UPSC ESE Mechanical Engineering Fluid Mechanics Syllabus

What is the relationship between absolute, gauge, and atmospheric pressure?

$$p_{abs} = p_{atm} + p_{gauge}$$ Gauge pressure is measured relative to atmospheric; a negative gauge pressure is a vacuum pressure: $p_{vacuum} = p_{atm} - p_{abs}$.

State Pascal's law.

Pascal's law states that pressure at a point in a fluid at rest is equal in all directions (isotropic): $p_{x} = p_{y} = p_{z}$. It also implies pressure applied to a confined fluid is transmitted undiminished throughout.

How does a simple piezometer measure pressure, and what is its limitation?

A piezometer is an open vertical tube tapped into the pipe; the liquid rises to a height $h$ giving gauge pressure $p = \rho g h$. Limitations: it cannot measure negative (vacuum) or gas pressures and is impractical for large pressures (very tall column).

For a simple U-tube manometer with a heavier gauge fluid, write the pressure at the pipe centre.

$$p_{A} = \rho_{m} g h_{m} - \rho_{1} g h_{1}$$ where $\rho_{m}$ is the manometer (gauge) fluid density rising height $h_{m}$ and $\rho_{1}, h_{1}$ refer to the fluid in the pipe limb. Pressures are balanced about a common horizontal datum.

What does a differential manometer measure and write its reading for two pipes A and B.

A differential manometer measures the pressure difference between two points. With a heavier manometer fluid: $$p_{A} - p_{B} = g\,[\rho_{m} h_{m} + \rho_{B} h_{B} - \rho_{A} h_{A}]$$ obtained by balancing pressures at the datum.

Convert standard atmospheric pressure into height of water and mercury columns.

Standard atmosphere $= 101.325\,\mathrm{kPa}$. In water: $h = \frac{p}{\rho g} = \frac{101325}{1000 \times 9.81} \approx 10.33\,\mathrm{m}$. In mercury ($\rho = 13600\,\mathrm{kg/m^{3}}$): $h \approx 0.760\,\mathrm{m}$ ($760\,\mathrm{mm\,Hg}$).

Why is mercury preferred over water in barometers and manometers?

Mercury has a high density ($13600\,\mathrm{kg/m^{3}}$), so it gives a short, manageable column for a given pressure; it has very low vapour pressure and does not wet glass, giving accurate readings.

State Archimedes' principle and the buoyant force formula.

Archimedes' principle: a body wholly or partly immersed in a fluid experiences an upward buoyant force equal to the weight of fluid displaced. $$F_{B} = \rho_{fluid}\, g\, V_{displaced}$$

What conditions determine whether a body floats, sinks, or is neutrally buoyant?

Compare body weight $W = \rho_{body} g V$ with buoyancy $F_{B} = \rho_{fluid} g V_{sub}$. If $\rho_{body} < \rho_{fluid}$ it floats; if $\rho_{body} > \rho_{fluid}$ it sinks; if equal, it is neutrally buoyant (suspended).

Define the centre of buoyancy.

The centre of buoyancy is the point of application of the buoyant force; it coincides with the centroid (centre of gravity) of the displaced fluid volume.

Define metacentre and metacentric height.

The metacentre $M$ is the point where the vertical line through the new centre of buoyancy (after a small tilt) intersects the original line through the centre of gravity. Metacentric height $GM$ is the distance between the centre of gravity $G$ and the metacentre $M$.

Give the formula for metacentric height $GM$ of a floating body.

$$GM = \frac{I}{V} - BG$$ where $I$ is the second moment of area of the waterline plane about the tilt axis, $V$ is the submerged volume, and $BG$ is the distance between centre of buoyancy $B$ and centre of gravity $G$.

State the conditions for stable, unstable, and neutral equilibrium of a floating body.

For a floating body: stable if $M$ is above $G$ ($GM > 0$); unstable if $M$ is below $G$ ($GM < 0$); neutral if $M$ coincides with $G$ ($GM = 0$). A larger $GM$ gives greater stability but more rapid (stiff) rolling.

State the condition for stability of a fully submerged body.

For a fully submerged body, stability requires the centre of buoyancy $B$ to be above the centre of gravity $G$. (Metacentre coincides with $B$ since the displaced volume does not change on tilting.)

Give the formula for total hydrostatic force on a submerged plane surface.

$$F = \rho g \bar{h} A$$ where $\bar{h}$ is the vertical depth of the centroid of the surface below the free surface and $A$ is the area. The force acts normal to the surface.

Locate the centre of pressure on a vertically submerged plane surface.

$$h_{cp} = \bar{h} + \frac{I_{G}\sin^{2}\theta}{A\bar{h}}$$ (for a vertical surface $\sin\theta = 1$). Here $I_{G}$ is the second moment of area about the centroidal axis. The centre of pressure always lies below the centroid.

Why does the centre of pressure lie below the centroid of a submerged surface?

