Software development for registration of queries for ATM

The total amount of heat generated by people power (formula 21):

Qn=n*Q', (21)

where n - the number of people,

Q '- the heat generated by one person, kJ / h in Table 2.3

Heat dissipation per person is 125 Kcal / h, given that 1 cal = 4.2 J, then Q '= 525 kJ / h Maximum number of people in the audience (number of employees) 7 people.

Q_n=7*525=3 675 (Кдж/ч).

Heat generation from solar radiation: since space is on the fifth floor, the heat will flow through glass openings. Heat dissipation from the window openings calculated by the formula 22:

Qпр=q0*F0*A0, (22)

where q0 - amount of radiation;

Ao = 1.15, as a double glazing;

Fo-glazing surface;

For t = +35 ° C, side, double glazed windows in plastic frames, we obtain:

q₀=125 кКал/м²*4,2=525 (кДж/м²);

Q_пр=525*(2*2,2*1,8) *1,15=4 781,7 (кДж/ч).

Heat dissipation from artificial light (Eq. 23):

Qo = * 3612 * Noy, (23)

where = 1 - coefficient taking into account the conversion of electric energy into heat;

Noy = 1,2 - the system power consumption of artificial lighting kW;

3612 - thermoelectric coefficient, kJ / kWh;

Qo = 1 * 3612 * 1.2 = 4334.4 (kJ / hr).

Heat dissipation from the computer power (Eq. 24)

Qdv = () * 3612 * 1 * 2 * 3 * 4,(24)

where 1 is the average coefficient of performance (COP) engines;

2 - utilization;

3 - factor simultaneously used an electric motor;

4 - coefficient characterizing the conversion of mechanical energy into heat.

For electric equipment without special cooling 1 * 2 * 3 * 4 = 0,26, thus:

Q_дв =(7*26) *3612*0,26= 170 919,84 (кДж/ч).

The total amount of heat coming from sources of Qmax = Qn + Qnp + Qo + EFA = 3675 4 170 781.7 4334.4 919.84 = 183,710.94 (kJ / h).

Then calculated in water generation area. For our data are the main source of water generation workers. According to the normative data generated amount of moisture for one person would be 115 g / h

Then the total number of employees released moisture equals:

G_вл = 115 * 7 = 0,805 (кг/ч).

Calculation of air produced by Id chart. To do this, we first define teplovlazhnoe ratio using the following formula 25 :

E = Q / Gвл , (25)

where Q - full excess heat in the room;

Gvl - amount of moisture released.

we obtain:

E = 183 710,94 / 0,805 = 228 212,35(кДж/кг).

From the point P, which expresses the initial state of the supply air (200C, 40%), transferred to the room to draw a line chart Id I = const (d = 6,5 g / kg). Assigning an arbitrary value of d (in our case, 0.1 g / kg), we determine the heat content (enthalpy) Supply air Ip = 35 kJ / kg, and the air in the work area - Ip.

In our case, we have the following formula 26:

(26)

where E - teplovlazhnoe attitude, кДж/кг;

d - arbitrary value, г/кг.

(ккал/кг) ~ 45,44 (кДж/кг).

Find the schedule Ip = 91.28 (kcal / kg) = 182 (kJ / kg).

The required amount of air for ventilation in the room simultaneously with the release of heat and moisture determined by the formula 27:

G = m*Q / (Iр - Iп), (27)

where m - coefficient taking into account the share of heat entering the work area; in the absence of experimental measurements take m = 1;

Q - the amount of excess heat to be removed;

Ip and Ip - enthalpy of supply air and in the working area defined by Id chart.

For our data, we obtain:

G = 1 * 228 212,35/ 91,28 - 17 = 2 483,14 кг/ч / 1,88 = 1320,82 (м³/ч).

On the basis of calculations assume that for air conditioning facilities can use SDA 0351 and 0501 SUA with the following characteristics : air capacity 3020 m3 / h and 4720 m3 / h , respectively. Characteristics of the selected air conditioner is more than the calculated values ??that satisfies the requirements of SNIP RK 2.02.04-2002 .

Based on the available heat and humidity characteristics ( temperature and humidity ) on the nomogram state of moist air (h, x - on the nomogram ) , shown in Figure 21 should find the moisture content (x) for the initial air and defined circumstances.



Figure 21. Nomogram state of moist air

The amount of moisture required for room humidification under specified conditions, is calculated using the following formula 28:

(28)

where V - volume of humidified air , m3; For areas where recycling is available only air , V = volume of the room (m3) . For rooms with outdoor air supply V = external supply air flow ( m3 );

1.2 - the proportion of air in kg/m3 ( at 21 ° C and atmospheric pressure of 1013 mbar) ;

X1 - moisture content ( absolute humidity ) air to be humidified , g / kg;

X2 - moisture (absolute humidity ) of the humidified air , g / kg;

Y - minor correction value , taking into account other factors that affect the performance of the required moisture, but not included in the formula used to increase the accuracy of calculations. Assumed to be 0 .

