In order to meet a certain octane number specification, it is necessary to produ
ID: 494436 • Letter: I
Question
In order to meet a certain octane number specification, it is necessary to produce a petrol containing 80% w/w i-octane and 20% w/w n-heptane. What volume flow rates (m3/h) of high-octane petrol (90% w/w i-octane and 10% w/w n-heptane) and low-octane petrol (65% w/w l-octane and 35% w/w n-heptane) must be blended to obtain 100 m3/h of the desired petrol? (N.B. You cannot assume that volume is conserved) Density of low-octane petrol = 696 kg/m^3 Density of high-octane petrol = 701 kg/m^3 Density of product petrol = 699 kg/m^3 What is the mass flow rate of the final product? Write down the component balances in terms of the flow rates of the low- and high-octane petrol? Solve the simultaneous equations to find the low- and high-octane petrol flow rates?Explanation / Answer
Given
Volumetric flow rate of Product = 100 m3/hr
Density of product = 699 kg/m3
so mass flow rate of product = 100 m3 /hr * 699 kg/m3 = 69900 kg/hr
Given
petrol 1
90 % i octane x1F1 = 0.9 10 % n heptane x2F1 = 0.1 mass flow rate F1
petrol 2
65 % i octane x1F2 = 0.65 35 % n heptane x2F2 = 0.35 mass flow rate be F2
Product petrol
80 % i octane x1P = 0.8 20 % n heptane x2P = 0.2 mass flow rate be P
Mass will be conserved in the mixing process
writing overall mass balance
F1 + F2 = P
F1 + F2 = 69900 kg/hr --------> (1)
writing individual mass balance for i octane
x1F1 * F1 + x1F2 * F2 = x1P * P
0.9 * F1 + 0.65 * F2 = 0.8 * 69900
0.9 * F1 + 0.65 * F2 = 55920 kg/hr --------> (2)
writing individual mass balance for n heptane
x2F1 * F1 + x2F2 * F2 = x2P * P
0.1 * F1 + 0.35 * F2 = 0.2 * 69900
0.9 * F1 + 0.65 * F2 = 13980 kg/hr --------> (3)
so component balance are (1), (2), (3) This will be answer for 2)
solving these equations (1), (2), (3)
F1 = 41940 kg/hr
F2 = 27960 Kg/hr
Volumetric flow rate = mass flow rate / density
For high octane petrol flow rate = 41940 Kg/hr / 701 Kg/m3 = 59.83 m3/hr Answer 3)
For low octane petrol flow rate = 27960 Kg/hr / 701 Kg/m3 = 40.17 m3/hr Answer 3)