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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 wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at \(T_{\infty}=1200^{\circ} \mathrm{C}\) and maintains a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) over the blade. The blades, which are fabricated from Inconel, \(k \approx 20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), have a length of \(L=50 \mathrm{~mm}\). The blade profile has a uniform cross-sectional area of \(A_{c}=6 \times 10^{-4} \mathrm{~m}^{2}\) and a perimeter of \(P=110 \mathrm{~mm}\). A proposed blade- cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of \(T_{b}=300^{\circ} \mathrm{C}\). (a) If the maximum allowable blade temperature is \(1050^{\circ} \mathrm{C}\) and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory? (b) For the proposed cooling scheme, what is the rate at which heat is transferred from each blade to the coolant?

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

A truncated solid cone is of circular cross section, and its diameter is related to the axial coordinate by an expression of the form \(D=a x^{3 / 2}\), where \(a=1.0 \mathrm{~m}^{-1 / 2}\). The sides are well insulated, while the top surface of the cone at \(x_{1}\) is maintained at \(T_{1}\) and the bottom surface at \(x_{2}\) is maintained at \(T_{2}\). (a) Obtain an expression for the temperature distribution \(T(x)\). (b) What is the rate of heat transfer across the cone if it is constructed of pure aluminum with \(x_{1}=0.075 \mathrm{~m}\), \(T_{1}=100^{\circ} \mathrm{C}, x_{2}=0.225 \mathrm{~m}\), and \(T_{2}=20^{\circ} \mathrm{C}\) ?

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?
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