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Heat Loss calculation for Insulation thickness?

I have not had thermodynamis in 10+ years so please excuse my ignorance. I want to calculate the required thickness of an insulated matierial like thinsulate (R= 5 / in.) to prevent a temperature fluctuation within a glass of suspended solids.

I have a 3" diameter by 13" tall glass cylinder filled with water and solids. I do not want my suspension temperature to vary by more than 2* C over a 24 hour period. My room temperature may span from 15* to 25* C during that 24 hour period.

How would I determine the thickness and corresponding R value needed to maintain a 2*C internal temperature range?

Thanks in advance

Reply to
gtslabs
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Dear gtslabs:

Is the suspension doing anything exothermic or endothermic? Is the suspension supposed to be a particular value, or just "in between 15 and 25 C"? Is the glass cylinder sealed, so that evaporation cannot occur?

David A. Smith

Reply to
N:dlzc D:aol T:com (dlzc)

David No the suspension is not reactive. Just soil and water. The cylinder is open at the top. The temp should stay constant within 2 deg. c over a 24 hour period. Typical temperature would be 21*c. Evaporation should be ignored. Thanks

Reply to
gtslabs

You give the following worst case temperature loss of 21degC to 19degC in an ambient of 15degC after 24 hours. Supposing the volume of slurry is 1506 cc the maximal loss of energy desired is 2 X 1506 calories =

12650 joules in 24 hrs = 0.01464 watts through a surface area of 2 end caps and one cylinder wall 2 X 46sq cm + 791 = 883 cm sq

So the question is, what thickness of insulation with R=5 lets 0.01464 watts through an area of 883 cm sq when theres a temp difference of

6 degC (from 15degC to 21 degC) i.e 0.28 watts / per degC meter squared which is U = 0.28 or R = 5 / 0.28 = 18 This looks like 3.5 inches of thinsulate at R = 5/inch for the worst case.

Brian Whatcott Altus OK

Reply to
Brian Whatcott

Thank you. Can you explain how you got 3.5 inches from R = 5/0.28=18?

Reply to
gtslabs

U is the metric kindly unit for thermal conductance in watts per degreeC of temp drop across a square meter of surface

R is the American-customary-friendly unit for thermal resistance in degreesF of temp drop across a square foot of surface per BTU per hr.

So 1 R is 1 X 5/9 (degC/degF) x 12/39.37 X 12/39.37 (sq m/ sq ft)

----------------------------------------------------------------- 0.293 (watts/BTU/hr)

= 1/5.68 ( U/R)

But the one is the reciprocal of t'other, and I used 5 as a multiplier - shoulda been 5.68 i.e I started with U = 0.28 took the reciprocal 1/U = 3.57 and multipied 3.57 times 5 = R18

But more accurately, R = 1/U X 5.68 So actually R = 3.57 X 5.68 = 20 is closer. but I approximated in various places without worrying too much...

You specified insulation at R = 5 per inch. That's a thickness of four inches for R = 20

Brian Whatcott Altus OK

Reply to
Brian Whatcott

Use ISOWTC Software for the Insulation Thickness Calculation -

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Reply to
isowtc

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