ON THE SELECTION OF A DRIVE TYPE FOR WORKING MACHINES

V. A. Bragin

Presented by Prof.-Doctor I. N. Butakov

At the present time, in many branches of industry—machine building, metallurgy, chemistry—alongside electric motors, steam drives are widely used for driving working machines. The latter, especially at low powers, presupposes the use of back-pressure steam engines. This reduces their cost and increases reliability of operation. But the economy of steam engines is determined mainly by the possibilities for utilizing the heat of the exhaust steam. These possibilities are usually limited, especially where there is a factory thermal power station. The use of steam exhausted in the drive engine for heating or technological needs is connected with a redistribution of the output of condensing and district-heating kWh at the thermal power station, which considerably lowers the effectiveness of the steam drive. It is therefore of interest to compare the electric drive with the steam drive under various schemes for utilizing the heat of the exhaust steam. Such a comparison was made by L. A. Melent'ev [1, 2]. As the basis of the comparison method he adopted the reduction of the compared schemes—with electric and steam drive—to an equal release of heat from the boiler house of the thermal power station, and their comparison according to useful release of electrical energy. In this case the economy of the type of drive is determined by the ratio \(\xi = \frac{W_\mathrm{e} - W_{\mathrm{pr}}}{W_\mathrm{p}}\). If \(\xi > 1.0\), then the electric drive is more effective, and conversely.

But in deriving the formula for determining the fuel saving from the use of an electric drive, a whole series of inaccuracies is allowed. For example, all efficiencies of the thermal power station and of the drive drop out of the comparison, with the exception of the boiler-house efficiency and the internal relative efficiency. Taking the indicated method as a basis considerably schematizes, for some operating regimes of the thermal power station, the comparison of the selected schemes. Further, it is approximately assumed that the steam consumption of the steam engine of the drive is equal to the steam consumption from turbine extraction for heat consumers. Finally, the formula obtained does not make it possible to judge the change in the relative effectiveness of the drive with a prospective change in the energy coefficient of the factory thermal power station or of the power system.

The indicated inaccuracies can for the most part be eliminated if the heat-balance method [3, 4] is used for comparing different types of drives.

The heat balance is drawn up for the system: factory thermal power station—drive. In doing so it is assumed that both with an electric drive and with a steam drive, as

and with steam drive, the specified system supplies external consumers with one and the same amount of output (electric power and heat energy). Then the difference in the heat expenditures of the fuel in the system will show which drive is more effective under the given conditions.

Let us consider this in the most complicated example, when steam is supplied to the steam drive from extractions of the main turbines of the CHP plant, and the heat of its exhaust steam is used partly to displace district-heating extractions of the CHP turbines and partly to displace steam from the PRDS.

With an electric drive present, the heat balance of the specified system will have the form:

\[ \frac{860{\mathcal E}'_{\mathrm{t}}+Q'_{\mathrm{otb}}}{\eta'_{\mathrm{t}}} + \frac{860{\mathcal E}''_{\mathrm{t}}+Q''_{\mathrm{otb}}}{\eta''_{\mathrm{t}}} + \frac{860\left({\mathcal E}-{\mathcal E}'_{\mathrm{t}}-{\mathcal E}''_{\mathrm{t}}\right)}{\eta_{\mathrm{k}}} + \frac{Q_{\mathrm{rou}}}{\eta_{\mathrm{ku}}\cdot\eta_{\mathrm{p}}} + \frac{860{\mathcal E}_{\mathrm{pr}}}{\eta_{\mathrm{k}}\cdot\eta_{\mathrm{s}}\cdot\eta_{\mathrm{dv}}} = \frac{860{\mathcal E}+Q_{\mathrm{otp}}+860{\mathcal E}_{\mathrm{pr}}}{\eta'} \tag{1} \]

and with a steam drive:

