### 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
m³ h^{-1} = 2.97 · 10^{-2}
m³ s^{-1}, a gas throughput of $Q$ = 4,285 *·* 2.97 · 10^{-2}
= 127 Pa m³ s^{-1} is specified. The DN 63 piping has an
inside diameter of 0.07 m and a length of 2 m. Two 90° 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
· 10^{-3} Pa · m for $\bar{I} \cdot p$ according
to Chapter 1, Table 1.5 we obtain a mean free path of 1.56 · 10^{-3}
m. We use the Knudsen number, Formula 1-13, to determine the flow
range and obtain:

$K_n=\frac{\bar{I}}{d}=$ 2.23 · 10^{-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 m s^{-1}

And we determine the density $\rho$ of the air at 4,285 Pa from the
air density $\rho_0$ = 1.293 kg m^{-3} at atmospheric pressure

$\rho=\frac{1.293 \cdot 4,285}{101,325}=$ 5.47 · 10^{-2}
kg m^{-3}

and according to Formula 1-14 with a dynamic viscosity for air of 18.2
· 10^{-6} Pa · s we obtain

$Re=\frac{\rho \cdot \nu \cdot l}{\eta}$= 1,820

i. e. laminar flow.

We use Formula 1-26 in Chapter 1 p_{1} to obtain the inlet
pressure for the piping:

$C_{Rohr,\,lam}=\frac{\pi\cdot d^4}{256\cdot\eta\cdot l}\cdot(p_1+p_2)=\frac{\pi\cdot d^4}{228\cdot\eta\cdot l}\cdot\bar p$

We multiply by $\Delta p=p_1-p_2$ 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 $p_2$ = 4,285 Pa and $Q$ = 127 Pa · m^{3}
· s^{-1} it is possible to directly determine $p_1$
from these values:

$p_1=q_{diff} \cdot A_d \cdot \sqrt{p_2^2+\frac{Q \cdot 256\cdot\eta\cdot l}{\pi \cdot d^4}}=$ 4,287.2 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 m^{3}s^{-1} or 58,000
l s^{-1}

The effective volume flow rate

$S_{eff}=\frac{S_v \cdot C_{Rohr,\,lam}}{S_v + C_{Rohr,\,lam}}=$2.9707
m^{3} s^{-1}

is only slightly lower than the volume flow rate without the piping:
$S_v$ of 2.9222 m^{3} s^{-1}.