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Chapter 2: Problem 77

A test technique for measuring the thermal conductivity of copper-nickel alloys is based on the measurement of the tip temperature of pin fins made from the alloys. The standard fin dimensions are a diameter of \(5 \mathrm{~mm}\) and a length of \(20 \mathrm{~cm}\). The test fin and a reference brass fin \((k=111 \mathrm{~W} / \mathrm{m} \mathrm{K})\) are mounted on a copper base plate in a wind tunnel. The test data include \(T_{B}=100^{\circ} \mathrm{C}, T_{e}=20^{\circ} \mathrm{C}\), and tip temperatures of \(64.2^{\circ} \mathrm{C}\) and \(49.7^{\circ} \mathrm{C}\) for the brass and test alloy fins, respectively. (i) Determine the conductivity of the test alloy. (ii) If the conductivity should be known to \(\pm 1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\), how accurately should the tip temperatures be measured?

Short Answer

Expert verified
The test alloy's thermal conductivity is approximately 50.73 W/mK. Tip temperatures should be measured within 0.3掳C accuracy.

Step by step solution

01

Define the Problem Parameters

First, let's organize all the given parameters from the problem. We have two fins: the test alloy fin and the reference brass fin, with known dimensions (diameter \(d = 5 \text{ mm}\) and length \(L = 20 \text{ cm}\)). The thermal conductivity of the brass is \(k_{brass} = 111 \text{ W/mK}\). The base temperature of the fins is \(T_B = 100^{\circ} \text{C}\) and the environment temperature is \(T_e = 20^{\circ} \text{C}\). The tested tip temperatures for the brass and alloy fins are \(T_{tip, brass} = 64.2^{\circ} \text{C}\) and \(T_{tip, alloy} = 49.7^{\circ} \text{C}\) respectively.
02

Calculate the Performance Ratio

The performance ratio \(R\) of the fins is defined as the tip temperature difference, normalized by the base and environment temperature difference for each fin. Calculate \(R_{brass}\) and \(R_{alloy}\) as follows:\[ R_{brass} = \frac{T_{tip, brass} - T_e}{T_B - T_e} \]\[ R_{alloy} = \frac{T_{tip, alloy} - T_e}{T_B - T_e} \] Substituting the given values:\[ R_{brass} = \frac{64.2 - 20}{100 - 20} = 0.5525 \]\[ R_{alloy} = \frac{49.7 - 20}{100 - 20} = 0.3725 \]
03

Relate Performance Ratios to Thermal Conductivities

Assuming that the geometric parameters and convective conditions are the same for both fins, the performance ratio is proportional to the square root of thermal conductivity:\[ \frac{R_{alloy}}{R_{brass}} = \left(\frac{k_{alloy}}{k_{brass}}\right)^{1/2} \]Rearranging gives:\[ k_{alloy} = k_{brass} \left( \frac{R_{alloy}}{R_{brass}} \right)^2 \]
04

Calculate the Thermal Conductivity of the Test Alloy

Use the formula derived in the previous step:\[ k_{alloy} = 111 \left( \frac{0.3725}{0.5525} \right)^2 \approx 50.73 \text{ W/mK} \]
05

Determine the Accuracy Required for Tip Temperature

To ensure that the thermal conductivity has a confidence interval of \(\pm 1.0 \text{ W/mK}\), the error in \(R_{alloy}\) needs to be calculated since it propagates to uncertainties in conductivity via:\[ \Delta k_{alloy} = 2k_{alloy} \frac{\Delta R_{alloy}}{R_{alloy}} \]Substituting values:\[ 1.0 = 2 \times 50.73 \times \frac{\Delta R_{alloy}}{0.3725} \]Solve for \(\Delta R_{alloy}\):\[ \Delta R_{alloy} \approx 0.0037 \]Convert this into a temperature difference knowing:\[ \Delta R_{alloy} = \frac{\Delta T_{tip, alloy}}{80} \] \[ \Delta T_{tip, alloy} \approx 0.2956^{\circ} \text{C} \]Thus, tip temperatures must be measured to within \(0.3^{\circ} \text{C}\) accuracy.

