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Chapter 4: Problem 28

In an experiment, the temperature of a hot gas stream is to be measured by a thermocouple with a spherical junction. Due to the nature of this experiment, the response time of the thermocouple to register 99 percent of the initial temperature difference must be within \(5 \mathrm{~s}\). The properties of the thermocouple junction are \(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). If the heat transfer coefficient between the thermocouple junction and the gas is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the diameter of the junction.

Short Answer

Expert verified
Answer: The diameter of the thermocouple junction required to achieve a response time of 5 seconds is approximately 5.88 mm.

Step by step solution

01

Find the time constant required to reach 99 percent of the initial temperature difference

Since the response time is given as 5 seconds to reach 99 percent of the initial temperature difference, we can use the exponential relation between the response time (t) and the time constant (蟿) to find the time constant: \(t=5 \tau\) Since 蟿 is the time required to reach 63.2% of the initial temperature difference, the relation between t and 蟿 can be expressed as: \(0.99 = 1 - e^{-t / \tau}\) We need to solve this equation for 蟿.
02

Calculate the time constant for the thermocouple (蟿)

Solving the equation from the previous step, we find: \(\tau = -\frac{t}{\ln(1-0.99)}\) Plugging in the given response time, t = 5 seconds, we have: \(\tau = -\frac{5}{\ln(1-0.99)} = 1 \mathrm{~s}\) The time constant for the thermocouple is 1 second.
03

Use the time constant and thermocouple properties to find the diameter of the junction

The time constant of the thermocouple with a spherical junction is given by: \(\tau = \frac{\rho c_{p} r}{3h}\) Here, r is the radius of the junction, and h is the heat transfer coefficient between the junction and the gas. We want to find the diameter, D, where D = 2r. Substituting D/2 for r, we have: \(\tau = \frac{\rho c_{p} D}{6h}\) We can now solve this equation for D: \(D = \frac{6h \tau}{\rho c_{p}}\) Substitute the given values for 蟻, c_p, h, and 蟿: \(D = \frac{6(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}) (1 \mathrm{~s})}{(8500 \mathrm{~kg} / \mathrm{m}^{3})(320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})} = 0.005882 \mathrm{~m}\)
04

Convert the diameter to appropriate units and report the final answer

Convert the diameter from meters to millimeters: \(D = 0.005882 \mathrm{~m} \times \frac{1000 \mathrm{~mm}}{1 \mathrm{~m}} = 5.88 \mathrm{~mm}\) The diameter of the junction required to achieve a response time of 5 seconds is approximately 5.88 mm.

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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 is a measure of how easily heat is transferred from one material to another. In this experiment, it's important because it defines how quickly the heat from the hot gas is transferred to the thermocouple junction. A higher heat transfer coefficient means that heat is transferred more efficiently.
In this case, the heat transfer coefficient is given as 250 W/m虏路K. This value helps determine the rate at which the thermocouple can respond to changes in the gas temperature. For accurate temperature measurements, understanding the heat transfer coefficient is crucial:
  • It affects the time it takes for the thermocouple to reach thermal equilibrium with the gas.
  • A higher coefficient leads to a faster response time.
  • It is essential for calculating the thermocouple's time constant and ultimate diameter.
To achieve quick response times, optimizing the heat transfer coefficient is key, especially in dynamic environments such as this experiment with moving gas.
Spherical Junction
A spherical junction in a thermocouple means that the sensor part, or the junction, is shaped like a sphere. This shape is beneficial because it ensures uniform heat transfer from all directions, leading to consistent temperature readings.
Here's why a spherical junction is important:
  • This uniform shape helps in evenly distributing the heat transfer, avoiding hotspots that could lead to measurement errors.
  • Spheres have minimal surface area for a given volume, which can be advantageous in minimizing aerodynamic drag or heat loss to the surroundings.
  • The spherical shape simplifies the mathematical modeling of thermal response, which is crucial for calculating the time constant.
For this experiment, the spherical junction ensures that the calculations for response time and diameter take into account the unique thermal dynamics of a uniform shape.
Time Constant Calculation
The time constant (\(\tau\)) is essential to understanding how quickly the thermocouple responds to changes in temperature. It is defined as the time needed for the sensor to register approximately 63.2% of the total temperature change.
In this exercise, the time constant is calculated as 1 second. This calculation is critical because:
  • The defined response time for the experiment to register 99% of the temperature change is 5 seconds, which relates to the time constant by the equation: \(t = 5 \tau\).
  • The time constant provides insight into the speed and efficiency of the thermocouple's response.
Calculating the time constant accurately is foundational for designing the thermocouple to meet experimental requirements, such as the desired fast response time.
Diameter Calculation
The diameter of the thermocouple junction must be calculated to ensure that the desired response time is achieved. This involves understanding the relationship between the time constant and physical properties of the thermocouple.
The calculation uses the formula:\[ D = \frac{6h\tau}{\rho c_{p}} \]where \(D\) is the diameter.Substituting the given values, the resulting diameter is approximately 5.88 mm.
This process highlights several critical points:
  • The diameter affects the thermocouple's thermal capacity and therefore its response time.
  • Smaller diameters usually mean faster response times due to reduced thermal mass.
  • The calculated diameter must ensure the thermocouple's responsiveness aligns with the 5-second response requirement.
By using the relevant equations and experimental parameters, the diameter calculation ensures the thermocouple is appropriately designed for the experiment's conditions and requirements.

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Most popular questions from this chapter

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

For heat transfer purposes, an egg can be considered to be a \(5.5-\mathrm{cm}\)-diameter sphere having the properties of water. An egg that is initially at \(8^{\circ} \mathrm{C}\) is dropped into the boiling water at \(100^{\circ} \mathrm{C}\). The heat transfer coefficient at the surface of the egg is estimated to be \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the egg is considered cooked when its center temperature reaches \(60^{\circ} \mathrm{C}\), determine how long the egg should be kept in the boiling water. Solve this problem using analytical one-term approximation method (not the Heisler charts).

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

What are the factors that affect the quality of frozen fish?

In Betty Crocker's Cookbook, it is stated that it takes \(2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\) to roast a \(3.2-\mathrm{kg}\) rib initially at \(4.5^{\circ} \mathrm{C}\) "rare" in an oven maintained at \(163^{\circ} \mathrm{C}\). It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min}\). If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using analytical one-term approximation method (not the Heisler charts).
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