The industrialization of the agro-industrial complex on the basis of intersectoral relations and increasing its efficiency will make it possible to eliminate the existing agriculture imbalances, as well as eliminate large losses of products during their production, transportation, storage, processing and sale. Under the conditions of perestroika, it is necessary to improve the form and organization of production, improve its planning and management.

Introduction 3
1. Calculation of herd structure ……… 6
2. Development of a master plan for the livestock complex. 6
2.1 Justification of type industrial premises and determining the need for them. 8
2.2 Calculation of the annual feed requirement. nine
2.3 Calculate the capacity of feed storage facilities and determine the need for them. 12
2.4 Calculation of the manure storage. 15
2.5 Calculation of water consumption. 17
3. Substantiation and selection of mechanization and automation tools for performing the main processes of the farm. 17
3.1 Milking cows. 17
3.2 Manure removal. twenty
3.3 Stall equipment. 21
4. Designing a flow-technological feed distribution line. 22
4.1 Determine the performance of PTL 22
4.2 Compiling constructively technological scheme PTL. 23
4.3 We make calculation and selection of equipment for PTL. 24
4.4 Daily schedule of machines and equipment. 32
4.5 Graph of electricity consumption by hours per day. 33
5. Analysis of indicators of the technological map. 34
Conclusion. 36
Literature 37

The work contains 1 file

4.Operational and energy calculation.

The operational and energy calculation includes the determination of energy costs for the performance of such technological operations as water supply, steam and heat consumption, lighting, heating, air exchange, drive of the working bodies of equipment for milking, processing and storage of milk.

Tab. : Approximate rates of water consumption for technological needs

4.1 Daily consumption of cold water defined as

,

where q 1 , q 2 ,…,q n- the average daily rate of water consumption by a given consumer;

m 1 , m 2 ,…,m n- the number of consumers of this type.

.

4.1.1 Hourly water consumption for the technological needs of PTL

,

where α - coefficient of daily unevenness of water analysis ( α = 3…4).

4.1.2 For some technological operations, water is used in a heated state. Such water is obtained by mixing hot water heated to 90 ° C with cold tap water. The daily consumption of water heated to 90 ° C is determined by the formula:

where Q c1 , Q c2 ,…,Q cn- daily amount of mixed water, l;

t c1 , t c2 ,…,t cn is the temperature of the mixed water, °С;

t G- hot water temperature, (t Г = 90 °С);

t X– temperature of cold water, (t X = 8…12 °С).

4.2 Steam consumption for technological needs PTL is determined by the formula:

,

where R P , R r-t , R from , R about- steam consumption, respectively, for pasteurization, steaming of the thermos tank, sterilization of milk pipes and for heating.

4.2.1 Steam consumption for pasteurization product (milk, cream) for steam pasteurizers is determined by the formula:

,

where M– performance of the pasteurizer, ;

FROM m is the heat capacity of milk, ;

i and λ– heat content of steam and condensate, ;

η T– thermal efficiency of the pasteurizer;

t n and t P– initial product temperature and pasteurization temperature, °C.

4.2.2Steam consumption for steaming cooling tank is defined as

where k f- the amount of steam for steaming one tank-thermos

k f = 0,2 kg;

Z f- the number of thermos tanks.

.

4.2.3 Steam consumption for sterilization milk pipes and fittings is:

where k c- steam consumption for sterilization after processing each batch

milk, k c = 25 kg;

n c– the number of individual processing cycles per day.

.

2.4) Steam consumption per space heating is defined as

where k 0 - specific steam consumption for heating, k 0 = 0,5…0,75 kg/m 3 ;

V P- the volume of the room, V P = a∙b∙h = 66∙150∙6 =60000 m 3 .

.

Then

4.3 calculation of farm water supply

The total average daily water consumption on the farm Q average day (m 3 / day) is determined by the formula

,

where g i– average daily water consumption by one consumer;

n i - the number of consumers.

Maximum daily water consumption.

Q max day =Q cf day *ά day

where ά day is the coefficient of daily unevenness.

ά day =1.3

Q max day \u003d 180 * 1.3 \u003d 234 m 3 / day

Maximum hourly water consumption, l\h

where ά h \u003d coefficient of hourly unevenness (on farms with auto-drinking ά h \u003d 2 ... .2.5; without auto-drinking ά h \u003d 4

Second calculation of water, l\s

L\s

Daily consumption pumping station, should be equal to max daily water consumption on the farm, and the hourly flow rate of the pump station is determined by the formula:

M 3\h

where: t is the duration of the operation of the pump or station per day hours.

t=7h

By the value of Q, we choose the type and brand of the pump 3V-27.

