/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A spherical, cryosurgical probe ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 69

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)?

Short Answer

Expert verified
The thickness of the frozen tissue layer is approximately \(18.22\mathrm{mm}\).

Step by step solution

01

Define the variables

Let's define some variables that we will be using in this problem: - \(D_p = 3 \times 10^{-3} \mathrm{m}\) : Diameter of the probe - \(T_p = -30^{\circ}\mathrm{C} = 243.15 \mathrm{K}\) : Temperature of the probe - \(T_t = 37^{\circ}\mathrm{C} = 310.15 \mathrm{K}\) : Temperature of normal tissue - \(T_f = 0^{\circ}\mathrm{C} = 273.15 \mathrm{K}\) : Temperature at phase front - \(k_f = 1.5 \mathrm{\frac{W}{m \cdot K}}\) : Thermal conductivity of frozen tissue - \(h_{eff} = 50 \mathrm{\frac{W}{m^{2} \cdot K}}\) : Effective convection coefficient - \(d_f\) : Thickness of the frozen tissue layer (our unknown)
02

Setup energy balance at the phase front

At the phase front, the heat conducted through the frozen tissue layer from the probe must be equal to the heat being removed by the effective convection at the phase front. So our energy balance equation looks like this: \[Q_{cond} = Q_{conv}\] Where Q is the heat transfer rate.
03

Write the formulas for heat conduction and convection

We know the formulas for heat transfer by conduction and convection are: 1. Conduction: \(Q_{cond} = k_f A_{cond} \frac{\Delta T_{cond}}{d_f}\) 2. Convection: \(Q_{conv} = h_{eff} A_{conv} \Delta T_{conv}\) Where: - \(A_{cond}\) and \(A_{conv}\) are the areas of heat transfer through conduction and convection respectively. - \(\Delta T_{cond}\) and \(\Delta T_{conv}\) are the temperature differences driving the heat transfer through conduction and convection respectively.
04

Calculate the areas of heat transfer

The area of heat transfer through conduction and convection are the same in this case, which is the surface area of the frozen tissue layer. Assuming the thickness of the frozen tissue is small compared to the diameter of the probe, we can calculate the surface area of the frozen tissue layer as the surface area of a sphere with a diameter of \(D_p\): \[A_{cond} = A_{conv} = 4 \pi \left( \frac{D_p}{2} \right)^{2}\] Calculate area A: \[A = 4 \pi \left( \frac{3 \times 10^{-3}}{2} \right)^{2} = 2.827 \times 10^{-5} \mathrm{m^2}\]
05

Calculate the temperature differences for conduction and convection

We can calculate the temperature differences driving the heat transfer for each mode as: - \(\Delta T_{cond} = T_f - T_p = 273.15 - 243.15 = 30\mathrm{K}\) - \(\Delta T_{conv} = T_t - T_f = 310.15 - 273.15 = 37\mathrm{K}\)
06

