Bernulli qonuni nima?

2-masala: suv favvorasi muhandisligi
Sizga katta bir mehmonxona tomonidan 15 cm15 cm15, start text, space, c, m, end text diametrli gorizontdan 8,00 m8,00 m8, comma, 00, start text, space, m, end text chuqurlikdagi quvur bilan taʼminlanadigan favvora qurib berish soʻraldi. Suv quvurning oxirida diametri 5,00 cm5,00 cm5, comma, 00, start text, space, c, m, end text va gorizontdan balandligi 1,75 m1,75 m1, comma, 75, start text, space, m, end text boʻlgan silindr teshikdan tik yuqoriga 32,0 m/s32,0 m/s32, comma, 0, start text, space, m, slash, s, end text tezlikda otilib chiqadi. Suvning zichligi 1000���31000m3kg​1000, start fraction, k, g, divided by, m, cubed, end fraction.
Ushbu favvora toʻgʻri ishlashi uchun quvurdagi gidrostatik bosim qanday boʻlishi kerak?
Bernulli qonuniga doir masalalar biroz murakkab boʻlgani uchun biz qiziqishimiz ostida boʻlgan har ikki nuqtaning chizmasini chizib olishimiz kerak (rasm oʻlchash uchun berilmagan).

Biz quvurning pastki qismidagi nuqtani 1-nuqta sifatida tanlaymiz, chunki biz ushbu nuqtadagi bosimni aniqlashimiz kerak va quvurning yuqori qismidan 2-nuqtani tanlaymiz, chunki masala shartida ushbu nuqtadagi suvning tezligi haqida maʼlumot berilgan.
�1+12��12+��ℎ1=�2+12��22+��ℎ2(yechishni Bernulli qonunidan boshlaymiz)P1​+21​ρv12​+ρgh1​=P2​+21​ρv22​+ρgh2​(yechishni Bernulli qonunidan boshlaymiz)P, start subscript, 1, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 1, end subscript, squared, plus, rho, g, h, start subscript, 1, end subscript, equals, P, start subscript, 2, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 2, end subscript, squared, plus, rho, g, h, start subscript, 2, end subscript, start text, left parenthesis, y, e, c, h, i, s, h, n, i, space, B, e, r, n, u, l, l, i, space, q, o, n, u, n, i, d, a, n, space, b, o, s, h, l, a, y, m, i, z, right parenthesis, end text
�1=�2+12��22+��ℎ2−12��12−��ℎ1(�1 bosimni algebraik hisoblab toping)P1​=P2​+21​ρv22​+ρgh2​−21​ρv12​−ρgh1​(P1​ bosimni algebraik hisoblab toping)P, start subscript, 1, end subscript, equals, P, start subscript, 2, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 2, end subscript, squared, plus, rho, g, h, start subscript, 2, end subscript, minus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 1, end subscript, squared, minus, rho, g, h, start subscript, 1, end subscript, start text, left parenthesis, P, start subscript, 1, end subscript, space, b, o, s, i, m, n, i, space, a, l, g, e, b, r, a, i, k, space, h, i, s, o, b, l, a, b, space, t, o, p, i, n, g, right parenthesis, end text
Bizga suvning 1-nuqtadagi tezligi nomaʼlum. Bernulli tenglamasini qoʻllab 1-nuqtadagi bosimni topish uchun biz oldin �1v1​v, start subscript, 1, end subscript tezlikni topishimiz kerak.
