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Chapter 9: Problem 112

A probe, used to measure the velocity of air in a lowspeed wind tunnel, is fabricated of an \(L=100 \mathrm{~mm}\) long, \(D=8\)-mm outside diameter horizontal aluminum tube. Power resistors are inserted into the stationary tube and dissipate \(P=1.5 \mathrm{~W}\). The surface temperature of the tube is determined experimentally by measuring the emitted radiation from the exterior of the tube. To maximize surface emission, the exterior of the tube is painted with flat black paint having an emissivity of \(\varepsilon=0.95\). (a) For air at a temperature and cross flow velocity of \(T_{\infty}=25^{\circ} \mathrm{C}, V=0.1 \mathrm{~m} / \mathrm{s}\), respectively, determine the surface temperature of the tube. The surroundings temperature is \(T_{\text {sur }}=25^{\circ} \mathrm{C}\). (b) For the conditions of part (a), plot the tube surface temperature versus the cross flow velocity over the range \(0.05 \mathrm{~m} / \mathrm{s} \leq V \leq 1 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The surface temperature of the tube (\(T_s\)) is found by setting up an equation for the total heat transfer rate, accounting for both convective and radiation heat transfer, and equating it to the power dissipation from the resistors. The equation for the total heat transfer rate is given by: \[ P = 1.5 W = hA(T_s - T_{\infty}) + \varepsilon \sigma A (T_s^4 - T_{sur}^4) \] To find \(T_s\) for the given conditions and velocity range (0.05 m/s to 1 m/s), first, determine the convective heat transfer coefficient (\(h\)) using the given problem parameters, then substitute it back into the equation. After simplification, find the expression for \(T_s\) as a function of cross-flow velocity (\(V\)). Finally, evaluate the expression for \(T_s\) for each value of \(V\) within the specified range, and plot the tube surface temperature against the cross-flow velocity.

Step by step solution

01

(Step 1: Determine heat transfer mechanisms and rates)

Given the problem description, we understand the two modes of heat transfer acting in this scenario: convection from the air to the tube and radiation from the tube to its surroundings. Thus, we can write the equation for the heat transfer rate within the tube as: \[ Q = Q_{conv} + Q_{rad} \] For the convection, we can use Newton's law of cooling, which states that the convection heat transfer rate is proportional to the temperature difference between the surface and the fluid: \[ Q_{conv} = hA(T_s - T_{\infty}) \] For the radiation, we can use the Stefan-Boltzmann law, which considers the surface temperature of the object (\(T_s\)) and the surroundings temperature (\(T_{sur}\)): \[ Q_{rad} = \varepsilon \sigma A (T_s^4 - T_{sur}^4) \] Now, we need to find the convective heat transfer coefficient (\(h\)).
02

(Step 2: Determine convective heat transfer coefficient)

In this case, we have a cross flow over a tube. It is common to use correlations to determine the average heat transfer coefficient. For example, we can use the Dittus-Boelter correlation, which requires the Reynolds number (\(Re\)) and Prandtl number (\(Pr\)) for air: \[ Nu = h \frac{D}{k} = C Re^m Pr^n \] However, the values of \(C\), \(m\) and \(n\) rely on experimental data, so we will use a simplified approach. From the problem description, we know that the total power dissipation \(P\) from the resistors is 1.5 W. For simplicity, we will assume that this entire power is accounted for through convection: \[P = Q_{conv}\] That means we can find \(h\) from this equation and the given problem parameters: \[ h = \frac{P}{A (T_\infty - T_s)} \]
03

(Step 3: Set up an equation for total heat transfer rate)

Combining the convective and radiation heat transfer equations, and using the power dissipation from the resistors as the total heat transfer rate, we find: \[ P = 1.5 W = hA(T_s - T_{\infty}) + \varepsilon \sigma A (T_s^4 - T_{sur}^4) \] Note that we know all parameters in the equation except \(T_s\). Hence, the above equation can be solved for the unknown \(T_s\).
04

(Step 4: Solve for \(T_s\) in part (a))

We can now solve for the surface temperature of the tube, given the initial conditions in part (a). Apply values to the equation (in Kelvin) as follows: \[T_{\infty} = 25 + 273.15 = 298.15 K \newline T_{sur} = T_{\infty} = 298.15 K \newline L = 100 \cdot 10^{-3} m, \: D = 8 \cdot 10^{-3} m \newline A = \pi D L, \newline V = 0.1 \: m/s \newline \varepsilon = 0.95\] To find the surface temperature of the tube, solve this nonlinear equation for \(T_s\) numerically: \[ P = 1.5 W = hA(T_s - T_{\infty}) + \varepsilon \sigma A (T_s^4 - T_{sur}^4) \]
05

