8

H6 - INTRODUCTION: Heat-exchangers are widely used in the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online