An oxidation reactor is being fed with air to aid in the combustion. The air must first be heated from 25°C to 300°C at 1 atm pressure and the specific enthalpy change associated with this temperature change for air is 8170 J/mol. The inlet velocity of air is 5 m/s in pipe of 0.5 m ID (inside diameter). You may consider the gas to be ideal.
a. What is the inlet flow rate of air (mol/min)?
b. Given that the outlet diameter from this heater is 0.5 m ID, then what is the outlet velocity (m/s)?
c. Calculate the flow rate of heat required to heat the gas
(kW)?
a. Flow = pipe area x velocity
inlet flow = 3.14 x (0.5 m/2)2 x 5 m/s = 0.9813 m3/s
or 0.98 m3/s x 60 s/min = 59 m3/min
The molar volume of air at 25°C is
22.4 L/mol x 298 K/273 K = 22.45 L/mol or 0.02245 m3/mol
Inlet flow = 59 m3/min / 0.02245 m3/mol = 2630 mol/min
b.
you have same pipe diameter at inlet and outlet.
By heating, the air volume is increasing (see gas law):
573 K/ 298K = 1.9 times
You have the same increase for the outlet velocity
5 m/s x 1.9 = 9.6 m/s
c.
8170 J/mol x 2360 mol/min / 60 sec/min = 321 kJ/s = 321 kW
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