Thermal Conductivity and Boundry Layer

by morrobay
Tags: boundry, conductivity, layer, thermal
 PF Patron P: 347 Thermal Conductivity , k , for water is .1455 cal/sec/ meter * C ( converted watts to cal/sec) Q/t = k* A delta T/ d A= area M2 d = thickness of water/ice boundary layer Suppose I want to calculate time for 100 grams of ice to melt ( 8000 calories absorbed from surroundings by conduction. I have delta T and initial surface area of ice/water interface . Two questions : What would be the boundary layer thickness. d And in the case that the surface area, A, is decreasing as the ice melts is it correct to use just the initial surface area in the calculation ? My reasoning on the area is that it would actually be : Integral A(initial) - A(final) and A(final ) = 0
PF Patron
P: 347
 Quote by morrobay Thermal Conductivity , k , for water is .1455 cal/sec/ meter * C ( converted watts to cal/sec) Q/t = k* A delta T/ d A= area M2 d = thickness of water/ice boundary layer Suppose I want to calculate time for 100 grams of ice to melt ( 8000 calories absorbed from surroundings by conduction. I have delta T and initial surface area of ice/water interface . Two questions : What would be the boundary layer thickness. d And in the case that the surface area, A, is decreasing as the ice melts is it correct to use just the initial surface area in the calculation ? My reasoning on the area is that it would actually be : Integral A(initial) - A(final) and A(final ) = 0
I just did this experiment with 100 cc ice in 1 liter water at 30 deg C , 30000 cal
so final temp 22000 cal /1100 gram water = 20 C , melting time was 6 minutes
In close agreement with Q/t = k*A delta T/ x
x = thickness of boundary layer between ice and water = .002 M
Reference Journal Physical Oceanography
A= surface area 100 cc ice cylinder = .01224 M^2
delta t = 30 deg
k, thermal conductivity water = .1455 cal/sec/Meter * Kelvin
So, Q/t = .1455 cal/sec/M*T ( 30 C) (.01224 M^2) / .002 M = 26.5 cal/sec
8000 cal/ 26.5cal/sec = 301 sec. = 5 min

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