Mechanika płynów – wzory
$\rho = \frac{m}{V}\ ,\ \ \ \ \gamma = \rho g\ \ \ ,\ \ p = \frac{F}{A}\text{\ \ \ \ \ \ }$

pV = nℝT ,    p = ρℝT

$$E_{v} = - \frac{\left( p_{2} - p_{1} \right)}{\frac{\left( V_{2} - V_{1} \right)}{V_{1}}},\ \ \ \ E_{v} = \rho\frac{\text{dp}}{\text{dρ}},\ \ \ \ \frac{p}{\rho^{\kappa}} = const.,\ \ \ \ \ = \frac{C_{p}}{C_{v}}.$$

$K = \frac{1}{E_{v}}$

$$C = \sqrt{\mathbb{R}T}$$

$$\tau = \mu\frac{u}{h} = \mu\frac{\text{du}}{\text{dy}}$$

$$Q = \frac{\pi\ p\ D^{4}}{128\ \mu\ L}$$

$$M = \frac{2\ \pi\ R^{3}\text{μ\ ω\ L}}{h}$$

$$P = M\omega = \frac{2\ \pi\ R^{3}\text{μ\ }\omega^{2}\text{\ L}}{h}$$

$$\mu = \frac{d^{2}\left( \gamma_{k} - \gamma_{p} \right)t}{18\ l}$$

$$U_{i + 1} = \sqrt{\frac{\alpha}{C_{D}(U_{i})}},\ \ \ \alpha = \frac{4\ D\ g\ (\rho_{k} - \rho_{p})}{3\ \rho_{p}},\ \ Re = \frac{\text{ρ\ D}}{\mu}U_{1}$$

$$Re = \frac{4\ Q}{\text{π\ μ\ D}}$$

V = βV₁(T₂−T₁)

$$\beta = - \frac{1}{\rho}\frac{\text{dρ}}{\text{dT}} = \ \frac{\frac{V}{V_{1}}}{T}$$

$$p = \sigma\left( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right)$$

$$\frac{\text{dp}}{\text{dz}} = \rho g$$

p = γ • z + p_A

P = γ H A

$$x_{s} = \frac{I_{y}}{S_{y}},\ \ \ y_{s} = \frac{I_{\text{xy}}}{S_{y}},\ \ \ \ \ S_{y} = x_{c}A$$

p_a = p₀ + h ρ_Hg g

p_A = p_a + ρ h g

p_A = p_a + g(h₁ρ_m − h₂ρ)

p_A − p_B = ρ(Hρ_m + h₂ρ₂ − h₁ρ₁)

$${m = \frac{I}{V_{z}} - n}{W = V\ \rho\ g}$$

$$\overrightarrow{a} = \frac{\partial\overrightarrow{v}}{\partial x}\overrightarrow{v_{x}} + \frac{\partial\overrightarrow{v}}{\partial y}\overrightarrow{v_{y}} + \frac{\partial\overrightarrow{v}}{\partial z}\overrightarrow{v_{z}} + \frac{\partial\overrightarrow{v}}{\partial t}$$

Q = vA

$$\ {\dot{m}}_{1} - {\dot{m}}_{2} = \frac{\text{dρ}}{\text{dt}}V,\ \ \ \ \dot{m_{1}} = \rho_{1}v_{1}A_{1}\ $$

$\text{\ \ }v_{2} = \sqrt{\frac{2\ g\ h}{1 - \left( \frac{A_{2}}{A_{1}} \right)^{2}}}$ , $v_{2} = \sqrt{2\ g\ h}$

$$\frac{1}{2}\rho v^{2} + p + \gamma z,\ \ \gamma z = const.$$

$$t = \frac{2}{\alpha}\left( \sqrt{H_{0}} - \sqrt{h} \right),\ \ \ \ \ \alpha = \left( \frac{d}{D} \right)^{2}\sqrt{2g}$$

$$v = \sqrt{\frac{2\left( p_{2} - p_{a} \right)}{\rho\left( 1 - \left( \frac{d}{D} \right) \right)^{4}}}$$

$$v = \sqrt{2g\left( H - h \right)}$$

$$u = \sqrt{2\frac{\rho_{m}\text{\ g\ H}}{\rho_{p}}}$$

$$Q = \pi\sqrt{\text{gH}\left( \rho_{m} - \rho \right)\frac{1}{8\rho\left( \frac{1}{d^{4}} - \frac{1}{D^{4}} \right)}}$$

$$Q = \frac{\pi\ p\ D^{4}}{128\ L\ \mu} = \frac{\pi\left( p \pm \rho\ g\ L\sin\theta \right)\text{\ D}^{4}}{128\ \mu\ L\ }$$

$$dla\ Re < 2300:\ \ \ \ p = \frac{128\ L\ \mu\ Q}{\text{π\ }D^{4}} = \frac{32\ \mu\ L\ v_{s}}{D^{2}}$$

$$dla\ dowolnego\ przeplywu:\ \ \ p = \frac{1}{2}\text{ρ\ f}\left( Re,\ \frac{e}{D} \right)\frac{L}{D}\ {v_{s}}^{2}$$

$${dla\ Re < 2300:\ \ \ f = \frac{64}{\text{Re}}\backslash n}{dla\ 2300 \leq Re \leq 4000:\ \ \ f = 2,87 \bullet 10^{- 7}\text{Re}^{1,5}\backslash n}{dla\ Re > 4000:\ \ \ \frac{1}{\sqrt{f}} = 1,14 - 2\log_{10}\left( \frac{e}{D} + \frac{9,35}{\text{Re}\sqrt{f}} \right)\backslash n}{wzor\ iteracyjny:\ f_{i + 1} = \frac{1}{\left\lbrack 1,14 - 2\log_{10}\left( \frac{e}{D} + \frac{9,35}{\text{Re}\sqrt{f_{i}}} \right) \right\rbrack^{2}}}$$

$$h_{\text{st}} = f\left( Re,\ \frac{e}{D} \right)\frac{L}{D}\ \frac{v^{2}}{2g}$$

$$f\left( v_{s} \right){v_{s}}^{2} = \frac{2\ g\ D}{L}\left( \frac{p_{1} - p_{2}}{\text{g\ ρ}} \pm l\sin\theta \right)$$

h_Lj = ζ_j ρ C_D v² S  ;

dla Re > 2 • 10⁵ :   C_D = 0, 4 

F = p_iA_i − p_eA_e + ρ_iv_i²A_i − ρ_ev_e²A_e

$$v_{s} = \frac{R_{h}^{\frac{2}{3}}\ \sqrt{s}}{n}$$

$${R_{h} = \frac{A}{P}}{\theta = 2\arccos\left( \frac{R - H}{R} \right),\ P = R\theta\ ,\ A = \frac{R^{2}}{2}(\theta - sin\theta)}$$