lecture-slides-body/lecture-03-body.tex:    \item Compressibility (\ensuremath{\beta}) is mathematically defined as \ensuremath{\beta} = -\frac{1}{V}\frac{dV}{dp} = \frac{1}{\ensuremath{\rho}}\frac{d\ensuremath{\rho}}{dp}.
lecture-slides-body/lecture-03-body.tex:    \item Bulk Modulus (K) is the reciprocal of compressibility: K = -V\frac{dp}{dV} = \ensuremath{\rho}\frac{dp}{d\ensuremath{\rho}}.
lecture-slides-body/lecture-03-body.tex:    \item Water: K \approx 2.2 \times 10\textsuperscript{9} Pa; Air (at standard conditions): K \approx 1.01 \times 10\textsuperscript{5} Pa.
lecture-slides-body/lecture-03-body.tex:    \item Acoustic waves propagate isentropically through gases, making K\textsubscript{s} the relevant parameter for the speed of sound (c = \sqrt{K\textsubscript{s} / \ensuremath{\rho}}).
lecture-slides-body/lecture-03-body.tex:    \item Acoustic wave speed equation: c = \sqrt{\frac{K}{\ensuremath{\rho}}}.
lecture-slides-body/lecture-03-body.tex:    \item In water: c = \sqrt{2.2 \times 10\textsuperscript{9} \text{ Pa} / 1000 \text{ kg/m}\textsuperscript{3}} \approx 1483 \text{ m/s}.
lecture-slides-body/lecture-03-body.tex:    \item In air (isentropic, 15\textsuperscript{\ensuremath}{\circ} C): c = \sqrt{1.4 \times 101325 \text{ Pa} / 1.225 \text{ kg/m}\textsuperscript{3}} \approx 340 \text{ m/s}.
lecture-slides-body/lecture-03-body.tex:    \item Newton's Law of Viscosity: \ensuremath{\tau} = \ensuremath{\mu} \frac{du}{dy}.
lecture-slides-body/lecture-03-body.tex:    \item Gas viscosity variation is commonly modeled using Sutherland's law: \ensuremath{\mu} = \ensuremath{\mu}\textsubscript{0} \left(\frac{T}{T\textsubscript{0}}\right)\textsuperscript{3/2} \frac{T\textsubscript{0} + S}{T + S}.
lecture-slides-body/lecture-18-body.tex:    \item Volumetric flow rate equation: Q = C\textsubscript{d} * A\textsubscript{a} * \sqrt{2 * g * V\textsubscript{f} * (\ensuremath{\rho}\textsubscript{f} - \ensuremath{\rho}) / (A\textsubscript{f} * \ensuremath{\rho})}.
lecture-slides-body/lecture-18-body.tex:    \item For small taper angles, the y\textsuperscript{2} term is negligible, yielding a linear relationship: A\textsubscript{a} \approx \ensuremath{\pi} * D\textsubscript{f} * y * \tan(\ensuremath{\theta}/2).
lecture-slides-body/lecture-37-body.tex:    \item Mathematical relation: \ensuremath{\eta}\textsubscript{o} = P\textsubscript{s} / P\textsubscript{w} = (P\textsubscript{s} / P\textsubscript{r}) \times (P\textsubscript{r} / P\textsubscript{w}) \times (Q\textsubscript{runner} / Q)
lecture-slides-body/lecture-37-body.tex:    \item Simplifying the product: \ensuremath{\eta}\textsubscript{o} = \ensuremath{\eta}\textsubscript{m} \times \ensuremath{\eta}\textsubscript{h} \times \ensuremath{\eta}\textsubscript{v}
lecture-slides-body/lecture-37-body.tex:    \item If leakage is neglected (\ensuremath{\eta}\textsubscript{v} \approx 1), overall efficiency becomes: \ensuremath{\eta}\textsubscript{o} = \ensuremath{\eta}\textsubscript{h} \times \ensuremath{\eta}\textsubscript{m}
lecture-slides-body/lecture-37-body.tex:    \item Step 1: Jet velocity, V\textsubscript{1} = C\textsubscript{v} \sqrt{2gH} = 0.98 \sqrt{2 \times 9.81 \times 300} = 75.18 m/s.
