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2 years ago

Are there any wastes associated with Wind Energy?

2 answers:

Taya2010 [7]2 years ago

8 0

Pollution costs

Wind turbines when isolated from the electric grid, produce negligible amounts of carbon dioxide, carbon monoxide, sulfur dioxide, nitrogen dioxide, mercury and radioactive waste when in operation, unlike fossil fuel sources and nuclear energy station fuel production, respectively.

Naily [24]2 years ago

6 0

Answer:

Pollution costs

Explanation:

Wind turbines when isolated from the electric grid, produce negligible amounts of carbon dioxide, carbon monoxide, sulfur dioxide, nitrogen dioxide, mercury and radioactive waste when in operation, unlike fossil fuel sources and nuclear energy station fuel production, respectively.

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1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

4 0

4 years ago

When you light a fireplace what energy transformation occurs?

snow_tiger [21]

<span>thermal energy

hope this helped</span>

4 0

3 years ago

Which element above has 6 protons in its nucleus?

Vinil7 [7]

B. carbon because it's atomic number is 6

5 0

3 years ago

Read 2 more answers

There is a 50 g sample of ra-229. It has a half-life of 4 minutes.how much will be left after 12 minutes? a. 3.13 g b. 6.25 g c.

polet [3.4K]

It's B. 6.25

first divide 50/2

then divide 25/2

after that divide 12.5/2

Then you get 6.25

Hope it helps!

3 0

3 years ago

A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press

earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

6 0

4 years ago