HYDRAULICS

$HL=\frac{0.0826\;f\;L\;Q^2}{D^5}

$HL=\frac{10.29\;n^2\;L\;Q^2}{D^\frac{16}{3}}

$HL=\frac{10.67\; L\;Q^{1.85}}{D^{4.87}C^{1.85}}

$HL=\ \frac{f\ L}{D'}\ \bullet\ \frac{v^2}{2\ g}$
$$
$D^\prime=4R^\prime$
$R^\prime\ =\ \frac{\ pipe\ {\rm AREA}_{cross-sect}}{pipe\ PERIMETER}$

$HL=\ \frac{6.35\ n^2\ L\ v^2}{\left(D'\right)^\frac{4}{3}}$
$$
$D^\prime=4R^\prime$
$R^\prime\ =\ \frac{\ pipe\ {\rm AREA}_{cross-sect}}{pipe\ PERIMETER}$

$y=\frac{r^2\;{\omega}^2}{2g}

$Q\ =\ A\ v$
$A\ \bot\ v$

$P_{out}\ =\ Q\ \gamma_{liq}\ \left(H.A.\right)$

$P_{in}\ =\ Q\ \gamma_{liq}\ \left(H.E.\right)$

$Eff.\ \left(\% \right)\ =\ \left(\frac{P_{out}}{P_{in}}\right)\bullet\ 100$

1hp
is equivalent to

746 watts
2544.43 BTU/h

$y\ =\ \frac{H^2}{2D\ }

$y_2\ =\left(\frac{D}{H\ }\right)\ \left(y\ -\frac{H^2}{2D\ }\right)

$V_p=\frac{1}{2}{\pi r}^2h$

The internal pressure inside the cylinder
is unit weight of liquid multiplied by
the distance below the surface of liquid
to the surface of liquid inside the cylinder

$S.G.\ \ =\frac{\rho_{liq.}}{\rho_{water}}\ \ \ \ \ or\ \ \ \ \ \frac{\gamma_{liq.}}{\gamma_{water}}\ $

$9.81\ kN/\ m^3$
$62.4\ lb/\ {\rm ft}^3$

$1000\ kg/m^3$
$1.94\ slugs/\ {ft}^3$
$1\ g/cc$

$9.81\ m/s^2$
$32.2\ ft/s^2$

$287\ \ J\ /\ kg-°K $
$1,\ 716\ \ lb-ft/\ slug-°R $

$P_{2(abs)}\ {V_2}^k=P_{1(abs)}\ {V_1}^k\ $
$\ \frac{T_2}{T_1}\ =\ \left(\frac{P_{2(abs)}}{P_{1(abs)}}\ \right)^\frac{k-1}{k}$

$\rho_{air\ }=\ \frac{P_{abs}}{R\ T}$

$\tau=\ \mu\ \frac{dv}{dy}$

$\nu=\ \frac{\mu}{\rho}$

$1\ Pa-s\ \ \ \ =\ \ \ \ 10\ poise$
$1\ m^2/s\ \ \ \ =\ \ \ \ 10,\ 000\ stokes$

$P=\frac{4\sigma}{d}$

$P=\frac{8\sigma}{d}$

$h=\frac{4\sigma\ cos\theta}{\gamma d}$

$E_B=\frac{stress}{strain}\ =\ \ \ \frac{\Delta p}{\frac{\Delta V}{V}}\ \ =\ \ -\frac{dp}{\frac{dV}{V}}$

$\beta=\frac{1}{E_B}$

$P_{abs}=\ P_{atm}+\ \ P_{gage}$

$\frac{P_{1(abs)}V_1}{T_1}\ =\frac{P_{2(abs)}V_2}{T_2}$

$\ \ \ °K = °C +273.15 $

$°R = °F +459.67$

$c=\sqrt{\frac{E_B}{\rho}\ }=\sqrt{\frac{1}{\beta\rho}}$

$760\ mmHg =760\ Torr =\ 101.325\ kPa\ \ \ $
$2,\ 116.22\ lb/{\rm ft}^2=14.7psi =\ 29.92\ inHg\ \ $

$100\ kPa$
$1,\ 000,\ 000 \ dynes/ cm^2$

$ P_2-\ P_1=\ \gamma h $

$\frac{P_1}{\gamma_w}+\ \sum{\left[(S.G.)h_i\right]}=\ \frac{P_2}{\gamma_w}$

$F=\ \gamma\bar{h}\ A$

$e\ =\frac{I_g}{\bar{y}A}$

$q_{max(+)\ }\ \&\ q_{min(-)\ }\ =\frac{R_y}{B}\ \left(1\pm\ \frac{6\ e}{B}\right)$

