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Ishlab chiqarishdagi baxtsiz xodisalar va baxtsiz xodisalarni tekshirish
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| bet | 9/9 | | Sana | 29.12.2023 | | Hajmi | 411,8 Kb. | | #128943 |
Bog'liq Ishlab chiqarishdagi baxtsiz xodisalar va baxtsiz xodisalarni te-fayllar.orgBerilgan
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Vatiantlar
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61
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62
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63
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64
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65
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66
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67
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68
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69
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70
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L (m)
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6
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7
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8
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9
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10
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11
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12
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13
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14
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15
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B (m)
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4
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3
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2
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5
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7
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8
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9
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10
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8
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9
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H (m)
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0,4
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0,6
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0,8
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1
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1,2
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1,4
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1,6
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1,8
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2
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2,2
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ФL (lk)
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2300
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2310
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2280
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2290
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2320
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2330
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2340
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2285
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2295
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2305
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ЕH (lk)
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450
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180
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100
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120
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150
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200
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250
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300
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350
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400
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Кzk (lk)
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0,5
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0,6
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0,7
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0,8
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0,9
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1
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1,1
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1,2
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1,3
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1,4
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Z
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1
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1,02
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1,04
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1,06
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1,07
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1,08
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1,09
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1,1
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1,12
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1,13
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Wsq
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5
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5,2
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5,4
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5,6
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5,8
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6
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6,2
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6,4
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6,6
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6,8
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Sn (m2)
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22
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210
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220
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230
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240
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250
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260
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270
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280
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280
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n
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8
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9
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5
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4
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5
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6
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7
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8
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9
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10
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η
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45%
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45%
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45%
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45%
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45%
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45%
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45%
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45%
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45%
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45%
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μ
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1,1
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1,2
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1,3
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1,4
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1,5
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1,6
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1,7
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1,8
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1,9
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2,0
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2.3. Hisoblash uslubi1
1. Yorug’lik oqimidan foydalanish koeffitsienti usuli2
Bino uskunasini aniqlaymiz:
, (1.1)
bunda L - xona uzunligi;
B - xona kengligi;
Hp – chiroqlarni ilish balandligi.
НР = Н-0,3, (1.2)
Chiroqning yorug’lik oqimi:
, (1.3)
bunda ФL – lampaning yorug’lik oqimi;
ЕН – yoritilganlik, лк;
S = L·B xonaning maydoni;
Кzk – zahira koeffitsienti;
z – yoritishning nochiziqli koeffitsienti;
N – chiroqlar soni.
Shundan chiroqlar sonini quyidagi formula bo’yicha topamiz:
, (1.4)
Фsv = 2·ФL –chiroqning yorug’lik oqimi;
EH – yoritilganlik.
2. Chiroqning solishtirma quvvati usuli.
Solishtirma quvvat usuli bo’yicha aniqlanadi:
P = 40 B, (1.5)
Shunday qilib, chiroqlar sonini topamiz:
, (6)
bunda Sn – xonaning maydoni, m2 (shart bo’yicha);
P – chiroqning nominal quvvati;
n – chiroqning soni;
N – chiroq SOL(sanoatdagi ocma lyuminetsentli) bo’lganligi sababli, chiroqdagi lampalar soni.
Wx=K Wsq, (1.7)
bunda WX – chiroqning xisobli almashtirma quvvati,vt/m2;
K – korreksiya koeffitsienti;
Wsq – jadvalda muvofiq solishtirma quvvat.
Chiroqning oynali tarqatgichsiz, teshikchalarsiz va panjarasiz LD-400 turi uchun chiroqning h va S0 osilgan balandligida Wj vt/m2 ELK uchun Sshift=50%, Sst=30%, yer=10% da Kz.
Negaki, shart bo’yicha Sshift=50%, Set=50%, Ser=10%, unda WJ ni 10%ga kamaytirish zarur ya’ni
W=0.9•W0.
Shart bo’yicha Kz W ni korreksiya koefsentlariga ko’paytirish zarur:
αkz = .
Shart bo’yicha Z, αkz, ga kupaytirish natijasida olingan W, korreksiya koefsientlariga kupaytirish zarur:
αZ = .
