# EJECTOR COMPRESSORS Division

Depending on the type of driving fluid ejector compressors are divided into:

**Operating principle:**

All ejector compressors operate on the principle of using high pressure energy of the driving fluid for pumping in and compressing pumped gases. The pressure of the fluid mixture at the ejector output depends on the input pressures of the driving and pumped fluid and their mass flow ratio.

## 6.1 Ejector hydro compressors (liquid-gas)

Liquid, most often water is used as the driving fluid for these compressors for pumping in and compressing gases (most often air). Depending on the pressure of the driving liquid and mass ratio between the gas-liquid flow it is possible to compress gases to pressures higher than 10 bars. The volume ratio between the gas-liquid flow can be realized within the limits of 0 – 3 (V_{g}/V_{t}, [m^{3}/m^{3}]).

Compressed gas exiting the ejector is practically washed and partly cleansed from various pollutants. This compression procedure practically removes dust and other mechanical pollutants from gas and also droplets and evaporates of other liquids (water, oil, etc.).

**Application:**

They are used for compression of gas requiring relatively low flows and large pressures (for example ozoning drinking water, figure 6.1), pumping in compressed air into a hydrofor (figure 6.2) and many other processes.

They are especially applied in the food industry that requires clean compressed air with no dirt and compressed oil content. The relative moisture of such compressed air is close to 100% that is also desirable in the food industry.

Diagram 6.1 shows the dependence between the output pressure p_{3} on the pump input pressure p_{1} = p_{p} and the volume ratio between the pumped in air and water V_{2}/V_{1}.

**Example 6.1**

Data: Air using water needs to be pumped in a vessel (figure 6.3) under pressure p_{3} = 1.6 bar. The pressure of driving water in front of the ejector is p_{1} = 10 bar.

How many cubic meters of air can be pumped in with one cubic meter of water?

Solution: For the reservoir pressure of p_{3} = 1.6 bar and water pressure in front of the ejector of p_{1} = 10 bar, from diagram 6.1 one reads that V_{2}/V_{1} = 1.1 that means that 1.1 m^{3 }of air can be pumped in with 1 m^{3} of water.

Diagram 6.2 shows the dependence of the ratio between manometer pressures at the ejector output and pump pressure p_{3}/p_{p} on the ratio between the volume flow of air and water, for operating conditions presented in figures 6.3 and 6.4.

The maximal pressure that can be reached in the vessel according to the connection scheme presented in figure 6.3 is 0.7 of the driving pressure of pump p_{p}, i.e. pipeline pressure p_{1} (p_{3} =0.7×p_{1 }= 0.7×p_{p}) and according to the connection scheme given in figure 6.4 is 2.44 times the pump pressure p_{p} (2.44×p_{p})

**Example 6.2**

Data: The pressure ratio between the reservoir and pump is p_{3}/p_{p} = 0.25. The flow for the connection schemes given in figures 6.3 and 6.4 is required?

Solution: For the pressure ratio of p_{3}/p_{p} = 0.25 the volume flow ratio is read from figure 6.3 as V_{2}/V_{1} = V_{2}/V_{p} = 0.75 and according to figure 6.4 V_{2}/V_{1} = 0.9.

## 6.2 Ejector gas compressors

These compressors are used for mixing and compressing gases. Compressed gas (most often air) is used as the driving fluid for these compressors. It is used for pumping, mixing and compressing the pumped gas. Using the Laval nozzle the output rates of the driving gas rise to speeds above the speed of sound, so they can be efficiently used for attaining relatively high output pressures.

Figure 6.5 Mixing combustible gas and air |

**Application:**

They are used for extraction of natural undergoing gases with low pressure using other or similar gases present at higher pressures, for mixing compressed gases with different pressure in order to obtain a mixture with a corresponding pressure, for obtaining gas sinter under very high pressures (several hundred bars), for example for obtaining ammoniac.

Figure 6.5 shows the scheme of pumping, mixing and compressing combustible gas and air.

Diagram 6.3 shows the dependence of the ratio between the absolute pressure of the driving and pumped fluid (p_{1}/p_{2}) on the ratio between mass flows (m_{2}/m_{1}) for t_{1} = t_{2}. The diagram can be used for different temperatures t_{1} and t_{2} though the read values of m must be multiplied with (T_{1}/T_{2})^{0.5} (T_{1} and T_{2}are absolute temperatures of the driving and pumped gas).

