Propane (C3H8) is burned in a combustion chamber at a flow rate of 10.50 m3/min
ID: 495482 • Letter: P
Question
Propane (C3H8) is burned in a combustion chamber at a flow rate of 10.50 m3/min with 20.0% excess air. If the propane and air streams enter the chamber at 30.0°C and 1.10 atm (absolute), and the products leave the chamber at 250.0°C and 2.50 atm (absolute), what is the volumetric flow rate of product gas from the chamber? Assume 100% conversion and ideal behavior. To solve for the flow rate, first label the process flow diagram below. Label any quantities as "unknown" if they are not given or implied above and "none" if the quantity is zero. Do not leave any blank spaces. Variable names are given for reference later.
n1 n2 F m3/min C3Ha OC atm m3/min air Combustion Chamber atm mol/min air mol O2/mol air mol N2/mol air 0.21 none 30.0 079 1.10 10.50 2.500 250.0 unknown ng n4 n5 n6 n8 m3/min product °C atm mol/min product mol/min CO mol/min H20 mol/min O mol/min N mol/min C3H 18Explanation / Answer
V1=10.50m3/min;T1=30;P1=1.10atm;n1=469 moles/min
V2=59.1L;T2=250;P2=2.5atm;n2=13434 moles/min
n3=3283 moles/min;
n4=3×469=1407 moles/min;n5=4×469=1876 mol/min;n6=5×469=2345mol/min;n7=. ;n8=469 mol/min
V3=22.4L;T3=273K;P3=1 atm
First let me see if I agree with your molar flow rate for propane.
(a)At 30and 1.10 atm, the molar volume should be
22.4 L (303/273) (1 atm/1.10 atm) = 22.4 L, so the 10.50 m^3 represents469moles per minute.
(a) The complete combustion requires
C3H8 + 5 O2 --> 3 CO2 + 4 H2O
so there must be 5 moles of O2 for every mole of propane. Bearing in mind that air is only 20.95% oxygen (by moles), you will need 5/0.2095 = 23.87 moles of air for every mole of propane. 469 moles of propane enter every minute, you need 11195moles of air for the combustion, plus the 30% excess, which makes 13434moles of air per minute.
(b) For every mole of propane burned, there are 7 moles of "products". Thus the molar flow rate of products leaving the chamber is 7*469= 3283moles/minute. I'm assuming the excess air is not considered a "product."
(c) Molar volume at 250 and 2.50 atm is
22.4 * (523/273) * (1/2.5) = 17.1136L,
so 3283moles per minute is 56183 L/min = 56.1m^3/min.
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