Isihloko mayelana nesimiso sikaBernoullis kanye nesibalo sikaBernoullis
Uma sigibela isithuthuthu, izingubo esizigqokayo ziyavuvukala emuva. Ngezinye izikhathi, uma umoya uvunguza ngamandla, umnyango ungazivala. Nakuba umoya uvunguza ngaphandle kwendlu, kuyilapho umnyango ungaphakathi endlini.
Lokhu kungachazwa kusetshenziswa isimiso sikaBernoulli. UDaniel Bernoulli (1700-1782) wathola isimiso esingasetshenziswa ukuchaza lesi simo esingenhla.
Isimiso sikaBernoulli
Isimiso sikaBernoulli sithi lapho ijubane lokugeleza koketshezi liphezulu, ingcindezi yoketshezi iphansi. Ngokuphambene nalokho, uma ijubane lokugeleza koketshezi liphansi, ingcindezi yoketshezi iphakeme. Lapho isithuthuthu sihamba ngokushesha, isivinini somoya ngaphambili nasohlangothini lomzimba wakho siphezulu. Ngakho-ke, ingcindezi yomoya iba phansi. Ingemuva lomzimba wakho livinjwe yingaphambili lomzimba wakho, ngakho ijubane lomoya ngemuva lomzimba wakho aliphenduki phezulu. Ngenxa yalokho, ingcindezi yomoya ngemuva komzimba wakho iba phezulu. Ngoba kunomehluko ekucindezelweni komoya, lapho ngemuva komzimba ingcindezi yomoya iphakeme kunomoya osunduza ihembe lakho emuva ukuze izingubo zakho zibukeke zivuvukele emuva.
Kuthiwani ngomnyango wendlu ozivalela wona uma umoya uvunguza ngaphandle kwendlu? Umoya ongaphandle kwendlu uhamba ngokushesha kunomoya ongaphakathi kwendlu. Ngenxa yalokho, umfutho womoya ongaphandle kwendlu mncane kunomfutho womoya ongaphakathi kwendlu. Ngenxa yokuthi kunomehluko womfutho, lapho umfutho womoya ongaphakathi kwendlu mkhulu khona, umnyango uyaphushelwa ngaphandle. Ngamanye amazwi, iqabunga lomnyango lisuka endaweni lapho umfutho womoya mkhulu khona liye endaweni lapho umfutho womoya mncane khona.
Isibalo sikaBernoulli
Ngaphambilini, sifunde ngesimiso sikaBernoulli. UBernoulli uphinde wathuthukisa lesi simiso ngokwezibalo. Ukuze sithole i-equation kaBernoulli, sicabanga ukuthi ukugeleza koketshezi kuzinzile futhi ku-laminar, akucindezelwanga, i-viscosity incane kangangokuthi inganganakwa.
Engxoxweni ye-equation of continuity, sifunde ukuthi izinga lokugeleza koketshezi lingahluka kuye ngendawo yokugeleza kwepayipi lokugeleza. Ngokusekelwe esimisweni sikaBernoulli esichazwe ngenhla, ingcindezi yoketshezi ingahluka kuye ngereyithi yokugeleza koketshezi. Ingcindezi yoketshezi ingahluka futhi kuye ngereyithi yokugeleza koketshezi. Ukucindezeleka koketshezi kungahluka kuye ngereyithi yokugeleza. Ubudlelwano phakathi kwengcindezi, izinga lokugeleza, kanye nokuphakama kokugeleza kungatholakala ku-equation kaBernoulli.
Isibalo sikaBernoulli sibaluleke kakhulu ngoba singasetshenziswa ukuhlaziya izindiza, izitshalo zamandla kagesi, izinhlelo zamapayipi, njll. Ukuze isibalo sikaBernoulli sitholakale ngokujwayelekile, sicabanga ukuthi uketshezi lugeleza ngepayipi lokugeleza elinendawo enqamulayo engafani futhi ukuphakama nakho kuhlukile. Ukuze sithole isibalo sikaBernoulli, sisebenzisa i-theorem yomsebenzi namandla kuketshezi olusepayipini lokugeleza.
Umbala ongacacile epayipini lokugeleza esithombeni esingezansi ukhombisa ukugeleza koketshezi, kanti umbala omhlophe awubonisi uketshezi.