**Example 6.3**

Data: The driving pressure at the ejector input p_{1} = 4 bar_{abs}, pumped pressure p_{2} = 0.9 bar_{abs}, pressure at the ejector output is atmospheric p_{3} = 1 bar_{abs}. What is the mass flow ratio?

Solution: For t_{1} = t_{2} and pressure ratios p_{1}/p_{2} = 4/0.9 = 4.4 and p_{3}/p_{2} = 1/0.9 = 1.11. From diagram 6.3 m = m_{2}/m_{1} = 2.4.

## 6.3 Ejector vapor compressors

Saturated or pre-heated water vapor is used as the driving fluid for these compressors. Driving vapor, passing through the nozzle expands to very low pressures and enters the ejector mixing chamber at the speed of sound or higher. In the mixing chamber particles of the driving liquid are taken up, compressed and take along vapor from the inlet port. The ratio between the output and input pressure is 1-10. Higher values are related to low pumping pressures of 0.01-0.1 bar_{abs}.

**Application:**

Using fresh vapor as the driving fluid it is possible to pump in and compress used vapor and return it back into the process. Ejector vapor compressors are used in the following processes:

- evaporation,
- cooling,
- crystallization,
- dezoxidation,
- degasification,
- drying,
- for compression of evaporated condensed water vapor and – coupler evaporate.

Figures 6.4-6.8 give schemes of some processes that use ejector compressors.

The diagram labeled y 6.4 marked the dependence relations 1 kg mass flow of energy mix at the exit from ejector and 1 kg mass flow driving energy at the entrance to ejector depending on the mass relations suctioned and drive the flow multiplied with the square root of the relationship of their absolute temperature y = (m_{2}/m_{1})·(T_{2}/T_{1})^{0,5}.

The symbolic R, T and k marked the gas constant, absolute temperature and exponent adiabate process. Index 1 refers to the drive gas, the index 2 refers to suction gas, and the index 3 to the mix driving and sunctioned gas to exit from ejector. Label h indicated a coefficient of utility. Diagram 6.4 can be used for all types of gases and water vapor and for all temperatures.

where is : T_{1,2,3} – absolute temperature of gas in idlea, R_{1,2,3} – gas constant.

**Example 6.4**

Data: With 1 kg of compressed air driving pressure p_{1} =11 bar_{aps} be suctioned 0,61 kg of air which is under pressure p_{2} =1,1 bar_{aps}.

Search is what you can get pressure on the exit ejector p_{3}:

a) when the drive temperature and air usisavanog equal t_{1} = t_{2} and

b) when the air temperature driving t = 800 °C, and suctioned air t_{2} = 200 °C?

Solution: The ratio of the mass flow m = m_{2}/m_{1} = 0,61/1 = 0,61 from diagram reading to y = 0,281 and the coefficient of usefulness m = 0,27.

a) How is the air k_{1}= k_{2} = k_{3} i R_{1} = R_{2} = R_{3} it from the equation 6.2-1 za t_{1} = t_{2} gets

p_{3} = p_{2}·1,66 = 1,1·1,66 = 1,83 bar_{aps}.

b) For m = 0,61 temperature at the exit from ejector is

**p _{3} = 1,7287·1,1 = 1,9 bar_{aps}.**

Diagram 6.5 for k = 1.4 gives the ratio between the absolute pressure at the ejector output and the pumped pressure (p_{3}/p_{2}) depending on the absolute pressure ratio between the driving and pumped air p_{1}/p_{2} and mass ratio of the pumped and driving flow (m_{1}/m_{2}). Diagram 6.5 is given for conditions when the driving and pumped gas have the same adiabatic exponent k = 1.4 and temperature t_{1} = t_{2}. The diagram can be used for different temperatures of the driving and pumped gas by multiplying the value of m from the diagram with

(T_{1}/T_{2})^{0,5 }[m = (m_{2}/m_{1})·(T_{2}/T_{1})^{0,5}].

Thus, for the same operating conditions, a temperature increase of the driving gas will increase the amount of pumped gas and vice versa. T_{1} denotes the temperature of the driving gas and T_{2} denotes the pumped gas temperature.

Ejectors that use compressed gas (air) for drive can be used for extraction and separation of water vapor and other evaporates.

**Example 6.5**

Data:Air present in a vessel under pressure of p_{2} = 1.4 bar_{abs} and temperature t_{2} = 200^{o}C should be pumped in and compressed to a pressure of p_{3} = 2.45 bar_{abs}. Air under pressure p_{3} = 12.5 bar_{abs} and temperature t_{1} = 200^{o}C is used to drive the ejector. How many kilograms of air can be pumped in with 1 kg of driving air?