lecture-slides-body/lecture-37-body.tex:    \item Step 2: Blade speed, u = \ensuremath{\pi} D N / 60 = \ensuremath{\pi} \times 1.2 \times 500 / 60 = 31.42 m/s.
lecture-slides-body/lecture-37-body.tex:    \item Step 3: Relative velocity at inlet, V\textsubscript{r1} = V\textsubscript{1} - u = 43.76 m/s. At outlet, V\textsubscript{r2} = k V\textsubscript{r1} = 0.9 \times 43.76 = 39.38 m/s.
lecture-slides-body/lecture-37-body.tex:    \item Step 4: Whirl velocity at outlet, V\textsubscript{w2} = V\textsubscript{r2} \cos \ensuremath{\beta} - u = 39.38 \cos 15\textsuperscript{\ensuremath{\circ}} - 31.42 = 6.62 m/s. Power = \ensuremath{\rho} Q (V\textsubscript{w1} + V\textsubscript{w2}) u = 1000 \times 2.0 \times (75.18 + 6.62) \times 31.42 = 5.14 MW.
lecture-slides-body/lecture-37-body.tex:    \item Step 5: Hydraulic efficiency, \ensuremath{\eta}\textsubscript{h} = \text{Power} / (\ensuremath{\rho} g Q H) = 5.14 \times 10\textsuperscript{6} / (1000 \times 9.81 \times 2 \times 300) = 87.3%.
lecture-slides-body/lecture-37-body.tex:    \item Step 1: Calculate discharge, Q = \ensuremath{\pi} / 4 \times (D\textsubscript{o\textsuperscript}{2} - D\textsubscript{b\textsuperscript}{2}) \times V\textsubscript{f} = \ensuremath{\pi} / 4 \times (4.5\textsuperscript{2} - 2.0\textsuperscript{2}) \times 8 = 102.1 m\textsuperscript{3}/s.
lecture-slides-body/lecture-37-body.tex:    \item Step 3: Since V\textsubscript{w2} = 0, \ensuremath{\eta}\textsubscript{h} = V\textsubscript{w1} u / (g H) \implies u = \ensuremath{\eta}\textsubscript{h} g H / V\textsubscript{w1} = 0.94 \times 9.81 \times 20 / 13.86 = 13.31 m/s.
lecture-slides-body/lecture-37-body.tex:    \item Step 4: Speed N = 60 u / (\ensuremath{\pi} D\textsubscript{m}) where D\textsubscript{m} = (D\textsubscript{o} + D\textsubscript{b})/2 = 3.25 m. N = 60 \times 13.31 / (\ensuremath{\pi} \times 3.25) = 78.2 rpm.
lecture-slides-body/lecture-37-body.tex:    \item Step 5: Overall efficiency, \ensuremath{\eta}\textsubscript{o} = \ensuremath{\eta}\textsubscript{h} \times \ensuremath{\eta}\textsubscript{m} = 0.94 \times 0.97 = 91.18%.
lecture-slides-body/lecture-37-body.tex:    \item Step 6: Shaft Power, P\textsubscript{s} = \ensuremath{\eta}\textsubscript{o} \ensuremath{\rho} g Q H = 0.9118 \times 1000 \times 9.81 \times 102.1 \times 20 = 18.26 MW.
lecture-slides-body/lecture-27-yt.tex:    \item Maximum efficiency \ensuremath{\eta}\textsubscript{max} = 2 * (V/3) * (2*V/3)\textsuperscript{2} / V\textsuperscript{3} = 8/27 \approx 29.63\%.