Suv siqilmasligi sababli biz buni �1�1=�2�2A1​v1​=A2​v2​A, start subscript, 1, end subscript, v, start subscript, 1, end subscript, equals, A, start subscript, 2, end subscript, v, start subscript, 2, end subscript oqim uzluksizlik tenglamasidan foydalanib amalga oshirishimiz mumkin. Bilamizki, silindrsimon quvurning koʻndalang kesim yuzini �=��2A=πr2A, equals, pi, r, squared formula bilan topish mumkin, natijani oqim uzluksizlik tenglamasiga keltirib qoʻyamiz,
(��12)�1=(��22)�2(πr12​)v1​=(πr22​)v2​left parenthesis, pi, r, start subscript, 1, end subscript, squared, right parenthesis, v, start subscript, 1, end subscript, equals, left parenthesis, pi, r, start subscript, 2, end subscript, squared, right parenthesis, v, start subscript, 2, end subscript
Endi ikkala tarafdan �πpi larni qisqartirib, tenglamadan �1v1​v, start subscript, 1, end subscript ni topamiz,
�1=(�22�12)�2v1​=(r12​r22​​)v2​v, start subscript, 1, end subscript, equals, left parenthesis, start fraction, r, start subscript, 2, end subscript, squared, divided by, r, start subscript, 1, end subscript, squared, end fraction, right parenthesis, v, start subscript, 2, end subscript
Quvurlarning radiusini keltirib qoʻyib, 1-nuqtadagi suvning tezligini topamiz,
�1=(2,50 cm)2(7,50 cm)2(32,0 m/s)=3,56 m/sv1​=(7,50 cm)2(2,50 cm)2​(32,0 m/s)=3,56 m/sv, start subscript, 1, end subscript, equals, start fraction, left parenthesis, 2, comma, 50, start text, space, c, m, end text, right parenthesis, squared, divided by, left parenthesis, 7, comma, 50, start text, space, c, m, end text, right parenthesis, squared, end fraction, left parenthesis, 32, comma, 0, start text, space, m, slash, s, end text, right parenthesis, equals, 3, comma, 56, start text, space, m, slash, s, end text 
[Nega radiuslarni metrga oʻtkazmadik?]
Nihoyat, 1-nuqtadagi suvning tezligini topgandan keyin biz kattaliklarni Bernulli qonuniga keltirib qoʻyamiz,
�1=�2+12�(32 m/s)2+��ℎ2−12�(3,56 m/s)2−��ℎ1(tezliklarning qiymatini keltirib qoʻydik)P1​=P2​+21​ρ(32 m/s)2+ρgh2​−21​ρ(3,56 m/s)2−ρgh1​(tezliklarning qiymatini keltirib qoʻydik)P, start subscript, 1, end subscript, equals, P, start subscript, 2, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, left parenthesis, 32, start text, space, m, slash, s, end text, right parenthesis, squared, plus, rho, g, h, start subscript, 2, end subscript, minus, start fraction, 1, divided by, 2, end fraction, rho, left parenthesis, 3, comma, 56, start text, space, m, slash, s, end text, right parenthesis, squared, minus, rho, g, h, start subscript, 1, end subscript, start text, left parenthesis, t, e, z, l, i, k, l, a, r, n, i, n, g, space, q, i, y, m, a, t, i, n, i, space, k, e, l, t, i, r, i, b, space, q, o, ʻ, y, d, i, k, right parenthesis, end text
Biz 1-nuqtani ℎ=0h=0h, equals, 0 sath deb olamiz, shunda ℎ1=0 mh1​=0 mh, start subscript, 1, end subscript, equals, 0, start text, space, m, end text va ℎ2=8,00 m+1,75 m=9,75 mh2​=8,00 m+1,75 m=9,75 mh, start subscript, 2, end subscript, equals, 8, comma, 00, start text, space, m, end text, plus, 1, comma, 75, start text, space, m, end text, equals, 9, comma, 75, start text, space, m, end text boʻladi. 
[Shoshmang, qanday?]
Biz buni soʻnggi Bernulli tenglamasiga keltirib qoʻysak, ��ℎ1ρgh1​rho, g, h, start subscript, 1, end subscript qisqarib ketadi va biz quyidagiga ega boʻlamiz,
�1=�2+12�(32 m/s)2+��(9,75 m)−12�(3,56 m/s)2(biz ℎ larning qiymatlarini keltirib qoʻydik)P1​=P2​+21​ρ(32 m/s)2+ρg(9,75 m)−21​ρ(3,56 m/s)2(biz h larning qiymatlarini keltirib qoʻydik)P, start subscript, 1, end subscript, equals, P, start subscript, 2, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, left parenthesis, 32, start text, space, m, slash, s, end text, right parenthesis, squared, plus, rho, g, left parenthesis, 9, comma, 75, start text, space, m, end text, right parenthesis, minus, start fraction, 1, divided by, 2, end fraction, rho, left parenthesis, 3, comma, 56, start text, space, m, slash, s, end text, right parenthesis, squared, start text, left parenthesis, b, i, z, space, h, space, l, a, r, n, i, n, g, space, q, i, y, m, a, t, l, a, r, i, n, i, space, k, e, l, t, i, r, i, b, space, q, o, ʻ, y, d, i, k, right parenthesis, end text