$q_{max\ \ }=\ \frac{2\ R_y\ }{3\bar{x}\ }$

$\bar{x}\ =\ \frac{RM\ -\ OM}{R_y}$

$e\ =\frac{B}{2}\ -\ \bar{x}$

${\rm FS}_s\ =\ \frac{{\mu\ R}_y}{R_{x\ }}\ \geq\ 1.0$

${\rm FS}_o\ =\ \frac{RM}{OM}\ \geq\ 1.0$

$\gamma_{obj}\ V_{obj\ }=\ \gamma_{liq\ }V_{sub\ }$

$\bar{MB_0}\ =\ \frac{I_g}{V_{sub\ }cos\ \theta}$
$\bar{MG}\ =\ \bar{MB_0}\ \pm\ \bar{GB_0}$

$\bar{MB_0}\ =\ \frac{I_g}{V_{sub\ }cos\ \theta}$
$\bar{MG}\ =\ \bar{MB_0}\ \pm\ \bar{GB_0}$

$\frac{rev}{min}\ \ \ \ \ \Rightarrow\ \ \ \ \frac{rev}{min}\ \bullet\ \ \frac{\pi}{30\ }\ \ \ \ \Rightarrow\ \frac{rad}{sec}$

$tan\theta\ =\ \frac{a}{g}

$tan\theta\ =\ \frac{a_H}{g\ \pm\ a_V\ }

$p=\ \gamma h\ \left(1\ \pm\frac{a}{g}\right)

$c\ =\ \sqrt{\frac{E_b}{\rho_w}}$

$T_c\ =\ \frac{2\ L}{c}$

Instantaneous Closure
$P_h=\ \rho\ c\ v_{(velocity)}$

Rapid Closure
$P_h=\ \rho\ c\ v_{(velocity)}$

Slow Closure
$P_h=\ \rho\ \left(\frac{2L}{t}\right)\ v_{(velocity)}$

$E_1=\ E_2$
$\frac{{v_1}^2}{2\ g}+\ \frac{P_{gage1}}{\gamma_{liq}}+z_1=\ \frac{{v_2}^2}{2\ g}+\ \frac{P_{gage2}}{\gamma_{liq}}+z_2$

$v_2=\ \sqrt{2g\ \left(E_1\right)}$

$C_v=\ \frac{v_{\ actl}}{v_{\ theor}}$

$C_D=\ \frac{Q_{\ actl}}{Q_{\ theor}}$
$C_D= {C_{\ v}} \bullet{C_{\ a}}$

$t=\int_{E_2(final)}^{E_1(initial)}\frac{A_s\ dh}{C_D\ \ A_o\sqrt{2g\ \left(E_1\right)\ }}$

$t=\ \frac{2\ A_s\ \left(\sqrt{E_1}-\sqrt{E_2}\right)\ }{C_D\ \ A_o\sqrt{2g\ }}$

$Q=Q_1=Q_2=Q_3=\ Q_{(n)}$
$HL=h_1+h_2+h_3+h_{(n)}$$

$Q=Q_1+Q_2+Q_3+Q_{(n)}$
$h_1=h_2=h_3=h_{(n)}$

${V\ }_{initial}^{\ air}=\ {V\ }_{final}^{\ air}$

Initial Volume
is Equal To
Final Volume

$Re\ =\ \frac{D\ v\ \rho}{\mu}$

VENTURI METER for FLOW (Q)

PITOT TUBE for VELOCITY (v)

CONVERSION
$\gamma_a\ h_a=\ \gamma_b\ h_b$
IN ORIFICE convert all the liquid present in EQUIVALENT HEIGHT
of the FIRST FLUID to EXIT outflow.

$F=\ Mv$

$M=\ \ \rho\ \bullet Q\ $

$W=\ \ \rho\ \bullet g\ \bullet Q$
$W=\ \ \gamma\ \bullet Q$

$F=\ Mu$

$V=\ C\sqrt{RS}$

$m\ =\ \frac{A}{B}$

$R\ =\ \frac{A}{P}$

$z=\ A\ \sqrt m$

$C=\ \frac{{R\ }^\frac{1}{6}}{n}$

$C=\ \sqrt{\frac{8g}{f}}$

$C=\ \frac{23\ +\ \frac{1}{n}\ +\ \frac{0.00155}{S} } {1+\ \dfrac{n}{\sqrt R}\left(23\ +\ \frac{0.00155}{S}\right)}$

$C=\ \frac{87}{1+\frac{m}{\sqrt R}}$

$Q=\ A\ {\bullet\left(\frac{1}{n}\right)\bullet\left(R\right)}^{\ \frac{2}{3}}\ \bullet\left(S\right)^{\ \frac{1}{2}}$

$F_N=\ \frac{v}{\sqrt{g\ m}}$

$v_c=\ \sqrt{g\ m}$

$S_c=\ \frac{{v_c}^2\ n^2}{R^{4/3}}$

$\frac{{Q_c}^2}{g}=\ \frac{A^3}{B}$

$q=\ \frac{Q}{B}$

Critical Stress (Tangential Stress)

$\sigma_t=\dfrac{pD}{2t\left(eff \right)}$

$S=\dfrac{2T}{pD}$