In our case, V = flow outdoor supply air ( m3 / h) , then :

V = G = 1 * 228 212,35/ 91,28 - 17 = 2 483,14 kg / h / 1.88 = 1320.82 (m3 / h).

Determine the nomogram that at 20 ° C and humidity of 40%:

X1 = 6.5 g / kg.

Determine the nomogram that at 25 ° C and 60% RH:

X2 = 12.5 g / kg.

As a result, we obtain the following:

Q = [1320,82 * 1,2 * (12.5 - 6.5) / 1000] = 9.51 (kg / h).

Conclusion: the results of calculations show that in order to maintain the level of humidity in the data center to the standards, use the following humidifier: series humiSteam firm CAREL, the performance of which is from 1.5 to 126 kg / h

Industrial lighting

Light plays an important role in maintaining health, ensuring high availability. Properly organized workstation lighting conditions can improve vision, eliminate fatigue the body, enhances productivity, reduces occupational injuries [7]. Norms of luminosity are shown in Table 2.

Table 2 Indoor lighting norm EC (for personnel engaged in computer operation) (SNIP 23-05-95)

Premises	Light plane rationing its height from the floor, m	Norma light	K n, - no more than%
		When combined light, Luke	When general lighting
Norma personnel operating	Г-0,8	750	400	15

Calculation of natural lighting is to determine the area of skylights. Under side illumination determine the area of light openings (windows), So, providing the normalized values of KEO, the formula 29:

So=Sn*en*-n0*k3fl*k3/(100*то*r1), (29)

where Sn - computer room floor area , m;

en - normalized value KEO ;

n0 - light characteristic windows;

CPAs - coefficient taking into account shading windows opposing buildings;

k3 - the safety factor ;

T0 - the total light transmittance ;

r1 - coefficient reflecting the increase KEO under side illumination by the light reflected from the surface facilities and the underlying layer adjacent to the building.

The hall has dimensions: length L = 5 m, width B = 5 m, height H = 3 m Working surface height above the floor - 1 m , windows begin with a height of 1 m, height windows - 1.8 m The company is in Almaty ie IV in the light zone.

In the hall there are 7 PC , the length of each of the tables 1.5-2 m in length passes from 2 to 4 pm , tables are located at a distance of 1 m from the outer wall where the window openings . Thus, the minimum illumination will be at a point 5 meters from the outer wall .

Determine the values ??of all components:

Sn = B * L = 5 * 5 = 25 m2 ;

en = en (III Time ) * m * c = 1,5 * 0,9 * 0,8 = 1,08;

зo = 11 , the value from the table.

As a transparent material using hollow glass double opening blocks form - bearing coatings concrete form. Accepts the following values:

ф1 = 0,5; ф2 = 0,6; ф3 = 0,8;

ф0 = ф1 * ф2 * ф3 = 0,5 * 0,6 * 0,8 = 0,24;

Accept k3 = 1,1;

Standing next to the building is not, hence kzd = 1; r1 = 1 ;

So = 1.08 * 25 * 11 * 1.1 * 1 / ( 0.24 * 100 * 1) = 13.6 m2.

Since unilateral lateral illumination, the area of skylights on one side will 13,6:1 = 13.6 m2.

Noise and vibration

Noise and vibration also require protective measures to reduce the impact on human health. In industrial premises where work on VDT and PC is an auxiliary noise levels in the workplace must not exceed permitted for this type of work regulated by GOST 12.1.003 - 03SSBT "Noise . General requirements for safety . " To reduce the noise level can be used for decoration sound absorbing materials with a maximum sound absorption in the frequency range 63-8000 Hz , confirming the effectiveness of the measures acoustic calculations. Additional sound absorber are plain curtains and thick fabric, blend in with the color of the walls and hanging from the crease at a distance of 15-20 cm from the fence . The width of the curtain should be 2 times greater than the width of the window. A vibration absent ( GOST 12.1.012 - 03SSBT ) .

Harmful radiation

Exposure to electromagnetic fields (the main source - the monitor is working in violation of the functional state of the nervous system, heart and circulatory system. Permissible norms intensity electromagnetic fields during the workday in Table 3.

Currently it is believed that short-term and long-term impact of all types of radiation monitors, especially in the presence of shielding, do not pose a health risk operator.

Recommend the use of monitors that meet the safety standard MPR P.

Maximum tension in the casing of the monitor Samsung SyncMaster 510s, which complies with MPR II, is on passport data of 3.6 V / m, which corresponds to the background level.