Specifications

Innings

Pressure

Suction lift 6.0 m

Wheel speed 1450 min -1

Weight 65 kg

Power

Power consumption of the electric motor to drive the pump, W

Required power el. motor to drive the pump, W.

where: Q us = volumetric water flow m 3 \h

p-density of water, kg \ m 3 (p \u003d 1000 kg \ m 3)

K z \u003d power reserve factor taking into account possible overloads during pump operation (K z \u003d 1.1 ... .20)

g-acceleration of gravity, m / s 2

Pump efficiency, two vortex pumps:

=0,4…..0,6

Efficiency of transmission from engine to pump

1 with direct connection to the pump

4.4 calculation of daily output of manure

Determination of the daily output of manure in winter:

,

where g uh – average daily excretion of solid excrement;

g m- average daily urine output;

g P- the average daily bedding rate.

During the pasture period, the daily output of manure is less

Annual output of manure

where T st is the duration of the stall period (230 day);

T p - the duration of the pasture period (135 day).

4.4.1 Calculation of manure storage

where h is the manure laying height. We accept h = 2 m;

G day - daily output of manure on the farm from the total livestock, kg. Let us take the daily output of manure corresponding to the maximum amount, i.e. in winter;

D HR - the duration of storage of manure. We accept D XP = 180 days;

ρ - manure density, ρ = 900 kg / m 3;

φ is the filling factor of the manure storage. We accept φ = 0.8.

We accept manure storage V= 50 24 2.5 = 3000 m 3 .

1. Calculation of ventilation.

To maintain the microclimate parameters in the optimal mode or close to optimal, for this it is necessary to remove harmful gases from the room and renew the air, i.e. to carry out air exchange in accordance with the norms.

We determine the hourly air exchange by the content of carbon dioxide:

where: C is the amount of carbon dioxide emitted by one animal.

We accept C \u003d 130 dm 3 / h

M is the number of animals in the room

Permissible CO content in indoor air,

2.5 dm 3 / m 3

C 1 \u003d carbon dioxide content in the outside air, C \u003d 0.3 .... 0.4 dm 3 / m 3

We check the correctness of the calculation by the frequency of air exchange:

where V P is the internal volume of the room m 3:

Room size c= ,b= , h= ,

In livestock buildings n=3….5 h

When the air exchange rate is n, we select natural ventilation, when n=3….5 is forced ventilation without heating the supply air, and when n is forced ventilation with heating the supply air.

Choose………………………..

Literature

1. Braginets N.V., Palishkin D.A. Course and diploma design on the mechanization of animal husbandry. – M.: Agropomizdat, 1991.
2. All-Union norms process design cattle enterprises. ONTP 1-89 - M .: Gosagroprom USSR, 1989.
3. Murusidze D.N., Levin A.B. Technology for the production of livestock products.
4. Chugunov A.I., Pronichev N.P. etc. Guidelines for implementation term paper in the discipline "Technology and mechanization of animal husbandry". – M.: MGAU, 1998.
5. Pronichev N.P. Methodological instructions for calculation technological maps. – M.: MGAU, 1999.
6. Bogdanov V.D., Golovatov Yu.P. etc. Album of diagrams and drawings of an agricultural object. – M.: MGAU, 1996.

Air ventilation at ambient temperature can remove only volatile liquid residues with a boiling point not higher than 300 ° C. Steaming is used to clean equipment from liquid residues with a high boiling point. Unlike air ventilation, steaming is a more complex process. The devices are heated to a temperature at which heavy product residues begin to soften, melt and evaporate.

The steaming temperature is usually assumed to be 80...90°C. The steam flow required to maintain such a temperature in the gas space of the apparatus can be calculated based on the heat balance equation, which has the form:

Q 1 \u003d Q2 + Q 3 + Q4, (6.26)

where Q 1 - heat content of steam; Q2- the heat expended on the evaporation of a liquid at a temperature T;"Q 3 - heat loss through the walls, roof and bottom; Q 4 - heat used to pre-heat the remaining liquid, gas space and apparatus body to the steaming temperature.

