All categories

2 years ago

What are the advantages and disadvantages of a mine heardgear

Engineering

1 answer:

Irina18 [472]2 years ago

7 0

Answer:

If there is a shaft with headgear, then mining can take place until that depth of the shaft. If it is accessed horizontal Adits, it can mine until the lowest Adit from upwards. If it is accessed decline, the development and mining can continue so long as economic exploitation is possible.

Explanation:

What are the disadvantages of mining headgear? They totally cut off your vision of anything above your head. They are hot, most of the time

You might be interested in

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl

Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

Explanation:

(a) Volume compression ratio $\frac{v_1}{v_2} = 10$

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

$\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n$ Therefore p₂ = p₁ ÷ $(\frac{v_2}{v_1} )^n$ = (1 bar) ÷ $(\frac{1}{10} )^{1.3}$ = 19.953 bar

Similarly for a polytropic process we have

$\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}$ or T₂ = T₁ ÷ $(\frac{v_2}{v_1} )^{n-1}$ = $\frac{310}{0.1^{0.3}}$ = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = $\frac{8.3145}{28.9628}$ = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = $\frac{R(T_2-T_1)}{n-1}$ = $\frac{0.287(618.531-310)}{1.3 - 1}$ = 295.16 kJ/kg

Q = $(\frac{n-\gamma}{\gamma - 1} )W$ = $(\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg$ = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = $\frac{R(T_4-T_3)}{n-1}$ = Where T₄ is given by $\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}$ or T₄ = T₃ × $0.1^{0.3}$

= 2200 × $0.1^{0.3}$ T₄ = 1102.611 K

W = $\frac{0.287(1102.611-2200)}{1.3 - 1}$ = -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

$\eta = 1-\frac{T_4-T_1}{T_3-T_2}$ = $1-\frac{1102.611-310}{2200-618.531}$ = 0.499 or 49.9 % Efficient

$p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}$ where r = compression ratio and $r_p$ = $\frac{p_3}{p_2}$

However p₃ = $\frac{p_2T_3}{T_2}$ = $\frac{(19.953)(2200)}{618.531}$ =70.97 atm

$r_p$ = $\frac{p_3}{p_2}$ = $\frac{70.97}{19.953} = 3.56$

Therefore $p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}$ = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

8 0

3 years ago

Vivian's Violins has sales of $326,000, contribution margin of $184,000 and fixed costs total $85,000. Vivian's Violins net oper

frosja888 [35]

Based on the calculations, Vivian's Violins net operating income is equal to: D. $99,000.

<h3>How to calculate net operating income?</h3>

Mathematically, the net operating income of an individual or business firm can be calculated by using this formula:

Net operating income = Contribution margin - Total fixed costs

<u>Given the following data:</u>

Sales = $326,000

Contribution margin = $184,000

Fixed costs total $85,000

Substituting the given parameters into the formula, we have;

Net operating income = $184,000 - $85,000

Net operating income = $99,000.

Read more on net operating income here: brainly.com/question/24165947

#SPJ1

5 0

2 years ago

Mnsdcbjksdhkjhvdskjbvfdfkjbcv hjb dfkjbkjfvvfebjkhbvefgjdf

Julli [10]

Answer:

ik true eedcggftggggfffff

8 0

3 years ago

Read 2 more answers

Which of the following sensors is used to provide suspension control module with feedback regarding vehicle cornering forces?

Readme [11.4K]

Answer:

Explanation:

4 0

2 years ago

Explain the difference between thermoplastics and thermosets giving structure property correlation.

Misha Larkins [42]

Answer:

Explanation:

Thermosetting polymers are infusible and insoluble polymers. The reason for such behavior is that the chains of these materials form a three-dimensional spatial network, intertwining with strong equivalent bonds. The structure thus formed is a conglomerate of interwoven chains giving the appearance and functioning as a macromolecule, which as the temperature rises, simply the chains are more compacted, making the polymer more resistant to the point where it degrades.

Macromolecules are molecules that have a high molecular mass, formed by a large number of atoms. Generally they can be described as the repetition of one or a few minimum units or monomers, forming the polymers. In contrast, a thermoplastic is a material that at relatively high temperatures, becomes deformable or flexible, melts when heated and hardens in a glass transition state when it cools sufficiently. Most thermoplastics are high molecular weight polymers, which have associated chains through weak Van der Waals forces (polyethylene); strong dipole-dipole and hydrogen bond interactions, or even stacked aromatic rings (polystyrene). Thermoplastic polymers differ from thermosetting polymers or thermofixes in that after heating and molding they can overheat and form other objects.

Thermosetting plastics have some advantageous properties over thermoplastics. For example, better resistance to impact, solvents, gas permeation and extreme temperatures. Among the disadvantages are, generally, the difficulty of processing, the need for curing, the brittle nature of the material (fragile) and the lack of reinforcement when subjected to tension. But even so in many ways it surpasses the thermoplastic.

The physical properties of thermoplastics gradually change if they are melted and molded several times (thermal history), these properties are generally diminished by weakening the bonds. The most commonly used are polyethylene (PE), polypropylene (PP), polybutylene (PB), polystyrene (PS), polymethylmethacrylate (PMMA), polyvinylchloride (PVC), ethylene polyterephthalate (PET), Teflon (or polytetrafluoroethylene, PTFE) and nylon (a type of polyamide).

They differ from thermosets or thermofixes (bakelite, vulcanized rubber) in that the latter do not melt when raised at high temperatures, but burn, making it impossible to reshape them.

Many of the known thermoplastics can be the result of the sum of several polymers, such as vinyl, which is a mixture of polyethylene and polypropylene.

When they are cooled, starting from the liquid state and depending on the temperatures to which they are exposed during the solidification process (increase or decrease), solid crystalline or non-crystalline structures may be formed.

This type of polymer is characterized by its structure. It is formed by hydrocarbon chains, like most polymers, and specifically we find linear or branched chains

4 0

3 years ago

What are the advantages and disadvantages of a mine heardgear​

What are the advantages and disadvantages of a mine heardgear