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

A long roll of 1 -m-wide and 0.5 -cm-thick \(1-\mathrm{Mn}\) manganese steel plate \(\left(\rho=7854 \mathrm{kg} / \mathrm{m}^{3}\right)\) coming off a furnace is to be quenched in an oil bath to a specified temperature. If the metal sheet is moving at a steady velocity of \(10 \mathrm{m} / \mathrm{min}\) determine the mass flow rate of the steel plate through the oil bath.

Short Answer

Expert verified
Answer: The mass flow rate of the steel plate through the oil bath is 131.0 kg/s.

Step by step solution

01

Convert dimensions from meters to centimeters

To calculate the volume, it is easier to work with the same units. Since the thickness is given in centimeters, we will convert other dimensions to centimeters: Width: \(1 \,\text{m} * 100 = 100 \,\text{cm}\) Velocity: \(10 \,\text{m/min} * 100 *\frac{1 \,\text{min}}{60 \,\text{s}}= 16.67 \,\text{cm/s}\)
02

Calculate the area of the cross-section of the steel plate

Using the width and the thickness given in the problem, we can determine the area of the cross-section: Cross-sectional area: \(A = \text{width} \times \text{thickness} = 100 \,\text{cm} \times 0.5 \,\text{cm} = 50 \,\text{cm}^2\)
03

Convert cross-sectional area to square meters

We will convert the cross-sectional area to square meters for further calculations: \(A = 50 \,\text{cm}^2 * \frac{1 \,\text{m}^2}{10000 \,\text{cm}^2} = 0.005 \,\text{m}^2\)
04

Calculate the volume flow rate of the steel plate

Using the cross-sectional area and the velocity, we can determine the volume flow rate of the steel plate: Volume flow rate: \(Q_v = Av = 0.005 \,\text{m}^2 \times 16.67 \,\text{cm/s} = 0.01667 \,\text{m}^3 \,\text{s}^{-1}\)
05

Calculate the mass flow rate

Using the density and the volume flow rate, we can find the mass flow rate. Recall that: \(\rho = \frac{m}{V} \implies m = \rho V\) Mass flow rate: \(Q_m = \rho Q_v = 7854\,\text{kg}/\mathrm{m}^3 \times 0.01667 \,\text{m}^3 \,\text{s}^{-1} = 131.0 \,\text{kg/s}\) The mass flow rate of the steel plate through the oil bath is \(131.0 \,\text{kg/s}\).

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.

Volume Flow Rate
Understanding the volume flow rate is paramount when examining the motion of fluids, or in our case, the movement of a solid object perceived as a continuous flow. The volume flow rate, represented by the symbol Qv, is defined as the volume of fluid (or material) that passes through a given surface per unit time.

For solid materials like the manganese steel plate in the exercise, we consider the continuous movement over the cross-sectional area, assuming the sheet's thickness remains uniform. By calculating the product of the cross-sectional area and the velocity of the steel sheet, we get a volume per second passing a given point, which in thermodynamics is critical for understanding heat and mass transfer during processes like quenching in an oil bath.

Qv = Area × Velocity
Where area is in square meters (m2) and velocity in meters per second (m/s). When these units are multiplied, we get cubic meters per second (m3/s), the unit for volume flow rate.
Thermodynamics
Thermodynamics is the study of heat, work, and energy within physical systems, and it plays a crucial role in industrial processes including the heat treatment of metals. The basic principles of thermodynamics allow us to understand how energy is transferred between the system (in this case, the steel plate) and its surroundings (the oil bath).

In our exercise, as the manganese steel plate moves through the oil bath, it loses heat energy due to the temperature difference between the hot steel and the cooler oil. This heat transfer must be precisely managed to avoid thermal stresses or unwanted microstructural changes in the metal. Engineers use principles of thermodynamics to calculate the rate of energy transfer and design cooling processes that result in the desired properties in the final product.
Density of Materials
Material density is central to many engineering equations, including the calculation of mass flow rate from volume flow rate. Density (ρ) is defined as the mass of a material per unit volume and is expressed in kilograms per cubic meter (kg/m3).

Density can vary greatly between different materials and even within the same material under different conditions. The density of the manganese steel used in the exercise, given as 7854 kg/m3, is essential to convert the volume flow rate into mass flow rate – which is more practical for applications where weight is more relevant than volume.

Understanding the density is also vital when designing components as it impacts the material's strength, durability, and how it interacts with other materials, such as in the oil quenching process where buoyancy and heat transfer rates are influenced by the steel's density.
Heat Treatment of Metals
Heat treatment of metals is a controlled process used to alter the microstructure of a material, thereby changing its mechanical properties such as strength, hardness, and ductility. The heat treatment process typically involves heating the metal to a specific temperature, holding it at that temperature, and then cooling it at a controlled rate.

Quenching, the process mentioned in the exercise, is one type of heat treatment where metal is rapidly cooled after being heated. By dunking the hot steel plate into an oil bath, the aim is to achieve specific material properties determined by the steel's composition and the cooling rate. The mass flow rate of the steel plate through the oil bath affects how quickly heat is removed from the material, influencing the hardness and strength of the final product.

An understanding of thermodynamics and material properties, such as density and specific heat capacity, is crucial in designing effective heat treatment processes to ensure the metal achieves the desired mechanical properties.

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

A scuba diver's \(2-\mathrm{ft}^{3}\) air tank is to be filled with air from a compressed air line at 120 psia and \(85^{\circ} \mathrm{F}\). Initially, the air in this tank is at 20 psia and \(60^{\circ} \mathrm{F}\). Presuming that the tank is well insulated, determine the temperature and mass in the tank when it is filled to 120 psia.

Refrigerant-134a at \(1.4 \mathrm{MPa}\) and \(90^{\circ} \mathrm{C}\) is throttled to a pressure of 0.6 MPa. The temperature of the refrigerant after throttling is \((a) 22^{\circ} \mathrm{C}\) \((b) 56^{\circ} \mathrm{C}\) \((c) 82^{\circ} \mathrm{C}\) \((d) 80^{\circ} \mathrm{C}\) \((e) 90^{\circ} \mathrm{C}\)

A 110 -volt electrical heater is used to warm \(0.3 \mathrm{m}^{3} / \mathrm{s}\) of air at \(100 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\) to \(100 \mathrm{kPa}\) and \(30^{\circ} \mathrm{C}\). How much current in amperes must be supplied to this heater?

A \(4-m \times 5-m \times 6-m\) room is to be heated by an electric resistance heater placed in a short duct in the room. Initially, the room is at \(15^{\circ} \mathrm{C}\), and the local atmospheric pressure is \(98 \mathrm{kPa} .\) The room is losing heat steadily to the outside at a rate of \(150 \mathrm{kJ} / \mathrm{min} .\) A \(200-\mathrm{W}\) fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of \(40 \mathrm{kg} / \mathrm{min} .\) The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 20 min for the room air to reach an average temperature of \(25^{\circ} \mathrm{C}\), find \((a)\) the power rating of the electric heater and ( \(b\) ) the temperature rise that the air experiences each time it passes through the heater.

A \(2-m^{3}\) rigid insulated tank initially containing saturated water vapor at \(1 \mathrm{MPa}\) is connected through a valve to a supply line that carries steam at \(400^{\circ} \mathrm{C}\). Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure in the tank rises to 2 MPa. At this instant the tank temperature is measured to be \(300^{\circ} \mathrm{C}\). Determine the mass of the steam that has entered and the pressure of the steam in the supply line.

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