/*! 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 102 A cryogenic probe is used to tre... [FREE SOLUTION] | 91Ó°ÊÓ

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A cryogenic probe is used to treat cancerous skin tissue. The probe consists of a single round jet of diameter \(D_{e}=2 \mathrm{~mm}\) that issues from a nozzle concentrically situated within a larger, enclosed cylindrical tube of outer diameter \(D_{o}=15 \mathrm{~mm}\). The wall thickness of the AISI 302 stainless steel probe is \(t=2 \mathrm{~mm}\), and the separation distance between the nozzle and the inner surface of the probe is \(H=5 \mathrm{~mm}\). Assuming the cancerous skin tissue to be a semiinfinite medium with \(k_{c}=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(T_{c}=37^{\circ} \mathrm{C}\) far from the probe location, determine the surface temperature \(T_{s}\). Neglect the contact resistance between the probe and the tissue. Cold nitrogen exits the jet at \(T_{e}=100 \mathrm{~K}, V_{e}=20 \mathrm{~m} / \mathrm{s}\). Hint: Due to the probe walls, the jet is confined and behaves as if it were one in an array such as in Figure \(7.18 c\).

Short Answer

Expert verified
The surface temperature of the cryogenic probe, T_s, is approximately 107°C.

Step by step solution

01

Calculate heat transfer coefficient (h)

We can use the Nusselt number correlation for a jet in an array (Figure 7.18c), as the problem suggests. The correlation is: Nu = 0.27 * (2 * H / D_e) * (V_e * D_e / ν)^(1/2) * (H / L)^(1/7) Where Nu is the Nusselt number, ν is the kinematic viscosity of air, and L is the length of the confined jet (equal to H in this case). First, let's calculate the Reynolds number (Re): Re = V_e * D_e / ν We can assume the kinematic viscosity of air to be ν = 15 * 10^(-6) m^2/s. Re = (20 m/s * 0.002 m) / (15 * 10^(-6) m^2/s) ≈ 2666.67 Now, calculate L (equal to H): L = H = 0.005 m Calculate the Nusselt number using the correlation: Nu = 0.27 * (2 * 0.005 m / 0.002 m) * (2666.67)^(1/2) * (0.005 m / 0.005 m)^(1/7) ≈ 14.56 Now, we can calculate the heat transfer coefficient, h, using the Nusselt number: h = Nu * k_air / D_e Considering the thermal conductivity of air, k_air = 0.025 W/m K, we have: h = 14.56 * 0.025 W/m K / 0.002 m ≈ 181.95 W/m^2 K
02

Calculate surface temperature, T_s

Using Newton's law of cooling for the convective heat transfer and the conduction equation in the probe and the skin tissue, we get the following equation: Q = h * A * (T_e - T_s) = k_c * A / H * (T_s - T_c) Where Q is the heat transfer rate and A is the surface area of the jet in contact with the tissue. Now, to find T_s, we can rearrange the equation: T_s = (h * T_e + k_c * T_c / H ) / (h + k_c / H) Substituting the known values, T_s = (181.95 W/m^2 K * 100 K + 0.20 W/m K * 37°C / 0.005 m) / (181.95 W/m^2 K + 0.20 W/m K / 0.005 m) ≈ 107°C The surface temperature of the cryogenic probe, T_s, is approximately 107°C.

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Key Concepts

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

Heat Transfer Coefficient
The heat transfer coefficient, denoted as \( h \), is an essential concept in understanding how efficiently thermal energy moves between a surface and a fluid. In this exercise, it helps us determine the rate of heat exchange between the cryogenic probe and the surrounding air.
The higher the heat transfer coefficient, the more efficient the transfer of heat. It is usually measured in units of \( \, W/m^2 \cdot K \).
To calculate \( h \), we need to involve another crucial dimensionless number, the Nusselt number. The heat transfer coefficient is directly proportional to the Nusselt number and can be calculated with the equation:
\[ h = \frac{\text{Nu} \times k_{\text{air}}}{D_{e}} \]
Here, \( k_{\text{air}} \) is the thermal conductivity of air, which in the original problem is given as \( 0.025 \text{ W/m } K \). This indicates how readily heat can flow through the air at a micro level. With \( D_{e} \) being the diameter of the jet, we can apply this relation to get the required heat transfer coefficient, \( h \).
Nusselt Number
The Nusselt number (Nu) is a dimensionless parameter that is crucial for assessing the convective heat transfer between a surface and a fluid. Named after Wilhelm Nusselt, it represents the enhancement of heat transfer through a fluid as compared to conduction alone.
  • The Nusselt number model used in the given exercise is specific for a jet in an array, taking the form:
\[ \text{Nu} = 0.27 \times \left(\frac{2H}{D_e}\right) \times (\text{Re})^{1/2} \times \left(\frac{H}{L}\right)^{1/7} \]
In this formula:
  • \( H \) is the separation distance,
  • \( D_{e} \) is the diameter of the jet,
  • \( Re \) stands for the Reynolds number, which we calculate later,
  • \( L \) corresponds to the length of the confined jet, equal to \( H \) here.
By determining the Nusselt number from this correlation, we can then calculate the heat transfer coefficient \( h \), as it provides insight into how much heat is transferred via convection.
Reynolds Number
The Reynolds number \( (Re) \) is another critical dimensionless quantity in fluid dynamics. It predicts flow patterns and indicates whether the flow will be laminar or turbulent. In the problem at hand, the Reynolds number helps us understand the behavior of the cold nitrogen as it exits the probe.
This number is calculated using the formula:
\[ \text{Re} = \frac{V_e \times D_e}{u} \]
where \( V_e \) is the velocity of the fluid, \( D_{e} \) is the jet diameter, and \( u \) is the kinematic viscosity.
  • The kinematic viscosity is essentially the fluid's resistance to flow. In this situation, it's assumed to be \( 15 \times 10^{-6} \text{ m}^2/\text{s} \).
  • A higher Reynolds number typically indicates turbulent flow, which influences the effectiveness of the cooling process.
When the Reynolds number is high, convective heat transfer efficiency increases, leading to a potentially quicker cooling time.
Thermal Conductivity
Thermal conductivity, symbolized as \( k \), is intrinsic to materials and signifies their ability to conduct heat. Materials with high thermal conductivity are effective at transferring heat, whereas those with low thermal conductivity act as insulators.
In our exercise, thermal conductivity plays a role in determining how heat moves through both the probe and the cancerous skin tissue.
The formula used for the surface temperature \( T_s \) involves the thermal conductivity of the tissue \( k_{c} \), which is given as 0.20 W/m K. This relates to how well the skin tissue itself can absorb and transfer heat away from the probed area.
  • High thermal conductivity in tissues or materials results in a more swift spread of temperature changes.
  • The property is vital for applications where temperature regulation and transfer are critical factors, as with the cryogenic probe.
Understanding and applying thermal conductivity allows for precise control of temperatures in medical therapies and technological processes.

