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1 year ago

Write the design brief (identify the problem) of mine headgear

1 answer:

3 0

The design brief that identified the problem of mine headgear is:

The structural structure above an underground mine shaft that facilitates the hoisting of machinery, persons, or supplies is known as a headframe (also called as a gallows frame, winding tower, hoist frame, pit frame, shafthead frame, headgear, headstock, or poppethead).

Mine headgear supports wheel systems that suspend winding cables that convey employees and ore up and down deep level shafts. These weird humanoid constructions have become the mining industry's defining emblem.

A miner's helmet consists of four major components:

Part 1: The hoist or winch is in a winding house. This component of the system is responsible for winding and unwinding the steel cable.

A motor and a control system are connected to the hoist.

When a steel cable unwinds from the winch, the mine cage and skips are lowered into the mine.

When the steel cable is wound up again, the mine cage and skips are elevated.

The sheave wheel is a pulley wheel that stands above the mining shaft in Part 2. The hoist rope travels over the sheave wheel and down the mine shaft.

The sheave wheel minimizes the mine cable's sliding friction.

Part 3: The head frame is the framework that holds the sheave wheel in place. When lifting the heavy mine cage, it must be robust enough to maintain the sheave wheel in place.

The head frame's left "legs" slope towards the hoist. This is due to the cable's strain dragging the entire frame in that direction. The sloping legs keep the head frame from tipping or collapsing.

Part 4: The cage and the jumps. Miners and equipment are transported up and down the mine in the cage. Skips are attached beside or beneath the cage.

Skips are used to transport ore and waste materials from mines.

<h3>What is a design brief?</h3>

A design brief, also called as a creative brief, is a program management document that identifies the scope, scale, and key aspects of your impending design project.

Learn more about design brief:

brainly.com/question/21422013

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1 year ago

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

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Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),

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Final temperature = 550 c

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calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

$\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}$

= 0.4167 = $1.0396e^{-0.4888*Fo}$

therefore Fo = 3.8264

Now determine how long the plate should be left in the furnace

Fo = $(\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )$

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L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

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1 year ago

About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa

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The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

$K.E = \frac{2KZe^2}{r}$

where;

Z is the atomic number of aluminium = 13
e is charge
r is distance of closest approach = thickness of aluminium
k is Coulomb's constant = 9 x 10⁹ Nm²/C²

<h3>For 2.5 MeV electrons</h3>

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

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1 year ago

Write the design brief (identify the problem) of mine headgear​

Write the design brief (identify the problem) of mine headgear