Bernulli qonuni nima?

Bernulli qonuni yordamida 2-qismdagi manometrik bosimni toping.

�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
�2=�1+12��12+��ℎ1−12��22−��ℎ2(algebraik jihatdan �2 ni toping)P2​=P1​+21​ρv12​+ρgh1​−21​ρv22​−ρgh2​(algebraik jihatdan P2​ ni toping)P, start subscript, 2, end subscript, equals, 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, minus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 2, end subscript, squared, minus, rho, g, h, start subscript, 2, end subscript, start text, left parenthesis, a, l, g, e, b, r, a, i, k, space, j, i, h, a, t, d, a, n, space, end text, P, start subscript, 2, end subscript, start text, space, n, i, space, t, o, p, i, n, g, end text, right parenthesis
Endi biz ℎ=0h=0h, equals, 0 nisbiy sathni tanlashimiz kerak. Biz 1-qismning balandligini ℎ=0h=0h, equals, 0 sath deb olamiz. Bunda ℎ1=0h1​=0h, start subscript, 1, end subscript, equals, 0 va ℎ2=1,2 mh2​=1,2 mh, start subscript, 2, end subscript, equals, 1, comma, 2, start text, space, m, end text boʻladi. Ushbu qiymatlarni balandlik oʻrniga keltirib qoʻysak,
�2=�1+12��12+��(0 m)−12��22−��(1,2�)( ℎ1 va ℎ2 ning qiymatlarini keltirib qoʻyamiz)P2​=P1​+21​ρv12​+ρg(0 m)−21​ρv22​−ρg(1,2m)( h1​ va h2​ ning qiymatlarini keltirib qoʻyamiz)P, start subscript, 2, end subscript, equals, 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, left parenthesis, 0, start text, space, m, end text, right parenthesis, minus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 2, end subscript, squared, minus, rho, g, left parenthesis, 1, comma, 2, m, right parenthesis, start text, left parenthesis, space, h, start subscript, 1, end subscript, space, v, a, space, h, start subscript, 2, end subscript, space, 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, a, m, i, z, right parenthesis, end text
Biz nol qatnashgan hadni tashlab yuborib, qolgan hadlarning son qiymatini keltirib qoʻyamiz,
�2=12300 Pa+12(1090���3)(3,00 m/s)2−12(1,090���3)(0,750 m/s)2−(1090���3)�(1,20 m)P2​=12300 Pa+21​(1090m3kg​)(3,00 m/s)2−21​(1,090m3kg​)(0,750 m/s)2−(1090m3kg​)g(1,20 m)P, start subscript, 2, end subscript, equals, 12300, start text, space, P, a, end text, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, 1090, start fraction, k, g, divided by, m, cubed, end fraction, right parenthesis, left parenthesis, 3, comma, 00, start text, space, m, slash, s, end text, right parenthesis, squared, minus, start fraction, 1, divided by, 2, end fraction, left parenthesis, 1, comma, 090, start fraction, k, g, divided by, m, cubed, end fraction, right parenthesis, left parenthesis, 0, comma, 750, start text, space, m, slash, s, end text, right parenthesis, squared, minus, left parenthesis, 1090, start fraction, k, g, divided by, m, cubed, end fraction, right parenthesis, g, left parenthesis, 1, comma, 20, start text, space, m, end text, right parenthesis
�2=4080 Pa(hisoblang va nishonlang)P2​=4080 Pa(hisoblang va nishonlang)P, start subscript, 2, end subscript, equals, 4080, start text, space, P, a, end text, 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: bizga maʼlumki, 2-qismdagi bosim absolyut bosimning emas, balki gidrostatik bosimning qiymati, chunki biz 1-qism uchun gidrostatik bosimni keltirib qoʻydik. Agar bizga absolyut bosimni topish kerak boʻlsa, unda natijaga (1,01×105 Pa)(1,01×105 Pa)left parenthesis, 1, comma, 01, times, 10, start superscript, 5, end superscript, start text, space, P, a, end text, right parenthesis ni qoʻshamiz.