\[ \begin{aligned} & \frac{860{\mathcal E}'_{\mathrm{t}}+Q'_{\mathrm{otb}}}{\eta'_{\mathrm{t}}} + \frac{860{\mathcal E}''_{\mathrm{t}}+Q''_{\mathrm{otb}}}{\eta''_{\mathrm{t}}} + \frac{860\left({\mathcal E}-{\mathcal E}'_{\mathrm{t}}-{\mathcal E}''_{\mathrm{t}}\right)}{\eta_{\mathrm{k}}} + \frac{Q_{\mathrm{rou}}}{\eta_{\mathrm{ku}}\cdot\eta_{\mathrm{p}}} \\[4pt] &\quad+ \frac{860\left({\mathcal E}_{\mathrm{pr}}-{\mathcal E}^{\mathrm{t}}_{\mathrm{pr}}\right)}{\eta^{\mathrm{v}}_{\mathrm{pr}}} + \frac{860{\mathcal E}^{\mathrm{t}}_{\mathrm{pr}}+Q'_{\mathrm{pr}}+Q''_{\mathrm{pr}}}{\eta^{\mathrm{t}}_{\mathrm{pr}}} - \frac{860\Delta{\mathcal E}'_{\mathrm{t}}+Q'_{\mathrm{pr}}}{\eta'_{\mathrm{t}}} + \frac{860\Delta{\mathcal E}'_{\mathrm{t}}}{\eta'_{\mathrm{k}}} \\[4pt] &\quad- \frac{860\Delta{\mathcal E}''_{\mathrm{t}}}{\eta_{\mathrm{k}}} + \frac{860\Delta{\mathcal E}''_{\mathrm{t}}}{\eta''_{\mathrm{t}}} - \frac{Q''_{\mathrm{pr}}}{\eta_{\mathrm{ku}}\cdot\eta_{\mathrm{p}}} = \frac{860{\mathcal E}+Q_{\mathrm{otp}}+860{\mathcal E}_{\mathrm{pr}}}{\eta''} \tag{2} \end{aligned} \]

Here the following notation is used, respectively:

\[ {\mathcal E}'_{\mathrm{t}} \text{ and } {\mathcal E}''_{\mathrm{t}} \]
— generation of electric power on the basis of district-heating and process extractions, kWh/year;

\[ {\mathcal E} \]
— total generation of electric power at the CHP plant, required for external consumers, kWh/year;

\[ \Delta{\mathcal E}'_{\mathrm{t}} \text{ and } \Delta{\mathcal E}''_{\mathrm{t}} \]
— those changes in the generation of electric power \(\left({\mathcal E}'_{\mathrm{t}}\right.\) and \(\left.{\mathcal E}''_{\mathrm{t}}\right)\) which occurred at the CHP plant as a result of the appearance of the steam drive, kWh/year;

\[ Q'_{\mathrm{otb}} \text{ and } Q''_{\mathrm{otb}} \]
— heat released from turbine extractions, kcal/year;

\[ Q_{\mathrm{rou}} \]
— heat released from the PRDS, kcal/year;

\[ {\mathcal E}_{\mathrm{pr}} \]
— amount of energy required for the drive of the working machine, kWh/year;

\[ {\mathcal E}^{\mathrm{t}}_{\mathrm{pr}} \]
— the same, but when using the heat of exhaust steam, kWh/year;

\[ Q'_{\mathrm{pr}} \text{ and } Q''_{\mathrm{pr}} \]
— heat of the exhaust of the steam drive, used to replace part of \(Q'_{\mathrm{otb}}\) and \(Q_{\mathrm{rou}}\), kcal/year;

in this case \(Q'_{\mathrm{pr}}+Q''_{\mathrm{pr}}=Q_{\mathrm{pr}}\), while \(Q'_{\mathrm{pr}}=\psi Q_{\mathrm{pr}}\) and \(Q''_{\mathrm{pr}}=(1-\psi)Q_{\mathrm{pr}}\).

Here the coefficient \(\psi\) shows what fraction of all the utilized exhaust heat went to replace part of \(Q'_{\mathrm{otb}}\).

\[ Q_{\mathrm{otp}} = Q'_{\mathrm{otb}}+Q''_{\mathrm{otb}}+Q_{\mathrm{rou}} = Q'_{\mathrm{otb}}+Q''_{\mathrm{otb}}+Q_{\mathrm{rou}}-Q'_{\mathrm{pr}}-Q''_{\mathrm{pr}}+Q_{\mathrm{pr}} \]
— the total amount of heat energy released for external consumers, kcal/year;

\[ \eta'_{\mathrm{t}},\ \eta''_{\mathrm{t}},\ \eta_{\mathrm{k}},\ \eta_{\mathrm{ku}},\ \eta_{\mathrm{p}} \]
— efficiencies for the production of heat for heating and process purposes, for the production of condensing kWh, of the boiler plant, and of the heat flow within the CHP plant [3, 4];

\(\eta_c,\ \eta_{dv}\) — efficiencies of the electrical network and the motor;

\(\eta_{pr}^{v},\ \eta_{pr}^{t}\) — efficiencies of the steam engine when operating with exhaust to atmosphere and with utilization of exhaust steam. They are determined in the same way as \(\eta_k\) and \(\eta_t\) [3], but instead of the product \(\eta_{ku}\cdot \eta_p\), \(\eta_t''\cdot \eta_p'\) is substituted.