Table 3 Permissible norms electro-magnetic field during the workday

Radiation frequency, Hz	The electric magnetic field V / m, not more than
According to the electric component	According to the magnetic component	According to the electric component	According to the magnetic component
60*103	60*103	50	5
30-30*106	60-50*103	20	0,3
30-50*106	-	10	-
50-300*106	-	5	-

The intensity of electromagnetic radiation at 5 cm from the screen is 64 V / m, but at a distance of 30 cm, does not exceed 2.4 V / m, which is lower than the permissible level. The same can be said about the intensity of ultraviolet and infrared radiation. Thus, when operating at the insistence of 40-50 cm from the harmful effects of the display screen is possible.

Safety (electrical)

There are different requirements and protective measures in the premises for the operation of the VDT and PC . All premises intended for use VDT and PC , PC and ES , must be equipped with a separate ground loop . Prohibited from posting jobs with VDT , PC , PC in the basement . You may not use the basement room to accommodate VDT , PC and PC pre-school , secondary and tertiary educational institutions. Premises with VDT and PC must have natural and artificial lighting . Premises with VDT and PC must be equipped with heating and air conditioning. If there is no air conditioning in the rooms should be provided natural ventilation [8].

Individual requirements are defined in covering premises and workplaces . Natural lighting must be made via skylights and adjusted so that the light levels in the workplace comply with the requirements . In the workplace, in the VDT and PC illumination normalized in the vertical plane ( the plane of the screen) and in the horizontal plane ( the plane of the section in the area of ??working with documents ) . Normalization is carried out in absolute units ( suites ) , regardless of whether natural or artificial light in a room. On the horizontal surface of a table in the area of ??document management combined illumination shall be not less than 500 lux. In the absence - of at least 400 lux. The screen illumination shall be 200 lux . Local lighting should not create glare on the increase and its more than 300 lux illumination .

Calculation of PE. Electronic equipment used in the hall, refers to closed installations or locations. Under the terms of electrical safety as it relates to electric appliances up to 1 kV operating voltage of 380/220V AC power frequency of 50 Hz with an installed capacity of up to 5MW . According to climatic design electrical shop belongs to the third category. The electrical network is made of three-phase four-wire with a solidly grounded neutral . By the degree of reliability of power supply , the network belongs to the third category.

Used for grounding angle steel section 50X50X5 mm and a height of 2.5- 3m. Hammer it into the ground in a pre- dug trench depth 0.6- 0.7m . from the planning point. These grounders interrelate steel strip thickness of 4-5 mm .

Of natural earthing used in the workshop building area of ??reinforced concrete foundation sf = 360 sq.m.

Soil at the site where the earthing should be built , clayey , the resistivity of which is given in Table 4 .

Table 4 Resistivity

Type of soil	Resistivity (OmHm)
Clay garden soil	40
humus	50
Loam, stony clay	100
Crushed sand, stony soil	200
Sandy loam	300
Sand with pebbles	800

Soil resistivity measured in sunny dry weather with ambient temperature of 20 0C.

Calculation of ground :

Determine the desired resistance of the grounding device. According to the EMP resistance grounding device with line voltages of 380V three-phase power at any time of the year should be less than 40m. Therefore zadamsya value RL = 4 0m.

In accordance with the climatic characteristics determine the design resistance of the soil (Eq. 30) :

PB = Pu * in = 40 * 1.5 = 60 (ohm m) (30)

where PB - estimated soil resistivity for vertical electrodes ohm m

Rea - measured value of soil resistivity

Шv - seasonality factor

where Rf - current spreading resistance of concrete foundation of the building;

о - coefficient taking into account the presence of the concrete layer and frost ( 1.5-1.8 );

p - resistivity of the soil ;

Sf - square concrete foundation .

Since the resulting current spreading natural resistance grounding satisfies : 1.16 < 4 ohms, then the calculation of artificial earthing optional ( SAE 96, PTEiPTB 2004 ) .

Fire and Explosion

Flammability class D project is estimated ( flammable ) .

At the approach to fire safety focuses on the presence of a conductive wiring in the room. This requirement is compliance with fire safety due to the fact that according to the statistics 85% of all fires occur precisely because of the poor performance of the wiring.

Fire prevention facilities ( SNIP 21-01-97 )

- Availability manual powder fire extinguisher brand 0L -2M

intended to extinguish the ignition at a distance of 2 m, at a temperature

40-50 ° C. Fire extinguisher is kept protected from sunlight and

heaters place an easily accessible in the event of

ignition;

- Used for heating water central

heating ;

- Installed electric fire alarm system, two

thermal emitter firefighters respond to an increase in temperature

environment to a value of 80 ° C or higher within 3 meters for premises datacenters recommended thermal fire emitters type LBD, ATP -ZM , and others;

- Room is situated in the building so that there are at least two means of escape , one of which can not be blocked by fire fire escape .

