Determination of Steam Consumption in a Steam Ejector
Abstract Generated abstract
This paper addresses the practical calculation of motive steam consumption in a steam ejector required to remove a unit mass of air or steam-air mixture. It criticizes existing tabular and thermodynamic methods as either ambiguous or computationally cumbersome, then derives a simplified energy-balance formulation that accounts for ideal compression work and losses in the nozzle, mixing chamber, and diffuser. The resulting equation for steam consumption is solved graphically as the intersection of a bisector and a hyperbola. An example calculation gives 1.68 kg of fresh steam per kg of steam-air mixture, close to a previously published result, suggesting that the method offers sufficient accuracy for engineering practice with reduced calculation effort.
Full Text
Determination of Steam Consumption for a Steam Ejector
E. N. Shadrin
When calculating a steam ejector, it is necessary to know the consumption of working steam required to remove 1 kg of air or a steam-air mixture. Most authors [1; 2; 3], when determining steam consumption, refer to the tables given in the book by Caillou and Robinson. These tables divide all ejectors into large and small ones, and for each type of ejector, for a certain suction pressure of the steam-air mixture, the steam consumption is given. At the same time, it remains entirely unclear which ejectors should be classed as large and which as small. An even greater arbitrariness is allowed in the question of distributing the total steam consumption, determined from the table, among the individual stages. Here it is simply assumed that the ratio between the consumptions of the first and second stages lies within the limits from 1:7 to 1:9 for ejectors without cooling and from 1:3.5 to 1:4.5 for ejectors with cooling [1; 2]. A. M. Kazanskii [4], on the basis of thermodynamic considerations, derives a formula for determining the steam consumption of an ejector. But the use of this formula is difficult because of its complexity. The problem has to be solved by a trial-and-error method. We present a simpler solution of the question, which gives accuracy sufficient for practical purposes.
Let us first determine what minimum quantity of steam must be expended on an ejector in order to remove 1 kg of air or a steam-air mixture.
Let 1 kg of air or steam-air mixture enter the mixing chamber of the ejector, the pressure and specific volume of which are denoted by \(p_k\) and \(v_{\mathrm{cm}.k}\). Suppose that, to compress 1 kg of steam-air mixture from the condenser, \(\alpha\) kg of live steam is required. Then in the diffuser, for each 1 kg of steam-air mixture coming from the condenser, \(1+\alpha\) kg of steam-air mixture will be compressed, containing more steam than the mixture entering the mixing chamber from the condenser.
The adiabatic work, in kcal, expended on compressing 1 kg of mixture from pressure \(p_k\) to pressure \(p_{\mathrm{pr}}\), will be expressed by the formula:
\[ AL=\frac{Ak}{k-1}p_k v_k\left[\left(\frac{p_{\mathrm{pr}}}{p_k}\right)^{\frac{k-1}{k}}-1\right](1+\alpha), \tag{1} \]
where \(k\) is the exponent of the compression adiabat. Its value may be taken equal to 1.135.
\(v_k\) is the specific volume of the mixture at the inlet to the diffuser.
If the specific volume of the working steam at the end of expansion is \(v_{p.k}\), then the specific volume of the mixture \(v_k\) at the inlet to the diffuser can readily be determined from the condition of equality of the volumes of the removed mixture and the working steam. This equality is written as follows:
\[ v_{\mathrm{cm}.k}+\alpha v_{p.k}=(1+\alpha)v_k. \]
Whence
\[ v_k=\frac{v_{\mathrm{sm}.k}+a v_{\mathrm{p}.k}}{1+a}. \tag{2} \]
If the value of \(v_k\) from equality (2) is substituted into equation (1), then after the simplest transformations we obtain:
\[ AL=\frac{Ak}{k-1}p_k v_{\mathrm{sm}.k} \left[\left(\frac{p_{\mathrm{pr}}}{p_k}\right)^{\frac{k-1}{k}}-1\right]+ \]
\[ +a\frac{Ak}{k-1}p_k v_{\mathrm{p}.k} \left[\left(\frac{p_{\mathrm{pr}}}{p_k}\right)^{\frac{k-1}{k}}-1\right]. \tag{3} \]
In the right-hand side of the equation obtained, the first term characterizes the expenditure of energy for compressing the steam-air mixture from the condenser pressure \(p_k\) to the end-of-compression pressure \(p_{\mathrm{pr}}\). The second term expresses the work that must be expended to compress the motive steam from the condenser pressure to the end-of-compression pressure.
