6 replies to this topic

[sgkim]

Hi Bradley,

If the jacket is baffled, the "equivalent diameter" may be calculated from the cross sectional area and wetted perimeter. Based on the equivalent diameter, the Reynolds number, friction factor, and pressure drop could be calculated. But if the jacket is not baffled, the jacket can be regarded as "double pipe". Then follow the same pocedure to estimate the pressure drop thru pipes. N.B., consider 4 elbows for each turn of spiral and the resistance coefficient with respect to R/De ratio.

For heat transfer coefficient calculatons, however, the "heat transfer perimeter" should be used for the calculation of "equivalent diameter". From the equation (h)jacket = f (Re, Pr, μ, k, De), jacket side coefficient could be calculated. Refer to Chem Eng Apr, 1993 or Jan, 1999 for details.

[sgkim]

Thanks very much for that - gives me a good starting point.

Would you have any idea how to calculate the effects of the hemispherical section of the jacket (the jacket covers the sides and bottom of the vessel).

If not, any idea how much error is going to be present by ignoring this factor??

Total head loss through unbaffled jacket, hemispherical jacket, and inlet & outlet nozzles may be estimated as 1.25 times the head loss through inlet and outlet nozzles only. Which is one of jacketed vessel vendors' comment, and implies that the head loss thru the hemispherical jacket areas can be neglected.
(Refer to Chem Eng Nov 17, 1971)

The velocity and turbulency through the hemispherical jacket area is slightly higher than the annular area. No problem would be expected in estimating the total heat transfer area by applying the same coefficient as in the annular area.

You can ignore the effects of the characteristic featuresaround hemispherical area in pressure drop and heat transfer calculations.

Hope the above help.

[tzarrin]

I am designing a jacketed vessel.the vessel which the jacket will surround has 3m inner dia and 8 m length using 12mm thick SHEET, the material is S/S.the inside of vessel has air .the surrounding jacket is under 45psig pressure and is filled with 100c water temp.i want to know the size of annulus space b/w vessel and jacket . i made some calculation may help u to understand my problem
plz reply quickly  Intro_new4.rtf   49.36KB   199 downloads

[sgkim]

Hi, tzarrin

To determine the equivalent diameters of spiral jackets, two approaches are to be made - one for heat transfer calculation and the other for pressure drop.

The shape of the cross sectional flow area through spiral jacket would be rectangular with width W and length L . (in your work sheet, W=0.033 m and L= 1.40/7 = 0.20 m, since every section has 7 spiral turns)

Then each equivalent diameter would be:
(1) for heat transfer calculation,
De= 4*(Flow Area)/(Heat Transfer Perimeter) = 4* (W*L)/(L) = 4W = 4*0.033 = 0.132 m

(2) for pressure drop calculaton,
D'e = 4*(Flow Area)/(Frictional Perimeter) = 4* (W*L)/(2W+2L)= 4*(0.033*0.20)/{2*(0.033+0.20)}=0.057 m.......

The above two are to be employed for the calculations of heat transfer and pressure drop respectively, but somewhat different definitions from the above could also be imagined as,

(3) the equivalent diameter for the same average flow velocity: D"e=(4WL/π)^0.5 from WL=(π/4)(D"e)^2. D"e=(4*0.33*0.2/π)^0.5=0.092 m and,

(4) the equivalent diameter for the same wetted perimeter: D"'e=2(W+L)/π
D"'e = 2*(0.033+0.20)/π = 0.148 m.

Traditionally, however, the equivalent (inside) diameter of a non-circular cross sectional flow path seems to be defined as the (inside)diameter of the circular pipe that has the same wetted perimeter of heat tranfer or frictional resistance as the non-circular path. (Refer to D.Q. Kern's "Process Heat Transfer" p.105 for details.)

I am sure all the pressure drop calculations for non-circular paths are based on the definition (2), but I do not think all of heat transfer model equations suggested by lots of authors employ consistently the definition like (1) above.

Stefano