Question

The indicator dilution method is a technique used to determine flow rates of fluids in channels for which devices like rotameters and orifice meters cannot be used (e.g., rivers, blood vessels, and large- diameter pipelines). A stream of an easily measured substance (the tracer) is injected into the channel at a known rate and the tracer concentration is measured at a point far enough downstream of the injection point for the tracer to be completely mixed with the flowing fluid. The larger the flow rate of the fluid, the lower the tracer concentration at the measurement point. 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.(a) Estimate the gas flow rate (k mol/min) upstream of the injection point.(b) Eighteen seconds elapses from the instant the additional CO2 is first injected to the time the CO2 concentration at the measurement point begins to rise. Assuming that the tracer travels at the average velocity of the gas in the pipeline (i.e., neglecting diffusion of CO2), estimate the average velocity (m/s), if the molar gas density is 0.123k mol/m3 what is the pipe diameter?

Answer 1

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.

Answer 2

A) The gas flow rate (k.mol/min) upstream of the injection point is;

55.51 k.mol/min

B) The average velocity and the diameter of the pipe are respectively;

8.33 m/s and 1.072 m

Gas Flow rate in thermodynamics

Due to the fact that there is no reaction or accumulation in the pipeline, we can say that the input of CO₂ is equal to it's output. Thus;

Upstream + Injection = Downstream

A) If we assume that x is the flow rate of gas (kmol/min), then there is the we can write the balance as:

0.015x + (20/44) = 0.023(x + 20/44)

Expanding gives us;

0.015x + 0.4545 = 0.023x + 0.01045

Solving for x gives;

x = 55.51 kmol/min

B) Formula for velocity is;

v = distance/time

Thus, since the length of the pipe is 150 m/s and time is 18 seconds, then velocity is;

v = 150/18

v = 8.33 m/s

Now, volumetric flow rate is calculated from;

Q = q' * ρ

where;

q' is gas flow rate

ρ is gas density

Thus;

Q = 55.51 * 0.123

Q = 451.3 m³/min = 7.522 m³/s

We can also express average velocity as;

v = Q/A

where A is area = πD²/4

Thus;

8.33 = 7.522/(πD²/4)

Making D the subject of the formula gives;

D = √((4 * 7.522)/(8.33π))

D = 1.072 m

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