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Thermal Conductivity and Boundry Layer

by morrobay
Tags: boundry, conductivity, layer, thermal
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morrobay
#1
Apr9-11, 09:03 PM
P: 376
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
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morrobay
#2
Nov24-11, 02:26 AM
P: 376
Quote Quote by morrobay View Post
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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