Aerodynamics Notes: Pressure, Force, Aircraft Performance PDF

Here is the text extracted and converted to markdown format: ### Aerodynamic force The force due to air flow. **Thrust** $mg$ $D-T_{ray} \; aerodynamic \\ efficiency = \frac{L}{D}$ (for normal aircrafts it is $\approx 10$) $e.g. \; W = 100 kg \\ L = 100 g \longrightarrow $ for equilibrium fli...

Here is the text extracted and converted to markdown format: ### Aerodynamic force The force due to air flow. **Thrust** $mg$ $D-T_{ray} \; aerodynamic \\ efficiency = \frac{L}{D}$ (for normal aircrafts it is $\approx 10$) $e.g. \; W = 100 kg \\ L = 100 g \longrightarrow $ for equilibrium flight cond. and if $\frac{L}{D} = 10 \longrightarrow D = 10g$ Thus engine required to power, should give a thrust of at least $10g$. * **Gliders** have a larger $\frac{L}{D}$ ratio, i.e. very high efficiency. ### Atmosphere * Ideal atmosphere $\longrightarrow$ neglecting **winds, clouds and objects** * Most aircrafts fly mid-range of troposphere and stratosphere. i.e. in **Tropopause** $\approx 11 km \longrightarrow 20 km$ and **stratosphere** $\approx 20 km \longrightarrow 47 km$ ### Pressure Variation Diagram with a rectangular block with height h and a small change in height denoted $dh$. The pressure at h + dh is $(P + dP)$ and the pressure at h is $P$. Since $P = \rho R T$ and $\rho = \frac{P}{R T}$, $\longrightarrow -dP = \rho gdh \longrightarrow - \frac{dP}{P} = \frac{g \cdot dh}{R \cdot T}= \frac{g}{R T(W)}$ --- ### Lapse Rate Rate of change Temp w.r.t the height $L = \frac{dT}{dh}$ The temperature varies as the atmosphere (green house gases) traps certain regions/wavelength from the electromagnetic radiations. Diagram with temperature increasing with increasing altitude. * Stratosphere * Tropopause * Troposphere $\longrightarrow$ Temp inc. due to presence of ozone. For each layer: $L_i = \frac{dT}{dh}$ $\frac{dP}{P} = - \frac{g \cdot dh}{R T}$ $\longrightarrow dP = - \frac{g}{R T} \cdot d T$ On integrating $\longrightarrow \int^{P_1}_{P_0} \frac{dP}{P} = -g \cdot \int^{T_1}_{T_0} \frac{dT}{R L_t T}$ $= ln \big (\frac{P}{-P_1} \big) = - \frac{g}{R L_i} ln (\frac{T}{-T_0})$ $\Longrightarrow \frac{P}{P_i} = \big( \frac{T}{T_0} \big) - \frac{g}{R L_i}$ $T(h) = T_0 + a h \; since \; \frac{\partial T}{\partial h} = a = L_i$ * As h $\uparrow T \downarrow$ $a = - ve$ $T < T_0 \Longrightarrow \frac{P_i}{P_0} \propto { \big( \frac{T}{T_0} \big) }^{-\frac{g}{R u}} + ve \Longrightarrow P_i < P_0$ * As h $\uparrow \, T\uparrow$ $a = + ve$ $T > T_0 \Longrightarrow \frac{P_i}{P_0} \propto { \big( \frac{T}{T_0} \big) }^{-\frac{g}{R u}} - ve \Longrightarrow P_i < P_0$ ### Aircraft Performance. * Max velocity $\longrightarrow$ actual ambient * Pressure Altitude $\longrightarrow$ By knowing pressure we associate the altitude to ISA model alt. $\hookrightarrow$ Not true Altitude --- Image containing two diagrams of airplane. We can use Radar,GPS for some flights. But they might not work at heights above 18000 ft. Nose hole ti measure pressure: Pressure is distorted around the body might not measure the actual ISA ht. There are tail wind and head wind. * Tail * Head Actual speed. Ground speed (GS). GS = AS + TWS GS = AS - HWS Air speed & Ground speed w.r.t. surrounding ground air Jetstreams happen during winter affecting the flow of air and wind. Altitude $\approx 9$ km. * In polar regions strong during winter * Jet streams are slower in subtropical regions. Approximately same as atmosphere and ground are relatively at rest. All commercial flights vel > 800 mph When wild speeds are high, air speed and ground speed is diff. Study of effects of relative motion between