A single light beam is split into two equal beams, denoted a and b. beam a travels through a medium with a higher index of refra
1 answer:
You might be interested in
The correct answer is
<span>c) very small and very large
Let's see this with a few examples:
1) if we have a very small number, such as
</span>
<span>we see that we can write it easily by using the scientific notation:
</span>
<span>2) Similarly, if we have a very large number:
</span>
<span>we see that we can write it easily by using again the scientific notation:
</span>
<span>
</span>
<span>c) very small and very large
Let's see this with a few examples:
1) if we have a very small number, such as
</span>
<span>we see that we can write it easily by using the scientific notation:
</span>
<span>2) Similarly, if we have a very large number:
</span>
<span>we see that we can write it easily by using again the scientific notation:
</span>
</span>
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:
= 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
The mass of the coal that is burned per sec = 187.97 lb/s
The mass of sulfur burned
= 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