lecture-slides-body/lecture-38-yt.tex:ode[red, font=\tiny, right] at (2.1, 0.4) {Micro-jet (v \approx 100  m/s)};
lecture-slides-body/lecture-21-body.tex:h\textsubscript{f} = \frac{p\textsubscript{1} - p\textsubscript{2}}{\ensuremath{\gamma}} = \frac{4 \ensuremath{\tau}\textsubscript{0} L}{\ensuremath{\gamma} d}
lecture-slides-body/lecture-21-body.tex:To relate this to flow velocity, we introduce Froude's definition of shear stress in turbulent flow: \ensuremath{\tau}\textsubscript{0} = f' \frac{\ensuremath{\rho} v\textsuperscript{2}}{2}, where f' is the coefficient of friction. Substituting this yields:
lecture-slides-body/lecture-21-body.tex:h\textsubscript{f} = \frac{4 f' L v\textsuperscript{2}}{2 g d}. Defining the Darcy friction factor f = 4f' gives the final Darcy-Weisbach equation:
lecture-slides-body/lecture-21-body.tex:h\textsubscript{f} = \frac{f L v\textsuperscript{2}}{2 g d}
lecture-slides-body/lecture-21-body.tex:Writing \frac{p\textsubscript{1} - p\textsubscript{2}}{\ensuremath{\gamma}} = h\textsubscript{f}, we get \ensuremath{\tau}\textsubscript{0} = \ensuremath{\gamma} \frac{A}{P} \frac{h\textsubscript{f}}{L}.
lecture-slides-body/lecture-21-body.tex:k v\textsuperscript{2} = \ensuremath{\rho} g m i \Rightarrow v = \sqrt{\frac{\ensuremath{\rho} g}{k}} \sqrt{m i}.
lecture-slides-body/lecture-21-body.tex:Defining Chezy's constant C = \sqrt{\frac{\ensuremath{\rho} g}{k}}, we obtain:
lecture-slides-body/lecture-21-body.tex:v = C \sqrt{m i}
lecture-slides-body/lecture-21-body.tex:ode[right, blue] at (4.5, 1.95) {\frac{v\textsuperscript{2}}{2g}};
lecture-slides-body/lecture-21-body.tex:From Darcy-Weisbach: h\textsubscript{f} = \frac{f L v\textsuperscript{2}}{2 g d} \Rightarrow v\textsuperscript{2} = \frac{2 g d h\textsubscript{f}}{f L}.
lecture-slides-body/lecture-21-body.tex:For a circular pipe, the hydraulic radius is: m = \frac{A}{P} = \frac{\ensuremath{\pi} d\textsuperscript{2} / 4}{\ensuremath{\pi} d} = \frac{d}{4} \Rightarrow d = 4m.
lecture-slides-body/lecture-21-body.tex:v\textsuperscript{2} = \frac{2 g (4m) i}{f} = \frac{8 g}{f} m i \Rightarrow v = \sqrt{\frac{8g}{f}} \sqrt{m i}.
lecture-slides-body/lecture-21-body.tex:Comparing this with Chezy's formula (v = C \sqrt{m i}), we establish:
lecture-slides-body/lecture-21-body.tex:C = \sqrt{\frac{8g}{f}}
lecture-slides-body/lecture-21-body.tex:2. The friction coefficient f' = 0.005. Note that Darcy friction factor f = 4f' = 4 \times 0.005 = 0.02.
lecture-slides-body/lecture-21-body.tex:3. Apply Darcy-Weisbach equation: h\textsubscript{f} = \frac{f L v\textsuperscript{2}}{2 g d}
lecture-slides-body/lecture-21-body.tex:4. Substitute values: h\textsubscript{f} = \frac{0.02 \times 1500 \times 2\textsuperscript{2}}{2 \times 9.81 \times 0.3}
lecture-slides-body/lecture-21-body.tex:5. Calculate: h\textsubscript{f} = \frac{120}{5.886} = 20.39\text{ m} of water.
lecture-slides-body/lecture-21-body.tex:2. For a pipe running half full: Area A = \frac{1}{2} (\frac{\ensuremath{\pi} d\textsuperscript{2}}{4}) = 0.5655\text{ m}\textsuperscript{2}.
lecture-slides-body/lecture-21-body.tex:Wetted perimeter P = \frac{1}{2} (\ensuremath{\pi} d) = 1.885\text{ m}.