Uketshezi endaweni enqamulayo 1 (ohlangothini lwesobunxele) lugeleza luze lufike ku-L1 futhi kuphoqa uketshezi olusesigabeni 2 (ohlangothini lwesokudla) ukuthi luhambe luye ku-L2Ngenxa yokuthi indawo enqamula ingxenye 2 ngakwesokudla incane, ijubane lokugeleza koketshezi ohlangothini lwesokudla lwepayipi lokugeleza likhulu (Khumbula isibalo sokuqhubeka). Lokhu kubangela umehluko wokucindezela phakathi kwesigaba 2 (uhlangothi lwesokudla lwepayipi lokugeleza) kanye nesigaba 1 (uhlangothi lwesobunxele lwepayipi lokugeleza) – Khumbula isimiso sikaBernoulli. Uketshezi olukwesobunxele sesigaba 1 lunikeza ingcindezi (P).1) kuketshezi olungakwesokudla futhi luyasebenza:
Bese kuba yi-W1 Isibalo singabhalwa kanje:
W 1 = p 1 A 1 L 1
Esigabeni 2 (uhlangothi lwesokudla lwepayipi lokugeleza), umsebenzi owenziwe kuketshezi yilona:
W2 = − p2 A2 L2
Uphawu olungeluhle lubonisa ukuthi amandla asetshenziswayo aphambene nesiqondiso sokunyakaza. Ngakho-ke, uketshezi lusebenza ngakwesokudla kwesigaba sesi-2. Ngaphezu kwalokho, amandla adonsela phansi ayasebenza oketshezini. Esimweni esingenhla, ezinye izinqwaba zoketshezi zidluliselwa kusuka esigabeni 1 kuze kufike ku-L1 kuya esigabeni 2 kuze kufike ku-L2, ikuphi ivolumu yoketshezi esigabeni 1 (A)1 L1) = umthamo woketshezi esigabeni 2 (A2 L2Umsebenzi owenziwe ngamandla adonsela phansi yilona:
W 3 = − mg (h 2 − h 1 )
W 3 = − mgh 2 + mgh 1 )
W 3 = mgh 1 − mgh 2
Isibonakaliso esingesihle sibangelwa uketshezi olugeleza phezulu, ngokungafani nendlela amandla adonsela phansi ahamba ngayo. Ngakho-ke, umsebenzi wenethi owenziwa oketshezini yilona:
W = W 1 + W 2 + W 3
W = P1 A1 L1 - P2 A2 L2 + mgh1 - mgh2
Ithiyori yamandla okusebenza ithi umsebenzi wenethi owenziwe ohlelweni ufana noshintsho lwamandla e-kinetic. Ngakho-ke, singawushintsha umsebenzi (W) ngezinguquko zamandla e-kinetic (EK).2 – EK1).
Isibalo esingenhla singabhalwa futhi:
W = P1 A1 L1 - P2 A2 L2 + mgh1 - mgh2
EK2 - EK1 =P1 A1 L1 - P2 A2 L2 + mgh1 - mgh2
1⁄2 mv22 – 1⁄2 mv12 =P1 A1 L1 - P2 A2 L2 + mgh1 - mgh2
Isisindo soketshezi esigeleza kuze kufike ku-L1 esigabeni A1 = isisindo soketshezi olugeleza kuze kufike ku-L2 (isigaba A esiphambeneyo2). Isisindo soketshezi, isibonelo m, sinomthamo ongu-A1 L1 kanye A2 L2 lapho u-A1 L1 =A2 L2 (L2 inde kune-L1).
Manje sifaka u-m ku-equation engenhla ngo-m = ρ AL:
Lesi yisibalo sikaBernoulli. Isibalo sikaBernoulli sithathwe ngokusekelwe esimisweni samandla okusebenza ukuze sibe uhlobo lokulondolozwa kwamandla.
P = ingcindezi, ρ = ukuminyana, v = isivinini soketshezi, g = ukusheshisa kwamandla adonsela phansi, h = ukuphakama kwepayipi ngaphezu komhlabathi.
Izingxenye zesobunxele nesokudla zesibalo sikaBernoulli ezingenhla zingabhekisela kumaphuzu amabili noma kuphi eduze kwepayipi lokugeleza ukuze sikwazi ukubhala kabusha lesi sibalo esingenhla sibe:
Manje ake sibukeze isibalo sikaBernoulli kwezinye izimo.
Isibalo sikaBernoulli ku-static fluids
Icala elikhethekile le-equation kaBernoulli limayelana noketshezi olumile, lapho uketshezi lungenaso isivinini. Ngakho-ke, v1 =v2 = 0. Uma kwenzeka uketshezi olumile, singakha i-equation kaBernoulli ibe yilokhu:
Uma u-h2 - h1 = h, lesi sibalo singabhalwa kanje:
p 1 − p 2 = ρ g (h 2 − h 1 )
p 1 − p 2 = ρ gh
Isibalo sikaBernoulli epayipini lokuphakama okufanayo
Uma ukuphakama kwepayipi kufana, isibalo sikaBernoulli sishintshwa sibe:
