REDUCED AIR INJECTION TIME DURING CONTAINMENT TESTING DUE TO THE USE OF AN EJECTOR

Reduced air injection time during containment testing due to the use of an ejector. When testing the tightness of the containment and elements of the accident localization system at the Ukrainian NPP, the "absolute pressure" method is used. Using this method, as a result of measuring air pressure, temperature, and humidity, the Mendeleev-Clapeyron equation determines the mass of air present in the containment. That is, the turn is determined indirectly by determining the change in the mass of air in the containment over time. The tests consist of five stages: vacuuming; air injection, to achieve the required pressure value; parameter stabilization; measurement; pressure relief, and last more than 25 hours. During the tests, no work is carried out in the containment. A necessary condition for testing is to ensure overpressure in the containment. This is done by the operation of the compressor. Given the large volume of containment, this requires a relatively long time, which affects the economic performance of nuclear power plants. In this paper, it is proposed to use an ejector to reduce the time of air injection, the working environment for which is the air after the compressor. The environment for injection is taken from the environment (from the "pure" volume). The article calculates the time of air injection by the compressor to the containment under today’s conditions and when using an ejector. The pressure in the containment changes from 0 to 0.0686 MPa during injection. The ejector outlet pressure is optimized for designing relative to the minimum discharge time. The optimal pressure at the ejector outlet during design is 0.45 bar. It is shown that due to the use of an ejector, the injection time can be reduced by about 30 %.


Introduction
The main equipment of the reactor unit, which is operated in Ukraine at NPP power units (mostly WWER-1000 reactors) is placed in cylindrical structures (containments) of prestressed reinforced concrete with a convex oval roof [1,2]. Containments are designed for internal pressures up to 5 bar. Because concrete is not a dense barrier to gaseous media, the inside of the containment is lined with a dense metal coating of 8 mm thick. Thus, metal cladding provides the tightness of the containment and elements of the accident localization system and the forces that may occur under emergency conditions under internal pressure on the metal cladding are perceived by reinforced concrete containment.
The strength of the containment is provided by pre-stressed steel cables, which pass in special channels in the wall and dome of containment [3]. Containment is also a barrier for radiation. The reliability of the system is checked by testing the tightness of the containment with excess internal air pressure and determine the integrated leakage from it, which is regulated [4].
Tests to determine the integrated leakage are carried out in several stages, namely [5]: 1. Vacuuming the containment to determine the location of leaks and eliminate damage -4 hours.
2. Injection of air into the contaminant to the control pressure. Determination of leaks of nodes (main and auxiliary locks, hatch of transport corridor, etc.) 4.5 hours (for Rivne NPP).
3. Stabilization of parameters to align the parameters throughout the volume -8 hours. 4. Measurements of temperature, pressure and humidity -8 hours. 5. Pressure release -1.5 hours.
The specified time of stages may be different for different NPPs and different power units for different reasons.
Thus, according to the method of "absolute pressure" to determine the integrated leakage, the value of which confirms the possibility of operation of the unit at power, it is necessary to spend more than a day. It should be added that no works in the containment during this period are not carried out. That is, the tests to determine the integrated leakage are on the critical path of the schedule of planned preventive work and should be reduced over time.

Analysis of literature data and statement of the problem
One of the stages of the test is the injection of air into the containment to a pressure of 0.0686+0.0196=0.0882 МPa (excess) [5]. To do this, a compressor carries out the air supply.
To reduce the time of air injection, it was proposed to use an ejector, which increases the flow of air at the outlet due to the suction of ambient air.
The aim of the work is to analyze the use of the ejector to accelerate the process of air injection during tests for the leak of the containment of NPPs with WWER. The calculations were performed on the example of Unit №3 of Rivne NPP.
From the experience of tests on the leak of the reactor unit with WWER-1000, located at the third and fourth power units of RNPP, it was found that: -the free volume of the containment is 60,000 m 3 ; the compressor of the ЦK-135/8 brand is used to provide tests; -at the beginning of operation of the compressor at pressure in containment of 0.101 MPa (abs.) pressure on a compressor pressure is equal to 0.245 MPa (excess).
According to [5], the rate of increase of pressure in containment should not exceed 0.039 MPa / hours). The tests shall be performed at a pressure of 0.0686 MPa. It follows that the minimum time of air injection can be equal to 0.0686 / 0.039=1.76 hours. This is 2.5 times less than the current injection time at RNPPs and significantly less than this value for other NPPs.
Given that the pressure behind the compressor at the beginning of the injection is 0.245 MPa (excess), ie at a pressure in the containment is 0.101 MPa (abs.), we can assume that the resistance of the path from the compressor to the containment is 0.245 MPa.
Nominal parameters of the compressor specified in the passport: Q = 135 m 3 /min. at a pressure of P =0.667 MPa. The characteristics of the compressor are in the range of 0.42...0.6 MPa and Q = 110...150 m 3 /min. [6]. Thus, during the period when the compressor increases the pressure in the containment from 0.101 MPa to 0.1696 MPa, the compressor operates out of operation.
Assuming that the resistance of the path from the compressor to the containment does not change when the pressure changes in the containment, we calculate the injection time according to the following algorithm.
According to the accepted pressure on the compressor P 1 determine the flow rate q (m 3 /min). We accept the time interval τ (for example 10 minutes), during which the air flow will not change. Then, during this time to the containment will be submitted: Find the mass of air in the containment: where m 0 -initial amount of air in containment; ∆m -mass of injected air: ρ -air density at 20 °С.
According to the Mendeleev-Clapeyron equation for an ideal gas we determine the corresponding pressure P 2 , which will take place in the containment at a new mass of air: If P 2 < 0.1696 MPa, repeat the calculation from the beginning, taking into account the new initial mass of air and the risk in containment.
Calculate the initial amount of air in the containment at atmospheric pressure. We assume air temperature = 293 K. The molar mass of air (oxygen -21%, nitrogen -78%, argon -1%) is 29 kg/mol. Pressure in containment = 101 MPa.
Universal gas constant for air: From the analysis of the characteristics of the compressor ЦК-135/8 it can be noted that from 0.42 to 0.6 MPa, this characteristic is almost a linear function. As a result of the approximation we obtain the equation of the dependence of the specific air flow on the pressure after the compressor at P = 0.10132 +0.245 = 0.3463 MPa: where P -compressor pressure, MPa; The results of further pressure changes in the containment with a time interval of 10 minutes are shown in Table 1.  The average feed of the compressor from table. 1 is equal to q = 153.14 m 3 /min. Thus, for the implementation of the injection will be required: / 40030 / 153.14 261.4 V q τ = = = min.
That is, a value is obtained that coincides with the previous value of the injection time.

