Determination of the estimated steam flow. Determination of the estimated steam consumption Method for calculating steam consumption
According to the obtained value of η oe, a preliminary estimated flow pair
which will be specified later.
For turbines with one controlled steam extraction (by order), the preliminary steam flow rate 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 by - the value of controlled (industrial, heating) selection at pressure R by (by task); H t 0pvd - heat drop of an ideal turbine from the initial pressure R 0 to extraction pressure R by (Fig. 6).
When calculating the flow path of a turbine with controlled extraction:
1) all steps up to regulated selection are calculated for the total steam flow, found by formula (13);
2) stages after controlled extraction are calculated for the flow rate in a purely condensing mode, determined by expression (12).
The low pressure stages must ensure the passage of steam when the turbine is operating at rated electric power with the controlled extraction off (condensation mode).
The calculation of the thermal scheme, the determination of steam flow rates in the turbine compartments and the reduction of the energy balance is carried out for two modes of operation of the turbine:
a) with controlled extraction at rated electric power (cogeneration mode);
b) without controlled extraction (condensation mode) at rated electric power.
Adjustment of the lengths of the nozzle and working blades of the stages before controlled extraction is carried out according to the steam flow rates through the compartments obtained in the heating mode, and the remaining stages ― on steam flow rates through the compartments in the condensing mode.
EXAMPLE OF CALCULATION OF A MULTISTAGE STEAM TURBINE
K-12-35 with three regenerative extractions for heating feed water up 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.
Steam consumption determination
We calculate the turbine for economic power. Accept
N eq =0.9 N e \u003d 0.9 ∙ 12000 \u003d 10800 kW.
Pressure in front of the nozzles of the control stage in the 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 VP /100) 2 ,
having accepted with vp = 120 m/s, λ = 0.07, we get
Δ R\u003d 0.006 ∙ 0.07 ∙ (120/100) 2 \u003d 0.0006 MPa,
steam pressure behind the blades of the last stage
R to =p" to + Δ R= 0.006 +0.0006 = 0.0066 MPa.
Roughly depict the process in h,s- diagram
(see Fig. 1), putting points A "0, A 0, A" to t, A to t.
Let's find h 0 = 3304 kJ/kg; h′ to t= 2143 kJ/kg; h to t= 2162 kJ/kg;
H t 0id \u003d 3304-2143 \u003d 1161 kJ / kg; H t 0 \u003d 3304-2162 \u003d 1142 kJ / kg;
η dr \u003d 1142/1161 \u003d 0.984.
We accept η vvr = 1.0, η ′ o i= 0.8, according to reference data
η m =0.98; η g = 0.97.
Thus, we have
η oe \u003d η dr ∙η ′ o i∙η vvr ∙η m ∙η g = 0.984∙0.8∙1.0∙0.98∙0.97=0.748.
Estimated steam flow rate for the turbine
All turbine stages will be designed for this steam flow rate.
Preliminary process line in h,s-diagram is applied according to the accepted value η " o i in the following way:
H t i= 1142∙0.8=913.6 kJ/kg.
postponing H t i in h,s-diagram, we get point A k on the isobar R to (Fig. 6).
The task of drawing an indicative line of change in the state of steam in h,s-diagram is only a search for the specific volume of steam at the outlet of the last stage. We find the state of the steam at the exit of this stage by laying down along the isobar R to from A to output loss
H in z =c 2 2 z/2000.
In preliminary calculation H in z is found from the expression
H in z = ζ id ∙N t 0id ,
where ζ id a is the output loss coefficient of the last stage.
When calculating, evaluate ζ id a and find H in z and with 2z.
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Fig.6. Steam expansion process in the condensing room(s)
and cogeneration (b) turbines in h,s-diagram
The smaller z id a, the smaller, therefore, with 2 z- the output speed of the steam in the last stage, but the longer the blade length will be.
The value of ζ id a should be set on the basis of 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 backpressure turbines ζ id a<0,015.
Let's take ζ id a = 0.0177. Then
H in z = 0.0177∙1161 = 20.55 kJ/kg.
