Calculate steam consumption. Calculation of steam consumption for an industrial building. Calculation of steam consumption for heating and ventilation

3.2.2 Calculation of steam consumption for heating and ventilation

Calculation of heat costs for heating and ventilation is determined by the formula:

Q=q · V · (t pom – t calculation ) · T year , kW/year, (3.11)

where q is the specific heat consumption for heating and ventilation of 1 m 3 room at a temperature difference of 1 ° C, kW/(m 3 deg).

The average value of this value can be taken: for heating - 0.45 · 10 -3 kW/(m 3 .deg), for ventilation 0.9 · 10 -3 kW/(m 3 .deg).

V – the total volume of the premises of the site without taking into account the volume drying chambers, m 3;

t room – room temperature, assumed to be 20°C;

t calc – design temperature for heating and ventilation;

T year - the duration of the heating season is determined by the formula:

T year = 24*τ from, h,

where τ from is the duration of the heating season, days.

T year = 24 · 205 = 4920 hours.

Q from = 0,45 · 10 -3 · 4456,872 · (20-(-26)) · 4920 = 453,9 · 10 3 kW/year.

Q vent = 0,09 · 10 -3 · 4456,872 · (20-(-12)) · 4920 = 63,15 · 10 3 kW/year.

Table 3.3 – Calculation of heat consumption for heating and ventilation

Calculation annual needs in pairs for heating and ventilation is determined by the formula:

3.2.3 Calculation of heat (steam) consumption for domestic needs

Calculation of heat (steam) consumption for household needs is determined by the formula:

where q is steam consumption per person per shift;

m – number of people working on the busiest shift;

n is the number of work shifts at the site (it is advisable to take 2);

τ – number of days of operation of the site per year.

3.2.4 Calculation of the total annual steam demand for technological and domestic needs, heating and ventilation

Calculation of the total annual steam demand for technological and domestic needs, heating and ventilation is determined by the formula:

D generally = D academic year + D from + D everyday life , t/year. (3.14)

D generally =8.13+891.47+2.6=902.2 t/year.

Based on the obtained value η oe determine the preliminary estimated flow rate pair

which will be clarified later.

For turbines with one controlled steam extraction (as specified), the preliminary steam flow is determined by an approximate formula (assuming that the relative internal efficiency of the high-pressure part and the turbine as a whole are the same):

(13)

Where G- value of regulated (industrial, district heating) extraction at pressure R according to (assigned); N t 0chvd - heat drop of an ideal turbine from the initial pressure R 0 to extraction pressure R according to (Fig. 6).

When calculating the flow path of a turbine with controlled extraction:

1) all stages up to the controlled extraction are calculated for the total steam consumption found using formula (13);

2) the stages after controlled extraction are calculated for the flow rate in pure condensation mode, determined by expression (12).

The low-pressure stages must ensure the passage of steam when the turbine operates at rated electrical power with the controlled extraction switched off (condensing mode).

Calculation of the thermal circuit, determination of steam flow rates in the turbine compartments and reduction of the energy balance are carried out for two turbine operating modes:

a) with controlled extraction at rated electrical power (cogeneration mode);

b) without controlled extraction (condensation mode) at rated electrical power.

Adjustment of the lengths of the nozzle and working blades of the stages to the controlled extraction is carried out according to the steam flow through the compartments obtained in the cogeneration mode, and the remaining stages ― by steam flow through the compartments in condensation mode.

EXAMPLE OF CALCULATION OF A MULTISTAGE STEAM TURBINE

K-12-35 with three regenerative selections for heating feed water to 145 °C according to the following initial data:

nominal electric power N e = 12000 kW;

rotation frequency n=50 s -1 ;

steam pressure in front of the turbine R" 0 = 3.5 MPa;

steam temperature in front of the turbine t"0 = 435 o C;

exhaust steam pressure R" k = 0.006 MPa;

nozzle steam distribution.

Determination of steam consumption

We calculate the turbine for economic power. Let's accept

N eq =0.9 N e =0.9∙12000 = 10800 kW.

Pressure in front of the nozzles of the control stage at design mode

R 0 = 0,95∙R"0 = 0.95∙3.5=3.325 MPa.

The pressure loss in the exhaust pipe is determined by the formula

Δ p = p" To ∙ λ∙( With ch /100) 2,

having accepted With VP =120 m/s, λ = 0.07, we get

Δ R=0.006∙0.07∙(120/100) 2 = 0.0006 MPa,

steam pressure behind the last stage rotor blades

R To =p" To + Δ R= 0.006 +0.0006 = 0.0066 MPa.

