Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration , so:
where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:
W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2)= 1.976 rpm
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