The state of steam at point a to z corresponds to the specific volume of steam v 2 z\u003d 20.07 m 3 / kg. Steam enthalpy behind the turbine h k =
2390.4 kJ/kg.
The first stage of the preliminary calculation ends with the determination of the approximate steam flow through the turbine and the approximate specific volume of steam at the outlet of the last stage.
The second stage consists in checking the possibility of constructive implementation of the last stage and tentative determination of the isentropic heat drop in it.
2. Preliminary calculation of the last step
For the preliminary calculation of the last stage, the following parameters are known:
H t 0id, H in z ,ζ id a, G,n.
In the further calculation, the index z discard.
Steam velocity at the outlet of the working grate of the last stage
To determine the diameter of the last step, it is necessary to set the ratio ν = d/l 2 , where d is the average diameter of the last step; l 2 - the 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 large condensing turbines, large values are typical for small capacity condensing turbines and turbines with back pressure. The blades of the last stages can be performed either with a constant or with a variable profile. The question of the transition from blades with a profile that is constant along the height to a swirling one should be decided on the basis of a comparison of the losses caused by the flow around the working blades with a change in the value of ν. For ν<8 лопатки приходится всегда выполнять закрученными. При ν >12, the use of spin does not give a tangible gain in efficiency.
Let be , for example, the ratio ν = 5.2. Then, assuming the axial exit of steam in the last stage, i.e. α 2 \u003d 90 ° (and, therefore, with 2a =c 2), we get:
Thus, the length of the rotor blades
l 2 =d/ν \u003d 1.428 / 5.2 \u003d 0.2746 m.
Circumferential speed at average step diameter
u=π ∙d∙n= 3.14∙1.428∙50 = 224.3 m/s.
Peripheral speed at the end of the blade
u in =u∙(d+l 2 )/d\u003d 224.3 ∙ (1.428 + 0.2746) / 1.428 \u003d 267.4 m / s .
Such speeds are quite acceptable.
When calculating turbines of small power, there is no need to check the strength of the rotor blades if u in does not exceed 300 m/s .
Root section diameter
d to \u003d d - l 2 \u003d 1.428 - 0.2746 \u003d \u003d 1.153 m .
The peripheral speed of the blades in the root section
u to = π ∙ d to ∙n=181.17 m/s.
The determination of the heat drop processed in the axial turbine stage is made for optimal operating conditions, which are expressed by the optimal ratio of speeds
(14)
where ρ – degree of step 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 average section of the step.
Since in any section along the height of the blade, the heat drop H 0 must be the same (the pressure in front of and behind the stage is constant in height), then it can be calculated by expression (15) for the root section of the last stage, where ρc ≈0 (all stages of chamber turbines are designed with a degree of reactivity in the root section ρc ≈0), u=u to, assuming approximately φ \u003d 0.95 and α 1 \u003d 15 about:
For a given heat drop H 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 of the step ρ k \u003d 0, φ \u003d 0.955, α 1 \u003d 15 o, we obtain the optimal diameter of the root section at H 0 =78 kJ/kg:
3. Control stage calculation
We select the control stage in the form of a two-crown Curtis disk. Let us take the heat drop in it equal to 30% of the total heat drop H t 0 , which will be
H 0 rs \u003d 0.3 1142 \u003d 342.6 kJ / kg.
From the preliminary calculation of the turbine, the following are known:
1) Estimated steam consumption G= 12.436 kg/s;
2) design pressure in front of the nozzles of the control stage p 0 = 3.325 MPa;
3) steam enthalpy in front of the control stage nozzles h 0 =3304 kJ/kg.
The method for calculating a two-row control stage practically does not differ from the above method for calculating a single-stage turbine with a two-row impeller.
We build in h,s-diagram of water vapor, the isentropic expansion process in this stage from the initial point A 0 (Fig. 7) to the point a to t pc, postponing the heat drop H 0 rs =
342.6 kJ / kg, and we find the pressure behind the control stage R k rs = 0.953 MPa.
Rice. 7. Determining the pressure behind the control stage and
available heat drop H 0(2-z )
We accept the degree of reactivity of the lattices
First working ρ p1 =0,
Guide ρ n \u003d 0.05,
The second working ρ p2 =0.
