On the Question of Selecting Air-Intake Points in the Fractional Method of Removing Air from Cooling Water and from the Condensers of Utilization Plants

E. N. Shadrin

A distinctive feature of utilization plants, as compared with ordinary power plants, is the large expenditure of energy for their own needs. This expenditure consists mainly of the energy required for supplying cooling water for steam condensation and for removing air from the condensers. The present work gives a theoretical investigation of the possibility of reducing the energy expenditure for air removal.

With mixing condensation, from \(1 \, m^3\) of cooling water in the condenser there is released, in kilograms, an amount of air

\[ G_v = a \cdot (p_n - p_k), \tag{1} \]

where \(a\) is the coefficient of the weight solubility of air in water, \(kg/m^3\);
\(p_n\) is the pressure above the water source, from which the latter is supplied to the condenser, \(ata\);
\(p_k\) is the pressure in the condenser of the utilization plant, \(ata\).

If all the air were removed from the condenser, then for degassing \(1 \, m^3\) of water, under adiabatic compression of pure air, it would be necessary to expend the work

\[ L = \frac{k}{k - 1} \cdot p_k \cdot v_k \left[ \left( \frac{p_n}{p_k} \right)^{\frac{k-1}{k}} - 1 \right] \cdot a \cdot (p_n - p_k). \tag{2} \]

As is evident, this work is the greater, the deeper the vacuum that must be maintained in the condenser.

If, for example, one takes the temperature of the cooling water for winter conditions as \(t_n = 2^\circ C\), \(p_k = 0.01 \, ata\), \(p_n = 1 \, ata\), then, in order to remove the air released from \(1 \, m^3\) of cooling water, it will be necessary to expend, in an ideal compressor, the work \(L = 2660 \, kgm\). Therefore, it must be considered natural to raise the question of reducing this item of expenditure for the own needs of utilization plants.

Prof. I. N. Butakov [1], in order to reduce the expenditure of energy for maintaining the vacuum in the condensers of utilization plants, proposed applying a fractional method of air removal. The essence of this method consists in the preliminary degassing of the water near the tops of its barometric column. The scheme of such preliminary degassing may be represented in the following way. The cooling water (Fig. 1) from the water-supply source (1) is supplied by a pump (2) to a degasser (3), in which the pressure \(p_1\) is maintained, with \(p_n > p_1 > p_k\). From the degasser the water then enters the condenser (4). Since the pressure \(p_1\) in the degasser is lower than the pressure \(p_n\) above the water-supply source, part of the air will be released in the degasser. The temperature of the water during the passage of the latter through the degassers all the time

remains constant; therefore the amount of air released from each \(m^3\) of water in the deaerator is

\[ G_{v1}=a\cdot(p_n-p_1). \tag{3} \]

Figure labels: Air-removing device; Air from the deaerator; Air from the condenser.

Fig. 1.

In the condenser of the utilization installation, with such an arrangement with one deaerator, air will be released in the amount

\[ G_{vk}=a\cdot(p_1-p_k). \tag{4} \]

With such an arrangement, not all the air will be compressed from the pressure in the condenser to atmospheric pressure, but only that part of it determined by formula (4). Consequently, the expenditure of energy for this compression will also be smaller. We shall not consider the question of the air-removing device. We shall only point out that this may be done either by means of ejectors installed separately both for removing air from the condenser and from the deaerator, or by means of one multistage ejector, into an intermediate stage of which the air from the deaerator may be directed, or, finally, by using a multistage centrifugal compressor.

Here we shall consider the following questions:

1. How will the magnitude of the work expended on air removal change when the fractional method described is used?

2. At what points of the barometric column of liquid should air be extracted in order to obtain the minimum compression work?

3. The expedient number of air-extraction points.

In investigating the questions indicated, we shall assume that 1) the temperature of the cooling water does not change as the latter passes through the deaerators and through the condenser, and 2) the air pump removes only air. For this purpose let us consider a scheme with one extraction point. The amount of air released in the deaerator and in the condenser from \(1\ m^3\) of cooling water is calculated by formulas 3 and 4.