In calculating the ejector, the parameters of the motive steam are specified; consequently, the adiabatic heat drop \(H\) of this steam is also known. Therefore, from the equality \(aH=AL\) it is possible to determine the steam consumption for removing 1 kg of the steam-air mixture, provided that there are no losses in the ejector.
This consumption will be expressed by the formula:
\[ a= \frac{1} {\dfrac{H} {\dfrac{Ak}{k-1}p_k v_{\mathrm{sm}.k} \left[\left(\dfrac{p_{\mathrm{pr}}}{p_k}\right)^{\frac{k-1}{k}}-1\right]} -\dfrac{v_{\mathrm{p}.k}}{v_{\mathrm{sm}.k}}}. \tag{4} \]
Under actual operating conditions of a steam ejector, the following losses occur:
1) energy loss in the nozzle;
2) energy loss in the mixing chamber;
3) energy loss in the diffuser.
The energy loss in the nozzle, in kcal/kg, is
\[ h_i=(1-\varphi^2)H, \tag{5} \]
where \(\varphi\) is the velocity coefficient of the nozzle.
The energy losses in the mixing chamber may be determined on the basis of the following considerations.
The motive steam enters the mixing chamber with a high velocity \(c_1\), whereas the mixture drawn from the condenser enters with an insignificant velocity. In the mixing chamber these two streams mix, so that the phenomenon of an inelastic impact is present. Equating to zero the velocity of entry of the steam-air mixture into the mixing chamber, we obtain the energy loss due to impact per 1 kg of the mixture entering the diffuser,
\[ h_{\mathrm{k.s.}}=\frac{A}{2g}\left[ac_1^2-(1+a)c_2^2\right]. \tag{5'} \]
In formula (5′), \(c_2\) is the velocity of the steam-air mixture at the entrance to the diffuser. Its magnitude can be determined from the condition of equality of the quantities of motion before and after mixing of the streams. This equality is written as
\[ ac_1+c'_0=(1+a)c_2. \]
As already indicated, the velocity \(c_0'\) of the steam-air mixture at the entrance to the mixing chamber is insignificant in comparison with the velocity \(c_1\) of the steam leaving the nozzle and may be taken as equal to zero, so that
\[ c_2=\frac{\alpha}{1+\alpha}c_1. \tag{5''} \]
Substituting \(c_2\) from \((5'')\) into \((5')\) and carrying out a number of simple transformations, we obtain
\[ h_{k.c.}=\frac{\alpha}{1+\alpha}H. \]
We shall estimate the energy loss in the diffuser, as is customary, by the efficiency of the diffuser \(\eta_0\). Then the expenditure of energy for removing \(1\ \mathrm{kg}\) of steam-air mixture, with allowance for losses, is expressed by the formula:
\[ \alpha H=(1-\varphi_i^2)H\alpha+\frac{\alpha}{1+\alpha}H+ \]
\[ +\alpha\,\frac{Ak}{(k-1)\eta_0}\cdot p_k \left[\left(\frac{p_{pr}}{p_k}\right)^{\frac{k-1}{k}}-1\right] \left(\frac{v_{c.m.k}}{\alpha}+v_{p.k}\right). \]
From this equation we determine the amount of fresh steam required for removing \(1\ \mathrm{kg}\) of steam-air mixture. Thus, after simple transformations, we obtain:
\[ \alpha= \frac{1}{ \varphi_i^2- \frac{Ak}{H(k-1)\eta_0}\cdot p_k \left[\left(\frac{p_{pr}}{p_k}\right)^{\frac{k-1}{k}}-1\right] \left(\frac{v_{c.m.k}}{\alpha}+v_{p.k}\right) } -1. \tag{6} \]
As can be seen, the required quantity \(\alpha\) enters both the left- and right-hand sides of equation (6); therefore such an equation can be solved either by trial or graphically. The latter method is the simplest and most convenient. This equation can be represented as a system of two equations:
\[ y=\alpha \]
and
