2.2.2 Condenser mode

Condenser Calculation

In many vacuum processes (drying, distillation), large volumes of vapor are released that have to be pumped down. Moreover, significant volumes of leakage air will penetrate into large vessels, and those substances that are being vaporized or dried will release additional air that is contained in pores or dissolved in liquids.

Excel spreadsheet templates for calculating the heat transfer area needed for a heat exchanger are presented and discussed in the next section. For more details about preliminary heat exchanger design, the parameters involved, and example heat exchanger calculations, see the articles: ' and '.'
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In drying processes, the vapor can always be displaced against atmospheric pressure by a vacuum pump having sufficient water vapor capacity and can then be condensed there. However, this process has the following disadvantages:

The pump must be very large
A large volume of gas ballast air will be entrained which, together with the vapor, will carry a great deal of oil mist out of the pump
It will be necessary to dispose of the resulting condensate from the water vapor and oil mist, which is a costly process

Distillation processes operate with condensers, and the object is to lose as little of the condensing distillate as possible through the connected vacuum pump.

Let us consider a vacuum chamber, or recipient, containing material to be dried, to which enough energy will supplied by heat that 10 kg of water will evaporate per hour.

In addition, 0.5 kg of air will be released per hour. The pressure in the chamber should be less than 10 hPa. A pumping station in accordance with Figure 2.3 is used for drying, enabling the steam to be condensed cost-effectively through the use of a condenser.

The material to be dried (2) is heated in the vacuum chamber (1). The Roots pump (3) pumps the vapor / air mixture into the condenser (4), where a major portion of the vapor condenses.

The condenser is cooled with water. The condensing water at a temperature of 25°C is in equilibrium with the water vapor pressure of 30 hPa. An additional vacuum pump (5) pumps the air content, along with a small volume of water vapor, and expels the mixture against atmospheric pressure. The first step is to calculate the gas flow from the chamber: $Q=p_{vc} cdot S_1$

With the ideal gas law according to Formula 1-15 we obtain

[Q=p_{vc} 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 down vapors

$T$	Intake gas temperature	[K]
$R$	General gas constant	[kJ kmol^-1 K^-1
$t$	time	[s]
$p_{vc}$	Pressure in vacuum chamber	[Pa]
$m_{water}$	Water vapor mass	[kg]
$M_{water}$	Molar mass of water	[kg mol^-1]
$m_{air}$	Air mass	[kg]
$M_{air}$	Molar mass of air	[kg mol^-1]

Where:

Tube Condenser Calculations

$T$	Intake gas temperature	300 K
$R$	General gas constant	8.314 kJ kmol^-1 K^-1
$t$	time	3600 s
$p_{vc}$	Pressure in vacuum chamber	1000 Pa
$m_{water}$	Water vapor mass	10 kg
$M_{water}$	Molar mass of water	0.018 kg mol^-1
$m_{air}$	Air mass	0.5 kg
$M_{air}$	Molar mass of air	0.0288 kg mol^-1

Air Cooled Condenser Design Calculation Excel

we obtain a gas throughput for air of 12 Pa m³ s^-1 and for water vapor of 385 Pa m³s^-1, together 397 Pa m³s^-1. Divided by the inlet pressure $p_{vc}$ von 1000 Pa we obtain a pumping speed $S_1$ von 0.397 m³s^-1 or 1429 m³h^-1.

When evacuating the condenser, the partial air pressure should not exceed 30 %, i. e. a maximum of 12.85 hPa. It follows from this that:

$S_2=frac{Q_{air}}{0.3 cdot p_{air}}$

With a gas throughput for air of 12 Pa m³ s^-1 and a pressure of 1285 Pa, a pumping speed $S_2$ of 0.031 m³ s^-1 or 112 m³ h^-1 is obtained.

We therefore select a Hepta 100 screw pump as the backing pump. Because its pumping speed is somewhat lower than the calculated one, this pump will achieve a slightly higher partial air pressure. And we select an Okta 2000 with the following values as the Roots pump:

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

We estimate the inlet pressure $p_a$ = 1000 Pa and calculate $S_1$ in accordance with Formula 2-7. Radimpex software keygen 2019.

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

We obtain 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 $p_a$ vof 785 Pa we obtain the inlet pressure in the drying chamber, and by using this figure again in Formula 2-7 we arrive at the more precise pumping speed $S_1$ = 1.736 m³ h^-1 for an inlet pressure $p_a$ = 823 Pa.

We calculate the condenser for a 10 kg h^-1 volume of vapor to be condensed. The following applies for the condensation surface area:

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

Formula 2-12: Calculation of the condensation surface area

$A_k$	Condensation surface area	[m²]
$Q_{water}$	Specific enthalpy of evaporation	[Ws kg^-1]
$m_{water}$	Water vapor mass	[kg]
$Delta T_m$	Temperature differential between vapor and condensation surface	[K]
$k$	Thermal transmission coefficient	[W m^-2 K^-1]

Where:

$Q_{water}$	2.257 ⋅ 10⁶ Ws kg^-1
$m_{water}$	10 kg
$t$	3600 s
$Delta T_m$	60 K
$k$	400 W m^-2 K^-1

we obtain a condensation surface area of $A_k$ 0.261 m².

The vapor is heated by more than 100 K through the virtually adiabatic compression, however it re-cools on the way to the condenser. So the assumption tha