Labarai

Lokacin da Zonel Filtech ke taimaka wa abokan ciniki don inganta masu tattara ƙura, wasu daga cikinsu sun koka da cewa tsarin tsabtace su.jakar tace gidajeba ya aiki da kyau ko da suna amfani da bututun da ke jagorantar iska akan bututun iska, har ila yau tare da venturi, da kuma tare da matsi daidai don matsewar iska, don haka ba za su iya samun mafita don inganta ayyukan tsaftacewa ba.

Bayan nazarin tsarin aikin su na tsaftacewa, an gano injiniyoyin Zonel babban dalilin shi ne tazarar da ke tsakanin bututun iskar su zuwa takardar buhu ba daidai ba ne. Idan nisa ya yi girma sosai, iska na iya busa wasu zuwa takardar bututun jaka maimakon cikin jakunkunan tacewa; akasin haka, idan ya yi ƙanƙanta, matsewar iska ba zai iya haifar da isasshiyar iska a waje zuwa cikin jakunkuna masu tacewa ba, tasirin tsarkakewa ba zai yi kyau ba.

Amma yadda za a ayyana wannan nisa (H1 a cikin zane mai zuwa)?

１. Mataki na farko, kuna buƙatar ayyana matsakaicin ƙimar Øp a cikin zane.
Kamar yadda aka saba, muna lissafin Øp tare da dabara mai zuwa:
Øp=(C*D^2/n) ^1/2
C = Coefficient, kamar yadda aka saba zaɓi 50% ~ 65%.
D= pulse jet bawul diamita, kamar yadda ya saba iri daya zuwa iska busa bututu.
n = lambar jakar tacewa a jere (tsaftacewa tare da bawul ɗin jet guda ɗaya)
Kamar yadda aka saba, C muna zaɓar 0.55.
Yawancin, bututun da ke jagorantar iskar diamita shine sau 2 ~ 3 na Øp.

 

2. Ƙayyade tsawon bututun da ke jagorantar iska.
Air gubar bututu kamar yadda aka saba amfani da dabara mai zuwa:
L=Ck* Øp/K
Ck = ƙididdiga, kamar yadda aka saba zaɓi 0.2 ~ 0.25
K=is jet turbulence coefficient, cylindrical zabi 0.076.
watau L= kimanin 0.2* Øp/0.076=2.65 Øp

 

３. Yana da sauqi don samun wannan tg digiri = (1/2 Øb)/H2
tg digiri = 3.4K = 0.272 (ana iya bi da shi azaman dindindin)
Don haka digiri ya zaɓi digiri 15.

 

Misali:
Idan zaɓin 3” bawul ɗin bugun bugun jini, jagorar bututu d = 30mm, diamita jakar tace shine 160mm, yadda ake samun H1.
Amsa:
Babu shakka, H1=H2-L
Don haka dole ne mu ayyana H2 da L.

tg digiri = (1/2 Øb)/H2=3.4K=0.272
watau H2=1.838 Øb

Øb = 160mm
Saboda haka H2 = 294 mm

2
Daga sakamakon da ya gabata, L = 2.65 Øp, don haka L = 2.65 * 15 = 40 mm
Don haka H1=294-40=254mm.

 

Ga Qp, gabaɗaya matsakaicin bayanai ana iya zaɓar kamar haka:
Girman bawul ɗin bawul ---- Qp
3/4" ----5 ~ 7mm
1" ---- 6 ~ 8mm
1 1/2" ----7-9mm
2"----8-11mm
2 1/2" ----9 ~ 14mm
3"----14-18mm
4"----16-22mm

 

Kamar yadda aka saba, lokacin da ƙirar Qp za ta raba zuwa ƙungiyoyin 3 ~ 4, kusa da bawul ɗin jet ɗin bugun jini, girman buɗewa ya fi girma, da rukuni don haɗa bambance-bambancen diamita game da 1mm.