Question

A diesel engine with an engine volume of 4.0 L and an engine speed of 2500 rpm operates on an air–fuel ratio of 18 kg air/kg fuel. The engine uses light diesel fuel that contains 800 ppm (parts per million) of sulfur by mass. All of this sulfur is exhausted to the environment, where the sulfur is converted to sulfurous acid (H2SO3). If the rate of the air entering the engine is 336 kg/h, determine the mass flow rate of sulfur in the exhaust. Also, determine the mass flow rate of sulfurous acid added to the environment if for each kmol of sulfur in the exhaust, 1 kmol sulfurous acid will be added to the environment. The molar mass of sulfur is 32 kg/kmol.

Answer 1

Answer:

mass flow rate of sulfur = 14.936*10^-3 kg/hr

mass flow rate of sulfurous acid = 38.23 * 10^-3 kg/hr

Explanation:

Please see attached images for detailed explanation.

Feel free to ask anything if it seems confusing. Thank you.

Answer 2

Answer:

Explanation:

i) (18 X 336) air/h = 6048 air/h

Engine speed = 2500 rpm; Capacity of diesel = 800 ppm

∴ 2500/800 = 3.125 kg/pm

Mass flow rate = (3.125 X 6048) Kg/h/pm =18900 Kg/h/pm

ii) Mass flow rate = 32 X 4/1000 X 2500 = 320 Kg/h/pm