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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

In a production facility, 3-cm-thick large brass plates \(\left(k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) are heated by passing them through oven maintained at \(700^{\circ} \mathrm{C}\). The plates remain in the oven for a period of \(10 \mathrm{~min}\). Taking the convection heat transfer coefficient to be \(h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be solved using lumped system analysis? Justify your answer.

An 18-cm-long, 16-cm-wide, and 12 -cm-high hot iron block \(\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(20^{\circ} \mathrm{C}\) is placed in an oven for heat treatment. The heat transfer coefficient on the surface of the block is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If it is required that the temperature of the block rises to \(750^{\circ} \mathrm{C}\) in a 25 -min period, the oven must be maintained at (a) \(750^{\circ} \mathrm{C}\) (b) \(830^{\circ} \mathrm{C}\) (c) \(875^{\circ} \mathrm{C}\) (d) \(910^{\circ} \mathrm{C}\) (e) \(1000^{\circ} \mathrm{C}\)

The Biot number during a heat transfer process between a sphere and its surroundings is determined to be \(0.02\). Would you use lumped system analysis or the transient temperature charts when determining the midpoint temperature of the sphere? Why?

A potato that may be approximated as a \(5.7-\mathrm{cm}\) solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes by the time the center temperature reaches \(100^{\circ} \mathrm{C}\) is (a) \(56 \mathrm{~kJ}\) (b) \(666 \mathrm{~kJ}\) (c) \(838 \mathrm{~kJ}\) (d) \(940 \mathrm{~kJ}\) (e) \(1088 \mathrm{~kJ}\)

A potato may be approximated as a 5.7-cm-diameter solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-1} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes during a 30-min period is (a) \(77 \mathrm{~kJ}\) (b) \(483 \mathrm{~kJ}\) (c) \(927 \mathrm{~kJ}\) (d) \(970 \mathrm{~kJ}\) (e) \(1012 \mathrm{~kJ}\)
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