3 replies to this topic

[jeffthechef]

Hey All,

Some guidance with a problem I have would be fantastic.

The problem: A 12ltr diver air bottle contains high temp air (say 80ºC). It is submerged in water and I am trying to estimate the time taken to cool.

Assuming a sea temp of 10ºC and knowing dimensions and properties of the bottle, I have simplified the bottle to be a cylinder and can use the conductive heat transfer equation* to work out the heat transfer in Watts/m (or just watts by multiplying by the length of cylinder). A lot of assumptions I know, but just after an order of magnitude of time to cool.

Trouble is, I am then struggling to understand what to do with this value of Q, the rate of heat transfer from the cylinder. My textbooks have little information on unsteady heat transfer, so I am hoping somebody can point me in the right direction!? It's 12ltr of air at 1bar and 80ºC cooling by Q Watts, I'm sure it's simple but I'm stuck!!!

Any help would be appreciated!

Cheers

Jeff

* q = 2 π k (t_i - t_o) / ln(r_o / r_i)

[kkala]

Mentioned equation concerns heat transfer by conduction through cylinder side wall; average wall area is (similarly to LMTD) Am=L*2π(Ro-Ri)/ln(Ro/Ri), L=cylinder height, so q=k*A*(ti-to)/(Ro-Ri)=L*2πk(ti-to)/ln(Ro-Ri). But is heat conduction the controlling stage?
Taking k=45 W/(moC) for steel(http://en.wikipedia...._conductivities) and 3 mm wall thickness for the cylinder, corresponding heat transfer coefficient (through steel wall) is 45/0.003=15000 W/(oCm2), which is much higher than heat transfer coefficient for natural convection to sea water; the latter can be h= 20 - 100 W/(oCm2), see http://www.engineeri...sfer-d_430.html.
Therefore external convection seems to be the controlling stage in this case of heat transfer, not conduction through the wall. Air inside cylinder is only about 12 grammes (thus of negligible heat content), so it is neglected for simplicity, despite of its small local heat transfer coefficient. Nevertheless advice on this assumption would be welcomed.
Taking U=0.1 kW/(oCm2) due to sea currents (even so it might be conservative), and arbitrarily assuming a cylinder of 2R=0.15 m - L=0.68 m - wall thickness=3 mm (to be revised according the actual data), order of magnitude estimate can procede as follows.
Total cylinder surface: 0.356 m2.
Total cylinder weight (3 mm thickness, steel s.g.=8 g/cm3) : 8.5 kg
Steel heat capacity : 0.51 kJ/(oCkg), see (http://www.engineers...s_of_metals.htm.
Sea water temperature tw=10 oC (probably a higher value is realistic, in Greece tw = 11 - 27 oC, see [url]http://www.sailingissues.com/seawater-temperatures-greece.html[url].
Variation of cylinder temperature ti=80 oC to to=30 oC (say) during cooling.
Total heat loss from cylinder : q= 8.5*0.51*(80-30)= 217 kJ.
U & heat capacity assumed constant during cooling.
Average heat transfer Δt during cooling = ((80-10)+(30-10))/2=45 oC.
Estimated cooling time: 217/(0.356*0.1*45)=135 sec

Note. A differential equation could be used instead of the last average, though approximations are such that actual precision would not be improved: dq/dx = 0.356*0.1*(t-tw) = 8.5*0.51*dt/dx, where x= time in sec & t=cylinder temperature in oC. Thus 0.0356(t-10)=4.34dt/dx.

Edited by kkala, 02 September 2011 - 10:54 AM.

[jeffthechef]

Kkala,

Thank you very much for the help with this problem. Your solution of 135s is in the ball park of what I was expecting and seems much more feasible than the 3s I had clumsily arrived at.

Think I have some reading to do to understand calc, but it's the guidance I needed to help me in the right direction.

Thank you!

Jeff

[kkala]

Please note that result is based on 10 oC sea water and 30 oC final cylinder temperature, apart from rest assumptions. Time would be much longer, if e.g. sea water temperature were 27 oC. Besides sea water close to surface is further heated by sun, so at noon it is locally warmer.