\(\eta_{ln}\) — efficiency taking into account the additional heat losses by the steam line from the CHP plant to the drive motor;

\(\eta',\ \eta''\) — overall efficiencies of the system for electric and steam drives.

The quantities denoted above by \(E_t'\), \(E_t''\), and \(E_{pr}^{t}\) are determined as
\(E_t'=\dfrac{Q_{otb}'}{q'}\),

\[ E_t''=\frac{Q_{otb}''}{q''}\quad \text{and}\quad E_{pr}^{t}=\frac{Q_{pr}}{q_{pr}}, \]

where \(q'\), \(q''\), and \(q_{pr}\) indicate how many kcal of heat are released from the extraction and back pressure of the drive motor per each kWh of energy generated on the basis of the corresponding extractions [3].

If now the corresponding parts of equation (1) are subtracted from the left- and right-hand sides of equation (2), the difference will show the effectiveness of applying one drive or another. With a positive difference the electric drive will be more economical, and with a negative difference, the steam drive.

Omitting simple intermediate transformations and denoting by

\[ A'=\frac{1}{\eta_t'}-\frac{860}{q'}\left(\frac{1}{\eta_k}-\frac{1}{\eta_t'}\right); \tag{3} \]

\[ A''=\frac{1}{\eta_t''}-\frac{860}{q''}\left(\frac{1}{\eta_k}-\frac{1}{\eta_t''}\right); \tag{3'} \]

\[ A_{pr}=\frac{1}{\eta_{pr}^{t}}-\frac{860}{q_{pr}}\left(\frac{1}{\eta_{pr}^{v}}-\frac{1}{\eta_{pr}^{t}}\right); \tag{3''} \]

\[ z=\frac{Q_{otp}}{860E} \]
— the energy coefficient;

\[ \varphi_{pr}=\frac{E_{pr}}{E} \]
— the share of the energy expended on the drive in the total kWh output at the CHP plant;

\[ \omega_{pr}=\frac{Q_{pr}}{860E_{pr}}=\frac{q_{pr}}{860}\cdot \delta,\quad \text{where} \tag{4} \]

\[ \delta=\frac{D_{pr}'}{D_{pr}} \]
— the share of the exhaust steam of the drive motor used for external consumers;

we obtain:

\[ (1+\varphi_{pr}+z)\left(\frac{1}{\eta''}-\frac{1}{\eta'}\right) =(1+\varphi_{pr}+z)\Delta \frac{1}{\eta} = \]

\[ =\varphi_{pr}\left[\left(A_{pr}\cdot A''\cdot \eta_t''-\psi\cdot A' -\frac{1-\psi}{\eta_{ku}\cdot \eta_p}\right)\omega_{pr} +\frac{1}{\eta_{pr}^{v}}\cdot A''\cdot \eta_t'' -\frac{1}{\eta_k\cdot \eta_c\cdot \eta_{dv}}\right], \]
or, finally,

\[ \Delta \frac{1}{\eta} =(A_0\cdot \omega_{pr}+B_0)\frac{\varphi_{pr}}{1+\varphi_{pr}+z}. \tag{5} \]

Here

\[ A_0=A_{pr}\cdot A''\cdot \eta_t''-\psi\cdot A' -\frac{1-\psi}{\eta_{ku}\cdot \eta_p} \tag{6} \]

\[ B_0=\frac{1}{\eta_{pr}^{v}}\cdot A''\cdot \eta_t''-\frac{1}{\eta_k\cdot \eta_s\cdot \eta_{dv}} . \tag{7} \]

The quantity \(\Delta 1/\eta\) in formula (5) indicates the saving or excess consumption of fuel heat in the system when one or another drive is used, per each kcal released in the form of electric power and heat.

If \(\Delta 1/\eta\) is positive, the electric drive is more effective; in the opposite case, the steam drive is more effective.

Formula (5) makes it possible to trace the change in \(\Delta 1/\eta\) when the energy coefficient is increased or decreased. In the general case, as \(z\) changes, the quantities \(\varphi_{pr}\) and \(\omega_{pr}\) may change, while \(A_0\) and \(B_0\) remain approximately constant.

