I am currently working on a hobby project, and I would like to count on a little help for the calculation below;
Suppose I have a water vessel (cubus) with a volume of 1m³. The wall of the vessel is made of steel plate, but we will neglect this wall in our calculation. The vessel is insulated with PUR plates with a total thickness of 30cm.
The temperature of the water in the vessel is 80°C, the ambient temperature is 10°C. How long will it take for the water to reach a temperature of 50°C?
What do we know:
volume of water: 1m³
mass of water: 1000kg
surface along which heat loss will occur: 10.14m² (insulation added)
specific heat capacity of water: 4178 J/KgK
80°C = 353°K
50°C = 323°K
heat capacity difference = 1000KG x 4178K/KgK x 30°K = 125340000J = 125.34MJ
Lambda value PUR= 0.03W/mK
R-value PUR= 10m²K/W
I have already found a number of formulas via google, but I still can’t figure it out… What are your findings?
Answer
Best,
using your data (provided I have not miscalculated) I arrive at about 44 days.
I have attached an excel file with the calculation.
to calculate this in a simple way I assume that the temperature of the vessel is approximately constant for 10000s (you can also reduce this step to increase the accuracy of the calculation, furthermore the ambient T also remains at 10C and is thus supposedly unaffected by the vessel). we calculate how much heat Qt1 flows away in that time and then we calculate how much the temperature drops when Qt1 heat flows away this gives us a new temperature of the vessel 352.89947343226400000 then you can repeat this process again with this new one temperature as input.
after about 387 reps or 3870000s or about 44 days you will reach about 323K or 50C.
this is mathematically not the most correct method but does give a good approximation of the solution.
Best regards
sam
Answered by
ir. Sam van der Heijden
material sciences (plastics and nanotechnology)
http://www.ugent.be
.