lecture-slides-body/lecture-21-body.tex:4. Velocity: v = C \sqrt{m i} = 55 \times \sqrt{0.3 \times 0.001} = 55 \times 0.01732 = 0.9526\text{ m/s}.
lecture-slides-body/lecture-21-body.tex:5. Discharge: Q = A \times v = 0.5655 \times 0.9526 = 0.5387\text{ m}\textsuperscript{3}/s.
lecture-slides-body/lecture-32-yt.tex:ode at (6.75, 2.0) [font=\bfseries] {Overall Efficiency: \ensuremath{\eta}\textsubscript{o} = \ensuremath{\eta}\textsubscript{h} \times \ensuremath{\eta}\textsubscript{v} \times \ensuremath{\eta}\textsubscript{m}};
lecture-slides-body/lecture-32-yt.tex:ode at (1.5, 4.0) [align=center, font=\small] {Pelton\\\\(Tangential Flow)\\High Head\H > 250m\N\textsubscript{s} \approx 10-35};
lecture-slides-body/lecture-32-yt.tex:ode at (5.25, 2.5) [align=center, font=\small] {Francis\\\\(Mixed/Radial Flow)\\Medium Head\H \approx 60-250m\N\textsubscript{s} \approx 60-300};
lecture-slides-body/lecture-32-yt.tex:ode at (9.75, 0.85) [align=center, font=\small] {Kaplan / Propeller\\\\(Axial Flow)\\Low Head\H < 60m\N\textsubscript{s} \approx 300-1000};
lecture-slides-body/lecture-35-body.tex:    \item Tangential speed relation: U = \ensuremath{\pi} D N / 60 \propto \sqrt{H}, which implies diameter D \propto \sqrt{H}/N.
lecture-slides-body/lecture-35-body.tex:    \item Substituting D into the power relation: P \propto (\sqrt{H}/N)\textsuperscript{2} H\textsuperscript{1.5} \propto H\textsuperscript{2.5}/N\textsuperscript{2}.
lecture-slides-body/lecture-35-body.tex:    \item Solving for speed: N\textsuperscript{2} \propto H\textsuperscript{2.5}/P, which gives specific speed N\textsubscript{s} = N \sqrt{P} / H\textsuperscript{1.25}.
lecture-slides-body/lecture-35-body.tex:    \item Mathematically defined as N\textsubscript{sp} = N \sqrt{Q} / H\textsuperscript{0.75}.
lecture-slides-body/lecture-35-body.tex:    \item For turbines: Dimensionless N\textsubscript{s}* = \ensuremath{\omega} \sqrt{P / \ensuremath{\rho}} / (g H)\textsuperscript{1.25}.
lecture-slides-body/lecture-35-body.tex:    \item For pumps: Dimensionless N\textsubscript{sp}* = \ensuremath{\omega} \sqrt{Q} / (g H)\textsuperscript{0.75}.
lecture-slides-body/lecture-34-yt.tex:    \draw[<->, red] (-0.8, 0.5) -- (0.8, 0.5) node[midway, above] {2\ensuremath{\theta} \approx 7\textsuperscript{\ensuremath{\ensuremath{\circ}}} - 8\textsuperscript{\ensuremath{\ensuremath{\circ}}}};
lecture-slides-body/lecture-43-yt.tex:  \draw[->, thick, >=stealth] (3.2, 0.75) -- (5.2, 0.75) node[midway, above] {\ensuremath{\eta}\textsubscript{h} \times \ensuremath{\eta}\textsubscript{v}} node[midway, below] {Runner}; 
lecture-slides-body/lecture-43-yt.tex:  \draw[->, thick, >=stealth] (8.4, 0.75) -- (10.4, 0.75) node[midway, above] {\ensuremath{\eta}\textsubscript{mano} \times \ensuremath{\eta}\textsubscript{v}} node[midway, below] {Water}; 
lecture-slides-body/lecture-12-body.tex:In fluid mechanics, analyzing a fixed mass of fluid (a system) is often impractical due to fluid motion. Instead, we use a Control Volume (CV) which is a defined region in space through which fluid flows. The mathematical bridge between system-based conservation laws (Lagrangian description) and control volume analysis (Eulerian description) is the Reynolds Transport Theorem (RTT). For an extensive property B and intensive property b = B/m, RTT is expressed as: d/dt(B\textsubscript{sys}) = d/dt \int\textsubscript{CV} b \ensuremath{\rho} dV + \int\textsubscript{CS} b \ensuremath{\rho} (V\textsubscript{r} \cdot n) dA, where V\textsubscript{r} is the relative velocity.