Solution: For pressure ratios p_{3}/p_{2} = 2.45/1.4 = 1.75 and p_{1}/p_{2} = 12.5/1.4 = 8.93 from diagram 6.5 [m = (m_{2}/m_{1})× (T_{2}/T_{1})^{0.5} = 0.51. With 1 kg of driving air about 0.51 kg of air can be pumped in from the vessel.

Diagram 6.6 for k = 1.135 (saturated water vapor) shows the ratio between the absolute pressure at the ejector output and the pumped pressure (p_{3}/p_{2}) depending on the absolute ratio between the pressure of the driving and pumped water vapor p_{1}/p_{2} and mass ratio of the pumped and driving flow (m_{2}/m_{1}). Diagram 6.6 is valid for conditions when the driving and pumped vapor have the same adiabatic exponent k = 1.135 and the same temperature t_{1} = t_{2}. The diagram can be used for different temperatures of driving and pumped water vapor, but the value of m read from the diagram must be multiplied by (T_{2}/T_{1})^{0.5}. From this follows that, for the same operating conditions, with increase in the temperature of the driving vapor the amount of pumped vapor will be higher and vice versa with increase in the temperature of pumped vapor the amount of pumped vapor will be lower. T_{1} denotes the temperature of the driving and T_{2} of the pumped water vapor.

Diagram 6.7 for k = 1.3 (saturated water vapor, gases, and all who have K = 1.3) shows the ratio of absolute pressure at the exit from ejektora and usisavanog pressure (p_{3}/p_{2}) depending on the relationship of absolute pressure and driving usisavane steam p1/p2 and mass relations usisavanog and driving the flow (m_{2}/m_{1}). Diagram 6.7 is given for conditions when the driving and usisavana pairs have the same izložitelj adijabate k = 1.3 and the same temperature t_{1} = t_{2}. Diagram can be used for different temperatures and driving and suctioned steam which, in the diagram guest value of m should be multiplied by (T_{1}/T_{2})^{0.5}. From here it follows that, in the same conditions, with increasing temperature steam drive sunctions the greater the amount of steam sunctioned and vice versa, or with increasing temperature sunctioned staeam sunction the smaller amount of steam sunctioned and vice versa. The T_{1} is a marked temperature driving, and with T_{2} sunctioned steam.

**Example 6.6**

Data:The pressure of saturated pumped vapor of p_{2} = 1.1 bar_{abs} and temperature t_{2} = 102.32^{o}C needs to be compacted to p_{3} = 2 bar_{abs.}

How many kilograms of pumped vapor m_{2} can be pumped with m_{1} = 1 kg of driving vapor if the pressure of driving saturated vapor is p_{1} = 11 bar_{abs} and the temperature is t_{1 }= 184.07^{o}C?

Solution: For k=1.134 and p_{3}/p_{2} = 2/1.1 = 1.81 and p_{1}/p_{2} = 11/1.1 = 10 m = (m_{2}/m_{1})× (T_{2}/T_{1})^{0.5} =0.61 is read from diagram 6.6. Due to unequal temperatures T_{1}1T_{2} of the driving and pumped vapor a correction will be made so m_{2}/m_{1} = m×(T_{1}/T_{2})^{0.5} = 0.61×[(273+184.07)/(273+102.32)]^{0.5} = 0.67.

For the set conditions with m_{1} = 1 kg of driving vapor m_{2} = 0.67 kg of vapor can be pumped.

**Example 6.7**

Data:With a driving pressure of pre-heated vapor p_{1} = 18 bar_{abs} and temperature t_{1} = 2800^{o}C one needs to pump in preheated vapor with a pressure of p_{2} = 2 bar_{abs} and temperature t_{2} = 1600^{o}C and compress it to a pressure of p_{3} = 4.5 bar_{abs}. What is the mass flow ratio m_{2}/m_{1}?

Solution: For pressure ratios p_{3}/p_{2} = 4.5/2 = 2.25 and p_{1}/p_{2} = 18/2 = 9 on diagram 6.7 m = (m_{2}/m_{1})× (T_{2}/T_{1})^{0.5} = 0.29. The mass flow ratio is m_{2}/m_{1} = m×(T_{1}/T_{2})^{0.5} = 0.29×[(273+280)/(273+160)]^{0.5} = 0.33.