Answer:
[a]. 55.5kmol/min.
[b]. The average velocity = 8.3m/s.
pipe diameter = 1.0954m.
Explanation:
So, from the question above we are given the following data or information or parameters which are going to help us in solving the above question;
=> ''A gas stream that contains 1.50 mole% CO2 flows through a pipeline''
=> ''Twenty (20.0) kilograms of CO2 per minute is injected into the line.''
=> '' . A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2''
=> ''molar gas density is 0.123k mol/m3''
[a].The gas flow rate (k mol/min) upstream of the injection point can be calculated as follows.
- Determine the mass balance= j₁ +[ 20 /.123 × 44] = j₃. Thus, the flow gas rate is given as;
w₁ × 1.50 mole% CO2 + 3.6954= 2.3 mole% CO2 × w₃.
-3.610 = - 0.008w₁.
w₁ = 451.3 m³/min.
Thus, the flow rate = 0.123 × 451.3 = 55.51 kmol/min.
[b]. The average velocity = 150/ 18 seconds = 8.3 m/s.
[c]. Pipe Diameter can be calculated as follows;
0.123 × 8.3 × [3.12 × (pipe diameter)² / 4] = 55.51/60.
[1.0209 × 3.12]/4 × [pipe diameter]² = 0.9252.
0.8014065 × [pipe diameter]² = 0.9252.
[pipe diameter]² = 0.9252/ .8014065.
[pipe diameter]² = 1.15443.
[pipe diameter] = √1.15443.
[pipe diameter] = 1.2 m² = 1.0954m.