If you do not take into account the preheating of liquid residues, gas space and apparatus body (Q 4 =0), and the steaming process is considered stationary, the heat balance equation will take the form:

Q 1 \u003d Q 2 + Qs. (6.27)

Expanding the values ​​of Q1...Q3, we get:

where α i and fi- heat transfer coefficients and corresponding surfaces i th elements of the apparatus design; T- average volume temperature; T in - outdoor air temperature; Go- the amount of evaporating product; r 0 - heat of evaporation of the product; G B- total consumption of water vapor; r c is the heat of vaporization.

From equation (6.28), given the flow rate and parameters of water vapor, it is possible to estimate the temperature in the steam-air space of the apparatus during its steaming:

. (6.29)

To solve the inverse problem (to find the flow rate and parameters of water vapor), the steaming temperature is set. The steaming of large-volume apparatuses without thermal insulation (for example, tanks with a capacity of more than 10,000 m 3) is extremely long and does not allow achieving the desired result.

It should be borne in mind that steaming, as well as ventilation, cannot remove solid and viscous combustible residues. In this case, the devices should be cleaned using safe methods of washing the devices with technical solutions. detergents or wash away residues with product circulating in the system.

When steam is used to purge flammable products from apparatus, precautions must be taken to avoid excessive pressurization of the apparatus (by removing weight plates from breather valves and covers from skylights and mounting hatches) and the accumulation of dangerous static electricity charges that can be generated in a fast a jet of water vapor, especially when it hits an obstacle. Therefore, in the initial period of steaming (until the combustible medium in the apparatus is phlegmatized), steam must be supplied slowly. If a fire occurs during the steaming process, it is dangerous to use water inside or outside the machine, as this will condense the steam; air from the atmosphere will penetrate into the apparatus, there will be a threat of formation of a combustible mixture inside the apparatus and an explosion.

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, for boiling raw materials in brewers or heating and mixing liquids by bubbling, for creating excess pressure in autoclaves, as well as for changing 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

from– specific heat capacity of the product at its average temperature, J/kg K;

t to, t n – initial and final temperature, °C;

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) is spent thermal energy, 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

The specific steam consumption is calculated using formula 76.

Task and initial data. Let's calculate how many kilograms of water are evaporated in each of the evaporation chambers per 100 kg of beets. Such a calculation is of great importance, since it allows one to determine the steam consumption for evaporation and, in addition, one can then calculate the amount of heat transferred in each vessel through the heating surface, and determine the size of the required heating surfaces and the dimensions of the vessels.
Let us calculate the five-fold residue as the simplest, although far from the best. It is used in the case when diffusion works with a large pumping of juice (USA), for example, 140% by weight of beets, and 100 kg of beets have to evaporate W = 120 kg of water. Let us accept for this case the following system for the use of residue vapors (Table 23).

So, E1 = 7.0; E2 = 9.5 and E3 = 21.0. A significant part of the steam consumption in the plant (17.0 kg) does not depend on evaporation: exhaust (return) steam is used to boil syrup in vacuum apparatuses.
Payment. Let's denote the amount of water evaporated in the fifth evaporation chamber per 100 kg of beets through x kg. As a basis for all calculations, we assume that 1 kg of heating steam evaporates 1 kg of water; this is close enough to reality for practical purposes.
Obviously, in order to evaporate x kg of water in the V building, x kg of steam must be directed there from the IV building, i.e. W4 = x kg of water is also evaporated in the IV building. In order to evaporate x kg of water in the IV building, it is necessary to direct x kg of juice heating steam from the III building. However, in the III building of the residue (see Fig. 135) not only these x kg of water are evaporated, which are sent in the form of steam to the IV building; Juice steam of the III building is also used as an extra steam, in the amount of E3 - 21.0 kg for heating some stations, a sugar factory. Consequently, in the III building,

W3 = (x + 21) kg.

W2 \u003d (x + 21 + 9.5) kg.

W1 = (x + 21 + 9.5 + 7.0) kg.

W1 + W2 + W3 + W4 + W5 = W

x + 21 + 9.5 + 7 + x + 21 + 9.5 + x + 21 + x + x = 120,

W5 = 6.2; W4 = 6.2; W3 \u003d 6.2 + 21 - 27.2;
W2 = 6.2 + 21 + 9.5 = 36.7;
W1 \u003d 6.2 + 21 + 9.5 + 7 \u003d 43.7 kg.

D = E1 + E2 + E3 + W5,

D \u003d 7 + 9.5 + 21 + 6.2 \u003d 43.7 kg,

E \u003d E1 + E2 + E3 \u003d 7 + 9.5 + 21 \u003d 37.5 kg,