Pressure drop in a pipeline Consider a pipeline made of commercial steel with an inner diameter of 40 cm for transporting CO2 at a flow rate of 3.5 Mega tons per year. Use CO2 properties under 11 MPa and 25 °C as follows: ρ = 877 kg/m³ and μ = 7.73 × 10^-5 Pa · s. Estimate the pressure drop over a

Question

Pressure drop in a pipeline
Consider a pipeline made of commercial steel with an inner diameter of 40 cm for transporting [latex]CO_{2}[/latex] at a flow rate of 3.5 Mega tons per year. Use [latex]CO_{2}[/latex] properties under 11 MPa and 25 °C as follows: ρ = 877 kg/m³ and μ = 7.73 × [latex]10^{-5}[/latex] Pa · s. Estimate the pressure drop over a distance of 100 km.

Accepted Answer

Convert the unit of the [latex]CO_{2}[/latex] flow rate as
[latex]\dot{m}=3.5\frac{Mt}{year}=\frac{3.5 imes 10^{6} imes 1000 kg}{365 imes 24 imes 3600 s}=111 kg/s[/latex]

The average speed in the pipe can be determined using
[latex]U=\frac{4\dot{m}}{ρ \pi d^{2}}=\frac{4 imes 111}{877 imes \pi imes 0.4^{2}}=1.007 m/s[/latex]

The corresponding Reynolds number is determined using Eq. (12.63).
[latex]Re=\frac{ ho Ud}{\mu }=\frac{4\dot{m}}{\mu \pi d}[/latex] (12.63)
[latex]Re=\frac{4\dot{m}}{μ \pi d}=\frac{4 imes 111}{7.73 imes 10^{-5} × \pi imes 0.4}=4.57 × 10^{6}[/latex]

With ε = 4.6 × [latex]10^{-5}[/latex] m, d = 0.4 m and Re = 4.57 × [latex]10^{6}[/latex], the Darcy friction factor can be determined using Eq. (12.62) as follows:
[latex]f^{-\frac{1}{2}}_{D}=-2\log _{10}\left(\frac{\varepsilon }{3.7d}+\frac{2.51}{Ref^{\frac{1}{2}}_{D}} ight)[/latex] (12.62)
[latex]=-2\log _{10}\left(\frac{4.6 × 10^{-5} }{3.7 × 0.4}+\frac{2.51}{4.57 × 10^{6}f^{\frac{1}{2}}_{D}} ight)[/latex]
[latex]=-2\log _{10}\left(3.108 imes 10^{-5}+\frac{5.492 imes 10^{-5}}{f^{\frac{1}{2}}_{D}} ight)[/latex]

By iteration, we can get [latex]f_{D} = 0.0217[/latex].
Then we can calculate the pressure drop over 100 km distance length using Eq. (12.61).
[latex]\frac{\Delta P}{\Delta L}=f_{D}\left(\frac{ ho U^{2}}{2d} ight)[/latex] (12.61)

[latex]\Delta P=f_{D}\left(\frac{ ho U^{2}}{2d} ight)\Delta L[/latex]
[latex]=0.0217\left(\frac{877 imes 1.007^{2}}{2 imes 0.4} ight) imes 100 imes 10^{3}=2.41 imes 10^{6} Pa[/latex]

So the pressure drop is about 2.41 MPa over a distance of 100 km. Due to this kind of great resistance, intermediate pumping (or booster) stations are required at certain intervals along the pipeline.

We have to be careful that the [latex]CO_{2}[/latex] properties are assumed constant in the above example in order to simplify the calculation. In reality, they cannot remain constant. Glilgen et al. [25] experimentally determined the pressure, density, and temperature relationship of [latex]CO_{2}[/latex] in the homogeneous region for pressures up to 13 MPa and temperatures in the range of 220–360 K. For P < 9 MPa and T > 298 K, [latex]CO_{2}[/latex] density varies considerably with P and T.
In the engineering practice, it is critical to ensure enough pressure to maintain above vapor–liquid equilibrium conditions to avoid liquid slugs and other operational problems resulted from a two-phase (gas–liquid) flow. A typical operating pressure is in the range of 84.9–207.3 atm where [latex]CO_{2}[/latex] is a dense-phase fluid over a wide range of temperatures. Nonetheless, the gas properties also depend on impurities such as [latex]SO_{x}, NO_{x}[/latex], and other air contaminants. A complex model is required to accurately predict the fluid property along the pipeline. Interested readers have to dig deeper into the literature for in-depth knowledge.

Question 12.5: Pressure drop in a pipeline Consider a pipeline made of comm...

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