Investigation shows that formula (5) is valid for the most diverse cases of using a steam drive. Indeed, in practice the following conditions may be encountered:

1) The drive engine receives steam from extractions of cogeneration-plant turbines, but the heat of the exhaust \(Q_{pr}\) is used only to displace part of \(Q_{otb}\). Then \(\psi=1.0\) and

\[ A_0=A_{pr}\cdot A''\cdot \eta_t''-A' . \tag{6'} \]

2) The same, but \(Q_{pr}\) displaces only part of \(Q_{rou}\). Then \(\psi=0\) and

\[ A_0=A_{pr}\cdot A''\cdot \eta_t''-\frac{1}{\eta_{ku}\cdot \eta_p}. \tag{6''} \]

3) The heat of the exhaust steam is not used at all. Then \(Q_{pr}=0\), \(\delta=0\), and \(A_0\cdot \omega_{pr}=0\), while

\[ \Delta \frac{1}{\eta}=B_0\cdot \frac{\varphi_{pr}}{1+\varphi_{pr}+z}. \tag{5'} \]

4) The steam drive is supplied with steam from a boiler. Let \(\psi=1.0\). Then the product \(A''\cdot \eta_t''=1.0\) and

\[ A_0=A_{pr}-A' . \tag{6'''} \]

\[ B_0=\frac{1}{\eta_{pr}^{v}}-\frac{1}{\eta_k\cdot \eta_s\cdot \eta_{dv}} . \tag{7'} \]

The equality of \(A''\cdot \eta_t''\) to unity is easy to prove:

\[ A''\cdot \eta_t'' = \eta_t'' \cdot \left[ \frac{1}{\eta_t''} - \frac{860}{q''} \left( \frac{1}{\eta_k} - \frac{1}{\eta_t''} \right) \right] = 1- \frac{h_a''\cdot \eta_{oi}''\cdot \eta_m''\cdot \eta_2'}{i_{otb}''-t_k} \cdot \left( \frac{\eta_t''}{\eta_k}-1 \right). \]

But when the drive is supplied with live steam, the heat drop used in the turbine on the basis of this steam, \(h_a''\cdot \eta_{oi}''=0\), and \(A''\cdot \eta_t''=1\). In addition, in the expressions for \(\eta_{pr}^{t}\) and \(\eta_{pr}^{v}\), instead of \(\eta_t''\) it is necessary to substitute the product \(\eta_{ku}\cdot \eta_p\).

5) The same, but the heat of the exhaust is used year-round. Then \(\delta=1.0\) and \(\eta_{pr}^{v}=\eta_{pr}^{t}\),

\[ A_0=\frac{1}{\eta_{pr}^{t}}-A' \tag{6^{IV}} \]

\[ B_0=\frac{1}{\eta_{pr}^{t}}-\frac{1}{\eta_k\cdot \eta_s\cdot \eta_{dv}} . \tag{7^{II}} \]

6) If a condensing steam engine is used for drive purposes, then, when it is supplied with live steam, we have \(A''\cdot \eta_t''=1.0\), \(\omega_{pr}=0\), and \(A_0\omega_{pr}=0\), while

\[ B_0=\frac{1}{i_{pr}^{k}}-\frac{1}{\eta_k\cdot \eta_{c}\cdot \eta_{dob}} . \tag{7'''} \]

7) If the exhaust steam of the drive engine is used to produce electric power in a back-pressure turbine, then it is easy to prove that all the quantities in formula (5) take the form:

\[ A_0=\frac{1}{\eta_b^{m}}-\frac{1}{\eta_k}, \tag{6'} \]

\[ \omega_{pr}=\frac{E_{km}}{E_{pr}}, \tag{4'} \]

where \(E_{km}\) is the number of kWh of energy obtained from a turbogenerator of type \(MK\) with efficiency \(\eta_k^m\), kWh/year.

\(B_0\) is determined by expression \((7'')\).

These are almost all the cases that may occur under the conditions of industrial enterprises.

Since the quantity \(\Delta^1 i_k\) in formula (5) is proportional to fuel consumption, it is easy to calculate the absolute saving of fuel, or its excess consumption, in the case of using one or another type of drive. To do this, we divide the right- and left-hand sides of expression (5) by

\[ \frac{\varphi_{pr}}{1+\varphi_{pr}+z} \]

and multiply by

\[ \frac{860}{Q_p^n}. \]

Then we obtain an expression for calculating the absolute saving of fuel (or its excess consumption) for each kWh of energy expended on the drive \((Q_p^n=7000\ \text{kcal/kg})\):

\[ b_{ek}^{y}=0.123\,(A_0\cdot \omega_{pr}+B_0)\ \text{kg/kWh}. \tag{8} \]

The obtained expressions (5) and (8) make it possible to compare not only the effectiveness of electric and steam drives, but also to establish the dependence of effectiveness on a whole series of factors (the initial and final parameters of the drive steam, the initial steam parameters at the thermal power station, the power of the drive engine, etc.).

