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

Consider a 1000-W iron whose base plate is made of \(0.5-\mathrm{cm}\)-thick aluminum alloy \(2024-\mathrm{T} 6\left(\rho=2770 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(\left.875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \alpha=7.3 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\). The base plate has a surface area of \(0.03 \mathrm{~m}^{2}\). Initially, the iron is in thermal equilibrium with the ambient air at \(22^{\circ} \mathrm{C}\). Taking the heat transfer coefficient at the surface of the base plate to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine how long it will take for the plate temperature to reach \(140^{\circ} \mathrm{C}\). Is it realistic to assume the plate temperature to be uniform at all times?

Short Answer

Expert verified
Answer: It takes approximately 225.5 minutes for the iron's base plate to reach a temperature of 140°C under the given conditions.

Step by step solution

01

Calculate the heat generated by the resistance wire and transferred to the plate.

First, we need to find the amount of heat generated by the resistance wire that is transferred to the plate. Using the given wattage of the iron and the percentage of heat transferred to the plate, we can calculate this as follows: Heat Generated: Q_gen = 1000 W (Wattage) Heat Transferred: Q_transfer = 0.85 * Q_gen = 0.85 * 1000 W = 850 W
02

Calculate the bioheat transfer equation and the lumped capacity parameter.

We are given the heat transfer coefficient, h = 12 W/m²K, the surface area, A = 0.03 m², and the volume of the aluminum base plate as V = A * thickness = 0.03 m² * 0.005 m = 1.5 × 10⁻⁴ m³. The mass and the specific heat capacity of the aluminum base plate are given as ρ = 2770 kg/m³ and c_p = 875 J/kgK, respectively. Let's calculate the bioheat transfer equation and the lumped capacity parameter: Mass of the base plate: m = ρ * V = 2770 kg/m³ * 1.5 × 10⁻⁴ m³ = 0.4155 kg Lumped Capacity Parameter: Bi = (h * A) / (ρ * V * c_p) = (12 W/m²K * 0.03 m²) / (0.4155 kg * 875 J/kgK) ≈ 0.097 Since the calculated Lumped Capacity Parameter is less than 0.1, we can assume that the base plate temperature is uniform at all times.
03

Calculate the temperature difference, time constant, and the required time to reach the desired temperature.

We need to solve for the time it takes for the base plate to reach the desired temperature of 140°C. We are already given the initial temperature as 22°C. Temperature difference: ΔT = (140 - 22) °C = 118 °C Time constant (τ): τ = (ρ * V * c_p) / (h * A) = (0.4155 kg * 875 J/kgK) / (12 W/m²K * 0.03 m²) ≈ 10,178 s Now we can use the lumped capacity analysis formula to find the time required to reach the desired temperature: ΔT(t) = ΔT_initial * exp(-t/τ) t = -τ * ln(ΔT(t) / ΔT_initial) Plugging in the values, we get: t = -10,178 s * ln(0/118) ≈ 13,529 s
04

Converting the time to minutes and providing the answer.

Now that we have the time in seconds, we can convert it to a more convenient unit like minutes: t(min) = 13,529 s * (1 min / 60 s) ≈ 225.5 min So, it will take approximately 225.5 minutes for the plate temperature to reach 140°C under the given conditions, and it's realistic to assume the plate temperature to be uniform at all times.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Equilibrium
Thermal equilibrium is a fundamental concept in heat transfer that occurs when two or more substances or systems have no net exchange of heat. They are at the same temperature, hence, there is no temperature gradient driving heat flow. For example, when the base plate of an iron is initially at the same temperature as the surrounding air at 22°C, the system is in thermal equilibrium. Changes occur only when the iron starts to heat and energy flows from a hotter area to a cooler one until equilibrium is reached again with a new environment or state.
Heat Transfer Coefficient
The heat transfer coefficient, often denoted by the symbol \( h \), plays a crucial role in convective heat transfer processes. It quantifies the rate of heat transfer per unit area per unit temperature difference between a solid and a fluid in contact with each other. For example, the coefficient given as 12 \( W/m^2 \cdot K \) describes the efficiency of heat transfer from the surface of the iron's base plate to the surrounding air. Higher coefficients indicate more efficient heat transfer.
Lumped Capacity Analysis
Lumped capacity analysis is a simplification method used in heat transfer analysis, which assumes that the temperature within a body changes uniformly over time. This holds true when the object's lumped capacity parameter is less than 0.1. For the iron's base plate, the Bio number calculated as 0.097 allows us to apply this analysis, assuming uniform temperature throughout. This greatly simplifies the mathematical modeling of heating processes. It means complex spatial temperature gradients are neglected, which is valid for small, highly conductive objects like thin metal plates.
Specific Heat Capacity
Specific heat capacity (\( c_p \)) is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. For the aluminum alloy used in the iron's base plate, this value is 875 \( J/kg \cdot K \). It helps determine how much energy is needed to raise the temperature of the material, given its mass. A high specific heat means more energy is needed for a temperature change, influencing calculations of heat required in thermal applications like the one in the iron's heating process.

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

An ordinary egg can be approximated as a \(5.5-\mathrm{cm}-\) diameter sphere whose properties are roughly \(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.14 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\). The egg is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is dropped into boiling water at \(97^{\circ} \mathrm{C}\). Taking the convection heat transfer coefficient to be \(h=\) \(1400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the center of the egg to reach \(70^{\circ} \mathrm{C}\). Solve this problem using analytical one-term approximation method (not the Heisler charts).

Plasma spraying is a process used for coating a material surface with a protective layer to prevent the material from degradation. In a plasma spraying process, the protective layer in powder form is injected into a plasma jet. The powder is then heated to molten droplets and propelled onto the material surface. Once deposited on the material surface, the molten droplets solidify and form a layer of protective coating. Consider a plasma spraying process using alumina \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) powder that is injected into a plasma jet at \(T_{\infty}=15,000^{\circ} \mathrm{C}\) and \(h=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The alumina powder is made of particles that are spherical in shape with an average diameter of \(60 \mu \mathrm{m}\) and a melting point at \(2300^{\circ} \mathrm{C}\). Determine the amount of time it would take for the particles, with an initial temperature of \(20^{\circ} \mathrm{C}\), to reach their melting point from the moment they are injected into the plasma jet.

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

An experiment is to be conducted to determine heat transfer coefficient on the surfaces of tomatoes that are placed in cold water at \(7^{\circ} \mathrm{C}\). The tomatoes \((k=0.59 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=\) \(\left.0.141 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \rho=999 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.99 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) with an initial uniform temperature of \(30^{\circ} \mathrm{C}\) are spherical in shape with a diameter of \(8 \mathrm{~cm}\). After a period of 2 hours, the temperatures at the center and the surface of the tomatoes are measured to be \(10.0^{\circ} \mathrm{C}\) and \(7.1^{\circ} \mathrm{C}\), respectively. Using analytical one-term approximation method (not the Heisler charts), determine the heat transfer coefficient and the amount of heat transfer during this period if there are eight such tomatoes in water.

Hailstones are formed in high altitude clouds at \(253 \mathrm{~K}\). Consider a hailstone with diameter of \(20 \mathrm{~mm}\) and is falling through air at \(15^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(163 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the hailstone can be modeled as a sphere and has properties of ice at \(253 \mathrm{~K}\), determine the duration it takes to reach melting point at the surface of the falling hailstone. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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