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Most popular questions from this chapter

Consider laminar, parallel flow past an isothermal flat plate of length \(L\), providing an average heat transfer coefficient of \(\bar{h}_{L^{-}}\)If the plate is divided into \(N\) smaller plates, each of length \(L_{N}=L / N\), determine an expression for the ratio of the heat transfer coefficient averaged over the \(N\) plates to the heat transfer coefficient averaged over the single plate, \(\bar{h}_{L, N} / \bar{h}_{L, 1}\).

Air at \(27^{\circ} \mathrm{C}\) with a free stream velocity of \(10 \mathrm{~m} / \mathrm{s}\) is used to cool electronic devices mounted on a printed circuit board. Each device, \(4 \mathrm{~mm} \times 4 \mathrm{~mm}\), dissipates \(40 \mathrm{~mW}\), which is removed from the top surface. A turbulator is located at the leading edge of the board, causing the boundary layer to be turbulent. (a) Estimate the surface temperature of the fourth device located \(15 \mathrm{~mm}\) from the leading edge of the board. (b) Generate a plot of the surface temperature of the first four devices as a function of the free stream velocity for \(5 \leq u_{s} \leq 15 \mathrm{~m} / \mathrm{s}\). (c) What is the minimum free stream velocity if the surface temperature of the hottest device is not to exceed \(80^{\circ} \mathrm{C}\) ?

An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches are \(S_{L}=S_{T}=24 \mathrm{~mm}\). There are 3 rows of elements in the flow direction \(\left(N_{L}=3\right)\) and 4 elements per row \(\left(N_{T}=4\right)\). Atmospheric air with an upstream velocity of \(12 \mathrm{~m} / \mathrm{s}\) and a temperature of \(25^{\circ} \mathrm{C}\) moves in cross flow over the elements, which have a diameter of \(12 \mathrm{~mm}\), a length of \(250 \mathrm{~mm}\), and are maintained at a surface temperature of \(350^{\circ} \mathrm{C}\). (a) Determine the total heat transfer to the air and the temperature of the air leaving the duct heater. (b) Determine the pressure drop across the element bank and the fan power requirement. (c) Compare the average convection coefficient obtained in your analysis with the value for an isolated (single) element. Explain the difference between the results. (d) What effect would increasing the longitudinal and transverse pitches to \(30 \mathrm{~mm}\) have on the exit temperature of the air, the total heat rate, and the pressure drop?

A circular pipe of 25 -mm outside diameter is placed in an airstream at \(25^{\circ} \mathrm{C}\) and 1 -atm pressure. The air moves in cross flow over the pipe at \(15 \mathrm{~m} / \mathrm{s}\), while the outer surface of the pipe is maintained at \(100^{\circ} \mathrm{C}\). What is the drag force exerted on the pipe per unit length? What is the rate of heat transfer from the pipe per unit length?

An array of 10 silicon chips, each of length \(L=10 \mathrm{~mm}\) on a side, is insulated on one surface and cooled on the opposite surface by atmospheric air in parallel flow with \(T_{\infty}=24^{\circ} \mathrm{C}\) and \(u_{\infty}=40 \mathrm{~m} / \mathrm{s}\). When in use, the same electrical power is dissipated in each chip, maintaining a uniform heat flux over the entire cooled surface. If the temperature of each chip may not exceed \(80^{\circ} \mathrm{C}\), what is the maximum allowable power per chip? What is the maximum allowable power if a turbulence promoter is used to trip the boundary layer at the leading edge? Would it be preferable to orient the array normal, instead of parallel, to the airflow?

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