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2.3 Piping conductivities

In calculating the pump-down times of vessels, we have left piping resistance out of consideration for both Roots pumping stations as well as for turbopumping stations. However this is usually an inadmissible simplification since the presence of piping between the vessel and the pump will also reduce the effective pumping speed.

2.3.1 Laminar conductance

Let us consider the pumping station for the drying system (Figure 2.3) and calculate the drop in pressure between the condenser and the backing pump. In this case, due to the pressure of 4,285 Pa and the pumping speed of 4,285 Pa, 𝑆𝑣 of the backing pump of 107 mΒ³ h-1 = 2.97 Β· 10-2 mΒ³ s-1, a gas throughput of 𝑄 = 4,285 Β· 2.97 Β· 10-2 = 127 Pa mΒ³ s-1 is specified. The DN 63 piping has an inside diameter of 0.07 m and a length of 2 m. Two 90Β° pipe bends having an equivalent length of 0.2 m each are also taken into consideration.

From a pump inlet pressure of 4,285 Pa and a value for air of 6.7 Β· 10-3 Pa Β· m for 𝐼¯⋅𝑝 according to Chapter 1, Table 1.5 we obtain a mean free path of 1.56 Β· 10-3 m. We use the Knudsen number, Formula 1-13, to determine the flow range and obtain:
  • K_{n}=\frac{\bar{l}}{d}=2.23\cdot10^{-2}

Since Kn is less than 0.01, this results in viscous flow. This can be either laminar or turbulent. For laminar flow the conductance are significantly higher than for turbulent flow, which means that significantly lower volume flow losses will occur. The Reynolds number Re must be less than 2,300 for laminar flow. To calculate the Reynolds number, we first determine the flow velocity 𝑣 in the piping:
  • v=\frac{4\cdot S_{v}}{d^2\cdot\pi}=8.66\,\text{m}\,\text{s}^{-1}

And we determine the density 𝜌 of the air at 4,285 Pa from the air density 𝜌0 = 1.293 kg m-3 at atmospheric pressure
  • \rho=\frac{1.293\cdot4.285}{101.325}=5.47\cdot10^{-2}\,\text{kg}\,\text{m}^{-3}

and according to Formula 1-14 with a dynamic viscosity for air of 18.2 Β· 10-6 Pa Β· s we obtain
  • \text{Re}=\frac{\rho\cdot\nu\cdot l}{\eta}=1{,}820

i. e. laminar flow.


We use Formula 1-26 in Chapter 1 p1 to obtain the inlet pressure for the piping:
  • C_{Rohr,\,lam}=\frac{\pi\cdot d^4}{256\cdot\eta\cdot l}\cdot(p_1+p_2)=\frac{\pi\cdot d^4}{128\cdot\eta\cdot l}\cdot\bar{p}

We multiply by Δ𝑝=𝑝1βˆ’π‘2 to obtain the gas throughput
  • Q=C_{Rohr,\,lam}\cdot\Delta p=\frac{\pi\cdot d^4}{256\cdot\eta\cdot l}\cdot(p_1^2-p_2^2)

Since 𝑝2 = 4,285 Pa and 𝑄 = 127 Pa Β· m3 Β· s-1 it is possible to directly determine 𝑝1 from these values:
  • p_1=q_{diff}\cdot A_{d}\cdot\sqrt{p_2^2+\frac{Q\cdot256\cdot\eta\cdot l}{\pi\cdot d^4}}=4{,}287.2\,\mathrm{Pa}

We have a pressure loss of merely 2.2 Pa, a very low value.


The conductivity of the piping is obtained from Chapter 1, Formula 1-18:
  • C=\frac{Q}{\Delta p}=58\,\mathrm{m^3\cdot s^{-1}} oder 58{,}000\,\mathrm{l\cdot s^{-1}}

The effective volume flow rate
  • S_{\text{eff}}=\frac{S_{v}\cdot C_{\text{Rohr, lam}}}{S_{v}+C_{\text{Rohr, lam}}}=2.9707\,\mathrm{m^3\cdot s^{-1}}

is only slightly lower than the volume flow rate without the piping: 𝑆𝑣 of 2.9222 m3 s-1.

2.3.2 Molecular conductance

Now let us also consider the conductance of the same piping in the molecular flow range. The piping has a diameter of 0.07 m and a length of 2 m. The elongated length of 0.235 each of the two 90Β° pipe bends, i. e. a total length of 𝑙 = 2.47 m. In accordance with Chapter 1, Formula 1-30, the piping resistance is:
  • L_{R_{m}}=\frac{\dot{c}\cdot\pi\cdot d^3}{12\cdot l}=1.71\cdot10^{-2}\,\text{m}^3\text{s}^{-1}

  • S_{eff}=\frac{S_{\nu}\cdot L_{R_{m}}}{S_{\nu}+L_{R_{m}}}=1.09\cdot10^{-2}\,\text{m}^3\text{s}^{-1} with S_{\nu}=2.97\cdot10^{-2}\,\text{m}^3\text{s}^{-1}

In the molecular flow range, the pumping speed of the backing pump would be reduced to nearly one third. In this range, it is absolutely necessary to pay strict attention to short runs and large piping cross sections between the pump and the vacuum chamber. This applies in particular for turbopumps that should ideally be flanged directly to the vacuum chamber.