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. The mass flow rate of sulfurous acid added to the environment is kg/h.

m_s=9.335*10^-3 kg/h

m_(H2SO3)=0.0239 kg/h

Explanation:

solution:

First the mass flow rate of fuel is determined from the mass flow rate of air and the air/fuel ratio:  

m_fuel=m_air/(A÷F)

          =336/18 kg/h

          =18.67 kg/h

The mass flow rate of sulfur is determined from the mass flow rate of the fuel and the amount of sulfur:  

m_s=n*m_fuel

      =800 *10^-6*18.67

      =9.335*10^-3 kg/h

The mass flow rate is obtained suing the molar masses of sulfur and the sulfurous acid:  

m_(H2SO3)=(M_(H2SO3)/M_s)*m_s

                   =(82.0791/32.065)*9.335*10^-3

                   =0.0239 kg/h

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