PLANETARY COOLER SURFACE LOSSES

HOW CAN I CALCULATE THE SAME? ONE WAY IS SU = K .D.L K = FACTOR DEPENDS ON NO. OF COOLER TUBES D = DIA. OF TUBE L = LENGTH OF TUBE I HAVE 10 COOLER TUBES EACH 1.5 M DIA , 12.2 M LONG & TAKING K= 26 THAN RADIATION SURFACE PER METER = 1.5*26*12.2 = 485 M2 SO , ACCORDINGLY = 485/6 M2 PER METER PER TUBE = 79 M2 I TAKE TUBE SHELL TEMP. AT EVERY METER ALONG THE LENGTH OF TUBE FROM INLET TO OUTLET AT 6 LOCATIONS. I CALCULATE IT LIKE THIS 1 500K 79 M2 = KCAL/HR/M2 *79 2 600K 79 M2 = KCAL/HR/M2 *79 3 550K 79 M2 =KCAL/HR/M2 *79 4 573K 79 M2 =KCAL/HR/M2 *79 5 589K 79 M2 =KCAL/HR/M2 *79 6 600K 79 M2 = KCAL/HR/M2 *79 SUM IT IT WILL BE KCAL /HR IS IT CORRECT? RAJ

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Re: PLANETARY COOLER SURFACE LOSSES

My understanding is that shell losses from planetary coolers are estimated by using the diameter of a circle round the outside of all the tubes.

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Re: PLANETARY COOLER SURFACE LOSSES

Mr Raj

1. Find total surface area of each tube

2.find the mean temperture of the each tube (radiation temp)

3.find the radiation cofficient in that temperture from the graph (I think available in all cement operation handbooks)

use the multipy value as per the ave radiation temp.

200 degree - 17

250 degree - 18

300 degree - 20.5

350 degree - 23.5

find radiation in Kcal/hr as

 Factor X Surface area X (ave temp - base temp) for each cooler

sum all the cooler losses to get kcal lost per hour

thanks

chari

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Re: PLANETARY COOLER SURFACE LOSSES

Dear Chari Can we have the radiation factors exactly according to the surface temp. factor 17 has to used upto 200 deg. or at 200 deg. What will happen for the surface temp. greater than 350 deg? What about the convection losses from the surface? Also radiation losses are not directly proportional to temp. so I think we can not use the formula what u have written. Thanks Raj

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