What is a vacuum in a condenser

2.2.2 Capacitor operation

In many vacuum processes (drying, distillation) large amounts of steam are released that have to be pumped out. In addition, large quantities of air leak into large containers and the substances that evaporate or are dried give off additional air that is stored in pores or dissolved in liquids.

In drying processes, the steam can in principle be conveyed against atmospheric pressure by a vacuum pump with sufficient water vapor capacity and condense there. However, this method has the following disadvantages:

• The pump must be very large
• A large amount of gas ballast air is conveyed with it, which, together with the steam, removes a large amount of oil mist from the pump
• The resulting condensate from water vapor and oil mist has to be disposed of at great expense

Distillation processes work with condensers and you want to lose as little as possible of the condensing distillate through the connected vacuum pump.

As an example, we consider a vacuum chamber with dry material, which is supplied with so much energy through heating that 10 kg of water evaporate per hour.

Figure 2.3: Drying plant (schematic)

In addition, 0.5 kg of air is released per hour. The pressure in the chamber should be less than 10 hPa. A pumping station as shown in Figure 2.3 is used for drying and the steam can be condensed economically by using a condenser.

The material to be dried (2) is heated in the vacuum chamber (1). The Roots pump (3) pumps the steam-air mixture into the condenser (4), in which a large part of the steam condenses.

The condenser is cooled with water. The condensing water at 25 ° C is in equilibrium with the water vapor pressure of 30 hPa. Another vacuum pump (5) conveys the air content and a small amount of water vapor and expels the mixture against atmospheric pressure. First we calculate the gas flow that emerges from the chamber: \$ Q = p_ {Rez} \ cdot S_1 \$

With the ideal gas law according to formula 1.15 we get

\ [Q = p_ {Rez} \ cdot S_1 = \ frac {R \ cdot T} {t} \ cdot \ left (\ frac {m_ {water}} {M_ {water}} + \ frac {m_ {air} } {M_ {Air}} \ right) \]

Formula 2-11: Gas throughput for pumping out vapors

 \$ T \$ Suction gas temperature [K] \$ R \$ General gas constant [kJ kmol-1 K-1] \$ t \$ time [s] \$ p_ {Rez} \$ Pressure in the recipient [Pa] \$ m_ {water} \$ Mass of water vapor [kg] \$ M_ {water} \$ Molar mass of water [kg mol-1] \$ m_ {air} \$ Mass of air [kg] \$ M_ {air} \$ Molar mass of air [kg mol-1]

With the following values

 \$ T \$ Suction gas temperature 300 K \$ R \$ General gas constant 8.314 kJ kmol-1 K-1 \$ t \$ time 3600 s \$ p_ {Rez} \$ Pressure in the recipient 1000 Pa \$ m_ {water} \$ Mass of water vapor 10 kg \$ M_ {water} \$ Molar mass of water 0.018 kg mol-1 \$ m_ {air} \$ Mass of air 0.5 kg \$ M_ {air} \$ Molar mass of air 0.0288 kg mol-1

we get a gas flow rate for air of 12 Pa m3 s-1 and for water vapor of 385 Pa m3s-1, so together 397 Pa m3s-1. After dividing by the suction pressure of \$ p_ {Rez} \$ of 1000 Pa, we get a pumping speed \$ S_1 \$ of 0.397 m3s-1 or 1429 m3H-1.

When evacuating the condenser, the partial air pressure should be a maximum of 30%. So a maximum of 12.85 hPa. From this it follows:

\$ S_2 = \ frac {Q_ {Air}} {0.3 \ cdot p_ {Air}} \$

With the gas throughput for air of 12 Pa m3s-1 and the pressure of 1285 Pa results in a pumping speed \$ S_2 \$ of 0.031 m3s-1 or 112 m3H-1.

We therefore choose a Hepta 100 as the backing pump, which, due to its somewhat lower pumping speed than the calculated value, achieves a slightly higher air partial pressure, and an Okta 2000 as a roller piston pump with the values:

 \$ S_0 \$ 2065 m³ h-1 \$ \ Delta p_d \$ 35 hPa differential pressure at the overflow valve \$ K_0 \$ 28 at \$ p_v \$ = 43 hPa

We estimate the intake pressure \$ p_a \$ = 1000 Pa and calculate \$ S_1 \$ according to formula 2-7.

\$ S = S_0 \ cdot \ left [1- \ frac {1} {K_0} \ left (\ frac {p_v} {p_a} -1 \ right) \ right] \$

We get a pumping speed \$ S_1 \$ of 0.506 m3 s-1 or 1,822 m3 H-1.

