Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a \(1.2\)-mm-diameter sphere. The properties of the 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}$, and the heat transfer coefficient between the junction and the gas is \(h=110 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine how long it will take for the thermocouple to read 99 percent of the initial temperature difference.

Short Answer

Expert verified
Question: Determine the time it takes for a thermocouple to read 99 percent of the initial temperature difference. Given: - Diameter of the thermocouple junction sphere (d) = 1.2 mm - Density (ρ) = 8500 kg/m³ - Specific heat capacity (cp) = 320 J/(kg·K) - Heat transfer coefficient (h) = 110 W/(m²·K) Following the steps in the provided solution: 1. Calculate the volume (V) and surface area (A) of the junction sphere. 2. Calculate the time constant (τ) using the lumped capacitance method. 3. Calculate the time (t) for the thermocouple to reach 99 percent of the initial temperature difference. Answer: The time it takes for the thermocouple to read 99 percent of the initial temperature difference is the calculated value for t.

Step by step solution

01

Calculate Volume and Area of the Junction Sphere

We can first find the volume and surface area of the junction sphere: $$ V=\frac{4}{3} \pi\left(\frac{d}{2}\right)^{3} \quad \text { and } \quad A=4 \pi\left(\frac{d}{2}\right)^{2} $$ where, \(d\) is the diameter of the thermocouple junction sphere (\(1.2\) mm).
02

Calculate Lumped Capacitance Time Constant

Now we can calculate the time constant in the lumped capacitance method: $$ \tau=\frac{\rho V c_{p}}{h A} $$ Using the given values for density (\(\rho = 8500 \frac{kg}{m^3}\)), specific heat capacity (\(c_p = 320 \frac{J}{kg \cdot K}\)), and heat transfer coefficient (\(h = 110 \frac{W}{m^2\cdot K}\)).
03

Calculate the Time for 99 Percent of Temperature Difference

Given that we want to find the time it takes to reach 99 percent of the initial temperature difference, we can use the following formula: $$ t = -\tau \ln (1 - 0.99) $$ Calculating the time using the obtained value for the time constant, \(\tau\).
04

Result

The time it takes for the thermocouple to read 99 percent of the initial temperature difference is the calculated value for \(t\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How does \((a)\) the air motion and \((b)\) the relative humidity of the environment affect the growth of microorganisms in foods?

How does immersion chilling of poultry compare to forced-air chilling with respect to \((a)\) cooling time, \((b)\) moisture loss of poultry, and (c) microbial growth?

It is claimed that beef can be stored for up to two years at $-23^{\circ} \mathrm{C}\( but no more than one year at \)-12^{\circ} \mathrm{C}$. Is this claim reasonable? Explain.

A hot dog can be considered to be a \(12-\mathrm{cm}\)-long cylinder whose diameter is \(2 \mathrm{~cm}\) and whose properties are $\rho=980 \mathrm{~kg} / \mathrm{m}^{3}\(, \)c_{p}=3.9 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.76 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\alpha=2 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\(. A hot dog initially at \)5^{\circ} \mathrm{C}\( is dropped into boiling water at \)100^{\circ} \mathrm{C}$. The heat transfer coefficient at the surface of the hot dog is estimated to be \(600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the hot \(\operatorname{dog}\) is considered cooked when its center temperature reaches \(80^{\circ} \mathrm{C}\), determine how long it will take to cook it in the boiling water. Solve this problem using the analytical one-term approximation method.

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=0.45\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at a uniform temperature of \)15^{\circ} \mathrm{C}$ are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be $0.95 \mathrm{~g} / \mathrm{cm}^{3}$ and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and $45 \mathrm{~min}$. Also, determine if any part of the chicken will freeze during this process. Solve this problem using the analytical one-term approximation method.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free