Fluid Mechanics

Fluid statics and viscous flow

In a fluid at rest, pressure increases linearly with depth. The absolute pressure at depth h is p_{abs} = p_{atm} + \rho g h, and the gauge pressure, measured relative to atmospheric pressure, is p_{gauge} = \rho g h. Always check whether a reading is absolute or gauge before substituting it into the formula. Hydrostatic pressure depends only on depth and density, not on the shape of the container. For viscous flow, kinematic viscosity \nu and dynamic viscosity \mu are related by \mu = \rho\nu, so the density must be used to convert between them before computing a shear stress. In plane Couette flow, where one plate moves and there is no pressure gradient, the velocity varies linearly across the gap and the shear stress is uniform, \tau = \mu U / h. A body hinged in a fluid is in equilibrium when the moments of buoyancy and weight about the hinge balance. A frictionless hinge reaction passes through the pivot and contributes no moment, so only the buoyant and gravitational forces matter. Specific gravity multiplies the density of water to give the material density, and the net buoyant force equals the density difference times the displaced volume times g. For a gate separating two fluids, the force on each side per unit area is \rho g h at depth h, and its moment about the hinge is obtained by integrating this pressure over the wetted height. Atmospheric pressure acting on both free surfaces cancels, so only gauge pressure produces a net moment.

Bernoulli equation for inviscid flow

Bernoulli's equation, p/\rho + V^2/2 + gz = \text{constant}, expresses energy conservation along a streamline for steady, inviscid, incompressible flow. It can be applied between any two points on the same streamline, even when the flow is rotational. The constant is the same for all streamlines only when the flow is irrotational, which is why the equation can relate arbitrary points in an irrotational field but not in a rotational one. At a stagnation point the velocity is zero, and the pressure there exceeds the free-stream static pressure by \rho V^2/2. A pitot tube with a manometer measures this difference, and for an incompressible fluid V = \sqrt{2\Delta p/\rho}, while for a U-tube manometer \Delta p = (\rho_m - \rho) g h. For example, a pitot reading of 2 kPa in air of density 1.2 kg/m^3 gives V \approx 57.7 m/s. When a stagnation inlet discharges to atmosphere through an outlet located above the free surface, the elevation of the outlet reduces the outlet velocity to \sqrt{V^2 - 2gh}, because the kinetic energy is converted into potential energy. In a diffuser the flow area grows, the velocity falls and the pressure rises. A combined Bernoulli and momentum analysis gives the net force that the fluid exerts on the diffuser walls, and that force depends on the inlet and outlet velocities and pressures.

Continuity equation for 2D incompressible flow

For two-dimensional incompressible flow, continuity reduces to \partial u/\partial x + \partial v/\partial y = 0, so the velocity divergence is zero. This relation is used to eliminate one derivative when simplifying expressions. A typical simplification expands products such as \partial(u^2)/\partial x + \partial(uv)/\partial y and replaces \partial v/\partial y with -\partial u/\partial x, which leaves the convective form u\,\partial u/\partial x + v\,\partial u/\partial y. To check a candidate velocity field, compute the divergence and the vorticity. The field is incompressible if the divergence vanishes, and irrotational if the z-component of vorticity, \partial v/\partial x - \partial u/\partial y, is zero. A field can satisfy both conditions even though its velocity components depend on both coordinates, so the flow may be two-dimensional and still be irrotational and incompressible. Also, at a point where both velocity components vanish, the steady convective acceleration (\vec V\cdot\nabla)\vec V is zero. To find an unknown velocity component, integrate continuity with respect to the coordinate. For example, if u = 3x and v(x,0) = 0, then \partial v/\partial y = -3, so v = -3y. An integration constant is a function of x alone and is fixed by the boundary value on the line y = 0.

Continuity equation for compressible flow

For steady one-dimensional compressible flow, the mass flow rate \dot m = \rho A V is constant along the duct. The momentum balance on the gas then relates the wall force to the pressure difference and the change in momentum flux, (p_1 - p_2)A + F_{wall} = \dot m (V_2 - V_1), with the sign of the wall force set by the chosen convention. The force the gas exerts on the pipe is equal and opposite to the wall force on the gas by Newton's third law. For unsteady flow with spatially uniform density, the continuity equation \partial\rho/\partial t + \nabla\cdot(\rho\vec V) = 0 becomes \partial u/\partial x = -(1/\rho)\,d\rho/dt, so the velocity gradient is set by the rate of change of density in time. For an isothermal ideal gas, p = \rho R T means density is proportional to pressure. Since the mass flux \rho V is constant in a uniform pipe, a fall in pressure along the pipe lowers the density and raises the velocity. For example, with \rho = 2 kg/m^3, A = 0.01 m^2 and V = 10 m/s, the mass flow rate is 0.2 kg/s.

Compressible flow in converging-diverging nozzle

In a converging-diverging nozzle, subsonic flow accelerates through the converging section, and the flow reaches sonic conditions at the throat. Beyond the throat, the flow can accelerate to supersonic speed only if the area increases. In isentropic supersonic flow in a diverging duct, the velocity rises and the pressure falls, which is the opposite of subsonic behaviour. A normal shock is a thin, adiabatic, irreversible discontinuity across which the flow decelerates from supersonic to subsonic. Across a normal shock, the static pressure, temperature and density increase, the Mach number decreases, and the specific entropy increases. A static pressure rise in a diverging supersonic passage therefore cannot occur isentropically and requires a normal shock, while viscous losses and boundary layer effects are not the standard mechanism for such a rise. In compressible flow the volume flow rate is not conserved, so the velocity cannot be found from volume continuity alone. The mass flow rate is the quantity that remains constant.