We roughly depict the process in h,s- diagram

(see Fig. 1), drawing points A" 0, A 0, A" to t, A to t.

We'll find h 0 = 3304 kJ/kg; h′ To t= 2143 kJ/kg; h To t= 2162 kJ/kg;

N t 0id = 3304-2143 = 1161 kJ/kg; N t 0 = 3304-2162 = 1142 kJ/kg;

η dr = 1142/1161 = 0.984.

We accept η вр = 1.0, η ′ o i= 0.8, according to reference data

η m =0.98; η g =0.97.

Thus we have

η oe = η dr ∙η ′ o i∙η vvr ∙η m ∙η g =0.984∙0.8∙1.0∙0.98∙0.97=0.748.

Preliminary calculated steam flow per turbine

All stages of the turbine will be designed for this steam flow.

Preliminary process line in h,s-the diagram is plotted according to accepted valueη " o i in the following way:

N T i= 1142∙0.8=913.6 kJ/kg.

Postponing N T i V h,s-diagram, we get point A k on the isobar R k (Fig. 6).

The task of drawing an approximate line of change in the state of steam in h,s-diagram is only to find the specific volume of steam at the exit from the last stage. We find the state of the steam at the exit from this stage by plotting down the isobar R to from A to output loss

N in z =c 2 2 z/2000.

In preliminary calculation N in z is found from the expression

N in z = ζ id a ∙H t 0id ,

where ζ id a is the output loss coefficient of the last stage.

When calculating, evaluate ζ id a and find N in z and With 2z.

Fig.6. The process of steam expansion in the condensation room(s)

and heating (b) turbines in h,s-diagram

The smaller ζ id a, the smaller, therefore, With 2 z – the steam output speed in the last stage, but the longer the blade length will be.

The value of ζ id should be set based on available data on similar turbine designs.

For small condensing turbines ζ id a = =0.015...0.03; for large condensing turbines ζ id a = 0.05 ... 0.08.

For turbines with back pressure ζ id a<0,015.

Let's take ζ id a =0.0177. Then

N in z = 0.0177∙1161 =20.55 kJ/kg.

State of steam at point a to z corresponds to the specific volume of steam v 2 z=20.07m 3 /kg. Enthalpy of steam behind the turbine h k =

2390.4 kJ/kg.

By determining the approximate steam flow through the turbine and the approximate specific volume of steam at the exit from the last stage, the first stage of the preliminary calculation ends.

The second stage consists of checking the possibility of constructively implementing the last stage and approximate determination of the isentropic heat difference in it.

2. Preliminary calculation of the last stage

For preliminary calculation of the last stage, the following parameters are known:

N t 0id, N in z ,ζ id a, G,n.

In further calculations, the index z discard.

Steam velocity at the outlet of the last stage working grate

To determine the diameter of the last step, it is necessary to set the ratio ν = d/l 2 where d– average diameter of the last stage; l 2 – output length of the blade of the last stage.

In existing turbines the value ν lies within 2.7 ... 50.0. Small values ​​apply to high-power condensing turbines, large values ​​are typical for low-power condensing turbines and turbines with back pressure. The blades of the last stages can be made with either a constant or variable profile. The issue of transition from blades with a constant height profile to a twisted one should be decided on the basis of a comparison of losses caused by the flow around the rotor blades as the value of ν changes. For values ​​of ν<8 лопатки прихо­дится всегда выполнять закрученными. При ν >12, the use of twisting does not provide a noticeable gain in efficiency.

Let , for example, ratioν =5.2. Then, assuming an axial steam exit in the last stage, i.e. α 2 = 90° (and therefore With 2a =c 2), we get:

Thus, the length of the working blades

l 2 =d/ν =1.428/5.2=0.2746 m.

Peripheral speed at the middle diameter of the stage

u =π ∙d∙n= 3.14∙1.428∙50 = 224.3 m/s.

Peripheral speed at blade tip
u V =u∙(d+l 2 )/d=224.3∙(1.428+0.2746)/1.428=267.4m/s .

Such speeds are quite acceptable.

When calculating small-power turbines, there is no need to test the strength of the rotor blades if u does not exceed 300 m/s .

Root section diameter

d To = d - l 2 = 1.428 - 0.2746 = =1.153 m .