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Key Concepts

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

Copper-Nickel Alloys
Copper-nickel alloys are known for their exceptional resistance to corrosion, especially in marine environments. These alloys are made by combining copper and nickel in varying percentages, which significantly influences their thermal and electrical properties.
One of the key advantages of copper-nickel alloys is their thermal conductivity, which is essential in applications where heat dissipation is necessary. For instance, these alloys are used in heat exchangers because they efficiently transfer heat.
  • Enhanced corrosion resistance, especially in seawater.
  • Good thermal and electrical conductivity.
  • Used in marine hardware, shipbuilding, and power generation.
Understanding the thermal properties of copper-nickel alloys, such as thermal conductivity, helps in designing systems that require efficient heat management.
Pin Fins
Pin fins are small, protruding features on a surface that increase the surface area for heat exchange. They're commonly used in electronic components to dissipate heat, preventing overheating.
In the context of thermal conductivity measurements, pin fins made from different materials, like copper-nickel alloys, provide insights into how well these materials can conduct heat. The efficiency of a pin fin in transferring heat directly relates to its material's thermal conductivity.
Some benefits of using pin fins include:
  • Increased surface area for heat transfer, leading to better cooling efficiency.
  • Used in a wide range of cooling applications, from electronics to automotive systems.
Pin fins also affect airflow patterns in systems like wind tunnels, further influencing the transfer of thermal energy.
Temperature Measurement
Temperature measurement is critical when analyzing the thermal performance of materials. It involves determining the temperature at the tip of pin fins to calculate the material's thermal conductivity.
Accurate temperature measurements are essential because small errors in temperature can lead to significant errors in calculated thermal properties. For precise temperature measurement:
  • Use calibrated temperature sensors or thermocouples.
  • Ensure consistent environmental conditions during measurements.
In the given exercise, the tip temperature of the pin fins signals how effectively the material conducts heat from the base to the environment.
Thermal Analysis
Thermal analysis involves studying how materials conduct heat and how this affects their overall performance. In the exercise, conducting a thermal analysis involves calculating the thermal conductivity of a test alloy by comparing it with a known reference.
The performance ratio obtained from temperature measurements is pivotal since it correlates with the thermal conductivities of the fins being tested.
Key points in thermal analysis include:
  • Comparing known and unknown thermal properties through measurable parameters like temperature.
  • Understanding the influence of material properties on heat transfer efficiency.
Conducting thorough thermal analysis allows engineers to optimize material selection in thermal management applications.

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Most popular questions from this chapter

Consider a heat barrier consisting of a \(2 \mathrm{~mm}\)-thick brass plate to which \(3 \mathrm{~mm}\) copper tubing is soldered. The tubes are spaced \(10 \mathrm{~cm}\) apart. Cooling water passed through the tubes keeps them at approximately \(315 \mathrm{~K}\). The underside of the brass wall is insulated with a \(1.5 \mathrm{~cm}\)-thick asbestos layer, which in turn contacts a hot wall at \(600 \mathrm{~K}\). Assuming that the heat transfer from the cold side of the brass plate is negligible, estimate the temperature of the hottest spot on the brass wall. Take \(k=0.16 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\) for the asbestos and \(111 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the brass.

A spherical metal tank has a \(2.5 \mathrm{~m}\) outside diameter and is insulated with a \(0.5\) \(\mathrm{m}\)-thick cork layer. The tank contains liquefied gas at \(-60^{\circ} \mathrm{C}\) and the ambient air is at \(20^{\circ} \mathrm{C}\). The inside heat transfer coefficient can be assumed to be large, and the combined convection and radiation outside heat transfer coefficient is estimated to be \(8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Atmospheric water vapor diffuses into the cork, and a layer of ice forms adjacent to the tank wall. Determine the thickness of the layer. Assume that the cork thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \mathrm{K}\) is unaffected by the ice and water, but comment on the validity of this assumption.

A \(2 \mathrm{~mm}\)-diameter resistor, for an electronic component on a space station, is to have a sheath of thermal conductivity \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). It is cooled by forced convection with \(\bar{h}_{c} \simeq 1.1 D^{-1 / 2} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), for diameter \(D\) in meters, and by radiation with \(q_{\mathrm{rad}}=\sigma \varepsilon\left(T_{s}^{4}-T_{e}^{4}\right)\). Determine the radius of the sheath that maximizes the heat loss when the resistor is at \(400 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). Take the value of the surface emittance \(\varepsilon\) as (i) \(0.9\). (ii) \(0.5\).

A \(2 \mathrm{~cm}\)-thick composite plate has electric heating wires arranged in a grid in its centerplane. On one side there is air at \(20^{\circ} \mathrm{C}\), and on the other side there is air at \(100^{\circ} \mathrm{C}\). If the heat transfer coefficient on both sides is \(40 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), what is the maximum allowable rate of heat generation per unit area if the composite temperature should not exceed \(300^{\circ} \mathrm{C}\) ? Take \(k=0.45 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the composite material.

Heat is generated at a rate \(\dot{Q}_{v}^{\prime \prime \prime}\) in a long solid cylinder of radius \(R\). The cylinder has a thin metal sheath and is immersed in a liquid at temperature \(T_{e}\). Heat transfer from the cylinder surface to the liquid can be characterized by an overall heat transfer coefficient \(U\). Obtain the steady-state temperature distributions for the following cases: (i) \(\dot{Q}_{v}^{\prime \prime \prime}\) is constant. (ii) \(\dot{Q}_{v}^{\prime \prime \prime}=\dot{Q}_{v 0}^{\prime \prime \prime}\left[1-(r / R)^{2}\right]\). (iii) \(\dot{Q}_{v}^{\prime \prime \prime \prime}=a+b\left(T-T_{e}\right)\).
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