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

Calculate the time constant for the thermocouple (τ)

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 ρ, c_p, h, and τ: \(D = \frac{6(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}) (1 \mathrm{~s})}{(8500 \mathrm{~kg} / \mathrm{m}^{3})(320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})} = 0.005882 \mathrm{~m}\)
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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Most popular questions from this chapter

Oranges of \(2.5\)-in-diameter \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(78^{\circ} \mathrm{F}\) are to be cooled by refrigerated air at \(25^{\circ} \mathrm{F}\) flowing at a velocity of \(1 \mathrm{ft} / \mathrm{s}\). The average heat transfer coefficient between the oranges and the air is experimentally determined to be \(4.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine how long it will take for the center temperature of the oranges to drop to \(40^{\circ} \mathrm{F}\). Also, determine if any part of the oranges will freeze during this process.

A 5-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=\) \(8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is being heat treated as it moves through a furnace at a speed of \(1 \mathrm{~cm} / \mathrm{s}\). The air temperature in the furnace is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the furnace length is \(3 \mathrm{~m}\) and the stainless steel strip enters it at \(20^{\circ} \mathrm{C}\), determine the temperature of the strip as it exits the furnace.

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

Thick slabs of stainless steel \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and copper \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) are subjected to uniform heat flux of \(8 \mathrm{~kW} / \mathrm{m}^{2}\) at the surface. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Determine the temperatures of both slabs, at \(1 \mathrm{~cm}\) from the surface, after \(60 \mathrm{~s}\) of exposure to the heat flux.

Consider a 7.6-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(\left.\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Such a meat chunk intially at \(2^{\circ} \mathrm{C}\) is dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the center temperature of the meat chunk to rise to \(75{ }^{\circ} \mathrm{C}\) is (a) \(136 \mathrm{~min}\) (b) \(21.2 \mathrm{~min}\) (c) \(13.6 \mathrm{~min}\) (d) \(11.0 \mathrm{~min}\) (e) \(8.5 \mathrm{~min}\)
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