Peripheral speed of the blades in the root section

u To = π ∙d To ∙n=181.17 m/s.

The heat drop processed in the axial turbine stage is determined for optimal operating conditions, which are expressed by the optimal speed ratio

(14)

where ρ – degree of stage reactivity.

The available heat drop processed in the turbine stage with the greatest efficiency can be determined from expression (14):

,

after transforming which we find

In this formula the quantities u,ρ , φ, α 1 refer to the middle section of the step.

Since in any section along the height of the blade the heat drop N 0 must be the same (the pressure in front and behind the stage is constant in height), then it can be calculated using expression (15) for the root section of the last stage, where ρ k ≈0 (all stages of chamber turbines are designed with a degree of reactivity in the root section ρ k ≈0), u=u k, taking approximately φ = 0.95 and α 1 = 15 o:

At a given heat difference N 0 optimal diameter of the root section of the step d k can be determined after transforming expression (15):

. (16)

Taking, for example, for the root section the steps ρ к =0, φ=0.955, α 1 =15 о, we obtain the optimal diameter of the root section at N 0 =78 kJ/kg:

3. Control stage calculation

We select a control stage in the form of a double-crown Curtis disk. Let us take the heat drop in it to be equal to 30% of the total heat drop N t 0, which will be

N 0 rs =0.3∙1142=342.6 kJ/kg.

From the preliminary calculation of the turbine we know:

1) approximate steam consumption G= 12.436 kg/s;

2) design pressure in front of the control stage nozzles p 0 =3.325 MPa;

3) enthalpy of steam in front of the nozzles of the control stage h 0 =3304 kJ/kg.

The method for calculating a two-row control stage is practically no different from the above method for calculating a single-stage turbine with a two-row impeller.

We build in h,s- diagram of water vapor is an isentropic process of expansion in this stage from the initial point A 0 (Fig. 7) to point a to t pc, setting aside the heat drop N 0 rs =

342.6 kJ/kg, and find the pressure behind the control stage R to rs =0.953 MPa.

Rice. 7. Determination of pressure behind the control stage and

available heat drop N 0(2- z )

We accept the degree of reactivity of the gratings

First working ρ р1 =0,

Guide ρ n =0.05,

Second working ρ р2 =0.

The heat difference processed in the nozzle grille is

N 011 =(1- ρ р1 -ρ n - ρ р2)∙ N 0 rs =0.95∙342.6=325.47 kJ/kg.

The pressure behind the first working grid, equal to the pressure behind the nozzles (since ρ р1 =0), is determined by h,s-diagram:

R 11 =p 21 =1.024 MPa.

The heat difference processed in the guide grid is

N 012 = ρ n ∙ N 0 rs =0.05∙432.6=17.13 kJ/kg.

The pressure behind the guide grid is equal to the pressure behind the stage (since ρ р2 =0):

R 12 =p 22 = p k p With=0.953 MPa .

Having previously specified the speed coefficient φ=0.965, we determine the loss in the nozzles:

N c =(1- φ 2) N 011 =(1-0.965 2)∙325.47 =22.384 kJ/kg.

Postponing the loss N from to h,s-diagram (see Fig. 2), we find on the isobar R 11 =p 12 point a 11, characterizing the state of the steam behind the nozzles. At this point we determine the specific volume of steam v 11 =0.24 m 3 /kg .

Isoentropic (conditional) velocity of steam outflow from the nozzle array

With from = .

Let's take the values u/c of equal to 0.2; 0.22; 0.24; 0.26; 0.28 and carry out variant calculations, the results of which are summarized in

table 2 (in all variants α 11 =12.5° is taken).

For the first option attitude u/c from = 0.2. Peripheral speed in this version

u=(u/c from)· c from = 0.2 827.8 = 165.554 m/s.

Average step diameter d=u/(π n)= 1.054 m.

Actual steam velocity at the exit from the nozzle array

778.57 m/s .

From the continuity equation for the exit section of the nozzle array

ε l 11 = Gv 11 / (π ·d·c 11 · sinα 11)=

12.436·0.24/(π·1.054·778.57·sin12.5°)= 0.00536 m .

Since ε l 11 <0,02 м, принимаем парциальный подвод пара к рабочим лопаткам и находим оптимальную степень парциальности

Output length of nozzle blades

l 11 = ε l 11 / ε opt =0.0243 m.