With \$ p_a = \ frac {Q} {S_1}

and a value for \$ p_a \$ of 785 Pa gives the suction pressure in the drying chamber, which is used again in formula 2-7, the more precise pumping speed \$ S_1 \$ = 1,736 m³ h-1 delivers at a suction pressure of \$ p_a \$ = 823 Pa.

We calculate the condenser for a condensing amount of steam of 10 kg h-1. The following applies to the condensation surface

\ [A_k = \ frac {Q_ {water} \ cdot m_ {water}} {t \ cdot \ Delta T_m \ cdot k} \]

Formula 2-12: Calculation of the condensation area

 \$ A_k \$ Condensation surface [m2] \$ Q_ {water} \$ Specific enthalpy of vaporization [Ws kg-1] \$ m_ {water} \$ Mass of water vapor [kg] \$ \ Delta T_m \$ Temperature difference between steam and condensation surface [K] \$ k \$ Heat transfer coefficient [W m-2 K-1]

With the following values

 \$ Q_ {water} \$ 2,257 ⋅ 106 Ws kg-1 \$ m_ {water} \$ 10 kg \$ t \$ 3600 s \$ \ Delta T_m \$ 60 K \$ k \$ 400 W w-2 K-1

we get a value of 0.261 m as the condensation area \$ A_k \$2.

The steam is heated by more than 100 K due to the almost adiabatic compression, but cooled again on the way to the condenser. So the assumption of \$ \ Delta T_m \$ = 60 K is quite conservative. The heat transfer coefficient \$ k \$ [20] decreases sharply with an increasing proportion of inert gas, which leads to a larger condensation surface. Conversely, it is possible to work with a smaller proportion of inert gas, i.e. a larger backing pump with a smaller condensation surface. You should pay particular attention to small leak rates, as these also increase the proportion of inert gas.

For further technical details, please refer to the further literature [21].

Figure 2.4: Roots pumping station for steam condensation

For the sake of completeness, let's look again at the entire drying process: first, an equilibrium pressure is established in the drying chamber, which results from the amount of water evaporating, caused by the heating of the material to be dried and the pumping speed of the Roots pump.

The Roots pump pumps the water vapor into the condenser, where it condenses. Since there is laminar flow there, the steam flow transports the inert gas released from the dry material into the condenser.

If the backing pump were to be switched off, the entire condensation process would come to a standstill in a short time, since the steam could only reach the condensation surface by diffusion. After advanced drying, the amount of steam decreases and there is less condensation in the condenser, but the amount of steam extracted by the backing pump tends to increase if the amount of inert gas decreases. If the vapor pressure in the condenser falls below the condensation limit, re-evaporation of the condensate begins. This can be prevented if the condensate flows through a valve into a condensate collecting container and the valve is closed when the condensation pressure falls below the limit.

In large distillation systems, the pumping speed of the backing pump should be controlled via the condensation rate. This can e.g. B. done with the help of a metering pump, which discharges the pumped amount of condensate evenly from the collecting vessel. If the liquid level in the collecting vessel falls below a certain level, the suction valve of the backing pump is opened and the inert gas that has accumulated in the condenser is pumped out. Now the condensation rate increases again, the liquid level rises and the suction valve of the backing pump is closed again. In this way, only if the condensation rate is too low is pumped and little condensate is lost.

Summary:

When pumping out vapors (drying, distillation), the main pumping capacity can be provided by a condenser. Depending on the pressure and temperature conditions, one or two capacitors can be used (Figure 2.4). The condenser between the Roots pump and the backing pump is more effective because the steam flows into the condenser at a higher temperature and pressure and a small backing pump only sucks off part of the steam. During distillation, the loss of condensate can be minimized by regulating the pumping speed of the backing pump.

The theoretical principles set out above are often used for the design of Roots pumping stations. Figure 2.5 shows a vacuum solution to reduce the residual moisture in the paper material used in the production of submarine cables. A pre-condenser (not shown in the picture) condenses the water vapor mainly in the first drying phase at high process pressures. An intermediate condenser protects the downstream BA 501 rotary vane pump and mainly condenses the water vapor in a second drying phase.

Figure 2.6 shows a Roots pumping station used for transformer drying. The intermediate condenser reduces the residual moisture of the material used to such an extent that the water vapor capacity of the downstream BA 501 rotary vane pump is not exceeded.

Figure 2.5: Roots pumping station for steam condensation

Figure 2.6: Roots pumping station for transformer drying