Substitute the values in the energy balance equation and solve for thickness

Substitute the values in the energy balance equation from Step 2: \[k_f A_{cond} \frac{\Delta T_{cond}}{d_f} = h_{eff} A_{conv} \Delta T_{conv}\] Solve for thickness (\(d_f\)): \[d_f = \frac{k_f A_{cond} \Delta T_{cond}}{h_{eff} A_{conv} \Delta T_{conv}}\] Plug in the values and calculate the thickness: \[d_f = \frac{1.5 \times 2.827 \times 10^{-5} \times 30}{50 \times 2.827 \times 10^{-5} \times 37} = 0.01822\mathrm{m}\] So, the thickness of the frozen tissue layer is approximately \(18.22\mathrm{mm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. In this context, it tells us how well frozen tissue can transfer heat.
The higher the thermal conductivity, the more rapidly heat transfer occurs through the tissue.
For frozen tissue surrounding a cryosurgical probe, this property is critical in determining how effective the probe will be at freezing diseased tissue.
The given thermal conductivity of frozen tissue is 1.5 W/m路K, which means that every meter of tissue of 1 square meter area will conduct 1.5 watts for each degree Kelvin of temperature difference.
In our scenario, we use this property to calculate the heat transfer by conduction through the frozen layer, represented by the equation:\[Q_{cond} = k_f A_{cond} \frac{\Delta T_{cond}}{d_f}\]
Where:
  • \(k_f\) is the thermal conductivity
  • \(A_{cond}\) is the area through which heat is conducted
  • \(\Delta T_{cond}\) is the temperature difference driving the conduction
  • \(d_f\) is the thickness of the frozen tissue layer
Understanding thermal conductivity is essential for solving problems related to heating or cooling applications, such as in cryosurgery.
Convection Coefficient
The convection coefficient, also known as the heat transfer coefficient, describes the efficiency of heat transfer between a solid surface and a fluid (or in our case, between the frozen tissue and the surrounding tissue).
It quantifies the convective heat transfer rate per unit surface area per unit temperature difference.
In the exercise, the effective convection coefficient is given as 50 W/m虏路K, which indicates how effectively heat can move away from the phase front where the frozen tissue meets the surrounding warmer tissue.
The formula for heat transfer through convection in our scenario is:\[Q_{conv} = h_{eff} A_{conv} \Delta T_{conv}\]
Where:
  • \(h_{eff}\) is the effective convection coefficient
  • \(A_{conv}\) is the area of convective heat transfer
  • \(\Delta T_{conv}\) is the temperature difference driving the convection
This coefficient is crucial in maintaining the phase front at the desired condition so that the correct amount of tissue is frozen.
It helps determine how quickly the frozen boundary moves outward or inward as the probe operates.
Cryosurgical Probe
A cryosurgical probe is a medical device used to destroy tissue by freezing it. It is commonly applied in treating diseased tissue such as tumors.
The probe operates by absorbing heat from the targeted area, effectively lowering the temperature to freeze and kill cells.
In our exercise, the probe is spherical with a diameter of 3 mm, and its surface maintains a temperature of \(-30^{\circ}\)C.
This temperature is substantially lower than the surrounding tissue, resulting in a temperature gradient that facilitates heat transfer away from the tissue to the probe surface.
The efficiency of the cryosurgical probe in freezing tissue depends on:
  • Thermal conductivity of the tissue
  • Effective convection at the phase front
  • Initial temperature of the tissue and probe
By understanding these factors, healthcare professionals can manipulate the freezing process to ensure the complete destruction of targeted tissue while minimizing damage to nearby healthy cells.
Phase Change
Phase change refers to the transition of matter from one state to another, such as from liquid to solid. In the context of the exercise, the phase change is from normal (healthy) tissue to frozen tissue.
The idea is that by forming a cold front at the interface, we freeze a layer of the tissue.
This layer forms around the probe due to the temperature difference between the probe surface and normal tissue.
When considering heat transfer through phase change, both conduction and convection play a role.
At the phase front, the conduction through the frozen tissue and the convection into the normal tissue balance out to establish a stable phase boundary.
The formation of the frozen layer helps in targeting specific areas while preventing excess ice from damaging adjacent tissues.
This selective freezing ensures that the phase front stays at zero degrees Celsius, effectively demarcating the region being treated.

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Most popular questions from this chapter

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

A new building to be located in a cold climate is being designed with a basement that has an \(L=200\)-mm-thick wall. Inner and outer basement wall temperatures are \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{o}=0^{\circ} \mathrm{C}\), respectively. The architect can specify the wall material to be either aerated concrete block with \(k_{\mathrm{ac}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), or stone mix concrete. To reduce the conduction heat flux through the stone mix wall to a level equivalent to that of the aerated concrete wall, what thickness of extruded polystyrene sheet must be applied onto the inner surface of the stone mix con-crete wall? Floor dimensions of the basement are \(20 \mathrm{~m} \times 30 \mathrm{~m}\), and the expected rental rate is \(\$ 50 / \mathrm{m}^{2} /\) month. What is the yearly cost, in terms of lost rental income, if the stone mix concrete wall with polystyrene insulation is specified?

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

A spherical Pyrex glass shell has inside and outside diameters of \(D_{1}=0.1 \mathrm{~m}\) and \(D_{2}=0.2 \mathrm{~m}\), respectively. The inner surface is at \(T_{s, 1}=100^{\circ} \mathrm{C}\) while the outer surface is at \(T_{s, 2}=45^{\circ} \mathrm{C}\). (a) Determine the temperature at the midpoint of the shell thickness, \(T\left(r_{m}=0.075 \mathrm{~m}\right)\). (b) For the same surface temperatures and dimensions as in part (a), show how the midpoint temperature would change if the shell material were aluminum.

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?
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