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Chapter 3: Problem 9

A technique for measuring convection heat transfer coefficients involves bonding one surface of a thin metallic foil to an insulating material and exposing the other surface to the fluid flow conditions of interest. By passing an electric current through the foil, heat is dissipated uniformly within the foil and the corresponding flux, \(P_{\text {elec }}^{\prime \prime}\), may be inferred from related voltage and current measurements. If the insulation thickness \(L\) and thermal conductivity \(k\) are known and the fluid, foil, and insulation temperatures \(\left(T_{\infty}, T_{s}, T_{b}\right)\) are measured, the convection coefficient may be determined. Consider conditions for which \(T_{\infty}=T_{b}=25^{\circ} \mathrm{C}, P_{\text {elec }}^{\prime \prime}=2000\) \(\mathrm{W} / \mathrm{m}^{2}, L=10 \mathrm{~mm}\), and \(k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) With water flow over the surface, the foil temperature measurement yields \(T_{s}=27^{\circ} \mathrm{C}\). Determine the convection coefficient. What error would be incurred by assuming all of the dissipated power to be transferred to the water by convection? (b) If, instead, air flows over the surface and the temperature measurement yields \(T_{s}=125^{\circ} \mathrm{C}\), what is the convection coefficient? The foil has an emissivity of \(0.15\) and is exposed to large surroundings at \(25^{\circ} \mathrm{C}\). What error would be incurred by assuming all of the dissipated power to be transferred to the air by convection? (c) Typically, heat flux gages are operated at a fixed temperature \(\left(T_{s}\right)\), in which case the power dissipation provides a direct measure of the convection coefficient. For \(T_{s}=27^{\circ} \mathrm{C}\), plot \(P_{\text {elec }}^{\prime \prime}\) as a function of \(h_{o}\) for \(10 \leq h_{o} \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What effect does \(h_{o}\) have on the error associated with neglecting conduction through the insulation?

Short Answer

Expert verified
In summary, for water flow over the surface (scenario a), the convection coefficient is \(h_{o_a} = 998.4\, \text{W/m}^2 \cdot \text{K}\) and the error incurred by assuming all of the dissipated power to be transferred to the water by convection is -0.16%. For air flow over the surface (scenario b), the convection coefficient is \(h_{o_b} = 147.2\, \text{W/m}^2 \cdot \text{K}\) and the error incurred by assuming all of the dissipated power to be transferred to the air by convection is -8%.

Step by step solution

01

Write down the given values.

We have the following values: - Fluid, foil, and insulation temperatures: \(T_\infty = T_b = 25^{\circ} C\) - Electric power flux: \(P_{\text {elec }}^{\prime \prime} = 2000\, \text{W/m}^2\) - Insulation thickness: \(L = 10\, \text{mm}\) - Thermal conductivity: \(k = 0.040\, \text{W/m} 路 \text{K}\) - Foil temperature with water flow: \(T_s = 27^{\circ} \mathrm{C}\)
02

Express the conduction power flux.

The conduction power flux through the insulation is given by \[P_{cond}^{\prime\prime} = \frac {k (T_s - T_b)}{L}\]
03

Calculate the convection coefficient with water flow (scenario a).

Using the formula for the conduction power flux, we will find the convection coefficient in scenario a: \[P_{cond}^{\prime\prime} = \frac {k (T_s - T_b)}{L} = \frac {0.040 (27 - 25)}{0.01} = 3.2\, \text{W/m}^2\] The convection power flux is given by \[P_{conv}^{\prime\prime a} = P_{\text {elec }}^{\prime \prime} - P_{cond}^{\prime\prime} = 2000 - 3.2 = 1996.8\, \text{W/m}^2\] Now, we can find the convection coefficient for water flow using Newton's law of cooling: \[h_{o_a} = \frac{P_{conv}^{\prime\prime a}}{T_s - T_\infty} = \frac{1996.8}{27 - 25} = 998.4\, \text{W/m}^2 \cdot \text{K}\]
04

Calculate the error incurred in scenario a.

We will calculate the error incurred by assuming all of the dissipated power to be transferred to the water by convection in scenario a: \[\text{Error}_a = \frac{P_{conv}^{\prime\prime a} - P_{elec}^{\prime\prime}}{P_{elec}^{\prime\prime}} = \frac{1996.8 - 2000}{2000} = -0.0016 = -0.16\%\]
05

Calculate the convection coefficient with air flow (scenario b).

For scenario b, we have a foil temperature of \(T_s = 125^{\circ} \mathrm{C}\). We will calculate the conduction power flux and convection power flux in scenario b: \[P_{cond}^{\prime\prime} = \frac {k (T_s - T_b)}{L} = \frac {0.040 (125 - 25)}{0.01} = 160\, \text{W/m}^2\] The convection power flux is given by \[P_{conv}^{\prime\prime b} = P_{\text {elec }}^{\prime \prime} - P_{cond}^{\prime\doubleprime} = 2000 - 160 = 1840\, \text{W/m}^2\] Now, we can find the convection coefficient for air flow using Newton's law of cooling: \[h_{o_b} = \frac{P_{conv}^{\prime\prime b}}{T_s - T_\infty} = \frac{1840}{125 - 25} = 147.2\, \text{W/m}^2 \cdot \text{K}\]
06

Calculate the error incurred in scenario b.