The heat drop processed in the nozzle grid,
H 011 \u003d (1- ρ p1 - ρ n - ρ p2) ∙ H 0 rs \u003d 0.95 342.6 \u003d 325.47 kJ / kg.
The pressure behind the first working grate, equal to the pressure behind the nozzles (because ρ p1 = 0), is determined by h,s-chart:
R 11 =p 21 = 1.024 MPa.
The heat drop processed in the guide grid,
H 012 = ρ n ∙ H 0 rs \u003d 0.05 432.6 \u003d 17.13 kJ / kg.
The pressure behind the guide grid is equal to the pressure behind the stage (because ρ p2 = 0):
R 12 =p 22 = p to r with=0.953 MPa .
Having previously set the velocity coefficient φ=0.965, we determine the loss in the nozzles:
H c \u003d (1- φ 2) H 011 \u003d (1-0.965 2) ∙ 325.47 \u003d 22.384 kJ / kg.
postponing the loss H 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 \u003d 0.24 m 3 / kg .
Isentropic (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
tab. 2 (in all cases, α 11 =12.5° is accepted).
For the first option attitude u/c out = 0.2. Peripheral speed in this variant
u=(u/c from)· c out \u003d 0.2 827.8 \u003d 165.554 m / s.
Average step diameter d=u/(π n)= 1.054 m
Actual steam velocity at the outlet of the nozzle array
778.57 m/s .
From the continuity equation for the outlet 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
Outlet length of nozzle vanes
l 11 = ε l 11 / ε opt \u003d 0.0243 m.
We accept the width of the nozzle blades b 11 = 0.04 m .
The refined velocity coefficient of the nozzle array is determined from fig. 4 at b 11 /l 11 \u003d 0.04 / 0.0243 \u003d 1.646 and the value of the angle α 11 \u003d 12.5 °:
The refined velocity coefficient of the nozzle array φ does not differ from that adopted earlier, therefore, the steam velocity at the outlet of the nozzle array c 11 and energy loss in the nozzle array H c is not specified.
The dimensions of the nozzle vanes remain unchanged. The dimensions of the working and guide vanes are taken to ensure the smooth opening of the flow path in this calculation option as follows:
l 21 = 0.0268 m, l 12 \u003d 0.0293 m, l 22 = 0.0319 m ,
b 21 \u003d 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. The formulas for determining all numerical values of the quantities are given above, in the example of calculating a turbine with speed steps.
From variant calculations (Table 2) it follows that the highest internal relative efficiency of the control stage η o i max =0.7597 with average diameter d pc =1.159 m (variant with velocity ratio u/c of =0.22). Steam enthalpy downstream of the control stage in this variant
h to r with =h 0 - H i pc \u003d 3304 -260.267 \u003d 3043.733 kJ / kg.
This enthalpy corresponds to the state of steam at the point a to p with on the isobar R to r with=0.953 MPa h,s-diagrams (see Fig. 7) and takes into account all the blade and additional losses of the control stage. From this point, the process of steam expansion in the unregulated turbine stages begins.
