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2 years ago

What type of power plant is virtually pollution free

1 answer:

6 0

A power plant that is virtually pollution free would be a power plant that does not consume gas, petrol, coal or any of the other resources that may cause pollution.

The virtually pollution-free plant will be mainly depending on natural renewable sources of energy

Examples:
1- geothermal plants (including dry steam plants, binary cycle plants and flash steam plants)
2- solar plant
3- wind turbines
4- water turbines

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A laser produces light at 5.32x10-7 m.

Blababa [14]

Answer:

Explanation:

speed = frequency x wavelength

speed of light in vacuum is 3.0 x 10^8

wavelength = 5.32 x10 ^-7

3.0 x 10 ^ 8 = 5.32 x 10^-7 x frequency

frequency = 5.63909 x 10^14

round it off = 5.64 x 10^14 Hz

thus the answer is D

hope this helps please mark it

6 0

3 years ago

In a building with 10.000 cubic feet where the air changes every two hours, what the rate of air change? A. 167.7 cfm B. 83.3 cf

Naya [18.7K]

Answer:

Flow rate of air is given as 83.33 cubic feet per minute

Explanation:

As we know that total volume of the air flow is given as

$V = 10,000 cubic feet$

also we know that total time is

$t = 2 hours = 120 min$

now we have flow rate given as

$Q = \frac{V}{t}$

$Q = \frac{10000}{120}$

$Q = 83.3 cf/m$

3 0

2 years ago

] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0

2 years ago

Explain the application of pascals law<br>

Marta_Voda [28]

Answer:

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6 0

3 years ago

Your car is traveling at 70/mi/h (1000ft/s) your sneeze and your eyes close for 0.33 seconds how far did you travel during your

Novay_Z [31]

If you're moving at 70 mi/hr, then you cover 33.88 feet in 0.33 sec. If you're moving at 1,000 ft/sec, then you cover 33 feet in 0.33 sec. 70 mph and 1,000 fps are not equivalent. 1,000 fps is about 682 mph, whereas 70 mph is about 103 fps.

6 0

3 years ago