Shart bo’yicha EH, αZ ga kupaytirish natijasida olingan W, koreksiya koeffitsientiga ko’paytirish zarur:
αE = .
3. Yaltirovchi chiziqlar usuli.
l = 0,5L, m, (1.8)
bunda l – devordan joylashish chizig’igacha masofa.
Chiziqli yorug’lik oqimining chiziqligi quydagiga teng:
, (1.9)
bunda K3 – zaxira koeffitsienti;
µ – akslangan yorug’lik va uzoqlashgan chiziq tasirini inobatga oluvchi koeffitsient;
LL = L=20 m;
EH = 400 lk;
Σе – hisoblash natijasidagi shartli yoritilganlik yig’indisini chiziqning barcha bo’laklarini 42.84 ga teng deb olamiz.
H’=H-0,3.
Natijaviy jadval: 2.2.
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Р
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P’=P/H’
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L
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L’=L/H’
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e
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1
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2
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3
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4
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5
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6
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Jami
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2.4. Mustaqil tayyorlanish uchun savollar:
1. Yoritilganlik nima?
2. Yoritilganliknin sifat ko’rsatkichlari sanab bering.
4. Suniy yoritishni me’yorlashtirishda hisobga olinadigan omillar.
5. Ko’zni qamashtiruvchi ta’sirni cheklash talabi.
6. Chiroqning solishtirma quvvati hisoblash usulini tushuntiring.
7. Stroboskopik effect nima.
III. Hisobot tayyorlash.
Hisobot tayyorlashda qo‘yidagilarga amal qidinadi:
1. Titul varog‘i (ilova 1 ga qarang).
2. Topshiriq shaklini o‘zgartirilmagan holatda to‘ldirish (word da yozish va tahrirlash qonun qoidalariga qatiy rioya etgan holda).
3. Nazariy savollarga javob berish.
4. Amaliy topshiriqni bajarish.
5. Bajarilgan ishlar bo‘yicha xulosa chiqarish. (ilova 2 ga qarang).
6. Foydalanilgan adabiyotlar va internet xavolalari ruyhati.
7. Baholash mezoni: №1 topshiriq bali 6 bal.
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Kerakli harakatlar ketma-ketligiga rioya qilgan holda ishni to‘liq bajarish
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0.5 ball
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Talaba mustaqil mushohada yuritsa berilgan topshiriq mavzularining mohiyatini tushunsa
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1 ball
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Rasmiylashtirish sifati (tartibliligi, mantiqliligi)
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1 ball
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Hisob qitoblarda o‘lchov birliklarining mavjudligi
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0.5 ball
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Mavzu bo‘yicha maqsad va asosiy tushunchalar va ta’riflarning mavjudligi
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1 ball
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Variant bo‘yicha vazifani bajarilishi
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0.5 ball
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Amaliy ishlarining nazariy himoyasi (talaba mustaqil xulosa va qaror qabul qilsa, ijodiy fikrlay olsa, mustaqil mushohada yuritsa, berilgan topshiriq mavzularining mohiyatini tushunsa, bilsa, ifodalay olsa, aytib bera olsa, hamda topshiriqlar bo‘yicha tasavvurga ega bo‘lsa)
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1.5 ball
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Fan bo‘yicha umumiy ballar: jami-100 bal
Topshiriq
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Maksimal ball
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Topshiriq-1. Ish ishlab chiqarish ergonomikasi (paramaetrlarni hisoblash)
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6
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Joriy nazorat bo‘yicha maksimal 18 ball
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Topshiriq-2. Tabiiy va texnogen tabiatning xavfli va zararli omillarni hisoblash.
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6
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Topshiriq-3. Ishlab chiqarishdagi baxtsiz xodisalar va baxtsiz xodisalarni tekshirish
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6
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Ma’ruzalarni o’zlashtirish
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27
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Davomat bo‘yicha maksimal ball
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5
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Yakuniy nazorat bo‘yicha maksimal ball
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50
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Jami:
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100
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100 ball
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2 ilova
Bajarilgan ishlar bo‘yicha xulosa chiqarish.
(xulosa qo‘lda yoziladi)
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Foydalanilgan adabiyotlar va internet xavolalari ruyhati
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http://fayllar.org
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