Scenario
1 answer:
Answer:
hello your question is poorly written and I have tried to understand it hence I will give a general diagram as related to your question
answer : attached below
Explanation:
Attached below is the required diagram of the situation you are trying to describe
let the wide receiver be ; W
Corner back = C
westward velocity = Vw
eastward velocity = Vc
at initial position X = Xw
x = 0 ( initial position )
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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