(Step 5: Determine \(T_s\) as a function of \(V\))

Determine the mathematical expression of the surface temperature of the tube by substituting the equation for the convective heat transfer coefficient into the total heat transfer equation: \[ 1.5 W = \frac{P}{A (T_\infty - T_s)} A(T_s - T_{\infty}) + \varepsilon \sigma A (T_s^4 - T_{sur}^4) \] Simplify the equation and find the expression of \(T_s\) depending on \(V\). Keep in mind that \(h\) relies on the air properties, which in turn are a function of the cross-flow velocity \(V\). This can be calculated by the Dittus-Boelter correlation mentioned before and also considering the properties of air in terms of density and viscosity.
06

(Step 6: Plot \(T_s\) vs \(V\) for the given range in part (b))

Having obtained the mathematical expression for \(T_s\) as a function of \(V\), we are now able to plot the tube's surface temperature against the given velocity range: \[ 0.05 \: m/s \leq V \leq 1 \: m/s \] Using numerical analysis tools, such as MATLAB or Python, evaluate the expression of \(T_s\) for each value of \(V\) within the specified range, and then plot the values to visualize the tube surface temperature versus the cross-flow velocity.

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Key Concepts

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

Radiation Heat Transfer
Radiation heat transfer occurs when energy is emitted by a surface or body and travels through space without the need for a medium. This kind of heat transfer takes place in the form of electromagnetic waves. For instance, the heat you feel from the sun is due to radiation. In the given exercise, the aluminum tube uses radiation to emit heat to its surroundings.
This happens because the tube surface is painted with flat black paint, maximizing its emissivity to
  • Emissivity (\( \varepsilon \)): It is a measure of a material's ability to emit thermal radiation. A perfect black body has an emissivity of 1, and the painted tube here approximates this condition with \( \varepsilon = 0.95 \).
  • Stefan-Boltzmann Law: Helps to calculate radiation heat transfer rate with the formula: \[ Q_{rad} = \varepsilon \sigma A (T_s^4 - T_{sur}^4) \] where \( \sigma \) is the Stefan-Boltzmann constant.
Essentially, this equation helps understand how much heat is lost or gained due to radiation between two bodies at different temperatures. By using this information, students can see how different surface finishes affect heat dissipation.
Cross-Flow Convection
In cross-flow convection, a fluid (in this case, air) flows across a solid object, leading to heat transfer. This is a critical aspect when analyzing heat exchangers or any scenario where fluid moves perpendicularly over the surface. For the aluminum tube, the air in the wind tunnel moves across the tube, causing a heat exchange.
  • Newton's Law of Cooling: Expresses the convection heat transfer rate as proportional to the difference in temperature between the fluid and the solid surface \[ Q_{conv} = hA(T_s - T_{\infty}) \]
  • Reynolds and Prandtl Numbers: These dimensionless numbers help determine if convection is laminar or turbulent. This information is crucial for finding the appropriate correlation for average heat transfer coefficients.
Understanding cross-flow convection is important for calculating the primary means of heat dissipation from the tube in this scenario. Observing how velocity changes affect the convection process also illustrates how dynamic systems like wind tunnels operate.
Heat Transfer Coefficient
The heat transfer coefficient (\( h \)) is a pivotal parameter in convective heat transfer calculations. It quantifies the effectiveness of heat being transferred between a solid surface and a fluid flowing over or past it. In our example, the heat transfer coefficient determines how effectively the air stabilizes the temperature of the tube in the wind tunnel.

Factors Influencing Heat Transfer Coefficient

Several elements affect \( h \):
  • Fluid properties: Viscosity, thermal conductivity, specific heat, etc.
  • Flow characteristics: Whether the flow is laminar or turbulent plays a role.

Practical Applications

Engineers estimate \( h \) using correlations like the Dittus-Boelter formula, which incorporates the Reynolds and Prandtl numbers. These correlations make it possible to predict how varying air speeds (cross-flow velocity) influence \( h \), and thus how you can project the surface temperature of structures like the aluminum tube.Therefore, understanding how to find and use the heat transfer coefficient allows for precise engineering of systems to ensure optimal thermal performance.