table 2
The main results of the calculation of the control stage of the turbine
| No. pp | Physical quantity and designation of its unit | Designation | Speed ratio u/with from | ||||
| 0,20 | 0,22 | 0,24 | 0,26 | 0,28 | |||
| Peripheral speed, m/s | u | 165,55 | 182,11 | 198,66 | 215,22 | 231,78 | |
| Average step diameter, m | d | 1,054 | 1,159 | 1,265 | 1,37 | 1,476 | |
| Angle of exit of the steam flow from the nozzle array, deg. | a11 | 12,5 | |||||
| Product ε l 11 , m | ε· l 11 | 0,00536 | 0,00487 | 0,00443 | 0,00414 | 0,00384 | |
| Degree of partiality | ε o pt | 0,2205 | 0,2094 | 0,2006 | 0,1929 | 0,1859 | |
| Nozzle blade length, m | l 11 | 0,0243 | 0,0233 | 0,0223 | 0,0214 | 0,0207 | |
| Width of nozzle blades, m | b 11 | 0,04 | 0,04 | 0,04 | 0,04 | 0,04 | |
| Nozzle velocity coefficient | φ | 0,965 | 0,965 | 0,964 | 0,963 | 0,963 | |
| Dimensions of blades of working and guide gratings, m | l 21 l 12 l 22 b 21 b 12 b 22 | 0,0268 0,0293 0,0319 0,025 0,03 0,03 | 0,0257 0,0282 0,0308 0,025 0,03 0,03 | 0,0247 0,0272 0,0298 0,025 0,03 0,03 | 0,0239 0,0263 0,0289 0,025 0,03 0,03 | 0,0231 0,0255 0,0280 0,025 0,03 0,03 | |
| Abs. steam velocity at the nozzle outlet, m/s | with 11 | 778,57 | 778,57 | 777,76 | 776,96 | 776,96 | |
| Energy loss in the nozzle array, kJ/kg | H with | 22,384 | 22,384 | 23,012 | 23,639 | 23,639 | |
| Rel. steam velocity at the entrance to the first working grate, m/s | w 11 | 617,98 | 602,07 | 585,39 | 568,75 | 552,96 | |
| Angle of flow entry into the first working grate, deg. | β11 | 15,82 | 16,25 | 16,71 | 17,20 | 17,71 | |
| Speed coefficient of the first working grate | Ψ p 1 | 0,947 | 0,946 | 0,946 | 0,945 | 0,945 | |
| Energy loss in the first working grate, kJ/kg | H l1 | 19,786 | 18,939 | 18,043 | 17,156 | 16,331 | |
| Rel. steam velocity at the outlet of the first working grate, m/s | w 21 | 585,09 | 569,75 | 553,71 | 537,74 | 522,59 | |
| Specific volume of steam behind the first working grate, m 3 /kg | v 21 | 0,2449 | 0,2448 | 0,2447 | 0,2446 | 0,2445 | |
| Steam flow exit angle from the first working grate, deg. | β21 | 15,44 | 15,80 | 16,18 | 16,59 | 17,01 | |
| Abs. steam velocity at the outlet of the first working grate, m/s | with 21 | 427,79 | 397,62 | 367,11 | 337,12 | 308,50 | |
| The angle of exit of the steam flow from the first working grate in absolute motion, deg. | a 21 | 21,28 | 22,96 | 24,85 | 27,09 | 29,71 | |
| Guide Grating Speed Coefficient | φ n | 0,946 | 0,945 | 0,945 | 0,944 | 0,944 | |
| Steam speed at the outlet of the guide grate, m/s | with 12 | 440,84 | 414,61 | 388,47 | 363,23 | 339,65 | |
| Energy loss in the guide grid, kJ/kg | H n | 11,459 | 10,231 | 9,060 | 7,985 | 7,036 |
Steam consumption for industrial consumers
To determine the enthalpy of steam in a steam collector, it is necessary to use the tables of thermodynamic properties of water and steam given in. Required reference materials are provided 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 O C(column 2);
The enthalpy of water on the saturation curve - , kJ / kg (column 5),
Steam enthalpy 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 is between the values given in the table, then it is necessary to interpolate between two adjacent values \u200b\u200bof the quantities between which the desired value is located.
The enthalpy of steam in a steam collector is determined by the steam pressure in it () according to Table B.1. Applications 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 (given).
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.
According to this temperature and pressure of the condensate, according to 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 industrial and housing and communal loads
Steam consumption for auxiliary needs of the boiler house is taken in the range of 15-30% of the external load, i.e. steam consumption to cover industrial and housing and communal loads. Steam supplied for own needs is used in the thermal scheme of the boiler house for heating additional and make-up water, as well as for their deaeration.
We accept steam consumption for own needs equal to 18%. Subsequently, this value is specified as a result of the calculation of the thermal scheme of the boiler house.
Steam consumption for own needs:
Steam losses in the thermal scheme of the boiler house are 2-3% of the external steam consumption, we accept 3%.
The amount of steam supplied through the steam manifold 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: - steam enthalpy after throttling, - steam enthalpy 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 are reduced during throttling, some superheating of the steam occurs. In order for the steam to remain saturated after the reduction-cooling plant, feed water is supplied to it.