The work expended on removing the air released from \(1\ m^3\) of cooled water in the deaerator and in the condenser with one stage of preliminary deaeration will be expressed by the formula:

\[ L=a\cdot\frac{k}{k-1}\cdot R\cdot T_1 \left[ \left(\frac{1}{p_1}\right)^{\frac{k-1}{k}}-1 \right](1-p_1)+ \]

\[ {}+a\cdot\frac{k}{k-1}\cdot RT_2 \left[ \left(\frac{1}{p_k}\right)^{\frac{k-1}{k}}-1 \right](p_1-p_k). \tag{5} \]

The right-hand side of (5) consists of two terms. The first term expresses the work expended on removing air from the degasser, and the second—the same for the air released in the condenser. Here it is assumed that \(p_k = 1\) ata. With a decrease in the pressure \(p_1\) in the degasser, the first term of equation (5) increases and the second decreases. The limiting cases should be considered those in which \(p_1 = p_n = 1\) ata and \(p_1 = p_k\). At \(p_1 = 1\) ata the first term in formula (5) is equal to 0. In this case the second term gives the amount of work for removing the entire quantity of air from the condenser. If, however, \(p_1 = p_k\), then the amount of work for removing the entire quantity of air is given by the first term, while the second becomes equal to zero. For all values \(p_k < p_1 < p_n\), the work will be less than \(L\) at \(p_1 = p_n\) and at \(p_1 = p_k\), and at some quite definite value \((p_1)^1\) there will be a minimum of the work required to remove air from \(1\,m^3\) of cooling water with one stage of preliminary degassing. This value \((p_1)\) is determined from the equation obtained by setting the first derivative \(\frac{dL}{dp_1}\) equal to zero. Then, after some transformations, taking \(T_1 \simeq T_2\), we obtain

\[ p_1 = k \left[\left(\frac{1}{p_k}\right)^{\frac{k-1}{k}} \cdot p_1^{\frac{2k-1}{k}} - 1\right] + 1\ \text{ata}. \tag{6} \]

From formula (6) it is evident that the pressure in the degasser \(p_1\) depends only on the vacuum in the condenser and can readily be found from equation (6) by trial. For a particular example, when \(p_n = 1\) ata, \(p_k = 0.01\) ata, \(k = 1.4\), the pressure in the degasser is \(p_1 = 0.18\) ata. To check the correctness of formula (6), Fig. 2 gives a graphical solution of equation (6).

The curve \(L_o\) shows the change in the work required to remove air from the degasser at different \(p_1\); the curve \(L_k\)—the same for the stream of air entering the condenser. The upper curve gives the dependence of the total work required to remove the air dissolved in \(1\,m^3\) of cooling water. From Fig. 2 one can see a decrease in the work \(L_o\) for removing air

Fig. 2.

from the degasser and an increase in the work \(L_k\) for removing air from the condenser as \(p_1\) increases. The curve of the dependence of the total work \(L\) clearly shows the presence of a minimum. The pressure \(p_1\) at \(L_{\min}\) on the graph agrees sufficiently well with that found from formula (6). If we now substitute into equation (5) the optimal value \(p_1 = 0.18\) ata, then for the previously consid-

for the example considered, we obtain \(L=964\) kgm. Consequently, with one stage of preliminary degassing, the work expended on removing air will be

\[ \frac{2660}{964}=2.76 \]

times less than without the use of such a fractional method. It follows from formula (6) that, as the pressure in the condenser of the utilization installation increases, the optimum pressure in the degasser increases.

