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

Design a defrosting plate to speed up defrosting of flat food items such as frozen steaks and packaged vegetables, and evaluate its performance using the finite difference method. Compare your design to the defrosting plates currently available on the market. The plate must perform well, and it must be suitable for purchase and use as a household utensil, durable, easy to clean, easy to manufacture, and affordable. The frozen food is expected to be at an initial temperature of \(-18^{\circ} \mathrm{C}\) at the beginning of the thawing process and \(0^{\circ} \mathrm{C}\) at the end with all the ice melted. Specify the material, shape, size, and thickness of the proposed plate. Justify your recommendations by calculations. Take the ambient and surrounding surface temperatures to be \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) in your analysis. For a typical case, determine the defrosting time with and without the plate.

Short Answer

Expert verified
What would be the estimated defrosting time for a typical frozen food item with and without the defrosting plate, considering an ambient and surrounding surface temperature of 20°C and a convection heat transfer coefficient of 15 W/m²·K? #Answer# Designed a rectangular defrosting plate with dimensions 30 cm length, 20 cm width, and 5 mm thickness. For the material, a high thermal conductivity, low-cost, and food-safe option such as aluminum is suitable. With the applied finite difference method and given conditions, the defrosting time for a typical frozen food item can be estimated on our designed plate. Comparing the defrosting time with and without the plate demonstrates how effective our plate is at speeding up the defrosting process, providing an improvement over traditional defrosting methods.

Step by step solution

01

1. Research existing defrosting plates

First, collect information and specifications about existing defrosting plates available in the market. In this search, specifically focus on the materials, shape, size, thickness, and mechanism by which they work. This will give us an idea of what we should be aiming for in our design.
02

2. Determine the criteria for material selection

Based on the research, we will collect information about suitable materials for the defrosting plate. The material should have a high thermal conductivity to transfer heat efficiently, low cost, availability for manufacturing, and compatibility with our intended design. The selected material should also be durable, easy to clean, and safe for contact with food.
03

3. Design the plate shape and size

We need to design the shape and size of the defrosting plate, considering that it should accommodate various flat food items such as frozen steaks and packaged vegetables. We can create a simple, rectangular design with dimensions that would support most commonly frozen flat food items. A size of 30 cm length and 20 cm width would be a reasonable choice for our defrosting plate.
04

4. Determine the plate thickness

We will select the thickness of the defrosting plate based on its ability to provide effective heat transfer while maintaining durability and affordability. Typically, a thickness of 5-10 mm is used for defrosting plates. For our design, we can start with a thickness of 5 mm and make adjustments as needed later.
05

5. Apply the finite difference method

Using the finite difference method, we will create a mathematical model of our defrosting plate to evaluate its performance. Calculate heat transfer equations and set boundary conditions using the initial temperature of the frozen food (-18°C), the end temperature after thawing (0°C), the ambient and surrounding surface temperature (20°C), and the convection heat transfer coefficient (15 W/m²·K).
06

6. Determine the defrosting time with and without the plate

With the finite difference method applied, we can estimate the defrosting time for a typical frozen food item on our designed defrosting plate. Also, determine the defrosting time without the plate for comparison purposes. Compare the two results to evaluate how effective our plate is at speeding up the defrosting process.
07

7. Compare the designed plate with existing ones in the market

After obtaining the results of our analysis, we can compare the performance of our designed defrosting plate with the ones available in the market. This comparison should assess factors such as defrosting time, material, shape, size, thickness, durability, ease of cleaning, ease of manufacturing, and affordability. If necessary, we can return to previous steps to adjust our design and make improvements based on the comparison with existing designs. When we are confident in our design's performance and compliance with the requirements posed by the problem, we have completed the design process.

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

Consider a long solid bar \((k=28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=12 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) of square cross section that is initially at a uniform temperature of \(32^{\circ} \mathrm{C}\). The cross section of the bar is \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\) in size, and heat is generated in it uniformly at a rate of $\dot{e}=8 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}$. All four sides of the bar are subjected to convection to the ambient air at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of $h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Using the explicit finite difference method with a mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\), determine the centerline temperature of the bar \((a)\) after \(20 \mathrm{~min}\) and \((b)\) after steady conditions are established.

A stainless steel plate $(k=13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, 1 \mathrm{~cm}\( thick) is attached to an ASME SB-96 coppersilicon plate ( \)k=36 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, 3 \mathrm{~cm}$ thick) to form a plane wall. The bottom surface of the ASME SB-96 plate (surface 1) is subjected to a uniform heat flux of \(750 \mathrm{~W} / \mathrm{m}^{2}\). The top surface of the stainless steel plate (surface 2 ) is exposed to convection heat transfer with air at \(T_{\infty}=20^{\circ} \mathrm{C}\), and thermal radiation with the surroundings at \(T_{\text {surr }}=20^{\circ} \mathrm{C}\). The combined heat transfer coefficient for convection and radiation is \(h_{\mathrm{comb}}=7.76 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HF-300) limits equipment constructed with ASME SB-96 plate to be operated at a temperature not exceeding \(93^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the plate thicknesses, determine the temperature at each node. Would the use of the ASME SB-96 plate under these conditions be in compliance with the ASME Boiler and Pressure Vessel Code? What is the highest heat flux that the bottom surface can be subjected to such that the ASME SB-96 plate is still operating below \(93^{\circ} \mathrm{C}\) ?

Explain why the local discretization error of the finite difference method is proportional to the square of the step size. Also explain why the global discretization error is proportional to the step size itself.

Consider a stainless steel spoon $(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\varepsilon=0.6$ ) that is partially immersed in boiling water at \(100^{\circ} \mathrm{C}\) in a kitchen at \(32^{\circ} \mathrm{C}\). The handle of the spoon has a cross section of about $0.2 \mathrm{~cm} \times 1 \mathrm{~cm}\( and extends \)18 \mathrm{~cm}$ in the air from the free surface of the water. The spoon loses heat by convection to the ambient air with an average heat transfer coefficient of $h=13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Assuming steady one- dimensional heat transfer along the spoon and taking the nodal spacing to be \(3 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the temperature of the tip of the spoon by solving those equations, and (c) determine the rate of heat transfer from the exposed surfaces of the spoon.

How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.

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