Because pressure increases with depth, the lower portion of the surface experiences greater force. This shifts the resultant (centre of pressure) below the centroid by $\frac{I_{G}\sin^{2}\theta}{A\bar{h}}$. As depth increases, the centre of pressure approaches the centroid.

How are the horizontal and vertical components of hydrostatic force on a curved surface determined?

Horizontal component $F_{H} = \rho g \bar{h} A_{proj}$ equals the force on the vertical projection of the curved surface. Vertical component $F_{V} = \rho g V$ equals the weight of fluid contained (or displaced) above the surface. Resultant $F = \sqrt{F_{H}^{2} + F_{V}^{2}}$.

Give the second moment of area $I_{G}$ about the centroid for a rectangle and a circle.

Rectangle (width $b$, depth $d$): $I_{G} = \frac{b d^{3}}{12}$. Circle (diameter $D$): $I_{G} = \frac{\pi D^{4}}{64}$. These are used in centre-of-pressure calculations.

State the continuity equation for one-dimensional steady incompressible flow.

$$A_{1} V_{1} = A_{2} V_{2} = Q = \text{constant}$$ The volume flow rate is constant; velocity is inversely proportional to cross-sectional area.

State the continuity equation for steady compressible one-dimensional flow.

$$\rho_{1} A_{1} V_{1} = \rho_{2} A_{2} V_{2} = \dot{m} = \text{constant}$$ The mass flow rate $\dot{m}$ is conserved; density variation is included for compressible flow.

Write the differential (general) continuity equation for compressible flow.

$$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{V}) = 0$$ For steady flow $\frac{\partial \rho}{\partial t}=0$; for incompressible flow it reduces to $\nabla \cdot \vec{V} = \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} = 0$.

State Bernoulli's equation and name each term's physical meaning.

$$\frac{p}{\rho g} + \frac{V^{2}}{2g} + z = \text{constant}$$ The terms are pressure head ($\frac{p}{\rho g}$), velocity (kinetic) head ($\frac{V^{2}}{2g}$), and elevation (potential) head ($z$). Their sum is the total head along a streamline.

List the assumptions underlying Bernoulli's equation.

The flow is steady, incompressible, inviscid (frictionless / non-viscous), irrotational, along a single streamline, and only gravity (no external work/heat) acts. It expresses conservation of mechanical energy.

How is Bernoulli's equation modified for real (viscous) flow between two sections?

$$\frac{p_{1}}{\rho g} + \frac{V_{1}^{2}}{2g} + z_{1} = \frac{p_{2}}{\rho g} + \frac{V_{2}^{2}}{2g} + z_{2} + h_{L}$$ where $h_{L}$ is the head loss due to friction between sections 1 and 2.

Derive the Torricelli velocity of efflux from a small orifice using Bernoulli's equation.

For a tank with free surface a height $H$ above the orifice, applying Bernoulli gives $$V = \sqrt{2gH}$$ the theoretical velocity of efflux (same as a freely falling body through height $H$).

Give the discharge formula for a venturimeter.

$$Q = C_{d}\,\frac{A_{1} A_{2}}{\sqrt{A_{1}^{2} - A_{2}^{2}}}\,\sqrt{2gh}$$ where $C_{d}$ is the coefficient of discharge, $A_{1}, A_{2}$ are inlet and throat areas, and $h$ is the manometer/pressure head difference.

How does a pitot tube measure flow velocity?

A pitot tube measures stagnation pressure; combined with static pressure it gives the velocity head. From Bernoulli: $$V = C_{v}\sqrt{2gh}$$ where $h$ is the difference between stagnation and static heads and $C_{v}$ is the coefficient of velocity.

Write the Navier-Stokes equation for incompressible, constant-viscosity flow in vector form.

$$\rho\left(\frac{\partial \vec{V}}{\partial t} + (\vec{V}\cdot\nabla)\vec{V}\right) = -\nabla p + \mu \nabla^{2}\vec{V} + \rho \vec{g}$$ It expresses Newton's second law (momentum balance) per unit volume for a viscous fluid.

Identify the physical meaning of each term in the Navier-Stokes equation.

$\rho\frac{\partial \vec{V}}{\partial t}$ = local (unsteady) inertia; $\rho(\vec{V}\cdot\nabla)\vec{V}$ = convective inertia; $-\nabla p$ = pressure force; $\mu\nabla^{2}\vec{V}$ = viscous (diffusion) force; $\rho\vec{g}$ = body (gravity) force per unit volume.

What do the Euler equations of motion represent and how do they relate to Navier-Stokes?

The Euler equations are the Navier-Stokes equations with the viscous term dropped ($\mu = 0$): $$\rho\frac{D\vec{V}}{Dt} = -\nabla p + \rho\vec{g}$$ They govern inviscid flow; integrating Euler's equation along a streamline yields Bernoulli's equation.