We take the width of the nozzle blades b 11 = 0.04 m .

The adjusted speed coefficient of the nozzle array is determined from Fig. 4 at b 11 /l 11 = 0.04/0.0243 = 1.646 and angle α 11 = 12.5°:

The adjusted nozzle array velocity coefficient φ does not differ from that adopted earlier, therefore the steam velocity at the exit from the nozzle array c 11 and energy loss in the nozzle array H c we do not specify.

The dimensions of the nozzle blades remain unchanged. To ensure smooth opening of the flow part in this calculation option, the dimensions of the working and guide blades are taken as follows:

l 21 = 0.0268 m, l 12 =0.0293 m, l 22 =0.0319 m ,

b 21 =0.025 m, b 12 = 0.03 m, b 22 = 0.030 m .

The main results of calculations of the turbine control stage for all five options are summarized in table. 2. Formulas for determining all numerical values ​​of quantities are given above, in the example of calculating a turbine with speed steps.

From the variant calculations (Table 2) it follows that the highest internal relative efficiency of the control stage η o i max =0.7597 at average diameter dрс =1.159 m (version with speed ratio u/s from =0.22). Enthalpy of steam behind the control stage in this embodiment

h k p With =h 0 - H i рс =3304 -260.267=3043.733 kJ/kg.

This enthalpy corresponds to the state of steam at point a to p With on the isobar R k p With=0.953 MPa h,s-diagrams (see Fig. 7) and takes into account all blade and additional losses of the control stage. From this point the process of steam expansion begins in the unregulated stages of the turbine.

table 2

Main results of calculating the turbine control stage

Steam consumption for industrial consumers

To determine the enthalpy of steam in a steam manifold, it is necessary to use the tables of the thermodynamic properties of water and steam given in. The necessary reference materials are given in Appendix B of this manual. According to Table B1, which shows the specific volumes and enthalpies of dry saturated steam and water on the saturation curve for a certain pressure, the following are given:

Saturation temperature - t ABOUT C(column 2);

Enthalpy of water on the saturation curve - , kJ/kg (column 5),

Enthalpy of steam on the saturation curve - , kJ/kg (column 6).

If it is necessary to determine the enthalpies of steam and water at a pressure whose value lies between the values ​​​​given in the table, then it is necessary to interpolate between two adjacent values ​​of the values ​​between which the required value is located.

The enthalpy of steam in the steam header is determined by the steam pressure in it () according to Table B.1. Appendixes B.

The enthalpy of condensate returned from production is determined by its temperature and condensate pressure according to Appendix A.

Amount of condensate returned from production

where is the return of condensate from production (specified).

Steam consumption to cover the heating and ventilation load

The temperature of the heating steam condensate at the outlet of the surface heater is assumed to be 10-15 o C higher than the temperature of the heated medium at the inlet to this heater. In the heater 8, the network water is heated, which enters it from the return pipeline of the heating network with a temperature of 70 o C. Thus, we take the temperature of the heating steam condensate at the outlet of the heater 8 equal to 85 o C.

Using this temperature and pressure of the condensate, using the table in Appendix A, we find the enthalpy of the condensate:

Steam consumption for hot water supply

Steam consumption for heating plant

Total steam consumption to cover production and housing and utility loads

The steam consumption for the boiler house's own needs is assumed to be in the range of 15-30% of the external load, i.e. steam consumption to cover production and housing and communal loads. Steam used for auxiliary needs is used in the thermal circuit of the boiler room to heat additional and makeup waters, as well as to deaerate them.

We take the steam consumption for our own needs to be 18%. Subsequently, this value is clarified as a result of calculating the thermal diagram of the boiler room.

Steam consumption for own needs:

Steam losses in the thermal circuit of the boiler house are 2-3% of external steam consumption, we assume 3%.

Amount of steam supplied through the steam header after the reduction-cooling unit:

When steam passes through narrowed sections, a throttling process occurs, accompanied by a decrease in pressure, temperature, and an increase in the volume and entropy of the steam. For the case of an adiabatic throttling process, the following condition is satisfied:

where: is the enthalpy of steam after throttling, is the enthalpy of steam before throttling.

Thus, the steam energy does not change during the throttling process. The temperature of saturated steam is equal to the saturation (boiling) temperature and is a direct function of pressure. Since steam pressure and saturation temperature decrease during throttling, some overheating of the steam occurs. In order for the steam after the reduction-cooling unit to remain saturated, feed water is supplied to it.

