Answer:
See step by step explanations for answer.
Explanation:
600 megawatts =
568 690.272 btu / second
thermal eficiency=work done/Heat supllied
0.38=568690.272/Heat supplied
Heat supplied=1496553.35btu /s
heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s
feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day
sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input
so
the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %
the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day
Question
Determine the minimum hole diameter for the spring to operate in
Answer:
11 mm
Explanation:
Mean spring coil diameter is given from the following relationship
D=cd where C is spring index and d is the diameter of oil-tempered wire
By substitution
D=10*1 mm=10 mm
To find the outer diameter, we add the d to D hence getting the minimum hole diameter as