air and bodies, can be due ti wind / motion if body. In air crafts reference, wind only moves, i.e. relative velocity is afflicted only, and $\rho,T, \phi$ remains consent according to local atm surrounding. But Pressure distribution is there. Bernoulli's eqn. $P + \frac{1}{2} \rho v^2 = const$ For rigid body $F = m \frac{d^2 r}{d t^2}$ But for a fluid con wont tell about the mech of whole fluid. --- ### Continuum Mech Diagrams showing the effect of viscous force. $F = m \frac{d^2 r}{d t^2}$ Streamlines ### Steady/ Un-steady flow steady - same element property / but not w.r.t Time position i.e. not uniform throughout represents steady flow Momentum eqn Properties: $(P, T, S, \overline{v})$ can be written as function of position, whereas mass and volume can't be written for a point. Pressure-stress = F / Area is used since F can't be determined for a pt. $F = m a \longrightarrow \, Euler \\ Navies Stokes$ V6 c Vi because of relative motion there will be friction forces, viscous stress Newton's law of viscosity: viscous stress $(\tau) = \mu \frac{d v}{d y}$ In Euler's equation, viscous effects are neglected, pressure and gravity effect are considered! It assumes the fluid to InviSCID. --- $F_4 = ma_i$ force along a dir depends on acc streamline co-ordinate system along that dir. $(2 \mathbb{Q} flow)$: assuming $Fs = m \hat{a} \hat{s} steady \, flow$ $= PdA - (P + dP)dA$ $= -dP dA \overline{aS} = \int dA ds dv \frac{ds}{dt}$ $\frac{- dP}{\partial s} dA \partial s = \int dA ds \overline{dv} \frac{ds}{dt}$ velocity of fluid particle which will change according to position $\frac{dp}{ds} = \int \frac{d \overline{d}v}{\partial s} \cdot \frac{v}{\rightarrow} S dP = Sv dv$ A stream line defined as a pass a mass-less particle would take when carried in a stream. Local velocity is always tangent to stream line, in which case there can be no mass flow across a stream Line. This means that fluid must be inviscid. $\frac{dP}{ds} + \int \frac{v dv}{ds}=0$ $\Rightarrow \frac(\left[P+\int \frac{V^{2}}{2}\right])=0$ *steady flow assumption is used as V(s) =$\frac{ds}{dt}$ steady $V(c,t)$* constant on a streamline $P+\mathbb{Q} P \cdot dm$ $\partial P \hat{n}$ $-dmm \longrightarrow$ $Pd A^{1}- (P+dp dm) d^{2} A^{1} =\operatorname{Am}\cdot \frac{V^{ 2}}{R}$ $-\Delta PA^{\dagger} dA^{\dagger} = -\int dom dA^{1}. \frac{v^{\partial}}{R}$ $\therefore -\frac{\Delta P}{\Delta M}$. (FVM) Keep P with Sign --- Free stream flow. (constant pressure and velocity Varr). Depending on the magnitude of Radius of curvature of stream lines the value of d P varies dn. Diagram exhibiting the effects of pressure Poo, Voo The pressure decreases along the direction of R. *Then by applying Bernoulli's eqn: $P++\mathbb{\Phi} \frac{Y^{2}}{2} 110= C$. We get the velocity at P minimum. at P, as R 1, IS maximum and & Curvature is highest * The Curvature reduces for B R p- - The curvatture of c incneases but the P onside preduces. P 4:00 0: pressure Medaces from 6-31. P++ V = ccan De applied even if we are not going along she dmeamline if Now is a free stream flow. * More/Higher che curvature & more is is The change in ce- coefficren't of. P-P80 (d) 1 max e 1:00+ 1- SV2. "stagnation point Vei reducered To And max Pressum --- $\hat{i}$ #Navier Stoke’s Eqn: Momentum equation with viscosity in the free stream there is no velocity gradient, although coefficient of viscosity is very much there Diagram representing Pressure Through BL and No slip condution ~ fluid element will have same velocity ad That of the Plare Boundary layer Layer of flud in centact, with the Body where friction us particularly dominant People Ignored This layer is It Is