Power plant efficiency/scrubber problem for intro environmental engineering class?

I'm having difficulty with this question...can anyone help?



"Consider a new 38 percent-efficient 600-MW power plant burning 9000 Btu/lb coal containing 1 percent sulfur. If a 70 percent-efficient scrubber is used, what would be the emission rate of sulfur (lb/hr)?



Thanks!Power plant efficiency/scrubber problem for intro environmental engineering class?W = 600 megawatts = 2.04728498 脳 10^9 Btu / hr
畏_pw = 38 % = 0.38
畏_sc = 70 % = 0.7
GHV = 9000 Btu/lb --%26gt; contained 1% sulphur for 1 lb Coal sulphur is 0.01 lb

Coal required every hour,
M_coal = W / 畏_pw*GHV = 2.04728498 脳 10^9 [Btu / hr] / 0.38*9000 [Btu/lb]
M_coal = 598621.3392 lb/hr

Sulphur rate,
S_total = 1% * M_coal
S_total = 5986.213392 lb/hr

Sulphur emission rate,
S_dot = 畏_sc*S_total
S_dot = 4190.35 lb/hr