Water consumption at the ROU is determined by the ratio:

The enthalpy of steam at the boiler exit is determined by the pressure in the boiler drum according to Table B.1. Appendix B,

We determined the enthalpy of steam in the steam header earlier, .

We take the feedwater pressure to be 10% higher than the pressure in the boiler drum:

The enthalpy of feed water at a pressure of 1.5 MPa is determined from the table in Appendix A,.

Full boiler room performance.

The article provides a fragment of a table of saturated and superheated steam. Using this table, the corresponding values ​​of the parameters of its state are determined from the value of steam pressure.

Column 1: Vapor pressure (p)

The table shows the absolute value of steam pressure in bar. This fact must be kept in mind. When we talk about pressure, we usually talk about excess pressure, which is shown by a pressure gauge. However, process engineers use absolute pressure in their calculations. In practice, this difference often leads to misunderstandings and usually with unpleasant consequences.

With the introduction of the SI system, it was accepted that only absolute pressure should be used in calculations. All pressure measuring instruments of technological equipment (except barometers) mainly indicate excess pressure, we mean absolute pressure. Normal atmospheric conditions (at sea level) mean barometric pressure of 1 bar. Gauge pressure is usually indicated in barg.

Column 2: Saturated steam temperature (ts)

The table, along with pressure, shows the corresponding temperature of saturated steam. The temperature at the corresponding pressure determines the boiling point of water and thus the temperature of the saturated steam. The temperature values ​​in this column also determine the steam condensation temperature.

At a pressure of 8 bar, the temperature of the saturated steam is 170°C. Condensate formed from steam at a pressure of 5 bar has a corresponding temperature of 152 °C.

Column 3: Specific volume (v”)

The specific volume is indicated in m3/kg. With increasing vapor pressure, the specific volume decreases. At a pressure of 1 bar, the specific volume of steam is 1.694 m3/kg. Or in other words, 1 dm3 (1 liter or 1 kg) of water during evaporation increases in volume by 1694 times compared to its original liquid state. At a pressure of 10 bar, the specific volume is 0.194 m3/kg, which is 194 times greater than that of water. The specific volume values ​​are used in calculating the diameters of steam and condensate pipelines.

Column 4: Specific gravity (ρ=rho)

Specific gravity (also called density) is given in kJ/kg. It shows how many kilograms of steam are contained in 1 m3 of volume. As the pressure increases, the specific gravity increases. At a pressure of 6 bar, steam with a volume of 1m3 has a weight of 3.17 kg. At 10 bar - already 5.15 kg and at 25 bar - more than 12.5 kg.

Column 5: Enthalpy of saturation (h’)

The enthalpy of boiling water is given in kJ/kg. The values ​​in this column show how much thermal energy is needed to bring 1 kg of water to a boil at a certain pressure, or how much thermal energy is contained in the condensate that condensed from 1 kg of steam at the same pressure. At a pressure of 1 bar, the specific enthalpy of boiling water is 417.5 kJ/kg, at 10 bar – 762.6 kJ/kg, and at 40 bar – 1087 kJ/kg. With increasing steam pressure, the enthalpy of water increases, and its share in the total enthalpy of steam is constantly growing. This means that the higher the steam pressure, the more thermal energy remains in the condensate.

Column 6: Total enthalpy (h”)

Enthalpy is given in kJ/kg. This column of the table shows the steam enthalpy values. The table shows that the enthalpy increases up to a pressure of 31 bar and decreases with a further increase in pressure. At a pressure of 25 bar the enthalpy value is 2801 kJ/kg. For comparison, the enthalpy value at 75 bar is 2767 kJ/kg.

Column 7: Thermal energy of vaporization (condensation) (r)

The enthalpy of vaporization (condensation) is indicated in kJ/kg. This column shows the amount of thermal energy required to completely evaporate 1 kg of boiling water at the appropriate pressure. And vice versa - the amount of thermal energy that is released during the process of complete condensation of (saturated) steam at a certain pressure.

At a pressure of 1 bar r = 2258 kJ/kg, at 12 bar r = 1984 kJ/kg and at 80 bar r = only 1443 kJ/kg. As pressure increases, the amount of thermal energy of vaporization or condensation decreases.

Rule:

As steam pressure increases, the amount of thermal energy required to completely evaporate boiling water decreases. And in the process of condensation of saturated steam at the appropriate pressure, less thermal energy is released.