Conclusion

In this thesis project was required to develop a system of accounting ATM network. It was necessary to automate the work of operators and engineers. The main objective, which was to form outfits for engineers operators, reporting through ATMs, database management containing information across the ATM network.

In the design phase established the basic functional requirements to system requirements for users. And then on the basis of these requirements were built case diagram, sequence diagram, classes that reflect the model of the system and give a clear understanding of the subsequent development.

Conducted database design. For this was built a conceptual data model. And then by using Visual Paradigme suite developed logical and physical models.

The result is a software product that allows you to updated database of ATMs, recharged and edited by using the user interface.

Annex 1

The text of the module

procedure TForm1.FormEdit(Sender: TObject);

begin

DBGrid1.Width:=Form1.Width-10;

end;

procedure TForm1.FormCreate(Sender: TObject);

begin

with query1 do begin

close;

sql.Clear;

sql.Add('select * from atmalmaty');

open;

end;

end;

procedure TForm1.N2Click(Sender: TObject);

begin

form1.Close;

end;

procedure TForm1.Button6Click(Sender: TObject);

var b:string;

begin

if ComboBox2.ItemIndex=0 then b:='Название';

if ComboBox2.ItemIndex=1 then b := 'Номер';

if ComboBox2.ItemIndex=2 then b:='Состояние';

with query1 do begin

close;

sql.Clear;

sql.Add('select * from atmalmaty order by atmalmaty."'+b+'"');

open;

end;

end;

procedure TForm1.N4Click(Sender: TObject);

begin

Form4.Show;

end;

procedure TForm1.N5Click(Sender: TObject);

begin

form5.show;

end;

procedure TForm1.N6Click(Sender: TObject);

begin

Form6.show;

end;

procedure TForm1.N8Click(Sender: TObject);

begin

ShowMessage('Created by Sl, on Deplhi 7 platform');

end;

procedure TForm1.FormShow(Sender: TObject);

var u:string;

begin

Passwordform.showmodal;

u:=Passwordform.ComboBox1.Text;

StatusBar1.Panels[0].Text:='Вы вошли как "'+ u + '"';

end;

procedure TForm1.N13Click(Sender: TObject);

begin

changepassword.show;

end;

procedure TForm1.N10Click(Sender: TObject);

begin

passwordtable.show;

end;

procedure TForm1.N11Click(Sender: TObject);

begin

changerule.show;

end;

procedure TForm1.N12Click(Sender: TObject);

begin

ATM_help.show;

end;

procedure TForm1.FormActivate(Sender: TObject);

var g:string;

begin g:=Passwordform.Query1.Fields[2].Value;

with Query2 do

if g='Пользователь' then

begin

Close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Добавить запись"');

open; n4.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Изменить запись"');

open; n5.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Удалить запись"');

open; n6.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Изменение пароля"');

open; n13.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Таблица пользователей"');

open; n10.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Таблица полномочий"');

open; n11.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Поиск банкомата"');

open; n16.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Изменение состояния"');

open; n17.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Формирование отчета"');

open; n18.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Создать наряд"');

open; n20.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Просмотреть наряд"');

open; n22.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from usergroup where usergroup."Privileges" = "Закрыть наряд"'); open;

n21.Visible:=Fields[0].AsBoolean;

end;

if g='Другие' then

with Query2 do

begin

Close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Добавить запись"');

open; n4.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Изменить запись"');

open; n5.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Удалить запись"');

open; n6.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Изменение пароля"');

open; n13.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Таблица пользователей"');

open; n10.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Таблица полномочий"');

open; n11.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Поиск банкомата"');

open; n16.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Изменение состояния"');

open; n17.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Формирование отчета"');

open; n18.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Создать наряд"');

open; n20.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Просмотреть наряд"');

open; n22.Visible:=Fields[0].AsBoolean;

close;

SQL.Clear;

SQL.Add('select access from othergroup where othergroup."Privileges" = "Закрыть наряд"');

open; n21.Visible:=Fields[0].AsBoolean;

end;

DBEdit1.Visible:=false;

DBEdit2.Visible:=false;

DBEdit3.Visible:=false;

DBEdit4.Visible:=false;

DBEdit5.Visible:=false;

end;

procedure TForm1.N16Click(Sender: TObject);

begin

find_form.show;

end;

procedure TForm1.N18Click(Sender: TObject);

begin

report_form.show;

end;

procedure TForm1.N17Click(Sender: TObject);

begin

Form3.show();

end;

procedure TForm1.N20Click(Sender: TObject);

begin

create_order_form.show;

end;

procedure TForm1.N21Click(Sender: TObject);

begin

close_order_form.show;

end;

procedure TForm1.N22Click(Sender: TObject);

begin

view_orders.show;

end;

end.

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