Water consumption at the ROU is determined by the ratio:
Steam enthalpy at the outlet of the boiler is determined by the pressure in the boiler drum according to Table B.1. Applications B,
The enthalpy of steam in the steam collector was determined by us earlier, .
The feed water pressure is assumed to be 10% higher than the pressure in the boiler drum:
The enthalpy of feed water at and a pressure of 1.5 MPa is determined from the table in Appendix A,.
Full performance of the boiler room.
At enterprises, water vapor is used for technological, household and power purposes.
For technological purposes, deaf and live steam is used as a coolant. Live steam is used, for example, to boil raw materials in boilies or to heat and mix liquids by bubbling, to create excess pressure in autoclaves, and also to change the state of aggregation of a substance (evaporation or evaporation of a liquid, drying materials, etc.). Deaf steam is used in surface heat exchangers with steam heating. The pressure of the steam used in meat processing enterprises ranges from 0.15 to 1.2 MPa (1.5 ÷ 12 kg / cm 2).
For each technological operation using water vapor, its consumption is determined according to the heat balance of each thermal process. In this case, the data of material balances of product calculations are used. For periodic processes, the heat treatment time for each cycle is taken into account.
In each particular case, the heat load of the apparatus (heat input) can be determined from the heat balance of the process. For example, the heat spent on heating the product from the initial ( t m) to the final ( t j) temperatures for a continuous apparatus are determined by formula 72:
Q = Gc (t k – t n)φ, (72)
where Q- heat spent on heating, J / s (W), i.e. thermal load of the device;
G
with– specific heat capacity of the product at its average temperature, J/kg K;
t to, t n – initial and final temperature, °С;
φ
- coefficient taking into account heat loss to the environment
Wednesday ( φ
= 1.03÷1.05).
The heat capacity of the product is chosen either from known directories, or calculated according to the principle of additivity for multicomponent systems.
To change the state of aggregation of a substance (solidification, melting, evaporation, condensation), thermal energy is consumed, the amount of which is determined by formula 73:
where Q is the amount of heat, J/s (W);
G is the mass flow rate of the product, kg/s;
r is the heat of phase transition, J/kg.
Meaning r determined according to reference data, depending on the type of product and the type of phase transition of the substance. For example, the heat of fusion of ice is taken to be r 0 \u003d 335.2 10 3 J / kg, fat
r w = 134 10 3 J / kg. The heat of vaporization depends on the pressure in the working volume of the apparatus: r = f (P a). At atmospheric pressure r= 2259 10 3 J/kg.
For continuous devices, heat consumption is calculated per unit of time (J / s (W) - heat flow), and for batch devices - for a cycle of operation (J). To determine the heat consumption per shift (day), it is necessary to multiply the heat flow by the operating time of the apparatus per shift, a day, or by the number of cycles of operation of a batch apparatus and the number of such apparatuses.
The flow rate of saturated water vapor as a heat carrier under the condition of its complete condensation is determined by the equation:
where D- the amount of heating water vapor, kg (or flow rate, kg / s);
Q total - total heat consumption or heat load of a thermal device (kJ, kJ / s), determined from the equation of the heat balance of the device;
– enthalpy of dry saturated steam and condensate, J/kg;
r is the latent heat of vaporization, kJ/kg.
The consumption of live steam for mixing liquid products (bubbling) is taken at the rate of 0.25 kg / min per 1 m 2 of the cross section of the apparatus.
Steam consumption for economic and domestic needs under this heading, steam is used to heat water for showers, laundry, washing floors and equipment, and scalding equipment.
Steam consumption for scalding equipment and inventory is determined by its outflow from the pipe according to the flow equation:
(75)
where D w – steam consumption for scalding, kg/shift;
d– inner diameter of the hose (0.02÷0.03 m);
ω – speed of steam outflow from the pipe (25÷30 m/s);
ρ - vapor density, kg / m 3 (according to Vukalovich's tables ρ = f(ρ ));
τ – scalding time, h (0.3÷0.5 h).
If we take in the equation τ = 1 h, then the steam consumption is determined in kg/h.