Pelton wheel turbine power and efficiency

A Pelton wheel is an impulse turbine in which a jet of velocity V_1 strikes buckets on a wheel of mean diameter D. The bucket speed is u = \pi D N/60, and the relative velocity at inlet is V_1 - u. When the bucket deflects the jet through an angle \theta, the tangential force on the buckets is F = \rho Q (V_1 - u)(1 - \cos\theta), and the power developed is P = F u. In the ideal case with no friction loss, the relative speed is unchanged across the bucket. Power is maximised when the bucket speed is half the jet speed, u = V_1/2, which is the condition for maximum power output in the ideal case. The rotational speed then follows from N = 60u/(\pi D). For example, with V_1 = 100 m/s and D = 1 m, the optimum speed is about 955 rpm.

Streamlines in 2D flow fields

A streamline is a curve tangent to the velocity vector at a given instant. In a two-dimensional flow at a fixed time, streamlines satisfy dx/u = dy/v, and integrating this relation gives the family of streamlines. The integration constant is then fixed by the point through which a particular streamline passes. A stream function \psi provides a compact way to describe a two-dimensional incompressible flow, with u = \partial\psi/\partial y and v = -\partial\psi/\partial x. Lines of constant \psi are streamlines, and once the velocity components are known from \psi, the convective acceleration at a point follows from a_x = u\,\partial u/\partial x + v\,\partial u/\partial y and a_y = u\,\partial v/\partial x + v\,\partial v/\partial y. For instance, the velocity field u = x, v = -y has streamlines xy = \text{constant}.

Tank draining through orifice

For a tank draining through a small orifice with no losses, Bernoulli's equation gives the exit velocity v = \sqrt{2gh}, where h is the head above the orifice, provided the free surface velocity is negligible. Mass conservation gives A_t\,dh/dt = -A_o\sqrt{2gh}. Integrating from an initial head H_1 to a final head H_2 gives t = (A_t/A_o)\sqrt{2/g}\,(\sqrt{H_1} - \sqrt{H_2}). The exit velocity depends only on the head and gravity, not on the fluid density. Two identical tanks with the same head therefore give the same exit velocity whether they hold water or oil. The orifice diameter affects the discharge rate but cancels out of the velocity relation. For example, with A_t/A_o = 100 and a drop from H_1 = 4 m to H_2 = 1 m, the drain time is about 45.2 s.

Francis turbine velocity triangles

In a Francis turbine, the flow enters the runner through guide vanes. At the runner inlet, the whirl component of absolute velocity follows from the guide vane angle, V_{w1} = V_{f1}/\tan\alpha_1, where V_{f1} is the flow velocity. The blade speed is U_1 = \pi D_1 N/60, and the relative velocity has \tan\beta_1 = V_{f1}/(V_{w1} - U_1), which determines the inlet vane angle. The hydraulic efficiency is the ratio of the Euler head H_u to the net head H available at the turbine inlet. The energy balance is H = H_u + h_{gv} + h_r + h_{dt} + V_3^2/(2g), where the losses are those in the guide vanes, runner and draft tube, and the last term is the kinetic head leaving at the draft tube exit. For example, with H = 100 m, total losses of 5 m and an exit kinetic head of 0.5 m, the Euler head is 94.5 m and the hydraulic efficiency is 94.5 percent.

Euler momentum equation for pressure gradient

For steady, inviscid flow with no body forces, the Euler equation states \rho(\vec V\cdot\nabla)\vec V = -\nabla p, so the pressure gradient is the negative of the density times the convective acceleration. The convective acceleration has components a_x = u\,\partial u/\partial x + v\,\partial u/\partial y and a_y = u\,\partial v/\partial x + v\,\partial v/\partial y. Once the acceleration is known at a point, the pressure gradient there is obtained directly by multiplying by -\rho. For example, if u = Ax and v = 0, then a_x = A^2 x and \partial p/\partial x = -\rho A^2 x. Integrating such gradients gives the pressure field, and the integration constant is fixed by a reference pressure at a known point.

Jet impingement force on flat plate

When a water jet strikes a flat plate normally and spreads radially, the fluid loses its normal momentum. For a jet of area A and velocity V, the force on the plate is F = \rho A V^2, since the mass flow rate \rho A V is turned from velocity V to zero in the normal direction. If the plate rests on a spring, a steady deflection x means the jet force balances the spring force, F = kx. Equating the two gives the jet velocity from the spring constant and the deflection. For example, a spring deflection of 1 cm under a force of 10 N gives the jet velocity from V = \sqrt{F/(\rho A)}.

Turbulent boundary layer and Reynolds stresses

In a turbulent boundary layer, velocity fluctuations produce the Reynolds stresses -\rho\overline{u'v'}, which act as additional shear stresses that transfer momentum across the layer. These stresses are the physical source of the enhanced momentum exchange in turbulent flow. The Boussinesq hypothesis models the Reynolds stresses as \tau_t = \varepsilon_m\,\partial\bar u/\partial y, where \varepsilon_m is the eddy viscosity. The eddy viscosity is a modelled coefficient that depends on the flow, while the Reynolds stresses are the quantities being modelled. Prandtl's mixing length theory provides a way to estimate \varepsilon_m, and Nikuradse is associated with experiments on flow in roughened pipes rather than with the boundary layer stress itself.