Question

] 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. If a scrubber with 75% removal efficiency is utilized, what is the net emissions rate of sulfur? Express answer in lbs/hr.

Answer 1

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