With two stages of preliminary degassing, the work expended on removing the air released from \(1\ \text{m}^3\) of cooling water is expressed by the formula:

\[ L=a\left\{\frac{k}{k-1}RT_1\left[\left(\frac{1}{p_1}\right)^{\frac{k-1}{k}}-1\right](1-p_1)+\frac{k}{k-1}RT_1\left[\left(\frac{1}{p_2}\right)^{\frac{k-1}{k}}-1\right](p_1-p_2)+\right. \]
\[ \left.+RT_2\frac{k}{k-1}\left[\left(\frac{1}{p_k}\right)^{\frac{k-1}{k}}-1\right](p_2-p_k)\right\}, \tag{7} \]

where \(p_2\) is the pressure in the second degasser, atm.

Here, as in formula (5), each term characterizes the work expended on compressing a separate air stream.

To determine the optimum values of \(p_1\) and \(p_2\), we take the partial derivatives \(\left(\dfrac{\partial L}{\partial p_1}\right)_{p_2}\) and \(\left(\dfrac{\partial L}{\partial p_2}\right)_{p_1}\) and set these derivatives equal to zero.

As a result, we obtain two equations:

\[ \left(\frac{\partial L}{\partial p_1}\right)_{p_2} =\frac{1-k}{k}\cdot p_1^{\frac{1}{k}-2} -\frac{1}{k}\cdot p_1^{\frac{1}{k}-1} +p_2^{\frac{1}{k}-1}=0, \tag{8} \]

\[ \left(\frac{\partial L}{\partial p_2}\right)_{p_1} =\frac{1-k}{k}\,p_1\,p_2^{\frac{1}{k}-2} -\frac{1}{k}\cdot p_2^{\frac{1}{k}-1} +\left(\frac{1}{p_k}\right)^{\frac{k-1}{k}}=0. \tag{9} \]

Solving this system of equations with respect to \(p_1\) and \(p_2\), we obtain

\[ p_2= \frac{ kp_2^{\frac{1}{k}} \left[(1-k)p_2-\dfrac{k}{1-k}\left(\dfrac{1}{p_k}\right)^{\frac{k-1}{k}}\cdot p_2^{\,2-\frac{1}{k}}\right]^{2-\frac{1}{k}} }{ k-1+(1-k)p_2-\dfrac{k}{1-k}\left(\dfrac{1}{p_k}\right)^{\frac{k-1}{k}}\cdot p_2^{\,2-\frac{1}{k}} }, \tag{10} \]

\[ p_1=\frac{p_2}{1-k}-\frac{k}{1-k}\left(\frac{1}{p_k}\right)^{\frac{k-1}{k}}\cdot p_2^{\,2-\frac{1}{k}}. \tag{11} \]

As in the case with one stage of preliminary degassing, the distribution of pressures over the stages with two air-intake points depends only on the vacuum in the condenser. Equation (10) is solved by the method of trial selection. Thus, for the particular example considered earlier, the optimum values for \(p_1\) and \(p_2\) should be taken as \(0.25\) atm and \(0.07\) atm. With such a distribution of pressures, the work \(L=703\) kgm per \(\text{m}^3\) of cooling water, i.e., the energy expenditure for removing air will be

\[ \frac{2660}{703}=3.73 \]

times less.

less than without the application of the fractional method, and

\[ \frac{964}{713} = 1.35 \]

times less than with one stage of preliminary degassing. In Fig. 3 a curve is given for the dependence of the work \(L\) on the number of air-extraction points. From

Fig. 3.

this figure it is evident that the use of three stages of preliminary degassing can hardly be considered expedient, because the gain in energy expenditure for air removal from the use of a third stage will already be so insignificant that it will not compensate for the additional capital costs of the installation caused by the application of this stage.

CONCLUSIONS

1. The application of the fractional method of removing air from cooling water and from the condensers of utilization plants with mixing condensation significantly reduces the work expended on maintaining the specified vacuum.

2. The most expedient scheme of preliminary degassing is a two-stage scheme. With a larger number of stages of preliminary degassing, the additional cost and complexity of the installation are not justified by the additional gain thereby obtained in the expenditure of work for air removal.

LITERATURE

1. Butakov I. N. The use of the heat of circulating water at power stations for obtaining electric power. Izvestiya TPI, vol. 63, 1944.