The calculation of steam consumption for all items is summarized in table 8.3.
Table 8.3 - Steam consumption, kg
| Expenditure | At one o'clock | On shift | Per day | In year |
| Total |
The specific steam consumption is calculated using formula 76.
The article contains a fragment of the table of saturated and superheated steam. With the help of this table, according to the value of steam pressure, the corresponding values of the parameters of its state are determined.
|
Steam pressure |
Saturation temperature |
Specific volume |
Density |
Steam enthalpy |
Heat of vaporization (condensation) |
|
Column 1: Steam pressure (p)
The table shows the absolute value of the steam pressure in bar. This fact must be kept in mind. When it comes to pressure, as a rule, they talk about the excess pressure that the pressure gauge shows. However, process engineers use absolute pressure in their calculations. In practice, this difference often leads to misunderstandings and usually backfires.
With the introduction of the SI system, it was accepted that only absolute pressure should be used in calculations. All process equipment pressure gauges (except barometers) basically show gauge pressure, we mean absolute pressure. Normal atmospheric conditions (at sea level) mean a barometric pressure of 1 bar. Gauge pressure is usually indicated in barg.
Column 2: Saturated steam temperature (ts)
In the table, along with the pressure, the corresponding saturated steam temperature is given. The temperature at the appropriate pressure determines the boiling point of water and thus the temperature of saturated steam. The temperature values in this column also determine the condensing temperature of the steam.
At a pressure of 8 bar, the saturated steam temperature is 170°C. The 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 given in m3/kg. As the vapor pressure increases, the specific volume decreases. At a pressure of 1 bar, the specific steam volume 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 the initial liquid state. At a pressure of 10 bar, the specific volume is 0.194 m3/kg, which is 194 times that of water. The specific volume value is used in calculating the diameters of steam and condensate pipelines.
Column 4: Specific Gravity (ρ=po)
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 1 m3 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 heat energy is needed to bring 1 kg of water at a certain pressure to a boiling state, or how much heat energy is contained in the condensate, which condenses 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 it is 762.6 kJ/kg, and at 40 bar it is 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 vapor 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 given in kJ/kg. This column gives 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 in the process of complete condensation of (saturated) steam at a certain pressure.
At 1 bar r = 2258 kJ/kg, at 12 bar r = 1984 kJ/kg and at 80 bar r = only 1443 kJ/kg. With increasing pressure, the amount of thermal energy of vaporization or condensation decreases.
Rule:
With increasing steam pressure, the amount of thermal energy required to completely evaporate boiling water decreases. And in the process of condensing saturated steam at the appropriate pressure, less thermal energy is released.
3.2.2 Calculation of steam consumption for heating and ventilation
The calculation of heat costs for heating and ventilation is determined by the formula:
Q=q · V · (t pom – t calc ) · T year , kW/year, (3.11)
where q is the specific heat consumption for heating and ventilation of 1m 3 of the 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, excluding the volume of drying chambers, m 3;
t pom is the temperature in the room, assumed to be 20°С;
t calc - design temperature for heating and ventilation;
T year - the duration of the heating season is determined by the formula:
T year \u003d 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
|
Name of steam consumers |
Specific consumption q, kW / (m 3 .deg). |
Room volume |
Temperature difference inside and outside the building (t room - t calculated), ° С |
Length of the heating season |
Annual heat consumption Q, |
|
Heating of the drying area |
453,9 · 10 3 |
||||
|
Ventilation |
63,15 · 10 3 |
||||
|
517,05 · 10 3 |
|||||
The calculation of the annual demand for steam for heating and ventilation is determined by the formula:
3.2.3 Calculation of heat (steam) consumption for domestic needs
The calculation of heat (steam) consumption for domestic needs is determined by the formula:
where q - steam consumption per 1 person per shift;
m is the number of people working on the busiest shift;
n is the number of shifts in the work of the section (it is advisable to take 2);
τ is the 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
The calculation of the total annual steam demand for technological and domestic needs, heating and ventilation is determined by the formula:
D common = D academic year + D from + D life , t/year. (3.14)
D common \u003d 8.13 + 891.47 + 2.6 \u003d 902.2 tons / year.