We will calculate the error incurred by assuming all of the dissipated power to be transferred to the air by convection in scenario b: \[\text{Error}_b = \frac{P_{conv}^{\prime\prime b} - P_{elec}^{\prime\doubleprime}}{P_{elec}^{\prime\doubleprime}} = \frac{1840 - 2000}{2000} = -0.08 = -8\%\]

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Key Concepts

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

Electric Power Flux
Electric power flux refers to the distributed power over a given area, specifically in the context of heat generated by electrical means. In the exercise, the foil is heated by passing an electric current through it, with a known electric power flux denoted as \( P_{\text {elec }}^{\prime \prime} \). Here, it is given as 2000 \( \mathrm{W} / \mathrm{m}^2 \). This represents the rate at which electrical energy is being converted into heat per unit area.

Understanding electric power flux is crucial in evaluating the heat transfer processes involved as it helps in determining how much of the generated heat is conducted through the material and how much is lost to the surrounding fluid. Calculating any deviations or errors in assumptions that all electrical power is transferred via convection is key to accurately measuring the convection heat transfer coefficient.
  • It helps quantify the energy conversion from electrical to thermal.
  • Important for calculating conduction or convection heat fluxes.
  • Exposes potential errors in heat transfer assumptions.
Thermal Conductivity
Thermal conductivity \( k \) is a property of a material that indicates its ability to conduct heat. In this exercise, the material's thermal conductivity is provided as \( 0.040 \, \text{W/m} \cdot \text{K} \). Understanding this concept helps determine how efficient the insulating material is in transferring heat through it.

The conduction power flux can be calculated using the formula:
\[ P_{cond}^{\prime \prime} = \frac{k (T_s - T_b)}{L} \]
where \( T_s \) is the surface temperature of the foil, \( T_b \) is the background temperature, and \( L \) is the insulation thickness.
  • Higher thermal conductivity indicates better heat conduction ability.
  • Contributes to the overall heat transfer calculations in scenarios.
  • Aids in determining the potential error in the convection-only approach.
Infrared Emissivity
Infrared emissivity is a measure of a material's effectiveness in emitting energy as thermal radiation. An emissivity of 0.15, as given for the foil in the exercise, indicates that the foil is not a very efficient emitter of radiation, meaning less energy is lost through radiation compared to conduction and convection.

When air flows over the foil and the temperature rises to \( 125^{\circ} C \), the errors arise due mainly to assumptions about heat transfer only via convection. The emissivity plays a crucial role because it dictates the amount of radiation lost, which can be significant at higher temperatures.
  • Lower emissivity means less energy radiated at any given temperature.
  • Affects the calculation of the heat lost through radiation.
  • Must be considered in high-temperature environments to prevent miscalculations of convection heat losses.

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Most popular questions from this chapter

A nuclear reactor fuel element consists of a solid cylindrical pin of radius \(r_{1}\) and thermal conductivity \(k_{f}\). The fuel pin is in good contact with a cladding material of outer radius \(r_{2}\) and thermal conductivity \(k_{c^{*}}\). Consider steady-state conditions for which uniform heat generation occurs within the fuel at a volumetric rate \(\dot{q}\) and the outer surface of the cladding is exposed to a coolant that is characterized by a temperature \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain equations for the temperature distributions \(T_{f}(r)\) and \(T_{c}(r)\) in the fuel and cladding, respectively. Express your results exclusively in terms of the foregoing variables. (b) Consider a uranium oxide fuel pin for which \(k_{f}=2\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{1}=6 \mathrm{~mm}\) and cladding for which \(k_{c}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{2}=9 \mathrm{~mm}\). If \(\dot{q}=2 \times 10^{8}\) \(\mathrm{W} / \mathrm{m}^{3}, h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty}=300 \mathrm{~K}\), what is the maximum temperature in the fuel element? (c) Compute and plot the temperature distribution, \(T(r)\), for values of \(h=2000,5000\), and 10,000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the operator wishes to maintain the centerline temperature of the fuel element below \(1000 \mathrm{~K}\), can she do so by adjusting the coolant flow and hence the value of \(h\) ?

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of \(300-\mathrm{mm}\) inner diameter and \(30-\mathrm{mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to ensure that the temperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss(b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length. per meter of tube length?

From Problem 1.71, consider the wire leads connecting the transistor to the circuit board. The leads are of thermal conductivity \(k\), thickness \(t\), width \(w\), and length \(L\). One end of a lead is maintained at a temperature \(T_{c}\) corresponding to the transistor case, while the other end assumes the temperature \(T_{b}\) of the circuit board. During steady-state operation, current flow through the leads provides for uniform volumetric heating in the amount \(\dot{q}\), while there is convection cooling to air that is at \(T_{\infty}\) and maintains a convection coefficient \(h\). (a) Derive an equation from which the temperature distribution in a wire lead may be determined. List all pertinent assumptions. (b) Determine the temperature distribution in a wire lead, expressing your results in terms of the prescribed variables.

Consider uniform thermal energy generation inside a one-dimensional plane wall of thickness \(L\) with one surface held at \(T_{s, 1}\) and the other surface insulated. (a) Find an expression for the conduction heat flux to the cold surface and the temperature of the hot surface \(T_{s, 2}\), expressing your results in terms of \(k, \dot{q}, L\), and \(T_{s, 1}\). (b) Compare the heat flux found in part (a) with the heat flux associated with a plane wall without energy generation whose surface temperatures are \(T_{s, 1}\) and \(T_{s, 2}\).

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?
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