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3 years ago

Do you think the mining process is faster when you know in advance that the land must be restored? Explain.

Engineering

1 answer:

vfiekz [6]3 years ago

3 0

Answer: No, the mining process isn't faster when you know in advance that the land must be restored.

Explanation:

The mining process would be slower when mining companies know in advance that the land must be restored because they have to be more careful about their impact on the environment & avoid taking unnecessary damage to ruin the land which they know they must restore.

So, it's evident how this would slow down the mining process. If there were no obligations to restore land, the mining companies would just come at the mining space and run their mining process fast without big concerns for the consequences.

But the caution definitely slows them down.

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What happens to a commercial airline at cruising altitude if the pilot does not touch the throttles?

jolli1 [7]

Answer:

it slows down

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3 years ago

The most important rating for batteries is the what

kifflom [539]

Answer:

I'm completely sure that the answer is: The most important rating for batteries is the ampere-hour rating. Ampere-hour is the battery discharge rating. It's used as a measure of charge in your device. It indicates how long your device will work without charging.

Explanation:

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2 years ago

The following is a list of metals and alloys:

viktelen [127]

Answer:

A) Gray cast iron

B) Aluminum

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2 years ago

A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl

bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1. Steady operating conditions exist.

2. Kinetic and potential energy changes are negligible.

Properties: The specific heat of geothermal water ( $c_{geo}$ [) is taken to be 4.18 kJ/kg.ºC.

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

P $P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }$

= $=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788$

b. Pump

$h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\$

$W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\$

c. $The thermal efficiency of the cycle n_{th} =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%$

7 0

3 years ago

Read 2 more answers

Two units for measuring magnetic flux density are:

romanna [79]

The correct answer is A
Faraday and Weber
:)

5 0

2 years ago