Answer:Could you give a picture or something so I could answer.
Explanation:
Hi
Acetylene and propane
I hope this help you!
Answer:
2.83 kg
Explanation:
Given:
Volume, V = 0.8 m³
gage pressure, P = 200 kPa
Absolute pressure = gage pressure + Atmospheric pressure
= 200 + 101 = 301 kPa = 301 × 10³ N/m²
Temperature, T = 23° C = 23 + 273 = 296 K
Now,
From the ideal gas equation
PV = mRT
Where,
m is the mass
R is the ideal gas constant = 287 J/Kg K. (for air)
thus,
301 × 10³ × 0.8 = m × 287 × 296
or
m = 2.83 kg
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
Answer:
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