Endi biz bajarishimiz kerak boʻladigan narsa – bu 2-nuqtadagi �2P2​P, start subscript, 2, end subscript bosimni aniqlash. 2-nuqtadagi bosim atmosfera bosimiga teng boʻlishi kerak deyishimiz mumkin, chunki suv ochiq atmosferada turibdi. Bu – Bernulli tenglamasiga doir koʻplab masalalarda eʼtiborga olinishi kerak boʻladigan gipoteza. Har qanday nuqta ochiq atmosferada boʻlganda, uning bosimi atmosfera bosimiga teng. Bernulli tenglamasida biz mutlaq bosim �2=1,01×105��P2​=1,01×105PaP, start subscript, 2, end subscript, equals, 1, comma, 01, times, 10, start superscript, 5, end superscript, P, a dan yoki manometrik bosim �2=0P2​=0P, start subscript, 2, end subscript, equals, 0 dan foydalanishimiz mumkin (chunki manometrik bosim atmosfera bosimidan yuqori bosim). Har doim nollarni kiritishimiz mumkin, bu bizning ishimizni osonlashtiradi, shuning uchun biz manometrik bosim �2=0P2​=0P, start subscript, 2, end subscript, equals, 0 dan foydalanamiz. Bu bizning oʻzgartirilgan Bernulli tenglamamizni quyidagicha koʻrinishga keltiradi:
�1=12�(32 m/s)2+��(9,75 m)−12�(3,56 m/s)2(�2=0 ni keltirib qoʻydik)P1​=21​ρ(32 m/s)2+ρg(9,75 m)−21​ρ(3,56 m/s)2(P2​=0 ni keltirib qoʻydik)P, start subscript, 1, end subscript, equals, start fraction, 1, divided by, 2, end fraction, rho, left parenthesis, 32, start text, space, m, slash, s, end text, right parenthesis, squared, plus, rho, g, left parenthesis, 9, comma, 75, start text, space, m, end text, right parenthesis, minus, start fraction, 1, divided by, 2, end fraction, rho, left parenthesis, 3, comma, 56, start text, space, m, slash, s, end text, right parenthesis, squared, start text, left parenthesis, P, start subscript, 2, end subscript, equals, 0, space, n, i, space, k, e, l, t, i, r, i, b, space, q, o, ʻ, y, d, i, k, right parenthesis, end text
Endi suvning zichligi �=1000���3ρ=1000m3kg​rho, equals, 1000, start fraction, k, g, divided by, m, cubed, end fraction va erkin tushish tezlanishi �=+9,8��2g=+9,8s2m​g, equals, plus, 9, comma, 8, start fraction, m, divided by, s, squared, end fraction ning son qiymatlarini keltirib qoʻyamiz,
�1=12(1000���3)(32 m/s)2+(1000���3)(+9,8��2)(9,75 m)−12(1000���3)(3,56 m/s)2P1​=21​(1000m3kg​)(32 m/s)2+(1000m3kg​)(+9,8s2m​)(9,75 m)−21​(1000m3kg​)(3,56 m/s)2P, start subscript, 1, end subscript, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 1000, start fraction, k, g, divided by, m, cubed, end fraction, right parenthesis, left parenthesis, 32, start text, space, m, slash, s, end text, right parenthesis, squared, plus, left parenthesis, 1000, start fraction, k, g, divided by, m, cubed, end fraction, right parenthesis, left parenthesis, plus, 9, comma, 8, start fraction, m, divided by, s, squared, end fraction, right parenthesis, left parenthesis, 9, comma, 75, start text, space, m, end text, right parenthesis, minus, start fraction, 1, divided by, 2, end fraction, left parenthesis, 1000, start fraction, k, g, divided by, m, cubed, end fraction, right parenthesis, left parenthesis, 3, comma, 56, start text, space, m, slash, s, end text, right parenthesis, squared
�1=6,01×105��(hisoblang va nishonlang)P1​=6,01×105Pa(hisoblang va nishonlang)P, start subscript, 1, end subscript, equals, 6, comma, 01, times, 10, start superscript, 5, end superscript, P, a, start text, left parenthesis, h, i, s, o, b, l, a, n, g, space, v, a, space, n, i, s, h, o, n, l, a, n, g, right parenthesis, end text
Yodda tuting: biz �2=0P2​=0P, start subscript, 2, end subscript, equals, 0 ni keltirib qoʻyganimiz uchun natijamiz gidrostatik bosimga teng boʻladi. Agar bosimga �2=1,01×105 PaP2​=1,01×105 PaP, start subscript, 2, end subscript, equals, 1, comma, 01, times, 10, start superscript, 5, end superscript, start text, space, P, a, end text ni qoʻyganimizda, 1-nuqtadagi absolyut bosimni topgan boʻlardik.