In Fig. 1, for a number of the cases analyzed above, the dependences are shown of \(b_{ek}^{y}\), the fuel saving from the use of electric drive as compared with a steam-turbine drive, on \(\delta\), the degree of utilization of the exhaust steam for various purposes \((N_{pr}=700\ \text{kW})\).

Let us consider individual curves. Curves 1 and 2 \((P_0^{pr}=29\ \text{ata};\ t_0^{pr}=425^\circ\text{C};\ P_k^{pr}=1.2\ \text{and}\ 7.0\ \text{ata};\ \psi=1.0;\ A''\cdot \eta_t''=1.0)\) show that, as the heat drop in the engine of the steam drive decreases, its effectiveness falls, and the more so the smaller \(\delta\) is.

Curves 4 and 5 represent the same thing, but for \(\psi=0\). Here, for certain values of \(\delta\) (greater than 5.5 for \(P_k^{pr}=1.2\ \text{ata}\) and 7.75 for \(P_k^{pr}=7.0\ \text{ata}\)), the steam drive already becomes more economical. This is explained by the fact that, in the case of electric drive, the heat consumers are supplied with heat through a pressure-reducing and desuperheating unit, and a certain heat drop is not used to perform work.

The ordinates, respectively, between curves 1–4 and 2–5 show the change in the effectiveness of the steam drive for different \(\delta\) as a function of the values of \(\psi\) (when it changes from zero to unity). The higher the value of \(\psi\), the lower the effectiveness of the steam drive. This is understandable, since with the use of

Fig. 1. Dependence of fuel savings from the use of an electric drive, as compared with a steam drive, on the degree of utilization of its exhaust steam.

As \(\psi\) increases, the cogenerated \(\text{kWh}\), produced with high efficiency at the plant CHP station, are increasingly displaced. Curve 3 shows the effectiveness of using a condensing engine for the drive \((P_o^{pr} = 29\ \text{ata};\ t_o^{pr} = 425^\circ\text{C};\ P_k^{pr} = 0.05\ \text{ata})\). For most values of \(\delta\), such a drive is more economical than back-pressure engines.

Curves 6 and 7, for different \(\delta\), show how the effectiveness of the drive changes with the initial steam parameters at the CHP station (for curve 6: \(p_0 = 29\ \text{ata};\ t_0 = 425^\circ\text{C}\); for curve 7: \(p_0 = 90\ \text{ata};\ t_0 = 500^\circ\text{C}\); in both cases \(\psi = 1.0;\ P_o^{pr} = 7\ \text{ata};\ P_k^{pr} = 1.2\ \text{ata};\ P_{otb}^{pr} = 8\ \text{ata}\)). It is evident that the economy of the steam drive increases as the initial steam parameters before the CHP turbines are raised.

As can be seen from Fig. 1, in most cases (even for \(\delta = 1.0\)) the electric drive is more advantageous. The fuel saving will be greater the more: a) the smaller the value of \(\delta\); b) the greater the value of \(\psi\); c) the lower the power required to drive the working machine; d) the lower the initial steam parameters of the CHP station; e) the smaller the heat drop processed in the steam engine of the drive, or, what is the same, the higher its back pressure; f) the farther the drive engine is located from the energy source.

From the thermal point of view, a steam drive becomes competitive with an electric drive only when heat consumers are supplied with heat through a ROU and at sufficiently high values of \(\delta\), and also at considerable drive powers. Under these conditions, the effectiveness of a steam drive will be the greater, the higher \(\delta\) is and the greater the drive power.

Naturally, in addition to the thermal comparison, it is always necessary to carry out a comparison in terms of capital investments and annual costs. But here as well, in most cases, the electric drive proves to be more advantageous [1].

However, in making the comparison it is necessary not to overlook one further circumstance, which is very important in many cases. When a steam drive is used (at its relatively large powers), a certain number of kWh of electric energy is freed up at the plant’s CHP station or in the power system for satisfying other consumers. This is especially important under strained operating conditions of a CHP station, if its expansion is not expected in the near future.

References

1. Melent’ev L. A.—Fundamental Problems of Industrial Heat Power Engineering, GEI, 1954.
2. Melent’ev L. A.—District Heating, Part II, Publishing House of the Academy of Sciences of the USSR, 1948.
3. Butakov I. N.—Thermal Power Installations, Parts I, II, TPI Publishing House, 1954.
4. Butakov I. N.—On discrepancies in determining the efficiency of CHP stations, Collection on the Efficiency of CHP Stations, GEI, 1953.