/*! 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

91影视

Log out

Learning Materials

Features

Discover

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 \(18.22\mathrm{mm}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The fin array of Problem \(3.142\) is commonly found in compact heat exchangers, whose function is to provide a large surface area per unit volume in transferring heat from one fluid to another. Consider conditions for which the second fluid maintains equivalent temperatures at the parallel plates, \(T_{o}=T_{L}\), thereby establishing symmetry about the midplane of the fin array. The heat exchanger is \(1 \mathrm{~m}\) long in the direction of the flow of air (first fluid) and \(1 \mathrm{~m}\) wide in a direction normal to both the airflow and the fin surfaces. The length of the fin passages between adjoining parallel plates is \(L=8 \mathrm{~mm}\), whereas the fin thermal conductivity and convection coefficient are \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (aluminum) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) If the fin thickness and pitch are \(t=1 \mathrm{~mm}\) and \(S=4 \mathrm{~mm}\), respectively, what is the value of the thermal resistance \(R_{t, o}\) for a one-half section of the fin array? (b) Subject to the constraints that the fin thickness and pitch may not be less than \(0.5\) and \(3 \mathrm{~mm}\), respectively, assess the effect of changes in \(t\) and \(S\).

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.

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

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.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Electromagnetism

Read Explanation

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Mechanics and Materials

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.