\[ y= \frac{1}{ \varphi_i^2- \frac{Ak}{H(k-1)\eta_0}\cdot p_k \left[\left(\frac{p_{pr}}{p_k}\right)^{\frac{k-1}{k}}-1\right] \left(\frac{v_{c.m.k}}{\alpha}+v_{p.k}\right) } -1. \]
The first equation in the axes \(y-\alpha\) is a bisector, and the second is a hyperbola. The bisector may be constructed once and for all, whereas the position of the hyperbola is determined by the initial conditions. Thus, for \(y=\infty\), the denominator of the first fraction must be equal to zero. Hence
\[ \alpha= \frac{1}{ \frac{\varphi_i^2(k-1)\cdot H\cdot\eta_0}{ Akp_kv_{c.m.k} \left[\left(\frac{p_{pr}}{p_k}\right)^{\frac{k-1}{k}}-1\right] } -\frac{v_{p.k}}{v_{c.m.k}} }. \]
If we put \(\alpha=\infty\), then
\[ y= \frac{1}{ \varphi_i^2- \frac{Ak}{H(k-1)\eta_0}\cdot p_k v_{p.k} \left[\left(\frac{p_{pr}}{p_k}\right)^{\frac{k-1}{k}}-1\right] } -1. \]
To construct the hyperbola it is necessary to know one more point belonging to this hyperbola. We determine this point by putting \(\alpha=1\). Then
\[ y=\frac{1}{c^{2}-\dfrac{Ak}{H(k-1)\eta_{0}}\cdot p_{k}\left[\left(\frac{p_{np}}{p_{2}}\right)^{\frac{k-1}{k}}-1\right]\left(v_{c.m.k}+v_{n.k}\right)}-1. \]
The intersection of the angle bisector and the hyperbola gives the required value of \(\alpha\). Let us determine the steam consumption for the ejector by the method considered by us for an example taken from the book by Kazanskii [4], in order to compare the results obtained.
Mass quantity of air being exhausted \(G_{v}=0.0103\ \text{kg/sec}\).
Temperature of the steam-air mixture at the ejector inlet \(t_{c.m}=35^{\circ}C\).
Pressure of fresh steam before the ejector nozzle \(p_{0}=9\ \text{ata}\).
Temperature of fresh steam before the ejector nozzle \(t_{0}=240^{\circ}C\).
Pressure of the steam-air mixture at the ejector inlet \(p_{k}=0.07\ \text{ata}\).
Pressure after the first stage of the ejector \(p_{np}=0.28\ \text{ata}\).
Under these conditions the quantity of steam in the steam-air mixture is \(G_{k}=0.0295\ \text{kg/sec}\), so that the total quantity of mixture is \(G_{c.m}=0.0398\ \text{kg/sec}\). The specific volume of the steam-air mixture is \(v_{c.m.k}=31.1\ \text{m}^{3}/\text{kg}\); the specific volume of the motive steam at the end of expansion is \(v_{n.k}=17.3\ \text{m}^{3}/\text{kg}\). The adiabatic drop is \(H=183\ \text{kcal/kg}\).
Figure 1 gives the graphical solution of equation (6) for this particular example. The intersection of the angle bisector \(y=\alpha\) with the hyperbola gives the value \(\alpha=1.68\), i.e., for 1 kg of steam-air mixture it is necessary to expend 1.68 kg of fresh steam, so that the per-second consumption of fresh steam is \(G_{n}=0.067\ \text{kg/sec}\).
Fig. 1.
Kazanskii A. M. [4] gives, for this same example, \(G_{n}=0.063\ \text{kg/sec}\). Thus, the proposed method for determining the steam consumption for an ejector is simpler and gives an accuracy sufficient for practical purposes. In this case it is not necessary to make numerous laborious calculations, determining the steam consumption for the ejector by the method of selection [4] or of successive approximation [5].
Literature
- Anatoliev F. N. Calculation of Auxiliary Devices of Steam-Power Installations. ONTI, 1936.
- Makeev V. A. Condensing Devices of Steam-Power Installations. ONTI, 1933.
- Radtsig A. A. Theory and Calculation of Condensing Installations. Gosenergoizdat, 1930.
- Kazanskii A. M. Condensing Devices. GONTI, 1939.
- Bliudov V. P. Condensing Devices of Steam Turbines. Gosenergoizdat, 1951.