2 replies to this topic

[sathish.chem]

Dear Members,

Have anyone of you experienced a heat loss calculation for the vessel. In my opinion the heat loss is less for the below case, however need to calculate to prove the assumptions.

My carbon steel vessel is vertical with elliptical heads, 1.5m diameters, 4m length & 10mm thickness. The vessel is filled with 4608kg of 99% Mono ethylene glycol (MEG), operating @ 1 barg. Fluid initial temperature is 141˚C duration of cooling is 2hr 30 min. I am trying to calculate outlet temperature at the end of 2hr 30 min. Vessel is uninsulated or with standoffs (PP) to allow natural cooling.

The Vessel surrounding temperature around 34˚C. Can any expert advice the solution to this problem?

Thanks a lot
Sathish Kumar

[shan]

Step 1.  To divide 150 min into N time intervals based the accuracy you need.

Step 2.  To use 141 C as the beginning temperature and to assume the tank temperature at the end of 1st time interval to calculate the heat loss during the 1st time interval.

Step 3.  To calculate tank temperature at end of 1st interval per the calculated heat loss, tank mass, steel specific heat, contained MEG mass and MEG specific heat.

Step 4.  If the calculated tank temperature is not matched with the assumed temperature, repeat Step 1 to Step 3 until they are matched.

Step 5.  When the calculated tank temperature is matched with the assumed the assumed tank temperature, calculate heat loss of the next time interval with verified the tank temperature at the end of the last time interval as the beginning temperature and to assume the tank temperature at the end of next time interval.

Step 6. To repeat Step 1 to Step 5 until reaching Nth time interval 150 min.   

[latexman]

 For no wind, no rain, no insulation and no agitation, I use 2 Btu/hr-ft²-^oF.