A spherical, cryosurgical probe may be imbedded in diseased...

A spherical, cryosurgical probe may be imbedded in diseased tissue for the purpose of freezing, and thereby destroying, the tissue. Consider a probe of $3-\mathrm{mm}$ diameter whose surface is maintained at $-30^{\circ} \mathrm{C}$ when imbedded in tissue that is at $37^{\circ} \mathrm{C}$. A spherical layer of frozen tissue forms around the probe, with a temperature of $0^{\circ} \mathrm{C}$ existing at the phase front (interface) between the frozen and normal tissue. If the thermal conductivity of frozen tissue is approximately $1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ and heat transfer at the phase front may be characterized by an effective convection coefficient of $50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, what is the thickness of the layer of frozen tissue (assuming negligible perfusion)?

Key Concepts

Spherical Heat Conduction

In spherical coordinates, steady?state heat conduction with radial symmetry leads to a temperature distribution that depends inversely on the radial distance. The general solution takes the form T(r) = A + B/r, and the corresponding heat flux is calculated using the gradient of this temperature profile. This concept is crucial when analyzing heat transfer from a spherical source or sink, as it accurately describes how conduction behaves in three-dimensional, radially symmetric systems.

Convective Heat Transfer at a Boundary

Convective heat transfer is characterized by an effective convection coefficient that relates the heat flux at a surface to the temperature difference between the surface and the ambient environment. At the phase change interface, where the tissue temperature abruptly changes, this coefficient provides a means to quantify the heat loss from the frozen layer to the warmer tissue. It bridges the gap between conduction inside the frozen region and the external boundary conditions imposed by the surrounding tissue.

Thermal Resistance Network in Series

Complex heat transfer problems that involve both conduction and convection can be simplified using the thermal resistance concept, where each mode of heat transfer (conduction through the frozen tissue and convection at the phase front) is represented as a thermal resistor. The overall temperature difference is then divided into drops across each resistance, allowing for an energy balance approach that facilitates the calculation of characteristic dimensions such as the thickness of the frozen tissue layer.

Steady-State Energy Balance at the Phase Front

At the phase change interface, an energy balance must be maintained, meaning that the conductive heat flux arriving at the interface equals the convective heat flux leaving it. This balance provides a crucial boundary condition, linking the temperature distribution from within the frozen tissue to the convective losses into the unfrozen tissue. By equating these fluxes, one can solve for unknown variables such as the radius of the phase change front, and hence determine the thickness of the frozen region.

Transcript

00:01 The setup in this question is that we have a dual flask in which the top layer is made of ice over here.

00:12 And we are given that the ice, the temperature of the ice is at negative 18 degrees celsius.

00:22 We want to find what is the rate at which the ice below is the water below is actually freezing.

00:32 So this part of the flask is still water, right? so the edge over here is actually still at 0 degree celsius, just starting to freeze.

00:44 To find the rate at which this ice is basically forming, we will first need to find what is the rate at which heat is being transferred out from the water.

01:01 The rate of the heat transfer and with this rate of heat transfer we can then find how much water is being converted to ice per unit time per second and from there we can then find what is this speed right at which the ice is moving right based on the area over here i'm going to call this area to be a right just an unknown so the rates average heat is being conducted out of there is equals to k times a times delta t divided by d where do you see thickness so the thickness that we're actually interested in is 0 .7 centimeters all right so the conductivity of ice k over here is given to be 2 .2 watts per meter per degree celsius and the delta t will basically be 18 degrees because from 0 to minus 18 so we have 2 .2 times 18 divided by the thickness which is 0 .7 cm is also equals to 0 .2 0 .007 meters.

03:04 Alright, so this is our rates of heat transfer.

03:10 From this rate of heat transfer, we want to find what is the total amount of ice that's being converted.

03:23 So this implies that 5657 joules per meter square times a of energy is being transferred per second.

03:33 Energy transferred per second.

03:40 Using this value of the latent heat of fusion, taking this value divided by the latent heat of fusion, you'll be able to get the amount of mass of water that's actually converted into ice per second.

04:01 So this is 5657 divided by 334 kilojouze so we're going to convert kilojouze into joules so it times 10 power 3 alright so we're not done yet we need to convert this kilograms into meter square or meter cube sorry because what we really want to find is the volume so the volume of ice that has been created per second.

05:15 We use the density of ice, just about 0 .92 grams per cm cube...

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