Calculation of injection time using an ejector
The calculation of the ejector [7] is performed at the specified pressure before and after the ejector. At once there is a question: at what pressure behind an ejector it is necessary to carry out calculation. In the process of injection, the pressure in the containment varies from 0.1013 to 0.1696 MPa. That is, in any case, the designed device will work in a wide range of parameters. Thus, to determine the operating time of the compressor with the ejector before reaching the pressure in the containment of 1.72 bar (abs.), you must select the initial parameters, calculate the ejector, and then determine the flow of the ejector already specified design as a function of ejector pressure. As it is not known at which parameters the ejector operating time will be minimal, it is necessary to perform appropriate calculations for several ejector variants, calculated for different initial ejector pressures.
Here are the calculations of the first selected option. Since the ejector pressure varies in the specified range, an average pressure of 0.135 MPa was chosen for the calculation. At a compressor, pressure of 0.7 MPa, the pressure in front of the ejector is 0.45 MPa. Consumption of working air on the compressor 8000 m 3 /h.
For the calculated design, the calculation of the change in parameters when changing the pressure in the containment (after the ejector) was performed. The results of the calculation -the dependence of air flow after the ejector on the pressure behind it (pressure in containment) are shown in Table 2. In the Table 2 in the last column shows the average flow in the pressure range given in the first column. That is, for example, when operating the ejector in the pressure range in the containment from 0.1279 to 0.123 MPa, the consumption of the ejector is 268.4 m 3 /min. Until the pressure in the containment of 0.1279 MPa is reached, the flow rate behind the ejector is constant and equal to 270.11 m 3 /min. After reaching the pressure in the containment of 0.1551 MPa, the flow rate of the ejector is equal to the flow rate of the compressor -133.3 m 3 /min. That is, you can bypass the ejector, which will even increase air consumption. With the help of the given data, the calculation of pressure change in containment over time was carried out. The calculation of the operating time of the ejector to achieve the desired pressure in the containment is given in Table 3.

Table 3
Calculation of the time of injection of air into the containment to achieve the required pressure of 0.1696 MPa using an ejector According to the calculation, it was obtained that the set pressure of 0.17 MPa when using an ejector, which was designed for a final pressure of 0.135 MPa, will be achieved in 209.7 / 60 = 3.49 hours which is 4.50 --3.49 = 1.01 hours less than without an ejector. Time savings of 23.5%.
Appropriate calculations were performed for the ejector designed for an outlet pressure of 0.125; 0.145; 0.156 and 0.17 MPa. The dependence of the injection time for different design pressure on the ejector is shown in Fig. 1. In a result of the calculations, it was found that the minimum time of air injection in the containment to reach a pressure of 0.169 MPa is 3.09 hours (186 minutes) and corresponds to the design ejector pressure of 0.145 MPa. Conclusion 1. When testing the tightness of the NPP containment, to reduce the time of air injection, it is proposed to use an ejector, which due to the working air from the standard compressor will suck air from the environment and thus increase air flow in the containment.  The calculation of the air injection time for today's conditions at the Rivne NPP was performed, the result of which coincided with the actual air injection time during the tests (4.5 hours). The characteristic of the compressor was used in the calculation, namely the dependence of the flow rate on the pressure in front of the compressor.
3. The program of calculation of an air ejector for work with the regular compressor of the ЦК-135/8T brand for which initial data are initial and final air pressure, consumption of working air is developed.
4. The mathematical model of work of an ejector of the set design at work in a variable mode is developed. That is, the dependence of air flow on the ejector depending on the change in pressure behind it is determined.
5. With the help of the developed mathematical models, calculations were performed regarding the operating time of the ejector paired with the compressor to achieve a given ejector pressure of 0.17 MPa. As a result of calculations to determine the injection time depending on the design pressure on the ejector, it was determined that the minimum injection time is 3.09 hours, which is 1.48 hours (32.4%) less than under existing conditions. The optimal pressure behind the ejector for its design is a pressure of 0.145 MPa.
6. Reducing the time of containment tests will provide an opportunity to increase electricity generation and earn NPPs an additional profit of approximately 1.48 h⋅10 6 kW⋅0.6 UAH / (kWh) = 888 thousand UAH / year. 7. Under other conditions, it can be assumed that the reduction of injection time due to the use of the ejector will be about 30% of the available time.