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Most popular questions from this chapter

A computer code is being developed to analyze a 12.5-mm-diameter, cylindrical sensor used to determine ambient air temperature. The sensor experiences free convection while positioned horizontally in quiescent air at \(T_{\infty}=27^{\circ} \mathrm{C}\). For the temperature range from 30 to \(80^{\circ} \mathrm{C}\), derive an expression for the convection coefficient as a function of only \(\Delta T=\) \(T_{s}-T_{\infty}\), where \(T_{s}\) is the sensor temperature. Evaluate properties at an appropriate film temperature and show what effect this approximation has on the convection coefficient estimate.

In the analytical treatment of the fin with uniform cross-sectional area, it was assumed that the convection heat transfer coefficient is constant along the length of the fin. Consider an AISI 316 steel fin of 6-mm diameter and 50-mm length (with insulated tip) operating under conditions for which \(T_{b}=125^{\circ} \mathrm{C}, T_{\infty}=27^{\circ} \mathrm{C}\), \(T_{\text {sur }}=27^{\circ} \mathrm{C}\), and \(\varepsilon=0.6\). (a) Estimate average values of the fin heat transfer coefficients for free convection \(\left(h_{c}\right)\) and radiation exchange \(\left(h_{r}\right)\). Use these values to predict the tip temperature and fin effectiveness. (b) Use a numerical method of solution to estimate the foregoing parameters when the convection and radiation coefficients are based on local, rather than average, values for the fin.

A biological fluid moves at a flow rate of \(\dot{m}=0.02 \mathrm{~kg} / \mathrm{s}\) through a coiled, thin-walled, 5 -mm-diameter tube submerged in a large water bath maintained at \(50^{\circ} \mathrm{C}\). The fluid enters the tube at \(25^{\circ} \mathrm{C}\). (a) Estimate the length of the tube and the number of coil turns required to provide an exit temperature of \(T_{m, o}=38^{\circ} \mathrm{C}\) for the biological fluid. Assume that the water bath is an extensive, quiescent medium, that the coiled tube approximates a horizontal tube, and that the biological fluid has the thermophysical properties of water. (b) The flow rate through the tube is controlled by a pump that experiences throughput variations of approximately \(\pm 10 \%\) at any one setting. This condition is of concern to the project engineer because the corresponding variation of the exit temperature of the biological fluid could influence the downstream process. What variation would you expect in \(T_{m, o}\) for a \(\pm 10 \%\) change in \(\dot{m}\) ?

A horizontal 100-mm-diameter pipe passing hot oil is to be used in the design of an industrial water heater. Based on a typical water draw rate, the velocity over the pipe is \(0.5 \mathrm{~m} / \mathrm{s}\). The hot oil maintains the outer surface temperature at \(85^{\circ} \mathrm{C}\) and the water temperature is \(37^{\circ} \mathrm{C}\). Investigate the effect of flow direction on the heat rate (W/m) for (a) horizontal, (b) downward, and (c) upward flow.

The maximum surface temperature of the \(20-\mathrm{mm}-\) diameter shaft of a motor operating in ambient air at \(27^{\circ} \mathrm{C}\) should not exceed \(87^{\circ} \mathrm{C}\). Because of power dissipation within the motor housing, it is desirable to reject as much heat as possible through the shaft to the ambient air. In this problem, we will investigate several methods for heat removal. (a) For rotating cylinders, a suitable correlation for estimating the convection coefficient is of the form $$ \begin{gathered} \overline{N u}_{D}=0.133 \operatorname{Re}_{D}^{2 / 3} \operatorname{Pr}^{1 / 3} \\ \left(\operatorname{Re}_{D}<4.3 \times 10^{5}, \quad 0.7<\operatorname{Pr}<670\right) \end{gathered} $$ where \(R e_{D} \equiv \Omega D^{2} / \nu\) and \(\Omega\) is the rotational velocity (rad/s). Determine the convection coefficient and the maximum heat rate per unit length as a function of rotational speed in the range from 5000 to \(15,000 \mathrm{rpm}\). (b) Estimate the free convection coefficient and the maximum heat rate per unit length for the stationary shaft. Mixed free and forced convection effects may become significant for \(R e_{D}<4.7\left(G r_{D}^{3} / P r\right)^{0.137}\). Are free convection effects important for the range of rotational speeds designated in part (a)? (c) Assuming the emissivity of the shaft is \(0.8\) and the surroundings are at the ambient air temperature, is radiation exchange important? (d) If ambient air is in cross flow over the shaft, what air velocities are required to remove the heat rates determined in part (a)?
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