lecture-slides-body/lecture-12-body.tex:Applying RTT with B = m (mass) and b = 1 yields the integral form of the Continuity Equation: d/dt \int\textsubscript{CV} \ensuremath{\rho} dV + \int\textsubscript{CS} \ensuremath{\rho} (V\textsubscript{r} \cdot n) dA = 0. For steady, incompressible flow with one-dimensional inlets and outlets, the equation simplifies to the conservation of volumetric flow rate: \sum A\textsubscript{in} V\textsubscript{in} = \sum A\textsubscript{out} V\textsubscript{out}, or Q = A\textsubscript{1} V\textsubscript{1} = A\textsubscript{2} V\textsubscript{2}. This states that fluid velocity is inversely proportional to cross-sectional area.
lecture-slides-body/lecture-12-body.tex:    \item Area-averaged velocity V is defined as (1/A) \\int V dA across the flow cross-section.
lecture-slides-body/lecture-12-body.tex:Applying the Divergence Theorem to the control surface integral of the continuity equation yields the differential form: \partial \ensuremath{\rho} / \partial t + \nabla \cdot (\ensuremath{\rho} V) = 0. In Cartesian coordinates, this expands to: \partial \ensuremath{\rho} / \partial t + \partial (\ensuremath{\rho} u)/\partial x + \partial (\ensuremath{\rho} v)/\partial y + \partial (\ensuremath{\rho} w)/\partial z = 0. This form represents conservation of mass at an infinitesimal point in space, which is critical for computational fluid dynamics (CFD) and local flow analysis.
lecture-slides-body/lecture-12-body.tex:    \item The term \\partial \\ensuremath{\rho} / \\partial t represents local rate of density change.
lecture-slides-body/lecture-12-body.tex:Applying Newton's second law (F = ma) to a fluid system using RTT, where B = m V (momentum) and b = V, yields the Linear Momentum Equation: \sum F = d/dt \int\textsubscript{CV} V \ensuremath{\rho} dV + \int\textsubscript{CS} V \ensuremath{\rho} (V\textsubscript{r} \cdot n) dA. The net force \sum F acting on the control volume includes body forces (gravity, electromagnetic) and surface forces (pressure, shear stress, and reaction forces from solid walls).
lecture-slides-body/lecture-12-body.tex:    \item Body forces act on the entire mass of the control volume (e.g., W = \\int\textsubscript{CV} \\ensuremath{\rho} g dV).
lecture-slides-body/lecture-12-body.tex:For rotating hydraulic machinery like pumps and turbines, torque and angular momentum are the primary variables of interest. The Moment of Momentum equation is derived by taking the cross product of the position vector r with the terms in Newton's second law: \sum T = d/dt \int\textsubscript{CV} (r \times V) \ensuremath{\rho} dV + \int\textsubscript{CS} (r \times V) \ensuremath{\rho} (V\textsubscript{r} \cdot n) dA. Under steady conditions with uniform inlets/outlets, this yields Euler's Turbine Equation: T = \ensuremath{\rho} Q (r\textsubscript{2} V\textsubscript{t2} - r\textsubscript{1} V\textsubscript{t1}), where V\textsubscript{t} is the tangential velocity component.
lecture-slides-body/lecture-07-yt.tex:  \draw[<->, thick] (0.6, 0.8) -- (0.6, 4.5) node[midway, right] {h \approx 760 mm};